Summary of the course

Nonlinear partial differential equations IStefan Muller

Bonn UniversityFall term 2013–2014

This is only a summary of the main results and arguments discussed in classand not a complete set of lecture notes. These notes can thus not replacethe careful study of the literature. As discussed in class, among others thefollowing two books are recommended:

[Ev] L.C. Evans, Partial differential equations, Amer. Math. Soc., 1998(2nd edition 2010).

[GT] D.Gilbarg and N.S. Trudinger, Elliptic partial differential equations ofsecond order, Springer, 1998 (reprinted as Classics in Mathematics).

These notes are based on the books mentioned above and further sourceswhich are not always mentioned specifically. These notes are only for theuse of the students in the class ’Nonlinear partial differential equations I’ atBonn University, Fall term 2013–2014.

Please send typos and corrections to sm@hcm.uni-bonn.de.

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Contents

1 Introduction 41.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 The Navier-Stokes equation . . . . . . . . . . . . . . . 41.1.2 Mean curvature flow . . . . . . . . . . . . . . . . . . . 51.1.3 Ricci flow . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.4 Minimal surfaces . . . . . . . . . . . . . . . . . . . . . 71.1.5 Harmonic maps . . . . . . . . . . . . . . . . . . . . . . 71.1.6 Nonlinear elasticity . . . . . . . . . . . . . . . . . . . . 8

1.2 Some general strategies . . . . . . . . . . . . . . . . . . . . . 91.2.1 Soft functional analytic and topological methods . . . 91.2.2 Weak convergence methods and compactness . . . . . 141.2.3 Variational methods . . . . . . . . . . . . . . . . . . . 161.2.4 Convex integration . . . . . . . . . . . . . . . . . . . . 18

1.3 Overview of a priori estimates . . . . . . . . . . . . . . . . . . 191.4 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Elliptic equations and systems 212.1 Energy methods: Lp and Cα estimates for elliptic equations

and systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.1.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . 222.1.2 Existence of weak solutions . . . . . . . . . . . . . . . 262.1.3 Reverse Poincare inequality and pointwise estimates . 312.1.4 Morrey and Campanato spaces . . . . . . . . . . . . . 332.1.5 Regularity theory in Campanato spaces, interior esti-

mates . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.1.6 Estimates up to the boundary . . . . . . . . . . . . . . 482.1.7 Flattening the boundary . . . . . . . . . . . . . . . . . 562.1.8 Global estimates . . . . . . . . . . . . . . . . . . . . . 602.1.9 Ck,α theory, Schauder estimates up to the boundary . 622.1.10 An alternative route to Schauder estimates by rescal-

ing and blow-up . . . . . . . . . . . . . . . . . . . . . 652.1.11 Lp theory . . . . . . . . . . . . . . . . . . . . . . . . . 712.1.12 More general elliptic systems and the complementing

condition . . . . . . . . . . . . . . . . . . . . . . . . . 892.1.13 A short look back on elliptic regularity . . . . . . . . . 91

2.2 Harnack inequality and the DeGiorgi-Moser-Nash theorem . . 922.3 BMO , exponential integrability, interpolation and the Hardy

space H1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1032.3.1 Exponential integrability of functions in BMO ,

John-Nirenberg estimate . . . . . . . . . . . . . . . . . 1032.3.2 Interpolation between BMO and L2 . . . . . . . . . . 1102.3.3 Maximal functions and the Hardy space H1 . . . . . . 113

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2.3.4 Meyers’ estimate for systems with L∞ coefficients . . 1222.4 Regularity theory for nonlinear pde . . . . . . . . . . . . . . . 128

2.4.1 Harmonic maps and Hardy space estimates . . . . . . 1282.4.2 Regularity for minimizers of variational problems . . . 1602.4.3 Monotone operators and the Minty-Browder trick . . . 175

2.5 Maximum principles and existence for quasilinear elliptic equa-tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1792.5.1 Classcial maximum principles for linear equations . . . 1792.5.2 Maximum and comparison principles for quasilinear

elliptic equations . . . . . . . . . . . . . . . . . . . . . 1872.5.3 Existence of classical solutions . . . . . . . . . . . . . 1952.5.4 C1,α estimates . . . . . . . . . . . . . . . . . . . . . . 2022.5.5 Gradient bounds at the boundary . . . . . . . . . . . 2072.5.6 Global gradient bounds . . . . . . . . . . . . . . . . . 213

3 A brief look back 2163.1 Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2163.2 Regularity for nonlinear PDE . . . . . . . . . . . . . . . . . . 2173.3 Existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2173.4 Some topics that were not covered . . . . . . . . . . . . . . . 218

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[17.10. 2013, Lecture 1]

1 Introduction

1.1 Examples

1.1.1 The Navier-Stokes equation

Let n ≥ 2, ν > 0 and let

v : [0,∞) × Rn → Rn ’velocity’, (1.1)p : [0,∞) × Rn → Rn ’pressure’. (1.2)

The Navier-Stokes equations are given by

∂

∂tvj +

n∑i=1

vi∂

∂xivj − ν∆vj +

∂

∂xjp = 0, ∀j = 1, . . . n, (1.3)

div v = 0, (1.4)

where div v :=∑n

i=1∂∂xi v

i and ∆ =∑n

i=1∂2

∂2xi , with initial condition

v(0, x) = v0(x). (1.5)

The sixth Clay Millennium problem states1: prove or disprove for n = 3that

If v0 ∈ C∞(Rn) with rapid decay, i.e., |∂αu(x)| ≤ CαK(1 + |x|)−K for all αand K, then there exists (v, p) ∈ C∞([0,∞) × Rn; R3 × R) which satisfies

(1.3)–(1.5).

What is known is that there exist solutions for a small time interval[0, T ) for general smooth initial data v0 with sufficiently rapid decay (whereT depends on v0) and solutions on an infinite time interval for small initialdata. These results are proved by a perturbation argument based on goodestimates for the linear heat equation ∂tv = ∆v and on the Helmholtzdecomposition of a vector field into a gradient and a divergence free part(this is used to eliminate the pressure)2.

An important breakthrough was J. Leray’s paper in 1934 where heshowed the global existence of weak solutions. This reduces the question

1See http://www.claymath.org/millennium/Navier-Stokes Equations/navierstokes.pdffor the precise statement

2Over time such results have been proved in various function spaces. A nearly optimalresult is H. Koch, D. Tataru, Well-posedness for the Navier-Stokes equations, Adv. Math.157 (2001), 22–35.

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of existence of smooth solutions to the question to prove that weak solu-tions are already regular. Leray himself already showed that his solutionsare smooth except on closed set in time of one dimensional Lebesgue measure0. The estimate on the size of the set of possible singularities in space-timewhere much later improved by Scheffer and Cafarelli-Kohn-Nirenberg.

1.1.2 Mean curvature flow

Let Mn be a compact n-dimensional manifold and consider a family of mapsF : [0,∞) ×M → Rn+1. We say that F evolves by mean curvature if

∂tF = −Hν (1.6)

where ν denotes the normal of the hypersurface F (t, ·) and H its meancurvature (more precisely the sum of the principal curvatures). If F0 is asphere then one easily sees that the mean curvature flow yields a family ofshrinking spheres whose radii satisfy the ODE

R(t) = − n

R(t)(1.7)

so that R2(t) = R(0)2 − 2nt and the flow has a point singularity at T =R2(0)/2n.

If the surfaces are (locally) graphs of a function u : [0,∞) × Ω → R theequation is equivalent (up to reparametrisation due to tangential motion)to

∂tu =√

1 + |∇u|2 div∇u√

1 + |∇u|2=(δij −

∂iu∂ju

1 + |∇u|2

)︸ ︷︷ ︸

=aij(x)

∂i∂ju. (1.8)

If |∇u| ≤M , i.e., if the slope of the graph is bounded then

∞∑i,j=1

aij(x)ξiξj ≥1

1 +M2|ξ|2, (1.9)

i.e., the equation (1.8) is parabolic. Moreover in the limit of vanishing slopethe equation reduces to the linear heat equation. This suggests that meancurvature flow at least for small times behaves similar to the heat equation.

The following global results are known for the mean curvature flow.Embedded curves shrink to a round point, see M. A. Grayson, J. Diff. Geom.26 (1987), 285–314. Convex hypersurfaces shrink to a round point, seeG. Huisken, Invent. Math. 84 (1986), 463–480. Dumbbells can developsingularities, see M. A. Grayson, Duke Math. J. 58 (1989), 555–558. Thefull classification of singularities is open despite important partial results,

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see, e.g., Huisken and Sinestrari, Acta Math. 183 (1999), 45–79 and Invent.Math. 175 (2009), 137–221.

In the analysis of the mean curvature flow maximum principles play acrucial role. The simplest one is a geometric maximum principle. If F ′

0 iscontained in the open bounded set enclosed by F0 then the relation holdsat all later times (as long as both flows exist and are smooth). There aremore subtle maximum principles for the curvatures of the surfaces. Roughlyspeaking these guarantee that the surfaces become rounder along the flow(i.e., the ratio of the largest and smallest principal curvature approaches 1).

1.1.3 Ricci flow

The mean curvature flow describes the evolution of a family of surfaces in anambient space. The Ricci flow describes the evolution of a family of metricson a fixed manifold. Under suitable conditions both make the geometrynicer as the flow evolves.

Consider a family of Riemannian metrics gt on an n-dimensional mani-fold M . Then the Ricci flow is given by the equation

∂tgt = −2Ricg (1.10)

where Ricg is the Ricci curvature of the metric g which depends affinely onthe second derivatives of g. The Ricci flow should be viewed as a kind of heatequation for the metric g. Nonetheless the equation (1.10) is in general nota parabolic equation because the right hand side is not a positive definiteexpression in the second derivatives of g (this is related to the fact thatthe flow is invariant under a change of coordinates). If one makes a goodchoice of coordinates (a ’gauge’) then the equation becomes parabolic. Inthe simplest case n = 2 one can always choose conformal coordinates, i.e.,one can choose coordinates for which the metric has the form

gαβ = e2pδαβ. (1.11)

Then (1.10) becomes∂tp = e−2p∆p = ∆gp, (1.12)

where ∆g denotes the Laplace-Beltrami operator for the metric g. TheGauss curvature K and the scalar curvature R are given by

R = 2K = −2e−2p∆p (1.13)

and thus∂tR = R2 + e−2p∆R = R2 + ∆gR. (1.14)

In this form one sees the competition between the ODE ∂tR = R2 whichwould lead to blow-up in finite time and which enhances differences betweendifferent spatial values of R and the heat equation ∂tR = ∆R which makes R

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smoother in space as time evolves and reduces the difference between valuesat R at different spatial points. Fundamental results about the Ricci flowcentered around the idea that after suitable rescaling the flow converges toa metric of constant (scalar) curvature and that possible singularities canbe controlled.

The most spectacular application of Ricci flow is certainly Perelman’sproof of the Poincare conjecture (see Tao’s paper arXiv:math/0610903 fora high-level overview of Perelman’s proof from a PDE perspective and thebook ’Ricci flow and the Poincare conjecture’ by J. Morgan and G. Tian fora full exposition).

Another striking recent application of the Ricci flow is the proof of thedifferentiable version of the Klingenberg sphere theorem (every simply con-nected compact manifold with section curvatures strictly between 1/4 and1 is diffeomorphic to the sphere3).

1.1.4 Minimal surfaces

These are by definition stationary points of the mean curvature flow, i.e.,surfaces which satisfy

H = 0. (1.15)

For graphs this reduces to the PDE

div∇u√

1 + |∇u|2= 0, (1.16)

which is elliptic as long as |∇u| ≤M .The equation (1.16) arises as the Euler-Lagrange equation of the area

functionalA(u) =

∫Ω

√1 + |∇u|2 dx. (1.17)

1.1.5 Harmonic maps

Let Ω ⊂ Rn be open, let M ⊂ Rm be a submanifold and consider mapsu : Ω → M . A harmonic map is a stationary point (i.e., a solution of theEuler-Lagrange equation) of the Dirichlet integral

E(u) =12

∫Ω|∇u|2 dx. (1.18)

If M = Rn the Euler-Lagrange equation reduces to −∆u = 0 (hence thename). In general a (sufficiently regular) map is harmonic if and only if

−∆u = A(u)(∇u,∇u), (1.19)3see S. Brendle, R. Schoen, Bull. Amer. Math. Soc. 48 (2011), 1–32 and J. Amer. Math.

Soc. 22 (2009), 287–307, C. Bohm and B. Wilking, Ann. Math. 167 (2008), 1079–1097.

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where A(p) is a certain quadratic form with values in the normal space(TpM)⊥. If M = Sm−1 then u is a harmonic map if and only if

−∆u = |∇u|2u. (1.20)

It has long been open whether weak solutions of this equation are smooth.Finally in the 1990’s F. Helein showed that this is indeed the case if n = 2(C.R. Acad. Sci. Paris Math. 311 (1990), 519–524, while Riviere con-structed a counterexample with a dense set of singularities for n = 3 (ActaMath. 175 (1995), 197–226). Under a mild additional assumption for n ≥ 3one can show that the set of singularities is at most a closed set whose n−2dimensional Hausdorff measure is zero. We will discuss these results later inthe course. An important feature which makes the analysis of the regularityof solutions very delicate is that (1.20) is invariant under rescaling. Moreprecisely if we define

ur(x) := u(rx) (1.21)

then ur satisfies the same equation as u. If instead we consider the equation

−∆v = |∇v|pv with 1 ≤ p < 2 (1.22)

then the rescaled functions vr satisfy the equation

−∆vr = r2−p|∇vr|pv, (1.23)

i.e., by rescaling small scales back to 1 the equation becomes more linear andhence easier. For geometric problems, however, often scaling invariance is anatural feature of the problem and one often needs to develop new analyticaltools since one is at the borderline of classical compactness and regularitymethods.

1.1.6 Nonlinear elasticity

Let Ω ⊂ Rn. To an elastic deformation u : Ω → Rn we assign an energy

E(u) =∫

ΩW (∇u) dx. (1.24)

The corresponding Euler Lagrange equation is

−divDW (∇u) = 0 (1.25)

or, in components,

−n∑

α=1

∂

∂xα∂W

∂Fiα(∇u(x)) = 0 for all i = 1, . . . , n. (1.26)

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This is a system of PDE and we will discuss in the next section when thissystem is elliptic (the harmonic map equations are also a system of PDEbut in this case the term in the second order derivatives is diagonal).

The corresponding time dependent equation is given by

ρ0(x)∂2t u(t, x) = divDW (∇u). (1.27)

If W (∇u) = 12 |∇u|

2 and ρ0 ≡ 1 this becomes a system of n decoupledwave equations ∂2

t uj = ∆uj for j = 1, . . . , n. In general the system (1.27)

is a nonlinear system of wave equations. We will consider such hyperbolicsystems in NPDE II in the summer term.

As a side remark note that the physical interpretation of (1.27) is justNewton’s third law. On the right hand side DW (∇u) is the Piola-Kirchhoffstress tensor and its divergence is a force density (force per unit volume).On the left hand side ρ0 is a mass density (mass per unit volume) and ∂2

t u isthe acceleration. Thus the equation is a continuum version of Newton’s law:force equals mass times acceleration. A systematic derivation of (1.27), theNavier-Stokes equation and related equations was discussed in the summerterm 2013 in the course ’PDE and modeling’. A good reference is the book’Introduction to continuum mechanics’ by M.E. Gurtin, Academic Press.

1.2 Some general strategies

1.2.1 Soft functional analytic and topological methods

The idea is to view the left hand side of a PDE as a map between Banachspaces and to use the inverse function theorem or topological fixed pointtheorems. The basic idea from the point of PDE is that good estimatesimply existence of solutions.

We begin with a simply result for a family for linear operators whichillustrates this idea.

Lemma 1.1. Let X be a Banach space and let Y be normed space and letL0 and L1 be bounded linear operators from X to Y . For each t ∈ [0, 1] set

Lt = (1 − t)L0 + tL1

and suppose that there exists a constant C such that

∥x∥X ≤ C∥Ltx∥Y ∀x ∈ X, t ∈ [0, 1].

Then L1 maps X onto Y if and only if L0 maps X onto Y .

Proof. See [GT], Theorem 5.2. Idea: Assume that Ls is onto for somes ∈ [0, 1]. Now rewrite the equation Ltx = y as

x = L−1s y + (t− s)L−1

s (L0 − L1)x

and use the Banach fixed point theorem to show that this equation is solvablefor |s−t| < δ. Finally subdivide [0, 1] into intervals of length less than δ.

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New we recall the inverse function theorem which shows that a nonlinearproblem can be solved (for small data) if we have a good estimate for thelinearized problem.

Theorem 1.2. Let X and Y be Banach spaces. Let U ⊂ X be open andassume that A ∈ Cr(U, Y ) with r ≥ 1, i.e., A is r times Frechet differentiablewith continuous derivatives. Let u0 ∈ U , f0 := A(u0) and assume that thedifferential

DA(u0) : X → Y is an invertible map. (1.28)

Then there exist ε, δ > 0 such that

For each f ∈ Bε(f0) ⊂ Y there exists a unique u ∈ Bδ(u0) ⊂ X with

Au = f (1.29)

Moreover the map f 7→ u is a Cr map.

Application to PDEs. Let Ω ⊂ Rn be bounded with smooth boundary.Let g ∈ C1(R) with bounded derivative and consider the boundary valueproblem

−∆u+ g(u) = f in Ω, (1.30)u = 0 on ∂Ω. (1.31)

We first make the choice

X = u ∈ C2(Ω) : u|∂Ω = 0, (1.32)

Y = C(Ω) (1.33)

and we setA(u) = −∆u+ g u. (1.34)

It is easy to check that A ∈ C1(X,Y ) and that the differential at u0 = 0 isgiven by

Lv := DA(0)v = −∆v + g′(0)v. (1.35)

The invertibility of L is equivalent to the following assertion for a linearboundary value problem. For every f ∈ Y there exists a unique v ∈ X suchthat

−∆v + g′(0)v = f in Ω, (1.36)v = 0 on ∂Ω. (1.37)

Moreover there exists a constant C such that for all f ∈ Y

∥v∥X ≤ C∥f∥Y . (1.38)

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Assume first thatg′(0) = 0. (1.39)

It turns out that with the above choice of X and Y the linear map L is notinvertible (see, e.g., the lecture notes [FAPDE], Proposition 7.9).

As we shall see in the next section the operator L is invertible if we workinstead in the Holder spaces

X = u ∈ C2,α(Ω) : u|∂Ω = 0, (1.40)

Y = C0,α(Ω) (1.41)

with 0 < α < 1 (note that a Holder continuous function in Ω has a uniqueextension to Ω so that the condition u|∂Ω = 0 makes sense). Thus thereexist ε, δ > 0 such that

If ∥f∥C0,α < ε then the problem (1.30)– (1.31) (with g′(0) = 0)has a unique solution u with

∥u∥C2,α < δ. (1.42)

If g′(0) = 0 the same conclusion holds by the Fredholm alternative forelliptic second order PDE (see [GT], Thm. 8.6 or [Ev], Chapter 6, Thm. 4and 5 or the lecture notes [FAPDE]), as long as g′(0) is not an eigenvalueof the operator ∆. Strictly speaking the Fredholm alternative first providesa solution in the Sobolev space W 1,2

0 . The fact that the solution is in C2,α

then follows from the regularity theory we will develop in the next section.[17.10. 2013, Lecture 1][21.10. 2013, Lecture 2]

In a similar way one can use the inverse function theorem to show that thereexist solutions for the equations of nonlinear elasticity (1.25) which in C2,α

are close to the identity. In fact until J.M. Ball’s paper on a variationalapproach in 1977 this was the only way to show the existence of solutionsunder physically realistic conditions on W .

Remark. 1. There is a natural version of the inverse function theoremfor Banach manifolds. This allows one to handle problems with constraintssuch as the harmonic map problem.2. It suffices that DA(u0) is bijective. Then the boundedness of the inversefollows from the open mapping theorem.3. In some cases the operator L := DA(u0) is not invertible as a map fromX to Y , but only has an inverse which maps into a weaker space than X (’loss of regularity’). Under additional assumptions it is still possible to solvethe nonlinear problem through the so called ’hard inverse function theorem’of Nash and Moser4.

4see R.S. Hamilton, Bull. Amer. Math. Soc. 7 (1982), 65–222 or the book S. Alinhac,P. Gerard, Operateurs pseudo-differentiels et theoreme de Nash-Moser.

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Warning The assumption A ∈ C1(U, Y ) is not always trivial. Note forexample that the map

u 7→ sin u (1.43)

is differentiable (and in fact C∞) as map from L∞(0, 1) to itself, but thatthis map is not differentiable as a map from L2(0, 1) to itself. In fact wehave

Proposition 1.3. Let E ⊂ Rn be measurable with 0 < Ln(E) < ∞, let1 ≤ p <∞, let f : R → R and consider the Banach space X = Lp(E). Thenthe map

u 7→ f u (1.44)

is Frechet differentiable as a map from X to X if and only if f is affine.

Remark. Similarly one sees that if Ln(E) = ∞ then f must be linear.

Proof. The ’if’ part is clear. For the ’only if’ part first note that by consid-ering constant functions u one sees that f must be a differentiable function.By subtracting an affine function we may assume that f(0) = f ′(0) = 0. Itremains to show that f ≡ 0. This is left as an exercise.

Roughly speaking, the inverse function theorem can only be used inspaces which are already sufficiently strong with respect to the nonlinearitiesinvolved.

The inverse function theorem gives existence and uniqueness of solutions(in the open set where it applies). This is good, but may be more than wecan expect in particular if we also want to look for ’large’ solutions. Inthis case topological fixed point theorems which only give existence are veryuseful. We start with a fixed point theorem in finite dimensional spaces.

Theorem 1.4 (Brouwer’s fixed point theorem). Let B ⊂ Rn be the closedunit ball. Assume that f : B → B is continuous. Then f has a fixed point,i.e., there exists x ∈ B such that

f(x) = x. (1.45)

Remark. Clearly B can be replaced by a set which is homeomorphic toB.

Proof. For a short PDE proof see, e.g., [Ev] Chapter 8.1, Thm. 3. Seealso Problems 4. and 5. in Chapter 8 for the definition of the topologicaldegree.

Now we look for fixed point theorems in an (infinite-dimensional) Banachspace X. Here in addition to continuity we need some compactness.

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Theorem 1.5 (Schauder). Let S ⊂ X be compact, convex, and not empty.Let T : S → S be continuous. Then T has a fixed point.

Proof. See [GT], Thm. 11.1

Remark. It suffices that S is not empty and homeomorphic to a compactand convex set.

Corollary 1.6. Let S ⊂ X be closed, convex and not empty. Let T : S → Sbe continuous and assume that T (S) is precompact. Then T has a fixedpoint.

Proof. See [GT], Cor. 11.2

Theorem 1.7 (Leray-Schauder, special case). Suppose that T : X → X iscompact, i.e., the for every bounded set in X the image under T is precom-pact. Suppose that there exists R > 0 with the following property.

∀σ ∈ (0, 1) if x = σTx then ∥x∥ < R. (1.46)

Then T has a fixed point.

Proof. Define

T (x) :=

Tx if ∥Tx∥ ≤ RR

∥Tx∥Tx if ∥Tx∥ > R.(1.47)

Then T is continuous and compact and maps the closed ball B(0, R) toitself. Hence T has a fixed point x. If ∥Tx∥ ≤ R then Tx = T x = x and weare done. If ∥Tx∥ > R then ∥T x∥ = R and

x = T x =R

∥Tx∥Tx. (1.48)

Hence by assumption (with σ := R/∥Tx∥ < 1)

R = ∥T x∥ = ∥x∥ < R. (1.49)

This contradiction finishes the proof.

Application to PDEs. Let Ω ⊂ Rn be open and bounded with smoothboundary and let φ ∈ C2,α(Ω) with 0 < α < 1. Let g ∈ C1(R) with boundedderivative. Consider the boundary value problem

−∆u+ g(u) = 0 in Ω, (1.50)u = φ on ∂Ω. (1.51)

For v ∈ C0,α(Ω) consider the linear PDE for u

−∆u+ g(v) = 0 in Ω, (1.52)u = φ on ∂Ω. (1.53)

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Assume for a moment that this linear PDE has a unique solution u. Wenow define a map T by Tv := u.

Then u is a solution of the nonlinear problem if and only if it is a fixedpoint of T , i.e., if

T u = u. (1.54)

It is easy to see that g v in C0,α(Ω) and that the map v 7→ g v isa continuous map from C0,α(Ω) to itself. It follows from the linear theorywhich we will develop in the next section that T is a continuous map fromC0,α(Ω) to C2,α(Ω). The embedding C2,α(Ω) → C0,α(Ω) is compact (Arzela-Ascoli !) and thus

T : C0,α(Ω) → C0,α(Ω) is compact. (1.55)

Thus it suffices to show that

∃R > 0 ∀σ ∈ (0, 1) Tu =u

σ=⇒ ∥u∥C0,α < R. (1.56)

By the definition of T we thus need to show that if

−∆u

σ+ g(u) = 0 in Ω, (1.57)

u

σ= φ on ∂Ω (1.58)

then ∥u∥C0,α < R. Or, equivalently: there exists R > 0 such that if σ ∈ (0, 1)and if u solves

−∆u+ σg(u) = 0 in Ω,u = σφ on ∂Ω

(1.59)

then ∥u∥C0,α < R. Thus the problem of existence is reduced to the problemof showing a uniform estimate if a solution exists. Such estimates are knownas a priori estimates. Proving a priori estimates will be a key element ofthis course.

1.2.2 Weak convergence methods and compactness

Here is a general strategy to solve nonlinear PDE written symbolically asA(u) = f

(i) Consider approximative problems (e.g. finite dimensional approxima-tion, regularization by higher order terms, . . . )

Ak(uk) = fk with fk → f, Ak → A in a suitable sense. (1.60)

(ii) Derive a priori estimates

∥uk∥X ≤ C. (1.61)

14 [March 17, 2014]

(iii) Deduce (weak) convergence of a subsequence in a suitable topology τ

ukτ→ u. (1.62)

(iv) Show A(u) = f (compactness/ stability argument).

The crucial difficulty in the last step is that usually weak convergencedoes not commute with the action of nonlinear functions. This problem canbe overcome in several ways.

We can try to prove a priori estimates in a strong enough space to ensurestrong converge of the quantities which enter in the nonlinearities, e.g., ifthe nonlinearities only involve terms with lower order derivatives.

We can also use the fact that uk is an approximate solution to get strongconvergence. Think of the result that in a Hilbert space weak convergenceuk u plus convergence of the norm ∥uk∥ → ∥u∥ implies strong conver-gence. This follows simply by expanding the ∥uk − u∥2. The deeper reasonbehind this result is that Hilbert spaces are uniformly convex and the as-sertion holds more generally in uniformly convex Banach spaces like Lp (for1 < p < ∞). Convexity arguments or closely related monotonicity argu-ments5, are very powerful tools to show that A(u) = f , but convexity ormonotonicity are often not available.

We can also try to use additional structures to show that not all, butcertain nonlinear quantities pass to the limit. In the course PDE and mod-eling we showed, e.g., that the weak convergence uk u in W 1,p(Rn,Rn)implies that det∇k det∇u in Lp/n if p > n. This is the idea of themethod of compensated compactness developed by F. Murat and L. Tartar(the original French name ’compacite par compensation’ describes the ideaof the theory better).

The proof of existence via subsolutions and supersolutions (Perron’smethod) can be seen in that framework, too. The approximation consistsin replacing the equation by an inequality (understood in a rather weaksense). The necessary bounds on the class of all subsolutions come from amaximum principle (or more generally from a comparison principle). Weakconvergence is replaced by taking the supremum of all subsolutions. Thenthe key point is to show that this yields actually a solution.

In this course we will see a number of strategies to implement weakconvergence methods. A beautiful short survey of a number of importantmethods can be found in the book L.C. Evans, Weak convergence methodsfor nonlinear partial differential equations, Amer. Math. Soc., 1990. Theclassic book J.L. Lions, Quelques methodes de resolution des problemes auxlimites non lineaires, Dunod; Gauthier-Villars, 1969, which also emphasizesthe aspect of compactness is still very enjoyable to read.

5see Section 2.4.3

15 [March 17, 2014]

1.2.3 Variational methods

Let X be a Banach space and let I : X → R be a functional on X. We saythat u ∈ X is a stationary point of I if I is (Frechet) differentiable at u andthe derivative satisfies

DI(u) = 0. (1.63)

If u0 is a minimizer of I (in X or at least in some open neighbourhood of u0)and if I is differentiable at u0 then u0 is a stationary point. Many interestingPDE arise as stationary points of suitable functionals (these PDE are thencalled variational).

Example. Let Ω ⊂ Rn be bounded and open, let X = W 1,20 (Ω) and let

G ∈ C1(Rn). For simplicity of notation let n ≥ 3 and assume that

|G(s)| ≤ C(1 + |s|2∗), where 2∗ =2nn− 2

(1.64)

is the critical exponent in the Sobolev embedding theorem W 1,20 (Ω) →

L2∗(Ω). Consider on X the functional

I(u) =∫

Ω

12|∇u|2 +G(u) dx. (1.65)

If we assume in addition that

|G′(s)| ≤ C(1 + |s|2∗−1) (1.66)

then one can show (exercise !) that I : X → R is Frechet differentiable andthe derivative is given by

DI(u)φ =∫

Ω∇u · ∇φ+G′(u)φdx ∀φ ∈W 1,2

0 . (1.67)

Thus if u is a minimum of I then u is a weak solution of the PDE

−∆u+G′(u) = 0. (1.68)

This equation is called the Euler-Lagrange equation of I.Here is a fine technical point. One can make a connection between

minima of I and the PDE even if we replace (1.66) by the weaker condition

|G′(s)| ≤ C(1 + |s|2∗). (1.69)

Under this condition one can still show that I has directional derivatives ingood directions. More precisely for all φ ∈ C1

c (Ω) the functions t 7→ I(u+tφ)are differentiable (exercise !) and

d

dt |t=0I(u+ tφ) =

∫Ω∇u · ∇φ+G′(u)φdx ∀φ ∈ C1

c (Ω). (1.70)

16 [March 17, 2014]

Thus under assumptions (1.69) and (1.64) a minimizer u of I is still a dis-tributional solution of (1.68).

What are the advantages of the variational formulation ? One point isthat the variational formulation is much more robust. It only involves firstderivatives and not second derivatives and the existence of minimizers (evenfor rather ’rough’ integrands) can often be established by applying the di-rect method of the calculus of variations in the Sobolev space W 1,p(Ω) (ora subspace thereof). Another point is that solutions of the Euler-Lagrangeequation which come from minimizers (or local minimizers) often enjoy im-portant additional properties. We will see this later, e.g., in the discussionof the regularity of harmonic maps.

There are other important stationary points beyond minimizer. If X isone dimensional then between two minima of a function there is always alocal maximum. In higher dimensions one can argue that between two localminima there is a saddle point or mountain-pass. We say that x is a strictlocal minimum of a function E : X → R if there exists an open ball B(x, r)such that E(x) ≤ E(y) for all y ∈ B(x, r) and inf∂B(x,r)E > E(x).

Theorem 1.8 (Mountain pass theorem). Suppose that E ∈ C1(Rn) is coer-cive, i.e., limx→∞E(x) = ∞. Suppose further that E has two distinct strictlocal minima x1 and x2. Then E possesses a third critical point x3 which isnot a local minimum and characterized by the minmax principle

E(x3) = infp∈P

maxx∈p

E(x) (1.71)

where

P := p ⊂ Rn : x1, x2 ∈ p, p is compact and connected. (1.72)

There is a version of the theorem for C1,1 functionals E (i.e. maps whosefirst Frechet derivative is Lipschitz continuous) on Banach spaces providedthat E satisfies a so-called Palais-Smale condition. We say that sequence aumm∈N ⊂ X is a Palais-Smale sequence if

|E(um)| ≤ C and limm→∞

∥DE(um)∥X∗ = 0 (1.73)

The Palais-Smale condition is:

every Palais-Smale sequence contains a convergent subsequence. (1.74)

Note that in finite dimensional spaces this follows from coercivity since thenthe closure of bounded sets is compact.

An excellent reference on minmax method and on variational methods ingeneral is the book M. Struwe, Variational methods, Springer, 4th edition,2008. This book also contains many beautiful applications to problems ingeometry.

17 [March 17, 2014]

1.2.4 Convex integration

Ideas related to what is now called convex integration first appear in J.Nash’s seminal work on ’wild’ isometric immersion with minimal regular-ity and were later developed into a systematic theory by Gromov who alsocoined the name6. The method was first applied mostly to geometric equa-tions but recently also led to deep new insights in PDEs coming from physics,including in particular the Euler equations of hydrodynamics. L. Szekely-hidi’s lecture notes7 provide an excellent short survey.

The idea is initially somewhat counterintuitive since it relies to someextend on a certain lack of compactness while we usually argue that com-pactness is crucial for existence in nonlinear PDE. One replaces the originalproblem first by a ’relaxed’ problem, usually some sort of convexificationof the original problem which is much easier to solve. Then one iterativelyadds the right (usually one-dimensional) oscillations to the solution of therelaxed problem to obtain a solution of the original problem. The initiallysurprising fact is that this sequence of iterations actually converges stronglyto a solution of the original problem (one might expect that due to the os-cillations one only gets weak convergence and everything is lost again in thelimit). For those who know probability this argument is similar to the proofthat martingales converge.

To explain the idea let us just consider a toy example. Let Ω ⊂ Rn beopen and bounded. We look for a (Lipschitz) function u : Ω → R such that

|∇u| = 1 a.e. in Ω, (1.75)u = 0 on ∂Ω. (1.76)

The first condition can be written as a partial differential relation

∇u(x) ∈ Sn−1 ⊂ Rn. (1.77)

If we replace Sn−1 by its convex hull, the closed unit ball, we obtain therelaxed problem

|∇u| ≤ 1 a.e. in Ω, (1.78)u = 0 on ∂Ω. (1.79)

This has the trivial solution u0 = 0. Now we want to add oscillations to u0

to obtain a solution of the original problem. There are many ways to dothat for this concrete problem. Here is one possibility. Consider for ease ofimagination the case n = 2. Let Ωk := x ∈ Ω : dist (x,Rn \ Ω) ≥ 2−k. In

6see the book M. Gromov, Partial differential relations, Springer, 1986, which is toughreading but contains a wealth of deep ideas.

7From isometric embeddings to turbulence, Preprint MPI MIS Leipzig,http://www.mis.mpg.de/preprints/ln/lecturenote-4112.pdf

18 [March 17, 2014]

the first step choose finitely many disjoint triangles (contained in Ω) whichcover the compact set Ω1. On each triangle T choose a piecewise linear hatfunction vT which satisfies |∇vT | = 1 and agrees with u0 on the boundaryof T . Replacing u0 by vT on all triangles we obtain the next iterationu1. For the next step add further triangles so that Ω2 is covered and dothe replacement on this triangles. In the limit this yields a function whichsolves the original problem |∇u∞| = 1 a.e. and u∞ = 0 on ∂Ω.

This toy problem is actually too simple. We could have done the replace-ment in one step, e.g. by taking u as the distance function u = dist (x, ∂Ω).What makes this problem easy is that we can do directly a multidimen-sional replacement and in one step replace an affine function on a triangleby a piecewise affine one with the same boundary conditions such that thenew function already solves the original problem. In typical applications ofconvex integration one can only add oscillation in one spatial direction attime. This makes it impossible to satisfy affine boundary conditions on thetriangle exactly. Thus some additional errors arise in a thin region near theboundary of the triangle. To make sure that these errors are at least com-patible with the relaxed problem (and thus can be eventually removed inlater steps) we can usually not choose the one-dimensional piecewise affinemaps so that the gradients are exactly on set where we want them to be(in our example Sn−1) but we have to stay a little inside the interior of therelaxed set (in our example the unit ball Bn). Thus in each step on eachtriangle the gradient is only pushed closer to our desired set and the trian-gles have to be refined for the next step. Then the final strong convergenceof the gradients (a.e.) is less obvious, but also not difficult to show if thenew oscillations are always chosen much faster than the previous ones.

There is also a variant of this construction where one adds smooth oscil-lations rather than piecewise affine ones (this is indeed what Gromov does)but the issues are very similar.

[21.10. 2013, Lecture 2][24.10. 2013, Lecture 3]

1.3 Overview of a priori estimates

Let A denote a partial differential operator. As we have seen a priori esti-mates of the form

Au = f, ∥f∥... ≤ R1 =⇒ ∥u∥... ≤ R2. (1.80)

play a crucial role in the theory.Below is a rough overview of some important spaces and techniques.

This list is be no means exhaustive. To prove estimates in the Holder spacesCα one can, e.g., also so the representation of the solution in terms of aGreen’s function (see, e.g., [GT]).

19 [March 17, 2014]

L2 use test functions, most robustL∞ max. principle or Sobolev embedding

Lp, 1 < p <∞ harmonic analysis orCampanato spaces + interpolation

Cα Campanato spaces,or blow-up & compactness

BMO substitute for L∞, Campanato spacesor harmonic analysis

H1 substitute for L1, harmonic analysisor duality with BMO

parabolic space-time elliptic + semigroup ideas,harmonic analysis,damping of high frequencies

hyperbolic space-time L2 + Strichartz est., wave packetsdisperse due speed differences

While a lot can be achieved for nonlinear PDE using mostly good esti-mates for linear PDE for the most difficult and interesting nonlinear PDEone often has to find and use additional estimates which depend on thenonlinear structure.

For the Navier-Stokes equation, e.g., multiplication of the equation byv, integration in x and integration by parts yields the crucial global energyestimate

12

∫Rn

|v|2(T, x) dx+∫ T

0

∫Rn

|∇v|2 dxdt ≤ 12

∫Rn

|v0|2(x) dx ∀T > 0.

(1.81)This estimate depends upon the fact that for the most dangerous term weget (after a clever integration by parts and using that div v = 0)∫

Rn

n∑j=1

n∑i=1

[vi

∂

∂xivj]vj dx = 0. (1.82)

There is no similar estimate for the derivatives of v. One only gets the weakerestimates d

dt12∥v∥

2Wk,2 ≤ C∥v∥3

Wk,2 − ν∥∇v∥2Wk,2 . These estimates are useful

to construct local-in-time solutions but they do not rule out blow-up of thehigher norms in finite time.

Important nonlinear estimates are often motivated by insights from phy-sics about the conservation or increase or decrease of certain quantities suchas energy or entropy. One should note, however, that the derivation ofestimates like (1.81) assumes a certain degree of regularity and that theydo not automatically hold for weak solutions. Indeed for the Euler equationfor which the same calculation shows that ∥v(t, ·)∥L2 should be constant int there exist weak solutions which very strongly violate the energy estimate.

20 [March 17, 2014]

For the Ricci flow one key insight of Perelman was the constructionof certain nonlinear expression (in the spirit of a physical entropy) whichdecrease along every solution. Using these he could in particular rule outcertain potential singularities by showing that to form these this quantitywould have to go to ∞.

1.4 Prerequisites

Some basic knowledge in Functional analysis and the theory of Sobolevspaces and weak solutions is required. For participants with no previousexperience in these areas it is recommend to take the course ’FunctionalAnalysis and PDE’ in parallel. The material needed is more or less coveredin

• [GT] Chapter 5, Chapter 7.1-7.7 and 7.10 and Chapter 8.1 and 8.2, or

• [Ev] Chapter 5.1-5.8, Chapter 6.1 and 6.2 and Appendices C, D andE.

Other useful books include

• H.W. Alt, Linear Funktionalanalysis, Springer, 5th edition, 2006.

• H. Brezis, Functional analysis, Sobolev spaces and partial differentialequations, Springer, 2011.

2 Elliptic equations and systems

2.1 Energy methods: Lp and Cα estimates for elliptic equa-tions and systems

This subsection follows very closely Chapter III of the book

• M. Giaquinta, Multiple integrals in the calculus of variations and non-linear elliptic systems, Princeton Univ. Press, 1983.8

In the following we always assume that

Ω ⊂ Rn is open and bounded. (2.1)

Sometimes additional assumptions on Ω are needed. These will be statedexplicitly.

8A more detailed exposition can be found in the original paper Sergio Campanato,Equazioni ellittiche del II0 ordine e spazi L2,λ, Annali di Matematica Pura Appl (4) 69(1965), 321–381.

21 [March 17, 2014]

2.1.1 Overview

In the following we disucss the approach of Campanato to the regularitytheory for elliptic systems. The whole approach is based on three simpleinequalities: the L2 energy estimate, the reverse Poincare inequality andthe Poincare inequality. Their combination turns out to be surprisinglypowerful. We first illustrate the idea by looking at the Poisson equation andharmonic functions.

Energy estimate for the boundary value problem Suppose that f ∈L2(Ω) and that u is a weak solution of the boundary value problem

−∆u = −div f in Ω,u = 0 on ∂Ω.

The weak form of this boundary value problem is u ∈W 1,20 (Ω) and∫

Ω∇u · ∇φdx =

∫Ωf · ∇φdx ∀φ ∈W 1,2

0 (Ω) (2.2)

and the choice φ = u yields, in combination with the Cauchy-Schwarz in-equality, ∫

Ω|∇u|2 dx =

∫Ωf · ∇u dx ≤ ∥f∥L2 ∥∇u∥L2 (2.3)

and thus the energy estimate∫Ω|∇u|2 dx ≤

∫Ω|f |2 dx. (2.4)

Reverse Poincare estimate Now let Br = B(x0, r) ⊂ Rn be an openball in Rn. We assume that v is weakly harmonic in Br , i.e., a weak solutionof

−∆v = 0 in Br, (2.5)

but we do not impose any boundary conditions on v. The weak form of theequation is ∫

Br

∇u · ∇φdx = 0 ∀φ ∈W 1,20 (Ω). (2.6)

Now we cannot choose φ = v since we do not now that v has zero boundaryconditions. Thus we consider s ∈ (0, r) and a cut-off function η ∈ C∞

c (Br)with

0 ≤ η ≤ 1, η = 1 in Bs, |∇η| ≤ 2r − s

. (2.7)

Let a ∈ R and let φ = η2(v − a). Since ∇v = ∇(v − a) we get

∇(η2v) = η2∇v + 2η(v − a)∇η. (2.8)

22 [March 17, 2014]

and the weak form of the equation yields∫Br

η2|∇v|2 dx = −∫Br

2∇v · η(v − a)∇η dx

≤ 2(∫

Br

η2|∇v|2 dx)1/2(∫

Br

|∇η|2|v − a|2 dx)1/2

.(2.9)

Using that η = 1 in Bs and thus ∇η = 0 in Bs it follows that∫Bs

|∇v|2 dx ≤ 16(r − s)2

∫Br\Bs

|v − a|2 dx ∀a ∈ R, s ∈ (0, r). (2.10)

This inequality is called reverse Poincare inequality because it provides anestimate of ∇v in terms of v whereas the usual Poincare inequality (seebelow) provides an estimate for v in terms of ∇v. Some authors also callthis inequality a Cacciopoli inequality.

Poincare inequality

Theorem 2.1 (Poincare inequality for zero mean). Let Ω ⊂ Rn be open,bounded and connected, with Lipschitz boundary. Let 1 ≤ p ≤ ∞. Thenthere exists a constant C(Ω, p) such that

∥u− uΩ∥Lp(Ω) ≤ C(Ω, p)∥∇u∥Lp(Ω) ∀u ∈W 1,p(Ω), (2.11)

whereuΩ :=

1Ω

∫Ωu dx . (2.12)

In particular for 1 ≤ p <∞ there exist constants C1(p) such that∫Br\Br/2

|u− uBr\Br/2|p dx ≤ C1(p)rp

∫Br\Br/2

|∇u|p dx (2.13)

∫Br

|u− uBr |p dx ≤ C1(p)rp∫Br

|∇u|p dx. (2.14)

Remark. This can be combined with the Sobolev embedding theorem.For p ≤ n, 1

q ≤ 1p − 1

n and q = ∞ we thus obtain the Sobolev-Poincareinequality

∥u− uΩ∥Lq(Ω) ≤ C(Ω, p)∥∇u∥Lp(Ω) ∀u ∈W 1,p(Ω), (2.15)

Notation: we write

ux0,r as a shorthand for uB(x0,r). (2.16)

23 [March 17, 2014]

Proof. We may assume that uΩ = 0 since otherwise we can consider thefunction v = u− uΩ. Assume that the estimate (2.11) does not hold. Thenfor k ∈ N there exist uk ∈W 1,p(Ω) such that

∥uk∥Lp ≥ k∥∇uk∥Lp ,

∫Ωuk dx = 0. (2.17)

Set vk = uk/∥uk∥Lp . Then

∥vk∥Lp = 1, ∥∇vk∥Lp → 0,∫

Ωvk dx = 0. (2.18)

In particular the functions vk form a bounded set in W 1,p(Ω). Hence by thecompact Sobolev theorem there exists a subsequence vkj

such that vkj→ v

in Lp(Ω) as j → ∞. Since we also know that ∇vkj→ 0 in Lp(Ω; Rn) it

follows that vkj→ v in W 1,p(Ω) and ∇v = 0. Thus we get

∥v∥Lp = 1, ∇v = 0∫

Ωv dx = 0. (2.19)

Since Ω is connected the condition ∇v = 0 implies that v is constant in Ω.The condition

∫Ω v dx = 0 then implies that v = 0. This contradicts the fact

that ∥v∥Lp = 1. This contradiction finishes the proof of (2.11).Thus (2.13) and (2.14) hold for r = 1. To obtain the result for general

r it suffices to apply the result for r = 1 the the rescaled function ur(z) =u(x0 + rz).

Consequences of the reverse Poincare inequality

Widman’s hole filler If we combine the inverse Poincare inequalitywith s = r/2, a = uBr\Br/2

with the Poincare inequality with zero mean weget ∫

Br/2

|∇u|2 dx ≤ C

r2

∫Br\Br/2

|u− uBr\Br/2|2 dx ≤ C

∫Br\Br/2

|∇u|2 dx

(2.20)Now we ’fill the hole’, i.e., we add C

∫Br/2

|∇u|2 dx to both sides. This yields

(C + 1)∫Br/2

|∇u|2 dx ≤ C

∫Br

|∇u|2 dx (2.21)

and thus∫Br/2

|∇u|2 dx ≤ θ

∫Br

|∇u|2 dx, where θ =C

C + 1< 1. (2.22)

In dimension n = 2 this inequality has very powerful consequences. Ifthis inequality (and the reverse Poincare inequality) holds for all balls thenwe have (see Homework sheet 1)

24 [March 17, 2014]

• Regularity: u is in C0,α for some α > 0 (which only depends on θ),

• Liouville theorem: If u is bounded then u is constant.

Remark. There is often a close relation between Liouville theorems andregularity.We shall see later that the assertion on regularity hold for solutions of scalarelliptic equations with L∞ coefficients also in dimensions n ≥ 3 (DeGiorgi,Nash, Moser theorem), but for systems the conclusion in general do not holdfor n ≥ 3

Reverse Holder inequality Let n ≥ 2 and 1p = 1

2 + 1n . Then 1 ≤ p <

2. Then the Sobolev Poincare inequality yields

∥u− uB1∥L2(B1) ≤ C∥∇u∥Lp(B1). (2.23)

If we combine this with the reverse Poincare inequality we get∫B1/2

|∇u|2 dx ≤ C

∫B)

|u− uB1 |2 dx ≤ C

(∫B1

|∇u|p dx)2/p

, (2.24)

i.e., we can bound the L2 norm of the gradient on a smaller ball, by aweaker Lp norm on a larger ball. Such an inequality is called a reverseHolder inequality.

By scaling by obtain the corresponding inequality for arbitrary radii:(1

Ln(Br/2)

∫Br/2

|∇u|2 dx

)1/2

≤ C

(1

Ln(Br)

∫Br

|∇u|p dx)1/p

, (2.25)

where 1p = 1

2 + 1n .

A nontrivial results states that the increase in integrability from p < 2to 2 can be lifted into an increase from 2 to some q > 2.

Theorem 2.2 (Higher integrability). Assume that u ∈ W 1,2loc (Rn) and that

there exists a C > 0 and p < 2 such that (2.25) holds for all balls inB(x0, r) ⊂ Rn. Then there exists a q > 2 (which only depends on p, Cand n such that u ∈W 1,q

loc (Rn) and for all balls(1

Ln(Br/2)

∫Br/2

|∇u|q dx

)1/q

≤ C ′(

1Ln(Br)

∫Br

|∇u|2 dx)1/2

(2.26)

Proof. See Giaquinta, [Gi], Chapter V, Theorem 1.1 and Theorem 1.2.

Such higher integrability estimates play an important role in the so calleddirect approach to regularity of nonlinear PDE, see Giaquinta [Gi], ChapterVI.

[24.10. 2013, Lecture 3][28.10. 2013, Lecture 4]

25 [March 17, 2014]

2.1.2 Existence of weak solutions

We now begin the study of general elliptic equations and system. To warmup we recall the results for scalar equations proved in the course FA andPDE. Let Ω ⊂ Rn be open and bounded. Let f : Ω → Rn and g : Ω → R begiven. We seek solutions u : Ω → R of the boundary value problem

Lu := −n∑

α,β=1

∂αaαβ∂βu = g −

n∑α=1

∂αfα in Ω, (2.27)

u = 0 on ∂Ω, (2.28)

where the coefficients aαβ are bounded and elliptic, i.e.,

aαβ ∈ L∞(Ω), (2.29)

∃ν > 0 ∀ξ ∈ Rn∑

aαβξαξβ ≥ ν|ξ|2. (2.30)

To simplify the notation we will use from now on the so called summa-tion convention, i.e.,

we always sum over indices which appear twice.

Definition 2.3. A function u ∈W 1,2 is a weak solution of (2.27) if∫Ωaαβ∂βu∂αφdx =

∫Ωfα∂αφ+ gφ ∀ φ ∈W 1,2

0 (Ω; Rn). (2.31)

Moreover u is a weak solution of the boundary value problem (2.27)–(2.28)if in addition u ∈W 1,2

0 (Ω).

Theorem 2.4. Assume that f ∈ L2(Ω,Rn), g ∈ L2(Ω) and assume thatthe coefficients satisfy (2.29) and (2.30) Then the boundary value problem(2.27)–(2.28) has a unique weak solution. Moreover this solution satisfies∫

Ω|∇u|2 dx ≤ 1

ν2

(∥f∥L2(Ω;Rn) + C(Ω)∥g∥L2(Ω)

)2 (2.32)

where C(Ω) is the constant in the Poincare inequality (2.35)

The proof is based on the Lax-Milgram theorem and the Poincare in-equality in W 1,2

0 (Ω).

Theorem 2.5 (Lax-Milgram). Let X be a real Hilbert space. Let a : X ×X → K be a bilinear form. Suppose that there exist constants c0, C0 > 0such that

(i) (continuity) |a(u, v)| ≤ C0∥u∥ ∥v∥ ∀u, v ∈ X;

(ii) (coercivity) Re a(u, u) ≥ c0∥u∥2.

26 [March 17, 2014]

Let T be a continuous linear functional on X. Then there exists one andonly one u ∈ X such that

a(u, v) = T (v) ∀v ∈ X. (2.33)

Moreover∥u∥X ≤ 1

c0∥T∥X∗ (2.34)

Theorem 2.6 (Poincare inequality for W 1,p0 (Ω)). For each bounded open

set Ω ⊂ Rn there exists a constant C(Ω) > 0 such that

∥u∥Lp(Ω) ≤ C(Ω)∥∇u∥Lp(Ω) ∀u ∈W 1,p0 (Ω). (2.35)

If Ω ⊂∏ni1

(ai, ai + li) we may take

C(Ω) = mini=1,...,n

li. (2.36)

Sketch of proof. Assume for simplicity p <∞. For n = 1 and Ω = (0, l) oneuses the fundamental theorem of calculus and the Holder inequality to get

|u(t)| ≤∫ t

0|u′(s)| ds ≤ t

p−1p ∥u′∥Lp . (2.37)

The assertion follows by taking the p-th power and integrating over t. Forn > 1 we extend u be zero to a function in W 1,p(Rn) and we use the onedimensional result and Fubini’s theorem. The proof for p = ∞ is similar.Details were discussed in the course FA and PDG.

Proof of Theorem 2.4. Consider the quadratic form

a(φ, u) :=∫

Ωaαβ∂βu∂αφdx (2.38)

and the linear functional

T (φ) :=∫

Ωfα∂αφ+ gφ (2.39)

Then a is bounded on W 1,20 (Ω) and T is a bounded linear functional on this

space. Indeed by the Poincare inequality and the Cauchy-Schwarz inequality

T (φ) ≤ ∥f∥L2∥∇φ∥L2 + C(Ω)∥g∥L2∥∇φ∥L2 (2.40)

It thus remains to show that a is coercive. By the ellipticity we have

a(u, u) ≥ ν

∫Ω|∇u|2 dx (2.41)

and the Poincare inequality yields

∥u∥2W 1,2 = ∥u∥2

L2 + ∥∇u∥2L2 ≤ (1 + C(Ω)2)∥∇u∥2

L2 . (2.42)

Thus a is coercive. The estimate for ∇u follows directly from (2.40) withφ = u and (2.41).

27 [March 17, 2014]

We now turn to elliptic systems. For i, j = 1, . . . , N and α, β = 1, . . . , nconsider functions fαi : Ω → R and gi : Ω → R and coefficients

Aαβij ∈ L∞(Ω). (2.43)

We look for solutions u : Ω → RN of the elliptic system

(Lu)i := −∂α(Aαβij ∂βuj) = gi − ∂αf

αi ∀i = 1, . . . , N. (2.44)

Here we use again the summation convention. The Greek indices α and βare summed from 1 to n and Latin index j is summed from 1 to N . Wewrite this system in concise notation as

Lu := −divA∇u = −div f + g (2.45)

where (AG)αi = Aαβij Gjβ , (∇u)jβ = ∂βu

j and where the divergence is takencolumnwise, i.e. (divG)i = ∂αG

αi . On the space RN×n of N × n matrices

we use the Euclidean scalar product an norm, i.e.,

(F,G) = F ·G = F iαFiα, |F |2 = F iαF

iα. (2.46)

We also define

∥A∥ := sup(F,AG) : |F | ≤ 1, |G| ≤ 1 (2.47)

We say that u ∈W 1,2(Ω; RN ) is a weak solution of (2.44) if∫ΩAαβij ∂βu

j∂αφi dx =

∫Ωgiφ

i+fαi ∂αφi dx = 0 ∀φ ∈W 1,2

0 (Ω; RN ). (2.48)

We say that the coefficients Aαβij are strongly elliptic (with ellipticityconstant ν) if

∃ν > 0 Aαβij(x)ξαξβζiζj ≥ ν|ξ|2 |ζ|2 ∀ξ ∈ Rn, ζ ∈ RN and a.e. x ∈ Ω.(2.49)

This condition is also called the Legendre-Hadamard condition.Sometimes we require a stronger condition, namely

∃ν > 0 Aαβij (x)F iαFjβ ≥ ν|F |2 ∀F ∈ RN×n and a.e. x ∈ Ω. (2.50)

This is condition is sometimes called super-strong ellipticity.To outline the difference between these two conditions assume for sim-

plicity that the coefficients are independent of x consider the bilinear formB : RN×n × RN×n → R given by

B(F,G) = Aαβij FiαG

jβ . (2.51)

Super-strong ellipticity is equivalent to the condition that the quadratic formB is positive definite. Strong-ellipticity is equivalent to the weaker conditionthat B is positive on matrices of the form F = ζ⊗ξ, i.e., on rank-1 matrices.

28 [March 17, 2014]

Theorem 2.7. Assume that f iα ∈ L2(Ω), gi ∈ L2(Ω) and that either

the coefficients Aαβ are strongly elliptic and constant (2.52)

orthe coefficients Aαβij satisfy (2.50) and (2.43) (2.53)

Then there exists a unique weak solution u ∈ W 1,20 (Ω; RN ) of the elliptic

system−divA∇u = g − div f (2.54)

Moreover the solution u satisfies∫Ω|∇u|2 dx ≤ 1

ν2

(∥f∥L2(Ω;Rn) + C(Ω)∥g∥L2(Ω)

)2. (2.55)

Proof. Consider the quadratic form

a(φ, u) =∫

ΩAαβij ∂βu

j∂αφi dx. (2.56)

We only need to show that

a(u, u) ≥ ν

∫Ω|∇u|2 dx ∀u ∈W 1,2

0 (Ω,RN ). (2.57)

Then the rest of the proof proceeds as in the scalar case. Under assump-tion (2.53) this estimate follows from the pointwise estimate (2.50). Underassumption we state this estimate is contained in the following lemma.

Lemma 2.8. (i) Assume (2.52). Then∫ΩAαβij ∂βu

j∂αui dx ≥ ν

∫Ω|∇u|2 dx ∀u ∈W 1,2

0 (Ω,RN ). (2.58)

(ii) Assume the estimate in (2.49) holds for some x0 ∈ Ω, i.e.,

∃ν > 0 Aαβij (x0)ξαξβζiζj ≥ ν|ξ|2 |ζ|2 ∀ξ ∈ Rn, ζ ∈ RN (2.59)

and assume thatsupx∈Ω

∥A(x) −A(x0)∥ ≤ ν

2(2.60)

Then∫ΩAαβij ∂βu

j∂αui dx ≥ ν

2

∫Ω|∇u|2 dx ∀u ∈W 1,2

0 (Ω,RN ). (2.61)

29 [March 17, 2014]

Proof. To prove the first assertion extend u by zero to Rn and let Fu denotethe Fourier transform of u defined by

Fu(ξ) :=∫

Rn

u(x)e−ix·ξ dx. (2.62)

Then

(F∂αuj)(ξ) = i ξα(Fuj)(ξ) and thus (F∇u)(ξ) = i (Fu)(ξ) ⊗ ξ. (2.63)

Thus using Plancherel’s theorem, strong ellipticity, the identity |a ⊗ b|2 =|a|2|b|2 and again Plancherel’s theorem we get

a(u, u) =1

(2π)n

∫Rn

Aαβij ξβ(Fuj)(ξ) ξα(Fui)(ξ) dξ (2.64)

≥ 1(2π)n

∫Rn

ν|ξ|2|Fu(ξ)|2 dξ (2.65)

=1

(2π)n

∫Rn

ν|F∇u|2 dξ (2.66)

=ν∫

Rn

|∇u|2 dx. (2.67)

The second assertion follows from the first and the pointwise estimate

Aαβij (x)∂βuj∂αui ≥ Aαβij (x0)∂βuj∂αui −ν

2|∇u|2. (2.68)

For general systems with non-constant strongly elliptic coefficients theconclusion does not hold9. If the coefficients are continuous one can obtain,however, a slightly weaker estimate which is almost as useful.

Theorem 2.9 (Garding inequality). Suppose that

the coefficients Aαβij are continuous in Ω and strongly elliptic. (2.69)

Then there exists a constant K > 0 such that∫ΩAαβij ∂βu

j∂αuj dx ≥ ν

2

∫Ω|∇u|2 dx−K

∫Ω|u|2 dx ∀u ∈W 1,2

0 (Ω; RN ).

(2.70)9see, H. Le Dret, An example of H1-unboundedness of solutions to strongly elliptic sys-

tems of partial differential equations in a laminated geometry, Proc. Roy. Soc. EdinburghSect. A, 105 (1987), 77–82.

30 [March 17, 2014]

Proof. Homework sheet 2. Hint: use Lemma 2.8 on balls with sufficientlysmall radius r0, a partition of unity

∑k φ

2k = 1 in Ω, where

φk ∈ C∞c (B(xk, r0)), and estimate the difference

(φk∇u,Aφk∇u) − (∇(φku), A∇(φku)). (2.71)

The proof shows that the constant K can be estimated as

K ≤ C

r20

∥A∥2

ν, (2.72)

where r0 is such that

|x− y| ≤ r0 =⇒ ∥A(x) −A(y)∥ ≤ ν

2, (2.73)

and where C only depends on the dimension n

It follows from the Garding inequality and the Lax-Milgram theoremthat the problem Lu + Ku = −div f + g has a unique weak solution inW 1,2

0 (Ω). Moreover the Fredholm alternative and its proof (see the courseFA and PDG or [GT], Thm. 8.6. or [Ev], Section 6.2.3, Thm. 4) imply thatthere exists an at most countable set Σ ⊂ (−K,∞) such that the problem

Lu− σu = −div f + g has a unique weak soln. in W 1,20 (Ω; Rn) if σ /∈ Σ.

(2.74)

2.1.3 Reverse Poincare inequality and pointwise estimates

Theorem 2.10. Suppose that the coefficients Aαβij are constant and stronglyelliptic, that fαi ∈ L2(Ω) and that u ∈W 1,2(Ω; RN ) is a weak solution of

−divA∇u = −div f in Ω. (2.75)

Assume that Br = B(x0, r) ⊂ Ω. Then for 0 < s < r∫Bs

|∇u|2 dx ≤ C

(r − s)2

∫Br\Bs

|u− a|2 dx+C

ν2

∫Br

|f |2 dx ∀a ∈ R.

(2.76)If moreover

f = 0 (2.77)

then

supBr/2

|∇u|2 ≤ C

rn+2

∫Br

|u− ur|2 dx ≤ C

rn

∫Br

|∇u|2 dx, (2.78)

supBr/2

|∇2u|2 ≤ C

rn+2

∫Br

|∇u− (∇u)r|2 dx. (2.79)

Moreover the constants C depend only on the dimension n and the ratio∥A∥/ν.

31 [March 17, 2014]

Proof. We may assume thatν = 1. (2.80)

Indeed if the result has been proved for ν = 1 the general case follows byreplacing A by A/ν. We first assume that f = 0.

Let η ∈ C∞c (Br) with 0 ≤ η ≤ 1, η = 1 in Bs and |∇η| ≤ 2/(r − s).

Then∇(ηu) = η∇u+ u⊗∇η. (2.81)

Using Lemma 2.8 and applying the definition of weak solution with φ = η2uwe thus get

∫Br

|∇(ηu)|2 dx ≤∫Br

(∇(ηu), A∇(ηu)) dx

=∫Br

(∇(ηu), Aη∇u) + (∇(ηu), A(u⊗∇η)) dx

=∫Br

(∇(η2u), A∇u)) dx︸ ︷︷ ︸=0

−∫Br

(ηu⊗∇η,A∇u) dx+∫Br

(∇(ηu), A(u⊗∇η)) dx

≤2∥A∥ ∥∇(ηu)∥L2 ∥u⊗∇η∥L2 + ∥A∥∥u⊗∇η∥2L2

≤12∥∇(ηu)∥2

L2 + 2∥A∥2 ∥u⊗∇η∥2L2 + ∥A∥ ∥u⊗∇η∥2

L2 , (2.82)

where we used Young’s inequality ab ≤ 12a

2 + 12b

2. Now the first term of theright hand side can be absorbed into the left hand side and we get∫

Br

|∇(ηu)|2 dx ≤ C(∥A∥) 1(r − s)2

∥u∥2L2(Br\Bs)

(2.83)

Since ∥η∇∥L2 ≤ ∥(∇(ηu)∥L2 + ∥u ⊗ ∇η∥L2 assertion (2.76) follows fora = 0. Now if u is a weak solution so, is u − a and hence the general formof (2.76) follows.

Now consider the case f = 0. The only difference is that the term withvanished for f = 0 is now estimated as∫

Br

(∇(η2u), A∇u)) dx =∫Br

(∇(ηηu), f) dx

≤∥∇(ηu)∥L2∥f∥L2 + ∥u⊗∇η∥L2∥f∥L2

≤14∥∇(ηu)∥2

L2 + 2∥f∥2L2 +

12∥u⊗∇η∥2

L2 +12∥f∥2

L2 , (2.84)

where we used that 2ab ≤ 12a

2 + 2b2. Now the additional term 14∥∇(ηu)∥2

L2

can still be absorbed into the left hand side of (2.82) and the argument isfinished as for f = 0.

[28.10. 2013, Lecture 4]

32 [March 17, 2014]

[31.10. 2013, Lecture 5]To prove the other two assertions we use the fact that the coefficients areconstant and hence the derivatives ∂γu solve the same elliptic equation.

The rigorous justification was discussed in the course FA and PDE: onefirst considers the difference quotients ∂hγu(x) := u(x+heγ)−u(x)

h . It is easy tosee that these satisfy the weak form of the equation (in the set Ωh := x ∈Ω,dist (x, ∂Ω) > h. Then (2.76) provides estimates for ∇∂hγu which areuniformly in h. Now by the relation between difference quotients and weakderivatives (see, e.g., [GT], Lemma 7.24) we see that for each i the functions∂hi ∂ju are bounded in L2(Bs) uniformly for h < h0 and thus u ∈W 2,2(Bs).Moreover passage to the limit h→ 0 yields∫

Bs

|∇∂γu|2 dx ≤ C

(r − s)2

∫Br\Bs

|∂γu|2 dx ∀s ∈ (0, r). (2.85)

Iterative application of this argument with radii ri = 1 − i2k shows that∫

B1/2

|∇ku|2 dx ≤ Ck

∫B1

|u− a|2 dx ∀a ∈ R. (2.86)

The Sobolev embedding theorem implies that W k,2(B1/2) → C1(B1/2) fork > n/2+1. Thus we get the first inequality in (2.78). The second inequalityin (2.78) follows by the Poincare inequality. The statement for general rfollows by considering the rescaled function ur(z) = u(x0 + rz). Finally toget (2.79) we use that the derivatives of u are also weak solutions and hence(2.78) can be applied with u replaced by ∇u.

2.1.4 Morrey and Campanato spaces

The main idea is that scaled L2 estimates can be used to derive Cα andeven Lp estimates.

Definition 2.11. Let λ > 0. The Morrey space L2,λ(Ω) consists of allfunctions f ∈ L2(Ω) for which the seminorm [f ]L2,λ defined by

[f ]2L2,λ = supx∈Ω

supr<diamΩ

1rλ

∫B(x,r)∩Ω

|f |2 dy (2.87)

is finite.The Campanato space L2,λ(Ω) consists of all functions f ∈ L2(Ω) for

which the seminorm [f ]L2,λ defined by

[f ]2L2,λ = supx∈Ω

supr<diamΩ

1rλ

∫B(x,r)∩Ω

|f − fx,r|2 dy (2.88)

is finite. Here we set

fx,r :=1

Ln(B(x, r) ∩ Ω)

∫B(x,r)∩Ω

f dy (2.89)

33 [March 17, 2014]

One can easily check these spaces are Banach spaces with the norms

∥f∥2L2,λ = ∥f∥2

L2 + [f ]2L2,λ , ∥f∥2L2,λ = ∥f∥2

L2 + [f ]2L2,λ , (2.90)

respectively.

Examples.

f ∈ L∞ =⇒ f ∈ L2,n,

f ∈W 1,∞ =⇒ f ∈ L2,n+2

f ∈ C0,α =⇒ f ∈ L2,n+2α

If Ω has Lipschitz boundary the Poincare inequality implies that

∇f ∈ L2,λ =⇒ f ∈ L2,λ+2.

Note also that∫B(x,r)∩Ω

|f − fx,r|2 dx ≤∫B(x,r)∩Ω

|f − a|2 dx ∀a ∈ R. (2.91)

Taking a = 0 we see in particular that L2,λ ⊂ L2,λ. Another consequence of(2.91) is the following monotonicity property

Ω ⊂ Ω′ =⇒∫B(x,r)∩Ω

|f − fΩ∩B(x,r)|2 dx ≤∫B(x,r)∩Ω

|f − fΩ′∩B(x,r)|2 dx

≤∫B(x,r)∩Ω′

|f − fΩ′∩B(x,r)|2 dx (2.92)

Here

fΩ∩B(x,r) :=1

Ln(B(x, r) ∩ Ω)

∫B(x,r)∩Ω

f dy,

fΩ′∩B(x,r) :=1

Ln(B(x, r) ∩ Ω′)

∫B(x,r)∩Ω′

f dy.

An immediate consequence of (2.92) is

Ω ⊂ Ω′ =⇒ [f ]L2,λ(Ω) ≤ [f ]L2,λ(Ω′). (2.93)

Theorem 2.12 (Characterization of Campanato spaces). Suppose that thereexists an A > 0 such that

Ln(B(x, r) ∩ Ω) ≥ Arn ∀x ∈ Ω, ∀r < diamΩ. (2.94)

Then

34 [March 17, 2014]

(i) L2,λ(Ω) = L2,λ(Ω) if 0 < λ < n,

(ii) L2,n+2α(Ω) = C0,α(Ω)

with equivalence of the norms involved.

As usual the second assertion has to be understood in the sense thatf has an Holder continuous representative. We will always identify f withthat representative.

Proof. The main idea is to compare the averages of f ∈ L2,λ over balls ofradius 2−kr. By scaling Ω we may assume without loss of generality thatdiamΩ = 1. We have for x ∈ Ω and for s < r∫

B(x,s)∩Ω|fx,r − fx,s|2 dx

≤2∫B(x,s)∩Ω

|f − fx,r|2 dx+∫B(x,s)∩Ω

|f − fx,s|2 dx

≤2(rλ + sλ) [f ]2L2,λ . (2.95)

If we take s = r/2 and use (2.94) we get

|fx,r − fx,r/2|2 ≤ Crλ−n[f ]2L2,λ . (2.96)

If λ < n this implies that

|fx,r − fx,1| ≤ Crλ−n

2 [f ]L2,λ . (2.97)

for all 0 < r < 1 (indeed let k be such that 12 < 2kr ≤ 1 and apply (2.96)

with 2l−1r and 2lr for l ≤ k, finally apply (2.95) with s = 2kr and 1). Thisyields the first assertion since∫

B(x,r)∩Ω|f |2 dy =

∫B(x,r)∩Ω

|f − fx,r|2 dy + Ln(B(x, r) ∩ Ω)|fx,r|2

≤Crλ[f ]2L2,λ + Crn|fx,1|2 (2.98)

and since by (2.94) we have |fx,1| ≤ C∥f∥L2 .Now assume that α ∈ (0, 1) and λ = n + 2α. By the Lebesgue point

theorem we havelimr→0

fx,r = f(x) for a.e. x ∈ Ω. (2.99)

On the other hand it follows from (2.96) that the continuous functions x 7→fx,2−k converge uniformly to some limit g. Hence g is continuous and g = fa.e. In the following we do not distinguish between f and g. By (2.96) wehave

|fx,r − f(x)| ≤ Crα[f ]L2,n+2α . (2.100)

35 [March 17, 2014]

Now consider two points x, y ∈ Ω and take r = |x− y|. To prove Holdercontinuity we write

|f(x) − f(y)| ≤ |f(x) − fx,2r| + |fx,2r − fy,2r| + |fy,2r − f(y)| (2.101)

The first and the last term on the right hand side are controlled by (2.100).To control the middle term note that

Ln(B(x, 2r) ∩B(y, 2r) ∩ Ω)|fx,2r − fy,2r|2

=∫B(x,2r)∩B(y,2r)∩Ω

|fx,2r − fy,2r|2 dz

≤2∫B(x,2r)∩B(y,2r)∩Ω

|f(z) − fx,2r|2 dz + 2∫B(x,2r)∩B(y,2r)∩Ω

|f(z) − fy,2r|2 dz

≤Crn+2α[f ]L2,n+2α (2.102)

Now by the choice of r we have B(x, 2r) ∩ B(y, 2r) ⊃ B(x, r) and by as-sumption Ln(B(x, r) ∩ Ω) ≥ Arn. Thus

|fx,2r − fy,2r|2 ≤ Cr2α[f ]L2,n+2α (2.103)

and the proof is finished.

The following lemma was added on Nov 4, 2013

Lemma 2.13. Let 0 < λ < n+ 2. Let Ω ⊂ Rn be open. Then

(i) C∞(Ω) is not dense in L2,λ(Ω).

(ii) Assume that f ∈ L2loc(Rn) and [f ]L2,λ(Rn) < ∞. Then there exist

fk ∈ C∞(Rn) with

fk → f in L2loc(Rn), lim sup

k→∞[fk]L2,λ(Rn) ≤ [f ]L2,λ(Rn). (2.104)

(iii) Assume in addition that Ω is bounded with Lipschitz boundary. Letf ∈ L2,λ(Ω). Then there exist fk ∈ C∞(Ω) with

fk → f in L2(Ω), lim supk→∞

∥fk∥L2,λ(Ω) ≤ C(Ω)∥f∥L2,λ(Ω).

(2.105)

(iv)

fk f in L2loc(Ω) =⇒ [f ]L2,λ(Ω) ≤ lim inf

k→∞[fk]L2,λ(Ω), (2.106)

where both sides are allowed to take the value +∞.

36 [March 17, 2014]

Notation. We say that f ∈ L2loc(E) if f ∈ L2(K) for all compact sets

K ⊂ E. We say that fk and converges weakly or strongly to f in L2loc(E) if

the convergence holds in all L2(K), where K ⊂ E is compact.

Proof. (i), (ii), (iv) see Homework 3. Hints:(i): Denote by Xλ the closure of C∞(Ω) in L2,λ(Ω). Show first that

f ∈ Xλ =⇒ limr→0

1rλ

∫B(x,r)

|f − fx,r|2 dy = 0 ∀ x ∈ Ω (2.107)

(consider first f ∈ C∞(Ω)). Then assume without loss of generality that0 ∈ Ω and show that g ∈ L2,λ(Ω) \Xλ where

g(x) =

|x|

λ−n2 if λ = n,

− ln |x| if λ = n.(2.108)

(ii): Let η ∈ C∞c (B(0, 1)) with η ≥ 0 and

∫Rn η dx = 1 and let ηk(x) =

knη(kx). Set fk = ηk ∗ f . Apply Jensens’s inequality (for the probabilitymeasure ηk(x) dx), i.e.,∣∣∣∣∫

Rn

g(z)ηk(z) dz∣∣∣∣2 ≤

∫Rn

|g(z)|2ηk(z) dz (2.109)

tog(z) = f(y − z) − 1

Ln(B(x, r))

∫B(x,r)

f(y − z) dy (2.110)

and integrate the left hand side of (2.109) to deduce that [fk]2L2,λ ≤ [f ]2L2,λ .(iv): Argue by contradiction. First, passing to a subsequence if necessary

we may assume that L := limk→∞[fk]L2,λ(Ω) exists and L < ∞ (since forL = ∞ the assertion is trivially true). If the assertion does not hold thenthere exist x ∈ Ω, r > 0 and δ > 0 such that

1rλ

∫B(x,r)∩Ω

|f(y) − fx,r|2 dy > L2 + δ. (2.111)

Now derive a contradiction to the assumption fk f in L2loc(Ω) (for sim-

plicity assume first that the convergence holds in L2(Ω)).Assertion (iii) can be deduced from (ii) and the following extension result.

There exists a constant C(Ω) such that for each f ∈ L2,λ(Ω) there exists anF ∈ L2,λ(Rn) with F|Ω = f and ∥F∥L2,λ(Rn) ≤ C(Ω)|f∥L2,λ(Ω). In additionone uses the following trivial estimates for x ∈ Ω and fk = Fk |Ω∫

B(x,r)∩Ω|fk − (fk)x,r|2 dy ≤

∫B(x,r)∩Ω

|fk − (Fk)B(x,r)|2 dy

≤∫B(x,r)

|Fk − (Fk)B(x,r)|2 dy.

37 [March 17, 2014]

2.1.5 Regularity theory in Campanato spaces, interior estimates

Theorem 2.14 (Interior estimates in L2,λ for constant coefficients). Let

B = B(x0, R), B′ = B(x0, θR), where 0 < θ < 1. (2.112)

Let 0 < λ < n + 2 and suppose that fαi ∈ L2,λ(B). Assume that the coeffi-cients Aαβij are constant and strongly elliptic and let u ∈ W 1,2(B; RN ) be aweak solution of the system

−divA∇u = −div f in B. (2.113)

Then∇u ∈ L2,λ(B′; RN×n). (2.114)

and

[∇u]2L2,λ(B′) ≤C

ν2[f ]2L2,λ(B) + C

1Rλ

∫B|∇u− (∇u)B|2, (2.115)

where C depends on θ, λ, ∥A∥/ν and the dimension n.

Remark. Taking λ = n+ 2µ we see in particular that f ∈ C0,µ(B(x0, R))implies that ∇u ∈ C0,µ(B(x0, θR) with a quantitative estimate of the Holderseminorms.

Proof. Note first it suffices to prove the result in the case

ν = 1, B(x0, R) = B(0, 1), (∇u)B = 0. (2.116)

Indeed to recover the result for general ν > 0 (it suffices replace A byA/ν and f by f/ν). To recover the result for general R > 0 and x0 defineuR(z) = R−1u(x0 + Rz) and fR(z) = f(x0 + Rz), observe that uR is aweak solution of −divA∇uR = fR in B(0, 1) and that [∇u]2

L2,λ(B(x0,R))=

Rn−λ[uR]2L2,λ(B(0,1))

and right hand side of (2.115) scales in the same way.To see that we can assume that (∇u)B = 0 note if u is a weak solution

of (2.114) then so is the function v defined by v(x) = u(x)− (∇u)Bx (sinceA is constant). Now ∇vB = 0 and once we have the estimate for v weimmediately deduce the estimate for u since [∇u]L2,λ = [∇v]L2,λ and ∇u−(∇u)B = ∇v.

Let x1 ∈ B(0, θ). Set Br := B(x1, r), gr = gx1,r. We establish a decayestimate for

ϕ(r) :=∫Br

|∇u− (∇u)r|2 dx (2.117)

38 [March 17, 2014]

for r ≤ 1 − θ. This is sufficient because for r ≥ 1 − θ we can use the trivialestimates

r−λ∫Br∩B(0,1)

|∇u− (∇u)r|2 dx ≤ r−λ∫Br∩B(0,1)

|∇u|2 dx

≤(1 − θ)−λ∫B(0,1)

|∇u|2 dx. (2.118)

We first derive a functional inequality for ϕ. Together with a simpleiteration lemma this will give the decay of φ. In order to derive the functionalinequality we write u = v + w, where v solves an elliptic system with zeroright hand side (and hence has good decay) and where w has zero boundaryconditions. More precisely we define w as the unique weak solution of thesystem

−divA∇w = −div (f − fr) in Br, w ∈W 1,20 (Br; RN ) (2.119)

and we set v = u − w. Note that fr is a constant and hence div fr = 0. Itfollows that v is a weak solution of

−divA∇v = 0 in Br. (2.120)

By Theorem 2.10 we have

supBr/2

|∇2v|2 ≤ C

rn+2

∫Br

|∇v − (∇v)r|2 dx (2.121)

This implies the following excess decay estimate∫Bρ

|∇v − (∇v)ρ|2 dx ≤ Cρn+2

rn+2

∫Br

|∇v − (∇v)r|2 dx (2.122)

for all 0 < ρ ≤ r/2, where C depends on ∥A∥/ν and the dimension n. Ifwe take C ≥ 2n+2 the estimate holds trivially also for ρ ∈ (r/2, r]. Here weused ρ 7→

∫Bρ

|∇v − (∇v)ρ|2 dx is monotone. Indeed by (2.91) we have forρ < r∫

Bρ

|∇v−(∇v)ρ|2 dx ≤∫Bρ

|∇v−(∇v)r|2 dx ≤∫Br

|∇v−(∇v)r|2 dx (2.123)

Regarding w, the energy estimate in Theorem 2.7 yields∫Br

|∇w|2 dx ≤∫Br

|f − fr|2 ≤ rλ[f ]2L2,λ . (2.124)

39 [March 17, 2014]

Since v = u−w this provides an estimate for the right hand side of (2.122)∫Br

|∇v − (∇v)r|2 dx ≤ 2∫Br

|∇u− (∇u)r|2 dx+ 2∫Br

|∇w − (∇w)r|2 dx

≤2∫Br

|∇u− (∇u)r|2 dx+ 2rλ[f ]2L2,λ (2.125)

The energy estimate (2.124) implies trivially that for all 0 < ρ < r∫Bρ

|∇w − (∇w)ρ|2 dx ≤∫Bρ

|∇w|2 dx ≤∫Br

|∇w|2 dx ≤ rλ[f ]2L2,λ . (2.126)

Combining this estimate with (2.122) and (2.125) we get

ϕ(ρ) ≤ 2∫Bρ

|∇v − (∇v)ρ|2 dx+ 2∫Bρ

|∇w − (∇w)ρ|2 dx

≤2Cρn+2

rn+2

(2ϕ(r) + 2rλ[f ]2L2,λ

)+ 2rλ[f ]2L2,λ). (2.127)

This gives the desired functional inequality for ϕ

ϕ(ρ) ≤ C1ρn+2

rn+2ϕ(r) + C2r

λ[f ]2L2,λ ∀ 0 < ρ ≤ r ≤ 1 − θ. (2.128)

where C1 and C2 depend only on ∥A∥ and n. It now follows from Lemma2.15 below that

ϕ(ρ) ≤ C

(ϕ(1 − θ)(1 − θ)λ

+ [f ]2L2,λ

)ρλ (2.129)

where C depends on λ, ∥A∥ and n. Since (1− θ)−λϕ(1− θ) is controlled by(2.118) with r = 1 − θ this finishes the proof (note that argued at the verybeginning of the proof that it suffices to consider the case (∇u)B = 0).

[31.10. 2013, Lecture 5][4.11. 2013, Lecture 6]

Remark. Theorem 2.14 can be extended to equations of the form

−divA∇u = −div f + g (2.130)

withg ∈ L2,λ−2(B; RN ) (2.131)

provided that we add a term C[g]2L2,λ(B)

to the right hand side of (2.115)Here we use the following conventions for λ ≤ 2

L2,λ−2(Ω) = L2(Ω), [g]2L2,λ(Ω) = (diamΩ)2−λ∫

Ω|g|2 dx if 0 < λ ≤ 2.

(2.132)

40 [March 17, 2014]

The only difference in the proof is that w is now defined by solving−divA∇u = −div f + g and the energy estimate becomes∫Br

|∇w|2 ≤ (∥f −fr∥L2 +C(Br)∥g∥L2)2 ≤ 2rλ[f ]2L2,λ +2C(Br)2rλ−2∥g∥2L2,λ

(2.133)where C(Br) = rC(B1) is the constant in the Poincare inequality with zeroboundary values.

Lemma 2.15 (Iteration lemma). Let 0 < β < α. Let ϕ : (0, r0) → [0,∞)be an increasing function and suppose that

ϕ(ρ) ≤ A

[ρα

rα+ ε

]ϕ(r) +Brβ , whenever 0 < ρ ≤ r ≤ r0 (2.134)

Then there exists a constant ε0 > 0 (which depends on A, α and β) suchthat if ε < ε0 then

ϕ(ρ) ≤ C

(ρβ

rβϕ(r) +Bρβ

)whenever 0 < ρ ≤ r ≤ r0, (2.135)

where C depends on A, α and β.

Proof. For 0 < τ < 1 and r ≤ r0 we have

ϕ(τr) ≤ Aτα[1 + ετ−α]ϕ(r) +Brβ . (2.136)

Now let γ ∈ (α, β) and choose τ and ε0 such that

Aτα ≤ 12τγ , ε0 = τα. (2.137)

Thenϕ(τr) ≤ τγϕ(r) +Brβ . (2.138)

By induction we get for all integers k ≥ 0:

ϕ(τk+1r) ≤ φ(τ τkr)≤ τγϕ(τkr) +Bτkβrβ

≤ τγ(τγϕ(τ (k−1)r) +Bτ (k−1)βrβ

)+Bτkβrβ

≤ τ (k+1)γϕ(R) +Bτkβrβk∑j=0

τ j(γ−β)

≤ C ′τ (k+1)β [ϕ(r) +Brβ ].

This gives the assertion for the discrete values ρ = τkr. Now the followingstandard argument shows that we get the estimate for all ρ ≤ r, at theexpense of a slightly larger constant.

41 [March 17, 2014]

Given ρ ≤ τr choose k such that τk+2r < ρ ≤ τk+1r. Then

ϕ(ρ) ≤ ϕ(τk+1r) ≤ C ′( ρτr

)β[ϕ(r) +Brβ ] =

C ′

τβ

(ρβ

rβϕ(r) + ρβ

)(2.139)

For τr < ρ ≤ r one uses the trivial estimate

ϕ(ρ) ≤ ϕ(r) ≤ τ−βρβ

rβϕ(r). (2.140)

Corollary 2.16 (Global estimates in Rn). Assume that the coefficients Aαβijare constant and strongly elliptic.

(i) Let f ∈ L2(Rn; Rn×N ). Then there exists a weak solutionu ∈W 1,2

loc (Rn; RN ) of

−divA∇u = −div f, with ∇u ∈ L2(Rn; RN×n). (2.141)

Moreover u is unique up to the addition of constants and∫Rn

|∇u|2 dx ≤ 1ν2

∫Rn

|f |2 dx. (2.142)

(ii) Let 0 < λ < n+2. Assume that f ∈ L2loc(Rn; Rn×N ) with [f ]L2,λ(Rn) <

∞. Then there exists a weak solution u ∈W 1,2loc (Rn; RN ) of

−divA∇u = −div f, with [∇u]L2,λ(Rn) <∞. (2.143)

Moreover u is unique up to the addition of affine functions and

[∇u]L2,λ(Rn) ≤ C[f ]L2,λ(Rn), (2.144)

where C only depends on n and ∥A∥/ν.

Proof. Homework 3. Hint: As before we may assume ν = 1. To showuniqueness up to constant and affine maps use (2.78) and (2.79), respectively,and take the limit r → ∞.(i): Apply the Lax-Milgram theorem to show that the problem

−divA∇uk +1ku = −div f (2.145)

has a unique solution in W 1,2(Rn; RN ) and ∥∇uk∥L2 ≤ ∥f∥L2 . Deducethat (a subsequence of) ∇uk converges weakly in L2 and hence uk − (uk)0,1converges weakly in W 1,2

loc to the desired solution u.

42 [March 17, 2014]

Alternatively observe that ∥∇u∥ is a norm on C∞c (Rn; RN ) (and not just

a seminorm) and define

D1,20 := closure of C∞

c (Rn; RN ) in ∥∇u∥. (2.146)

Then apply the Lax-Milgram theorem in D1,20 .

(ii): Step 1. Assume in addition that f = 0 in Rn \B(0, R0).Then f ∈ L2. Let u be the solution in (i). Deduce the estimate for[∇u]L2,λ(Rn) from the interior estimate and (2.142) by exploiting that theinterior estimate holds for every R0 > 0.

Step 2. General f .Let η ∈ C∞

c (B(0, 1) with η = 1 on B(0, 1/2) and let ηk(x) = η(2−kx). Letfk = ηk(f − f0,2k) and let uk be the solution of −divA∇u = fk which existsby Step 1. Show that

[fk]L2,λ(Rn) ≤ C[f ]L2,λ(Rn). (2.147)

Then Step 1 yields[∇uk]L2,λ(Rn) ≤ C[f ]L2,λ(Rn). (2.148)

Deduce that (for ρ ≥ 1)∫B(0,ρ)

|∇uk − (∇uk)0,ρ|2 dx ≤ Cρλ[f ]2L2,λ(Rn),

|(∇uk)0,ρ − (∇uk)0,1| ≤ C(ρ)[f ]L2,λ(Rn). (2.149)

Deduce that there exists a subsequence of ∇uk − (∇uk)0,1 which convergesweakly in L2

loc(Rn) to some v. Then show that uk = uk− (uk)0,1− (∇uk)0,1xconverges weakly in W 1,2

loc (Rn) to some u. It is then easy to see that u isa weak solution of the PDE in Rn and the estimate for [u]L2,λ(Rn) followsfrom (2.106).

Remark: in fact it is easy to see that the whole sequence ∇uk− (∇uk)0,1converges strongly in L2

loc by using that

−divA∇(uk − uk+1) = 0 in B(0, 2k−1)

and the decay estimate for solutions of the homogeneous equation.

Theorem 2.17 (Interior estimates with continuous coefficients). Let

B = B(x0, R), B′ = B(x0, θR), where 0 < θ < 1. (2.150)

Let 0 < λ < n and suppose that fαi ∈ L2,λ(B). Assume that the co-efficients Aαβij are uniformly continuous in B and strongly elliptic. Letu ∈W 1,2(B; RN ) be a weak solution of the system

−divA∇u = −div f in B. (2.151)

43 [March 17, 2014]

Then∇u ∈ L2,λ(B′; RN×n), (2.152)

and[∇u]2L2,λ(B′) ≤

C

ν2[f ]2L2,λ(B) +

C

Rλ

∫B|∇u|2, (2.153)

where C depends on θ, λ, sup ∥A∥/ν, the dimension n and the modulus ofcontinuity of the rescaled function z 7→ A(x0 +Rz)/ν.

Remark. For 0 < λ < n we have L2,λ = L2,λ with equivalent norms. Wehave written [u]L2,λ and [f ]L2,λ because these are the seminorms we actuallyuse in the proof.

Notation The modulus of continuity of a function g is the function ωggiven by

ωg(r) := sup|g(x) − g(y)| : x, y ∈ Ω, |x− y| ≤ r. (2.154)

A function g is uniformly continuous if and only if limr↓0 ωg(r) = 0.

Proof. As in the proof of Theorem 2.14 we may assume that

ν = 1, B = B(x0, R) = B(0, 1). (2.155)

We set

ω(r) = sup∥A(x) −A(y)∥ : x, y ∈ Ω, |x− y| ≤ r. (2.156)

Let x1 ∈ B′. We set Br = B(x1, r). We have to prove decay of thequantity

ϕ(r) =∫Br

|∇u|2 dx. (2.157)

It suffices to control ϕ for r ≤ r0 ≤ 1− θ since for r > r0 we have the trivialestimate

r−λ0

∫Br∩B

|∇u|2 ≤ r−λ0

∫B|∇u|2 dx. (2.158)

The main idea is to ’freeze the coefficients’, i.e., we observe that u is alsoa weak solution of the constant coefficient PDE

−divA(x1)∇u = −divF, where F = (f − fr) + [A(x1) −A]∇u.

We write again u = v + w where w ∈ W 1,20 (Br) is the weak solution of the

constant coefficient system

−divA(x1)∇w = −divF, (2.159)

44 [March 17, 2014]

and where v solves −divA(x1)∇v = 0. Now using (2.78) we get∫Bρ

|∇v|2 dx ≤ Cρn

rn

∫Br

|∇v|2 dx (2.160)

for ρ ≤ r/2. By taking C ≥ 2n we can assume that this estimate holds forall ρ ≤ r ≤ 1 − θ. The energy estimate for w gives∫Br

|∇w|2 dx ≤ 2∫Br

|f |2 dx+ 2ω2(r)∫Br

|∇u|2 ≤ 2[f ]2L2,λrλ + 2ω2(r)ϕ(r).

(2.161)

This yields

ϕ(ρ) ≤ C

(ρn

rn+ ω(r)2

)ϕ(r) + C[f ]2L2,λr

λ ∀ ρ ≤ r ≤ 1 − θ, (2.162)

where C depends on ∥A(x1)∥ and n.Now let ε0 > 0 be as in the iteration lemma. Since A is uniformly

continuous there exists r0 > 0 such that ω2(r) ≤ ε0 for all r ≤ r0. Thusthe iteration lemma implies that there exists a constant (depending on λ,∥A(x1)∥ and n) such that

ϕ(ρ) ≤ Cρλ

rλϕ(r) + C[f ]2L2,λr

λ ∀ ρ ≤ r ≤ r0. (2.163)

Taking r = r0 and using the trivial estimate (2.158) we obtain the as-sertion.

Theorem 2.18 (Interior estimates with Holder continuous coefficients). Let0 < µ < 1, 0 < λ ≤ n+ 2µ and let

B = B(x0, R), B′ = B(x0, θR), 0 < θ < 1. (2.164)

Assume that the coefficients Aαβij are in C0,µ(B) and strongly elliptic. Letfαi ∈ L2,λ(B) and let u ∈W 1,2(B; RN ) be a weak solution of

−divA∇u = −div f in B. (2.165)

Then∇u ∈ L2,λ(B′,RN×n) (2.166)

and

[∇u]2L2,λ(B′) ≤C

ν2[f ]2L2,λ(B)+

C

Rλ

∫B|∇u−(∇u)B|2 dx+

C[A]2C0,µR2µ

ν2Rλ

∫B|∇u|2 dx,

(2.167)where C depends on supB ∥A∥/ν, λ, n and on Rµ[A]C0,µ/ν.

45 [March 17, 2014]

Remark. Since L2,n+2µ(B) = C0,µ(B) this shows in particular that

A, f ∈ C0,µ(B) =⇒ ∇u ∈ C0,µ(B′). (2.168)

The corresponding estimate is called a Schauder estimate.

Proof. Step 1. Simplifications.We claim that again it suffices to show the result for

ν = 1, B = B(0, 1), (∇u)B = 0, λ ≥ n. (2.169)

The restriction to ν = 1 is clear. As regards the restriction to R = 1 notethat the function uR(z) = R−1u(x0 +Rz) solves

−divAR∇uR = −div fR, (2.170)

whereAR(z) = A(x0 +Rz), fR(z) = f(x0 +Rz). (2.171)

Now [AR]C0,µ(B(0,1)) = Rµ[A]C0,µ(B(x0,R)) so that all the terms in (2.167)scale in the right way. Finally to see that we may assume (∇u)B = 0 takeB = B(0, 1) and note that v(x) = uB(x) − (∇u)Bx solves

−divA∇v = −div f , f := f −A(∇u)B. (2.172)

Now

[A(∇u)B]L2,λ ≤ [A]L2,λ |(∇uB)| ≤ [A]L2,n+2µ |(∇uB)|

≤C[A]C0,µ

(∫B|∇u|2 dx

)1/2

(2.173)

Thus [f ]2L2,λ is controlled by twice the sum of the first and the last term onthe right hand side of (2.167).

[4.11. 2013, Lecture 6][7.11. 2013, Lecture 7]

We finally note that the assertion for 0 < λ < n follows from Theorem 2.17(in this case the last term on the right hand side of (2.167) is not needed).

To obtain the result for λ ≥ n will use a so called bootstrap argument,i.e., we will successively improve the regularity. More precisely we will provethe estimates first for λ < n, then for n ≤ λ < n + 2µ and finally forλ = n+ 2µ on a sequence of decreasing balls

B(0, 1 − τ) ⊃ B(0, 1 − 2τ) ⊃ B(0, 1 − 3τ) = B(0, θ), (2.174)

where τ = (1 − θ)/3.

46 [March 17, 2014]

Step 2. Estimate of [∇u]L2,λ(B(0,1−2τ)) for n ≤ λ < n+ 2µ.By Theorem 2.17 applied with λ− 2µ < n instead of λ we have

[∇u]2L2,λ−2µ(B(0,1−τ)) ≤ C[f ]2L2,λ−2µ(B) + C

∫B|∇u|2 dx. (2.175)

We want to prove decay of

ϕ(r) =∫Br

|∇u− (∇u)r|2 dx

where Br = B(x1, r), x1 ∈ B(0, 1− 2τ) and r ≤ τ (for r ≥ τ we can use thetrivial estimate ϕ(τ) ≤

∫B |∇u|2).

We freeze the coefficients, i.e., we observe that u is also a weak solutionof the constant coefficient PDE

−divA(x1)∇u = −divF, where F = (f − fr) + [A(x1) −A]∇u.

We argue as in the proof of Theorem 2.17 but now use (2.79) for v insteadof (2.78) and get

ϕ(ρ) ≤ Cρn+2

rn+2ϕ(r) + C

∫Br

|F |2 dx (2.176)

and thus

ϕ(ρ) ≤ Cρn+2

rn+2ϕ(r) + Cr2µ

∫Br

|∇u|2 dx+ C[f ]2L2,λrλ (2.177)

≤Cρn+2

rn+2ϕ(r) + C[∇u|2L2,λ−2µ(Br)r

λ + C[f ]2L2,λrλ ∀ ρ ≤ r ≤ τ (2.178)

Then the desired estimate follows from the iteration lemma (with ε = 0)and the bound (2.175) on [∇u|2

L2,λ−2µ(B(0,1−τ)).

Step 3. λ = n+ 2µ.From the previous step we know that ∇u ∈ L2,n+2µ−δ(B(0, 1 − 2τ)), for allδ > 0 and that the corresponding norm is bounded by the right hand side of(2.167). Now we take δ = µ and use that L2,n+µ = C0,µ/2 (with equivalentseminorms) and

supB′′

|∇u| ≤ [∇u]C0,µ/2(B′′) +1

Ln(B′′)

∫B′′

∇u dx

≤ [∇u]C0,µ/2(B′′) + C∥∇u∥L2(B′′), (2.179)

where B′′ = B(0, 1 − 2τ). This shows that supB′′ |∇u|2 is controlled by theright hand side of (2.167).

47 [March 17, 2014]

Returning to (2.178) (now for x1 ∈ B′ = B(0, 1−3τ) and r ≤ τ) we thusget

ϕ(ρ) ≤ Cρn+2

rn+2ϕ(r) + C sup

Br

|∇u|2rλ + C[f ]2L2,λrλ (2.180)

and the iteration lemma shows that ∇u ∈ L2,λ and that (2.167) holds.

2.1.6 Estimates up to the boundary

We now prove estimates near the boundary. We first consider functionsu which vanish on the boundary of the half-space Rn

+ and we show thatthe estimates for balls which were derived in the previous section can beextended to half-balls. Later we will map a neighbourhood of a boundarypoint of a general open set with C1,α boundary to a half ball (’flattening theboundary’). We consider the half space

Rn+ := x ∈ Rn : xn > 0 = Rn−1 × (0,∞) with ∂Rn

+ = Rn−1 × 0(2.181)

and the half balls

B+(x0, r) = B(x0, r) ∩ Rn+, where x0 ∈ Rn−1 × 0. (2.182)

We want to define the space of Sobolev functions which vanish on the lowerboundary of B+, i.e., on the set x ∈ B(x0, r) : xn = 0. On possibility isto first set

V+ := u ∈W 1,2(B+(x0, r)) : ∃δ > 0 u = 0 for xn < δ (2.183)

and then to define

W 1,2+ (B+(x0, r)) = closure of V+ in W 1,2(B+(x0, r)). (2.184)

Note in particular that (assuming still x0 ∈ ∂Rn+)

u ∈W 1,2+ (B+(x0, r)), η ∈ C∞

c (B(x0, r)) =⇒ ηu ∈W 1,20 (B(x0, r))

(2.185)Functions u ∈W 1,2

+ (B+(x0, r)) satisfy the Poincare inequality∫B+(x0,r)

|u|2 dx ≤ Cr2∫B+(x0,r)

|∂nu|2 dx. (2.186)

This can be easily deduced from the one-dimensional Poincare inequality(applied in direction xn) or by a compactness argument as in the proofof the Poincare inequality with zero mean, see Homework 1 for a generalabstract version of the Poincare inequality. In this case on first considersr = 1 and then argues by scaling.

Theorems 2.10, 2.17 and 2.18 can easily be adapted to half-balls.

48 [March 17, 2014]

Theorem 2.19. Let B+r = B+(0, r). Suppose that the coefficients Aαβij are

constant and strongly elliptic and that u ∈W 1,2+ (B+

r ) is a weak solution of

−divA∇u = f in B+r . (2.187)

Then for 0 < s < r∫B+

s

|∇u|2 dx ≤ C

(r − s)2

∫B+

r \B+s

|u− a|2 dx+C

ν2

∫B+

r

|f |2 dx ∀a ∈ R.

(2.188)If moreover

f = 0 (2.189)

then

supB+

r/2

|∇u|2 ≤ C

rn+2

∫B+

r

|u|2 dx ≤ C

rn

∫B+

r

|∂nu|2 dx, (2.190)

supB+

r/2

|∇2u|2 ≤ C

rn+2

∫Br

|∂nu− λ|2 dx ∀ λ ∈ R. (2.191)

for all λ ∈ R. Moreover the constants C depend only on the dimension nand the ratio ∥A∥/ν.

Proof. In view of (2.185) the first estimate is proved in the same way as forballs.

For the second an third estimate we need control of higher derivatives.On the half-ball we can still use that all the tangential derivatives ∂iu,i = 1, . . . , n− 1 also solve the PDE. This gives the estimate∫

B+s

|∇∂iu|2 dx ≤ C

(r − s)2

∫Br

|∇iu|2 dx (2.192)

The left hand side gives control of all second derivatives ∂j∂iu and ∂n∂iufor i, j = 1 . . . , n − 1. The only missing second derivative is ∂n∂nu. Thisderivative can be controlled by the equation. Consider first scalar equations∑

α,β aαβ∂α∂βu = 0. With the choice ξ = en the ellipticity conditions gives

ann ≥ ν. Thus we get from the equation

∂n∂nu = − 1ann

∑(α,β) =(n,n)

aαβ∂α∂βu. (2.193)

For systems define the matrices Bαβ = Aαβij . Then

Bnn∂n∂nu = −∑

(α,β) =(n,n)

Bαβ∂α∂βu (2.194)

49 [March 17, 2014]

The choice ξ = en in the strong ellipticity condition implies that

Bnnij ζ

iζj ≥ ν|ζ|2 ∀ ζ ∈ RN . (2.195)

Hence the matrix Bnn is invertible (this is true, even if Bnn is not symmetric;use, e.g., the Lax-Milgram theorem on RN ). Thus we can obtain the estimatefor ∂n∂nu from (2.194)

For the higher derivative one argues by induction. First one gets∫B+

1/2

|∇∂i1 . . . ∂ik−1u|2 dx ≤ Ck

∫B+

1

|u|2 dx (2.196)

for ij = 1, . . . , n − 1. Thus gives control of all k-th order derivatives whichcontain at most one derivative in direction xn. By taking k − 2 tangentialderivatives of (2.194) we get those partial derivatives which contain at most2 derivatives in direction xn. Now we inductively take l derivatives of or(2.194) in direction xn and k − 2 − l derivatives in tangential directions forl = 1, . . . , k−2 to obtain and L2 estimates for all k-th order derivatives. Nowthe Poincare inequality for functions in W 1,2

+ and the Sobolev embeddingtheorem (2.190) and (2.191) for r = 1 and λ = 0. As usual the assertion forgeneral r follows by scaling.

Assume finally that λ = 0 and let v(x) = u(x)−λxn. Then v ∈W 1,2+ (B+

r )and v solves the same PDE as u. Thus we can apply (2.191) to v. Since∇2v = ∇2u and ∂nv = ∂nu− λ the estimate (2.191) holds for all λ ∈ R.

The regularity results in L2,λ(B+) follow from the estimates in Theorem2.19 and two simply ingredients:

• a decay estimate for half-balls and

• a simple geometric observation which allows one to combine the esti-mates and half-balls and the interior estimates.

We directly state the results for variable coefficients.

Lemma 2.20 (Decay on half balls with continuous coefficients). Let B+R =

B+(0, R), let 0 < λ < n, f ∈ L2,λ(B+R), Aαβij ∈ C0(B+

R) and let u ∈W 1,2

+ (B+R) be a weak solution of

−divA∇u = f in B+R . (2.197)

Then for 0 < r ≤ R∫B+

r

|∇u|2 dx ≤ C

ν2[f ]L2,λ(B+

R)rλ + C

rλ

Rλ

∫B+

R

|∇u|2 dx (2.198)

50 [March 17, 2014]

Proof. This is very similar to the proof of Theorem 2.17. As before we mayassume that ν = 1 and R = 1. We have to prove decay of

φ(r) :=∫B+

r

|∇u|2 dx. (2.199)

It suffices to control φ(r) for r ≤ r0. As before set

ω(r) := sup∥A(x) −A(y)∥ : |x− y| ≤ r (2.200)

We again freeze the coefficients, i.e., we note that u is a weak solution of

−divA(0)∇ = −divF, where F = (f − fr) + [A(0) −A]∇u. (2.201)

As before write u = v + w, where w ∈ W 1,20 (B+

r ) is the weak solution of−divA(0)∇w = −divF and where −divA(0)∇v = 0. Then∫

B+r

|∇w|2 dx ≤∫B+

r

|F |2 dx ≤ 2∫B+

r

|f |2 dx+ 2ω(r)2φ(r). (2.202)

and by (2.190) ∫B+

ρ

|∇v|2 dx ≤ Cρn

rn

∫B+

r

|∇v|2 dx (2.203)

for ρ ≤ r/2. As before, taking C ≥ 2n we may assume that this estimateholds for all ρ ≤ r. As before this yields

φ(ρ) ≤ C

[ρn

rn+ ω2(r)

]φ(r) + C[f ]L2,λrλ. (2.204)

Let ε0 be the constant in the iteration lemma and choose r0 such thatω2(r0) ≤ ε0. Then the iteration lemma yields

φ(ρ) ≤ Cρλ

rn0φ(r0) + C[f ]L2,λρλ ∀ ρ ≤ r0. (2.205)

This implies the assertion since for r0 ≤ ρ ≤ 1 we can use the trivial estimateρ−λϕ(ρ) ≤ r−λ0 ϕ(1).

Theorem 2.21 (Boundary estimates with continuous coefficents). Let

B+ = B+(0, R), B′,+ = B+(0, θR), where 0 < θ < 1. (2.206)

Let 0 < λ < n and suppose that fαi ∈ L2,λ(B+). Assume that the co-efficients Aαβij are uniformly continuous in B+ and strongly elliptic. Letu ∈W 1,2

+ (B+; RN ) be a weak solution of the system

−divA∇u = −div f in B+. (2.207)

51 [March 17, 2014]

Then∇u ∈ L2,λ(B′,+; RN×n). (2.208)

and[∇u]2L2,λ(B′,+) ≤

C

ν2[f ]2L2,λ(B+) +

C

Rλ

∫B+

|∇u|2, (2.209)

where C depends on θ, λ, sup ∥A∥/ν,the dimension n and the modulus ofcontinuity of z 7→ A(Rz)/ν.

Proof. As before we may assume ν = 1, R = 1. Let 2δ = 1 − θ,

a ∈ B+(0, θ) = B+(0, 1 − 2δ). (2.210)

We have to estimate1ρλ

∫B(a,ρ)

|∇u|2 (2.211)

for ρ ≤ δ/2. We distinguish three geometric situations

Table 1: Case 1

.

Case 1: an ≥ δ (i.e., a is far away from ∂Rn+)

Then B(a, δ) ⊂ B+ and we can use the interior regularity result (Theorem2.17) with x0, R = δ and θ = 1

2 and get

1ρλ

∫B(a,ρ)

|∇u|2 ≤ [u]L2,λ(B(a,δ/2))

≤C[f ]L2,λ(B(a,δ)) + C1δλ

∫B(a,δ)

|∇u|2 dx. (2.212)

Case 2: an ≤ ρ (i.e., a is very close to ∂Rn+)

Then B(a, ρ) ⊂ B+(a′, 2ρ), where a′ = (a1, . . . , an−1, 0) and we can use the

52 [March 17, 2014]

Table 2: Case 2

.

decay estimate on half balls. Since |a′| ≤ |a| < 1 − 2δ we have B+(a′, 2δ) ⊂B+ and thus∫

B(a,ρ)|∇u|2 ≤

∫B+(a′,2ρ)

|∇u|2 dx

≤C[f ]L2,λ(B+)(2ρ)λ + C

(2ρ)λ

(2δ)λ

∫B+(a′,2δ)

|∇u|2 dx. (2.213)

Table 3: Case 3

.

Case 3: ρ < an < δ (i.e., a has an intermediate distance from theboundary).

53 [March 17, 2014]

Set s = an. In this case we use first the interior estimate for the ballsB(a, s/2) ⊂ B(a, s) ⊂ B+. This gives for ρ ≤ s/2

1ρλ

∫B(a,ρ)

|∇u|2 dx ≤ C[f ]L2,λ +C

sλ

∫B(a,s)

|∇u|2 dx. (2.214)

For s/2 ≤ ρ ≤ s the estimate holds trivially as long as C ≥ 2λ. Then wenote that B(a, s) ⊂ B+(a′, 2s) ⊂ B+(a′, 2δ) ⊂ B+ and we use the decayestimate on half balls∫

B(a,s)|∇u|2 ≤

∫B+(a′,2s)

|∇u|2 dx

≤C[f ]L2,λ(B+)(2s)λ +

(2s)λ

(2δ)λ

∫B+(a′,2δ)

|∇u|2 dx. (2.215)

Together with (2.214) this concludes the proof.

Theorem 2.22 (Boundary estimates with Holder continuous coefficients).Let 0 < µ < 1, 0 < λ ≤ n+ 2µ and let

B+ = B+(x0, R), B′,+ = B+(x0, θR), 0 < θ < 1. (2.216)

Assume that the coefficients Aαβij are in C0,µ(B+) and strongly elliptic. Letfαi ∈ L2,λ(B+) and let u ∈W 1,2

+ (B+; RN ) be a weak solution of

−divA∇u = −div f in B+. (2.217)

Then∇u ∈ L2,λ(B′,+,RN×n) (2.218)

and[u]2L2,λ(B′,+) ≤

C

ν2[f ]2L2,λ(B+) +

C

Rλ

∫B+

|∇u|2 dx (2.219)

where C depends on supB ∥A∥/ν, λ, n and on Rµ[A]C0,µ/ν.

Remark. This remark was added on Nov 21, 2013.From this local estimate on can again derive a global estimate for systemsin Rn

+ with constant coefficients analogous to the estimate in Corollary 2.16for system in Rn. More precisely f ∈ L2(Rn

+). Then one can again passto the limit in the approximations −divA∇uk + 1

kuk = −div f that thereexists a unique distributional solution of

−divA∇u = −div f in Rn+ with u = 0 on ∂Rn

+ (2.220)

and ∇u = 0. Moreover

∥∇u∥L2(Rn+) ≤ C∥f∥L2(Rn

+) (2.221)

54 [March 17, 2014]

If in addition f ∈ L2,λ then we can apply (2.219) with x0 = 0, and letR→ ∞. This yields the global estimate

[∇u]L2,λ(Rn+) ≤ C[f ]L2,λ(Rn

+) (2.222)

Proof. This proof was not discussed in detail in classAgain we may assume that ν = 1, R = 1. If λ < n the assertion follows

from Theorem 2.21. Thus assume that λ ≥ n.As in the proof of the corresponding interior estimate we set 3τ = 1− θ

and we prove increasingly better estimates on the decreasing half balls

B+(0, 1 − τ) ⊃ B+(0, 1 − 2τ) ⊃ B+(0, 1 − 3τ) = B(0, θ). (2.223)

Step 1. n ≤ λ < n+ 2µ, estimate in L2,λ(B+(0, 1 − 2δ).By Theorem 2.21 applied with λ− 2µ < n instead of λ we have

[∇u]L2,λ−2µ(B+(0,1−δ)) ≤C

ν2[f ]2L2,λ−2µ(B+) +

C

Rλ−2µ

∫B+

|∇u|2 dx, (2.224)

The main point is to establish an improved version of the decay estimatefor half-balls, namely∫

B+(a′,s)|∇u− (∇u)a′,s|2 dx

≤ C[f ]2L2,λ(B+(0,1))sλ + C[A]2C0,µ [u]2L2,λ−2µ(B+(0,1−δ))s

λ

∀ a′ ∈ ∂Rn+, |a′| ≤ 1 − 2τ, 0 < s < τ. (2.225)

To prove this let

ϕ(s) =∫B+(a′,s)

|∇u− (∇u)a′,s|2 dx (2.226)

We have

−divA(a′)∇u = −divF, F = (f − fa′,r) + (A(a′) −A)∇u. (2.227)

Using Theorem 2.19 we can write u = v + w where∫B+(a′,r)

|∇w|2 dx ≤∫B+(a′,r)

|F |2 dx

≤2∫B+(a′,r)

|f − fa′,r|2 dx+ 2[A]2C0,µr2µ

∫B+(a′,r)

|∇u|2 dx

≤2[f ]L2,λrλ + 2[A]2C0,µrλ[u]2L2,λ−2µ(B+(0,1−δ)) (2.228)

and∫B+(a′,ρ)

|∇v− (∇v)a′,ρ|2 dx ≤ Cρn+2

rn+2

∫B+(a′,r)

|∇v− (∇v)a′,r|2 dx. (2.229)

55 [March 17, 2014]

This yields in the usual way

ϕ(ρ) ≤ Cρn+2

rn+2ϕ(r) + C[A]2C0,µ [u]2L2,n−δ(B+(0,1−δ))r

λ (2.230)

and (2.225) follows from the iteration lemma.Now the estimate for [∇u]2L2,λ(B(0,1−2δ)

follows by combining the half-balldecay estimate (2.225) with the interior estimate in the same way as in theproof for Theorem 2.21.

Step 2. λ = n+ 2.Let 0 < β < µ. From the estimates in Step 1 we get in particular the bound

[∇u]2C0,β(B+(0,1−2τ)) ≤ C[f ]L2,λ(B+) + C

∫B+(0,1)

|∇u|2 dx (2.231)

where the constant depends also on [A]C0,µ . Since u = 0 on ∂Rn+ ∩ B(0, 1)

we havesup

B(0,1−2τ)|u|2 ≤ [∇u]2C0,β(B+(0,1−2τ)). (2.232)

Now revisiting the derivation of (2.225) we easily obtain the half-balldecay estimate∫

B+(a′,s)|∇u− (∇u)a′,s|2 dx ≤ C[f ]2L2,λ(B+(0,1))s

λ + C[A]2C0,µ supB(0,1−2τ)

|u|2 sλ

∀ a′ ∈ ∂Rn+, |a′| ≤ 1 − 3τ, 0 < s < τ (2.233)

for λ = n+ 2µ.Combining this estimate with the interior estimate as before we finally

get (2.219)

[7.11. 2013, Lecture 7][11.11. 2013, Lecture 8]

2.1.7 Flattening the boundary

Definition 2.23. Let Ω ⊂ Rn be open and bounded. We say that Ω hasa Ck,µ boundary if for each point x0 ∈ ∂Ω there exists a r > 0 and aCk,µ function γ : Rn−1 → R such that - upon relabeling and reorienting thecoordinates if necessary - we have

Ω ∩B(x0, r) = x ∈ B(x0, r) : xn > γ(x1, . . . , xn−1). (2.234)

Likewise we define open bounded sets with Lipschitz, Ck, C∞ or analyticboundary.

56 [March 17, 2014]

Remark. For x ∈ Rn let x′ = (x1, . . . , xn − 1). It suffices to assume thatγ is defined on the n − 1 dimensional ball x′ : |x′ − x0| < r. Then γ canbe extended to Rn−1 as a C1,α function.

To ’flatten out’ the boundary we define a map Φ : Rn → Rn by

Φi(x) = xi if i = 1, . . . , n− 1, (2.235)Φn(x) = xn − γ(x1, . . . , xn−1). (2.236)

and writey = Φ(x). (2.237)

Note that

Φ(B(x0, r) ∩ Ω) ⊂ Rn+, Φ(B(x0, r) ∩ ∂Ω) ⊂ ∂Rn

+ (2.238)

Similarly we define a map Ψ : Rn → Rn by

Ψi(x) = xi if i = 1, . . . , n− 1, (2.239)Ψn(x) = xn + γ(x1, . . . , xn−1). (2.240)

and writex = Ψ(y). (2.241)

Note that Ψ and Φ are Lipschitz and invertible with Lipschitz inverse, indeedΨ Φ = Φ Ψ = id Rn . In particular

Ψ(Φ(B(x0, r) ∩ Ω)) = B(x0, r) ∩ Ω, Ψ(Φ(B(x0, r) ∩ ∂Ω) = B(x0, r) ∩ ∂Ω.(2.242)

Moreover we havedetDΦ = detDΨ = 1. (2.243)

The Campanato spaces are invariant under a composition with Bilips-chitz maps.

Lemma 2.24. Let U,Ω ⊂ Rn be open.

(i) Assume that Φ : U → Rn is a C1 diffeomorphism from U to Φ(U) and

1L|x− y| ≤ |Φ(x) − Φ(y)| ≤ L|x− y| ∀x, y ∈ U. (2.244)

Let 0 < λ < n+2. Then f ∈ L2,λ(Φ(U)) if and only if f Φ ∈ L2,λ(U)and

[f Φ]L2,λ(U) ≤ L(n+λ)/2[f ]L2,λ(Φ(U)), ∥f Φ∥L2(U) ≤ Ln/2∥f∥L2(Φ(U)).(2.245)

If detDΦ = 1 then the constants in these estimates can be improvedto Lλ/2 and 1 respectively.

57 [March 17, 2014]

(ii) Assume that k ∈ N, 0 < β < 1 and f, g ∈ Ck,β(Ω). Then fg ∈ Ck,β(Ω)and

∥fg∥Ck,β(Ω) ≤ C∥f∥Ck,β(Ω) ∥g∥Ck,β(Ω), (2.246)

where ∥f∥Ck,β(Ω) := [∇kf ]C0,β(Ω) +∑k

l=0 supΩ |∇lg|.

(iii) Assume in addition that Ω is bounded and that

Ln(B(x, r) ∩ Ω) ≥ Arn ∀r < diam(Ω). (2.247)

Let 0 < µ < 1 and 0 < λ ≤ n + 2µ. Suppose that f ∈ L2,λ(Ω) andg ∈ C0,µ(Ω). Then fg ∈ L2,λ(Ω) and

∥fg∥L2,λ(Ω) ≤ C∥f∥L2,λ(Ω) ∥g∥C0,µ(Ω), (2.248)

where ∥g∥C0,µ(Ω) := [g]C0,µ(Ω) + supΩ |g|.

Remark. For part (i) the assumption that Φ ∈ C1 is not needed. Itsuffices that Φ is Bilipschitz, i.e., (2.244). To see this one uses that Lipschitzmaps are differentiable almost everywhere and that for Bilipschitz maps thechange of variables formula still holds, if one uses the a.e. derivative in thisformula.Part (iii) is obvious for λ < n (indeed in this case it suffices to assume thatg ∈ L∞(Ω). For λ = n+ 2β with 0 < β ≤ µ Part (iii) follows from Part (ii)with k = 0 since Ln+2β(Ω) = C0,β(Ω).

Proof. (i): It suffices to show that f ∈ L2,λ(Φ(U)) =⇒ f Φ ∈ L2,λ(U)since Φ−1 has the same properties as Φ. Details: Homework. Hint: use thechange of variables formula.

(ii): This is a short calculation for k = 0 and the general case follows byinduction and the product rule.

(iii): By the remark after the lemma it suffices to consider the case λ = n.See Campanato’s original paper.

Lemma 2.25. Let 0 < µ < 1 and 0 < λ ≤ n + 2µ and assume that Ω isan open and bounded set with C1,µ boundary. Then for each x0 ∈ ∂Ω thereexists an r1 > 0 with the following property. If the coefficients Aαβij belongto C0,µ(Ω), if f ∈ C0,µ and if u ∈W 1,2

0 (Ω) is weak solution of

−divA∇u = −div f in Ω (2.249)

Then ∇u ∈ C0,µ(B(x0, r1)) and

[∇u]C0,µ(B(x0,r1)) ≤ C(∥f∥C0,µ(Ω) + ∥∇u∥L2(Ω), (2.250)

where C depends on Ω (and in particular on r1), µ, λ, n, sup ∥A∥/ν and[A]C0,µ(Ω).

58 [March 17, 2014]

Proof. Let Φ be the flattening map defined above. Then

Ω := Φ(B(x0, r) ∩ Ω) ⊂ Rn+. (2.251)

Defineu : Ω → RN by u(y) = u Ψ. (2.252)

Step 1. Derivation of the PDE for u. We have∫ΩAαβij (x)∂βuj(x)∂αφi(x) dx =

∫Ωfαi (x)∂αφi(x) dx (2.253)

for all φ ∈W 1,20 (Ω; RN ). Let φ ∈ C∞

c (Ω; RN ) and set

φ = φ Φ. (2.254)

Then φ ∈W 1,20 (Ω; RN ). Since u = u Φ we get from the chain rule and the

transformation formula (with x = Ψ(y), y = Φ(x))∫ΩAαβij (x) (∂βu)(y)∂βΦ

β(x) (∂αφ)(y)∂αΦα(x) dx

=∫

Ωfαi (x) (∂αφ)(y)∂αΦα(x) dx (2.255)

or ∫Φ(Ω)

Aαβij (y) ∂βu(y) ∂αφ(y) dy =∫

Φ(Ω)fαi (y) ∂αφ(y) dy (2.256)

for all φ ∈ C∞c (Ω; RN ), where

Aαβij (y) =[Aαβij ∂β Φβ ∂αΦα

](Ψ(y)) detDΨ(y),

f αi =[fαi ∂α Φα

](Ψ(y)) detDΨ(y). (2.257)

By density we see that (2.256) also holds for all φ ∈W 1,20 (Ω; RN ).

Step 2. Regularity of u.Let a0 = Φ(x0) = (x′0, 0). Since Ψ is continuous there exist a radius r0 suchthat Ψ(B(a0, 2r0) ⊂ B(x0, r). Then Ψ(B+(a0, 2r0) ⊂ B(x0, r) ∩ Ω. UsingLemma 2.24 we see that f ∈ L2,λ(Ω; Rn×N ) and A ∈ C0,µ(Ω). Moreover u ∈W 1,2(Ω; RN ) implies that u ∈W 1,2

+ (B+(a0, 2r0); RN ). We finally verify thatthat the transformed coefficients are still elliptic. Indeed, since detDΨ = 1,

Aαβij (y)ξαξβζiζj = [Aαβij ∂αΦα∂βΦβ ](Ψ(y)) detDΨ(y)ξαξβζ

iζj

= Aαβij (Ψ(y))(∂αΦα(Ψ(y))ξα

) (∂βΦβ(Ψ(y))ξβ

)ζiζj

≥ ν∣∣((∇Φ)T (Ψ(y)))ξ

∣∣2 |ζ|2≥ ν

∣∣∣((∇Φ)T (Ψ(y)))−1∣∣∣−2

|ξ|2 |ζ|2 ,

59 [March 17, 2014]

where in the last step we used that |ξ| ≤∣∣∣((∇Φ)T (Ψ(y))

)−1∣∣∣ |∇Φ(Ψ(y))ξ|.

Thus by Lemma 2.22 we see that ∇u ∈ L2,λ(B+(a0, r0)).

Step 3. Regularity of u.Since Φ is continuous there exists r1 ∈ (0, r) such that Φ(B(x0, r1)) ⊂B(a0, r0). Then Φ(B(x0, r1)∩Ω) ⊂ B+(a0, r0). Using again Lemma 2.24 wesee that ∇u ∈ L2,λ(B+(a0, r0)) implies that ∇u ∈ L2,λ(B(x0, r1) ∩ Ω).

2.1.8 Global estimates

Theorem 2.26. Let 0 < µ < 1. Assume that Ω is bounded and openwith C1,µ boundary. Assume that the coefficients Aαβij are in C0,µ(Ω; RN )and strongly elliptic, that f ∈ L2,λ(Ω; Rn×N ), g ∈ L2,λ−2(Ω; RN ) and theu ∈W 1,2

0 (Ω) is a weak solution of

−divA∇u = −div f + g in Ω. (2.258)

Then ∇u ∈ L2,λ(Ω; RN ) and

[∇u]L2,λ(Ω) ≤ C[f ]L2,λ(Ω) + C[g]L2,λ−2(Ω) + C∥∇u∥L2(Ω). (2.259)

Remark. This remark was added on Nov 22, 2013.If k ≥ 2 and if in addition ∂Ω ∈ Ck,µ, A ∈ Ck−1,µ, ∇lf ∈ L2,λ for alll ≤ k and ∇lg ∈ L2,λ for l ≤ k − 1 one can similarly obtain estimates for[∇ku]L2,λ by locally differentiating the equation (see the proof of Theorem2.30 below).In particular for k = 2, A ∈ C1,µ and f = 0 and g ∈ L2,λ one obtains

[∇2u]L2,λ(Ω) ≤ C[g]L2,λ(Ω) + C∥g∥2L2 + C∥∇u∥L2(Ω). (2.260)

Proof. This follows by combining the interior estimates and the boundaryestimates through a covering argument. By Lemma 2.27 we may assumeg = 0.

For each x ∈ ∂Ω there exists r1(x) > 0 such that the conclusion ofLemma 2.25 holds. Then the collection of balls B(x, r1(x)) for x ∈ ∂Ω covers∂Ω. Since ∂Ω is compact there exists finitely many sets Vi = B(ai, ri) withri = r1(ai) such that

∂Ω ⊂m∪i=1

Vi. (2.261)

Set

Ω0 = Ω \m∪i=1

Vi. (2.262)

60 [March 17, 2014]

Then Ω0 is a compact subset of Ω and thus

dist (Ω0, ∂Ω) = infx∈Ω0

dist (x, ∂Ω) = minx∈Ω0

dist (x, ∂Ω) > 0. (2.263)

Setr :=

12

min(dist (Ω0, ∂Ω), mini=1,...M

ri). (2.264)

If x ∈ Ω0 we can use the interior estimate with R0 = 2r, θ = 1 toestimate

ϕ(x, r) :=∫B(x,r)∩Ω

|∇u− (∇u)x,r|2 dz (2.265)

for r ≤ r.If x ∈ Vi we use the boundary estimate, Lemma 2.25 to estimate ϕ(x, r)

for r ≤ r ≤ 12ri.

Finally if r > r we use the trivial estimate ϕ(x, r) ≤ r−λ∥∇u∥2L2(Ω).

Lemma 2.27. Let 0 < λ < n + 2 and assume that there exists an A > 0such that

Ln(B(x, r) ∩ Ω) ≥ Arn ∀ x ∈ Ω, r < diam(Ω). (2.266)

Assume that g ∈ L2,λ−2(Ω) (where we use the convention (2.132) if λ < 2).Then there exists f ∈ L2,λ(Ω; Rn) such that

−div f = g (2.267)[f ]L2,λ(Ω) ≤ C(A)[g]L2,λ−2(Ω). (2.268)

Proof. Let R = 2diam(Ω). Then there exists x0 ∈ Ω such that Ω ⊂B(x0, R/2). Extend g by zero to B(x0, R), let v ∈ W 1,2

0 (B(x0, R)) be theweak solution of

−∆v = g in B(x0, R) (2.269)

and set f = ∇v.Then −div f = g and by the energy estimate∫

B(x0,R)|∇v|2 dx ≤ C1R

2

∫Ω|g|2 dx ≤ C1R

λ[g]2L2,λ(Ω), (2.270)

where C1 is the constant Poincare constant for the unit ball with zero boun-dary conditions.

The interior estimate for equations with constant coefficients (see theremark after Theorem 2.14) implies that ∇u ∈ L2,λ(B(x0, R),RN ) and

[∇v]2L2,λ(Ω) ≤ [∇v]2L2,λ(B(x0,R/2)) ≤ C[g]2Lλ−2(B(x0,R)) +C

Rλ

∫B(x0,R)

|∇v|2 dx.

(2.271)

61 [March 17, 2014]

Together with the energy estimate this yields the assertion since

[g]2Lλ−2(B(x0,R)) ≤ max(2λ−2, 42−λ)[g]2Lλ−2(Ω). (2.272)

Indeed for λ ≤ 2 this follows directly from the definition of the seminorm in(2.132) since diam(B(x0, R)) = 2R = 4diam(Ω). If λ > 2 let x ∈ B(x0, R)and 0 < ρ < R. If B(x, ρ) ∩ Ω = ∅ then g = 0 on B(x, ρ) and there isnothing to show. If x1 ∈ B(x, ρ) ∩Ω then B(x, ρ) ∩Ω ⊂ B(x1, r) ∩Ω wherer ≤ min(2ρ,diam(Ω)). Thus∫

B(x,ρ)|g|2 dy =

∫B(x,ρ)∩Ω

|g|2 dy ≤∫B(x1,r)∩Ω

|g|2 dy

≤ rλ−2[g]2L2,λ−2(Ω) ≤ 2λ−2ρλ−2[g]2L2,λ−2(Ω). (2.273)

2.1.9 Ck,α theory, Schauder estimates up to the boundary

Theorem 2.28. Let 0 < µ < 1. Assume that Ω is bounded and open withC1,µ boundary. Let φ ∈ C1,α(Ω). Assume that the coefficients Aαβij are inC0,µ(Ω; RN ) and strongly elliptic and that u ∈ W 1,2(Ω) is a weak solutionof

−divA∇u− divBu+ C∇u+Du = −div f + g in Ω (2.274)

withu = φ on ∂Ω, (2.275)

where

(Bu)αi (x) = Bαij(x)u

j(x), (C∇u)i(x) = Cαij(x)∂αu(x)j , (Du)i = Dij(x)uj(x)

(2.276)and

Bαij ∈ C0,µ(Ω), Cαij ∈ L∞(Ω), Dij ∈ L∞(Ω), (2.277)

f ∈ C0,µ(Ω), g ∈ L∞(Ω). (2.278)

Then u ∈ C1,µ(Ω) and

∥u∥C1,µ(Ω) ≤ C(∥f∥C0,µ + ∥g∥L∞ + ∥φ∥C1,µ(Ω) + ∥u∥L2). (2.279)

Remark. (i) The boundary condition is understood in the sense that u−φ ∈W 1,2

0 (Ω; RN ).(ii) It suffices to assume that C,D, g ∈ L2,n+2µ−2.(iii) If we have an a priori estimate in W 1,2 then the term with u on theright hand side can be dropped. Example: if B = C = D = 0, φ = 0and N = 1 the energy estimate and the Poincare inequality imply that∥u∥L2 ≤ C∥∇u∥L2 ≤ C(∥f∥L2 + ∥g∥L2).

62 [March 17, 2014]

(iv) In general the term ∥u∥L2(Ω) cannot be dropped because there can benontrivial solutions for f = g = 0 (i.e. eigenfunctions of the operator on theleft with eigenvalue zero) . Consider e.g. the case n = 1 and the equation−u′′ − π2u = 0 on (0, 1) which has the solution u(x) = sinπx.

[11.11. 2013, Lecture 8][14.11. 2013, Lecture 9]

Proof. We may assume that φ = 0. Indeed if φ = 0 we use that v := u− φsatisfies a similar equation (with f replaced by f−A∇φ−Bφ and g replacedby g − C∇φ−Dφ), use the estimate for v and deduce the estimate for u.

We first show the slightly weaker estimate

[∇u]C1,µ(Ω) ≤ C(∥f∥C0,µ + ∥g∥L∞ + +C supΩ

|u| + supΩ

|∇u|). (2.280)

Then the full estimate follows from a simple interpolation estimate, seeLemma 2.29 below. Indeed given (2.280) it suffices to take ε = 1

2C in theLemma. Then the assertion follows.

Let

G = g − C∇u−Du (2.281)F = f −Bu. (2.282)

Then−divA∇u = −divF +G. (2.283)

Now we can apply Theorem 2.26 with λ = n + 2µ. Since L2,λ+2µ = C0,µ

this yields

[∇u]C0,µ ≤ C([F ]C0,µ + [G]L2,n+2µ−2 + ∥ ∇u∥L2). (2.284)

Since [G]L2,n+2µ−2 ≤ Cdiam(Ω)2−2µ∥G∥L∞ and [u]C0,µ ≤ C[u]C0,1 ≤≤ C supΩ |∇u| (for the last estimate see Lemma 2.29(i)) this implies (2.280).

To absorb the lower order terms we have used the following result.

Lemma 2.29. Let Ω be a bounded domain with Lipschitz boundary.

(i) [u]C0,1(Ω) ≤ C(Ω) supΩ |∇u|.

(ii) Let 0 < µ < 1, let k ∈ N, k ≥ 1. Then for each ε > 0 there exists aCε > 0 (which also depends on µ and k) such that

k∑l=0

supΩ

|∇lu| ≤ ε[∇ku]0,µ + Cε∥u∥L2 . (2.285)

63 [March 17, 2014]

Proof. (i) : Homework sheet 5: If the line segment from [x, y] is containedin Ω this follows by calculating d

dtu((1 − t)x + ty). If x and y are close tothe boundary one can first map the region containing x and y to a half-balland use the same argument there.

(ii) This can be proved in the usual way by contradiction and compact-ness (which in this case comes from the Arzela-Ascoli theorem). Details:exercise.

Theorem 2.30. Let k ≥ 2. Let µ, Ω, A, B, C, D, f , g and u be as inTheorem 2.28 and assume in addition that

φ ∈ Ck,µ(Ω), A,B, f ∈ Ck−1,µ(Ω), C,D, g ∈ Ck−2,µ(Ω), ∂Ω ∈ Ck,µ.(2.286)

Then ∇u ∈ Ck−1,µ(Ω) and

∥u∥Ck,µ(Ω) ≤ C(∥f∥Ck−1,µ(Ω) + ∥g∥Ck−2,µ(Ω) + ∥φ∥Ck,µ(Ω) + ∥u∥L2). (2.287)

Proof. Again it suffices to consider the case φ = 0. Let V1, . . . , Vm and Ω0

and r > 0 be as in the proof of Theorem 2.26.Assume first that B = C = D = 0. We will show that estimate

[∇ku]L2,n+2µ(Ω) ≤ C([f ]Ck−1,µ(Ω) + ∥g∥Ck−2,µ(Ω) + ∥φ∥Ck,µ(Ω) + ∥u∥Ck(Ω)).(2.288)

If x ∈ Ω0 the B(x0, 2r) ⊂ Ω and by induction we see thatv := ∂γ1 . . . ∂γk−1

u is a weak solution of

−divA∇v = −div f + ∂γ1(∂γ2 . . . ∂γk−1

g)− div f (2.289)

where f is a sum of products of l derivatives of A and k− l derivatives of uwith l ≥ 1. Thus

[f ]C0,µ ≤ C∥A∥Ck−1,µ∥u∥Ck−1,µ ≤ C∥A∥Ck−1,µ∥u∥Ck . (2.290)

Thus the interior estimate yields the desired estimate for r−n−2µ∫B(x,r) |∇v−

(∇v)x,r|2 dx and hence for ∇ku.If x ∈ Vi we argue similarly with the following modifications. We first

show the estimate for the transformed function u = u Ψ. We differen-tiate the equation for u (k − 1) times in tangential direction. This yieldsthe estimates for all k-th partial derivatives of u which contain at most onederivative in normal direction. To recover the higher order normal deriva-tives we use the transformed PDE. This yields (with the same notation asbefore)

Bnn∂n∂nu = ∂n( Bnn ∂nu) − (∂nBnn)∂nu,

∂n(Bnn∂nu) = −∑

(α,β) =(n,n)

∂αBαβ∂βu (2.291)

64 [March 17, 2014]

Taking inductively tangential and normal derivatives of these identities weobtain control of all partial derivatives of u of order k in L2,n+2µ. Finally thechain rule yields control of all partial derivatives of order k of u = uΦ.

2.1.10 An alternative route to Schauder estimates by rescalingand blow-up

Here we give a different argument, only based on scaling, which can be usedto prove an estimate which only depends on the data of the problem (i.e.,f , A and Ω) under the assumption (!) that we already have a C1,µ solution.

By slight additional arguments one can also show the existence of C1,µ

solutions and that every weak solutions is already in C1,µ. This approach isbeautifully developed in the paper

• L. Simon, Schauder estimates by scaling, Calculus of Variations andPartial Diff. Equations 5 (1997), 391–407.

The notation is at first glance a bit overwhelming, because Simon treatssimultaneously elliptic equations like −∆u = −div f or ∆2u = div div f andparabolic equations like ∂t −∆u = −div f . For a first reading it completelysuffices to take κ = (1, . . . , 1) and I = J = 1 so that each term Dγ and Dδ

just contains one partial derivative.

Theorem 2.31. Let 0 < µ < 1. Assume that Ω has C1,µ boundary, thatthe coefficients Aαβij are in C0,µ(Ω) and strongly elliptic, that fαi ∈ C0,µ(Ω)and that u ∈ C1,µ(Ω; RN ) is a weak solution of

−divA∇u = −div f + g (2.292)

withu = 0 on ∂Ω. (2.293)

Then there exists a constant C which depends on ν, ∥A∥C0,µ, and on thenumber and the C1,µ norms of the functions describing ∂Ω such that

[∇u]C0,µ(Ω) ≤ C([f ]C0,µ(Ω) + [g]L2,n+2µ−2(Ω) + supΩ

|∇u| + supΩ

|u|). (2.294)

Remark. (i) Using the interpolation lemma we can replace C supΩ |∇u|+C supΩ |u| by C∥u∥L2 on the right hand side and [∇u]C0,µ(Ω) by ∥u∥C1,µ(Ω)

on the left hand side.(ii) Using a similar argument one can also prove a priori estimates in Cam-panato spaces with 0 < λ ≤ n.

Proof. For ease of notation we only show the estimate for a fixed Ω. Assumethat the conclusion of the theorem does not hold. Then there exist sequences

65 [March 17, 2014]

Ak, fk, uk such that the coefficients Ak are strongly elliptic with an ellipticityconstant ν > 0 which is independent of k such that uk = 0 on ∂Ω and

−divAk∇uk = −div fk + gk in Ω, (2.295)

with∥Ak∥C0,µ ≤M, (2.296)

and[∇uk]C0,α = 1, [∇fk]C0,α → 0, [gk]Ln+2µ−2 → 0, (2.297)

supΩ

|∇uk| + |uk| → 0. (2.298)

Thus there exist xk, yk such that

|∇u(xk) −∇u(yk)||xk − yk|

≥ 12. (2.299)

Since sup |∇uk| → 0 this implies that

rk := |xk − yk| → 0. (2.300)

Passing to a subsequence we may assume that

xk → x and yk → x. (2.301)

Case 1. lim supk→∞ dist (xk, ∂Ω)/rk = ∞.Pass to a subsequence (not relabeled) such that limk→∞ dist (xk, ∂Ω)/rk =∞. Define

vk(z) = r−1−µ[uk(xk + rkz) − uk(xk) −∇u(xk) · rkz] (2.302)fk(z) = r−µ[fk(xk + rkz) − f(xk)], (2.303)gk(z) = r1−µg(xk + rkz), (2.304)Ak(z) = Ak(xk + rkzk). (2.305)

Then

[∇vk]C0,α(Ωk) = [∇uk]C0,α(Ω) = 1, (2.306)

[fk]C0,α(Ωk) = [fk]C0,α(Ω) → 0, (2.307)[gk]Ln+2µ−2(Ωk) = [gk]Ln+2µ−2(Ω) → 0, (2.308)

sup ∥Ak∥ = sup ∥Ak∥, (2.309)[Ak]C0,µ = rµk [Ak]C0,µ → 0, (2.310)

whereΩk := z ∈ Rn : xk + rkz ∈ Ω (2.311)

66 [March 17, 2014]

Moreover the coefficients Ak are strongly elliptic with the same ellipticityconstant ν.

The equation for uk becomes

−div Ak(z)(∇uk(xk) + rµk∇vk(z)) = −div rµkfk(z) + rkrµ−1k gk. (2.312)

Here the divergence is taken with respect to z. This can be rewritten as

−div Ak(z)∇vk = div fk(z) + gk + div

[Ak(z) − Ak(0)

rµk∇uk(xk)

]. (2.313)

Here we used that Ak(0)∇uk(xk) is independent of z and hence its divergenceis zero.

We now want to pass to the limit in (2.313). We first observe that theassumption limk→∞ dist (xk, ∂Ω)/rk = ∞ implies that for every R > 0 thereexists a k0 such that B(0, R) ⊂ Ωk for all k ≥ k0. Fix R > 0. We claim that

fk → 0,Ak(z) − Ak(0)

rµk∇uk(xk) → 0 in C0(B(0, R)). (2.314)

Indeed the first assertion follows from the fact the fk(0) = 0 and [fk]C0,µ →0. We the second assertion we use that (Ak(z) − Ak(0))r−µk is bounded by[Ak]C0,µ and that supΩ |∇uk| → 0. Moreover we have∫

B(0,R)|gk|2 dx ≤ Rn+2µ−2[gk]2L2,n+2µ−2 → 0. (2.315)

This show that the right hand side of (2.313) goes to zero in a suitableway (in particular in the sense of distribution and in the dual space ofW 1,2

0 (B(0, R)).We now look at the left hand side. First, the estimate of the Holder

norm of Ak implies that

supB(0,R)

|Ak −Ak(x)| → 0. (2.316)

Thus there exists a constant A such that

Ak → A (2.317)

and A is strongly elliptic with ellipticity constant ν.Regarding the convergence of vk we have vk(0) = 0, ∇vk(0) = 0 and

thus (for k ≥ k0(R)).

|∇vk| ≤ |z|µ, |vk(z)| ≤ |z|1+µ, ∀z ∈ B(0, R). (2.318)

67 [March 17, 2014]

Since [∇vk]C0,µ = 1 by the Arzela-Ascoli theorem there exists a subsequence(not relabeled) such that10

vk → v in C1(B(0, R),RN ) ∀R > 0. (2.319)

It follows that v solves the constant coefficient PDE

−div A∇v = 0 in Rn. (2.320)

Now the first inequality in (2.318) and the (local) convergence in C1 implythat

|∇v(z)| ≤ |z|µ ∀ z ∈ Rn. (2.321)

Hence taking the limit r → 0 in the decay estimate (2.79) for solutions inRn we get ∇2v ≡ 0. Since ∇v(0) = 0 it follows that

∇v ≡ 0. (2.322)

How does this lead to a contradiction ? By construction we have

|∇vk(zk)| ≥12, where zk =

yk − xkrk

and hence |zk| = 1. (2.323)

Thus there exists a further subsequence such that zk → z. Now by theuniform convergence of vk in B(0, 2) we get

12≤ lim inf

k→∞|∇vk(zk)| = |∇v(z)| = 0. (2.324)

This contradiction finishes the proof.

Case 2. lim supk→∞ dist (xk, ∂Ω)/rk <∞.This was only sketched very briefly in class.One defines fk, gk and Ak and obtains again a limiting equation with con-stant coefficients and zero right hand side. In this case, however, the limitingequation does not hold in Rn, but only on a half space and we have to usethe zero boundary condition on ∂Ω to show that the limit map v vanisheson the boundary of that half space. To do so it is convenient to slightlychange the definition of vk.

Let xk ∈ ∂Ω be such that |xk − xk| = dist (xk, ∂Ω). Let

vk(z) = r−1−µk (uk(xk + rkz) −∇uk(xk) · z) (2.325)

Then∇vk(z) = r−µk (∇uk(xk + rkz) −∇uk(xk)) . (2.326)

10More precisely one first considers R = j and for each j ∈ N one obtains a subsequence.Then a diagonalization argument shows that there exists a single subsequence which suchthat converges holds for all j.

68 [March 17, 2014]

Thus vk(0) = uk(xk) = 0 (since xk ∈ ∂Ω) and ∇vk(0) = 0. It followsthat |∇vk(z)| ≤ |z|α and |vk(z)| ≤ C|z|1+α. Now assuming (after a possiblerenumbering of the coordinates) we we assume that

Ω ∩B(x, s) = x ∈ B(x, s) : xn > γ(x′) (2.327)

Hence vk is at least defined the the set

Yk := z ∈ B(0, s/(2rk)) : rk(zk)n > γ(x′k + rkz′k) − γ(x′k) (2.328)

Set

γk(z′) = r−1−µk

[γ(xk + rkz

′) − γxk−∇γ(xk) · zk

]νk = (∇γ(x′k),−1).

(2.329)Hence

Yk := z ∈ B(0, s/(2rk)) = −νk · z > rµk γk(z′). (2.330)

Let ν = (∇γ(x′),−1) Then for each δ > 0 and R > 0 there exists a k0

such that for k ≥ k0 we have

Yk ⊃ z ∈ B(0, R) : −ν · z ≥ δ = Hδ,R. (2.331)

It follows that for a subsequence vk → v in C1(Hδ,R). Thus v is defined onthe half-space

H = z ∈ Rn : −z · ν ≥ 0 (2.332)

and−divA(x)∇v = 0 in H, |∇v(z)| ≤ |z|α (2.333)

Moreover the boundary condition for u can be rewritten as

uk(x′k + rkz′k, γ(x

′k) + rkzn) = 0 if − z · νk = rµk γk(z

′). (2.334)

This implies that

vk(z) = 0 − r−µ∇uk(xk) · z if − z · νk = rµk γk(z′) (2.335)

Now ∇uk is parallel to νk (since u = 0 on ∂Ω), thus ∇uk = λkνk and λk → 0since supΩ |∇uk| → 0. Thus

vk(z) = 0 − r−µλkz · νk if − z · νk = rµk γk(z′) (2.336)

Since λk → 0 this implies that

v(z) = 0 if − z · ν = 0. (2.337)

By the half-space estimate (2.191) (with λ = 0) for systems with constantcoefficients we deduce again that ∇v = const and hence ∇v = 0 which leadsto a contradiction as in Case 1.

69 [March 17, 2014]

Remark. Already argument in Case 1 shows the main idea of the blow-up method. By blow-up one obtains a much simpler equation which can beeasily analyzed. The heart of the matter is then a compactness argumentwhich finally yields the contradiction. In our case the compactness comesfrom the Arzela-Ascoli theorem which gives compactness in C0 for sequenceswhich are bounded in C0,µ. For the proof of the analogous result with theL2,n norm rather then the C0,µ norm one uses a compactness argumentbased on the reverse Poincare inequality (and the compact embedding ofW 1,2(B(0, R) into L2(B(0, R)).

The following arguments until the end of this subsubsection were notdiscussed in detail in class.The result above assumes that a C1,µ solution exists. With slight additionalarguments one can develop also a good existence theory.

First possibility: smoothing From the result above one can develop atheory which also give existence of C1,µ, e..g. as follows. First observe theabove argument can also be applied to the equation −divAu+Ku = −div f ,where K ∈ R is a constant (which will later help to get a unique weaksolution). Then note that for A ∈ C∞(Ω) and ∂Ω in C∞ every weak solutionis in C∞(Ω). This can be proved by differentiating the equation in theinterior and by flattening and tangential differentiation at the boundary.

Now consider a sequence of coefficients A(k) ∈ C∞ converge to A uni-formly such that the Holder norm of A(k) is uniformly bounded. Similarlyone approximates Ω by smooth sets Ωk and f by f (k). Now by Gardings’sinequality there exists a K > 0 (independent of k) such that for all k theproblem

−divA(k)∇u(k) +Ku(k) = −div f (k) (2.338)

has a unique solution in W 1,20 (Ω; RN ) and ∥u(k)∥W 1,2 is uniformly bounded

and indeed u(k) → u in W 1,2(Ω; RN ), where u is the (unique) weak solutionof the limit problem.

Theorem 2.31 in combination with the interpolation inequality showsthat u(k) is uniformly bounded in C1,µ(Ω) and hence u ∈ C1,µ(Ω). Usingthe Fredholm alternative (or more precisely repeating its proof) we can nowsee that the problem

−divA∇u− σu = −div f (2.339)

has a unique C1,µ solution for all σ /∈ Σ where Σ is a countable set.This show the existence of C1,µ solution but it does not yet show that

every weak solution is C1,µ. We just sketch this argument. Note that if u isa weak solution of −divA∇u = −div f + g then u is also a solution of

Lu = −divA∇u+Ku = −div f + g (2.340)

70 [March 17, 2014]

where g = g + Ku. Now one needs a version of Theorem 2.31 in all L2,λ

spaces with λ ≤ n, i.e., an estimate of the form

[∇u]L2,λ(Ω) ≤ C([f ]L2,λ(Ω) + [g]L2,λ−2(Ω) + ∥∇u∥L2). (2.341)

which can be proved in a similar way. Then one improve the regularity of thesolutions of (2.340) by a bootstrap argument. One starts with u ∈W 1,2

0 (Ω)which gives (by the Poincare inequality) u ∈ L2,2. If 2 ≥ n+ 2µ− 2 we aredone. If 2 < n + 2µ − 2 we can apply (2.341) to get ∇u ∈ L2,4 and henceu ∈ L2,6. If 6 ≥ n+2µ− 2 we are done. If not we repeat the argument untilwe get u ∈ L2,n+2µ−2 = L2,n+2µ−2 and then use (2.341) to conclude.

Second possibility: interior and boundary estimate Using the sameargument as in the proof of Theorem 2.31 (in fact only Case 1) one can showan interior estimate

Rµ[∇u]C0,µ(B(x0,θR)

≤Cε(Rµ[f ]C0,µ(B(x0,R)) +R−n/2∥∇u∥L2(B(x0,R)) + ε( supB(x0,R)

|∇u| +R|u|)).

(2.342)

Using a nice geometric lemma one can then remove the termε(supB(x0,1) |∇u| + R|u|)) on the right hand side and obtain the full localestimate (see Theorem 2 in Simon’s paper and the lemma following it).

Similarly one can prove estimates for half balls. Then these estimatecan be combined in the same way as above by flattening the boundary andusing a covering of the boundary to show regularity of every weak solution.

[14.11. 2013, Lecture 9][18.11. 2013, Lecture 10]

2.1.11 Lp theory

Definition 2.32. Let Ω ⊂ Rn be open. We say that f belong to the spaceBMO(Ω) (the space of function of bounded mean oscillation) if f ∈ L1(Ω)and

[f ]BMO(Ω) := sup 1Ln(Q)

∫Q|f − fQ| dx : Q ⊂ Ω cube <∞. (2.343)

HerefQ :=

1Ln(Q)

∫Qf dx. (2.344)

The space BMO(Ω) is a Banach space with the norm

∥f∥BMO(Ω) := ∥f∥L1(Ω) + [f ]BMO(Ω). (2.345)

71 [March 17, 2014]

The Cauchy-Schwarz inequality yields

1Ln(Q)

∫Q|f − fQ| dx ≤

(1

Ln(Q)

∫Q|f − fQ| dx

)1/2

(2.346)

and it follows easily that

L2,n(Ω) ⊂ BMO(Ω), [f ]BMO (Ω) ≤ C[f ]L2,n(Ω). (2.347)

The difference between cubes and balls in the definition is irrelevant sincea cube is always contained in a ball of comparable volume and vice versa.We shall see later that for nice domains Ω (e.g. a cube or more generallya bounded domain with Lipschitz boundary) we actually have L2,n(Ω) =BMO(Ω) and that the norms are equivalent.

Theorem 2.33 (Lp estimate from L2 and BMO estimates). (i) Let Ω bean open bounded set in Rn and suppose that Q ⊂ Rn is a cube. Sup-pose that the linear operator T : L2(Ω; Rd1) → L2(Q; Rd2) is a linearoperator such that

∥Tf∥L2(Q) ≤ M2∥f∥L2(Ω) ∀f ∈ L2, (2.348)[Tf ]L2,n(Q) ≤ M∞∥f∥L∞(Ω) ∀f ∈ L∞. (2.349)

Then for all 2 < q <∞

∥Tf∥Lq(Ω) ≤ C(q, n,M2,M∞,Ln(Ω)/Ln(Q)) ∥f∥Lq(Ω) ∀ f ∈ Lq.(2.350)

(ii) Let Ω = U = Rn or Ω = U = Rn+ and suppose that T : L2(Ω; Rd1) →

L2(U ; Rd2) is a linear operator such that

∥Tf∥L2 ≤ M2∥f∥L2 ∀f ∈ L2, (2.351)[Tf ]L2,n ≤ M∞∥f∥L∞ ∀f ∈ (L2 ∩ L∞). (2.352)

Then for all 2 < q <∞

∥Tf∥Lq ≤ C(q, n,M2,M∞) ∥f∥Lq ∀ f ∈ (L2 ∩ Lq) (2.353)

and T has a unique extension to a bounded operator from Lq(Ω; Rd1)to Lq(U ; Rd2).

(iii) The assertion in (i) holds if the cube Q is replaced by a bounded openset with Lipschitz boundary.

Proof. A short proof of (i) (with L2,n replaced by BMO ) can be found in[Ca66]. It uses Lemma 3 of [JN]. In the form stated assertion (i) then followsfrom (2.347).

72 [March 17, 2014]

If Ω = U = Rn then assertion (ii) follows directly from the proof of (i)in [Ca66]. This proof shows that ∥Tf∥Lq(Rn) ≤ C∥u∥Lq(Rn) if u ∈ Lq(Rn) ∩L2(Rn). Since this space is dense in Lq the estimate T has a unique extensionto a bounded operator from Lq(Rn) to Lq(Rn). To treat the case Ω = U =Rn

+ we define an extension operator E : Lp(Rn+) → Lp(Rn) by Ef(x′, xn) =

f(x′, |xn|), i.e., by reflection. Then it is easy to show that [Ef ]L2,n(Rn) ≤2[f ]L2,n(Rn

+) (see Homework 4). Now let R : Lp(Rn) → Lp(Rn+) be the

restriction operator, i.e., Rf = f|Rn+

and let T = ETR. Then T satisfies theassumptions for Ω = U = Rn and hence is a bounded operator from Lq(Rn)to Lq(Rn). Now T = RTE and hence T is also a bounded operator fromLq(Rn

+) to Lq(Rn+)

For part (iii) one uses that there exists a linear extension operator Ewhich maps L2,n(U) to L2,n(Q) where Q is a cube which contains U . In factone can even extend to BMO (Rn), see Lemma 2.34. Then one can apply (i)to the operator T = ET and we conclude that

∥T f∥Lp(Q) ≤ Cp∥f∥Lp(Ω). (2.354)

Now T f = ETf = Tf in U and thus we have the trivial estimate

∥Tf∥Lp(U) ≤ ∥T f∥Lp(Q). (2.355)

Lemma 2.34. Let U ⊂ Rn be a bounded domain with Lipschitz boundary.Then there exists a linear bounded operator E : L2,n(Ω) → L2,n(Rn) with(Ef)|U = f . Moreover E is also bounded as an operator from L2(Ω) toL2(Rn).

Proof. This proof was not discussed in detail in class.To construct this operator one first considers the situation in the half spaceand then uses a local flattening of the boundary and a partition of unity.First, if v ∈ L2,n(Rn

+) and if one extends v by reflection to Rn, i.e., if onedefines (ERn

+v)(x′, xn) = v(x′, |xn|) then ∥ERn

+v)∥L2,n(Rn) ≤ 2∥v∥L2,n(Rn

+)

(see Homework 4).Now assume that x ∈ ∂Ω and ∂Ω∩B(x, r) is a Lipschitz graph and Ω∩

B(x, r) lies above this graph (see Definition 2.23). Assume that u ∈ L2,n(Ω)and suppu ⊂ B(x, r). Then by Lemma 2.24 we have uΨ ∈ L2,n(Rn

+) (notethat by the remark after that lemma it suffices that Ψ is Bilipschitz). Hencewe define an extension operator by

EB(x,r)u := [ERn+(u Ψ)] Φ. (2.356)

Then another application of Lemma 2.24 shows that

∥EB(x,r)u∥L2,n(Rn) ≤ C∥u∥L2,n(Ω) if suppu ⊂ B(x, r). (2.357)

73 [March 17, 2014]

Here C may depend on x and r.Now by compactness of ∂Ω there exist finitely many balls B(xi, ri) as

above such that∪mi=1B(xi, ri/2) ⊃ Ω. Set

Ω0 = Ω \m∪i=1

B(xi, ri/2). (2.358)

Now we consider a partition of unity subordinate to these sets. Let ηi ∈C∞c (B(xi, ri) with ηi = 1 in B(xi, ri/2) and η0 ∈ C∞

c (Ω) with η0 = 1 in Ω0.Then set

ηi =ηi∑mj=0 ηj

. (2.359)

Then∑m

i=0 ηi = 1 in Ω and the ηi are smooth since∑m

j=0 ηj ≥ 1 in Ω. Nowdefine

Eu :=m∑i=0

Ei(ηiu), (2.360)

where Ei = EB(xi,ri) for i ≥ 1 and where E0 denotes the extension by zerooutside Ω, i.e. (E0v)(x) = v(x) for v ∈ Ω and (E0v)(x) = 0 if x ∈ Rn \ Ω.Let

r :=12

min(dist (Ω0, ∂Ω), mini=1,...m

ri) (2.361)

Then one can easily check that ∥ηiu∥L2,n(Ω) ≤ C(r)∥u∥L2,n(Ω) for i ≥ 0and ∥E0(η0u)∥L2,n(Rn) ≤ ∥η0u∥L2,n(Ω). The estimates in L2 are similar buteasier.

Theorem 2.35. Suppose that the coefficients Aαβij are constant and stronglyelliptic. Let 1 < p <∞.

(i) Assume that f ∈ Lp(Rn; Rn×N ). Then the problem

−divA∇u = −div f (2.362)

has a distributional solution u ∈ W 1,ploc (Rn) with ∇u ∈ Lp(Rn; RN×n)

and u is unique up to the addition of constants. Moreover

∥∇u∥Lp ≤ C(p, n,A)∥f∥Lp . (2.363)

(ii) Let g ∈ Lp(Rn; RN ). Then the problem

−divA∇u = g (2.364)

has a distributional solution u ∈W 2,ploc (Rn; RN ) with ∇2u ∈ Lp(Rn; RN×n×n)

and u is unique up to the addition of affine functions. Moreover

∥∇2u∥Lp ≤ C(p, n,A)∥g∥Lp . (2.365)

74 [March 17, 2014]

Remark. The constant C(p, n,A) depends on A only through ∥A∥ andthe ellipticity constant ν.We sometimes write T1 for the solution operator f 7→ ∇u and T2 for thesolution operator g 7→ ∇2u.

Proof. Step 1. Proof of (i) for 2 < p <∞.This is a direct consequence of Corollary 2.16 and the interpolation estimateTheorem 2.33 (ii).

In more detail one argues as follows. Let f ∈ L2. Then Corollary 2.16states there exists a unique (up to constants) weak solution u (with ∇u ∈ L2)of −divA∇u = −div f . Define

Tf = ∇u. (2.366)

Then by Corollary 2.16 (and the trivial inclusion L2,n ⊂ L2,n = L∞)

∥Tf∥L2 ≤ C∥f∥L2 ∀f ∈ L2, (2.367)[Tf ]L2,n ≤ C∥f∥L∞ ∀f ∈ L2 ∩ L∞ (2.368)

Hence∥Tf∥Lp ≤ Cp∥f∥Lp ∀f ∈ L2 ∩ Lp. (2.369)

If f ∈ Lp there exist fk ∈ Lp ∩L2 such that fk → f in Lp. Thus ∇uk = Tfkis a Cauchy sequence in Lp. After subtracting a suitable constant from ukwe may assume that uk → u in W 1,p

loc . It follows that u is a distributionalsolution of −divA∇u = −div f and

∥∇u∥Lp ≤ Cp∥f∥Lp (2.370)

It remains to show uniqueness. Assume that u1 and u2 are distributionalsolutions with ∇ui ∈ Lp. Then u = u1 − u2 is a distributional solution of−divA∇u = 0. Since p > 2 we have u ∈ W 1,2(B(0, r) for every R > 0 andu is a weak solution in B(0, r). The decay estimate (2.78) in combinationwith the Holder inequality implies that

supB(0,r/2)

|∇u|2 ≤ Cr−n∫B(0,r)

|∇u|2 dx ≤ Cr−nrn(1− 2

p)

(∫B(0,r)

|∇u|p dx

) 2p

.

(2.371)Taking the limit r → ∞ we see that ∇u ≡ 0.

Step 2 Proof of (i) for 1 < p < 2.This follows by duality. Let f ∈ Lp and assume that u with ∇u ∈ Lp is adistributional solution of −divA∇u = f , i.e.,∫

Rn

A∇u · ∇φdx =∫

Rn

f · ∇φdx ∀φ ∈ C∞c (Rn; RN ). (2.372)

75 [March 17, 2014]

Let (AT )αβij = Aβαji , let g ∈ Lp′. By Step 1 there exists a distributional

solution v of −divAT∇v = −div g with ∇v ∈ Lp′, i.e.,∫

Rn

AT∇v · ∇ψ dx =∫

Rn

g · ∇ψ dx ∀ψ ∈ C∞c (Rn; RN ). (2.373)

Now by density (see Lemma 2.37) the identity in (2.372) also holds withφ = v and in (2.373) we may take ψ = u. Since

AT∇v · ∇u = A∇u · ∇v (2.374)

this yields the duality relation∫Rn

g · ∇u dx =∫

Rn

f · ∇v dx. (2.375)

From the estimate in Step 1 we deduce that∫g · ∇u dx ≤ ∥f∥Lp ∥v∥Lp′ ≤ C(p′, n, A)∥f∥Lp ∥g∥Lp′ . (2.376)

Taking g = ∇u|∇u|p−2 and noting that p′ = pp−1 we deduce that

∥∇u∥Lp ≤ C(p′, n, A)∥f∥Lp . (2.377)

This implies in particular uniqueness (up to constants) since f = 0 impliesthat ∇u = 0.

To show existence we first consider f ∈ Lp ∩ L2. Then by Corollary2.16 there exists a solution u of (2.372) with ∇u ∈ L2 and u is unique upto the addition of constants. As before we write ∇u = Tf . Similarly forg ∈ Lp

′∩L2 by Step 1 there exists a solution v of (2.373) with ∇v ∈ Lp′∩L2.

Thus using Lemma 2.37 with q = 2 we get again the duality relation (2.375)and the estimate in Step 1 yields∫

g · Tf dx ≤ C(p′, n, A)∥f∥Lp ∥g∥Lp′ ∀ g ∈ Lp′ ∩ L2. (2.378)

Now Lemma 2.36 implies that

∥Tf∥Lp ≤ C(p′, A)∥f∥Lp ∀ f ∈ Lp ∩ L2. (2.379)

To show that for each f ∈ Lp there exists a solution u of (2.372) with∇u ∈ Lp such that (2.379) holds we use an approximation argument asabove. Let f ∈ Lp. Then there exist fk ∈ Lp ∩ L2 such that fk → f in Lp.For each fk there exists a unique solution uk of (2.372) with ∇uk ∈ Lp ∩L2

and by (2.379) the maps ∇uk are a Cauchy sequence in Lp. Hence (afterpossible subtraction of constants) uk → u in W 1,p

loc , ∇uk → ∇u in Lp, themap u is a solution of (2.372) and estimate in (2.379) holds.

76 [March 17, 2014]

Step 3. Proof of (ii).The estimate follows directly from (i). If u is a distributional solution of−divA∇u = g and ∇2u ∈ Lp then ∂γu is a distributional solution of−divA∇∂γu = ∂γg and thus by (i) (applied with f δi = gi and fαi = 0 ifα = δ)

∥∇∂γu∥Lp ≤ C(p, n,A)∥g∥Lp (2.380)

This yields the estimate for ∇2u. This estimate shows in particular thatdifference of two solutions must satisfy ∇2u = 0 and hence be an affinefunction.

To show existence of u we note that by (i) for each γ = 1, . . . , n theequation

−divA∇vγ = ∂γg (2.381)

has a unique distributional solution with ∇vγ ∈ Lp. Moreover

∥∇vγ∥Lp ≤ C(p,A)∥g∥Lp (2.382)

We want to show that there exists a u such that ∂γu = vγ and u is a solutionof the original problem.

First assume p ≥ 2. The function w : ∂γvδ − ∂δvγj satisfies −divA∇w =0. Since w ∈ Lp this implies that w = 0. Indeed, for p ≥ 2 one has w ∈ L2

loc

and by taking difference quotients one deduces that ∇w in W 1,2loc . Then the

decay estimate and the estimate ∥w∥L2(B(0,R)) ≤ CRn/2−n/p∥w∥Lp(B(0,R)

imply that ∇w = 0 as in Step 1 and hence w = 0 since w ∈ Lp. Since w = 0for all γ, δ it follows that there exists a u ∈W 2,p

loc such that vγ = ∂γu. Hence∇2u ∈ Lp. Moreover

∂γ(−divA∇u− g) = 0 (2.383)

in the sense of distributions and hence

−divA∇u− g = const (2.384)

in the sense of distributions. Since ∇2u ∈ Lp and g ∈ Lp we must haveconst = 0. This finishes the proof of p ≥ 2.

The same argument works for 1 < p < 2 as long as we can show againthat w = 0. We argue by duality. We have by definition of a distributionalsolution ∫

Rn

w · (−divAT∇φ) dx = 0 ∀φ ∈ C∞c (Rn; RN ). (2.385)

Similar to Lemma 2.37 one can show that the identity also holds for allφ ∈ W 2,p′

loc such that ∇2φ ∈ Lp′. Now let g ∈ Lp

′and let v be the solution

of −divAT∇v = g with ∇2v ∈ Lp′. Then we get∫

Rn

w · g = 0 ∀g ∈ Lp′

(2.386)

and thus w = 0.

77 [March 17, 2014]

Alternative proof of (ii). This was added on Nov 21, 2013.The estimate for ∥∇u∥Lp is proved as before. The main point is to showthat for g ∈ L2 there exists a solution in W 2,2

loc . To show the existence ofsuch a solution consider the problems

−divA∇uk +1kuk = g. (2.387)

By Fourier transform we have∫Rn

(A∇u,∇u) dx ≥ ν

∫Rn

|∇u|2 dx, (2.388)

where ν > 0 is the ellipticity constant. Thus by the Lax-Milgram theoremthe equation (2.387) has a unique solution in W 1,2(Rn) and by using 1

kuk asa test function we see that

1kν∥∇uk∥2

L2 +1k2

∥uk∥2L2 ≤ ∥g∥L2

1k∥uk∥L2 ≤ 1

2∥g∥2

L2 +1

2k2∥uk∥2

L2 . (2.389)

By taking difference quotients and passing to the limit we see that ∂γvk isa weak solution of

−divA∇∂γuk +1kuk = ∂γg. (2.390)

Using ∂γu as a test function we get the estimate

ν∥∇∂γuk∥2L2 +

1k∥∂γuk∥2

L2 ≤ ∥g∥L2∥∂2γuk∥L2 ≤ 1

2ν∥g∥2

L2 +ν

2∥∇∂γuk∥2

L2 .

(2.391)Hence

∥∇2uk∥2L2 +

1k∥∇uk∥2

L2 +1k2

∥uk∥L2 ≤ C∥g∥2L2 . (2.392)

If we letuk(x) = uk(x) − (uk)0,1 − (∇uk)0,1x (2.393)

Then ∇2uk = ∇2uk is uniformly bounded in L2 and by the Poincare inequal-ity there exists a subsequence such that uk u in W 2,2

loc and ∇2uk ∇2uin L2. Moreover taking a further subsequence we may assume that

1kuk w in L2, ∇1

kuk → 0 in L2, (2.394)

where we used that k−1/2∇uk is bounded in L2. It follows that ∇w = 0, i.e.,w = const. Passing to the limit in (2.387) we see that u is a distributionalsolution of

−divA∇u+ const = g (2.395)

Since ∇2u ∈ L2 and g ∈ L2 it follows that the constant must be zero, sinceno non-trivial constant function is in L2(Rn).

78 [March 17, 2014]

Now let g ∈ L2 ∩ Lp and let u ∈ W 2,2loc with ∇2u ∈ L2 be the solution

just constructed. Then vγ := ∂γu is the unique distributional solution of

−divA∇vγ = ∂γg (2.396)

with ∇vγ ∈ L2. Thus ∇vγ = Tf (γ) where (f (γ))γi = 1 and (f (γ))αi = 0 ifα = γ. Thus by (2.369) and (2.379)

∥∇vγ∥Lp ≤ C(p, n,A)∥g∥Lp . (2.397)

And this shows that ∥∇2u∥Lp ≤ C(p, n,A)∥g∥Lp . Finally for g ∈ Lp asolution u is constructed as before by approximating g by gk ∈ L2 ∩ Lp.

Lemma 2.36. Let 1 ≤ p < 2. Let f ∈ Lploc(Rn; Rd) and assume f · g is

integrable for each g ∈ (L2 ∩ Lp′)(Rn; Rd) that there exists a constant M

such that∫Rn

f · g dx ≤M∥g∥Lp′ ∀ g ∈ (L2 ∩ Lp′)(Rn; Rd). (2.398)

Then f ∈ Lp(Rn; Rd) and∥f∥Lp ≤M. (2.399)

Proof. This proof was only sketched briefly in class.If f ∈ Lp and if (2.398) holds forall g ∈ Lp

′then one can simply take

g = f |f |p−2. To prove the result under the assumptions stated consideran increasing sequence of compact sets Kk ⊂ Rn with

∪kKk = Rn. Let

Ek := x ∈ Kk : |f(x)| ≤ k. Let fk = χEkf . Then fk ∈ L1 ∩ L∞ it follows

from (2.398) that ∫Rn

fkg ≤M∥g∥Lp′ (2.400)

for all g ∈ L2 ∩ Lp′ . Taking g = fk|fk|p−2 we see that ∥fk∥pLp ≤ Mp. Nowthe assertion follows by taking k → ∞ and using the monotone convergencetheorem.

The following lemma was slightly rewritten on Nov 21, 2013. The proofis the same as before.

Lemma 2.37. Let 1 ≤ q < ∞. Assume that ψ ∈ W 1,qloc (Rn) with ∇ψ ∈

Lq(Rn; Rn). Then there exist ψk ∈ C∞c (Rn) such that

∇ψk → ∇ψ in Lq(Rn; Rn). (2.401)

Proof. Since C∞c (Rn) is dense in W 1,q(Rn) it suffices to show that there

exist ψk ∈W 1,q(Rn) such that (2.401) holds.

79 [March 17, 2014]

Let η ∈ C∞c (B(0, 1)) with η = 1 on B(0, 1/2) and 0 ≤ η ≤ 1 and let

ηk(x) = η(2−kx). (2.402)

Let

Ak := B(0, 2k) \B(0, 2k−1), ak :=1

Ln(Ak)

∫Ak

ψ dx (2.403)

andψk := ηk(ψ − ak). (2.404)

Then∇ψk = ηk∇ψ + (ψ − ak)∇ηk. (2.405)

and thus∇ψk −∇ψ = (ηk − 1)∇ψ + (ψ − ak)∇ηk (2.406)

Now by the Poincare inequality with zero mean in the annulus Ak

∥(ψ − ak)∇ηk∥Lq ≤ C2k∥∇ψ∥Lq(Ak) sup |∇ηk| ≤ C∥∇ψ∥Lq(Rn\B(0,2k−1)).

(2.407)

Moreover∥(ηk − 1)∇ψ∥Lq(Rn) ≤ ∥∇ψ∥Lq(Rn\B(0,2k−1)). (2.408)

Thus the assertion follows.

[18.11. 2013, Lecture 10][21.11. 2013, Lecture 11]

Theorem 2.38. Suppose that the coefficients Aαβij are constant and stronglyelliptic. Let 1 < p <∞.

(i) Assume that f ∈ Lp(Rn+; Rn×N ). Then the problem

−divA∇u = −div f in Rn+ with u = 0 on ∂Rn

+ (2.409)

has a unique distributional solution u ∈W 1,ploc (Rn

+) with ∇u ∈ Lp(Rn+; RN×n).

Moreover∥∇u∥Lp ≤ C(p, n,A)∥f∥Lp . (2.410)

(ii) Let g ∈ Lp(Rn+; RN ). Then the problem

−divA∇u = g in Rn+ with u = 0 on ∂Rn

+ (2.411)

has a distributional solution u ∈W 2,ploc (Rn

+; RN ) with ∇2u ∈ Lp(Rn+; RN×n×n)

and u is unique up to the addition of linear functions of the form cxn.Moreover

∥∇2u∥Lp ≤ C(p, n,A)∥g∥Lp . (2.412)

80 [March 17, 2014]

Remark. Note that the uniqueness statement is slightly stronger thanfor solutions in Rn since the boundary condition rules out the addition ofconstants or of linear functions of x1, . . . , xn−1.

Proof. This is proved like Theorem 2.35 using (2.221) and (2.222) instead ofCorollary 2.16. In this case the alternative proof of part (ii) applies directly.The other proof can be modified by first only looking that the equation fortangential derivatives vγ = ∂γu. This defines u uniquely up to the additionof a function h(xn). Then the PDE for u becomes an ODE for h and onecan easily see that this ODE has solution (which is unique up to addingterms of the form cxn).

Corollary 2.39. Let 1 < p < ∞. Then there exists a constants Cp,n andCp,q,n with the following properties.

(i) If g ∈ Lp(B(0, R)) then there exists f ∈W 1,p(B(0, R); Rn) such that

−div f = g in B(0, R) (2.413)1R∥f∥Lp + ∥∇f∥Lp ≤ Cp,n∥g∥Lp . (2.414)

(ii) Assume in addition that ∫B(0,R)

g dx = 0. (2.415)

Let G = g in B(0, R) and G = 0 in Rn \B(0, R). Assume that q <∞and 1

q ≥ 1p −

1n Then there exist f ∈ Lq(Rn) such that

−div f = G in Rn, (2.416)1

R1−n/p+n/q ∥f∥Lq ≤ Cp,q,n∥g∥Lp . (2.417)

Proof. By scaling we may assume that R = 1. Define G as in (ii).(i): There exist u ∈W 1,p

loc (Rn) such that

−∆u = G, ∥∇2u∥Lp ≤ Cn,p∥G∥Lp . (2.418)

Set f = (∇u−(∇u)B(0,1) then div f = G and the estimate for f inW 1,p(B(0, 1))follows from the Poincare inequality. The Lq estimate for f then follows fromthe Sobolev embedding theorem.

(ii): This was discussed briefly in class in the lecture on Nov 25, 2013The main point is to show that the solution u decays sufficiently rapidly at∞. To do so we use a different representation of the solution from Intro-duction to PDE. Assume first that G ∈ C∞

c (Rn) with suppG ⊂ B(0, 1) andset

v(x) =∫B(0,1)

Kn(x− y)G(y) dy (2.419)

81 [March 17, 2014]

where Kn is the Green’s function given by

Kn(z) =

cn|z|−n+2 if n ≥ 3,− 1

2π ln |z| if n = 2.(2.420)

Then G ∈ C∞c (Rn) implies that v ∈ C∞(Rn). We also see easily that for

|x| ≥ 2 we have |∇2v(x)| ≤ C|x|−n∥G∥L1 . Thus ∇2v ∈ L2(Rn). Finally inIntroduction to PDE we have shown that −∆v = G. It follows from theuniqueness result in Theorem 2.35 that ∇2u = ∇2v and thus

∥∇2v∥Lp ≤ C(p, n)∥G∥Lp . (2.421)

Now we look at the decay of ∇v. For |x| > 1 we have

∇v(x) =∫B(0,1)

cnx− y

|x− y|nG(y) dy =

∫B(0,1)

cn

(x− y

|x− y|n− x

|x|n

)G(y) dy

where in the last step we used that∫G(y) dy = 0. One easily checks (e.g.,

by computing the derivative with respect to y) that for |x| ≥ 2 and |y| ≤ 1∣∣∣∣ x− y

|x− y|n− x

|x|n

∣∣∣∣ ≤ C

|x|n. (2.422)

It follows that

|∇v(x)| ≤ C

|x|n∥G∥L1(B(0,1) ≤

C

|x|n∥G∥Lp(B(0,1) ∀x ∈ Rn\B(0, 2). (2.423)

Since C∞c (B(0, 1) is dense in Lp(B(0, 1) it follows by approximation that

this estimate and (2.421) hold also without the additional assumption G ∈C∞c (B(0, 1)). We now can use a Poincare-Sobolev inequality of the form

∫B(0,3)

|h|q dx ≤ C

∫B(0,3)\B(0,2)

|h|q dx+ C

(∫B(0,3)

|∇h|p dx.

)q/p(2.424)

This inequality can be proved in the usual way by contradiction using thecompact Sobolev embedding W 1,p(B(0, 3)) → Lp(B(0, R)). Taking h = ∇vwe get ∫

B(0,3)|∇v|q dx ≤ C∥G∥q/pLp . (2.425)

and (2.423) implies that∫Rn\B(0,3)

|∇v|q dx ≤ C∥G∥q/pLp . (2.426)

Now taking f = ∇v we get the assertion.

82 [March 17, 2014]

We will make the following general assumptions for the rest of the sub-subsection

The coefficients Aαβij are strongly elliptic (2.427)

Ω ⊂ Rn is bounded and open. (2.428)

We consider the differential operator

Lu = −divA∇u+ div (Bu) + C∇u+Du (2.429)

Lemma 2.40 (Improvement of regularity). Let 1 < s ≤ q < ∞. Assumethat ∂Ω ∈ C1, A ∈ C0(Ω). Assume that u ∈W 1,s

0 (Ω; RN ) is a distributionalsolution of

Lu = −div f + g (2.430)

Iff ∈ Lq(Ω), g ∈ Lt (2.431)

with t > 1 and 1t ≤

1q + 1

n then u ∈W 1,q0 (Ω) and

∥u∥W 1,q ≤ C(∥f∥Lq + ∥g∥Lt + ∥u∥Lq). (2.432)

Proof. Steps 1 and 2 were briefly discussed in the lecture on Nov 25, 2013.Step 3 was discussed in the lecture on Nov 21.We first note that it suffices to consider the case g = 0. Indeed let B(0, R) bea ball which contains Ω and extend g by zero to that ball. Then by Corollary2.39 and the Sobolev embedding theorem there exists an f ∈ Lq(B(0, R)such that g = div f and ∥f∥Lq ≤ C∥g∥Lt .

Step 1. A priori estimate for q = s and B = C = D = 0.We first derive an interior estimate for functions u ∈W 1,s

0 (Ω) which satisfy

−divA∇u = −div f. (2.433)

Let B(x0, R) ⊂ Ω. Then we claim that

∥∇u∥Ls(B(x0,R/2)) ≤C∥f∥Ls(B(x0,R)) +C

R∥u∥Ls(B(x0,R)

+ C supB(x0,R)

|A−A0| ∥∇u∥Ls(B(x0,R)) (2.434)

where A0 = A(x0). To see this first note that

−divA0∇u = −div f , where f = f + (A0 −A)∇u. (2.435)

Now let η ∈ C∞c (B(x0, R) with η = 1 on B(x0, θR) and set v = ηu. Then

−divA0(ηu) = −∂αAαβ0 (η∂βu+ u∂βη)

= −η∂αAαβ0 ∂βu−Aαβ0 ∂αη∂βu− ∂αAαβ0 u∂βη

= −ηdiv f − ∂βAαβ0 u∂αη +Aαβ0 u∂α∂βη − ∂αA

αβ0 u∂βη.

83 [March 17, 2014]

Since −ηdiv f = −div (ηf)+ fα∂αf we get (after exchanging α and β in thesecond term on the right hand side)

−divA0(ηu) = −div f + g (2.436)

where

fα = ηfα +Aβα0 u∂βη +Aαβ0 u∂βη,

g = fα∂αη +Aαβ0 u∂α∂βη. (2.437)

Note that f and v have compact support in B(x0, R). Hence if we extendv, f and g by zero we see that v is a distributional solution of

−divA0∇v = −div f + g in Rn. (2.438)

Note that∫

Rn g = 0. Indeed for every φ ∈ C∞c (Rn) with φ = 1 onB(x0, R/2)∫

Rn

g dx =∫

Rn

φg dx =∫

Rn

∇φ · (A0(∇v) − f) dx = 0. (2.439)

By Corollary 2.39(ii) there exist f such that g = −div f and ∥f∥Ls ≤CR∥g∥Ls . By Theorem 2.35 we get

∥∇(ηu)∥Ls ≤ ∥f + f∥Ls (2.440)

and together with (2.437) this implies the estimate (2.434).Now for x0 ∈ ∂Ω one can obtain a similar estimate in B(x0, R/2)∩Ω (in

this case one get supB(x0,R)∩Ω |A− A0| instead of supB(x0,R) |A−A0| whereA are the coefficients in the flattened domain).

Finally since A is uniformly continuous and ∂Ω is C1 for each ε > 0there exists an R0 such that for all R ≤ R0 we have supB(x0,R) |A−A0| < ε

if B(x0, R) ⊂ Ω and supB(x0,R)∩Ω |A− A0| < ε if x0 ∈ ∂Ω and R ≤ R0.We first cover ∂Ω by finitely many balls B(yi, R0/2) with yi ∈ ∂Ω. Let

Ω0 = Ω \∪iB(yi, R0/2) and R1 := min(R0,dist (Ω0, ∂Ω)) and cover Ω0 by

finitely many balls B(yj , R1/2) (then B(yj , R1) ⊂ Ω). This can be done insuch a way that each point is only contained in a finite number M of theballs. Then taking the local estimates to the power s and summing over allballs we get∫

Ω|∇u|s dx ≤ 3sCM

(∫Ω|f |s dx+R−s

∫Ω|u|s dx+ εs

∫Ω|∇u|s dx

).

(2.441)By choosing ε small enough we get the estimate for q = s and B = C =D = 0, i.e., we get the implication

−divA∇u = −div f =⇒ ∥∇u∥Ls ≤ C(∥f∥Ls + ∥u∥Ls), (2.442)

provided that u ∈W 1,s0 (Ω).

84 [March 17, 2014]

Step 2. A priori estimate for q = s.We write the equation Lu = −div f as

−divA∇u = −div f + div (Bu) − C∇u−Du. (2.443)

For g ∈ Lp(Ω) we define ∇−1g = ∇u − (∇u)Ω, where u is the solution of−∆u = g in Rn (with g extended by zero). Note that u is in general onlydefined up to the addition of an affine function but that ∇u − (∇u)Ω doesnot change if we add an affine function to u. Let

Tu = −∇−1(C∇u+Du). (2.444)

Then the equation Lu = −div f can be written as

−divA∇u = −div f , where f = f −Bu+ Tu. (2.445)

By Corollary 2.39(i) the operator T is bounded from W 1,s(Ω) to itself.Since 1 < s <∞ the space W 1,s

0 (Ω) is reflexive and therefore we have11

uk 0 in W 1,s0 =⇒ Tuk 0 in W 1,s

0 . (2.446)

We claim that this implies that for each ε > 0 there exists a Cε such that

∥Tu∥Ls ≤ ε∥∇u∥Ls + Cε∥u∥Ls . (2.447)

Indeed if this was not true there exists an ε > 0 and a sequence such that∥Tuk∥Ls = 1, ∥∇uk∥ ≤ ε−1 and ∥uk∥Ls → 0. Thus uk u in W 1,s

0 andhence Tuk → 0 in W 1,s

0 . By the compact Sobolev embedding this impliesTuk → 0 in Ls and this is a contradiction. Now we can apply (2.442) to(2.445). This yields

∥∇u∥Ls ≤ C(∥f∥Ls +C∥u∥Ls + ∥Tu∥Ls) ≤ C(∥f∥Ls +Cε∥u∥Ls + ε∥∇u∥Ls).(2.448)

By choosing ε > 0 sufficiently small we can absorb the last term on the righthand side and this finishes the proof of the estimate for q = s.

Step 3. Higher regularity.Again we first prove interior regularity. Let B(x0, R) ⊂ Ω, letη ∈ C∞

c (B(x0, R) and letv = ηu. (2.449)

11See Functional Analysis and PDE. Sketch: let X = W 1,s0 and define the dual operator

by T ∗ : X ′ → X ′ by T ∗φ(u) = φ(Tu). Then T ∗ is bounded. Since X is reflexivethere exists a subsequence such that Tukj v in X. Thus φ(v) = limj→∞ φ(Tukj ) =limj→∞ T ∗φ(uk) = 0. It follows that v = 0 and by the usual argument one shows thatthe whole sequence Tuk converges weakly to zero.

85 [March 17, 2014]

We may extend v by zero to Rn. Using the abbreviations

(u⊗∇η)jβ = uj∂βη, [A(∇η ⊗∇u)]i = Aαβij ∂α∇η∂βuj (2.450)

we get−divA0∇v = −div (A0 −A)∇v − divA(∇v) (2.451)

and

−divA∇v = −divAη∇u− divA(u⊗∇η)= −ηdivA∇u−A(∇η ⊗∇u) − divA(u⊗∇η). (2.452)

Now

−divA∇u = −div (f −Bu) − C∇u−Du,

−ηdivA∇u = −div (ηf − ηBu) + (fα − (Bu)α) ∂αη − ηC∇u− ηDu,

and thus−divA∇v = −div f + g, (2.453)

where

f = η(f −Bu) +A(u⊗∇η),gi = fα∂αη − (Bu)α∂αη − ηC∇u− ηDu−A(∇η ⊗∇u) (2.454)

Let1r

= max(1q,1s− 1n

). (2.455)

Then by the Sobolev embedding theorem f ∈ Lr(B(x0, R)). Moreover g ∈Ls(B(x0, R)). Note also that f and v have compact support in B(x0, R).Thus

∫B(x0,r)

g dx = 0 (see Step 1). Hence by Corollary 2.39(ii) there existsf ∈ Lr(Rn) ∩ Ls(Rn) with div f = g in Rn where g is extend by zero.Extending v and f by zero to Rn we thus get

−A0∇v = −div (A0 −A)∇v − div h (2.456)

with h = f + f ∈ Lr(Rn) ∩ Ls(Rn).Let T = T1 be the solution operator in Theorem 2.35(i) and set Sg =

T (A−A0)g. Then (2.456) implies that

∇v = S∇v + Th. (2.457)

We now show that S is a contraction on Lr and on Ls. By Theorem 2.35

∥Sg∥Lr ≤ C∥(A−A0)g∥Lr ≤ C supB(x0,R)

|A−A0| ∥g∥Lr . (2.458)

86 [March 17, 2014]

and the same estimate holds in Ls. Since A is uniformly continuous thereexists R0 such that for R ≤ R0

∥Sg∥Lr ≤ 12∥g∥Lr and ∥Sg∥Ls ≤ 1

2∥g∥Ls . (2.459)

Again by Theorem 2.35 we have ∥Th∥Lr ≤ C∥h∥Lr and ∥Th∥Ls ≤ C∥h∥Ls .Thus one can use (2.457) to express ∇v by the Neumann series of (Id −S)−1,i.e.,

∇v =∞∑k=0

Sk Th (2.460)

where the series converges in Ls. Since ∥Sg∥Lr ≤ 12∥g∥Lr and Th ∈ Lr the

series on the right hand side also converges in Lr and thus ∇v ∈ Lr and

∥∇v∥Lr ≤ 2∥∇Th∥Lr ≤ C∥h∥Lr ≤ C(∥f∥Lq + ∥u∥W 1,s) (2.461)

and be Step 1 we get

∥∇v∥Lr ≤ C(∥f∥Lq + ∥u∥Ls) ≤ C(∥f∥Lq + ∥u∥Lq) (2.462)

Now if η = 1 on B(x0, θR) we get u ∈W 1,r(B(x0, θr).If r = q the interior estimate is proved. Otherwise define iteratively

1ri+1

= max(1q ,

1ri

− 1n). Then we get u ∈ W 1,ri(B(x0, θ

iR) and the corre-sponding estimate. If i is the smallest integer that 1

s −in <

1q . Then ri = q

and the interior estimate is established.The boundary estimate is proved in the same way and the estimates

can be combined to a global estimate by covering Ω with sufficiently smallballs.

Theorem 2.41. Assume that Ω is a bounded open set with C1 boundary.and asssume that the coefficients A are uniformly continuous and stronglyelliptic and that B,C,D ∈ L∞. Assume that 0 is not an eigenvalue of L,i.e.,

for all for f ∈ L2(Ω) there existsa unique weak solution u ∈W 1,2

0 (Ω) of Lu = −div f .

Then for all p ∈ (1,∞) and all f ∈ Lp(Ω) the problem

Lu = −div f (2.463)

has a unique solution and there exists a constant C (depending on p, n, Ω,ν, sup ∥A∥, ∥B∥L∞, ∥C∥L∞ , ∥D∥L∞ and the modulus of continuity of A)such that

∥u∥W 1,p ≤ C∥f∥Lp (2.464)

87 [March 17, 2014]

Proof. For p > 2 the existence and the estimate follow directly from Lemma2.40. The uniqueness in W 1,p

0 follows from the uniqueness in W 1,20 since

W 1,p0 ⊂W 1,2

0 .For 1 < p < 2 we first consider uniqueness. If u1 and u2 are solutions

then u = u1 − u2 ∈ W 1,p0 and Lu = 0. By Lemma 2.40 (applied with

s = p, q = 2) we get u ∈W 1,20 and hence u = 0 by assumption.

To prove existence we use uniqueness to show that solutions of Lu =−div f satisfy the improved estimate

∥u∥W 1,p ≤ C∥f∥Lp . (2.465)

By Lemma 2.40 (with q = p) is suffices to show that

∥u∥Lp ≤ C∥f∥Lp . (2.466)

If this was not true there would exist sequences fk → 0 in Lp and uk with∥uk∥Lp = 1 and Luk = −div fk. By Lemma 2.40 it follows that ∥uk∥W 1,p ≤C. Hence by the compact Sobolev embedding there exists a subsequence(not relabeled) such that uk → u in Lp and uk u in W 1,p

0 . Then Lu = 0and ∥u∥Lp = 1. This contradicts uniqueness.

Now let f ∈ Lp and let fk ∈ Lp ∩ L2 such that fk → f in Lp. Byassumption there exists uk ∈ W 1,2

0 with Luk = −div fk. It follows from(2.465) that uk is a Cauchy sequence in W 1,p and hence uk → u in W 1,p

0 andu is a (distributional) solution of Lu = −div f .

[21.11. 2013, Lecture 11][25.11. 2013, Lecture 12]

We finally consider Lp estimates for higher derivatives.

Lemma 2.42. Let 1 < p ≤ q < ∞. Assume that Ω is a bounded open setwith C1 boundary, that the coefficients A ∈ C1(Ω) are strongly elliptic thatB ∈ C0,1, C,D ∈ L∞ and that u ∈ (W 2,p ∩W 1,p

0 ()Ω; RN ) is a distributionalsolution of Lu = g. If

g ∈ Lq(Ω) (2.467)

then u ∈ (W 2,q ∩W 1,q0 )(Ω) and

∥u∥W 2,q ≤ C(∥g∥Lq + ∥u∥Lq + ∥∇u∥Lq). (2.468)

Remark. The term ∥∇u∥Lq) can be removed by using an interpolationinequality ∥∇u∥Lp ≤ ε∥∇2u∥Lp + Cε∥u∥Lp .

Proof. In the interior one considers the equation for the derivatives ∂γu andargues as in the proof of Lemma 2.40. At the boundary one first flattens theboundary, then considers the equation for the tangential derivatives ∂γ u for

88 [March 17, 2014]

the transformed function and applies the reasoning in Lemma 2.40. Finallythe double normal derivative is recovered in the usual way by using theequation.

Theorem 2.43. Assume that Ω is a bounded open set with C2 boundary.and asssume that the coefficients A are strongly elliptic and satisfy A ∈C1(Ω) and that B ∈ C0,1, C,D ∈ L∞. Assume that 0 is not an eigenvalueof L, i.e.,

for all for f ∈ L2(Ω) there existsa unique weak solution u ∈W 1,2

0 (Ω) of Lu = −div f .

Then for all p ∈ (1,∞) and all g ∈ Lp(Ω) the problem

Lu = g (2.469)

has a unique solution in u ∈W 2,p ∩W 1,p0 and there exists a constant C (de-

pending on p, n, Ω, ν, ∥A∥C1, ∥B∥, ∥C∥, ∥D∥ and the modulus of continuityof A) such that

∥u∥W 2,p ≤ C∥g∥Lp . (2.470)

Proof. By the global W 2,2 estimates for g ∈ L2 there exists a solution inW 2,2 ∩W 1,2

0 . Thus for p > 2 one can use Lemma 2.42. For p < 2 one arguesas in the proof of Theorem 2.41.

Remark. One can obtain slightly better results by looking at operatorsin non-divergence form, i.e.,

(Lu)i := Aαβij (x)∂α∂βuj +Bαij(x)∂αu

j + Cij(x)uj . (2.471)

Then for the ∥∇2u∥Lp estimate it suffices to assume that A ∈ C0(Ω), B,C ∈L∞(Ω) and ∂Ω ∈ C1,1. This can be proved in essentially the same way, see[GT], Chapter 9.2 to 9.6 for the details.

Remark. Similarly one can obtain W k,p estimates if g ∈ W k−2,p and thecoefficients lie in the appropriate spaces.

2.1.12 More general elliptic systems and the complementing con-dition

In this course we have focused on second order elliptic systems. One canconsider more general systems for arbitrary order and even systems wheredifferent variables and different equations appear naturally with differentorders of derivatives.

There is a systematic way to determine which terms in such a systemare the leading order terms and which terms can be considered as lower

89 [March 17, 2014]

order terms. In the general setting one requires that there exists weightss1, . . . , sN ∈ Z and t1, . . . tN ∈ Z such that the system can be written as

N∑j=1

∂Lij(x, ∂)uj = gi ∀i = 1, . . . , N. (2.472)

where the Lij are polynomials in the second variable and

degLij ≤ si + tj (2.473)

We denote by L′ij the sum of the terms in the polynomial which are homo-

geneous of degree si + tj (the leading order term) We say that the systemis elliptic if

L0(x, ξ) := detL′(x, ξ) = 0 ∀ξ ∈ Rn \ 0 (2.474)

Note that L0 is a homogeneous polynomial of degree∑

i si+∑tj = 0. Note

that the condition does not change if we add a constant to all the tj andsubtract the same constant from all the si. Hence we may assume thatmax si = 0. By Fourier transform one sees that for homogeneous ellipticsystems in Rn with constant coefficients (i.e., if Lij(x, ξ) = Lij(ξ) = L′

ij(ξ))one can find a solution with ∇tjuj ∈ L2 provided that ∇−si

fi ∈ L2.One can also consider general boundary conditions involving arbitrary

combinations of partial derivatives of the variables on the boundary. TheSchauder and Lp regularity theory of such systems is systematically devel-oped in the two papers

• S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundaryfor solutions of elliptic partial differential equations satisfying generalboundary conditions. I, Comm. Pure Appl. Math. 12 (1959), 623–727.

• S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundaryfor solutions of elliptic partial differential equations satisfying generalboundary conditions. II, Comm. Pure Appl. Math. 17 (1964), 35–92.

. At about the same time similar results and results for the correspondingparabolic systems were obtained by the Russian School, in particular by O.A. Ladyzenkaja, V.A. Solonnikov and N.N. Uralceva. A discussion of theLp and Schauder estimates can also be found in Morrey’s book [Mo].

The so called complementing condition which connects the boundaryconditions and the operator plays a crucial role. It can be loosely stated asfollows. Consider the blow-up (of the leading order part) at any boundarypoint. This leads to a problem with constant coefficients in a half-space.Assume for ease for notation that this half space is Rn

+. Then the comple-menting condition requires that all solutions u(x′, xn) = v(xn)eik

′·x′ with

90 [March 17, 2014]

k′ = 0 for which v is bounded in (0,∞) are already zero. One can showthat this condition is equivalent to a purely algebraic condition involvingthe coefficients of the operator and the boundary conditions. This is closelyrelated to the notion of hypoellipticity of the half space problem (see Simon’spaper on the blow-up approach [Si]).

One can easily check that for the scalar equation −∆u = f the Dirichletboundary condition is complementing in the above sense. Similarly one cancheck that the Neumann boundary condition is complementing (the shortcalculation was done in class).

Similarly one can check that second order strongly elliptic systems in thesense considered above (i.e. Aαβij ξαξβζ

iζj ≥ ν|ξ|2|ζ|2) are elliptic in the senseof (2.474) and that the Dirichlet boundary conditions are complementing,see [ADN2], pp. 43-44 (even for higher order systems).

2.1.13 A short look back on elliptic regularity

We have studied solutions of the second order systems in an open set Ω ⊂ Rn

Lu := −div (A∇u) − div (Bu) + C∇u+Du = −div f + g (2.475)

under the assumption that A are bounded and strongly elliptic, i.e., thereexists a ν > 0 such that

(A(x)(ζ ⊗ ξ), ζ ⊗ ξ) = A(x)αβij ζiζjξαξβ ≥ ν|ζ|2|ξ|2 ∀ ζ ∈ RN , ξ ∈ Rn, x ∈ Ω(2.476)

The basic results are

• In the Holder spaces Ck,α, with 0 < α < 1 and the Sobolev spacesW k,p

with 1 < p < ∞ the solution u is as good as the data f, g, A,B,C,D(and the regularity of ∂Ω for global estimates) allow.

• There is no such theory for α = 0, α = 1, p = 1 or p = ∞.

• All the estimates can be derived from L2 type estimates in the Cam-panato spaces L2,λ (and suitable perturbation/ compactness argu-ments)

One subtlety is that one usually needs A ∈ C0 in cases where one mightthink that A ∈ L∞ is sufficient. The reason is that one usually has toapproximate at least the leading order term −divA∇ by an operator withconstant coefficients and then uses a fixed point argument (or an iterationargument for the Campanato estimates) to treat the case of variable coeffi-cients. This requires that the oscillation of the coefficient is small in smallballs. The only exception are the W 1,2 estimates for scalar equations (orfor systems which satisfy the super strong ellipticity condition). For scalarequations there are some deep results which still hold for L∞ coefficients.We will discuss these in the next section.

91 [March 17, 2014]

We have derived different though closely related results. For illustrationwe consider the W 1,p estimates.

(i) Local or global a priori estimates. If u is a solution and if u ∈W 1,s

0 (Ω) then

∥u∥W 1,s ≤ C(∥f∥Ls + ∥g∥Lt) + C∥u∥Ls (2.477)

In these estimates usually a weaker norm of u still appears on the righthand side. In local estimates the norm on the right hand is usually ona slightly large set then the norm of the left hand side. The norm of uon the right hand side can in general not be avoided since the problemLu = 0 can have nontrivial solutions.

(ii) Improvement of regularity estimates. If u ∈W 1,s0 (Ω) (or in an even

weaker space) and if f ∈ Lq with q > s then u ∈ W 1,q0 . Again there

are local and global versions.

(iii) Existence and uniqueness of solutions (for problems with boundaryconditions)

(iv) Global estimates for problems with boundary conditions. Thesecan often be deduced from the apriori estimate using a compactnessargument to eliminate the weaker norm on the right hand side.

[25.11. 2013, Lecture 12][28.11. 2013, Lecture 13]

2.2 Harnack inequality and the DeGiorgi-Moser-Nash theo-rem

This subsection follow closely [GT], Sections 8.5–8.9So far we have developed a theory which tells us that if the coefficients

in the leading order term are sufficiently regular then in appropriate spacesthe solution of the PDE is as good as the coefficients and the right handside allows.

In this subsection we consider elliptic problems where the coefficients inthe leading order term are only in L∞ but not continuous. In this case thereis a big difference between scalar equations and systems. For scalar equationthe DeGiorgi-Nash-Moser theorem states that we have C0,α estimates forsome small α > 0. The corresponding results for systems fails, even if weassume super strong ellipticity, see [Gi] for the history of the subject andvarious counterexamples. More recent counterexamples can be found the inpapers [SY00] and [SY02].

Apart from its intrinsic interest C0,α estimates for solutions of linearequations with L∞ coefficients are of crucial importance for the regularity

92 [March 17, 2014]

theory for nonlinear problems. Consider for example the problem of mini-mizer the functional

I(u) =∫

Ωf(∇u) dx (2.478)

where f is C∞ and uniformly convex, i.e., there exists ν > 0 such that

∂2f

∂ξi∂ξjξiξj ≥ ν|ξ|2 ∀ξ ∈ Rn, (2.479)

and that the second derivatives of f are uniformly bounded, i.e.,∣∣∣∣ ∂2f

∂ξi∂ξj

∣∣∣∣ ≤ C. (2.480)

By the direct method of the calculus of variations one sees that for givenboundary data u0 ∈ W 1,2(Ω) the functional I has a (unique) minimizer uwith u−u0 ∈W 1,2

0 (Ω). Moreover u is a weak solution of the Euler-Lagrangeequation

−∂i∂f

∂ξi(∇u) = 0. (2.481)

If we formally take a derivative in direction xk we obtain that the functionv = ∂ku is a solution of the linear PDE

−∂i∂2f

∂ξi∂ξj(∇u(x))︸ ︷︷ ︸

aij(x)

∂jv = 0. (2.482)

The assumptions that f is uniformly convex and that the second derivativesare bounded imply that the coefficients aij are elliptic and in L∞. The factthat v ∈ W 1,2

loc (Ω) and that v is a weak solution of (2.481) can be justifiedby the difference quotient method (Homework).

Now if we can show that solutions of (2.481) are in C0,αloc (Ω) then we can

derive full regularity of v from the theory developed so far by a bootstrapargument. Indeed v ∈ C0,α

loc implies that u ∈ C1,αloc . Hence the coefficients aij

are in C0,αloc and the Schauder theory implies that v ∈ C1,α

loc . Hence aij ∈ C1,αloc

and thus v ∈ C2,αloc , etc.

To focus on the essential points we consider only the leading order termsand we look at the operator

Lu := −∂iaij(x)∂ju. (2.483)

in a bounded and open set Ω ⊂ Rn. We write Ω′ ⊂⊂ Ω if Ω′ is open, and ifΩ′ is compact with Ω′ ⊂ Ω. We assume that there exist Λ > 0, and ν > 0

93 [March 17, 2014]

such that ∑i,j

a2ij(x) ≤ Λ2 for a.e. x, (2.484)

∑i,j

aij(x)ξiξj ≥ ν|ξ|2 for all ξ ∈ Rn and a.e. x. (2.485)

Theorem 2.44 (DeGiorgi-Nash-Moser). Assume that (2.484) and (2.485)hold and that u ∈W 1,2(Ω) is a weak solution of

Lu = 0. (2.486)

Then for any open set Ω′ ⊂⊂ Ω we have

∥u∥C0,α(Ω′) ≤ C∥u∥L2(Ω) (2.487)

where

α = α(n,Λ/ν) > 0, C = C(n,Λ, ν, d′) with d′ = dist (Ω′,Ω). (2.488)

Remark. (i) As mentioned above such an estimate does not hold for sys-tems.(ii) For each α ∈ (0, 1) there exists a solution of an elliptic equation whichis C0,α but not in C0,β for any β > α. Indeed for n = 2 and z = x1 + ix2 thefunction u(z) = Re z

|z|1−α is the solution of an elliptic equation (see Home-work 1).(iii) For a version with lower order terms and solutions of Lu = −div f + gwith f ∈ Lq(Ω), g ∈ Lq/2(Ω) for q > n see Gilbarg-Trudinger [GT].

We follow Moser’s proof and will derive the Cα estimate from a Harnackinequality for weak solutions. In fact it is convenient to state separate Har-nack type inequalities for subsolutions and supersolutions. We say that u isa weak subsolution in Ω and write

Lu ≤ 0 (2.489)

if u ∈W 1,2(Ω) and∫Ω

∑i,j

aij∂ju ∂iφ ≤ 0 ∀φ ∈W 1,20 (Ω) with φ ≥ 0. (2.490)

Similar we define a weak supersolution, Lu ≥ 0. An easy way to rememberthe signs is the implication

u convex =⇒ u subsolution.

Theorem 2.45 (Weak Harnack inequality). Assume that (2.484) and (2.485)hold, let u ∈W 1,2(Ω) and assume that B(y, 4R) ⊂ Ω.

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(i) IfLu ≤ 0 and u ≥ 0 (2.491)

then for all p > 1

supB(y,R)

u ≤ CR−n/p∥u∥Lp(B(y,2R)) (2.492)

(ii) IfLu ≥ 0 and u ≥ 0 (2.493)

then for all p ∈ [1, nn−2)

R−n/p∥u∥Lp(B(y,2R)) ≤ C infB(y,R)

u. (2.494)

Here C = C(n,Λ/ν, p).

Proof. For ease of notation we assume that Lu = 0. By scaling and trans-lation we may assume that R = 1. We write Br := B(y, r).

Key idea. To prove (i) show an estimate for the form

∥u∥Lp(Bs) ≤ C(s, r, p, p)∥u∥Lp(Br)

for p > p and s < r. Then iterate this estimate and finally take the limitp → ∞. The procedure is know as Moser iteration. For (ii) one uses that

1inf u = sup 1

u and estimates negative powers of u.We also note that Lu = 0 implies L(u + δ) = 0. Hence we may assume

u ≤ δ > 0 on B(y, 4) and in the end let δ → 0. Thus we can negative powersof u without further discussion.

Step1: the basic estimate. Let β = 0 and use the test function

v = η2uβ , where η ∈ C∞c (B4) with 0 ≤ η ≤ 1.

Assume first that u is a weak subsolution and β > 0. Then

0 ≥∫B4

∑i,j

aij∂ju ∂iv dx

=∫B4

∑i,j

aij∂ju η2βuβ−1∂iu dx+

∑i,j

aij∂ju 2η∂iη βuβ dx

Ellipticity and the bound on the coefficients imply that

ν

∫B4

η2uβ−1|∇u|2 dx ≤ 2Λ|β|

∫B4

η|∇η|uβ |∇u| dx. (2.495)

95 [March 17, 2014]

The same estimate holds if u is a weak supersolution and β < 0.Now we want to write the term uβ−1|∇u|2 as |∇w|2 (up to a constant

factor). Assuming for the moment

β = −1

we setw = u

β+12

and obtain

∇w =β + 1

2u

β−12 ∇u,

uβ−1|∇u|2 =(

2β + 1

)2

|∇w|2,

uβ |∇u| =2

|β + 1|u

|β+1|2 |∇w| =

2β + 1

w|∇w|

Thus with C(β) = C |β+1||β|∫

B4

η2|∇w|2 ≤ C(β)∫B4

η|∇η|w|∇w|

≤ C(β)∥∇η w∥L2 ∥η∇w∥L2

and therefore

∥η∇w∥L2 ≤ C(β) ∥∇η w∥L2

∥∇(ηw)∥L2 ≤ (C(β) + 1) ∥∇η w∥L2 .

Roughly speaking we have transferred the gain in derivatives for solutions ofelliptic PDE from u to w = u(β+1)/2. Now the Sobolev embedding theoremgives

∥ηw∥L2κ ≤ (C(β) + 1)∥∇η w∥L2 , where κ =n

n− 2(2.496)

(for n = 2 we can take any κ <∞).Now let 0 < s < r < 4 and choose η such that

η ∈ C∞c (Br), η = 1 in Bs, sup |∇η| ≤ C

r − s. (2.497)

Then∥w∥L2κ(Bs) ≤

C

r − s(C(β) + 1)∥w∥L2(Br) (2.498)

Finally set γ = β + 1. Then we obtain the basic improvement estimate(∫Bs

|u|κγ dx) 1

2κ

≤ C

r − s(|γ| + 1)

(∫Br

|u|γ) 1

2

if |β| = |γ − 1| ≥ d > 0.

(2.499)

96 [March 17, 2014]

This estimate holds for

subsolutions if γ − 1 = β > 0,supersolutions if γ − 1 = β < 0.

Step 2: iteration. Set

Φ(p, r) :=(∫

Br

|u|p dx) 1

p

. (2.500)

Then

supBr

u = limp→∞

Φ(p, r) =: Φ(∞, r),

infBr

u = limp→−∞

Φ(p, r) =: Φ(−∞, r).

With this notation the basic estimate (2.499) becomes

Φ(κγ, s) ≤(

C

r − s(|γ| + 1)

) 2γ

Φ(γ, r) if γ > 0 (2.501)

and

Φ(κγ, s) ≥(

C

r − s(|γ| + 1)

) 2γ

Φ(γ, r) if γ < 0, (2.502)

as long as 0 < s < r and |γ−1| ≥ d > 0. Here (2.502) holds for subsolutions,while (2.501) holds for subsolutions if 0 < γ < 1 and for supersolutions ifγ > 1.

To prove assertion (i) of Theorem 2.45 take

1 < p ≤ 2, γm = κmp, rm = 1 + 2−m

and apply (2.501) with γ = γm−1, r = rm−1, s = rm. Not that |γ|+1| ≤ 2|γ|and the factor 2 can be included in the constant C in (2.501). Thus (2.501)yields

lnΦ(γm, rm) ≤ 2γm

(C +m+ ln γm) + lnΦ(γm−1, rm−1). (2.503)

Now∑m

2γm

(C +m+ ln γm) =∑m

2κmp

(C +m+m lnκ+ ln p) ≤ C. (2.504)

HencesupB1

u = Φ(∞, 1) ≤ CΦ(p, 2). (2.505)

This proves assertion (i).

97 [March 17, 2014]

Proof of assertion (ii) Let 1 < p < κ = nn−2 . We will show that for each

p0 = κ−M ∈ (0, 1) we have

Φ(p, 2) ≤ CΦ(p0, 3), (2.506)Φ(−∞, 1) ≥ Φ(−p0, 3). (2.507)

Since infB1 u = Φ(−∞, 1) assertion (ii) then follows if we can show that forsufficiently small p0 > 0 we have

Φ(p0, 3) ≤ CΦ(−p0, 3) (2.508)

To prove (2.506) we apply the estimate (2.501) with

γm = κmp0, rm = 2 + 2−m

and we deduce (2.506) after M iterations. Note that γm < 1 if m ≤M−1 soat each step in the iteration (2.501) can be applied since u is a subsolution.

To prove (2.507) we apply (2.502) with

γm = −κmp0, rm = 1 + 2−m+1

and the convergence follows again from (2.504), now with γm replaced by|γm| and p replaced by p0.

Step 3: closing the gap: Φ(p0, 3) ≤ CΦ(−p0, 3). We first show that thefunction

w := lnu

satisfies ∇w ∈ L2,n−2(B3) where L2,n−2 denotes the Morrey space.We use (2.495) with β = −1. Since ∇w = u−1∇u this gives∫

B4

η2|∇w|2 dx ≤ C

∫B4

η|∇η| |∇w| dx

≤ C

(∫B4

|∇η|2 dx) 1

2(∫

B4

η2|∇w|2 dx) 1

2

.(2.509)

Now let z ∈ B3, r ≤ 12 and η ∈ C∞

c (B(z, 2r) with η = 1 in B(z, r) and∥∇η| ≤ C

r . Then (2.509) implies that∫B(z,r)

|∇w|2 dx ≤∫B(z,2r)

|∇η|2 dx ≤ Crn−2. (2.510)

Similarly for η ∈ C∞c (B4) with η = 1 in B3 we get∫

B3

|∇w|2 dx ≤ C. (2.511)

98 [March 17, 2014]

Thus ∇w ∈ L2,n−2(B3) and ∥∇w∥L2,n−2 ≤ C(n,Λ/ν). The Poincare in-equality implies that

[w]L2,n(B3) ≤ C∥∇w∥L2,n−2 ≤ C(n,Λ/ν). (2.512)

Now it follows from the John-Nirenberg theorem (see Theorem 2.51 be-low) that ∫

B3

exp(σ|w − a|[w]L2,n

)≤ C (2.513)

where a = wB(0,3) . Let p0 = σ/[w]L2,n . Then using that ±(w−a) ≤ |w−a|we deduce from (2.513) that∫

B3

up0e−p0a dx ≤ C,

∫B3

u−p0ep0a dx ≤ C

Hence ∫B3

up0 dx

∫B3

u−p0 dx ≤ C

and taking this inequality to the power 1/p0 we obtain the assertionΦ(p0, 3) ≤ CΦ(−p0, 3).

Step 4: Justification of the choice of test function This step wasonly sketched in class.The main point of Step 1 is the implication

uγ ∈ L1(Br), supp η ⊂ Br =⇒ (ηuγ/2)2κ ∈ L1 and (2.499) holds.

To explain the main line of the argument we used the test function η2uβ

with β = γ − 1 without further discussion. This is no problem for β ≤ 1,since by our assumption u ≥ δ we then have v = η2uβ ∈ W 1,2

0 (B4) so wecan use v in the definition of sub- oder supersolution.

If, however, β > 1 it is initially not clear whether v ∈ W 1,2. We nowgive a rigorous argument which shows that

uγ ∈ L1(Br), supp η ⊂ Br

=⇒∫B4

η2|u|γ−2|∇u|2 dx <∞, ηuγ/2 ∈W 1,20 (Br) and (2.495) holds.

(2.514)

From this it follows that all the remaining calculation in Step 1 are justified.To prove (2.514) we consider the test function v = η2hM (u) where, for

γ > 2 the function hM (t) is a truncation of tγ−1, namely,

hM (t) =

tγ−1 if 0 ≤ t < M ,(2 − γ)Mγ−1 + (γ − 1)Mγ−2t if t ≥ nM .

99 [March 17, 2014]

Then hM is a C1 function on (0,∞) and h′M is bounded. Thus hM (u) ∈W 1,2. We have

h′M (t) =

(γ − 1)tγ−2 if 0 ≤ t < M ,(γ − 1)Mγ−2 if t ≥ nM .

Note that h′M is monotonically increasing in M . Using η2hM (u) as a testfunction and using ellipticity and boundedness of the aij we get∫

B4

η2h′M (u)|∇u|2 dx ≤ C

∫B4

η|∇η|hM (u)|∇u| dx (2.515)

Now one can check directly from the formulae for hM and h′M that

hM (t) ≤ th′M (t).

Using this estimate and the Cauchy-Schwarz inequality we get∫B4

η|∇η|hM (u)|∇u| dx ≤∫B4

η|∇η|uh′M (u)|∇u| dx

≤(∫

B4

|∇η|2h′M (u)u2 dx

) 12(∫

B4

η2h′M (u)|∇u|2 dx) 1

2

Thus (2.515) implies that∫B4

η2h′M (u)|∇u|2 dx ≤ C

∫B4

|∇η|2h′M (u)u2 dx

Now h′M (u)|u|2 ≤ (γ−1)|u|γ . Thus the right hand side is bounded indepentlyof M . By the monotone convergence theorem we can take the limit M → ∞on the left hand side and we get∫

B4

η2(γ − 1)uγ−2|∇u|2 dx <∞ (2.516)

From this we deduce that

ηuγ−1∇u = ηuγ2 u

γ2−1∇u ∈ L1(B4)

since uγ2 ∈ L2(Br) by assumption. Thus the right hand side of (2.515) is

bounded independently of M . Using monotone convergence we can pass tothe limit M → ∞ and we obtain (2.495).

Finally to see that ηuγ/2 ∈W 1,2 we approximate again uγ/2 by functionsgM (u) where gM has linear growth at infinity and we use the estimate 2.516and monotone convergence to pass to the limit M → ∞.

100 [March 17, 2014]

[28.11. 2013, Lecture 13][2.12. 2013, Lecture 14]

We will first show that the weak Harnack inequality implies C0,α regularity.Then we will note that the weak Harnack inequality implies the classicalHarnack inequality. Finally we prove the John-Nirenberg result on expo-nential integrability of BMO functions which was crucial to ’close the gap’between Φ(−p0, 3 and Φ(p0, 3) in the proof of the weak Harnack inequality.

Proof that Theorem 2.45 =⇒ Theorem 2.44.

Step 1. L∞ estimate.Let u ∈W 1,2(Ω) be a weak solution of Lu = 0. Assume that B(y, 4R) ⊂ Ω.We first show that u is bounded in B(y,R) by showing that convex functionsof u are subsolutions.12

Assume thatf ∈ C2(R), sup

R|f ′| + |f ′′| <∞. (2.517)

We claim thatf u ∈W 1,2(Ω), L(f u) ≤ 0. (2.518)

Choosing f(t) =√t2 + δ2 and using Theorem 2.45 (i) for f u and p = 2

we then get

supB(y,R)

|u| ≤ supB(y,R)

f u ≤ CR−n/2∥f u∥L2(B(y,2R).

Taking the limit δ → 0 we conclude that

supB(y,R)

|u| ≤ CR−n/2∥u∥L2(B(y,2R)). (2.519)

and thus u ∈ L∞loc(Ω).

The first assertion in (2.518) follows from the chain rule. To prove thatf u is a subsolution we first consider φ ∈ (W 1,2

0 ∩ L∞)(Ω) with φ ≥ 0.Then ∫

Ω

∑i,j

aij∂jf(u) ∂iφdx =∫

Ω

∑i,j

aijf′(u)∂ju ∂iφ

=∫

Ω

∑i,j

aij∂ju ∂i(f ′(u)φ) dx−∫

Ω

∑i,j

aij∂ju f′′(u)∂iu︸ ︷︷ ︸

≥0

φdx ≤ 0.

Here we used that the first term in the last line is zero since u is a weaksolution of Lu = 0 and since v = f ′(u)φ is in W 1,2

0 (Ω). Since φ ∈ W 1,20 ∩

12Alternatively the estimate (2.519) below can be proved by using Moser iteration withthe test functions |u|β−1u or more precisely a suitable truncation of these functions as inStep 4 of the proof of the weak Harnack inequality.

101 [March 17, 2014]

L∞(Ω) : φ ≥ 0 is dense in φ ∈ W 1,20 (Ω) : φ ≥ 0 it follows that f u is a

subsolution.13

Step 2. C0,α estimate.Set

ω(r) = oscB(y,r)

u := supB(y,r)

u− infB(y,r)

u. (2.520)

We claim that there exist a 0 < θ < 1 such that

B(y, 4r) ⊂ Ω =⇒ ω(r) ≤ θω(4r). (2.521)

From this one easily deduces that there exist α > 0 and C > 0 such forB(y, r) ⊂ B(y,R) ⊂ Ω we have

oscB(y,r)

u ≤ C( rR

)αosc

B(y,R)u. (2.522)

and together with the local estimates for supu and − inf u this implies theconclusion of Theorem 2.44 To prove the claim set Bρ = B(y, ρ) and define

M4 = supB4ru, M1 = supBr

u,m4 = infB4r u, m1 = infBr u.

ThenM4 − u ≥ 0, L(M4 − u) = 0 in B4r

u−m4 ≥ 0, L(u−m4) = 0 in B4r.

Thus Theorem 2.45 (ii) with p = 1 yields

r−n∫B2r

(M4 − u) ≤ C infBr

(M4 − u) ≤ C(M4 −M1),

r−n∫B2r

(u−m4) ≤ C infBr

(u−m4) ≤ C(m1 −m4).

Adding these two inequalities we get

2nLn(B1)(M4 −m4) ≤ C(M4 −M1) + C(m1 −m4)

and thus(M1 −m1) ≤ (1 − 1

C2nLn(B1))(M4 −m4).

Taking θ = 1 − 1C 2nLn(B1) we get (2.521).

13To prove density note that there exist φk ∈ C∞c (Ω) such that φk → φ in W 1,2 and

let φ+k = max(φk, 0). Then φ+

k → φ+ = φ in W 1,2.

102 [March 17, 2014]

Theorem 2.46 (Harnack inequality). Assume that (2.484) and (2.485) holdand that u ∈W 1,2(Ω) is a weak solution of

Lu = 0. (2.523)

Assume further thatu ≥ 0 in B(y, 4R) ⊂ Ω. (2.524)

ThensupB(y,R)

u ≤ C infB(y,R)

u (2.525)

where C = C(n,Λ/ν).

Proof. This follows directly from the weak Harnack inequality, Theorem 2.45.

Remark. The Harnack inequality can be seen as a quantitative version ofthe strong maximum principle. The strong maximum principle implies thatif u(x) = 0 for some x ∈ B(y,R) then u ≡ 0 in B(y,R). In other words thestrong maximum principle implies that

infB(y,R)

u = 0 =⇒ supB(y,R)

u = 0. (2.526)

The Harnack inequality provides a quantitative estimate between the supre-mum and the infimum.

2.3 BMO , exponential integrability, interpolation and theHardy space H1

2.3.1 Exponential integrability of functions in BMO ,John-Nirenberg estimate

Now we prove exponential integrability of the functions BMO . In the fol-lowing by a cube we mean a bounded cube parallel to the coordinate axis.We say that Q is an open cube of size 2l if Q = a+(−l, l)n for some a ∈ Rn.

Let Q0 be an open cube and recall that f ∈ BMO(Q0) if u is integrablein Q0 and

[f ]BMO(Q0) := sup

1Ln(Q)

∫Q|f − fQ| dx : Q ⊂ Q0 cube

<∞. (2.527)

Here as usualfQ :=

1Ln(Q)

∫Qf dx. (2.528)

The key estimate is contained in the following lemma.

103 [March 17, 2014]

Lemma 2.47. Let Q0 ⊂ Rn be a cube and assume that u ∈ BMO(Q0). Fort > 0 let

St := x ∈ Q0 : |f(x) − fQ0 | > t. (2.529)

ThenLn(St) ≤ Ae−σt/[f ]BMO Ln(Q0), (2.530)

where A and σ depend only on n. Indeed we may take A = e and σ =1/(2ne).

Remark. The estimate is sharp (up to the optimal choice of σ) since thefunction x 7→ ln |x| is in BMO (Q0) for every cube Q0.

Lemma 2.48. Let Q0 ⊂ Rn be a cube and let h ∈ L1(Q0). Let s > 0 besuch that

s >1

Ln(Q)

∫Q0

|h| dx. (2.531)

Then there exist at most countable many disjoint open cubes Ik such that

(i) |h| ≤ s a.e. in Q0 \∪k Ik,

(ii)∣∣∣ 1Ln(Ik)

∫Ikh∣∣∣ < 2ns

(iii)∑

k Ln(Ik) ≤1s

∫Q0

|h| dx.

Proof. This is taken literally from [JN], pp. 418–419 (with a few extra wordsfor the proof of (i)).DivideQ0 (by halving each edge) into 2n cubes of equal size and let I11, I12, . . .be those open cubes over which the average value of of |h| is greater than orequal to s. Then in view of the assumption (2.531) we have on those cubes

sLn(I1k) ≤∫I1k

|h| dx < 2nsLn(I1k).

Next subdivide each remaining cube, over which the average of |h| is strictlyless than s, into 2n cubes of equal size, and denote by I21, I22, . . . those cubesover which the average of |h| is greater than or equal to s. Again subdividethe remaining cubes, etc. In this way we obtain a sequence of cubes Iik,which we rename Ik, such that

sLn(I1k) ≤∫I1k

|h| dx < 2nsLn(I1k)

Clearly property (ii) is satisfied. Furthermore, summing the left inequalityover k we obtain (iii).

104 [March 17, 2014]

Finally let x ∈ Q0 \∪Ik. Then there exist closed cubes Ji such that

x ∈ Ji, the length li of Ji satisfies li = 2−il0 where l0 is the length of Q0 and

1Ln(Ji)

∫Ji

|h| dy < s. (2.532)

Now by the Lebesgue point theorem we have, for a.e. x ∈ Q0 \∪Ik,

limr→0

r−n∫B(x,r)

|h(x) − h| dy = 0. (2.533)

Since Ji ⊂ B(x,√nli) we get

|h(x)| =1

Ln(Ji)

∫Ji

|h(x)| dy ≤ 1Ln(Ji)

∫Ji

|h| + |h(x) − h| dy

≤s+1

Ln(Ji)

∫B(x,

√nli)

|h(x) − h| dy

where we used (2.532). Thus taking the limit i → ∞ and using (2.533) weget property (i).

Proof of Lemma 2.47. It suffices to prove the result if [f ]BMO = 1. Indeed,for a general function g ∈ BMO we can first consider f = g/[g]BMO . Thenthe assertion for g follows from the assertion for f since |g| > t if and onlyif |f | > t/[g]BMO . Moreover we may assume that fQ0 = 0.

For t > 0 let F (t) be the smallest number such that

Ln(St) = Ln(x ∈ Q0 : |f(x)| > t) ≤ F (t)Ln(Q0) (2.534)

for all cubes Q0 and all f ∈ L1(Q0) with

[f ]BMO(Q0) ≤ 1, fQ0 = 0. (2.535)

Note that trivially F (t) ≤ 1 for all t > 0.The key estimate is: if t ≥ 2n then

F (t) ≤ 1sF (t− 2ns) ∀s ∈ [1, 2−nt]. (2.536)

To prove this estimate first note that by (2.535)∫Q0

|f | dx =∫Q0

|f − fQ0 | dx ≤ Ln(Q0). (2.537)

Now apply the Calderon-Zygmund decomposition with h = f . Since s < twe have |f | ≤ t almost everywhere in Q0 \

∪k Ik. Hence

Ln(St) ≤∑k

Ln(x ∈ Ik : |f(x)| > t). (2.538)

105 [March 17, 2014]

Now by assertion (ii) in Lemma 2.48 we have |fIk | < 2ns. Thus |f(x)| > timplies that |f − fIk | > t− 2ns and therefore

Ln(x ∈ Ik : |f(x)| > t) ≤ Ln(x ∈ Ik : |f − fIk | > t− 2ns)≤F (t− 2ns)Ln(Ik) (2.539)

where we used the definition of F in the second inequality.Using first (2.538), (2.539) and then property (iii) in Lemma 2.48 and

(2.537) we get

Ln(St) ≤ F (t− 2ns)∑k

Ln(Ik) ≤1sF (t− 2ns)Ln(Q0).

Since Q0 and u were arbitrary (subject to (2.535)) we get (2.536).Now set s = e in (2.536) and let σ = 1/(2ne). Then

F (t) ≤ Ae−σt =⇒ F (t+ 2ne) ≤ 1eAe−σt = e−σ(t+2ne).

It follows that if the inequality F (t) ≤ Ae−σt holds for t in some interval oflength 2ne it holds for all larger t. Using the trivial estimate

F (t) ≤ 1 ≤ e e−σt for 0 ≤ t ≤ 2ne (2.540)

we see that (2.530) holds with A = e and σ = 1/(2ne).

Theorem 2.49 (John-Nirenberg estimate for functions in BMO). Let Q0 ⊂Rn be a cube.

(i) If f ∈ BMO (Q0) then∫Q0

exp(σ1|f − fy,r|

[f ]BMO

)≤ A1Ln(Q0) (2.541)

for some σ1 > 0 and A1 > 0 which depend only on the space dimensionn. We may take σ1 = 1/(2n+1e) and A1 = e+ 1.

(ii) If f ∈ BMO (Q0) then f ∈ Lp(Q0) for all 1 < p <∞ and1

Ln(Q)

∫Q|f − fQ|p ≤ C(p, n)[f ]pBMO(Q) for all cubes Q ⊂ Q0.

(2.542)Moreover the Lp norms grow at most linearly in p, i.e.,

lim supp→∞

C(p, n)1/p

p≤ C ′(n). (2.543)

(iii) We have L2,n(Q0) = BMO (Q0) with equivalent norms. More preciselythere exists a constant B which only depends only on n such that

B−1[u]L2,n ≤ [u]BMO ≤ B[u]L2,n , (2.544)

Ln(Q0)−1/2∥u∥L2 ≤ B

([u]BMO + Ln(Q0)

−1∥u∥L1

), (2.545)

Ln(Q0)−1∥u∥L1 ≤ Ln(Q0)

−1/2∥u∥L2 . (2.546)

106 [March 17, 2014]

Remark. The example x 7→ ln |x| shows that linear growth of the Lp normin p is optimal.

To deduce (2.541) from Lemma 2.47 we use the following general formulawhich was proved in Analysis 3.

Lemma 2.50. Let E ⊂ Rn be measurable and let f : E → [0,∞) be mea-surable. Let g ∈ C0[0,∞) ∩C1((0,∞) and assume that g ≥ 0 and g(0) = 0.Let

µf (t) := Ln(x ∈ E : f(x) > t. (2.547)

Then ∫Eg(f(x)) dx =

∫ ∞

0g′(t)µf (t) dt (2.548)

in the sense that if one of the integrals exists then the other exists andequality holds.

Proof. By the fundamental theorem of calculus

g(f(x)) =∫ f(x)

0g′(t) dt (2.549)

LetF := (x, t) ∈ Rn+1 : f(x) > t

Then F is a measurable subset of Rn+1 since the function f(x) − t is mea-surable in Rn+1. Thus by Fubini’s theorem∫

g(f(x)) dx =∫

Rn

∫RχF (x, t)g′(t) dt dx =

∫R

∫Rn

χF (x, t) dx g′(t) dt,

=∫

Rµf (t)g′(t) dt.

Proof of Theorem 2.49. (i): We may assume without loss of generality that[f ]BMO ≤ 1 and fQ0 = 0. Then by Lemma 2.47 we have

µf (t) ≤ ALn(Q0)e−σt.

If we apply Lemma 2.50 with g(t) = eσ1t − 1 and 0 < σ1 < σ we get∫Q0

(eβf − 1

)dx =

∫ ∞

0σ1e

σ1tALn(Q0)e−σt dt ≤ ALn(Q0)σ1

σ − σ1.

Hence we may take σ1 = 12σ = 1/(2n+1e) and A′ = A+ 1 = e+ 1.

(ii): It suffices to consider the case Q = Q0 since otherwise we can applythe result to f = f|Q and use that [f ]BMO(Q) ≤ [f ]BMO(Q0). By scaling wecan also assume that Ln(Q0) = 1. Then the assertion follows again from

107 [March 17, 2014]

Lemma 2.50. To get the asymptotics for C(p, n) we can use the change ofvariables τ = σt and the identities (for k ∈ N, k ≥ 1)∫ ∞

0ktk−1e−σt dt = σ−k

∫ ∞

0kτk−1e−τ dτ = σ−kk! ≤ σ−kkk. (2.550)

(iii): All the estimates are invariant under the scaling u(x) 7→ u(λx).Hence we may assume thatQ0 = (0, 1)n. The inclusion L2,n(Q0) ⊂ BMO(Q0)and the second inequality in (2.544) were already established in (2.347). Es-timate (2.546) is just the Cauchy-Schwarz inequality.

For the reverse inclusion L2,n(Q0) ⊃ BMO(Q0) and the remaining es-timates we use assertion (ii) and the following observation. If x ∈ Q0 =(0, 1)n and r ≤ 1 then there exist an open cube Q of size ≤ 2r such thatB(x, r) ∩ Q0 ⊂ Q ⊂ Q0. Indeed by reflection on the symmetry planes ofQ0 we may assume that 0 < xi ≤ 1

2 . If r ≤ 12 take Q = a + (−r, r)n where

ai = max(xi, r). If r > 12 take Q = Q0. Thus

r−n∫B(x,r)∩Q0

|f − fB(x,r)|2 dy ≤ r−n∫B(x,r)∩Q0

|f − fQ|2 dy

≤r−n∫Q|f − fQ|2 dy ≤ 2nC(n, 2)[f ]2BMO

where we used (2.91) in the first inequality and assertion (ii) and the estimateLn(Q) ≤ (2r)n in the last. This proves the first inequality in (2.544).

Finally we have∫Q0

|f |2 dx = |fQ0 |2 +∫Q0

|f − fQ0 |2 dx ≤ ∥f∥2L1 + C(n, 2)[f ]2BMO

and this implies (2.544). Thus BMO (Q0) ⊂ L2,n(Q0).

[2.12. 2013, Lecture 14][5.12. 2013, Lecture 15]

By using the extension operator in Lemma (2.34) the John-Nirenbergtheorem can be extended to general open and bounded sets with Lipschitzboundary.

Theorem 2.51. Let Ω ⊂ Rn be an open and bounded set with Lipschitzboundary. Then exists constants σ2 > 0 and A2 > 0 which only depend onthe dimension n and on Ω with the following property. If f ∈ L2,n(Ω) then∫

Ωexp

(σ2|f − fΩ|

[f ]L2,n

)≤ A2(diamΩ)n (2.551)

108 [March 17, 2014]

Remark. (i) By scaling one sees that σ2 and A2 do not change if Ω isreplaced by a dilation λΩ. In particular if Ω is a ball then the constants σ2

and A2 can be chosen indepent of the radius of the ball.(ii) The result holds even under the weaker condition that Ω is bounded andopen and that there exists an a > 0 such that

Ln(B(x, r) ∩ Ω) ≥ arn ∀x ∈ Ω, r < diamΩ. (2.552)

The constants σ2 and A2 depend only on n and a. In this case one use theextension

Ef(x) :=1

Ln(B(x, 2d(x)) ∩ Ω)

∫B(x,2d(x))∩Ω

f dy

for x ∈ Rn \ Ω where d(x) = dist (x,Ω).

Note that the condition (2.552) implies that Ln(∂Ω) = 0. Indeed otherwiseby the Lebesgue point theorem there exists x ∈ Ω such that

limr→0

Ln(∂Ω ∩B(x, r))Ln(B(x, r))

= 1 and thus limr→0

r−nLn(Ω ∩B(x, r)) = 0

and this leads to a contradiction with (2.552).

Proof of Theorem 2.51. It suffices to prove the Theorem for function withfΩ = 0 (if the result is know for such function it can be applied to thefunction g = f − fΩ). Moreover by scaling we may assume that diamΩ = 1

2and by translation we may assume that

Ω ⊂ Q0 := (0, 1)n.

Finally we may assume[f ]L2,n(Ω) = 1. (2.553)

By the definition of the L2,n seminorm we have for any point y ∈ Ω∫Ω|f |2 dx =

∫Ω|f − fΩ|2 dx =

∫Ω∩B(x, 1

2

|f − fΩ|2 dx ≤ 2−n. (2.554)

since [f ]L2,n(Ω) = 1. Thus ∥f∥L2,n ≤ 2.Now let E be extension operator in Lemma (2.34) and let

g = Ef|Q0(2.555)

Then∥g∥L2,n(Q0) ≤ CE∥f∥2

L2,n ≤ 2CE , (2.556)

where CE = CE(Ω) is the norm of the extension operator. In particular wehave

|gQ0 | ≤∫Q0

|g| dx ≤(∫

Q0

|g|2 dx)1/2

≤ 2CE . (2.557)

109 [March 17, 2014]

Finally there exist B = B(n) ≥ 1 such that

[g]BMO(Q0) ≤ B∥g∥L2,n(Q0) ≤ 2BCE . (2.558)

Thus

|g| ≤ |g − gQ| + |gQ| ≤ 2BCE|g − gQ|[g]BMO

+ 2CE . (2.559)

Now setσ2 =

σ1

2BCE. (2.560)

Then by Theorem 2.49∫Q0

exp(σ2|g|) dx ≤∫Q0

exp(σ1

|g − gQ|[g]BMO

)eσ22CE dx

≤ A1eσ1 .

Since σ22CE = σ1/B ≤ σ1. Since Ω ⊂ Q0 and g = f on Ω this implies theassertion with σ2 = σ1/(2BCE) and A2 = A1e

σ1 .

2.3.2 Interpolation between BMO and L2

Lemma 2.52 (Weak Lp estimate for BMO p). Let Q0 be a cube in Rn. Wesay that ∆ is a subdivision of Q0 if ∆ is a finite or countable collectionof subcubes of Q0 which have no common interior point and Q0 ⊂

∪Q∈∆.

Assume that f ∈ L1(Q0) and that there exists a constant K > 0 such that∑Q∈∆

[1

Ln(Q)

∫Q|f − fQ| dx

]pLn(Q)

1p

≤ K (2.561)

for all subdivisions ∆. Then

Lnx ∈ Q0 : |f − fQ0 | > t ≤ AKp

tp(2.562)

where A only depends only on n and p.

Remark. Lemma 2.52 and Lemma 2.47 also hold for functions with valuesin Rd if | · | denotes a norm in Rd. The constant A is independent of d.

Proof. This proved by induction, very similar to the proof of Lemma 2.47.See [JN], pp. 423–425.

We note that if f ∈ Lp(Q0) then condition (2.561) holds with K =2∥f∥Lp . Indeed

∫Q |f −fQ| dx ≤ 2

∫Q |f | dx and thus by Jensens’s inequality[

1Ln(Q)

∫Q|f − fQ| dx

]p≤ 2p

1Ln(Q)

∫Q|f |p dx

110 [March 17, 2014]

and thus∑Q∈∆

[1

Ln(Q)

∫Q|f − fQ| dx

]pLn(Q) ≤ 2p

∑Q∈∆

∫Q|f |p dx = 2p

∫Q0

|f |p dx.

(2.563)

We now use Lemma 2.52 to prove the interpolation result Theorem 2.33which was the basis of our approach to the Lp estimates. We first recall theMarcinkiewicz interpolation theorem.

Let E and E′ be measurable subsets of Rn. Assume that T maps mea-surable function on E to measurable functions on E′.We say that T is of type (p, q) if there exists a constant A such that

∥Tf∥Lq ≤ A∥f∥Lp ∀f ∈ Lp(E). (2.564)

Similarly, for q ∈ [1,∞), we say that T is of weak-type (p, q) if there existsa constant A such that

Ln(x ∈ E′ : |Tf(x)| > t) ≤(A∥f∥Lp

t

)q∀f ∈ Lp(E), t > 0. (2.565)

Note that if T is of type (p, q) it is in particular of weak-type (p, q) (forq <∞) since for any g ∈ Lq(E′) we have

Ln(x ∈ E′ : |g| > t) ≤(∥g∥Lp

t

)q. (2.566)

We say that T is sub-additive if |T (f + g)(x)| ≤ |Tf(x)| + |Tg(x)|. ByL1(E) + Lr(E) we denote that space of functions which can be written inthe form f + g with f ∈ L1(E) and g ∈ Lr(E).

Theorem 2.53 (Marcinkiewicz interpolation theorem). Let 1 ≤ p < r ≤ ∞.Suppose that T is a subadditive mapping from Lp(E)+Lr(E) to the space ofmeasurable functions on E′. If T is simultaneously of weak type (p, p) andweak type (r, r) then T is of type (q, q) for all q ∈ (p, r). More precisely if Tsatisfies

(i) |T (f + g)(x)| ≤ |Tf(x)| + |Tg(x)| ∀f, g ∈ Lp(E) + Lr(E),

(ii) Ln(x ∈ E′ : |Tf(x)| > t) ≤(Ap∥f∥p

t

)p∀f ∈ Lp(E).

(iii) If r <∞ then

Ln(x ∈ E′ : |Tf(x)| > t) ≤(Ar∥f∥r

t

)r∀f ∈ Lr(E).

If r = ∞ then ∥Tf∥L∞ ≤ A∞∥f∥L∞ ∀f ∈ L∞(E).

Then∥Tf∥Lq ≤ Aq∥f∥Lq ∀f ∈ Lq(E) (2.567)

for all q ∈ (p, r) where Aq depends only on Ap, Ar, p, q and r.

111 [March 17, 2014]

Remark. The result also holds for vector-valued functions. More preciselyif T maps Lp(E; Rd1) +Lr(E; Rd1) into measurable functions with values inRd2 then the result holds if we interpret | · | as the norm in Rd1 and Rd2 ,respectively. The constant Ap does not depend on d1 and d2.

Proof. See Homework 1, Problem 4 and [St70], Theorem 4 in Section I.4,pp. 21-22 and Appendix B, pp. 272–274. Note that it suffices to considerthe case E = E′ = Rn since we can extend all functions by zero to Rn.

Theorem 2.54 (Lp estimate from L2 and BMO estimates). Let Ω be ameasurable set in Rn with finite measure and suppose that Q0 ⊂ Rn is acube. Assume that T : L2(Ω; Rd1) → L2(Q; Rd2) is a linear operator suchthat

∥Tu∥L2(Q) ≤ M2∥u∥L2(Ω) ∀u ∈ L2(Ω; Rd1) (2.568)

[Tu]BMO(Q) ≤ M∞∥u∥L∞(Ω) ∀u ∈ L∞(Ω;Rd1) (2.569)

Then for all 2 < q <∞

∥Tu∥Lq(Ω) ≤ C∥u∥Lq(Ω) ∀ f ∈ Lq(Ω; Rd1) (2.570)

where C depends on q, n,M2,M∞ and Ln(Ω)/Ln(Q).

Proof. This proof is taken from [Ca66].Let ∆ be a subdivision of the cube Q0. For f ∈ L2(Ω) define T u as thefollowing function which is constant on each Q ∈ ∆

(T u)(x) :=1

Ln(Q)

∫Q|Tu− (Tu)Q| ∀x ∈ Q. (2.571)

Then T is sublinear. Moreover

∥T (u)∥L∞ ≤ [Tu]BMO ≤M∞∥f∥L∞

and (use that∫Q |f − fQ|2 ≤

∫Q |f |2 and compare also (2.563))

∥T u∥2L2 =

∑Q∈∆

[1

Ln(Q)

∫Q|Tu− (Tu)Q| dx

]2

Ln(Q)

≤∑Q∈∆

∫Q|Tu|2 dx =

∫Q0

|Tu|2 dx ≤M2∥u∥2L2

Thus T is of type (2, 2) and of type (∞,∞). By the Marcinkiewicz interpo-lation theorem

∥T u∥Lr ≤Mr∥u∥Lr (2.572)

112 [March 17, 2014]

for all r ∈ (2,∞) were Mr depends only on M2, M∞ and r.14 In particularthe constant Mr is independent of the partition ∆.

Using the definition of T we get∑Q∈∆

[1

Ln(Q)

∫Q|Tu− (Tu)Q| dx

]rLn(Q) = ∥T u∥rLr ≤M r

r ∥u∥rLr (2.573)

Now Lemma 2.52 implies that

Ln(x ∈ Q0 : |Tu− (Tu)Q0 | > t ≤ Ar

(Mr∥u∥Lr

t

)r. (2.574)

Therefore the map u 7→ Tu− (Tu)Q0 is of weak type (r, r) for all r <∞. Byassumption this map is also of type (2, 2) since ∥Tu− (Tu)Q0∥L2 ≤ ∥Tu∥L2 .

Now let q ∈ (2,∞) and take r = 2q. It follows from the Marcinkiewiczinterpolation theorem that the map u 7→ Tu− (Tu)Q0 is of type (q, q) and

∥Tu− (Tu)Q0∥Lq ≤M ′q∥u∥Lq , (2.575)

whereM ′q = M ′

q(q, n,M2,M∞). (2.576)

Finally we need to estimate (Tu)Q0 . For this we use the obvious esti-mates

1Ln(Q0)

∫Q0

|(Tu)Q0 |q dx ≤ |(Tu)Q0 |q ≤(

1Ln(Q0)

∫Q0

|Tu|2)q/2

≤M q2

(Ln(E)Ln(Q0)

1Ln(E)

∫E|u|2)q/2

≤M q2

(Ln(E)Ln(Q0)

)q/2 1Ln(E)

∫E|u|q dx

(2.577)

where we used Jensen’s inequality in the second and in the last step. Itfollows that

∥(Tu)Q0∥Lq ≤M2

(Ln(E)Ln(Q0)

) 12− 1

q

∥u∥Lq

and together with (2.575) this finishes the proof.

2.3.3 Maximal functions and the Hardy space H1

The part about the Hardy-Littlewood maximal function is standard and canbe found, e.g., in [St70]. For an introduction to Hardy spaces, see Semmes’primer on Hardy spaces (Section 1 in [Se]), Chapters 3 and 4 in the bookby Stein [St93] or the original paper [FS].

14Actually the Riesz-Thorin interpolation theorem gives the more precise estimate Mr ≤M1−θ

2 Mθ∞ where 1/r = (1 − θ)/2, but we do not need this.

113 [March 17, 2014]

Another important application of the Marcinkiewicz interpolation theo-rem are the Lp estimates for the maximal function. For f ∈ L1

loc(Rn) theHardy-Littlewood maximal function Mf is defined as

Mf(x) := supr>0

1Ln(B(x, r))

∫B(x,r)

|f | dy. (2.578)

Note that∥Mf∥L∞ ≤ ∥f∥L∞ . (2.579)

Theorem 2.55. There exists a constant A1 such that

Ln(x ∈ Rn : |Mf(x)| > t ≤ A1

t∥f∥L1 . (2.580)

Moreover for p ∈ (1,∞) there exist Ap such that

∥Mf∥Lp ≤ Ap∥f∥Lp . (2.581)

Proof. The second estimate follows from the first estimate, (2.579) and theMarcinkiewicz interpolation theorem (note that the maps f 7→Mf is sublin-ear). The first estimate follows from Vitali covering theorem (see Analysis3 or [St70], Chapter 1, Paragraph 1).

Sketch of proof: Let At := x : Mf(x) > t. For each x ∈ At thereexists a ball B(x, rx) such that∫

B(x,rx)|f | ≥ t

2Ln(B(x, rx)). (2.582)

By the Vitali covering theorem there exist a subfamily of disjoint ballsB(xi, ri) such that

∪iB(xi, 5ri) ⊃ At. Thus

Ln(At) ≤ 5n∑i

Ln(B(xi, ri) ≤ 5n2t

∫S

i B(xi,ri)|f | dy ≤ 5n

2t∥f∥L1 .

The map f 7→ Mf is not of type (1, 1). Take for example B = B(0, 1),f = χB. Then for |x| > 1 we have B(0, 1) ⊂ B(|x|, |x| + 1) and thusMf(x) ≥ (|x|+1)−n. ThusMf /∈ L1(Rn). If we now take fk(x) = knfk(x) =knχB(0,1/k) then ∥fk∥ = 1 but ∥Mfk∥L1(B) ∼ ln k → ∞. One can show that

Mf ∈ L1(B) ⇐⇒ |f | ln(2 + |f |) ∈ L1(B) if f = 0 on Rn \B, (2.583)

see [St70], Chapter 1, Paragraph 5.2.[5.12. 2013, Lecture 15][9.12. 2013, Lecture 16]

114 [March 17, 2014]

Let

φ ∈ C∞c (B(0, 1)) with φ ≥ 0,

∫B(0,1)

φdx = 1, φ =1ω n

in B(0, 1/2)

(2.584)where

ωn := Ln(B(0, 1))

and setφr(z) =

1rnφ(zr

)(2.585)

Definition 2.56. Let f ∈ L1loc(Rn). We define the regularized maximal

function by15

f∗(x) := supr>0

|(φt ∗ f)(x)| = supr>0

∫Rn

1rnφ

(x− y

r

)f(y) dy. (2.586)

We say that f belongs to the Hardy space H1(Rn) if

f∗ ∈ L1(Rn) (2.587)

and we set∥f∥H1 := ∥f∗∥L1 (2.588)

Remark. By the Lebesgue point theorem we have limr→0(φr ∗ f)(x) =f(x) for a.e. x. Thus |f | ≤ f∗ a.e. and ∥f∥L1 ≤ ∥f∥H1 .

Lemma 2.57 (Elementary properties of H1).

(i) Scaling. Let ft(x) = t−nf(xt ). Then

(ft)∗ = (f∗)t and thus ∥ft∥H1 = ∥f∥H1 . (2.589)

(ii) Cancellation.

f ∈ H1(Rn) =⇒∫

Rn

f dx = 0. (2.590)

(iii) Local cancellation. Let a ∈ Rn, a = 0 and suppose that 0 < ρ <|a|/10. Let

f(x) = ρ−nχB(0,ρ) − ρ−nχB(a,ρ).

Then∥f∥H1 ∼ ln

|a|ρ.

15Alternatively one can define a ’grand maximal function’ by additionally taking asupremum over a class of test function φ, e.g., φ ∈ T := ψ ∈ C∞

c (B(0, 1)) : |∇ψ| ≤ 1.This is the original approach of Fefferman. Semmes [Se] explains why this is more natural.In the end, however, this yields the same space (with equivalent) norms as the use of asingle function φ.

115 [March 17, 2014]

(iv) ’Atoms’. Suppose that p > 1. Then

f ∈ Lp(Rn), supp f ⊂ B(0, R),∫f dx = 0

=⇒ f ∈ H1(Rn) and R−n∥f∥H1 ≤ R−n/p∥f∥Lp . (2.591)

Proof. (i) This is a straightforward calculation using the linear change ofvariables ξ = tx, η = ty, ρ = r/t. We have∫

Rn

1rnφ

(x− y

r

)ft(y) dy =

∫Rn

1rnφ

(x− y

r

)t−nf

(yt

)dy

=∫

Rn

1tn

tn

rnφ

(t(ξ − η)

r

)f(η) dη =

1tn

∫Rn

1ρnφ

(ξ − η

ρ

)f(η) dη

≤ 1tnf∗(ξ) =

1tnf∗(xt

).

Taking the supremum over r we get f∗t (x) ≤ t−nf∗(xt

). Since f = (ft)1/t

we also obtain the reverse inequality.(ii): Assume that α :=

∫Rn f dx = 0. We claim that then there exist an

R0 such that

f∗(x) ≥ |α|2 4nωn

1|x|n

if |x| ≥ R0. (2.592)

This immediately implies that f∗ /∈ L1(Rn).Let

ω(R) :=∫

Rn\B(0,R)|f | dx.

Then limR→∞ ω(R) = 0 since f ∈ L1(Rn). To prove (2.592) we use r = 4|x|in the definition of the regularized maximal function. If |y| ≤ |x| thenφ((x− y)/4|x|) = ωn and thus∣∣∣∣∫

Rn

φ

(x− y

4|x|

)f(y) dy

∣∣∣∣≥

∣∣∣∣∣ 1ωn

∫B(0,|x|)

f(y) dy

∣∣∣∣∣−∫

Rn\B(0,|x|)supφ |f(y)| dy

≥ 1ωn

(|α| − ω(|x|)) − supφ ω(|x|).

The right hand side is ≥ |α|/(2ωn) if |x| ≥ R0. This implies (2.592).(iii): Homework. HInt: By (i) one may assume that ρ = 1. For a lower

bound on the H1 norm show that f∗(x) ≥ c|x|−n for 2 < |x| < |a|/4 by usingr = 4|x| in the definition of f∗. For the upper bound use the argument inthe proof of (iv) for large r.

116 [March 17, 2014]

(iv): By (i) and the usual scaling of the Lp norm we may assume thatR = 1. We have f∗(x) ≤ CMf(x) (with C ≤ supφ/ωn). Thus

∥f∗∥L1(B(0,2)) ≤ C∥f∗∥Lp(B(0,2) ≤ C∥Mf∥Lp(Rn) ≤ C∥f∥Lp(Rn).

Hence it suffice to estimate f∗(x) for |x| ≥ 2. In this case (φr ∗ f)(x) = 0 ifr ≤ |x|/2. For r > |x|/2 we have∣∣∣∣∫

Rn

φ

(x− y

r

)f(y) dy

∣∣∣∣ = ∣∣∣∣∫Rn

[φ

(x− y

r

)− φ

(xr

)]f(y) dy

∣∣∣∣≤∫Rn

sup |∇φ| |y|r

|f(y)| dy ≤ C

rn+1∥f∥L1(B(0,1) ≤

2n+1C

|x|n+1∥f∥Lp(B(0,1).

Here in the first identity we used that∫

Rn f(y) dy = 0 and in the secondinequality we used that |y| ≤ 1 and the set where f is not zero. Thus itfollows that f∗ ∈ L1(Rn \ (B(0, 2)) and the proof is finished.

Lemma 2.58 (Approximation in H1).

(i) For η ∈ L1(Rn) and f ∈ H1(Rn) we have

(η ∗ f)∗(x) ≤ (|η| ∗ f∗)(x). (2.593)

(ii) (Lp ∩H1)(Rn) is dense in H1(Rn) for all p ∈ (1,∞).

Remark. One can also show that C∞c (Rn) is dense in H1(Rn). In this

case the proof is easier if one defines f∗ as the ’grand maximal function’,see [Se] Proposition 1.42 and in particular Lemma 1.51. Alternatively onecan apply the standard convolution argument to finitely many terms in theatomic decompostion (introduced below).

Proof. (i) We have

|(φr ∗ η ∗ f)(x)| = |(η ∗ φr ∗ f)(x)| ≤ (|η| ∗ |φr ∗ f |)(x) ≤ |η| ∗ f∗(x).

Taking the supremum over r > 0 we get the assertion.(ii) Homework. Hint:

Let ηε(x) = ε−nη(x/ε) denote the usual family of mollifiers. Show first that

h(x) := lim supε→0

(ηε ∗ f − f)∗(x) = 0 a.e.

To do so note that h(x) ≤ aδ(x) + bδ(x) where

aδ(x) = lim supε→0

sup0<t≤δ

|(ηε ∗ φt ∗ f)(x) − (φt ∗ f)(x)|,

bδ(x) = lim supε→0

supt>δ

|(ηε ∗ φt ∗ f)(x) − (φt ∗ f)(x)|.

117 [March 17, 2014]

Then show that bδ(x) = 0 for every x and every δ > 0 and that limδ→0 aδ(x) =0 for every Lebesgue point x, i.e., for a.e. x. Finally use (i) and a suitableversion of the Lebesgue dominated convergence theorem to conclude thatlimε→0 ∥(ηε ∗ f − f)∗∥L1 = 0.

A fundamental result of Fefferman states that the dual of H1(Rn) isBMO(Rn) and that H1(Rn) is the dual of VMO(Rn) (’functions of vanishingmean oscillations’), the closure of Cc(Rn) in BMO (Rn). Here BMO (Rn) isunderstood as a space of equivalence classes with f ∼ g if f − g = const andwe use the BMO seminorm on this space of equivalence classes. Since theH1 functions have integral zero the expression

∫fg dx depends only on the

equivalence class of g if f ∈ H1.

Theorem 2.59 (Fefferman; H1 − BMO duality). For f ∈ H1(Rn) andg ∈ L∞(Rn) we have ∣∣∣∣∫

Rn

fg dx

∣∣∣∣ ≤ C∥f∥H1 [g]BMO . (2.594)

Moreover if f ∈ L1loc then

∥f∥H1 ≤ C sup∫

Rn

fg dx : g ∈ Cc(Rn), [g]BMO ≤ 1. (2.595)

Here C depends only on n (and the choice of the function φ).

Warning. If f ∈ H1 and g ∈ BMO then the product fg is in general notin L1.

One way to prove (2.594) is to show that each f ∈ H1 has an atomicdecomposition, i.e., f(x) =

∑k∈N αkak(x) with

∑k |αk| <∞ and each ak is

an atom, i.e.,

ak is supported in a ball Bk, |ak| ≤1

Ln(Bk)a.e.,

∫Bk

ak = 0.

The construction of the atomic decomposition is similar, but more subtle,than the construction of the Calderon-Zygmund decomposition, see [St93],Chapter III.2, pp. 101–112. Note that if a is an atom supported on a ballB then that∫

Bag dx =

∫Ba(g − gB) dx ≤ C∥a∥L∞Ln(B)[g]BMO ≤ C[g]BMO .

118 [March 17, 2014]

Theorem 2.60 (Elliptic estimates in H1). Assume that the coefficients Aαβijare constant and strongly elliptic.

(i) Let f ∈ H1(Rn; Rn×N ). Then there exists a distributional solutionu ∈W 1,1

loc (Rn; RN ) of

−divA∇u = −div f (2.596)

with∥∇u∥H1 ≤ C∥f∥H1 . (2.597)

Moreover u is unique up to the addition of constants.

(ii) Similarly if g ∈ H1(Rn) then there exists a distributional solution u ∈W 2,1loc (Rn; RN ) of

−divA∇u = g, (2.598)

with∥∇2u∥H1 ≤ C∥g∥H1 . (2.599)

Moreover u is unique up to the addition of affine functions.

Here C only depends on n,N, ∥A∥ and the ellipticity constant ν.

Proof. (i): This follows directly from the corresponding BMO estimate inCorollary 2.16 by duality (recall that BMO = L2,n with equivalent semi-norms).

Let f ∈ (H1 ∩ L2)(Rn). Then there exist u ∈ W 1,2loc such that ∇u ∈ L2

and u is a distributional solution of −divA∇u = f , i.e.,∫Rn

A∇u · ∇φdx =∫

Rn

f · ∇φdx ∀φ ∈ C∞c (Rn; RN ). (2.600)

Let (AT )αβij = Aβαji , let g ∈ C∞c (Rn). Then in particular g ∈ (BMO ∩

L2)(Rn) and by Corollary 2.16 there exists a distributional solution v of−divAT∇v = −div g with ∇v ∈ (BMO ∩ L2)(Rn). Thus∫

Rn

AT∇v · ∇ψ dx =∫

Rn

g · ∇ψ dx ∀ψ ∈ C∞c (Rn; RN ). (2.601)

For future reference we note that in addition ∇v ∈ L∞(Rn). Indeed, byusing the equations for the derivatives of v we see that ∇v ∈ W k,2(Rn) forall k ∈ N. Hence by the Sobolev embedding

∥∇v∥2L∞(B(x,1)) ≤ C(k, n)∥∇v∥2

Wk,2(B(x,1)) ≤ C(k, n)∥∇v∥2Wk,2(Rn)

for k > n2 and all x ∈ Rn.

Now by density (use Lemma 2.37 with q = 2) the identity in (2.372) alsoholds with φ = v and in (2.601) we may take ψ = u. Since

AT∇v · ∇u = A∇u · ∇v (2.602)

119 [March 17, 2014]

this yields the duality relation∫Rn

g · ∇u dx =∫

Rn

f · ∇v dx. (2.603)

Now ∇v ∈ L∞(Rn) and we deduce from (2.594) (applied to each productfi∂vi), that ∫

Rn

g · ∇u dx ≤ C[f ]H1 [∇v]BMO ≤ C[f ]H1 [g]BMO

Now it follows from (2.595) that

∥∇u∥H1 ≤ C[f ]H1 ∀f ∈ (H1 ∩ L2)(Rn) (2.604)

Since (H1∩L2)(Rn) is dense in H1(Rn) (see Lemma 2.58) we easily concludeby approximation that for each f ∈ H1(Rn) there exists a distributionalsolution u ∈ W 1,1

loc (Rn) of −divA∇u = −div f such that ∇u ∈ H1(Rn) andthe bound in (2.604) holds.

To prove uniqueness it suffices to consider the case f = 0. Assumetherefore that u ∈W 1,1

loc with ∇u ∈ H1(Rn) and∫Rn

A∇u · ∇φdx = 0 ∀φ ∈ C∞c (Rn; RN ). (2.605)

We have in particular ∇u ∈ L1. Thus one easily sees by approximation thatthis identity holds for all φ ∈ C∞(Rn; RN ) with ∇φ ∈ L∞. Thus we maytake φ = v where v is the solution of the dual problem constructed above.This yields ∫

Rn

g · ∇u dx = 0 ∀g ∈ C∞c (Rn).

Since ∇u ∈ L1(Rn) it follows that ∇u = 0.(ii): If g ∈ (H1∩L2) then there exists a solution in u ∈W 2,2

loc with ∇2u ∈L2 and the derivative vk := ∂ku are distributional solutions of −divA∇vk =−∂kg. Thus for g ∈ (H1∩L2) the assertion follows from (i). Now the generalresult follows by approximation.

[9.12. 2013, Lecture 16][12.12. 2013, Lecture 17]

Another breakthrough came in 1990 when Coifman, Lions, Meyer andSemmes showed that certain quantities which arise naturally in PDEs arenot only in L1 but in H1.

Theorem 2.61 (Special quantities are in H1). Assume that u ∈ W 1,2(Rn)and v ∈ L2(Rn; Rn) with

div v = 0 (2.606)

in the sense of distributions. Then ∇u · v ∈ H1(Rn), with the bound

∥∇u · v∥H1 ≤ C∥∇u∥L2 ∥v∥L2 . (2.607)

120 [March 17, 2014]

Proof. This proof follows Evans [Ev]. For additional proofs and more generalstatements see the original paper [CLMS].Fix x ∈ Rn. Set ψr(y) = r−nφ

(x−yr

). We have∫

Rn

∇u · v ψr dy = −∫B(x,r)

(u− (u)x,r) v · ∇ψr dy

because v is divergence free. Thus∣∣∣∣∫Rn

∇u · v ψr dy∣∣∣∣ ≤ C

rn+1

∫B(x,r)

|u− (ux,r)| |v| dy.

Choose any p ∈ (2, 2∗) where 2∗ = 2nn−2 and let q = p

p−1 ∈ (1, 2) be the dualexponent. Then∣∣∣∣ 1

rn

∫Rn

∇u · v ψr dy∣∣∣∣ ≤ C

rn+1

(∫B(x,r)

|u− (ux,r)|p dy

) 1p(∫

B(x,r)|v|q dy

) 1q

≤Cr

(1

Ln(B(x, r))

∫B(x,r)

|u− (ux,r)|p dy

) 1p(

1Ln(B(x, r))

∫B(x,r)

|v|q dy

) 1q

≤ C

(1

L(B(x, r))

∫B(x,r)

|∇u|s dy

) 1s(

1Ln(B(x, r))

∫B(x,r)

|v|q dy

) 1q

where s∗ = p, i.e., 1s = 1

p + 1n > 2. Thus s ∈ (1, 2). Thus∣∣∣∣∫

Rn

∇u · v ψr dy∣∣∣∣ ≤CM(|∇u|s)

1s (x) M(|v|q)

1q (x).

This implies that the regularized maximal function satisfies

(∇u · v)∗(x) ≤ C[M(|∇u|s)2s (x) +M(|v|q)

2q (x)]

Now |∇u|s ∈ L2s and |v|q ∈ L

2q . Thus by the Lp estimate by the Hardy-

Littlewood maximal function

∥M(|∇u|s)2s ∥L

2s≤ C∥|∇u|s∥

L2s

and therefore ∫Rn

(M(|∇u|s)2s dx ≤ C

∫Rn

|∇u|2 dx.

Similarly ∫Rn

(M(|v|q)2q dx ≤ C

∫Rn

|v|2 dx.

It follows that ∇u · v ∈ H1(Rn) and

∥∇u · v∥H1 ≤ C(∥∇u∥2

L2 + ∥v∥2L2

)121 [March 17, 2014]

Thus the proof is finished if ∥∇u∥L2 = ∥v∥L2 = 1. To obtain the full resultapply this assertion to u = u/∥∇u∥L2 and v = v/∥v∥L2 . Note that (2.607)holds trivially if ∇u = 0 or v = 0.

Historical comment: Both the invention of BMO and a precursor toTheorem 2.61 have their origins in the theory of elastiticity. The spaceBMO appears in the paper [Jo] by Fritz John. In linear elasticity onedeals with displacements u : Ω ⊂ Rn → Rn and due to linearized frameindifference the energy only depends on the symmetric part of the gra-dient sym∇u = [∇u + (∇u)T ]/2. Korn’s second inequality states thatfor a Lipschitz domain Ω one has ∥∇u − (∇u)Ω∥L2 ≤ C(Ω)∥ sym∇u∥L2 .If sym∇u ∈ L∞ this (and the natural scaling by dilation) implies that∇u ∈ BMO .Motivated by question in nonlinear elasticy in [Mu] it was shown that ifΩ ⊂ Rn, if u ∈W 1,n(Ω; Rn) and if det∇u ≥ 0 then det∇u ln(2 + det∇u) ∈L1loc(Ω) which is by logarithm better than the obvious estimate coming from

|detF | ≤ C|F |n. To prove this one shows that maximal function of det∇uis in L1

loc and uses (2.583). In [CLMS] the deep connections with harmonicanalysis were revealed. In particular the result holds without any sign re-striction if one uses the regularized maximal functions and the authors deriveHardy space estimates for a very general class of ’compensated compactnessquantities’.

2.3.4 Meyers’ estimate for systems with L∞ coefficients

The DeGiorgi-Nash-Moser theorem shows that for scalar elliptic equationswith L∞ coefficients the solution u is much better than L2, namely C0,α, butgives no information about better properties of ∇u. Meyers’ Theorem showsthat ∇u ∈ Lp where p is slightly large than 2. This results even holds forsystems with super-strongly elliptic L∞ coefficients. The text below followsclosely the original paper [Me63]. The discussion of Riesz-Thorin theoremis taken from [StW].

Theorem 2.62 (Meyers’ estimate). Suppose that Ω ⊂ Rn is a boundedopen set with C1 boundary. Suppose that the Aαβij ∈ L∞(Ω) and satisfy thesuper-strong ellipticity condition, i.e., there exists a ν > 0 such that

(AF,F ) ≥ ν|F |2. (2.608)

Then there exists a p0 > 2 such that for p ∈ (p′0, p0) and f ∈ Lp the problem

−divA∇u = −div f, u ∈W 1,p0 (Ω) (2.609)

has a unique solution and

∥∇u∥Lp ≤ C∥f∥Lp . (2.610)

Here p0 and C depend only on n, ∥A∥L∞/ν and Ω.

122 [March 17, 2014]

Remark. (i) One easily obtains the following local version. If −divA∇u =−div f in Ω, if u ∈ W 1,2(Ω) and f ∈ Lp(Ω) with 2 < p < min(p∗, 2∗) thenu ∈W 1,p

loc (Ω). To see this note that by the Sobolev embedding theorem andby Corollary 2.39 the function v = ηu solves −divA∇v = −div f − div hwhere h ∈ L2∗(Ω).(ii) The result is optimal, even for n = 2, scalar equations and f = 0. Indeedfor each α > 0 the function u = x1

|x|1−α is the solution of an elliptic equation(see Homework 1) and u /∈ Lp(B(0, 1)) if p ≥ 1/(1 − α).(iii) This result allows one to extend the regularity results for minimizers ofthe integral functionals (2.478) to systems if n = 2. Indeed if the integrandf is uniformly convex and has bounded second derivatives we get by thedifference quotient method and Meyers estimate ∇2u ∈ Lp for some p > 2.Now for n = 2 the Sobolev embedding theorem implies that ∇u ∈ C0,α forsome α > 0 and then the previous bootstrap argument in Schauder spacesworks.(iv) The paper by Meyers’ also covers the case of complex coefficients (thisrequires only obvious changes in the argument in Step 3 of the proof).

Proof. This is done by a clever perturbation argument. We only show theestimate for 2 < p < p′0. The other case can be handled by duality andapproximation (see [Me63] for the details).

Step 1. Linear algebra.Recall that ∥A(x)∥ := sup∥AF∥ : |F | ≤ 1 denotes the operator norm. Bydividing the equation by ∥A∥L∞ we may assume that

∥A(x)∥ ≤ 1 a.e. (2.611)

To focus on the main idea we assume for simplicity in addition that

A(x) is symmetric,

i.e., (A(x)F,G) = (F,A(x)G) for all F,G ∈ RN×n and a.e. x. Now the PDEcan be written in the equivalent form

−∆u = −div (Id −A)∇u− div f (2.612)

Using ellipticity and the bound ∥A(x)∥ ≤ 1 we get

((Id −A(x))F, F ) ≤ (1 − ν)|F |2 and ((Id −A(x))F, F ) ≥ 0.

Thus the eigenvalues of the symmetric operator Id −A(x) lie in the interval[0, (1 − ν)]. Therefore we obtain the crucial estimate

∥Id −A(x)∥ ≤ 1 − ν for a.e. x. (2.613)

123 [March 17, 2014]

Step 2. Fixed point argument in W 1,p0 (Ω).

On W 1,p0 (Ω) we can consider the norm

∥u∥ep := ∥∇u∥Lp . (2.614)

It follows from the Poincare inequality for W 1,p0 that this is indeed a norm

and that this norm is equivalent to the usual W 1,p norm. Theorem 2.41implies that for every p ∈ (1,∞) and every h ∈ Lp(Ω) the problem

−div∇v = −div h, v ∈W 1,p0 (Ω)

has a unique solution. We write define the operator R by Rh := v. Then,again by Theorem 2.41, we have

∥∇Rh∥Lp ≤Mp∥h∥Lp . (2.615)

Now (2.612) is equivalent to

u = Su+Rf where Su = R[(Id −A)∇u].

Thus∥Su∥ep ≤Mp∥(A− Id )∇u∥Lp ≤Mp(1 − ν)∥u∥ep. (2.616)

We are done if we can show that S is a contraction. For this it sufffices that

∥Mp(1 − ν)∥ < 1. (2.617)

We know that M2 = 1 (test the equation for v with v). Let p ∈ (2,∞), letθ ∈ (0, 1) and let

1p

= (1 − θ)12

+ θ1p.

By the Riesz-Thorin Theorem (stated below)

Mp ≤M1−θ2 M θ

p = M θp . (2.618)

If Mp ≤ 1/(1 − ν) then (2.617) holds for all p ∈ (2, p). If Mp > 1/(1 − ν)define θ∗ by the relation M θ∗

p = 1/(1 − ν) and set 1p∗

= (1 − θ∗)12 + θ − 81

p .Then (2.617) holds for all p ∈ (2, p∗).

Step 3. Nonsymmetric A. This part was not discussed in classLet 0 < ν ≤ 1 and assume that

min|F |=1

(AF,F ) ≥ ν and ∥A∥ ≤ 1.

We show that there exist θ > 0, λ > 0 and M > 0, a symmetric operator Sand an operator R such that

A = S +R, inf|F |=1

(SF, F ) ≥ λ, ∥S∥ ≤M, ∥R∥ ≤ (1 − θ)λ. (2.619)

124 [March 17, 2014]

Here θ, λ and M depend only on ν. For fixed ν the operators S and R de-pends linearly on A. To prove this define the symmetric and skew symmetricpart of A as

symA :=A+AT

2, skwA :=

A−AT

2.

Then A = symA+ skwA. We claim that

(symAF, skwAF ) = 0 and thus ∥AF∥2 = ∥ symAF∥2 + ∥ skwAF∥2.

To prove the first identity note that

(symAF, skwAF ) =(AF,AF ) + (ATF,AF ) − (AF,ATF ) − (ATF,ATF )

=([ATA−AAT ]F, F ) + ([ATAT −AA]F, F )

Now the operators ATA − AAT and ATAT − AA are skew symmetric andhence the right hand side vanishes. In particular we have

∥ symA∥ ≤ ∥A∥, ∥ skwA∥ ≤ ∥A∥ (2.620)

For c ≥ 0 write

A = (symA+ cId ) + (skwA− cId )

By (2.620)∥ symA+ cId ∥ ≤ ∥A∥ + c ≤ 1 + c

Since (skwAF,F ) = 0 we have

∥(skwA− cId )F∥2 =(skwAF, skwAF ) − 2c (skwAF,F )︸ ︷︷ ︸=0

+c2|F |2

≤∥A∥2 + c2 ≤ 1 + c2

where we used (2.620) in the last inequality. Now define θ by

1 − θ = infc≥0

g(c), where g(c) :=√

1 + c2

ν + c(2.621)

For c ≥ 1/ν we have (c+ ν)2 > c2 + 1. Hence infc≥0 g(c) < 1 and one easilysees that the infimum is a minimum since the limits of g(c) at 0 and ∞ are≥ 1. Let c be such that g(c) = min g and set

S = A+ c Id , R = skwA− c Id . (2.622)

Then

min|F |=1

(SF, F ) ≥ (ν + c), ∥S∥ ≤ 1 + c, ∥R∥ ≤√

1 + c2 = (1 − θ)(ν + c)

125 [March 17, 2014]

and the assertion follows with λ = ν + c, M = 1 + c.

To finish the proof we rewrite the PDE as

−M∆u = −div (M Id − S −R)∇u− div f

or

−∆u = −div(

Id − 1MS − 1

MR

)∇u− 1

Mdiv f.

Let λ′ = λ/M . Then the estimate (SF, F ) ≥ λ|F |2 implies that∥∥∥∥Id − 1MS

∥∥∥∥+∥∥∥∥ 1MR

∥∥∥∥ ≤ 1 − λ′ + (1 − θ)λ′ ≤ 1 − θλ′.

Now we can conclude as in Step 1.

Theorem 2.63 (Riesz-Thorin interpolation theorem). Suppose that T is alinear map which maps measurable functions on E to measurable functionson E′. Assume that T is of type (pi, qi) for i = 0, 1 and

∥Tf∥Lq0 ≤ k0∥f∥Lp0

∥Tf∥Lq1 ≤ k1∥f∥Lp1

Then T is of type (pt, qt) and

∥Tf∥Lqt ≤ k1−t0 k1t∥f∥Lpt

provided

1pt

= (1 − t)1p0

+ t1p1

and1qt

= (1 − t)1q0

+ t1q1

with the usual convention 1/∞ = 0.

Remark. This theorem is also known as M. Riesz’s convexity theorembecause it shows that:(i) The set of points (1

p ,1q ) for which T is of type (p, q) is a convex subset of

[0, 1]2

(ii) If φ(t) denotes the logarithm of the operator norm of T as an operatorfrom Lpt to Lqt then φ is a convex function on [0, 1].

Lemma 2.64 (Three line lemma). Suppose that F : [0, 1] × R → C isbounded and continuous and that F is holomorphic in (0, 1) × R. If

|F (iy)| ≤ m0 and |F (1 + iy)| ≤ m1 ∀y ∈ R

then|F (x+ iy)| ≤ m1−x

0 mx1 ∀x ∈ [0, 1], y ∈ R.

Thus if function k(x) := supy∈R |F (x + iy)| then x 7→ ln k(x) is a convexfunction on [0, 1].

126 [March 17, 2014]

Proof. The proof was only discussed very briefly in class.It suffices to prove the result for m0 > 0 and m1 > 0. Indeed if theresult holds in this case the assertion that mθ = 0 if m0 = 0 followsby letting m0 ↓ 0. Similarly one argues if m1 = 0. Replacing F (z)by F (z) = F (z)/(m1−z

0 mz1) we see that it suffices to consider the case

m0 = m1 = 1.Note that the real and imaginary part of F are harmonic functions.

Thus |F |2 = (ReF )2 + (ImF )2 is subharmonic. If we assume in additionthat lim|y|→∞ F (x+iy) = 0 uniformly for x ∈ [0, 1] then the assertion followsfrom the maximum principle for |F |2.

For general F we consider the functions

Fk(z) := F (z)e(z2−1)/k.

with z = x + iy. Since z2 = x2 − y2 + 2ixy it follows that |Fk(z)| ≤e(1−y

2)/k sup |F | and hence we can apply the previous reasoning to Fk andget |Fk(x + iy)| ≤ 1 for all (x, y) ∈ [0, 1] × R. Let k → ∞ we obtain theassertion.

Proof of Theorem 2.63. This proof was only sketched in class.We have to show∫

E′Tf g dx ≤ k1−t

0 kt1 if ∥f∥Lpt ≤ 1 and ∥g∥Lq′t

≤ 1 (2.623)

where 1q′t

= 1 − 1qt

. To do so we embed f and g in an holomorphic familyand use the three line theorem lemma. We set

fz = |f |α(z) f

|f |, gz = |g|β(z) g

|g|

where

α(z) = pt[(1 − z)1p0

+ z1p1

], β(z) = q′t[(1 − z)1q′0

+ z1q′1

].

Then α(t) = 1 = β(t) and thus f = ft and g = gt. Moreover α(0) = pt/p0

and α(1) = pt/p1. For a > 0 we have |aiy| = | exp(iy ln a)| = 1 and we thusget

|fiy|p0 = |f |pt , |f1+iy|p1 = |f |pt

and similarly|giy|q

′0 = |g|q′t , |g1+iy|q

′1 = |g|q′t .

SetF (z) =

∫E′Tfz gz dx

127 [March 17, 2014]

Then the conditions ∥f∥Lpt ≤ 1 and ∥g∥Lq′t

≤ 1 and the assumptions on Timply that

|F (iy)| ≤ k0 and |F (1 + iy)| ≤ k1 ∀y ∈ R.

Since ft = f and gt = g the expression on the left hand side of (2.623) isjust F (t) and the desired estimate follows from the three line lemma.

To be sure that F is really bounded on the strip [0, 1] × R on can firstconsider functions f and g which a finite linear combination of characteristicfunctions of sets with finite measure (and hence in all Lp spaces), prove(2.623) for these function and then use that these functions are dense in Lpt

and Lq′t .

[12.12. 2013, Lecture 17][16.12. 2013, Lecture 18]

2.4 Regularity theory for nonlinear pde

2.4.1 Harmonic maps and Hardy space estimates

Let M be a submanifold of RN . A harmonic map is a critical point of theDirichlet integral

I(u) =12

∫Ω|∇u|2 dx (2.624)

among all maps u : Ω ⊂ Rn → RN with u(x) ∈M for a.e. x.To keep the notation as simple as possible we focus on the case that

M is the N − 1 dimensional unit sphere SN−1. To motivate the precisedefinition of weakly harmonic maps we derive the usual necessary first ordercondition (or Euler-Lagrange equations) for a map u which minimizes theDirichlet integral (subject to suitable boundary conditions). Thus let u bea minimizer of the Dirichlet integral among maps in W 1,2(Ω; RN ) whichsatisfy the constraint

|∇u|2 = 1 a.e. in Ω. (2.625)

To derive the Euler-Lagrange equations we usually consider variationsu+ tφ. Here such variations are in general not admissible since they do notsatisfy the constraint |u+ tφ| = 1 a.e. Thus we start from the relation

I(u) ≤ I(ut) where ut =u+ tφ

|u+ tφ|with φ ∈ (W 1,2

0 ∩ L∞)(Ω). (2.626)

The condition φ ∈ L∞ guarantees that |u+ tφ| ≥ 12 for |t| ≤ 1

2∥φ∥L∞ so thatut is well defined for small t and ut ∈W 1,2(Ω; RN ). We have

∂iut =∂iu+ t∂iφ

|u+ tφ|− u+ tφ

|u+ tφ|3(u+ tφ) · (∂iu+ t∂iφ). (2.627)

128 [March 17, 2014]

Using that |u| = 1 and

u · ∂iu =12∂i|u|2 = 0 (2.628)

we deduce that

d

dt |t=0∂iut = ∂iφ− ∂iu (u · φ) − u (φ · ∂iu) − u (u · ∂iφ).

Thus using again (2.628) we see that t 7→ I(ut) is differentiable and

0 =d

dt |t=0I(ut) =

∫Ω

n∑i=1

∂iu · ∂iφ− |∂iu|2u · φdx. (2.629)

With the Euclidean scalar product on matrices we have ∇v ·∇φ =∑n

i=1 ∂iv ·∂iφ and we say that u is a weakly harmonic map to SN−1 if u ∈W 1,2(Ω; RN )with |u| = 1 a.e and∫

Ω∇u · ∇φ− |∇u|2u · φdx = 0 ∀φ ∈ (W 1,2

0 ∩ L∞)(Ω; RN ). (2.630)

In particular a weakly harmonic map is a distributional solution of the sys-tem of following system of PDEs

−∆u = u|∇u|2 (2.631)

Even though a lot of deep results have been obtained for geometric equa-tion of this type in the 1960’s and 70’s the question whether weakly harmonicmaps are smooth has been open until 1990. We will see the answer shortly.Let us first see why this question is delicate. A key point is that the equation(2.631) is critical in at least two ways.

(i) The equation is invariant under the natural rescaling. If u solves(2.631) and ur(z) = u(rz) then ur is a solution of the same equation.Thus the nonlinearity does not get weaker under blow-up.

(ii) The usual elliptic estimates do not give an improvement of the initialregularity u ∈ W 1,2 ∩ L∞. Indeed take n = 2. The initial regularityimplies that the right hand side is in L1. Hence the best we couldhope for is that u is in W 2,1. By the Sobolev embedding theorem thiswould give u ∈W 1,2 which is no improvement.In fact the situation is slightly worse because if v is a distributionalsolution −∆v = f with f ∈ L1 then this does not imply that ∇v ∈ L2

loc

(see Homework 8). One only gets that ∇v is (locally) in weak-L2, i.e.,|∇v| > t ≤ Ct−2. For n ≥ 3 the situation is even worse.

If there is a regularity result for weakly harmonic maps it must use somespecial structure of the equation. It is not enough to use that the right

129 [March 17, 2014]

hand side is a quadratic term in ∇u. Indeed, the seemingly simpler scalarequation

−∆v = |∇v|2 in B(0, 1/2) ⊂ R2

has the distributional solution

v = ln ln1|x|

and v ∈W 1,2(B(0, 1/2)) (see Homework 7).

A general principle for critical elliptic equations is that one gets regu-larity if a suitable scale invariant quantity is small. In this spirit we firstshow the classical result that weakly harmonic maps which are continuousare already smooth.

Theorem 2.65. Let u be a weakly harmonic map from Ω to SN−1. Ifu ∈ C0(Ω; RN ) then u ∈ C∞(Ω; RN ).

Proof.

Step 1. ∇u ∈ L2,λloc (Ω; RN ) for 0 < λ < n.

Let B(x, r) ⊂ Ω. As before decompose u as u = v + w where ∆w = ∆uand w has zero boundary conditions and ∆v = 0. More precisely we definew ∈W 1,2

0 (B(x, r)) as the unique weak solution of∫B(x,r)

∇w · ∇φdx =∫B(x,r)

∇u · ∇φdx ∀φ ∈W 1,20 (B(x, r); RN ) (2.632)

(existence and uniqueness follow from the Lax-Milgram theorem) and we set

v = u− w.

Then ∫B(x,r)

∇v · ∇φdx = 0 ∀φ ∈W 1,20 (B(x, r)), (2.633)

i.e., v is weakly harmonic.If we can show that in addition w is in L∞(B(x, r); RN ) then by the

definition of weakly harmonic maps we can use φ = w in (2.632) and we get∫B(x,r)

|∇w|2 dy =∫B(x,r)

u|∇u|2 · w dy ≤ supB(x,r)

|w|∫B(x,r)

|∇u|2 dy (2.634)

For this to be useful we need that supB(x,r) |w| is small. To show thiswe use the weak maximum principle (see Lemma 2.66) for each componentvi. Let Mi = maxB(x,r) u

i. Then ui −Mi ≤ 0 and thus vi −Mi = ui −Mi − wi belongs to the space W 1,2

− (B(x, r)) of functions which are ≤ 0 on

130 [March 17, 2014]

the boundary in a weak sense (see (2.641) and (2.642)). Thus vi ≤ Mi =maxB(x,r) u

i a.e. Similarly vi ≥ minB(x,r) ui. Since w = u− v it follows that

|wi| ≤ maxB(x,r)

ui − minB(x,r)

ui a.e. (2.635)

Since u is continuous it follows that for each ε > 0 there exists an r0 > 0such that ∫

B(x,r)|∇w|2 dy ≤ ε

∫B(x,r)

|∇u|2 dy (2.636)

Moreover we have by (2.78)∫B(x,ρ)

|∇v|2 dy ≤ C(ρr

)n ∫B(x,r)

|∇v|2 dy.

for all ρ ≤ r (note that for ρ ≥ r/2 the estimate holds trivially with C = 2n).Now we can combine the w and the v estimate in the usual way and use

the iteration lemma to conclude. Here are the details: assume that ε ≤ 1and set

ϕ(r) :=∫B(x,r)

|∇u|2 dy.

Then ∫B(x,r)

|∇v|2 dy ≤ 2∫B(x,r)

|∇u|2 dy + 2∫B(x,r)

|∇w|2 dy ≤ 4ϕ(r)

and thus

ϕ(ρ) ≤2∫B(x,ρ)

|∇v|2 dy + 2∫B(x,ρ)

|∇w|2 dy

≤C(ρr

)n ∫B(x,r)

|∇v|2 dy + 2εϕ(r)

≤4C[(ρr

)n+ ε]ϕ(r)

(here we assume without loss of generality that C ≥ 1). Now by Lemma2.15 for each λ < n there exists a constant C ′ which only depends on C, λ,n such that

ϕ(ρ) ≤ C ′(ρ

r0

)λϕ(r0),

for all ρ ≤ r0. Thus ∇u ∈ L2,λloc (Ω).

131 [March 17, 2014]

Step 2. ∇u ∈ L2,µloc (Ω; RN ) for all µ < n+ 1.

By Step 1 and the Poincare inequality we have u ∈ Lλ+2loc (Ω; RN ) for all

λ < n. Hence u ∈ C0,αloc (Ω; RN ) for all α < 1. Let n < µ < n + 1 and let

α ∈ (0, 1) be such that λ := µ− α < n.Define v and w as in Step 1 and set

ϕ(r) :=∫B(x,r)

|∇u− (∇u)x,r|2 dy.

The Holder continuity of u and (2.635) imply that

sup |wi| ≤ Crα. (2.637)

Using that u ∈ L2,λ(B(x, r); RN ) and (2.634) we get∫B(x,r)

|∇w|2 dy ≤ Crα rλ = Crµ.

By (2.79) we have∫B(x,ρ)

|∇v − (∇v)x,ρ|2 dy ≤ C(ρr

)n+2∫B(x,r)

|∇v − (∇v)x,r|2 dy.

Combining the v and the w estimates as above and using that∫B(x,ρ)

|∇w − (∇w)x,ρ|2 dy ≤∫B(x,ρ)

|∇w|2 dy

we get

ϕ(ρ) ≤ A(ρr

)n+2ϕ(r) +Brµ.

Since µ < n+ 2 it follows from Lemma 2.15 that there exists a constant Cwhich only depends on A, B, µ and n such that

ϕ(ρ) ≤ C(ρr

)µϕ(r)

for all ρ ≤ r. Thus ∇u ∈ L2,µloc (Ω).

Step 3. u ∈ C∞(Ω; RN ).By Step 2 and Theorem 2.12 we have ∇u ∈ C0,β

loc (Ω) for all β < 12 . We show

by induction that ∇u ∈ Ck,βloc (Ω; RN ) for all k ∈ N. Indeed the Schaudertheory gives

∇u ∈ Ck,βloc =⇒ u|∇u|2 ∈ Ck,βloc =⇒ u ∈ Ck+2,βloc =⇒ ∇u ∈ Ck+1,β

loc .

132 [March 17, 2014]

In the argument above we have used the following form of the maximumprinciple for weak subsolutions. Recall that for a ∈ R we use the notationa+ = max(a, 0).

Lemma 2.66 (Weak maximum principle). Assume that Ω ⊂ Rn is boundedand open that the coefficients aij ∈ L∞(Ω) are elliptic. Define

W 1,2− (Ω) := v ∈W 1,2(Ω) : v+ ∈W 1,2

0 (Ω). (2.638)

If v ∈W 1,2− (Ω) is weakly subharmonic, i.e., if∫Ω

∑i,j

aij(x)∂jv ∂iφdx ≤ 0 ∀φ ∈W 1,20 (Ω) with φ ≥ 0 a.e.

thenv ≤ 0 a.e.

Proof. See Homework 7. Hint: Take φ = v+ and use that ∇v+ = χv>0 ∇va.e. (see Functional Analysis and PDE or [GT]). Then

ν

∫Ω|∇v+|2 dx ≤

∫Ωaij∂jv

+ ∂iv+ =

∫Ωaij∂jv ∂iv

+ ≤ 0.

Hence ∇v+ = 0. Since v+ ∈ W 1,20 (Ω) by assumption it follows that v+ = 0

a.e.

Remark. The space W 1,2− (Ω) should be viewed as the space of function

which satisfy the condition v ≤ 0 on ∂Ω in a weak sense. One has thefollowing characterization (see Homework 7).

v ∈W 1,2− (Ω) ⇐⇒

∃ Kk ⊂ Ω compact, vk → v in W 1,2(Ω), vk ≤ 0 in Ω \Kk (2.639)

and from this one easily derives that (see Homework 7)

v ∈W 1,2(Ω) ∩ C0(Ω), v ≤ 0 on ∂Ω =⇒ v ∈W 1,2− (Ω), (2.640)

and

W 1,20 (Ω) ⊂ W 1,2

− (Ω), (2.641)

v, w ∈W 1,2− (Ω)− =⇒ v + w ∈W 1,2

− (Ω). (2.642)

We now come to the heart of the matter. The key observation is thatsince 2u · ∂ku = ∂k|u|2 = 0 the nonlinearity can be written in the form

ui|∇u|2 =∑j

(ui∇uj − uj∇ui) · ∇uj (2.643)

133 [March 17, 2014]

and that

div (ui∇uj − uj∇ui) = 0 in the sense of distributions.

Thus the nonlinearity is not only in L1 but in (a suitable local version of)the Hardy space H1. From this we will easily conclude that u is continuousand hence smooth.

Lemma 2.67. Assume that u is a weakly harmonic map to SN−1. Thenfor each i, j ∈ 1, . . . , N the vectorfield ui∇uj−uj∇ui is divergence free inthe sense of distributions, i.e.,∫

Ω(ui∇uj − uj∇ui) · ∇η dx = 0 ∀ η ∈ C∞

c (Ω).

Proof. Let ei be the i-th unit vector and use the test function φ = ujη eiin the definition of a weakly harmonic map. Since ∇(ujη) = ∇uj η + uj∇ηthis yields∫

Ω∇ui · ∇uj η dx+

∫Ωuj∇ui · ∇η dx =

∫Ωui|∇u|2 ujη dx.

Exchanging i and j we also get∫Ω∇uj · ∇ui η dx+

∫Ωui∇uj · ∇η dx =

∫Ωuj |∇u|2 uiη dx.

The assertion follows by substracting the two equations.

[16.12. 2013, Lecture 18][19.12. 2013, Lecture 19]

Lemma 2.68 (Localization of H1 − BMO estimates). Let Br = B(x, r) ⊂Rn.

(i) Assume thatb ∈ L2(B2r; Rn), div b = 0

in the sense of distributions in B2r. Then there exists an extension

b ∈ L2(Rn; Rn), b = b in Br, b = 0 in Rn \B2r

∥b∥L2 ≤ C∥b∥L2 , div b = 0

in the sense of distributions in Rn. Here C depends only on n.

(ii) Suppose that a ∈ W 1,2(B2r), b ∈ L2(B2r; Rn) and div b = 0 in thesense of distributions. Then there exists f ∈ H1(Rn) such that

f = b · ∇a in Br, ∥f∥H1 ≤ C∥∇a∥L2(B2r) ∥b∥L2(B2r),

where C only depends on Rn.

134 [March 17, 2014]

(iii) Suppose that a ∈ W 1,20 (Br), b ∈ L2(B2r; Rn), g ∈ L2,n(B2r) with

div b = 0 in the sense of distributions. Then∫Br

b · ∇a g dx ≤ C∥∇a∥L2(Br) ∥b∥L2(B2r) [g]L2,n(B2r) (2.644)

Remark. In part (i) it is important that the ball is simply connected.The function b(x) = x

|x|2 is smooth and divergence free in the annulus A =B(0, 2) \ B(0, 1) ⊂ R2 and it is not difficult to show that there exists nodivergence free extension b ∈ L2(R2; R2) (Exercise. Hint: first assume thatb is smooth and show that

∫∂B(0,R) b·ν dH

1 = 2π and use Jensen’s inequality.Then approximate).

Proof. By scaling and we easily see that it suffices to consider the case r = 1.(i): Let φ ∈ C∞

c (B2) with φ = 1 on B1 and |∇φ| ≤ 2. Assume first inaddition that b is C1 on suppφ. Then div (φb) = b · ∇φ in B2 \ B1. Let Ube the weak solution of the Neumann problem

∆U = b · ∇φ in B2 \B1,∂U

∂ν= 0 on ∂(B2 \B1),

i.e., ∫B2\B1

∇U · ∇ψ dx = −∫B2\B1

b · φψ dx ∀ψ ∈W 1,2(B2 \B1).

Note that since div b = 0 we have∫B2\B1

b · ∇φdx =∫∂(B2\B1)

b · ν φ dH1 = −∫∂B1

b · ν dH1 =∫B1

div b = 0.

Here we used that on ∂B1 the outer normal of B2 \ B1 is the inner normalof B1. Thus the Neumann problem has a solution, the solution is unique upto the addition of constants, and using the test function ψ = U we get∫B2\B1

|∇U |2 dx = −∫B2\B1

b · φ(U − (U)B2\B1

)dx ≤ 2CP ∥b∥L2∥∇U∥L2 ,

where CP is the Poincare constant of B2 \ B1. Thus ∥∇U∥L2 ≤ C∥b∥L2 .Finally extend h = ∇U by zero from B2 \B1 to Rn, extend φb by zero fromB2 to Rn and set

b = φb− h.

Then ∥b∥L2 ≤ C∥b∥L2 and b is divergence free in the sense of distributions.Indeed for every η ∈ C∞

c (Rn) we have∫Rn

φb · ∇ηdx = −∫B2

div (φb) dx = −∫B2\B1

b · φdx,∫Rn

h∇η dx =∫B2\B1

∇U · ∇η = −∫B2\B1

b · φdx

135 [March 17, 2014]

and the assertion follows by substracting the two equations.If b is only in L2 we can consider approximations bk = ρk ∗ b where

ρk(x) = knρ(kx) are the usual mollifying kernels. For large enough k themaps bk are smooth on suppφ. Define bk as before. Then ∥bk − bl∥L2 ≤C∥bk − bl∥L2 . Thus bk is a Cauchy sequence in L2 and the limit b has thedesired properties.

(ii): Let b and φ as in (i), let a = φa and extend a by zero from B2 toRn. Then

b · ∇a = b · ∇a, in B1

We may asssume that∫B2a dx = 0 since the quantity we are interested

in and the right hand side of the estimate depend only on ∇a. Thus by thePoincare inequality

∥∇a∥L2(Rn) ≤ C∥a∥W 1,2(B2) ≤ C∥∇a∥L2(B2)

Now by Theorem 2.61 the function f = ∇a · b belongs to H1 and

∥∇a · b∥H1 ≤ C∥∇a∥L2 ∥b∥L2 ≤ C∥∇a∥L2(B2) ∥b∥L2(B2).

(iii): Let b be as in (i), let a denote the extension of a by zero to fromB1 to Rn and let g = φg where φ is as in (i) and extend g by zero from B2

to Rn. Then ∫B1

b · ∇a g dx =∫

Rn

b · ∇a g dx

We may assume that g has average zero on B2 since∫B1

b · ∇a dx =∫B1

b · ∇a dx = 0.

Then it is easy to show that there exists a constant C (which only dependson n) such that

[φg]L2,n(Rn) ≤ C[g]L2,n(B2)

Now the assertion follows from Theorem 2.61 and the H1 − BMO dualityapplied to a, b and g and the equivalence of the BMO seminorm and theL2,n seminorm on Rn.

Theorem 2.69. (Helein) Let u : Ω ⊂ R2 → RN be a weakly harmonic mapto SN−1. Then u ∈ C∞(Ω).

Proof. In view of Theorem 2.65 it suffices to show that u is continuous. Forthis it suffices to show the following: if B(x, 2r) ⊂ Ω then u is continuousin B(x, r).

By (2.643) the component ui is a weak solution of

−∆ui =∑j

(ui∇uj − uj∇ui) · ∇uj

136 [March 17, 2014]

and by Lemma 2.67div (ui∇uj − uj∇ui) = 0

in the sense of distributions in Ω. Assume that B(x, 2r) ⊂ Ω. Then byLemma 2.68 (ii) there exist functions fj ∈ H1(R2) such that

−∆ui =∑j

fij in B(x, r)

By Theorem 2.60 there exists U i ∈W 2,1loc (R2) such that

−∆U i =∑j

fij in Rn

By Lemma 2.70 function in W 2,1loc (R2) are continuous. Finally ui − U i is

weakly harmonic in B(x, r) and hence smooth. Thus ui is continuous inB(x, r).

Lemma 2.70. Suppose that u ∈ W 2,1loc (R2) and ∇2u ∈ L1(R2). Then u is

continuous and there exist constants a ∈ R and b ∈ R2 such that

supx∈Rn

|u(x) − a− b · x| ≤∫

R2

|∂1∂2u| dy (2.645)

Proof. See Homework 8. Hint: First show that the assertion holds for v ∈W 2,1(R2). To see this note that for v ∈ C∞

c (R2)

v(x) =∫ x1

−∞

∫ x2

−∞∂1∂2v(y1, y2) dy2 dy1.

and argue by density. To show that u is continuous at x consider a cut-off function ψ ∈ Cc∞(Rn) which is 1 in a neighbourhood of x. Then ψu ∈W 2,1(Rn) and hence ψu is continuous. Thus u is continuous at x. To provethe full result argue in a similar way as in Lemma 2.37 to approximate u byfunctions uk ∈W 2,1(R2) such that ∇2uk → ∇2u.

We now consider the regularity of weakly harmonic maps Ω ⊂ Rn →SN−1 for n ≥ 3. In this case singularities can arise. Indeed

u(x) :=x

|x|is a weakly harmonic map

from B(0, 1) ⊂ Rn to Sn−1 for all n ≥ 3.

This can be seen by a straightforward calculation (see Homework 9). Amuch more subtle argument shows that this map u is actually a minimizerof the Dirichlet integral among all W 1,2 maps v with v = u = id on ∂B(0, 1)and |v(x)| = 1 a.e. (see [AL]). In general the situation is much worse.

137 [March 17, 2014]

Theorem 2.71 (Riviere). There exists a weakly harmonic map from B(0, 1) ⊂R3 to S2 for which the set of points of discontinuity is dense.

Proof. See [Ri95].

At this point we could give up. Alternatively we can argue that thenotion of a weakly harmonic map is simply too weak to select ’good’ har-monic maps. In fact one important feature of the harmonic map equationis invariance an rescaling of the independent variable: if u is a solution andut(x) = u(x/(1 + t)) is also a solution.

This suggests to consider inner variations which are obtained by varia-tions of the independent variable. Let

ut = u Φt, i.e., ut(x) = u(Φt(x)).

Here Φ0 = Id and Φt is a smooth one-parameter family of diffeomorphismsfrom Ω to itself, i.e.,

(x, t) 7→ Φt(x) is a C∞ map andΦt : Ω → Ω is a diffeomorphism for each t ∈ (−δ, δ).

We setΨt = Φ−1

t and η(x) =d

dt |t=0Ψt(x).

In the following we assume that the first and second derivatives of Φt andΨt in x and t are uniformly bounded.

The additional condition we want to impose on u is that

d

dt |t=0

∫Ω|∇ut|2 dx = 0.

To express this in a simpler form we compute

∇ut(x) = ∇ut(Φt(x))∇Φt(x)

and we use the change of variable y = Φt(x) and thus x = Ψt(y) and thechain rule ∇Ψ(y)∇Φ(x) = Id to get∫

Ω|∇ut(x)|2 dx =

∫Ω∇u(y)(∇Ψt(y))−1 · ∇u(y)(∇Ψt(y))−1 det∇Ψt dy.

Now if t 7→ A(t) is a C1 map into the space of invertible n× n matrices wehave

d

dtA−1(t) = −A−1(t)

d

dtA(t)A−1(t),

d

dtdetA(t) = detA(t) trA−1(t)

d

dtA(t)

Since ∇Ψ0 = Id we get

d

dt |t=0

∫Ω|∇ut(x)|2 dx =

∫Ω−(∇u∇η) ·∇u − ∇u · (∇u∇η)+ |∇u|2Id ·∇η dy

138 [March 17, 2014]

Now we use the following relation16 for n× n matrices F and G

FG · F = F TF ·G.

This yields

d

dt |t=0

∫Ω|∇ut(x)|2 dx =

∫Ω

(−2(∇u)T∇u+ |∇u|2Id

)· ∇η dy

Finally we note that if η ∈ C∞c (Ω; Rn) and if we define Ψt(x) as the

solution of the ODEd

dtΨt(x) = η(Ψt(x)), Ψ0(x) = x

then Ψt is a smooth one parameter family of diffeomorphisms and Ψt(x) = xon Ω\ supp η and thus Ψ maps Ω to itself and is the identity in a neighbour-hood of ∂Ω.

Definition 2.72 (Stationary harmonic map). A weakly harmonic map iscalled a stationary harmonic map if∫

Ω

(2(∇u)T∇u− |∇u|2Id

)· ∇η dx = 0 ∀η ∈ C∞

c (Ω; Rn). (2.646)

Here [(∇u)T∇u

]αβ

=N∑i=1

∂αui ∂βu

i = ∂αu · ∂βu

and Id denotes the identity map on Rn.

Remark. If u is sufficiently regular (e.g. W 2,2 ∩ W 1,∞) then (2.646)follows from the fact that u is weakly harmonic. It suffices to use the testfunction φ = η · ∇u =

∑α η

α∂αu. If u is only in W 1,2 ∩ L∞, however, ingeneral φ is not an admissible test function because the assumptions do notsuffice to ensure that φ ∈W 1,2.

Lemma 2.73 (Monotonicity formula). Assume that B(x,R) ⊂ Ω and thatu ∈W 1,2(Ω; RN ) satisfies (2.646). Then

1ρn−2

∫B(x,ρ)

|∇u|2 dy ≤ 1rn−2

∫B(x,r)

|∇u|2 dy ∀ 0 < ρ < r ≤ R.

Proof. For simplicity of notation assume that x = 0 and write Br = B(0, r).First note that by Fubini’s theorem (and transformation to polar coor-

dinates) the maps r 7→∫B(x,r) |∇u|

2 dx is absolutely continuous (i.e., theprimitive of an L1 function) and

d

dr

∫Br

|∇u|2 dx =∫∂Br

|∇u|2 dx for a.e. r < R

16Proof:P

i,β,α FiβG

βα F i

α =P

i,β,α(FT )βi F

iα Gβ

α

139 [March 17, 2014]

Hence it suffices to show that

d

dr

(r2−n

∫Br

|∇u|2 dx)

=r1−n∫Br

(2 − n)|∇u|2 dx+ r2−n∫∂Br

|∇u|2 dHn−1 ≥ 0 for a.e. r < R.

(2.647)

Now observe that by density (2.646) holds for all η ∈W 1,∞0 (Ω; RN ).17

Let 0 < s < r and apply (2.646) with

η(x) = ψ(|x|)x

with

ψ(t) =

1 if t < s,r−tr−s if s ≤ t ≤ r,0 if t > r.

Then ∇η(x) = Id for |x| < s and ∇η = 0 for |x| > r. Moreover

∇η(x) = ψ(|x|) Id − 1r − s

x⊗ x

|x|if s ≤ |x| ≤ r.

Thus (2.646) and the identities F TF · Id = |F |2 and F TF · z ⊗ z = |Fz|2imply that18

0 =∫Br

ψ(|x|)(2 − n)|∇u|2 − 1r − s

∫Br\Bs

|x|

(2∣∣∣∣∇u x

|x|

∣∣∣∣2 − |∇u|2)dx.

Taking the limit s ↑ r we get, for a.e. r < R,

0 =∫Br

(2 − n)|∇u|2 +∫∂Br

r

(−2∣∣∣∣∇u x|x|

∣∣∣∣2 + |∇u|2)dx

≤∫Br

(2 − n)|∇u|2 + r

∫∂Br

|∇u|2 dx

Multiplying by r1−n we see that (2.647) holds.

Lemma 2.74 (ε-regularity lemma). Let n ≥ 3. There exists ε0 > 0 (whichonly depends on n) with the following property. If u is a stationary harmonicmap from Ω to SN−1, if B(a, 4R) ⊂ Ω and if

1(4R)n−2

∫B(a,4R)

|∇u|2 dy ≤ ε20 (2.648)

17Note that C∞c (Ω; RN ) is not dense in W 1,∞

0 (Ω; RN ), but for each η ∈ W 1,∞0 (Ω; RN )

there exist ηk ∈ C∞c (Ω; RN ) such that ∥ηk∥W1,∞ is uniformly bounded and ∇ηk → ∇η

a.e. By the Lebesgue dominated convergence theorem this is enough to pass to the limitin (2.646).

18Proof of the second identity: FTF ·z⊗z =P

i,α,β FiαF

iβz

αzβ =P

i(Fz)i(Fz)i = |Fz|2

140 [March 17, 2014]

thenu ∈ C∞(B(a,R); RN ).

Proof.

Step 1. Let B′ = B(a, 3R). Then u ∈ L2,n(B′,RN ) and [u]L2,n(B′) ≤ Cε0where C depends only on n.Let x ∈ B′, 0 < r ≤ R. Then B(x, r) ⊂ B(x,R) ⊂ B(a, 4R). By (2.92), thePoincare inequality, the monotonicity formula and the inclusion B(x,R) ⊂B(a, 4R) we get

1rn

∫B(x,r)∩B′

|u− uB(x,r)∩B′ |2 dy

≤ 1rn

∫B(x,r)

|u− uB(x,r)|2 dy

≤ CPrn−2

∫B(x,r

|∇u|2 dy

≤ CPRn−2

∫B(x,R

|∇u|2 dy ≤ 4n−2CP ε20

where CP denotes the constant in the Poincare estimate for the unit ball.For the r ≥ R use (2.92) and the Poincare inequality to obtain

1rn

∫B(x,r)∩B′

|u− uB(x,r)∩B′ |2 dy

1rn

∫B(a,4R)

|u− uB(a,4R)|2 dy

1Rn

(4R)2CP∫B(a,4R)

|∇u|2 dy ≤ 4nCP ε20.

Step 2. ∇u ∈ L2,λ(B(a,R); RN×n) for all λ < n.This implies that u ∈ L2,λ+2(B(a,R); RN ) for all λ < n. Hence u is Holdercontinuous in B(a,R) and thus C∞ by Theorem 2.65.

The proof is very similar to Step 1 in the proof of Theorem 2.65. Themain point is that we use the local H1 − BMO estimate (2.644) to replacesmallness of w in L∞ by smallness in L2,n.

Let x ∈ B(a,R). Then B(x, 2R) ⊂ B′ = B(a, 3R). Let Br := B(x, r)and

ϕ(r) :=∫Br)

|∇u|2 dy. (2.649)

As before write u = v + w where w ∈W 1,20 (B(x, r),RN ) is the solution of∫

B(x,r)∇w · ∇φdy =

∫B(x,r)

∇u · ∇φdy∀φ ∈W 1,20 (B(x, r),RN ).

141 [March 17, 2014]

Then v is weakly harmonic. Moreover |u| = 1 and hence the weak maximumprinciple shows that |vi| ≤ 1 in B(x, r). Thus w is in L∞ and taking φ = wand using the definition of weakly harmonic maps we get

∥∇w∥2L2(Br) =

∫Br

u|∇u|2 · w dy =∑i,j

∫Br

(ui∇uj − uj∇ui

)· ∇ujwi dy

= −∑i,j

∫Br

(ui∇uj − uj∇ui

)· ∇wiuj dy

where in the last step we used that ∇ujwi = ∇(ujwi) − ∇wiuj and thediv ui∇uj − uj∇ui = 0. Now we apply (2.644) to each term in the last sumand using that |uk| ≤ 1 we get

∥∇w∥2L2(Br) ≤ CN2∥∇u∥L2(B2r

∥∇w∥L2(Br)[u]L2,n(B2r).

By Step 1 (and (2.93)) it follows that∫Br

|∇w|2 dy ≤ Cε20

∫B2r

|∇u|2 dy = Cε20ϕ(2r) (2.650)

where C depends on n and N . Together with the usual estimate∫Bρ

|∇v|2 dy ≤ C(ρr

)n ∫Br

|∇v|2 dy

we deduce that

ϕ(ρ) ≤ C(( ρ

2r

)n+ ε20

)ϕ(2r) ∀0 < ρ ≤ r ≤ R.

Replacing C by 2nC we see that this estimate also holds for ρ ≤ 2r. Thus itfollows from the iteration lemma (Lemma 2.15) (with r replaced by 2r)that

ϕ(ρ) ≤ C ′( ρ

2R

)λϕ(2R)

provided that Cε20 is sufficiently small. Here C ′ depends on λ, n and C.Now

ϕ(2R) =∫B(x,2R)

|∇u|2 dy ≤∫B(a,4R)

|∇u|2 dy

Since the right hand side is independent of the point x ∈ B(a,R) it followsthat ∇ ∈ L2,λ(B(a,R)).

[19.12. 2013, Lecture 19][6.1. 2014, Lecture 20]

142 [March 17, 2014]

Theorem 2.75 (Partial regularity of stationary harmonic maps for n ≥ 3).Let n ≥ 3 and let u : Ω ⊂ Rn → RN be a stationary harmonic map to SN−1.Then there exists a closed set S such

u ∈ C∞(Ω \ S; RN ) and Hn−2(S) = 0 (2.651)

where Hs denotes the s dimensional Hausdorff measure.

Proof. Let ε0 be as in Lemma 2.74 and define the set U by

U := x ∈ Ω : lim supr→0

r2−n∫B(x,r)

|∇u|2 < ε20.

We claim that U is open. Let a ∈ U . Then there exists an r0 > 0 such thatB(a, r0) ⊂ Ω and

r2−n0

∫B(a,r0)

|∇u|2 dy < ε20.

Thus by the ε-regularity Lemma u ∈ C∞(B(a, r0/4); RN ). In particular wehave for every x ∈ B(a, r0/4)

lim supr→0

r2−n∫B(x,r)

|∇u|2 dy = 0.

Thus B(a, r0/4) ⊂ U . Hence U is open and u ∈ C∞(U ; RN ). Finally letS = Ω \U . Then S is closed and by Lemma 2.76 we have Hn−2(S) = 0.

Lemma 2.76. Let Ω ⊂ Rn be open and let f ∈ L1(Ω).

(i)

Ln(x ∈ Ω : lim supr→0

r−n∫B(x,r)

|f | dy = ∞) = 0

(ii) Let 0 < α < n and let ε > 0. Then

Hα(x ∈ Ω : lim supr→0

r−α∫B(x,r)

|f | ≥ ε) = 0

where Hα denotes the α-dimensional Hausdorff measure.

Proof. (i) This follows from the Lebesgue point theorem.(ii) Let

S = x ∈ Ω : lim supr→0

r−α∫B(x,r)

|f | ≥ ε).

By (i) we have Ln(S) = 0. Let σ > 0. We claim that there exists U ⊂ Ωsuch that

S ⊂ U, U open,∫U|f | dy < σ.

143 [March 17, 2014]

Indeed there exists η > 0 such that for every set E ⊂ Ω with Ln(E) < η wehave

∫E |f | dy < σ. Since Ln(S) = 0 there exists an open set U ⊃ S with

Ln(U) < η.For δ > 0 consider the following family of balls

Fδ = B(x, r) : x ∈ S, 0 < r < δ, B(x, r) ⊂ U,

∫B(x,r)

|f | dy > 12εrα.

Then Fδ is a fine cover of S, i.e., for every x ∈ S and every r0 there existsB(x, r) ∈ Fδ with r < r0. By the Vitali covering theorem there existcountable many disjoint balls Bi = B(xi, ri) ∈ Fδ such that

S ⊂∞∪i=1

B(xi, 5ri).

Note that diamB(xi, 5ri) ≤ 10ri ≤ 10δ. Thus the Hausdorff premeasure ofS satisfies

Hα10δ(S) ≤

∑C(α)(5ri)α ≤ C(α)5α

ε

∞∑i=1

∫Bi

|f | dy

≤ C(α)5α

ε

∫U|f | dy ≤ C(α)5α

εσ.

Taking δ → 0 we get

Hα(S) ≤ C(α)5α

εσ

Since σ > 0 was arbitrary this yields the assertion.

The role of symmetries A key ingredient in the proof of (partial) reg-ulartiy of stationary harmonic maps is the identity

div (ui∇uj − uj∇ui) = 0.

in Lemma 2.67. The proof was short but gave little insight why such anestimate should be true.

In fact this identity follows from Noether’s theorem that every symmetryof a variational problem leads to a conservation law (i.e., the statement thatthe divergence of a suitable quantity vanishes). Here the symmetry is therotational symmetry of SN−1 or more precisely the invariance of SN−1 andthe Dirichlet integral under the action of SO(N). If u : Ω → SN−1 and ifQ ∈ SO(N) then Qu : Ω → SN−1 and∫

Ω|∇(Qu)|2 dx =

∫Ω|∇u|2 dx.

144 [March 17, 2014]

We now look for the infinitesimal version of this symmetry. Considervariations of the form

ut = Q(t)u, where Q(t) ∈ SO(N) and Q(0) = Id .

Set W := Q′(0). Then W is skew-symmetric (i.e, Wji = −Wij) and

d

dt t=0ut = Wu

This motivates to consider more generally variations of the form

ut(x) = u(x) + tW (x)u(x) where W ∈ C∞c (Ω; RN×N ) and Wji = −Wij .

Since u · (Wu) = 0 we get

|ut|2 = |u(x)|2 + t2|Wu(x)|2 = 1 + t2|Wu(x)|2.

Thus we should really divide ut by 1 + t2|Wu(x)|2 but for calculating firstorder necessary conditions the contributions from this term will drop out.Thus we get the following necessary first order condition for a minimizer u

0 =∫

Ω

∑k

(∂ku · ∂k(Wu) ) dx

=∫

Ω

∑k

(∂ku ·W∂ku) dx+∫

Ω

∑k

(∂ku · (∂kW )u) dx

We have (∂ku ·W∂ku) = 0 since W is skew symmetric. Using the skewsymmetry of W the other term can be rewritten as

∂ku · (∂kW )u =∑i,j

∂kuj (∂kWji)ui =

∑i<j

(ui∂kuj − uj∂ui)∂kWji.

Taking the sum over k we thus get

0 =∫

Ω

∑i<j

(ui∇uj − uj∇ui) · ∇Wji.

Since the n(n−1)2 functions Wji ∈ C∞

c (Ω) which correspond to indices i < jcan be chosen independently we thus get div (ui∇uj − uj∇ui) = 0 for alli < j (and hence by symmetry for all i, j).

Similarly the monotonicity formula is related to the symmetry underrescaling. Set

ut(x) = ut(x

1 + t).

Thend

dtut(x) = −x · ∇u(x) = −

∑α

xα∂αu.

145 [March 17, 2014]

If u is sufficiently regular one obtains the monotonicity formula multiplyingthe strong form of the equation by x · ∇u(x) or equivalently by using thetest function φ(x) = ψ(|x|) x · ∇u(x), where ψ is an approximation of thecharacteristic function of (0, r).

Harmonic maps with general targets Let M ⊂ RN be a smooth N−1dimensional submanifold of RN . Let Ω ⊂ Rn be open. We look for criticalpoints of the Dirichlet integral ∫

Ω|∇u|2 dx

among maps u : Ω → RN which satisfy u(x) ∈M a.e.To define admissible variations we consider the closest point projection

π : U →M which is defined on a sufficiently small neighbourhood U of M .We consider the variations

ut := π (u+ tφ).

Thend

dt |t=0ut = Dπ(u)φ = φ− (φ · ν u) ν u,

where ν(p) denotes for normal at a point p ∈ M . Thus u is a weaklyharmonic map to M if∫

Ω∇u · [φ− (φ · ν u) ν u] dx = 0 ∀φ ∈ (W 1,2

0 ∩L∞)(Ω; Rm). (2.652)

One easily sees that equivalently u is a weakly harmonic map if∫Ω∇u · ψ dx = 0 ∀φ ∈ (W 1,2

0 ∩ L∞)(Ω; Rm) with ψ(x) ∈ Tu(x)M a.e.,

(2.653)where TpM denotes the tangent space of M at p. Indeed (2.652) implies(2.653) since for the ψ in (2.653) we have ψ · ν = 0 and hence we cantake φ = ψ in (2.652). For the converse implication it suffices to note thatν u ∈ W 1,2 ∩ L∞, that φ − (φ · ν u) ν u ∈ Tu(x)M a.e. and thatf, g ∈W 1,2 ∩ L∞ implies that fg ∈W 1,2 ∩ L∞.

Now in contrast to the sphere which is invariant under rotations a generalsubmanifold M has in general no global symmetries. Hence we can not hopeto obtain a conservation law like div (ui∇uj − uj∇ui) = 0 in a obvious wayfrom global symmetries.

Helein realized that one can instead use local (or infinitesimal) symme-tries of the tangent space of M . To explain his idea we assume for simplicity

146 [March 17, 2014]

that the manifoldM is parallelizable, i.e., that there existN−1 smooth fieldsεi : M → RN such that for each p ∈M the vectors ε1(p), . . . , εN−1(p) are anorthonormal basis of the tangent space TpM . Define the maps ei : Ω → RN

byei(x) = εi(u(x)) (2.654)

Then ei ∈W 1,2∩L∞ and for a.e. x ∈ Ω the vectors e1(x), . . . , eN−1 form anorthonormal basis of Tu(x). For brevity we call e1, . . . , em−1 an orthonormalframe of the M (more precisely these vectors form an orthonormal frame ofthe pull-back bundle u∗(TM)). Using test functions of the form ψ = ηieiwe easily see that (2.653) is the weak form of the equation

∆u · ei = 0 for i = 1, . . . , N − 1. (2.655)

[6.1. 2014, Lecture 20][9.1. 2014, Lecture 21]

Since we are interested in local properties we may assume that

and Ω is a ball B

To express this equation in a more convenient form we assume from now on

n = 2 (2.656)

and expand all tangent vector fields in the basis e1, . . . eN−1. Since ∂αu(x) ∈Tu(x)M we have

∂αu =N1∑j=1

(∂αu · ej)ej . (2.657)

We want to derive a system of PDE for the coefficients ∂α · ei. First, by theproduct rule,

∂β(∂αu · ei) = (∂β∂αu · ei) + (∂αu · ∂βei) = (∂β∂αu · ei) + (∂αu · ej)(ej · ∂βei).

We now introduce the abbreviations

∇u · ei :=(∂1u · ei∂2u · ei

), ωij := ∇ei · ej :=

(∂1ei · ej∂2ei · ej

)With the notation we get

div (∇u · ei) = (∆u · ei)︸ ︷︷ ︸=0

+∑j

ωij · (∇u · ej) (2.658)

where the · after ωij denotes the scalar product in R2. Recall that forv : Ω ⊂ R2 → R2 we define

curl v = ∂2v1 − ∂1v2, v⊥ =(−v2v1

).

147 [March 17, 2014]

Thus

curl (∇u · ei) = (∂1∂2u− ∂2∂1u · ei)︸ ︷︷ ︸=0

+∑j

ω⊥ij · (∇u · ej). (2.659)

Now (2.658) and (2.659) is a first order system for the 2(N − 1) coefficients∂αu · ei. It is convenient to rewrite this as a second order system using theHodge (or Helmholtz) decomposition.

Before we do so we not that for ease of notation we have done the cal-culation using the strong form of the equation. The same assertion can bedirectly derived in the weak form by using the test functions ψ = ηei withη ∈ (W 1,2

0 ∩ L∞)(B) in (2.653). With the product rule and the expansion(2.657) this yields

0 =∫B∇u · (ei ⊗∇η) + ∇u · ∇ei η dx

=∫B

(∇u · ei) · ∇η +∑j

(∇u · ej)(ej · ∇ei), η dx,

i.e., the weak form of (2.658). Similarly to get the weak form of (2.659) onestarts from the identity

0 =∫B∂1u · ∂2ψ − ∂2u · ∂1ψ dx.

This identity original clearly holds for ψ ∈ C2c (B; RN ) since ∂1∂2ψ = ∂2∂1ψ =

0. Since ∇u ∈ L2 the identity also holds for ψ ∈W 1,20 (B; RN ). Now we take

again ψ = ηei and we get

0 =∫B

(∂1u · ei)∂2η − (∂2u · ei)∂1η dx

+∫B

∑j

(∂1u · ej)(ej · ∂2ei)η − (∂2u · ej)(ej · ∂1ei)η dx.

This is the weak form of (2.659).

Lemma 2.77 (Hodge decomposition in two dimensions). Let Ω ⊂ R2 beconvex and bounded. Let v ∈ L2(Ω; R2). Then

(i) If curl v = 0 in the sense of distributions the there exists a C ∈W 1,2(Ω)such that

v = ∇V.

(ii) If div v = 0 in the sense of distributions the there exists a D ∈W 1,2(Ω)such that

v = ∇⊥D.

148 [March 17, 2014]

(iii) There exist C and D such that

v = ∇C + ∇⊥D

and∆C = div v, ∆D = curl v

in the sense of distributions.

The assumption that Ω is convex can be replaced by the assumption thatΩ is simply connected.

Proof. (i): If v ∈ C1(Ω) we have shown this in Analysis 2. The general casefollows by approximation.(ii): If div v = 0 and curl v⊥ = 0 and the result follow from (i).(iii): Let C ∈ W 1,2

0 (Ω) be the weak solution of ∆C = div v. Then div (v −∇C) = 0 and the assertion follows from (ii).

By Lemma 2.77 there exist Ci, Di for i = 1, . . . , N − 1 such that

∇u · ei = ∇Ci + ∇⊥Di (2.660)

and

−∆Ci = −div (∇u·ei) = −∑j

ωij ·(∇Cj+∇⊥Dj) =∑j

(−ωij ·∇Cj+ω⊥ij ·∇Dj)

Similarly

−∆Di = −curl (∇u·ei) = −∑j

ω⊥ij ·(∇Cj+∇⊥Dj) =

∑j

(−ω⊥ij ·∇Cj−ωij ·∇Dj)

Finally we set W = (C1, D1, . . . , CN−1, DN−1). Then we can write thesecond order system in the compact form

−∆Wi =2N−2∑j=1

Aij · ∇Wj

or even more concisely as

−∆W = A · ∇W. (2.661)

where the Aij are linear functions of the ωij .This is an interesting reformulation of the harmonic map equation be-

cause the only geometric information which enters are the connection co-efficients ωij , but from the analytical point of view we so far have gainednothing. We only know that the coefficients ωij (and hence the Aij) are inL2 and that ∇W ∈ L2. Thus the equation (2.661) is still critical. Indeed,if ∇W is in L2 then the right hand side is in L1 and the best we can hopefor is that the solution W is in W 2,1. By the Sobolev embedding this wouldgive again ∇W ∈ L2. In fact without further structure we only get ∇2W inweak L1 and ∇W in weak L2 (see below for the second assertion).

149 [March 17, 2014]

Infinitesimal symmetries and Helein’s moving frame method Thekey idea of Helein was that there are many possible frames e1, . . . , eN−1 andthat one improve the properties of the connection coefficients by choosingan optimal frame. To focus on the main idea we assume now in additionthat the manifold M is two dimensional, i.e.

N − 1 = 2.

Since ∇(ei · ej) = 0 we have ωji = −ωij . Thus for N − 1 = 0 there is wehave

ω11 = ω22 = 0, ω12 = −ω21 and we set ω = ω12.

Now if e1, e2 is an orthonormal frame and R : Ω → SO(2) is a W 1,2 map(when considered as a map with values in R2×2) then the pair e1, e2 definedby

ei =∑j

Rijej

is also an orthonormal frame.A key observation is that ω = ∇e1 · e2 can easily be computed from ω.

Every element of SO(2) is of the form

R =(

cosψ − sinψsinψ cosψ

)and thus

e1 = cosψ e1 − sinψ e2, e2 = sinψ e1 + cosψ e2.

A short calculation shows that

(ω)α = (∂αe1) · e2 = ωα − ∂αψ

or, written more compactly,

ω = ω −∇ψ. (2.662)

Now Helein’s idea is to look for an optimal frame, i.e., one which mini-mizes the Dirichlet energy∫

Ω

∑i,j

(∇ei · ej)2 dx = 2∫

Ω|ω|2 dx = 2

∫Ω|ω −∇ψ|2 dx.

Since ϕ is an arbitrary W 1,2 function an optimal frame e1, e2 must satisfy∫Ωω · ∇η dx =

∫Ω(ω −∇ϕ) · ∇η dx = 0 ∀η ∈W 1,2

0 (Ω).

150 [March 17, 2014]

Thusdiv ω = 0 (2.663)

in the sense of distributions.We now show that this implies that the optimal frame has slightly better

regularity than obvious property ω ∈ L2. By Hodge decomposition thereexists ϕ ∈W 1,2(Ω) such that

∇e1 · e2 = ω = ∇⊥ϕ (2.664)

and

−∆ϕ = − curl ω = −curlω = −∂1(∂2e1 · e2) + ∂2(∂1e1 · e2)

= − ∂2e1 · ∂1e2 + ∂1e1 · ∂2e2 =N∑k=1

−∂2ek1 ∂1e

k2 + ∂1e

k1 ∂2e

k2. (2.665)

Now for each k we have

div(∂2e

k1

−∂1ek1

)= 0.

Thus each term in the sum has the div-grad structure in Lemma 2.68 Itfollows that there exists a function f ∈ H1(R2) such that −∆ϕ = f in B.Hence ϕ ∈ W 2,1(B′) for any ball B′ strictly contained in B. In particularwe have

ω ∈W 1,1(B′).

This is an improvement over the trivial estimate ω ∈ L2. Note that forn = 2 we have the Sobolev embedding W 1,1 → L2 so on the level of theSobolev embedding there seems to be no improvement. We will see belowhowever, that W 1,1 functions are in a space which is slightly better than L2.

The regularity now follows from the formulation (2.661) and the followingresult.

Theorem 2.78. Let B ⊂ R2 be a ball, let Aij ∈ W 1,1(B; Rm) for i, j ∈1, . . . ,m and let W ∈W 1,2(B; Rm) be a weak solution of

−∆W = A ·W in B. (2.666)

Let B′ ⊂ B be a concentric subball. Then W ∈ C0,α(B′,Rm) and ∇W ∈L2,2α(B′; Rm) for any α ∈ (0, 1).

Note that this result implies that ∇u =∑N−1

i=1 (∇u·ei)ei ∈ L2,2α and thusu ∈ L2,2+2α = C0,α (which in turns yields u ∈ C∞ by the same reasoning asin Theorem 2.65).

Sketch of proof. Only steps 1–3 were sketched in class.

151 [March 17, 2014]

Step 1. L2,∞ estimate.Let f ∈ L1(Ω) and 1 < p < 2. We claim that there exists one and only onedistributional solution w ∈W 1,p

0 of

−∆w = f in B

and that ∇w is in the weak L2 space L2,∞ and

∥∇w∥L2,∞(B) ≤ C∥f∥L1(B) (2.667)

where for a measurable function b we define

∥b∥L2,∞(B) := infM : L2(x ∈ B : |b(x)| > λ) ≤ M2

λ2

and C =

√8π/π. Note that ∥v∥L2,∞(B) is not really a norm (it satisfies the

triangle inequality only up to a factor), but it is equivalent to a norm.Proof: By scaling vr(z) = rv(rz) it suffices to show the result for the

unit ball. The main point is to show the estimate for sufficiently regular f .Let f ∈ C∞

c (B). From Introduction to PDE we know that a classical (andhence distributional) solution is given by

w(x) =∫BG(x, y)f(y) dy.

Here Greens function G is defined as

G(x, y) =12π

ln1

|x− y|− 1

2πln

1|y|2|x− y|

, y =y

|y|2

For |x| < 1 and |y| < 1 we have 19 |x− y| ≥ |x− y|. Thus

|∇xG(x, y)| ≤ 12π

1|x− y|

+12π

1|x− y|

≤ 1π

1|x− y|

.

LetK(x, y) :=

1|x− y|

.

Then the assertion (for smooth f) follows from the estimate

∥K ∗ f∥L2,∞(R2) ≤ C∥f∥L1(R2). (2.668)

To prove (2.668) let M = ∥f∥L1 and define

K1(z) :=

1|z| if |z| ≤ 2M

λ

0 else, K2 := K −K1.

19Proof: |y|2|x− y|2 = |y|2|x|2 − 2x · y + 1 = |x− y|2 + (1 − |x|2)(1 − |y|2) ≥ |x− y|2

152 [March 17, 2014]

Then |K2| ≤ λ/(2M) and hence |K2 ∗ f | ≤ λ/2. Now

x : |K ∗ f |(x) > λ ⊂ x : |K1 ∗ f |(x) > λ/2 ∪ x : |K2 ∗ f |(x) > λ/2

and thus

L2(x : |K ∗ f |(x) > λ) ≤ L2(x : |K1 ∗ f |(x) > λ/2)

≤ 2λ∥K1 ∗ f∥L1 ≤ 2

λ∥K1∥L1∥f∥L1 ≤ 2

λ∥K1∥L1M.

Finally, using polar coordinates we get∫Rn

|K1| dx = 2π∫ 2M/λ

0

1rr dr = 2π

2Mλ

and thus (2.668) holds with C =√

8π.The proof of the rest of the assertion is a standard approximation and

duality argument (this argument was not discussed in class).20

Step 2. A slight improvement of the embedding W 1,1(R2) → L2(R2).Assume that a ∈W 1,1(R2) and a = 0 outside a compact set21. Then

∞∑k=−∞

2k(L2(x ∈ R2 : |a(x)| > 2k)

)1/2≤ C∥∇a∥L1 (2.669)

The space of functions a for which the left hand side is bounded is theLorentz space L2,1. For the general definition of the Lorentz spaces Lp,q

20For f ∈ L1 we consider fk ∈ C∞c (B) with fk → f in L1. Since ∥g∥Lp(B) ≤

Cp∥g∥L2,∞(B) for any p < 2 it follows that the corresponding solutions wk are a Cauchy

sequence in W 1,p0 (Ω) and hence converge to v in W 1,p

0 (Ω). One easily checks that v isa distributional solution of the equation and that ∥∇v∥L2,∞ ≤ lim infk→∞ ∥∇wk∥L2,∞ .Finally to show uniqueness it suffices to consider f = 0 and we argue by duality. Let p′

be the dual exponent of p. By assumption we haveZ

B

∇v · ∇ψ dx = 0

Z

B

g · ∇η dx ∀ψ ∈ C∞c (B).

Since ∇w ∈ Lp this identity holds for all ψ ∈ W 1,p′

0 (B) Now let g ∈ Lp′(B; R2). Since

p′ ≥ 2 there exists a weak solution φ ∈W 1,20 (B) of −∆φ = −div g, i.e.,

Z

B

∇φ · ∇η dx =

Z

B

g · ∇η dx ∀η ∈W 1,20 (B).

For the Lq regularity theory we have ∇w ∈ W 1,p′(B) and hence this identity holds for

η ∈W 1,p0 (Ω). Taking η = w and ψ = φ we get

Z

B

g · ∇w dx =

Z

B

φ · ∇v = 0.

Since this holds for all g ∈ Lp′(B,R2) we deduce that ∇w = 0 and hence w = 0.

21this condition is not really needed and can easily be removed by approximation

153 [March 17, 2014]

with p ∈ [1,∞) and q ∈ [1,∞] and their properties see, e.g., [Zi89]. Tocompare the left hand side of (2.669) with the usual Lp,q norm compare thesum to an integral (using the monotinicity of λ 7→ L2(|a| > λ) and usethe change of variables of variables t = 2k.

To show (2.669) we may assume that a ≥ 0 since |a| ∈ W 1,1(R2) and|∇|a| | = |∇a| a.e. Define

ak =

a− 2k−1 if 2k−1 < a < 2k+1,0 if a ≤ 2k−1,2k+1 − 2k−1 if a ≥ 2k+1.

Note that ak can also be written as ak = min(max(a, 2k−1), 2k+1

)− 2k−1.

Thus

∇ak = ∇a χFka.e. where Fk = x ∈ Rn : 2k−1 ≤ a(x) < 2k+1.

The sets Fk are not disjoint but each point x is contained in at most two ofthe sets Fk. Thus ∑

k

χFk≤ 2. (2.670)

Moreover ak has compact support (since a has compact support anda = 0 implies that ak = 0). Hence by the Sobolev embedding theorem

∥ak∥L2 ≤ C

∫R2

|∇ak| dx ≤ C

∫R2

χFk|∇a| dx.

Now we have

L2(x ∈ R2 : a(x) > 2k) = L2(x ∈ R2 : ak(x) > 2k−1) ≤ 122(k−1)

∫R2

|ak|2 dx.

Taking the square root, multiplying by 2k−1, summing over k and using(2.670) we get

∞∑k=−∞

2k−1(L2(x ∈ R2 : a(x) > 2k)

)1/2≤∑

∥ak∥L2 ≤∫

R2

∑k

χFk︸ ︷︷ ︸≤2

|∇a| dx.

This proves (2.669).

Step 3. L2,1-L2,∞ duality.Let Br be a ball of radius r. Let a ∈ W 1,1(Br) and assume that b is inL2,∞(Br), i.e.,

L2(x ∈ Br : |b(x)| ≥ λ) ≤ B2

λ2.

154 [March 17, 2014]

We claim that ab ∈ L1(Br) and∫Br

|ab| dx ≤ CB(∥∇a∥L1(Br) + ∥a∥L2(Br)). (2.671)

It suffices to prove the result for r = 1. Indeed one can define ar(z) = ra(rz),br(z) = rb(rz) and one easily sees that the assertion for ar and br impliesthe assertion for a and b.

Thus assume r = 1 and set B = Br. We can extend a to a functiona ∈W 1,1(R2) which vanishes outside a ball of radius 2 and which satisfies

∥a∥W 1,1(Rn) ≤ C∥a∥W 1,1(B) ≤ C(∥∇a∥L1(B) + ∥a∥L2(B)).

SetEk := x ∈ B : 2k ≤ |a(x)| < 2k+1.

Since a = a in B we get from Step 2 applied to a∞∑k=0

2k(L2(Ek))1/2 ≤ C(∥∇a∥L1(B) + ∥a∥L2(B)). (2.672)

Now we claim that for any measurable set E ⊂ B∫E|b| dx ≤ 2∥b∥L2,∞(E)(L2(E))1/2 (2.673)

(in other words functions in weak L2 satisfy the same Holder inequality asfunctions in L2). To see this recall that the left hand side equals∫ ∞

0L2(x ∈ E : |b(x)| ≥ λ) dλ ≤

∫ ∞

λ∗

∥b∥2L2,∞

λ2dλ+ λ∗L2(E)

≤∥b∥2

L2,∞

λ∗+ λ∗L2(E)

for any λ∗ > 0. The choice λ∗ = ∥b∥L2,∞(L2(E))−1/2 gives the assertion.Thus∫

B|ab| dx ≤

∞∑k=−∞

∫Ek

2k+1|b| dx ≤∞∑

k=−∞2k+2∥b∥L2,∞(L2(Ek))1/2

≤ C∥b∥L2,∞(∥∇a∥L1(B) + ∥a∥L2(B)).

Step 4. Decay of the L2,∞ norm.This is proved in the usual way by decomposing W into a harmonic functionand a function with zero boundary conditions and using the iteration lemma.

Let B′ ⊂ B be a ball which is compactly contained in B. Let x ∈ B′

and let B(x,R) ⊂ B and let r ∈ (0, R). Write W = v + w where w is theunique solution of

−∆w = A · ∇W in B(x, r)

155 [March 17, 2014]

with zero boundary conditions in the sense of Step 1.By Step 3 we have

∥A · ∇W∥L1(B(x,r) ≤ Cµ(r)∥∇W∥L2,∞ ,

where

µ(r) := sup∥∇A∥L1(B(z,r) + ∥A∥L2(B(z,r) : B(z, r) ⊂ B.

Since A ∈W 1,1(B) ⊂ L2(B) we have

limr→0

µ(r) = 0.

Thus by using Step 1 for each component of w we get

∥∇w∥L2,∞(B(x,r)) ≤ Cµ(r)∥∇W∥L2,∞(B(x,r)). (2.674)

Now ∆v = 0 and thus for ρ ≤ r/2∫B(x,ρ)

|∇v|2 dx ≤ C(ρr

)2∫B(x,r/2)

|∇v|2 dx

≤ C(ρr

)2 1r2

∫B(x,r)

|v − (v)x,r|2 dx.

Now we claim that for any function g with ∇g ∈ L2,∞(B(x, r)∫B(x,r)

|g − (g)x,r|2 dx ≤ Cr2∥∇g∥2L2,∞(B(x,r). (2.675)

Indeed by the Poincare-Sobolev estimate and (2.673) we have∫B(x,r)

|v−(v)x,r|2 dx ≤ C

(∫B(x,r

|∇v| dx

)2

≤ CL2(B(x, r)∥∇v∥2L2,∞(B(x,r).

Together with the trivial estimate ∥∇v∥L2,∞(B(x,ρ)) ≤ ∥∇v∥L2(B(x,ρ)) we con-clude that

∥∇v∥L2,∞(B(x,ρ)) ≤ C(ρr

)2∥∇v∥2

L2,∞(B(x,r)). (2.676)

Combining this with (2.674) in the usual way we deduce that

∥∇W∥2L2,∞(B(x,ρ)) ≤ C

[µ2(r) +

(ρr

)2]∥∇W∥2

L2,∞(B(x,r)).

Since limr→0 µ(r) = 0 it follows from the iteration lemma (Lemma 2.15)that for each λ ∈ (0, 2) there exists a Cλ which depends on B′ and µ (butnot on x ∈ B′) such that

∥∇W∥2L2,∞(B(x,ρ)) ≤ Cλρ

λ∥∇W∥2L2,∞(B). (2.677)

156 [March 17, 2014]

Step 5. C0,α and L2,λ estimates.It follows from (2.675) and (2.677) that∫

B(x,r)|W − (W )x,r|2 dx ≤ Cr2∥∇W∥2

L2,∞(B(x,r)) ≤ Cλr2+λ∥∇W∥2

L2,∞(B).

Thus W ∈ L2,2+λ(B′) = C0,λ/2(B′) for all λ < 2.Finally to get the L2,λ bound for ∇W we argue as in the proof of Theorem

2.65. Using the weak maximum principle as in that proof and the C0,α

estimate for W we see that the function w defined in Step 4 satisfies

supB(x,r)

|w| ≤ Crα.

Thus using w as a test function we get∫B(x,r)

|∇W |2 dx ≤ Crα∥A∥L2∥∇W∥L2(B(x,r) ≤C

εr2α + ε∥∇W∥2

L2(B(x,r)).

Together with the estimate for ∥∇v∥L2(B(x,ρ)) this yields

∥∇W∥2L2(B(x,ρ)) ≤

[ε+ C

(ρr

)2]∥∇W∥2

L2(B(x,r)) +C

εr2α.

By the iteration lemma this implies that ∇W ∈ L2,2α(B′) for all α ∈ (0, 1).

The following generalizations to higher dimensions were not discussed inclass.

The case N − 1 > 2. In this case we can still define a new frame by

ei =N−1∑k=1

Rikek

if R(x) ∈ SO(N − 1) for a.e. x. In this case two matrices in SO(N − 1) nolonger commute and the new connection coefficients are given by

ω = (∇R)R−1 +RωR−1. (2.678)

The frame which minimizes∫Ω

∑i,j |ωi,j |2 dx still satisfies

div ωij = 0. (2.679)

Using this we still get ωij ∈W 1,1 and obtain regularity using Theorem 2.78.For the proof of (2.678) and (2.679) see Homework 11, Problem 1.

157 [March 17, 2014]

Partial regularity for stationary harmonic maps to M for n ≥ 3.If M is sufficiently regular this can also be proved by the moving framemethod, see [Be93]. One uses a higher dimensional counterpart of the Hodgedecomposition 22 and shows a decay estimate for the BMO norm of u. Theargument is similar to Step 2 in the proof of the corresponding result forM = SN−1. In this setting one does not control the L2 norm of all relevantgradients but one can use that in view of the Poincare inequality the BMOnorm of f in a ball B is also controlled by[

supB(x,r)⊂B

1rn−p

∫B(x,r

|∇f |p dx

]1/p

.

Another symmetry This was not discussed in class.For n ≥ 3 Rivere and Struwe recently found a very elegant proof of partialregularity of stationary harmonic maps which only requires minimal regu-larity of the target manifold M (namely M ∈ C2). This proof uses anothersymmetry (see [RiSt] which follows the earlier work [Ri07]). Returning tothe form (2.652) of the harmonic map equation and setting w = ν u wehave∫

Ω∇u · ∇[φ− (φ · w)w] dx = 0 ∀φ ∈ (W 1,2

0 ∩ L∞)(Ω; Rm). (2.680)

Now using ∇u · w = 0 since ∇u is tangential and w is normal to M , thiscondition for u be rewritten as∫

Ω∇u · ∇φ−∇u · ∇w (φ · w) dx = 0.

The strong form of this equation is

−∇ui =∑j

wi∇wj · ∇uj .

Using again that∑

j wj∂αu

j = 0 (because ∂αu is a tangent vector) we canrewrite the harmonic map equation as

−∆ui =∑j

Ωij · ∇uj (2.681)

whereΩij = wi∇wj − wj∇wi.

22A natural way to express this higher dimensional Hodge decomposition in higherdimensions is the language of differential forms. One can, however, also write it in astandard way.

158 [March 17, 2014]

Now the only symmetry which is used is that

Ωji = −Ωij , (2.682)

i.e., the skew symmetry of Ω. In fact under this assumption alone can showthe following ε regularity result.

Theorem 2.79. Let B be a ball. Then there exists an ε0 > 0 with thefollowing properties. If u ∈W 1,2(B; RN ) is a weak solutions of (2.681) andif

∥∇u∥2L2,n−2(B) < ε20 and ∥Ω∥L2,n−2(B) < ε20

then u is locally Holder continuous in B.

For stationary harmonic maps u : Ω → M one can then use the mono-tonicity formula as before to obtain regularity in Ω \ S where S is a closedset with Hn−2(S) = 0.

To exploit the skew symmetry of Ω one uses again a change of coordinatesor a ’gauge transformation’ (similar to the choice of the optimal frame inHelein’s approach). First one shows if u solves the equation −∆u = Ω · ∇u(with a skew-symmetric Ω) in a ball B and if P : B → SO(N) then

−div (P−1∇u) = [P−1∇P + P−1ΩP ] · P−1∇u (2.683)

Here · denotes the scalar product in Rn and the other operations a matrixtimes vector or matrix times matrix multiplication23 in RN . Note thatthis formula is exactly analogous to (2.678) if one makes the substitutionR = P−1.

Then one proves (this similar to but more subtle argument than thechoice of the optimal frame) that there exists a map P : B → SO(n) forwhich

div [P−1∇P + P−1ΩP ] = 0. (2.684)23Without the · notation the equations (2.681) reads

−n

X

α=1

∂2αu

i =

nX

α=1

NX

j=1

Ωijα∂αuj

or

−n

X

α=1

∂2αu =

nX

α=1

Ωα∂αuj

where Ωα is an N ×N matrix with entries Ωijα. Then (2.683) and (2.684) become

−n

X

α=1

∂α(P−1∂αu) =

nX

α=1

[P−1∂αP + P−1ΩαP ]P−1∂αu,

nX

α=1

∂α[P−1∂αP + P−1ΩαP ] = 0.

159 [March 17, 2014]

The choice of this P is called the Coloumb gauge. Now the right hand side of(2.683) almost has a div-grad structure (except for the extra P−1 in front of∇u). One uses a Hodge decomposition of P−1∇u and then one can handlethe extra term can by use Morrey space and BMO estimates, duality andthe usual iteration lemma.

[9.1. 2014, Lecture 21][13.1. 2014, Lecture 22]

2.4.2 Regularity for minimizers of variational problems

We now study the regularity of minimizers of variational integrals

I(v) =∫

Ωf(∇v) dx.

Here Ω ⊂ Rn is open and bounded and v : Ω → RN . We say that

u ∈W 1,2(Ω; RN ) is a minimizer of I if

I(u) ≤ I(v) for all v with v − u ∈W 1,20 (Ω; RN ). (2.685)

We assume throughout this subsection that

f ∈ C2(RN×n). (2.686)

Now we assume first that f is uniformly convex and D2f is bounded,i.e., we assume that there exists ν > 0 and M > 0 such that

D2f ≥ ν Id , (2.687)|D2f | ≤ M. (2.688)

More explicitly these conditions read

D2f(A)(F, F ) ≥ ν|F |2 ∀A,F ∈ RN×n,D2f(A)(F,G) ≤ M |F | |G| ∀A,F,G ∈ RN×n.

Differentiating I(u + tφ) at t = 0 we see that minimizers of I are weaksolutions of the Euler-Lagrange equations

−divDf(∇u) = 0. (2.689)

Using the uniform convexity of f and the difference quotient method wesee that u ∈ W 2,2

loc (Ω; RN ) and that each derivative w = ∂γu satisfies thelinearized equation24

−divD2f(∇u)∇w = 0.24In index notation we have for each i = 1, . . . , N

−X

α

∂α∂f

∂F iα

(∇u) = 0.

160 [March 17, 2014]

For N = 1, i.e., scalar problems, the DeGiorgi-Nash-Moser theoremimplies that w ∈ C0,α

loc (Ω) and thus ∇u ∈ C1,αloc (Ω; Rn) for some α > 0. Then

a bootstrap argument implies that u ∈ C∞(Ω) if f ∈ C∞(Rn). (see thebeginning of Section 2.2).

If N > 1 then the DeGiorgi-Nash-Moser theorem cannot be applied. Forn = 2, there is a different argument which yields regularity. Indeed, byMeyers’ theorem, Theorem 2.62, we get ∇w ∈ Lploc(Ω; RN ) for some p > 2and then the Sobolev embedding theorem gives again w ∈ C0,α

loc (Ω; RN ).For n ≥ 3 in general singularities arise may arise for systems. The first

counterexample (for n andN large) is due Necas [Ne], see also Chapter II.3 in[Gi]. More recently Sverak and Yan have constructed new counterexamples,including one for n = 3 and N = 4 [SY00, SY02].

Thus for n ≥ 3 and the case of systems (N ≥ 2) it is natural to look forpartial regularity, i.e., regularity outside a ’small’ closed set. In a numberof interesting applications of RN valued map (e.g., in nonlinear elasticity)convexity is a too restrictive condition and quasiconvexity is a more nat-ural condition (see, e.g., the lecture notes of PDE and modelling, [Mu99]or [Da08] for a discussion of the various convexity notions). For (partial)regularity it is natural to assume that f is uniformly quasiconvex, i.e., thereexists a ν > 0 such that∫

B(x,r)f(A+ ∇φ) − f(A) dy ≥ ν

2

∫B(x,r)

|∇φ|2 dy

∀A ∈ RN×n, φ ∈ C1c (B(x, r); RN ), (2.690)

and all x ∈ Rn, r > 0 (indeed if the condition holds for one open andbounded set U with Ln(∂U) = 0 it holds for all such sets). It is easy tosee that (2.687) implies (2.690). Under the assumption (2.688) it follows bydensity that ∫

B(x,r)f(A+ ∇φ) − f(A) dy ≥ ν

2

∫B(x,r)

|∇φ|2 dy

∀A ∈ RN×n, φ ∈W 1,20 ((B(x, r); RN ). (2.691)

Lemma 2.80 (Uniform quasiconvexity implies strong ellipticity). Assumethat f ∈ C2(RN×n) satisfies (2.690) then

D2f(A)(a⊗ b, a⊗ b) ≥ ν|a|2 |b|2 ∀A ∈ RN×n, a ∈ RN , b ∈ Rn. (2.692)

and differentiating with respect to xγ yields

−X

α

∂α

X

β,j

∂2f

∂F iα ∂F

jβ

(∇u) ∂γ(∇u)jβ = 0.

Now ∂γ(∇u)jβ = ∂γ∂βu

j = ∂β∂γuj = ∂βw

j = (∇w)jβ .

161 [March 17, 2014]

In particular for N = 1 and f ∈ C2 the uniform quasiconvexity condition(2.690) is equivalent to the uniform convexity condition (2.687).

Proof. The main point is to use highly oscillatory test functions. Let η ∈C∞c (B(x, r)) and let

φk(x) = εka

ksin(kb · x)η(x), with εk → 0.

Then∇φk = εk(a⊗ b) cos(kb · x)η +

εkk

(a⊗∇η) sin(kb · x)

and a short calculation, which uses that∫B(x,r) ∇φdy = 0, shows that

limk→∞

1ε2k

∫B(x,r)

f(A+ ∇φk) − f(A) dy

= limk→∞

∫B(x,r)

12D2f(A) [(a⊗ b) cos(kb · y)η, (a⊗ b) cos(kb · y)η] dy

= limk→∞

12D2f(A)(a⊗ b, a⊗ b)

∫B(x,r)

cos2(kb · y)η2 dy

=12D2f(A)(a⊗ b, a⊗ b)

∫B(x,r)

12η2 dy.

Similarly we get

limk→∞

1ε2k

∫B(x,r)

|∇φk|2 dy = |a⊗ b|2∫B(x,r)

12η2 dy.

Now |a⊗ b|2 = |a|2 |b|2 and the assertion follows from (2.690).

We use the following notation for the average.

–∫E

f dx :=1

Ln(E)f dx.

Theorem 2.81 (Evans 1986). Assume that f ∈ C2(RN×n) satisfies (2.690)and (2.688) and that D2f is uniformly continuous. Let u be a minimizer ofI. Then there exists a closed set S such that

u ∈ C1,αloc (Ω \ S; RN ), Ln(S) = 0. (2.693)

If, in addition, f ∈ C∞ then u ∈ C∞(Ω \ S; RN ). We may take

S := x ∈ Ω : lim supr→0

–∫

B(x,r)

|∇u− (∇u)x,r|2 dy > 0. (2.694)

162 [March 17, 2014]

Remark. (i) The assumption that D2f is uniformly continuous is notneeded for partial regulartiy (see Corollary 2.86 below). Without this as-sumption one may take

S := x ∈ Ω : lim supr→0

–∫

B(x,r)

|∇u−(∇u)x,r|2 dy > 0 or lim supr→0

–∫

B(x,r)

|∇u| dy = ∞.

(ii) It is important that u is a minimizer. There exist Lipschitz continuousweak solutions of the Euler-Lagrange equations −divDf(∇u) = 0 which arenowhere C1 (see [MS03]).(iii) If f is uniformly convex the estimate for the singular set can be improvedto Hn−p(S) = 0 for some p > 2 (see below).(iv) It is not known whether the estimate for the singular set can be improvedfor uniformly quasiconvex integrands. The best known result states that ifwe assume in addition that u ∈ W 1,∞

loc then there exists an ε > 0 such thatHn−ε(S) = 0, see [MiKr].

We first show that every minimizer satisfies a reverse Poincare inequalityor Cacciopoli inequality.

Lemma 2.82 (Cacciopoli inequality/ reverse Poincare inequality). Assumethat f ∈ C2(Rn×N ) satisfies (2.690) and (2.688). Then there exists a con-stant C such for every minimizer u of I, for every ball B(x, r) ⊂ Ω andevery A ∈ RN×n and every a ∈ RN

–∫

B(x,r/2)

|∇u−A|2 dy ≤ C

r2–∫

B(r)

|u− a−Ay|2 dy. (2.695)

Here C depends only on n, µ and M .

Proof. We may assume x = 0 and we write B(r) = B(0, r). To illustrate howthe combination of uniform quasiconvexity and the minimization propertyleads to an estimate for ∇u − A let us first consider a toy problem andassume for a moment that u is affine on the boundary ∂B(r), i.e., thereexists A and a such that u(y) = Ay + a on ∂B(r).

Then we can apply the uniform quasiconvexity condition with φ = u −a−Ay and we get

ν

2

∫B(r)

|∇u−A|2 dy ≤∫B(r)

f(∇u) − f(A) dy

On the other hand we can apply the minimizing property of u with v =a+Ay = u− φ. This yields∫

B(r)f(∇u) dy ≤

∫B(r)

f(A) dy.

163 [March 17, 2014]

Combining the two estimates we get

ν

2

∫B(r)

|∇u−A|2 dy = 0,

i.e., u is affine in B(r).In general u does not equal an affine function on the boundary so we

need to truncate u − a − Ay near the boundary. Let 0 < t < s < r and letζ ∈ C∞

c (Ω) be a cut-off function with

ζ = 1 on B(t), ζ = 0 on Ω \B(s),

0 ≤ ζ ≤ 1, |∇ζ| ≤ C

s− t.

Setφ = ζ(u− a−Ay), ψ = (1 − ζ)(u− a−Ay).

Then∇φ+ ∇ψ = ∇(φ+ ψ) = ∇u−A.

Now φ ∈W 1,20 (B(s); RN ) and thus uniform quasiconvexity implies that

ν

2

∫B(s)

|∇φ|2 dy ≤∫B(s)

f(A+ ∇φ) − f(A) dy.

Since u is a minimizer and u− φ = u on Ω \B(s) we get∫B(s)

f(A+ ∇ψ + ∇φ︸ ︷︷ ︸=∇u

) dy ≤∫B(s)

f(A+ ∇ψ︸ ︷︷ ︸=∇(u−φ)

) dy.

Adding −f(A + ∇ψ + ∇φ) + f(A + ∇φ) − f(A) to the integrand in thesecond inequality and using the first inequality we get

ν

2

∫B(s)

|∇φ|2 dy ≤∫B(s)

f(A+∇ψ)− f(∇ψ+∇φ) + f(A+∇φ)− f(A) dy.

Now by Taylor expansion and the assumption |D2f | ≤M we have

|f(A+ ∇ψ) − f(A+ ∇ψ + ∇φ) +Df(A+ ∇ψ)∇φ| ≤ 12M |∇ψ|2,

|f(A+ ∇φ) − f(A) −Df(A)∇φ| ≤ 12M |∇ψ|2,

|Df(A+ ∇ψ) −Df(A)| ≤ M |∇ψ|.

Thus

|f(A+ ∇ϕ) − f(A+ ∇ψ + ∇φ) + f(A+ ∇φ) − f(A)|

≤M |∇ψ|2 +M |∇ψ| |∇φ| ≤(M +

M2

ν

)|∇ψ|2 +

ν

4|∇φ|2.

164 [March 17, 2014]

where we used Young’s inequality ab ≤ 12εa

2 + ε2b

2 with ε = ν/2. It followsthat ∫

B(s)|∇φ|2 dy ≤ C

∫B(s)

|∇ψ|2 dy, (2.696)

where C = 4Mν + 4M2

ν2 . Now ∇ψ = 0 in B(t) and

|∇ψ| ≤ |∇u−A| + C

s− t|u− a−Ay|.

Thus (2.696) implies that∫B(t)

|∇u−A|2 dy ≤ C

∫B(s)\B(t)

|∇u−A|2 dy+ C

(s− t)2

∫B(s)\B(t)

|u−a−Ay|2 dy.

Now ’fill the hole’, i.e., add C∫B(t) |∇u− A|2 dy to both sides an divide by

C + 1 to conclude∫B(t)

|∇u−A|2 dy ≤ θ

∫B(s)\B(t)

|∇u−A|2 dy+ C

(s− t)2

∫B(s)\B(t)

|u−a−Ay|2 dy

for all r/2 ≤ s < t ≤ r where

θ =C

C + 1< 1.

Now the assertion follows from Lemma 2.83 below with A′ = C∫B(r) |u −

a−Ay|2 dy, B = 0, α = 2 and ρ = r0 = r/2, r1 = r.

Remark. One can simplify the algebra leading to (2.696) by observingthat it suffices to consider A = 0, a = 0 and that we may assume thatW (0) = 0 and W (0) = 0. Indeed set

W (F ) = W (A+ F ) −W (A) −DW (A)F, u = u− a−Ay,

I(v) =∫

ΩW (∇v) dx.

Then u is a minimizer of I if and only if u is a minimizer of I, and W satisfies(2.688) and (2.4.2). Moreover W (0) = 0, DW (0) = 0. Then the estimatesreduce to

|W (∇ψ + ∇φ) − W (∇ψ)| ≤ |DW (∇ψ)∇φ| + 12M |∇ψ|2 ≤M |∇φ| |∇ψ| + 1

2M |∇ψ|2,

|W (∇ψ) −W (0)| = |W (∇ψ)| ≤ 12M |∇ψ|2.

165 [March 17, 2014]

Lemma 2.83. Let 0 < r0 < r1 and let f : [r0, r1] → [0,∞) be a boundedfunction. Assume that there exists θ ∈ [0, 1), α > 0 and A′, B ≥ 0 such thatfor all r0 ≤ s < t ≤ r1

f(t) ≤ θf(s) +A′(s− t)−α +B.

Then there exists a constant c which depends only on α and θ such that

f(ρ) ≤ c[A′(R− ρ)−α +B] forall r0 < ρ < R ≤ r1.

Proof. Let τ ∈ (0, 1), let r0 < ρ < R ≤ r1 and consider the sequence tidefined by

t0 = ρ, ti+1 − ti = (1 − τ)τ i(R− ρ).

Then ti ∈ (ρ,R) and ti → R as i→ ∞. By iteration we get

f(t0) ≤ θkf(tk) +[

A′

(1 − τ)α(R− ρ)α +B

] k∑i=0

θiτ−iα.

Now choose τ ∈ (0, 1) such that τ−αθ < 1. Taking the limit k → ∞ we getthe assertion with c(α, τ) = (1 − τ)−α(1 − τ−αθ)−1.

[13.1. 2014, Lecture 22][16.1. 2013, Lecture 23]

Now we come to the heart of the matter. Define

U(x, r) := –∫

B(x,r)

|∇u− (∇u)x,r|2 dy, where (∇u)x,r := –∫

B(x,r)

∇u dy

The quantity U(x, r) measures how close u is to an affine function and isoften called the excess.

Lemma 2.84 (Excess decay). Assume that f satisfies (2.688) and (2.690)and that D2f is uniformly continuous. Let u be a minimizer of I. Thenthere exists a constant C2 (which only depends on n, M and ν) with theproperty that for each τ ∈ (0, 1

4) there exists an ε(τ) such that for everyB(x, r) ⊂ Ω the relation

U(x, r) ≤ ε(τ) (2.697)

implies thatU(x, τr) ≤ C2τ

2U(x, r). (2.698)

Proof. Let C2 be a constant, which will be determined later, and fix τ ∈(0, 1

4). To show that (2.697) =⇒ (2.698) we argue by contradiction.

166 [March 17, 2014]

Step 1. Blow-up.If the implication (2.697) =⇒ (2.698) was false, there would exist for m =1, 2, . . . balls B(xm, rm) ⊂ Ω such that

U(xm, rm) = λ2m → 0 (2.699)

butU(xm, τrm) ≥ C2τ

2λ2m. (2.700)

Define

am := (u)xm,rm , bm := (u)xm,2τrm ,Am := (∇u)xm,rm , Bm := (∇u)xm,2τrm .

(2.701)

and set

vm(z) =u(xm + rmz) − am − rmAmz

λmrmfor z ∈ B = B(0, 1). (2.702)

Then∇vm(z) =

∇u(xm + rmz) −Amλm

(2.703)

and(vm)1 = 0, (∇vm)1 = 0. (2.704)

Here we write (vm)r for the average over the ball B(r) = B(0, r). Define

dm := (vm)2τ , em := (vm)τ ,Dm := (∇vm)2τ , Em := (∇vm)τ ,

(2.705)

Now (2.699) implies that

–∫B

|∇v|2 dz = 1. (2.706)

Since (vm)1 the Poincare inequality yields

–∫B

|v|2 dz ≤ C. (2.707)

On the other hand (2.700) gives

–∫

B(τ)

|∇v − Em|2 ≥ C2τ2. (2.708)

By assumption we have|D2f(Am)| ≤M. (2.709)

167 [March 17, 2014]

Together with (2.706) and (2.707) and the compact Sobolev embeddingit follows that there exists a subsequence, which upon relabeling we alsoindex by m, such that

vm → v in L2(B; RN ),∇vm ∇v in L2(B; RN×n),∂2f

∂F iα∂F

jβ

→ Lαβij .

for some v ∈W 1,2(B; RN ) and Lαβij ∈ R. For future reference we set

d := (v)2τ , e := (v)τ ,D := (∇v)2τ , E := (∇v)τ ,

(2.710)

We now claim that v is a weak solution to the linear PDE25

−divL∇v = 0 (2.711)

and

|L| := supLF ·G : |F | ≤ 1, |G| ≤ 1 ≤M, (2.712)

L(a⊗ b) · (a⊗ b) ≥ ν|a|2|b|2 ∀a ∈ RN , b ∈ Rn. (2.713)

These estimates follow directly from the corresponding estimatesD2f(Am)(F,G) ≤M |F ||G| and D2f(Am)(a⊗ b, a⊗ b) ≥ ν|a|2|b|2. To prove(2.711) we note that the weak form of the equation for u implies that∫

B(xm,rm)Df(∇u) · ∇ψ dx = 0 ∀ψ ∈ C1

c (B(xm, rm); RN ).

By (2.702) it follows that

1λm

∫B

[Df(Am + λm∇vm) −Df(Am)] · ∇φdz = 0 ∀φ ∈ C1c (B; RN ).

Hence ∫B

∫ 1

0D2f(Am + sλm∇vm)∇vm · ∇φds dz = 0. (2.714)

We have λm∇v → 0 in L2. Thus for a further subsequence we may assumethat λm∇vm → 0 a.e. in B. Since D2f is uniformly continuous we deducethat

sups∈[0,1]

D2f(Am + sλm∇vm) −D2f(Am) → 0 a.e. in B. (2.715)

25In index notation this reads −P

α ∂α

P

β,j Lαβij ∂βv

j = 0.

168 [March 17, 2014]

Boundedness of D2f and the dominated convergence theorem imply that

D2f(Am + sλm∇vm) − L =

=(D2f(Am + sλm∇vm) −D2f(Am)

)+ (D2f(Am) − L) → 0

in L2(B;L(RN×n; RN×n)), uniformly in s. Thus after exchanging the inte-gration in s and z in (2.714) we get∫

BL∇v · ∇φ = 0 ∀φ ∈ C1

c (B; RN ).

By density this also holds for all φ ∈ W 1,20 (B; RN ) and hence v is a weak

solution of (2.711).

Step 2. Decay estimate for v.By the regularity estimate (2.79) for linear equations with constant, stronglyelliptic coefficients we have

supB( 1

2)

|D2v| ≤ C

∫B|∇v|2 dz ≤ C

and thus

–∫

B(τ)

|∇v − (∇v)τ |2 dz ≤ Cτ2. (2.716)

Step 2. Transfer of the decay from v to vm.Since vm → v strongly in L2(B; RN ) we have, using the Poincare inequalityand (2.716),

limm→∞

–∫

B(2τ)

|vm − dm −Dmz|2 dz = –∫

B(2τ)

|v − d−Dz|2

≤Cτ2 –∫

B(2τ)

|∇v −D|2 dz ≤ Cτ4. (2.717)

The Cacciopoli inequality, Lemma 2.82, provides the estimate

–∫

B(xm,τrm)

|∇u−Bm|2 dy ≤ C

(2τrm)2–∫

B(xm,2τrm)

|u− bm −Bm(y − xm)|2 dy.

Dividing by λ2m and rescaling yields

–∫

B(τ)

|∇vm −Dm|2 dz ≤C1

τ2–∫

B(2τ)

|vm − dm −Dmz|2 dz.

169 [March 17, 2014]

Combining this with (2.717) we get

lim supm→∞

–∫

B(τ)

|∇vm − Em|2 dz ≤ lim supm→∞

–∫

B(τ)

|∇vm −Dm|2 dz

≤C1

τ2lim supm→∞

–∫

B(2τ)

|vm − dm −Dmz|2 dz ≤ C3τ2.

This contradicts (2.708), provided we chose C2 > C3.

Remark on the structure of the proof. There are three key ingredi-ents:

• Blow-up and weak convergence to a function v which satisfies a linearequation with constant coefficient

• Decay estimate for solutions of such equations

• Transfer of the decay from v to vm

The first step, which may look complicated initially, is actually usually easy.In particular the first two steps still work if instead of uniform quasiconvex-ity we just assume the strong elliptictiy condition D2f(A)(a ⊗ b, a ⊗ b) ≥ν|a|2 |b|2.

For the last step we used the Cacciopoli inequality. More abstractly forthe last step one can use a compactness argument: if vm are minimizers(of the rescaled problem) and vm v in W 1,2(B; RN ) then vm → v inW 1,2(B(τ); RN ) (see [EvG] for a proof of this compactness result). Thusexcess decay and hence (partial) regularity is very closely related to com-pactness of minimizers or of solutions to the PDE.

Proof of Theorem 2.81. Let

Ω0 = Ω \ S = x ∈ Ω : limr→0

–∫

B(x,r)

|∇u− (∇u)x,r|2 dy = 0.

Let α ∈ (0, 1). We will show that

x0 ∈ Ω =⇒ ∇u ∈ C0,α(B(x0, ρ); RN ) for some ρ > 0. (2.718)

This implies that

limr→0

–∫

B(x,r)

|∇u− (∇u)x,r|2 dy = 0 ∀x ∈ B(x0, ρ).

Hence Ω0 is open and ∇u ∈ C0,αloc (Ω0). Since α ∈ (0, 1) was arbitrary this

proves (2.693).

170 [March 17, 2014]

To prove (2.718) let C2 be the constant in Lemma 2.84 and choose τ ∈(0, 1

4) such thatC2τ

2 ≤ τ2α.

By definition of Ω0 there exists r > 0 (with r < dist (x0, ∂Ω)) such that

U(x0, r) = –∫

B(x0,r)

|∇u− (∇u)x0,r|2 dy < ε2(τ).

Since x 7→ U(x, r) is continuous there exist ρ > 0 such that

U(x, r) < ε2(τ) ∀x ∈ B(x0, ρ).

Let x ∈ B(x0, ρ). Then by Lemma 2.84 we have U(x, τr) ≤ τ2αU(x, r) <ε(τ)2. Hence by induction

U(x, τkr) ≤ τ2kαε(τ)2 ∀k ∈ N.

Since s →∫B(x,s) |∇u − (∇u)x,s|2 dy is nondecreasing (see (2.92)) it follows

that

U(x, s) ≤ 1τn

(sr

)2αε(τ)2 = s2α

ε(τ)2

τnr2α.

Thus ∇u belongs to the Campanato space L2,n+2α(B(x0, ρ); RN×n) whichagrees with C0,α(B(x0, ρ); RN×n).

The higher regularity follows from Lemma 2.85 below and Schauder the-ory (see the beginning of Section 2.2).

Lemma 2.85. Assume that f ∈ C3(RN×n) and

|D2f | ≤M,

D2f(a⊗ b, a⊗ b) ≥ ν|a|2 |b|2 ∀A ∈ RN×n, a ∈ RN , b ∈ Rn

with ν > 0. Let Ω ⊂ RN be open and assume that u ∈ C1,αloc (Ω; RN ) is a

distributional solution of

−divDf(∇u) = 0.

Then u ∈ W 2,2loc (Ω; RN ) and each derivative w = ∂γu is a distributional

solution of−divD2f(∇u)∇w = 0. (2.719)

171 [March 17, 2014]

Remark. The assumptions f ∈ C3 and u ∈ C1,αloc can be weakened to

f ∈ C2 and u ∈ C1, respectively.

Proof. Homework.Hint: Let B(x, r) ⊂ Ω. It suffices to show that u ∈W 2,2(B(x, r/4); RN )

and that w is a weak solution of (2.719) in B(x, r/4).To prove this you may use that the identity

Df(G) −Df(F ) =∫ 1

0D2f((1 − s)F + sG)(G− F ) ds

implies that for h < r/2 the difference quotients

wh(x) :=w(x+ heγ) − w(x)

h

are weak solutions of

−div∫ 1

0D2f((1 − s)∇u(x) + s∇u(x+ heγ)) ds︸ ︷︷ ︸

=Ah(x)

∇wh = 0 (2.720)

in B(x, r/2) (see Homework 6, Problem 1). Now show that [Ah]C0,α(B(x,r/2)

is bounded independent of h, use the test function η2wh whereη ∈ C∞

c (B(x, r/2)) with η = 1 on B(x, r/4) and apply Garding’s inequality,Theorem 2.9 to obtain a uniform bound for ∥wh∥W 1,2(B(x,r/4);RN ). Finallypass to the limit h→ 0 in (2.720).

Removal of the condition that D2f is uniformly continuous

Corollary 2.86. Assume that f ∈ C2(RN×n) satisfies (2.690) and (2.688).Let u be a minimizer of I. Then there exists a closed set S such that

u ∈ C1,αloc (Ω \ S; RN ), Ln(S) = 0. (2.721)

Here we may take

S := x ∈ Ω : lim supr→0

–∫

B(x,r)

|∇u−(∇u)x,r|2 dy > 0 or lim supr→∞

–∫

B(x,r)

|∇u| dy = ∞.

(2.722)If, in addition, f ∈ C∞ then u ∈ C∞(Ω \ S; RN ).

To show this we use the following variant of the excess decay lemma.

Lemma 2.87. Assume that f satisfies (2.688) and (2.690). Then thereexists a constant C2 (which only depends on n, M and ν) with the property

172 [March 17, 2014]

that for each τ ∈ (0, 14) and each L > 0 there exists an ε(τ, L) such that for

every B(x, r) ⊂ Ω and every minimizer u of I the relations

U(x, r) ≤ ε(τ) and |(∇u)x,r| ≤ L (2.723)

imply thatU(x, τr) ≤ C2τ

2U(x, r). (2.724)

Proof. The proof is the same as for Lemma 2.84. Uniform continuity wasonly used to show (2.715). The assumptions now guarantee that |Am| ≤ L.SinceD2f is uniformly continuous on compact subsets we obtain (2.715).

Iteration of the excess decay lemma gives the following result.

Lemma 2.88. Let f and C2 be as in Lemma 2.87. Let α ∈ (0, 1) and letτ ∈ (0, 1

4) be such thatτ2α = C2τ

2.

For L > 0 letε(L) = min(ε(τ, 2L), (1 − τα)τnL)

Then for any minimizer u of I and any ball B(x, r) ⊂ Ω the relation

U(x, r) ≤ ε(L)2 and |(∇u)x,r| ≤ L (2.725)

imply that

U(x, τkr) ≤ τ2kαU(x, r), |(∇u)x,τkr| ≤ (2 − τ−kα)L. (2.726)

Proof. First note that for any ball B(x, ρ) ⊂ Ω

|(∇u)x,τρ − (∇u)x,ρ| ≤ –∫

B(x,τρ)

|∇u− (∇u)x,ρ| dy

≤ 1τn

–∫

B(x,ρ)

|∇u− (∇u)x,ρ| dy ≤ 1τnU1/2(x, ρ).

In particular the condition

U(x, ρ) ≤ τ2kαε(L)2

implies that

|(∇u)x,τρ − (∇u)x,ρ| ≤ τkα(1 − τα)L = [τkα − τ (k+1)α]L.

Now the lemma follows by induction from Lemma 2.87.

173 [March 17, 2014]

Proof of Corollary 2.86. Let x0 ∈ Ω \ S. Then

L := 2 lim supr→0

–∫

B(x0,r)

|∇u| dy <∞.

Thus there exists r0 > 0 such that

|(∇u)x0,r| < L ∀r ∈ (0, r0).

Moreover there exists r1 ∈ (0, r0) such that

U(x0, r1) < ε(L).

By continuity there exists a ρ > 0 such that

|(∇u)x,r1 | < L, U(x, r1) < ε(L) ∀x ∈ B(x0, ρ).

Now Lemma 2.88 implies that ∇u ∈ C0,α(B(x0, ρ); RN ). Hence Ω \ S isopen and ∇u ∈ C0,α

loc (Ω \ S; RN ).

Better estimates on the singular set of uniformly convex f

Corollary 2.89. Assume that f ∈ C2(RN×n) satisfies (2.687) and (2.688).Let u be a minimizer of I. Then there exists a closed set S and a q > 2 suchthat

u ∈ C1,αloc (Ω \ S; RN ), Hn−q(S) = 0. (2.727)

If, in addition, f ∈ C∞ then u ∈ C∞(Ω \ S; RN ).

Proof. By Meyers’ theorem there exists a p > 2 with the following property.If U ⊂⊂ Ω is open and compactly contained in Ω then ∇u ∈W 1,p(U ; RN×n).Now consider Uj ⊂⊂ Ω with

∪j Uj = Ω. Corollary 2.86 implies that ∇u ∈

C1,αloc (Uj \ Sj) where

Sj := x ∈ Uj : lim supr→0

–∫

B(x,r)

|∇u−(∇u)x,r|2 dy > 0 or lim supr→∞

–∫

B(x,r)

|∇u| dy = ∞.

Now by Lemma 2.90 we have Hq(Sj) = 0 for all q < p. Hence Hq(∪j Sj) =

0.

Lemma 2.90. Let U ⊂ Rn be open. Let g ∈ L2(U) and let ∇g ∈ Lp(U ; Rn)with 2 ≤ p ≤ n. Let

S1 := x ∈ U : lim supr→0

–∫

B(x,r)

|g − gx,r|2 dy > 0,

S2 := x ∈ U : lim supr→0

–∫

B(x,r)

|g| dy = ∞.

ThenHn−p(S1) = 0, Hn−q(S2) = 0 ∀q < p.

174 [March 17, 2014]

Proof. Homework.Hint: To estimate S1 for p < n use the Poincare inequality, Holder’s in-equality and Lemma 2.76 with ε = 1

k . For p = n the argument is slightlysimpler.To estimate S2 first show an estimate for gx,r/2 − gx,r in terms of ∇g andargue like in the proof that L2,n+2α = C0,α.

Generalizations This was not discussed in class. As we have seen inthe course PDE and modelling for systems (e.g., in nonlinear elasticity) oneoften wants to consider integrands f which satisfy

f(F ) ≥ |F |q − C

with q > n. This is incompatible with the assumption |D2f | ≤M . One canshow that partial regularity also holds under the weaker condition

|D2f(A)| ≤M(1 + |A|q − 2)

with q > 2 if one strengthens the quasiconvexity to condition to∫B(x,r)

f(A+ ∇φ) − f(A) dy ≥ ν

2

∫B(x,r)

(1 + |∇φ|q−2)|∇φ|2 dy

for all φ ∈ C1c (B(x, r); RN ), and all A, x, r > 0, see [Ev86]

The results can also be generalized to minimizer of functionals of theform ∫

Ωf(x, v(x),∇v(x)) dx,

see, e.g., [FH86, GM86] and there is a large literature on the regularity ofglobal and local minimizers.

2.4.3 Monotone operators and the Minty-Browder trick

This section was not discussed in class. Theorem 2.91 and Lemma 2.92 werethe subject of Homework 11, Problems 3 and 4.

Let Ω ⊂ Rn be bounded and open, let g ∈ L2(Ω; RN ) and assume that

f is convex

and that there exist c > 0 and C > 0 such that26

f(F ) ≥ c|F |2 − C ∀F ∈ RN×n. (2.728)26If f is C2 and Df2 ≥ νId with ν > 0 then (2.728) holds with c = ν/4. To see

this one can use that the function f(F ) = f(F ) − ν2|F |2 is convex, which implies that

f(F ) ≥ f(0) + L · F , and use Young’s inequality |F | ≤ ε2|F |2 + 1

2ε.

175 [March 17, 2014]

Consider the functional

I(v) =∫

Ωf(∇v) − g · v dx

It follows from the Poincare inequality that infW 1,2

0 (Ω;RN )I > −∞ and by

the direct method of the calculus of variations I has minimizer u among allmaps in W 1,2

0 (Ω; RN ). If, in addition, f ∈ C1(RN×n) and the exist M > 0and C > 0 such that

|Df(F )| ≤M |F | + C ∀F ∈ RN×n (2.729)

then the minimizer u is a weak solution of the PDE27

−divDf(∇u) = g, u ∈W 1,20 (Ω; RN ).

We now look for weak solutions of the equation

−div a(∇u) = g, u ∈W 1,20 (Ω; RN ).

and we want to identify sufficient condition on a which guarantee the exis-tence of a solutions.

One hint is the following observation. A function f ∈ C2(RN×n) isconvex if and only if Df is monotone

(Df(F ) −Df(G)) · (F −G) ≥ 0.

We now show that monotonicity and combination with suitable upper andlower bounds on a is sufficient for the existence of a solutions.

Theorem 2.91 (Existence for monotone operators). Assume thata : RN×n → RN×n is continuous and there exist c > 0, M > 0 and C > 0such that

(a(F ) − a(G)) · (F −G) ≥ 0 ∀F,G ∈ RN×n, (2.730)

a(F ) · F ≥ c|F |2 − C ∀F ∈ RN×n, (2.731)

|a(F )| ≤M |F | + C ∀F ∈ RN×n, (2.732)

Let Ω ⊂ Rn be open and bounded, let g ∈ L2(Ω).For k ∈ N let Xk be a finite dimensional subspaces of W 1,2

0 (Ω; RN ) suchthat Xk ⊂ Xk+1 for all k and

∪kXk is dense in W 1,2

0 (Ω; RN ). Equip W 1,20

(and the spaces Xk) with the scalar product (u, v) :=∫Ω ∇u · ∇v dx. Note

that by the Poincare inequality the corresponding norm is equivalent to theusual W 1,2 norm. Then

27If f ∈ C2 and |D2f(F )| ≤ M for all F then (2.729) follows with C = |Df(0)|. Ifwe only assume that f is convex then (2.729) can be deduced from the bound f(F ) ≤M ′|F |2 + C′ for all F

176 [March 17, 2014]

(i) there exists uk ∈ Xk such that∫Ωa(∇uk) · ∇φdx =

∫Ωg · φdx ∀φ ∈ Xk. (2.733)

Moreover ν∥uk∥ ≤ C∥f∥L2.

(ii) There exists u ∈W 1,20 (Ω; RN ) such that∫

Ωa(∇u) · ∇φdx =

∫Ωg · φdx ∀φ ∈W 1,2

0 (Ω). (2.734)

To prove (i) we use the following finite dimensional result.

Lemma 2.92. Let X be a finite dimensional Hilbert space and assume thatF : X → X is continuous and satisfies the following coercivity conditon:and that there exist c > 0 and C > 0 such that

F (x) · x ≥ c|x|2 − C. (2.735)

where a · b denotes the scalar product in H. Then for every y ∈ X theequation F (x) = y has a solution which satisfies ∥x∥2 ≤ ∥y∥2

c2+ 2C

c .

Proof. Exercise (this is very similar to Homework 11, Problem 3).Hint: Rewrite the equation as x = x− F (x) + y and use Young’s inequalityx · y ≤ c

2 |x|2 + 1

2c |y|2 to show that T (x) := x − F (x) + y satisfies the

assumptions for Theorem 1.6 if R2 = ∥y∥2

c2+ 2C

c .

Proof of Theorem 2.91. Homework 11, Problem 4.Hint for (i): Let u ∈ Xk. Use the Riesz representation theorem to showthat there exist gk ∈ Xk and Fk(u) ∈ Xk such that (gk, φ) =

∫Ω g ·φdx and

(F (uk), φ) =∫Ω a(∇u) · ∇φdx for all φ ∈ Xk. Then apply Lemma 2.92

Hint for (ii): To show existence note that a subsequence of the uk in (i)converges weakly to a limit u. Let v ∈W 1,2

0 (Ω). First show that∫Ωa(∇v) · (∇v −∇u) ≥

∫Ωg · (v − u) (2.736)

in the following way. Let vj ∈ Xj with vj → v in W 1,20 (Ω). Monotonicity

of a implies that∫Ω (a(∇vj) − a(∇(uk))) · (∇vj − ∇uk) ≥ 0. If j ≤ k that

yields ∫Ωa(∇vj) · (∇vj −∇uk) ≥

∫Ωg · (vj − uk) (2.737)

Now first take the limit k → ∞ and then the limit j → ∞. Finally let t > 0,consider v = u± tφ in (2.736), divide by t and let t ↓ 0.

To show that a(∇vj) → a(∇v) in L2 as j → ∞, one can use that for asubsequence ∇vj → ∇v a.e. and then use (2.732) and a suitable version ofthe dominated convergence theorem.

177 [March 17, 2014]

Alternatively note that to pass to the limit in (2.736) one can argueas follows. By (2.732) the sequence a(∇vj) is bounded in L2 and hence asubsequence has a weak limit a. Now by monotonicity for every w ∈ W 1,2

0

we have ∫Ω(a(∇w) − a(∇vj)) · (∇w −∇vj) dx ≥ 0.

Since ∇vj converges strongly we can pass to the limit and get∫Ω(a(∇w) − a) · (∇w −∇v) dx ≥ 0.

Now take again w = v ± tφ divide the inequality by t and take the limitt ↓ 0 to conclude that ∫

Ω(a(∇v) − a) · ∇φdx = 0

for all φ ∈W 1,20 (Ω; RN ).

Remark. (i) The Minty Browder trick consists in replacing the nonlinearequation (2.734) for u by the linear inequalities (2.736) for u. Thus one canpass to the limit in (2.736) using only weak convergence of u.(ii) If we assume the uniform monotonicity condition (a(F ) − a(G)) · (F −G)ν|F −G|2 with ν > 0 then the solution is unique.

The existence result can be easily extended to more general boundaryconditions. Let u0 ∈W 1,2(Ω; RN ). Then there exists a weak solution of

−div a(∇u) = g, u− u0 ∈W 1,20 (Ω; RN ).

Regularity Solutions of monotone PDE have the same regularity as min-imizers of convex integrands. Let u be a weak solution of

−div a(∇u) = 0

and assume a ∈ C1 and that

|a(F ) − a(G)| ≤ L|F −G|, (a(F ) − a(G)) · (F −G) ≥ ν|F −G|2 (2.738)

for all F,G ∈ RN×n where ν > 0.Then u ∈ C1,α

loc (Ω\S) where S is closed and Hn−q(S) = 0 for some q > 2.For the Cacciopoli inequality one starts from the identity∫

B(x,r)[a(∇u) − a(A)] · φ = 0

178 [March 17, 2014]

and uses the choice φ = ζ2(u − a − Ay) and monotonicity to obtain theestimate

ν

∫B(x,r)

ζ2|∇u−A|2 ≤∫B(x,r)

[a(∇u) − a(A)] · ζ2(∇u−A) dx

= −∫B(x,r)

[a(∇u) − a(A)] · 2ζ(u− a−Ay) ⊗∇ζ dx

and estimates the integrand on the right hand side by

2Lζ|∇u−A| |u− a−Ay||∇ζ| ≤ ν

2ζ2|∇u−A|2 +

2νL2|u− a−Ay|2|∇ζ|2.

The proof of excess decay by blow-up and the estimate for the size of thesingular set can then be carried out in the same way as for convex f .

Beyond Hilbert spaces The results so far for convex variational inte-grals and monotone elliptic PDE are modelled on functionals with quadraticgrowth ∫

Ω|∇u|2 dx

and linear equations−∆u = 0

and the natural setting is the Hilbert space W 1,20 (Ω).

In order to treat functionals like∫Ω|∇u|p dx

for 1 < p < ∞ one can extend the setting to reflexive Banach spaces andagain monotonicity easily leads to existence results, see [Li69], Chapitre 2.2and [KS], Chapter III.1.

[16.1. 2013, Lecture 23][20.1. 2014, Lecture 24]

2.5 Maximum principles and existence for quasilinear ellip-tic equations

2.5.1 Classcial maximum principles for linear equations

This subsection follows [GT], Chapter 3.1 and 3.2, but note that in contrastto [GT] we put a minus sign in front of the leading order

∑aij∂i∂ju.

So far we mostly considered estimates which can be derived by choosingsuitable test functions, scaling and interpolation. These estimates are ratherrobust and hold for higher (even) order equations and for systems (with the

179 [March 17, 2014]

exception of the results in Section 2.2). The maximum principle only appliesto scalar equations of second order, but for those it is a very powerful tool andin particular immediately give L∞ estimates. Before we study the maximalprinciple for general linear elliptic second order operators, let us recall twoversions of the maximum principle for −∆ which were already proved in thecourse Einfuhrung in PDG.

Let Ω be open and bounded and u ∈ C2(Ω) ∩ C(Ω). The maximumprinciple states that

−∆u ≤ 0 in Ω =⇒ u ≤ max∂Ω

in Ω. (2.739)

The strong maximum principle provides a strict inequality if u is nontrivial

−∆u ≤ 0 in Ω, Ω connected =⇒ u < max∂Ω

in Ω or u = const.

(2.740)The proof of the maximum principle was easy. If the conclusion is false

then M := maxΩ > max∂Ω. Thus u must have a maximum at x0 ∈ Ω. Sinceu in C2 this implies that

D2u(x0) ≤ 0,

i.e, the symmetric matrix D2u(x0) is negative semidefinite. Hence

∆u(x0) = trD2u(x0) ≤ 0.

This is not yet a contradiction to the hypothesis −∆u ≤ 0. To get a con-tradiction we let ε > 0 and consider

u(x) = u(x) + ε|x|2.

If ε is small enough then we still have maxΩ u > max∂Ω. Thus u has amaximum at a point xε ∈ Ω and

D2u(xε) ≤ 0. (2.741)

Now we have−∆u = −∆u− 2ε < 0

which leads to a contradiction with (2.741). This argument is very robust.It works without change if

∑ij aij(x)∂i∂ju(x) ≤ 0 in Ω as long as aij = aji

and the matrix aij is positive definite for each x ∈ Ω.The proof of the strong maximum principle was more delicate. We de-

fined the set

V = x ∈ Ω : u(x) : M, where M = maxΩ

u.

180 [March 17, 2014]

The set V is (relatively) closed in Ω since u is continuous. We then used themean value inequality for subharmonic functions

−∆u ≤ 0 in Ω =⇒ u(x) ≤ –∫

B(x,r)

u(y) dy for all balls B(x, r) ⊂ Ω.

(2.742)Since u ≤ M it follows that x ∈ V implies B(x, r) ⊂ V . Hence V is alsoopen. Now Ω is connected and therefore either V = ∅ (then u < M in Ω)or V = Ω (then u = M in Ω). This argument looks much more fragile since(2.742) does not hold for functions which satisfy −

∑ij aij(x)∂i∂ju(x) ≤ 0.

A more subtle replacement which serves the same purpose (for equations indivergence form) is the weak Harnack inequality in Theorem 2.45 (ii) appliedto v = M − u. Below we follow instead the classical argument via the Hopfboundary lemma which historically preceeds Theorem 2.45.

We now consider general second order operators of the form

Lu = −∑i,j

aij∂i∂ju+∑i

bi∂iu+ cu.

Since ∂i∂ju = ∂j∂iu we may and will assume that the coefficients of theleading part are symmetric

aij(x) = aji(x).

Remark. In [GT] the operator is written as Lu =∑

i,j aij∂i∂ju+

∑bi∂iu+

cu. I preferred to keep the minus sign in front of the highest order termbecause then subsolutions are given by Lu ≤ 0 and because of consistencywith earlier sections in the notes. To compare the results below with thosein [GT] one has to

replace L by −L, c by −c and b by −b.

We use the following notation.

• λ(x) is the smallest eigenvalue of the symmetric matrix (aij(x))

• Λ(x) is the largest eigenvalue of the symmetric matrix (aij(x)).

Thenλ(x)|ξ|2 ≤

∑i,j

aij(x)ξiξj ≤ Λ(x)|ξ|2 ∀ξ ∈ Rn

and λ(x) and Λ(x) are the largest and smallest numbers, respectively, forwhich this inequality holds. We say that

• L is elliptic at x ∈ Ω if λ(x) > 0;

181 [March 17, 2014]

• L is elliptic in Ω if λ > 0 in Ω;

• L is strictly elliptic if λ ≥ λ0 > 0 in Ω;

• L is uniformly elliptic if λ > 0 and Λ/λ is bounded.

Note that this terminology slighty differs from our earlier terminolgy. Whatwe called ’elliptic’ before now corresponds to strictly and uniformly ellip-tic. The finer distinctions of ellipticity will also be useful when we considernonlinear equations.

We usually need some mild control of the coefficients of the lower orderterms in terms of the aij . In particular we will always assume that thereexists a constant C such that

|b(x)|λ(x)

≤ C <∞ ∀x ∈ Ω. (2.743)

When L is strictly elliptic this conditions is in particular satisfied if b =(b1, . . . , bn) is bounded.

Theorem 2.93 (Maximum principle). Let L be elliptic in the bounded andopen set Ω and assume (2.743). Assume

c = 0.

Let u ∈ C2(Ω) ∩ C(Ω). Then

Lu ≤ 0 in Ω =⇒ maxΩ

u ≤ max∂Ω

u.

Remark. Added Jan 23, 2014Instead of (2.743) it suffices to assume that b/λ is bounded on every compactset K ⊂ Ω. Indeed we need to show that L > max∂Ω u implies u ≤ L. LetΩL := x ∈ Ω : u(x) > L. If ΩL is not empty the closure of ΩL is compactand contained in Ω. Hence we can apply the maximum principle in ΩL. ThusmaxΩL

u ≤ max∂ΩLu = L. This gives the desired contradiction ΩL = ∅.

Proof. Assume first Lu < 0 in Ω. If maxΩ u > max∂Ω u and u attains amaximum at x0 ∈ Ω, then

D2u(x0) < 0, ∇u(x0) = 0.

We now use the following fact about symmetric matrices. If A and B are

182 [March 17, 2014]

symmetric n× n matrices then28

A ≥ 0, B ≥ 0 =⇒∑i,j

AijBij ≥ 0.

Since L is elliptic and c = 0 this implies that Lu(x0) ≥ 0. This contradictsthe assumption.

Now we would like to reduce that case Lu ≤ 0 to the case Lu < 0. ForL = −∆ we could simply consider u = u + ε|x|2 because −∆|x|2 = −n.Now we no longer know that L|x|2 < 0 because we have no control on theterm |

∑i bi∂i|x|2| for large x. We thus seek a function φ such that Lφ < 0.

Consider first the one dimensional problem. Then the inequality reduces to

−aφ′′ + bφ′ < 0.

Setting ψ = φ′ this becomes −aψ′ + bψ < 0. Assuming ψ > 0 and theellipticity condition a > 0 this is equivalent to

ψ′

ψ>b

aor (lnψ)′ >

b

a. (2.744)

Now let us return to the multidimensional case. By the assumption (2.743)there exists a constant C such that |b1| ≤ Cλ and we have λ ≤ a11. Moti-vated by (2.744) we take φ = eγx1 . Then

Lφ = (−γ2a11 + γb1) eγx1 < 0 if γ ≥ 2C.

Finally let u = u+εφ with ε > 0. Then Lu < 0 and (since Ω is bounded)maxΩ u > max∂Ω u. Then we obtain a contradiction as shown above.

Corollary 2.94. [Upper bound for c ≥ 0] Let L be elliptic in the boundedand open set Ω and assume (2.743). Assume

c ≥ 0.

Let u ∈ C2(Ω) ∩ C(Ω). Then

Lu ≤ 0 in Ω =⇒ maxΩ

u ≤ max(max∂Ω

u, 0).

28Proof: If C is a symmetric matrix with C ≥ 0 then Cii ≥ 0 for each diagonal element.Thus the assertion is clear if A is diagonal. In general there exists Q ∈ SO(n) such thatD = QAQT is diagonal. Then A = QTDQ as QT = Q−1. Moreover

X

i,j

AijBij = trAB = trQTDQB = trDQBQT ≥ 0

since B′ := QBQT is symmetric and B′ ≥ 0.

183 [March 17, 2014]

Proof. This will be reduced to Theorem 2.93 by using that cu ≥ 0 if u ≥ 0.Let M := maxΩ u and assume that M > 0. Let m = max(max∂Ω u, 0) andlet U = x ∈ Ω : u(x) > 0. Then u ≤ m on ∂U . Let

L′u = −∑ij

aij∂i∂ju+∑i

bi∂iu.

ThenL′u = Lu− cu ≤ Lu ≤ 0 in U.

By the maximum in principle applied to L′ in the set U we get M =maxU u ≤ max∂U u ≤ m.

The maximum principle guarantees that u attains its maximum at aboundary point x0. To work towards the strong maximum principle we nowshow that (under suitable conditions) the normal derivative at x0 does notvanish.

Definition 2.95. We say that an open set Ω satisfies an interior spherecondition at x0 ∈ ∂Ω if there exists an open ball B ⊂ Ω with x0 ∈ ∂B.

Lemma 2.96 (Hopf boundary lemma). Assume L is uniformly elliptic with(2.743) and c = 0. Let u ∈ C2(Ω) and assume that Lu ≤ 0 in Ω. Letx0 ∈ ∂Ω be such that

(i) u is continuous at x0;

(ii) u(x0) > u(x) for all x ∈ Ω;

(iii) Ω satisfies an interior sphere condition at x0.

Then the outer normal derivative of u at x0, if it exists, satisfies the strictinequality

∂u

∂ν(x0) > 0.

Remark. (i) If ∂Ω does not have a tangent space at x0, the positivity ofthe outer normal derivative should be understood as follows. If B(y, r) ⊂ Ωis a ball with x0 ∈ ∂B(y, r) then

limt↓0

u(x0) − u(x0 − tv)t

> 0, where v =x0 − y

|x0 − y|.

If ∂Ω does not have a tangent space there may be more than one ’normal’vector v to which this reasoning can be applied but the importance is thatthere exists at least one.(ii) The interior sphere condition is important. Consider the first quadrantin R2, i.e., Ω = (x1, x2) : x1 > 0, x2 > 0 then u(x) = −x1x2 is harmonicin Ω, u < 0 in Ω, u(0) = 0 but ∇u(0) = 0. More generally let α > 1, letu = −Re zα. Then u is harmonic and u < 0 in Ωα = (r cosβ, r sinβ) : 0 <r <∞, 0 < β < π/α and ∇u(0) = 0.

184 [March 17, 2014]

Proof. We may assume without loss of generality that

u(x0) = 0. (2.745)

We try to ’squeeze in’ a function w for which we already know that ∂w∂ν > 0.

More precisely it suffices to find a function w with the following properties

(i) w(x0) = 0,

(ii) ∂w∂ν > 0,

(iii) u ≤ w on the segment (x0 − εν, x0) := x0 − tν : 0 < t < ε.

Then u− w ≤ 0 on (x0 − εν, x0) and hence

∂(u− w)∂ν

(x0) ≥ 0.

Thus ∂u∂ν (x0) ≥ ∂w

∂ν (x0) > 0.To satisfy (iii) we use the interior sphere condition. There exists a ball

B(y,R) ⊂ Ω with x0 ∈ ∂B(y,R). Instead of (iii) it will actually be easier toshow the stronger condition

u ≤ w in A := B(y,R) \B(y, ρ).

We first try to satisfy this relation on

∂A = ∂B(y,R) ∪ ∂B(y, ρ).

On ∂B(y, ρ) it is easy to satisfy the condition since u < 0 on ∂B(y, ρ) andthus

max∂B(y,ρ)

u < 0.

We also haveu ≤ 0 on ∂B(y,R).

Assume we have found w with u ≤ w on ∂A and w(x0) = 0. How do weget u ≤ w in A ?

The idea is to use the maximum principle. It suffices to show thatLw ≥ 0. Then L(u− w) = Lu− Lw ≤ 0 and hence u− w ≤ 0 in A.

Thus it suffices to find a function W ∈ C2(A) ∩ C(A) such that

W = 0 on ∂B(y,R), LW ≥ 0,∂W

∂ν> 0. (2.746)

Then min∂B(y,ρ)W > −∞ and we can take w = εW where

ε = max∂B(y,ρ)

u/ min∂B(y,ρ)

W.

185 [March 17, 2014]

It is natural to look for radially symmetric W . Take y = 0 and W (x) =ψ(|x|2) with

−ψ′′ ≥ 0, ψ′(R2) > 0, ψ(R2) = 0. (2.747)

Then with r = |x|

∂iW = 2ψ′(r2)xi, ∂j∂iW = 4ψ′′(r2)xixj + 2ψ′(r2)δij

LW = −∑i,j

4aijψ′′(r2)xixj − 2∑i

aiiψ′(r2) + 2ψ′(r2)∑i

bixi

≥− 4λr2ψ′′(r2) − 2nΛ|ψ′(r2)| − 2ψ′(r2)r|b|.

Try ψ(s) = −e−αs + e−αR2. If α is large enough then LW ≥ 0. Moreover

the conditions (2.747) are satisfied. Explicitly W is given by29

W (x) = −e−α|x|2 + e−αR2.

[20.1. 2014, Lecture 24][23.1. 2014, Lecture 25]

Theorem 2.97 (Strong maximum principle). Let Ω be open and connected(not necessarily bounded), let L be uniformly elliptic in Ω with (2.743) and

c = 0.

Assume that u ∈ C2(Ω) satisfies Lu ≤ 0. If u has an interior maximum inΩ then u is constant.

Proof. Assume that u is not constant and attains its maximum M at aninterior point of Ω. Let

Ω− := x ∈ Ω : u(x) < M.

Then Ω− is open and non-empty. Moreover ∂Ω− ∩ Ω = ∅ (otherwise Ω−

would be also relatively closed in Ω and thus Ω− = Ω contradicting theassumption that u(x) = M for some x in Ω).

Let y ∈ Ω− be a point with r = dist (y, ∂Ω−) < dist (y, ∂Ω). Then

B(y, r) ⊂ Ω−, B(y, r) ⊂ Ω and ∂B(y, r) contains a point x0 ∈ ∂Ω−.

Then u(x0) = M (otherwise x0 ∈ intΩ−).Now we can apply the Hopf boundary lemma to the open set Ω−. Thus

∂u∂ν (x0) > 0. But u attains a maximum at x0. Hence ∇u(x0) = 0. Thiscontradiction finishes the proof.

29[GT] takes v = −W and shows that u − u(x0) + εv ≤ 0 in A which yields again theassertion.

186 [March 17, 2014]

2.5.2 Maximum and comparison principles for quasilinear ellipticequations

This subsubsection follows [GT], Chapter 10. Again we take a minus signin front of the leading order part. Thus to compare the statements in thisand the following subsubsections with [GT] one needs to

replace L by −L and b by −b.

Notation and definitions We consider second order quasilinear opera-tors Qu of the form

Qu = −∑i,j

aij(x, u,∇)∂i∂ju+ b(x, u,∇u) with aji = aij (2.748)

where x is contained in an open set Ω ⊂ Rn and, unless otherwise stated, thefunction u belongs to C2(Ω). The coefficients of Q, namely the functions aij

and b are assumed to be defined for all values (x, z, p) in the set Ω×R×Rn.Two operators Q1 and Q2 of the form (2.748) will be called equivalent ifthere is a fixed positive function h on Ω × R × Rn such that aij2 = haij1 andb2 = hb1. Equations Qu = 0 for equivalent operators will also be calledequivalent.

For (x, z, p) ∈ Ω × R × Rn we define:

• λ(x, z, p) is the smallest eigenvalue of the matrix (aij)(x, z, p),

• Λ(x, z, p) is the largest eigenvalue of the matrix (aij)(x, z, p).

Then

λ(x, z, p)|ξ|2 ≤∑i,j

aij(x, z, p)ξiξj ≤ Λ(x, z, p)|ξ|2 ∀ξ ∈ Rn.

Let U ⊂ Ω × R × Rn. We say that

• Q is elliptic in U if λ > 0 in U ;

• Q is uniformly elliptic in U if, in addition, Λ/λ is bounded in U .

If Q is elliptic in Ω × R × Rn we simply say that Q is elliptic in Ω andwe use the same convention for uniform ellipticity. If u ∈ C1(Ω) andλ(x, u(x),∇u(x)) > 0 for all x ∈ Ω we say that Q is elliptic with respect tou.

We define a scalar function E by

E(x, z, p) =∑i,j

aij(x, z, p)pipj . (2.749)

187 [March 17, 2014]

If Q is elliptic in U then

0 < λ(x, z, p)|p|2 ≤ E(x, z, p) ≤ Λ(x, z, p)|p|2 ∀(x, z, p) ∈ U with p = 0.

The operator Q is of divergence form if there exists a differentiable func-tionB(x, z, p) and a differentiable mapA(x, z, p) = (A1(x, z, p), . . . , An(x, z, p)such that

Qu = −divA(x, u,∇u) +B(x, u,∇u) for u ∈ C2(Ω), (2.750)

i.e., in (2.748),

aij(x, z, p) =12

(∂Aj

∂pi(x, z, p) +

∂Ai

∂pj(x, z, p)

).

Unlike the case of linear operators, a quasilinear operator with smooth co-efficients is not necessarily expressibly in divergence form.

The operator Q is variational if the equation Qu = 0 is the Euler-Lagrange equation of a variational integral∫

ΩF (x, u,∇u) dx

where F : Ω × R × Rn → R is a differentiable function. In this case Q is ofdivergence form and

Ai(x, z, p) =∂F

∂pi(x, z, p), B(x, z, p) =

∂F

∂z(x, z, p).

The ellipticity of of Q is equivalent to strict convexity of F as a function ofp (more precisely to the conditions that D2

pf is positive definite.

Examples (i) The minimal surface equation.This is the Euler-Lagrange operator for the area integral∫

Ω

√1 + |∇u|2 dx.

Thus

Qu = − div∇u√

1 + |∇u|2= −

∑i

∂i∂iu√

1 + |∇u|2

=1√

1 + |∇u|2

−∆u+∑ij

∂iu∂ju

1 + |∇u|2∂i∂ju

Thus Q is equivalent to the operator

Q = −∆u+∑ij

∂iu∂ju

1 + |∇u|2∂i∂ju

188 [March 17, 2014]

Here

aij(x, z, p) = δij − pipj1 + |p|2

=1

1 + |p|2[(1 + |p|2)δij − pipj

].

Thus the minimal and maximal eigenvalue of aij are given by

λ(x) =1

1 + |p|2, Λ(x) = 1.

Hence Q (and thus Q) is elliptic in Ω, but not uniformly elliptic. MoreoverQ is uniformly elliptic on every set Ω × R ×B(0, R). We have

E(x, z, p) =|p|2

1 + |p|2

Geometrically div ∇u√1+|∇u|2

is the sum of the principle curvatures κi of

the graph of u, i.e., of the n dimensional surface (x, u(x) : x ∈ Ω. Wedefine the mean curvature30 of an n dimensional hypersurface as

H =1n

n∑i=1

κi

Surfaces with H = 0 are called minimal surfaces and thus the equationQu = 0 is called the minimal surface equation.

Here the sign convention is as follows: we choose the sign of the normalsuch that νn+1 > 0. For a codimension 1 surface Γ with unit normal ν theprincipal curvatures are defined as follows. For a point x ∈ Γ and η in thetangent space TxΓ define the operator Bx by Bxη = −η · ∇ν. Then oneeasily sees that Bx maps TxΓ to itself and is symmetric (with respect to thescalar product on Tx which is obtained by restricting the Euclidean scalarproduct in Rn+1 to Tx). The principal curvatures κi(x) are the eigenvaluesof Bx. The sign convention is such that if Γ is the boundary of a boundedand open convex set U then the principle curvatures of Γ are non-negativeif we n choose ν as the inner normal. For further details see, e.g., [GT],Chapter 14.6.

Example Let Γ = (x, u(x)) : x ∈ B(0, 1) be the graph of a scalarfunction u. Let

ν =1√

1 + |∇u|2(−∇u, 1)

be the unit normal of Γ. If ∇u(0) = 0 then TxΓ = Rn×0 andBx(0) = D2u.In particular the principle curvatures are the eigenvalue ofD2u and the meancurvature at 0 is H(0) = 1

ntrD2u = 1n∆u.

30Some authors also use the definition H =Pn

i=1 κi even though the word mean cur-vature suggests to divide by n

189 [March 17, 2014]

(ii) Prescribed mean curvature equation.Let H : Ω → R be given. The equation of prescribed mean curvature (ormean curvature equation for short) is

−div∇u√

1 + |∇u|2= −nH (2.751)

This is equivalent to

Mu := −(1 + |∇u|2)∆u+∑ij

∂iu∂ju ∂i∂ju = −nH(1 + |∇u|2)3/2

For the operator M we have

λ(x, z, p) = 1, Λ(x, z, p) = 1 + |p|2, E(x, z, p) = |p|2

The mean curvature equation is the Euler-Lagrange equation of the func-tional

I(v) =∫

Ω

√1 + |∇u|2 dx+ n

∫ΩHudx.

The first integral is the area of the graphu and for constant H the secondterm is H times the (oriented) volume of the set bounded by graphu andΩ × 0. If H is positive and u = 0 on ∂Ω then the maximum principleimplies that u < 0 and∫

Ω(−u) dx = Ln+1(x, s) : x ∈ Ω, u(x) < s < 0.

Comparison principle

Theorem 2.98 (Comparison theorem for quasilinear elliptic operators). LetΩ be open and bounded. Assume that u, v ∈ C2(Ω) ∩ C(Ω) satisfy

Qu ≤ Qv in Ω, u ≤ v on ∂Ω, (2.752)

where

(i) the operator Q is elliptic with respect to u or v;

(ii) the coefficients aij are independent of z;

(iii) the coefficient b is non-decreasing in z;

(iv) the coefficients aij and b are continuously diffentiable with respect tothe p variable in Ω × R × Rn.

Thenu ≤ v in Ω. (2.753)

If Qu < Qv in Ω and conditions (i)–(iii) hold, (but not necessarily (iv)), wehave strict inequality u < v in Ω.

190 [March 17, 2014]

Proof. Let us assume that Q is elliptic with respect to u. The idea is toshow that L(u − v) ≤ 0 whenever u − v ≥ 0 where L is a suitable linearelliptic operator. We have

0 ≥ Qu−Qv =∑i,j

−aij(x,∇u)∂i∂j(u− v) − [aij(x,∇u) − aij(x,∇v)]∂i∂jv

+ b(x, u,∇u) − b(x, u,∇v) + b(x, u,∇v) − b(x, v,∇v)︸ ︷︷ ︸≥0 if u−v≥0

.

(2.754)

For a continuously differentiable function g : Rn → R we have the followingversion of the mean value theorem

g(p) − g(q) = l · (p− q), with l =∫ 1

0∇g((1 − t)q + tp) dt. (2.755)

Thus writing

w = u− v,

aij(x) = aij(x,∇u(x)),

−∑i,j

[aij(x,∇u) − aij(x,∇v)]∂i∂jv + b(x, u,∇u) − b(x, u,∇v) =∑i

bi(x)∂iw

we see that

Lw := −aij(x)∂i∂jw + bi∂iw ≥ 0 on Ω+ = x ∈ Ω : w(x) > 0

andw ≤ 0 on ∂Ω.

Note that the functions bi(x) are bounded on every compact set K ⊂ Ω.This follows from (2.755) and assumption (iv) because u, v, ∇u, ∇v arebounded on compact subsets of Ω. Similar the continuity of λ implies thatminK λ > 0. Therefore we can apply the linear maximum principle, Theorem2.93, and the remark following to L and Ω+ and we get w ≤ 0 in Ω+ andhence w ≤ 0 in Ω.

This paragraph was corrected on Jan 27, 2014. If Qu < Qv andu − v has a maximum at x0 with u(x0) ≥ v(x0) then ∇u(x0) = ∇v(x0),D2(u− v)(x0) ≤ 0 and thus by conditions (i) and (ii)

−∑i.j

aij(x0,∇u(x0)) (∂i∂ju)(x0) ≥ −∑i.j

aij(x0,∇u(x0)) (∂i∂jv)(x0)

= −∑i.j

aij(x0,∇v(x0))(∂i∂jv)(x0).

191 [March 17, 2014]

Condition (iii) implies that

b(x0, u(x0),∇u(x0)) ≥ b(x, v(x0),∇u(x0)) = b(x0, v(x0),∇v(x0)).

Thus (Qu)(x0) ≥ (Qv)(x0) which contradicts the assumption Qu < Qv inΩ.

If Q is elliptic at v then we exchange the roles of u and v and similarlyfind a linear elliptic operator L such that L(v − u) ≥ 0 if v − u ≤ 0. Thuswe get again L(u− v) ≤ 0 if u− v ≥ 0 and we conclude as before.

Theorem 2.99. Let u, v ∈ C2(Ω) ∩ C(Ω) satisfy

Qu = Qv in Ω, u = v on ∂Ω

and suppose that conditions (i) to (iv) in Theorem 10.1 hold. Then u = v.

[23.1. 2014, Lecture 25][27.1. 2014, Lecture 26]

Maximum principles Assume that Q is elliptic, u ∈ C2(Ω) ∩ C(Ω) and

Qu ≤ 0.

We look for a bound for an upper bound for u of the form

maxΩ

u ≤ max∂Ω

u+ + C.

A natural idea is to prove an upper bound for u is to use the comparisontheorem, Theorem 2.98, and to construct a function v with Qv ≥ 0 in Ω andv ≥ u on ∂Ω. Then u ≤ v if the assumptions of the comparison theoremholds.

One drawback of the comparison is that the coefficients aij are not al-lowed to depend on z and b has to be non-decreasing in z. To free ourselvesfrom these restrictions we consider a new operator where the z dependenceis ’frozen’. We define an quasilinear operator

Qv := −∑ij

aij(x, u,∇v)∂i∂jv + b(x, u,∇v). (2.756)

More explicitly the coefficients aij and b of Q are defined by

aij(x, p) = aij(x, u(x), p), b(x, z, p) = b(x, u(x), p).

Then Q satisfies the assumptions (i)–(iii) of the comparison theorem. Notethat

Qu = Qu ≤ 0.

192 [March 17, 2014]

Thus we look for functions v with

0 < Qv = −∑ij

aij(x, u(x),∇v(x))∂i∂jv + b(x, u(x),∇v).

In the proof of the maximum principle for linear operators we have seenthat the function v(x) = −eξ·x is a good candidate (there we looked forsubsolutions and took the + sign rather than the minus sign). With thenotation q = −eξ·xξ this gives

Qv =∑ij

aij(x, u(x),−eξ·xξ)eξ·xξiξj + b(x, u(x),−eξ·xξ)

=∑ij

aij(x, u(x), q)e−ξ·xqiqj + b(x, u(x), q)

= E(x, u(x), q)e−ξ·x + b(x, u(x), q)

= e−ξ·x(E(x, u(x), q) + b(x, u(x), q)

|q||ξ|

)This suggests to impose the following structural condition

E(x, z, p) ≥ −µb(x, z, p)|p| if z ≥ 0

for some µ > 0. Then taking |ξ| > µ we get Qv > 0. The following theoremprovides an upper bound for u under a slightly more general condition.

Theorem 2.100 (Maximum principle). Let Q be elliptic in Ω and supposethat there exist µ1 ≥ 0 and µ2 ≥ 0 such that

b(x, z, p)E(x, z, p)

≥ −µ1|p| + µ2

|p|2∀(x, z, p) ∈ Ω × [0,∞) × Rn. (2.757)

Then if u ∈ C2(Ω) ∩ C(Ω) satisfies

Qu ≤ 0 in Ω,

we havemax

Ωu ≤ max

∂Ωu+ + Cµ2

where u+ = max(u, 0) and C = C(µ1,diamΩ).

Remark. By applying this argument also to −u we obtain the followingresult. If

b(x, z, p)sgn zE(x, z, p)

≥ −µ1|p| + µ2

|p|2∀(x, z, p) ∈ Ω × R × Rn

thenQu = 0 =⇒ max

Ω|u| ≤ max

∂Ω|u| + Cµ2

193 [March 17, 2014]

Remark. The following example suggests31 that some restriction on thegrowth of b is needed. For the simple choice aij(x, z, p) = δij and b(x, z, p) =−|p|2 we get the equation

−∆u− u|∇u|2 = 0

and we have seen that this equation unbounded solution ln ln 1|x| inB(0, 1/2) ⊂

R2.

Proof. As above set

Qv = aij(x, u,∇v)∂i∂jv + b(x, u,∇v).

Assume thatΩ ⊂ x ∈ Rn : 0 < x1 < d

and µ2 > 0 and choose the comparison function

v(x) = max∂Ω

u+ + µ2(eαd − eαx1)

with α ≥ µ1 + 1. Using (2.757) we get in Ω+ = x ∈ Ω : u(x) > 0

Qv = µ2α2a11(x, u,∇v)eαx1 + b(x, u,∇v)

≥ e−αx1

µ2E(x, u,∇v)

(1 − µ1

α− e−αx1

α2

)> 0 ≥ Qu.

Hence, by Theorem 2.98, we have u ≤ v in Ω+ and hence in Ω. The resultfor µ2 = 0 follows be letting µ2 tend to zero.

The mean curvature equation Mu = −nH(x)(1 + |∇u|2)3/2 does notsatisfy the structure condition (2.757). Indeed in this case

E(x, z, p)p = |p|2, b(x, z, p) = nH(x)(1 + |p|2)3/2 ∼ |p|3 as p→ ∞.

Indeed, one can obtain an bound on supu only under a natural smallnesscondition on H (which arises by comparison with half-spheres).

Theorem 2.101. Let Ω be an open and bounded set and assume that Ω ⊂B(y,R) for some y ∈ Rn and R > 0. Let H ∈ C0(Ω) and assume that

sup |H| ≤ 1R.

Assume that u ∈ C2(Ω) ∩ C(Ω) satisfies

−div∇u√

1 + |∇u|2= −nH. (2.758)

Thenmax

Ω|u| ≤ max

Ω|u| +R.

31I say ’suggests’ and not ’proves’ because the singular solution below is not C2 so doesnot strictly fall under the theory considered here

194 [March 17, 2014]

Remark. (i) The result is optimal in the following sense. If Ω = B(y,R)and if H ≤ −A or H ≥ A with A > 1

R then there exists no C2 solutionof (2.758) which is bounded in Ω, see Homework 12, Problem 3 (in fact byconsidering a slight smaller ball B(y,R′) with 1

A < R′ < R we see that thereexist no C2 solution of (2.758) in Ω).(ii) For domains which are not balls L∞ bounds for u can be obtained underthe weaker condition

supΩ

|H| <(Ln(B(0, 1)Ln(Ω)

)1/n

,

see [GT], Theorem 10.10 and (10.32). It even suffices to assume that∫Ω|H|n dx < Ln(B(0, 1)),

see [GT], (10.35) or Corollary 10.6.

Proof. It suffices to show the upper bound u ≤ max∂Ω u+R. Then the fullassertion follows by applying this estimate also to −u. For the proof of theupper bound see Homework 12, Problem 2. Hint: Let

v(x) =√R2 − |x|2 + sup

∂Ωu.

A short calculation (or a geometric argument) shows that

−div∇v√

1 + |∇v|2=n

R

Now use the comparison principle.

2.5.3 Existence of classical solutions

Let α ∈ (0, 1) Assume that Ω is a bounded open set and

∂Ω is of class C2,α, φ ∈ C2,α(Ω). (2.759)

We seek a solution u ∈ C2,α(Ω) of the boundary value problem

Qu = 0 in Ω, u = φ on ∂Ω.

We assume that the coefficients satisfy

aij ∈ C0,α(Ω×(−R,R)×B(0, R)), b ∈ C0,α(Ω×(−R,R)×B(0, R)) ∀R > 0.(2.760)

In particular the coefficients can be extended to continuous functions onΩ ×R× Rn. We assume that

Q is elliptic in Ω, (2.761)

195 [March 17, 2014]

and that i.e. the lowest eigenvalue of the matrix of the extended coefficients(aij) satisfies λ > 0 in Ω × R × Rn.

To find such a solution we rewrite the boundary value problem for uas a fixed point problem involving the solution of a linear boundary valueproblem. Fix the boundary datum φ ∈ C2,α(Ω). Given u ∈ C1,β(Ω) forsome β > 0 define the linear partial differential operator

Lv = −∑i,j

aij(x, u,∇u)∂i∂jv + b(x, u,∇u). (2.762)

Then the coefficients aij(x) := aij(x, u(x),∇u(x)) and b(x) := b(x, u(x),∇u(x))are in C0,αβ(Ω). Moreover the lowest eigenvalue λ(x) of (aij) satisfies λ > 0in Ω. Since λ is continuous this implies that there exists νu > 0 such thatλ ≥ νu in Ω. In fact it follows from (2.760) that there exists νR such that

λ ≥ νR > 0 in Ω ∀u ∈ C1(Ω) with ∥u∥C1(Ω) ≤ R.

We now use the following existence and uniqueness results for linearelliptic equations in non divergence form.

Theorem 2.102. Let Ω ⊂ Rn be a open and bounded with C2,α boundary,let f ∈ C0,α(Ω) and φ ∈ C2,α(Ω). Assume that aij ∈ C0,α(Ω) and that thereexists ν > 0 and M > 0 such that

∥aij∥C0,α(Ω) ≤M,∑ij

aij(x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

Then there exists one and only one u ∈ C2,α(Ω) such that

−∑ij

aij(x)∂i∂ju = f in Ω, u = φ on ∂Ω

Moreover there exists a constant C(Ω, ν,M, n) such that

∥u∥C2,α(Ω) ≤ C(Ω, ν,M, n)(∥f∥C0,α(Ω) + ∥φ∥C2,α(Ω)

).

A proof is given below. The corresponding result for divergence formoperators is Theorem 2.30.

If u in C1,β(Ω) and L is given by (2.762) then this result (with α replacedby γ = αβ) yields the existence and uniqueness of a solution v of the problem

Lv = 0 in Ω, v = φ on ∂Ω (2.763)

Define the operator T by Tu := v. (the solution also depends on φ but wesuppress this dependence in the notation because φ has been fixed).

Then u is a solution of the quasilinear boundary value problem if andonly if

u = Tu

196 [March 17, 2014]

Moreover the equationu = tTu

with 0 < t < 1 is equivalent to u/t = Tu or, more explicitly to

Lu

t= 0 in Ω,

u

t= φ on ∂Ω.

This can be rewritten as∑i.j

aij(x, u,∇u)∂i∂ju+ tb(x, u,∇u) = 0 in Ω, u = tφ on ∂Ω. (2.764)

Now we will use the Leray-Schauder fixed point theorem, Theorem 1.7, toshow the following result.

Theorem 2.103. Let α ∈ (0, 1), let Ω ⊂ Rn be an open and bounded set.Assume that (2.759), (2.760) and (2.761) hold. Assume further that thereexist β ∈ (0, 1) and C > 0 such that

∥ut∥C1,β(Ω) ≤ C

whenever ut is a solution of the boundary value problem (2.764). Then thereexists a solution of the boundary value problem

Qu = 0 in Ω, u = φ on ∂Ω.

Proof. We first show that the operator T is a continuous and compact mapfrom C1,β(Ω) to itself. Indeed if u is in a bounded subset B of C1,β(Ω)the coefficients aij and b are in a bounded set of C0,αβ(Ω). By assumptionλ > 0 in Ω × R × Rn. Since λ is a continuous function it is strictly positiveon compact subsets of Ω × R × Rn and thus there exists a ν > 0 suchthat λ(x, u(x),∇u(x) ≥ ν > 0 for all u ∈ B, x ∈ Ω. Thus the set TB isbounded in C2,αβ(Ω) and hence precompact in C2(Ω) and C1,β(Ω) by theArzela-Ascoli theorem. Thus T is compact.

To see that T is continuous assume that uk → u in C1,β and set vk = Tuk.Then

−∑i,j

aij(x, uk,∇uk)∂i∂jvk + b(x, uk,∇uk) = 0

and vk = 0 on ∂Ω By compactness there exists a subsequence such thatvk → v in C2(Ω). Thus

−∑i,j

aij(x, u,∇u)∂i∂jv + b(x, u,∇u) = 0.

and v = 0 on ∂Ω. Thus v = Tu. Since v is uniquely determined it followsthat the whole sequence vk converges to Tu. Hence T is continuous.

197 [March 17, 2014]

Thus the Leray-Schauder fixed point theorem, Theorem 1.7, implies thatthere exists a u ∈ C1,β(Ω) such that

u = Tu.

Above we have shown that Tu ∈ C2,αβ(Ω). Thus u ∈ C2,αβ(Ω). It followsthat aij , b ∈ C0,α(Ω). Thus, using Theorem 2.102 for the linear boundaryvalue problem (2.763) we get Tu ∈ C2,α(Ω).

Classical solutions for linear non divergence form equationsThis was not discussed in class. Here we prove Theorem 2.102.

Proof. Replacing u by u− φ and f by f +∑

ij aij(x)∂i∂jφ we may assume

that φ = 0. Now the estimate for u follows from Lemma 2.104 below. Thisimplies in particular uniqueness. To show existence we consider the affinefamily for linear differential operators Lt

Ltv = −∑ij

aijt (x)∂i∂ju, with at(x) = (1 − t)δij + taij(x).

Then L0 = −∆ is an invertible map from X = u ∈ C2,α(Ω) : u = 0 on ∂Ωto Y = C0,α(Ω). Indeed existence of a weak solution follows from the Lax-Milgram theorem and the C2,α regularity was obtained in Theorem 2.30.By Lemma 2.104 below we have

∥v∥X ≤ C(Ω, ν ′,M ′, n)∥Ltv∥Y ∀t ∈ [0, 1].

where ν ′ = min(ν, 1), M ′ = max(M, 1). Then by Lemma 1.1 the operatorsLt are invertible for all t ∈ [0, 1]. Invertibility of L1 implies the existence ofa solution u ∈ X for every f ∈ Y .

Lemma 2.104. Let Ω ⊂ Rn be a open and bounded with C2,α boundary, letν > 0, M > 0. Then there exists a constant C(Ω, ν,M, n) with the followingproperty. If f ∈ C0,α(Ω) and if u ∈ C2,α(Ω) satisfies

−∑ij

aij(x)∂i∂ju = f in Ω, u = 0 on ∂Ω

where

∥aij∥C0,α(Ω) ≤M,∑ij

aij(x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

Then∥u∥C2,α(Ω) ≤ C(Ω, ν,M, n)∥f∥C0,α(Ω).

198 [March 17, 2014]

Proof. By a blow-up argument we get the key estimate

[D2u]0,α ≤ C ′(Ω, ν, ∥aij∥0,α, n)(∥u∥C2(Ω) + [f ]0,α

). (2.765)

The proof follows exactly the same lines as the proof of Theorem 2.31.For the convenience of the reader we outline some of the details. If theestimate does not hold there exist uk, fk, a

ijk such that

[D2uk]α = 1, ∥fk∥α → 0, ∥uk∥C2 → 0,

∥aijk ∥α ≤M,∑i,j

aijk ξiξj ≥ ν|ξ|2.

In particular there exist xk, yk with yk − xk → 0 such that

|D2uk(xk) −D2uk(yk)| ≥12|xk − yk|α.

Case 1. Assume that lim supk→∞ r−1k dist (xk, ∂Ω) = ∞.

Upon passage to a subsequence we may assume that r−1k dist (xk, ∂Ω) → ∞.

We have

−∑i,j

aijk (xk)∂i∂juk =∑i,j

[aijk (x) − aijk (xk)

]∂i∂juk + fk.

Since −∑

i,j aijk (xk)∂i∂juk(xk) = fk(xk) this can be rewritten as

−∑i,j

aijk (xk) [∂i∂juk − ∂i∂juk(xk)] =∑i,j

[aijk (x) − aijk (xk)

]∂i∂juk+[fk−fk(xk)].

(2.766)Define rk = |yk − xk| and

Uk(z) =1

r2+αk

(uk(xk + rkz − uk(xk) − rk(∇uk)(xk) · z − r2k

12D2uk(xk)(z, z)

).

Aijk (z) = aijk (xk + rkz), Fk(z) =1rαk

(f(xk + rkzk) − f(xk)).

ThenD2Uk(z) =

1rαk

(D2uk(xk + rkz) −D2u(xk)

),

[D2Uk]α ≤ 1, Uk = 0, ∇Uk(0) = 0, D2Uk(0) = 0, (2.767)

[Aij ]C0,α ≤Mrαk , Fk(z) ≤ [fk]C0,α |z|α.

199 [March 17, 2014]

Dividing (2.766) by rαk we get

−∑i,j

aijk (xk)∂i∂jUk =∑i,j

Aijk (z) −Aijk (0)rαk

[(∂i∂ju)(xk) + rαk ∂i∂jUk] + Fk

(2.768)Now by assumption D2uk → 0 uniformly and it follows from the Arzela-Ascoli theorem and (2.767) that (for a subsequence) Uk → U in C2

loc(Rn).Moreover |Aijk (z) − Aijk (0)| ≤ Mrαk |z|α and aijk (xk) → aij . Finally Fk → 0locally uniformly an thus

−∑i,j

aij∂i∂jU = 0 in Rn. (2.769)

Moreover it follows from (2.767) that |D2U(z)| ≤ C|z|α. Now every secondderivative ∂k∂lU satisfies the same equation and thus application of theestimate (2.78) with u = ∂k∂lU and passage to the limit r → ∞ imply thatD3U = 0 and thus D2U = 0 since D2U(0) = 0. On the other hand wehave (again for a subsequence) zk := r−αk (yk − xk) → z ∈ Sn−1 and the C2

loc

convergence of Uk implies that

|D2U(z)| = limk→∞

|D2Uk(zk)| = limk→∞

r−αk |D2u(yk) −D2u(xk)| ≥12.

This contradiction finishes the proof in Case 1.

Case 2. Assume that lim supk→∞ r−1k dist (xk, ∂Ω) <∞.

The xk be the point on ∂Ω which is closest to xk (for large enough k this pointis even uniquely determined since ∂Ω is of class C2). Then by assumption|xk − xk| ≤ Crk. Define as before rk = |yk − xk| and set

Uk(z) =1

r2+αk

(uk(xk + rkz − uk(xk) − rk(∇uk)(xk) · z − r2k

12D2uk(xk)(z, z)

),

Aijk (z) = aijk (xk + rkz), Fk(z) =1rαk

(f(xk + rkzk) − f(xk)).

By a translation and rotation we may assume that xk = 0 and that theouter normal to Ω at xk is −en. Since ∂Ω is of class C2,α there exists aψ ∈ C2,α(Rn−1) such that

Ω ∩B(0, ε) = (x′, xn) ∈ B(0, ε) : xn > ψ(x′).

and ∇ψ(0) = 0. Let

Ψk(z′) =1rkψ(rkz′), then [D2Ψ]C0,α ≤ Cr1+αk .

200 [March 17, 2014]

Now we can again use the Arzela-Ascoli theorem to obtain again a subse-quence such that

Uk → U in C2(Ωk ∩B(0, R)) for every R > 0

whereΩk = (z′, zn) : zn > Ψk(z′).

In particular C2loc(Rn

+) since Ψk → 0 locally uniformly. Moreover as beforeFk → 0 in Ωk ∩B(0, R)) and we get

−∑i,j

aij∂i∂jU = 0 in Rn+.

We now show thatU = 0 on ∂Rn

+. (2.770)

Indeed we have

Uk(z′,Ψk(z′)) =1

r2+αk

uk(x′, ψ(x′))︸ ︷︷ ︸=0

−Tk(z′,Ψ(z′))

,

where Tk is the second order Taylor polynomial of uk at xk = 0. To estimatethe Taylor polynomial we differentiate the relation uk(x′, ψ(x′)) twice anduse that ∇ψ(0) = 0. This gives the relations

∇uk(0) · (z′, 0) = 0, D2uk(0)(z′, z′) +∂uk∂xn

(0)D2ψ(0)(z′, z′) = 0. (2.771)

We also have ∇Ψ(0) = 0, |D2Ψ| ≤ Crk and [D2Ψ]C0,α ≤ Cr1+αk . Thisimplies

|Ψ(z′)| ≤ Crk|z′|2, |Ψ(z′) − 12D2Ψ(0)(z′, z′)| ≤ Cr1+αk |z′|2. (2.772)

Thus

Tk(z′,Ψ(z′)) = rk∂uk∂xn

(0)Ψ(z′) +12r2kD

2uk(0)(z′, z′)

+ r2kD2uk(0)(z′,Ψ(z′)en) +

12r2kD

2uk(0)(Ψ(z′)en,Ψ(z′)en)︸ ︷︷ ︸≤C(R)r3k if |z′|≤R

Using the second identity in (2.771) and the second estimate in (2.772) weget for |z′| ≤ R

Tk(z′,Ψ(z′)) ≤ C

∣∣∣∣∂uk∂xn(0)∣∣∣∣ r2+αk + Cr3k.

201 [March 17, 2014]

Now by assumption sup |∇uk| → 0 and it follows that

Uk(z′,Ψk(z′)) → 0 uniformly for z′ ∈ B(0, R)

Since Uk converges uniformly in B(0, R)∩Ωk and Ψk → 0 in B(0, R) we get(2.770).

Now we know that |DU(z)| ≤ C|z|1+α (since DUk(0) = D2Uk(0) = 0and [D2Uk]C0,α ≤ 1). Let k ∈ 1, . . . , n − 1. Then we can apply the halfspace estimate (2.191) to any tangential derivative u = ∂kU with λ = 0.This yields shows that D2∂kU = 0 and thus ∂j∂kU is constant for all j ≤ nand all k ≤ n− 1. By the PDE for U we get that ∂n∂nU is constant, too.Since D2Uk converges uniformly to D2U in B(0, R)∩Ωk this again leads toa contradiction with the assumption |D2uk(yk) −D2uk(xk)| ≥ 1

2 |xk − yk|2.This finishes the proof of (2.765)

Given (2.765) it thus suffices to show that

∥u∥C2(Ω) ≤ C ′(Ω, ν, ∥aij∥0,α, n)∥f∥0,α. (2.773)

Assume that this estimate does not hold. Since the equation for u is linearthen there exist uk, fk, a

ijk such that

∥uk∥C2(Ω) = 1, ∥fk∥0,α → 0,

and∥aijk ∥0,α ≤M,

∑ij

aijk (x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

By (2.773) and the Arzela-Ascoli theorem it follows that for a subsequenceuk → u in C2(Ω). Moreover by assumption and the Arzela-Ascoli theoremfk → 0 uniformly and aijk → aij uniformly. In particular

∑ij a

ij(x)ξiξj ≥ν|ξ|2. Hence ∑

i,j

aij∂i∂ju = 0 in Ω, u = 0 on ∂Ω

Now by the maximum principle applied to ±u we get u = 0 in Ω. Hence∥uk∥C2(Ω) → 0. This contradicts the assumption ∥u∥C2(Ω) = 1.

[27.1. 2014, Lecture 26][3.2. 2014, Lecture 27]

2.5.4 C1,α estimates

We aim at proving global Holder estimates for the gradient of solutions ofquasilinear equations in divergence form. We follow [GT, Chapter 12] andassume that Ω ⊂ Rn is open, ∂Ω ∈ C2,α, and that Q is (equivalent to) an

202 [March 17, 2014]

elliptic operator in divergence form. Precisely, there are A ∈ C1(Ω × R ×Rn; Rn) and B ∈ C0(Ω × R × Rn) such that for u ∈ C2(Ω),

Qu := −divA(x, u,∇u) +B(x, u,∇u). (2.774)

Assume that Q is elliptic in Ω, i.e., λ > 0, where λ(x, z, p) is the smallesteigenvalue of aij(x, z, p),

aij = aji =12

(∂Ai

∂pj+∂Aj

∂pi

). (2.775)

Plan: Given a solution u ∈ C1 ∩W 2,2, use that its derivatives satisfylinear elliptic equations, and apply DeGiorgi-Nash-Moser’s theorem.

Example 2.105. Let u ∈ C1,α(Ω) be a solution to the minimal surfaceequation. For fixed 1 ≤ k ≤ n, we test the equation with ∂kψ for arbitraryψ ∈ C2

0 (Ω), and obtain that w := ∂ku is a solution of the linear equationLw := −

∑i,j ∂iaij∂jw = 0 with aij(x) := ∂pjA

i(∇u(x)). For arbitraryΩ′ ⊂⊂ Ω (cf. (2.762) and its discussion),

(i) the operator L is uniformly elliptic on Ω′, and

(ii) the coefficients aij are bounded.

We note that u ∈ W 2,2loc (Ω) (cf. the discussion of homework 11). Hence the

assumptions of Theorem 2.44 (DeGiorgi-Nash-Moser) are satisfied, and weobtain local Holder regularity of ∂ku.

To derive global Holder estimates, we need the following generalizationof Theorem 2.44 (see also the Remark following it). For a proof see [GT,Theorem 8.29].

Theorem 2.106. Let Ω := B+R ⊂ Rn, R > 0, be a half ball, and let T

be contained in the flat part of its boundary, i.e., T ⊂ ∂Ω ∩ xn = 0.Suppose that the operator L satisfies (2.484) and (2.485), f ∈ Lq(Ω,Rn)and g ∈ Lq/2(Ω) for some q > n. Suppose that u ∈W 1,2(Ω) satisfies

Lu := −∑i,j

∂iaij(x)∂ju = g + div f in Ω,

and suppose that there exist κ and α0 > 0 such that for all x0 ∈ T and allR > 0,

osc∂Ω∩B(x0,R)

u ≤ κRα0 .

Then u ∈ Cα(Ω ∪ T ) for some α > 0, and for every Ω′ ⊂⊂ Ω ∪ T ,

∥u∥Cα(Ω′) ≤ C(∥u∥L2(Ω) + κ+ λ−1(∥f∥Lq(Ω;Rn) + ∥g∥Lq/2(Ω))),

where α and C depend on n, Λ/λ, dist (Ω′, ∂Ω \ T ) and α0.

203 [March 17, 2014]

We prove the following global Holder regularity result for the gradient,given that K = sup(|u| + |∇u|) is bounded.

Theorem 2.107. Suppose Ω ⊂ Rn is open and bounded with ∂Ω ∈ C2.Suppose that u ∈ W 2,2(Ω) ∩ C1,α0(Ω) satisfies Qu = 0 in Ω and u = φ on∂Ω with φ ∈ C2(Ω), where Q is elliptic in Ω and of divergence form (2.774)with A ∈ C1(Ω × R × Rn; Rn) and B ∈ C0(Ω × R × Rn). Then there areα > 0 and C > 0 such that

[∇u]C0,α(Ω) ≤ C,

where C and α depend on n, K := supΩ(|u| + |∇u|), A, B, ∥φ∥C2 and Ω.(For explicit bounds see proof.)

Remark 2.108. To derive uniform Holder estimates, it suffices to assumeφ ∈ W 2,q(Ω) for some q > n, ∂Ω ∈ C1,α0, α0 > 0, and u ∈ W 2,2(Ω) ∩C0,1(Ω).

Proof. By considering u− φ, we may assume φ = 0.Step 1. Consider Ω = B+

1 , and let r < 1, i.e., B+r ⊂ B+

1 , and set T := ∂B+r ∩

∂Rn+. Let k ∈ 1, . . . , n. Testing the equation with ∂kψ for ψ ∈ C2

0 (Ω), wefind that w := ∂ku satisfies

−∂i(aij∂jw) = −∂if ik in Ω, (2.776)aij := ∂pjA

i, f ik(x, z, p) := −pk∂zAi − ∂kAi + δikB.

In particular, in B+r , the coefficients aij are bounded and strongly elliptic

since u and A are continuously differentiable, and similarly f is bounded.Further, by the boundary condition u = 0 on ∂Ω, we have w = ∂ku = 0 on∂Ω ∩ Rn

+ for k = 1, . . . , n − 1. In particular oscB(x0,R)∩∂Ωw = 0 for everyx0 ∈ T and R < 1 − r. Hence, by Theorem 2.106,

[∂ku]C0,α(B+r ) ≤ C, k = 1, . . . , n− 1. (2.777)

It remains to bound ∂nu, for which we use the equation and Morrey’s lemma(cf. proof of Theorem 2.19 and homework 4). Let R ≤ 1−r

3 . Then for everyy ∈ B+

r , we have B(y, 2R) ⊂ B(0, 13(2 + r)) ⊂⊂ B(0, 1), and thus,

[∂ku]C0,α(B+(y,2R)) ≤ C, k = 1, . . . , n− 1 (2.778)

uniformly for y ∈ B+(0, r) and R ≤ 1 − r

3.

For y0 ∈ B+r and R ≤ (1 − r)/3, choose a standard cut-off function η ∈

C10 (B(y0, 2R)) with 0 ≤ η ≤ 1, η = 1 on B(y0, R) and |∇η| ≤ 2/R. Set

c :=

w(y0), if B(y0, 2R) ⊂ B+

r ,

0, if B(y0, 2R) ∩ ∂Rn+ = ∅.

204 [March 17, 2014]

Then (since w = 0 on ∂B+ ∩∂Rn+) we have ζ := η2(w− c) ∈W 1,2

0 (B+1 ), i.e.,

ζ is an admissible test function. We use that ∂iζ = 2η∂iη(w− c) + η2∂iw toderive∫B+

1

η2aijw∂jw∂iw ≤∫B+

1

|2η(w−c)aij∂iη∂jw|+|η2f ik∂iw|+|2η(w−c)f ik∂iη|,

which by ellipticity and Young’s inequality implies that∫B(y0,R)

|∇w|2 ≤ C

∫B+

1

(η2 + |∇η|2(w − c)2) dy

≤ CRn +Rn−2 supB(y0,2R)

(w − c)2.

By (2.778), we have sup(w − c)2 ≤ C|x− y|2α, and thus∫B(y0,R)

|∂i∂ju|2 ≤ CRn−2+2α j = n. (2.779)

Recall that aij and f ik are uniformly bounded and elliptic if K <∞. Then,solving the equation (2.776) for ∂n∂nu, we obtain (cf. (2.193) and its discus-sion)

∂n∂nu = bij∂i∂ju+ b, i = 1, . . . , n, j = 1, . . . , n− 1,

where bij and b are bounded in terms of K, ΛK/λK , and µK/λK with

0 < λK ≤ infΩλ(x, u,∇u), ΛK ≥ sup

Ω|aij |, µK ≥ sup

Ω|f ik|.

Thus by (2.779) ∫B(y0,R)

|∂n∂nu|2 ≤ CRn−2+2α,

which by Morrey’s lemma (see homework 4, problem 4) finally implies that

[∂nu]C0,α ≤ C.

Step 2. Assume now that Ω ⊂ Rn is a general bounded domain with C2-boundary. Then for every x0 ∈ ∂Ω, there exists a ball B := B(x0) and abijective map ψ : B → D, with D ⊂ Rn open, and

ψ(B ∩ Ω) ⊂ Rn+, ψ(B ∩ ∂Ω) ⊂ ∂Rn

+, ψ ∈ C2(B), ψ−1 ∈ C2(D).

We refer to the independent variables on the reference domain as y = ψ(x)(and x = ψ−1(y)), and to the dependent ones as v(y) := u(ψ−1(y)). Furtherset B+ = B ∩ Ω, and D+ = ψ(B+). We may assume that ψ(x0) = 0and B+

1 ⊂⊂ D+. Then, since u = 0 = const on ∂Ω, we have ∂ykv =

205 [March 17, 2014]

0 on ∂D+ ∩ ∂Rn+ for k = 1, . . . , n − 1. By chain rule, since ψ is a C2-

diffeomorphism, it follows from the equation Qu = 0 in B+ that

Qv = −∂yiAi(x, u,∇u) +B(x, u,∇u) = 0, x = ψ−1(y) in D+,

where

Ai :=

∑i

∂yi∂xr

Ar, B :=∑i,r

∂

∂yi

(∂yi∂xr

)Ar +B.

Consequently, the derivatives w = ∂ykv, k = 1, . . . , n, are generalized

solutions in D+ to the linear equation

Lw := −∂i(aij∂jw) = −∂if ik,

with

aij(y) :=∂yi∂xr

∂yj∂xs

∂psAr,

f ik(x, z, p) :=∂yi∂xr

∂xs∂yk

(∂2yℓ∂xj∂xs

∂pjAr∂yℓ

u+ ps∂zAr + ∂sA

r

)+ δikB +

+(

∂

∂yk

(∂yi∂xr

)− δik

∂

∂yj

(∂yj∂xr

))Ar,

where the arguments are x = ψ−1(y), u(x) = v(y) and ∇u(x) = ∇u(ψ−1(y)).We have:

(i) For the Jacobian, λI ≤ (∂yi/∂xj)i,j=1,...,n ≤ ΛI in ψ−1(B1). In partic-ular, L is strictly elliptic in B+

1 with bounded coefficients aij . Further,f ik are bounded.

(ii) ∂ykv, k = 1, . . . , n− 1 satisfy the assumptions of Theorem 2.106.

Consequently, for every D′ ⊂⊂ B1,

[∂ykv]C0,α(D′∩D+) ≤ C for all k = 1, . . . , n− 1.

The remaining derivative ∂ynv is bounded following the lines of Step 1. Sinceψ−1 is C2, the Holder estimates for ∇v can be carried over to estimates for∇u. The estimate is done for every x0 ∈ ∂Ω, i.e., preimages of the ballsB1 ⊂⊂ ψ(B(x0)) build an open cover of ∂Ω. Extracting a finite subcover(∂Ω compact since Ω bounded), and combining with an interior estimate,we conclude the proof.

Remark 2.109. (i) If n = 2, then the proof works for arbitrary ellipticequations (not necessarily in divergence form). Precisely, an arbitraryquasilinear elliptic equation is equivalent to

−a11

a22∂1∂1u− 2a12

a22∂1∂2u− ∂2∂2u+

b

a22= 0,

206 [March 17, 2014]

and thus the derivative w1 := ∂1u of a solution u ∈ C2 satisfies

∂1

(−a11

a22∂1w1 −

2a12

a22∂2w1

)− ∂2∂2w1 = −∂1

b

a22,

and similarly, for w2 := ∂2u,

∂2

(−a22

a11∂2w2 −

2a12

a11∂1w2

)− ∂1∂1w2 = −∂2

b

a11,

and we proceed as before.

(ii) To treat general equations in higher dimensions, one shows that certaincombinations of derivatives are subsolutions of linear elliptic equationsin divergence form, and uses weak Harnack inequalities. For detailssee [GT, Sections 12.3. and 12.4].

[3.2. 2014, Lecture 27][6.2. 2014, Lecture 28]

2.5.5 Gradient bounds at the boundary

As before we consider a quasilinear operator

Qu = −∑i,j

aij(x, u,∇u)∂i∂ju+ b(x, u,∇u), with aij = aji,

we denote by λ(x, z, p) and Λ(x, z, p) the smallest and largest eigenvalue ofthe symmetric matrix aij(x, z, p) and we assume that Q is elliptic, i.e.,

λ(x, z, p) > 0 ∀(x, z, p) ∈ Ω × R × Rn.

Theorem 2.110. Let Ω ⊂ Rn be open and bounded. Let φ ∈ C2(Ω) andassume that u ∈ C2(Ω) satisifies

Qu = 0 in Ωu = φ on ∂Ω

Assume in addition that

(i) Ω is convex and ∂Ω is of class C1;

(ii) aij and b depend on only on x and p;

(iii) There exists C1 > 0, p0 > 0 such that

Λ(x, p) ≤ C1λ(x, p)|p|2 if |p| ≥ p0,|b(x, p)| ≤ C1λ(x, p)|p|2 if |p| ≥ p0,

(2.780)

Thenmax∂Ω

|∇u| ≤ C(n,C1, p0, ∥φ∥C2(Ω), supΩ

|u|)

207 [March 17, 2014]

Remark. (i) The result applies to the minimal surface operator M. Inthis case b = 0 and Λ(p)/λ(p) = 1 + |p|2. For the minimal surface operatorand n = 2 convexity of Ω is necessary to obtain an estimate for max∂Ω |∇u|for all boundary data φ. In higher dimensions the optimal condition is thatthe ∂Ω is mean convex, i.e., that the mean curvature of ∂Ω is ≥ 0.(ii) The result does not apply to the equation of prescribed mean curvatureMu = −nH(x)(1 + |∇u|2)3/2 since in this case λ(p) = 1, Λ(p) = 1 + |p|2,but b(x, p) ∼ |p|3 as p→ ∞. In this case on needs more stringent conditionson the (mean) curvature of ∂Ω, see [GT], Chapter 14.(iiI) One can obtain similar results for operators for aij and b depend alsoon z by applying the result above to the operator Q defined by

Qw = −∑i,j

aij(x, u,∇w)∂i∂jw + b(x, u,∇w),

compare the proof of the L∞ estimate, Theorem 2.100 and see [GT], Chapter14.

Added Feb 7, 2014. The proof is now correct. The reason I got stuckin class was that I tried to construct immediately a global barrier with w =ψ(dist (x, P )). This leads to a problem because one cannot have simultane-ously −ψ′′/(ψ′)2 large and |∇w| large in all of Ω. As the proof shows, onecan, however, have −ψ′′/(ψ′)2 large and |∇w| large in a neighbourhood Nof x0. I first tried the obvious choice N = Ω∩B(x0, r) with small r, but thisdoes not work since we cannot guarantee that w ≥ u− φ on Ω ∩ ∂B(x0, r).A better choice of N is made below. In order to emphasize the main idea,the proof is first done for zero boundary condition and the general case isthen deduced by a change of variables.

Proof. We first show the result for homogeneous boundary data

φ = 0.

In this case it suffices to assume that the second structure condition

|b(x, p)| ≤ C1λ(x, p)|p|2 if |p| ≥ p0, (2.781)

holds. Note that |∇u| = |∂u/∂ν| on ∂Ω since u = 0 on ∂Ω and hence thetangential derivatives of u are zero. We will show that

max∂Ω

|∇u| = max∂Ω

∣∣∣∣∂u∂ν∣∣∣∣ ≤ p0e

2C1M where M := supΩ

|u|. (2.782)

208 [March 17, 2014]

Step 1. Reduction to the construction of barriers.We argue similar as in the proof of the Hopf boundary lemma. If sufficesto show that for each point x0 ∈ ∂Ω there exists an open set N ⊂ Ω withB(x0, r)∩Ω ⊂ N for some small r and local barriers u± ∈ C1(N) such that

(i) u±(x0) = 0

(ii) u− ≤ u ≤ u+ in N ;

(iii) ∣∣∣∣∂u±∂ν (x0)∣∣∣∣ ≤ S (2.783)

where S only depends on C1, p0, and M = supΩ |u|. The first two propertiesimply that the directional derivative in direction of the inner normal satisfies

−∂u−

∂ν≤ −∂u

∂ν≤ −∂u

+

∂ν

Hence the third property implies that∣∣∣∣∂u∂ν∣∣∣∣ ≤ S

Step 2. Ansatz for the barrier.In the following we fix a point x0 ∈ ∂Ω and we focus on u+. The argumentfor u− = −u+ is analogous. We write w = u+. If first condition (i) in Step1 is

w(x0) = 0. (2.784)

To obtain the inequality w ≥ u in N it suffices to show that

w ≥ u on ∂N (2.785)Qw > 0 in N (2.786)

Then the comparison principle, Theorem 2.98, implies that w > u in N .Condition (2.785) is equivalent to the following two conditions

w ≥ 0 on ∂Ω ∩N , (2.787)w ≥ u on ∂N ∩ Ω. (2.788)

To summarize we are looking for a function w which satisfies

(2.784), (2.787), (2.788), (2.786)

and ∣∣∣∣∂w∂ν (x0)∣∣∣∣ ≤ S.

209 [March 17, 2014]

Step 3. First attempt: distance from a touching hyperplane.Since Ω is convex the conditions (2.784) and (2.787) can be easily satisfiedby a linear function. Let P be a hyperplane which touches Ω at x0 (such aplane exists since Ω is convex and it is unique since ∂Ω is of class C1) andlet

v(x) = dist (x, P ).

Then v is an affine and nonnegative function in Ω. Indeed, v(x) = −(x −x0) ·ν where ν is the outer normal at x0. Thus if we take w = v then (2.784)and (2.787) hold. It is also easy to satisfy (2.788) if we set

N := x ∈ Ω : v(x) < a.

Then ∂N ∩ Ω = x ∈ Ω : v(x) = a and thus the choice w = αv(x)with α = supΩ |u|/a satisfies (2.787) and (2.788). If a is large then the set∂N ∩Ω can be empty, but in this case (2.788) holds trivially. Unfortunatelythis choice of w does in general not satisfy the remaining condition (2.786)since Qv = b(x,∇v).

Step 4. A monotone function of the distance function.Let ψ ∈ C2[0,∞) with ψ(0) = 0 and ψ′ > 0. If w = ψ(v) then

∂jw = ψ′(v)∂jv, ∂i∂jw = ψ′′(v)∂iv ∂jv + ψ′(v)∂i∂jv

∑i,j

−aij∂i∂jw = −∑i,j

ψ′aij ∂i∂jv︸ ︷︷ ︸=0

− ψ′′

(ψ′)2∑i,j

aij∂iw∂jw

Assume now in addition that ψ′′ < 0. Then for v = dist (x, P ) the functionw = ψ(v) satisfies

Qw = − ψ′′

(ψ′)2∑i,j

aij(x,∇w)∂iw∂jw + b(x,∇w)

≥− λ(x,∇w)ψ′′

(ψ′)2|∇w|2 + b(x,∇w).

Now the structure conditions (2.781) implies that Qw > 0 if

|∇w| ≥ p0 in N ,

− ψ′′

(ψ′)2= ν with ν > C1. (2.789)

Since |∇v| = |ν| = 1 we have |∇w| = ψ′(v) and since ψ′ is decreasing thefirst condition is equivalent to

ψ′(a) ≥ p0. (2.790)

210 [March 17, 2014]

Using that ∂N ∩ Ω = x ∈ Ω : v(x) = a we see that the condition (2.788)is satisfied if

ψ(a) ≥M := supΩ

|u|. (2.791)

Thus w is a barrier if ψ(0) = 0, ψ′ > 0, and conditions

(2.789), (2.790) and (2.791)

hold.

Step 5. Solution of the ODE − ψ′′

(ψ′)2 = ν.Let ν > 0 and let η = ψ′. Then

ν = − η′

(η)2=(

1η

)′

and thus 1η (t) = νt + c and ψ(t) = ν−1 ln(νt + c) + c′. The conditions

w(x0) = 0 is equivalent to ψ(0) = 0 which yields ln c + νc′ = 0. Settingk := ν/c this finally yields

ψ(t) =1ν

ln(kt+ 1), (2.792)

where ν > 0 and k > 0. Then

ψ′(t) =k

ν

11 + kt

, ψ′′(t) = − k2

(1 + kt)2.

Step 6. Verification of the conditions (2.789), (2.790) and (2.791).Let ψ be given by (2.792) and let ν = 2C1. Then (2.789) holds. The tworemaining conditions reduce to the inequalities

k

ν

11 + ka

≥ p0 and1ν

ln(1 + ka) ≥M.

To see that these two conditions can be satisfied simultaneously let

ka = eνM′ − 1

to ensure that the second condition holds. Note that the first condition canbe rewritten as

ap0 ≤ 1ν

ka

1 + ka=

1ν

(1 − e−νM )

Thus all conditions are satisfied if

a =1νp0

(1 − e−νM′), k = νp0e

νM .

Then|∇w(x0)| = ψ′(0) =

k

ν= p0e

νM = p0e2C1M .

Hence we may take S = p0e2C1M in (2.783). This finishes the proof of

(2.782) under the assumption (2.781) for φ = 0.

211 [March 17, 2014]

Step 7. Extension to φ = 0.This is done by a change of variables. We set u = u− φ. Then

u = 0 on ∂Ω.

We will show that u is the solution of quasilinear PDE whose coefficientssatisfy the structure condition (2.781). We have

Qu = Q(u+ φ) = −∑i,j

aij(x,∇u+ ∇φ)(∂i∂j u+ ∂i∂jφ) + b(x,∇u+ ∇φ)

Set

aij(x, p) = aij(x, p+ ∇φ(x)),

b(x, p) = −∑i,j

aij(x, p+ ∇φ(x))∂i∂jφ(x) + b(x, p+ ∇φ(x)).

and define the quasilinear operator Q by

QU = −∑i,j

aij(x,∇U)∂i∂jU + b(x,∇U(x)).

ThenQu = Qu = 0 in Ω.

We will show that the coefficient functions aij and b satisfy the structurecondition

|b(x, p)| ≤ C1λ(x, p)|p|2 if |p| ≥ p0 (2.793)

where λ(x, p) is the smallest eigenvalue of the symmetric matrix aij(x, p).Then by what we have already shown we get the estimate

max∂Ω

|∇u| ≤ p0e2C1M where M := sup

Ω|u| = sup

Ω|u− φ|

and the assertion follows since |∇u| ≤ |∇u| + |∇φ|.To prove (2.793) set

p0 := max(2p0, 2 sup |∇φ|).

Then

|p| ≥ p0 =⇒ |p+ ∇φ(x)| ≥ |p| − supΩ

|∇φ| ≥ 12|p| ≥ p0, (2.794)

|p| ≥ p0 =⇒ |p+ ∇φ(x)| ≤ 2|p|. (2.795)

212 [March 17, 2014]

For two symmetric matrices A and B we have |A ·B| ≤ Λ|B| where A ·B :=∑i,j AijBij , |A|2 := A · A and where Λ is the maximum of the absolute

values of the eigenvalues of A.32 Thus

|b(x, p)| ≤ Λ(x, p+ ∇φ(x))∥φ∥C2 + |b(x, p+ ∇φ(x))|≤

(2.780),(2.794)C1(∥φ∥C2 + 1)λ(x, p+ ∇φ(x)) |p+ ∇φ(x)|2

≤(2.795)

C1(∥φ∥C2 + 1)λ(x, p+ ∇φ(x)) 4|p|2

= 4C1(∥φ∥C2 + 1) λ(x, p) |p|2

where in the last step we used the relation λ(x, p) = λ(x, p+ ∇φ(x)) whichfollows from aij(x, p) = aij(x, p+ ∇φ(x)). Thus (2.793) holds with

C1 = 4C1(∥φ∥C2 + 1)

and the proof is finished

2.5.6 Global gradient bounds

Theorem 2.111. Let A : Rn → Rn be C1 and elliptic, i.e., the lowesteigenvalue λ(p) of the symmetric matrix with coefficients

12

(∂Ai

∂pj+∂Aj

∂pi

)satisfies λ(p) > 0 for all p ∈ Rn. Suppose that u ∈ C2(Ω) satisfies

−n∑i=1

∂iAi(∇u) = 0

in Ω. Thenmax

Ω|∇u| ≤ max

∂Ω|∇u|2

Remark. (i) Note that the assumption λ(p) > 0 for all p is equivalent to∑i,j

∂Ai

∂pj(p)ξiξj > 0 ∀p ∈ Rn, ξ ∈ Rn \ 0

(ii) The assumptions are satisfied for the minimal surface equation where

A(p) =p√

1 + |p|2

32This is clear if A is diagonal. For the general case observe that for any Q ∈ SO(n)we have A ·B = (QAQT ) ·QBQT and hence |QBQT | = |B|. Now use that there exists aQ ∈ SO(n) such that QAQT is diagonal.

213 [March 17, 2014]

Proof. First assume that u ∈ C3(Ω).General recipe: differentiate the PDE for u with respect to xk multiply

by ∂ku and sum over k to get a PDE for w = 12 |∇u|

2 to which the maximumprinciple can be applied.

Differentiation of the PDE with respect to xk gives by the chain rule

−∑i,j

∂i

[∂Ai

∂pj(∇u)∂j∂ku

]= 0

Multiplying this expression by ∂ku and using the product rule we get

−∑i,j

∂i

[∂Ai

∂pj(∇u)∂j∂ku ∂ku

]+∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂i∂ku (2.796)

Now apply the ellipticity condition with ξ = ∂ku. This shows that thesecond summand in (2.796) is non negative. Set

w =12|∇u|2.

Then ∂jw =∑

k ∂ku ∂j∂ku. Thus summing (2.796) over k we get

−∑i,j

∂i∂Ai

∂pj(∇u)∂jw ≤ 0.

Hence w is a subsolution of linear equation

−∑i,j

∂iaij∂jw ≤ 0

where

aij(x) =∂Ai

∂pj(∇u(x))

Since u ∈ C1(Ω) the gradient ∇u(x) takes values in a compact set andtherefore the coefficients aij are bounded and elliptic in the sense that∑

i,j

aij(x)ξiξj ≥ ν|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

withν = minλ(p) : p ∈ ∇u(Ω) > 0.

Hence by the weak maximum principle

maxΩ

w ≤ max∂Ω

w.

This yields the assertion under the additional assumption u ∈ C3(Ω).

214 [March 17, 2014]

If we only assume u ∈ C2(Ω) then we can do the same calculation in theweak setting. Letting ψ ∈ C∞

c (Ω) then using η = −∂kψ in the weak for ofthe equation we get, using integration by parts,

0 = −∫

Ω

∑i

Ai(∇u) ∂i∂kψ dx

=∫

Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂iψ dx

Now u ∈ C1(Ω) implies that ∂Ai

∂pj (∇u) ∈ C(Ω). Thus[∂Ai

∂pj (∇u)∂j∂ku]∈

C(Ω) and by density we have

0 =∫

Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂iψ dx ∀ψ ∈W 1,1

0 (Ω).

Now let ζ ∈ W 1,20 (Ω) and take ψ = ζ∂ku. Since ∂iψ = (∂iζ) ∂ku + ζ ∂i∂ku

we deduce that ∫Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku ∂ku

]∂iζ dx

= −∫

Ω

∑i,j

[∂Ai

∂pj(∇u)∂j∂ku

]∂i∂ku ζ dx

As before ellipticity implies that the right hand side is ≤ 0 if ζ ≥ 0. Summingover k and defining w as above we thus get∫

Ω

∑i,j

[∂Ai

∂pj(∇u)∂jw

]∂iζ dx ≤ 0 ∀ζ ∈W 1,2

0 (Ω) with ζ ≥ 0.

Thus w is a weak subsolution of a linear elliptic PDE and the assertionfollows again be the weak maximum principle.

215 [March 17, 2014]

3 A brief look back

3.1 Estimates

We essentially used three methods to derive estimates

(i) (L2 estimates) Multiply the equation by η2u and integrate by parts;

(ii) (Max. principle) If u has an interior maximum at x0 thenD2u(x0) ≤ 0;

(iii) (Moser iteration) Multiply the equation by η2|u|β−1u and let β → ±∞.

Strategy (i) also works for systems and higher order equations while (ii)and (iii) work for scalar second order equations; (i) and (iii) are most usefulfor equations in divergence form while (ii) also works for equations in non-divergence form and even fully nonlinear equations.

Consequences of (i):

(i) W 1,2 estimates and reverse Poincare inequality

(ii) Decay estimates for constant coefficients

(iii) Estimates in scaled L2 spaces (Morrey, Campanato, BMO ) by freezingthe coefficients, (i), (ii) and the iteration lemma

(iv) Schauder estimates by C0,α = L2,n+2α or decay estimate and blow-up

(v) Lp estimates by interpolation between L2 and BMO and duality

(vi) Meyers’ Lp estimate for |p− 2| << 1 and L∞ coefficients

(vii) H1 estimates by duality with BMO

These estimates usually come in two very closely related variants. Localestimates (estimate of a stronger norm in a ball by the right hand sideand the solution in a weaker norm in a larger ball) and global estimates inRn. For the passage from local to global estimates see Corollary 2.16 forthe reverse passage see Lemma 2.40. For estimates up to the boundary oneusually uses a partition of unity and maps the neighbourhood of a boundarypoint to a half-ball (’flattening the boundary’). For estimates in a half-ballor halfspace one can still use tangential derivatives of the equation. Toestimate the second derivative in normal direction one uses the equation.Note:

L1 and L∞ are ’bad’ spaces for elliptic equations.The good replacements are the Hardy space H1 and BMO .

For right hand side in L1 one can obtain estimates in the weak-Lp spaces.

Consequences of (ii):

216 [March 17, 2014]

(i) Maximum principle: c = 0, Lu ≤ 0 =⇒ maxΩ u ≤ max∂Ω u

(ii) Hopf boundary lemma by squeezing in a supersolution on a touchingannulus

(iii) Strong maximum principle

(iv) Comparison principle for quasilinear or fully nonlinear equations

(v) Boundary gradient estimates by barrier functions

Consequences of (iii):

(i) Harnack inequality for weak sub- and supersolutions

(ii) C0,α estimates for equations in divergence form with L∞ coefficients

3.2 Regularity for nonlinear PDE

(i) From C0 or C1 to C∞ by linear theory

(ii) Initial (partial) regularity for harmonic maps by H1 estimates for div−curl quantities, H1 − BMO duality and the monotonicity formula

(iii) Role of symmetries and good gauge to get div − curl structure

(iv) Partial C1,α regularity for minimizers of quasiconvex integrals by ex-cess decay

(v) Excess decay by blow-up and Cacciopoli inequality/ compactness

3.3 Existence

(i) For elliptic equations without lower order terms by Lax-Milgram

(ii) For general elliptic equations by the Fredholm alternative

(iii) For elliptic systems with continuous coefficients by Garding’s inequal-ity and the Fredholm alternative

(iv) Existence in W 1,p for p > 2 by regularity, for 1 < p < 2 by duality.

(v) For Euler-Lagrange equations by of variational problems by the directmethod of the calculus of variations / weak lower semicontinuity (forconvex or quasiconvex integrands)

(vi) Existence for monotone equations by the Minty-Browder trick andweak convergence

(vii) Existence of classical solutions for quasilinear elliptic equations by theLeray-Schauder fixed point theorem and C1,α a priori estimates.

217 [March 17, 2014]

3.4 Some topics that were not covered

(i) Neumann boundary conditions for elliptic systems

(ii) The Stokes system −∆u+∇p = f, div u = 0 (not an ADN system)

(iii) The Alexandrov-Bakelmann-Pucci maximum principle (see [GT],Chapter 9.1)

(iv) C0,α estimates for equations in non divergence form (see [GT], Chapter9.8., 9.9)

(v) Classical solutions for fully nonlinear equations (see [GT], Chapter17).

(vi) Viscosity solutions for fully nonlinear equations (a notion of solutionfor C0 functions based on the maximum principle and Perron’s methodto obtain solutions as the supremum of subsolutions, see [CIL])

218 [March 17, 2014]

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