Partial Diﬀerential Equations

Partial Differential Equationsversion: December 4, 2019

Armin SchikorraEmail address: [email protected]

Contents

Bibliography 6.1. Introduction and some basic notation 8

Chapter 1. Model equations and special solutions 11I.1. Transport equation 11I.2. Laplace equation 12

Chapter 2. Second order linear elliptic equations, and maximum principles 40II.1. Linear Elliptic equations 40II.2. Maximum principles 41

Chapter 3. Sobolev Spaces 51III.1. Basic concepts from Functional Analysis 51III.2. Philosophy of Distributions and Sobolev spaces 53III.3. Sobolev Spaces 56

Chapter 4. Existence and Regularity for linear elliptic PDE 92

Chapter 5. Fractional Sobolev spaces as trace spaces 96V.1. The Fractional Sobolev space W s,p 96V.2. Its a trace space! 97V.3. Cool things one can do with this: Integration by parts revisited 101V.4. W s,p is not a gradient to the power p 103V.5. W s,p becomes W 1,p as s goes to one 103V.6. “The” other fractional Sobolev space Hs,p 103V.7. Embedding theorems, Trace-theorems etc. hold true 104

Chapter 6. Parabolic PDEs 106VI.1. The heat equation: Fundamental solution and Representation 106VI.2. Mean-value formula 109VI.3. Maximum principle and Uniqueness 109VI.4. Harnack’s Principle 114VI.5. Regularity and Cauchy-estimates 116

Chapter 7. linear parabolic equations 118VII.1. Definitions 118VII.2. Maximum principles 119

3

CONTENTS 4

Chapter 8. Principles of Semi-group theory 126VIII.1. m-dissipative operators 127VIII.2. Semigroup Theory 133VIII.3. An example application of Hille-Yoshida 140VIII.4. Formal regularity theory (a priori estimates) 147

Chapter 9. Linear Hyperbolic equations: waves 150IX.1. Halfwave-decomposition 150IX.2. D’Alambert’s formula in one dimension 151IX.3. Second-order hyperbolic equations 152IX.4. Existence via Galerkin approximation 153IX.5. Propagation of Disturbances 158

Chapter 10. Introduction to Navier–Stokes 160X.1. The Navier–Stokes equations 160X.2. Weak formulation 162X.3. Existence of weak solutions 170X.4. Uniqueness of weak solutions in dimension two 170X.5. Some Regularity-type theory for n = 2 (global), n = 3 (domains) 173

Chapter 11. Short introduction to Calderon-Zygmund Theory 177XI.1. Calderon-Zygmund operators 178XI.2. W 1,p-theory for the Laplace equation 181

Chapter 12. Short introduction to Viscosity solutions 185XII.1. Uniformly elliptic equations 189XII.2. Perons Method 190

CONTENTS 5

In Analysisthere are no theorems

only proofs

A large part of these notes are based on [Evans, 2010] and lectures by Heiko von derMosel (RWTH Aachen).

Bibliography

[Adams and Fournier, 2003] Adams, R. A. and Fournier, J. J. F. (2003). Sobolev spaces, volume 140 ofPure and Applied Mathematics (Amsterdam). Elsevier/Academic Press, Amsterdam, second edition.

[Bourgain et al., 2001] Bourgain, J., Brezis, H., and Mironescu, P. (2001). Another look at Sobolev spaces.In Optimal control and partial differential equations, pages 439–455. IOS, Amsterdam.

[Brezis and Nguyen, 2011] Brezis, H. and Nguyen, H. (2011). The Jacobian determinant revisited. Invent.Math., 185(1):17–54.

[Bucur and Valdinoci, 2016] Bucur, C. and Valdinoci, E. (2016). Nonlocal diffusion and applications, vol-ume 20 of Lecture Notes of the Unione Matematica Italiana. Springer, [Cham]; Unione MatematicaItaliana, Bologna.

[Bui and Candy, 2015] Bui, H.-Q. and Candy, T. (2015). A characterisation of the Besov-Lipschitz andTriebel-Lizorkin spaces using Poisson like kernels. arXiv:1502.06836.

[Cazenave and Haraux, 1998] Cazenave, T. and Haraux, A. (1998). An introduction to semilinear evolutionequations, volume 13 of Oxford Lecture Series in Mathematics and its Applications. The Clarendon Press,Oxford University Press, New York. Translated from the 1990 French original by Yvan Martel and revisedby the authors.

[Chen, 1999] Chen, Z.-Q. (1999). Multidimensional symmetric stable processes. Korean J. Comput. Appl.Math., 6(2):227–266.

[Coifman et al., 1993] Coifman, R., Lions, P.-L., Meyer, Y., and Semmes, S. (1993). Compensated com-pactness and Hardy spaces. J. Math. Pures Appl. (9), 72(3):247–286.

[Di Nezza et al., 2012] Di Nezza, E., Palatucci, G., and Valdinoci, E. (2012). Hitchhiker’s guide to thefractional Sobolev spaces. Bull. Sci. Math., 136(5):521–573.

[Evans, 2010] Evans, L. C. (2010). Partial differential equations, volume 19 of Graduate Studies in Math-ematics. American Mathematical Society, Providence, RI, second edition.

[Evans and Gariepy, 2015] Evans, L. C. and Gariepy, R. F. (2015). Measure theory and fine properties offunctions. Textbooks in Mathematics. CRC Press, Boca Raton, FL, revised edition.

[Galdi, 2011] Galdi, G. P. (2011). An introduction to the mathematical theory of the Navier-Stokes equa-tions. Springer Monographs in Mathematics. Springer, New York, second edition. Steady-state problems.

[Gazzola et al., 2010] Gazzola, F., Grunau, H.-C., and Sweers, G. (2010). Polyharmonic boundary valueproblems, volume 1991 of Lecture Notes in Mathematics. Springer-Verlag, Berlin. Positivity preservingand nonlinear higher order elliptic equations in bounded domains.

[Giaquinta and Martinazzi, 2012] Giaquinta, M. and Martinazzi, L. (2012). An introduction to the regular-ity theory for elliptic systems, harmonic maps and minimal graphs, volume 11 of Appunti. Scuola NormaleSuperiore di Pisa (Nuova Serie) [Lecture Notes. Scuola Normale Superiore di Pisa (New Series)]. Edizionidella Normale, Pisa, second edition.

[Gilbarg and Trudinger, 2001] Gilbarg, D. and Trudinger, N. S. (2001). Elliptic partial differential equa-tions of second order. Classics in Mathematics. Springer-Verlag, Berlin. Reprint of the 1998 edition.

[Grafakos, 2014a] Grafakos, L. (2014a). Classical Fourier analysis, volume 249 of Graduate Texts in Math-ematics. Springer, New York, third edition.

[Grafakos, 2014b] Grafakos, L. (2014b). Modern Fourier analysis, volume 250 of Graduate Texts in Math-ematics. Springer, New York, third edition.

6

BIBLIOGRAPHY 7

[Haj�lasz and Liu, 2010] Haj�lasz, P. and Liu, Z. (2010). A compact embedding of a Sobolev space is equiv-alent to an embedding into a better space. Proc. Amer. Math. Soc., 138(9):3257–3266.

[Iwaniec and Sbordone, 1998] Iwaniec, T. and Sbordone, C. (1998). Riesz transforms and elliptic PDEswith VMO coefficients. J. Anal. Math., 74:183–212.

[John, 1991] John, F. (1991). Partial differential equations, volume 1 of Applied Mathematical Sciences.Springer-Verlag, New York, fourth edition.

[Koike, 2004] Koike, S. (2004). A beginner’s guide to the theory of viscosity solutions, volume 13 of MSJMemoirs. Mathematical Society of Japan, Tokyo.

[Kuznetsov, 2019] Kuznetsov, N. (2019). Mean value properties of harmonic functions and related topics(a survey). Preprint, arXiv:1904.08312.

[Lenzmann and Schikorra, 2019] Lenzmann, E. and Schikorra, A. (2019). Sharp commutator estimates viaharmonic extensions. Nonlinear Analysis (accepted).

[Lieberman, 1996] Lieberman, G. M. (1996). Second order parabolic differential equations. World ScientificPublishing Co., Inc., River Edge, NJ.

[Littman et al., 1963] Littman, W., Stampacchia, G., and Weinberger, H. F. (1963). Regular points forelliptic equations with discontinuous coefficients. Ann. Scuola Norm. Sup. Pisa (3), 17:43–77.

[Llorente, 2015] Llorente, J. G. (2015). Mean value properties and unique continuation. Commun. PureAppl. Anal., 14(1):185–199.

[Maz’ya, 2011] Maz’ya, V. (2011). Sobolev spaces with applications to elliptic partial differential equations,volume 342 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of MathematicalSciences]. Springer, Heidelberg, augmented edition.

[Mironescu and Sickel, 2015] Mironescu, P. and Sickel, W. (2015). A Sobolev non embedding. Atti Accad.Naz. Lincei Rend. Lincei Mat. Appl., 26(3):291–298.

[Robinson et al., 2016] Robinson, J. C., Rodrigo, J. L., and Sadowski, W. (2016). The three-dimensionalNavier-Stokes equations, volume 157 of Cambridge Studies in Advanced Mathematics. Cambridge Univer-sity Press, Cambridge. Classical theory.

[Runst and Sickel, 1996] Runst, T. and Sickel, W. (1996). Sobolev spaces of fractional order, Nemytskij op-erators, and nonlinear partial differential equations, volume 3 of De Gruyter Series in Nonlinear Analysisand Applications. Walter de Gruyter & Co., Berlin.

[Samko, 2002] Samko, S. G. (2002). Hypersingular integrals and their applications, volume 5 of AnalyticalMethods and Special Functions. Taylor & Francis, Ltd., London.

[Schikorra et al., 2017] Schikorra, A., Spector, D., and Van Schaftingen, J. (2017). An L1-type estimatefor Riesz potentials. Rev. Mat. Iberoam., 33(1):291–303.

[Stein, 1993] Stein, E. M. (1993). Harmonic analysis: real-variable methods, orthogonality, and oscillatoryintegrals, volume 43 of Princeton Mathematical Series. Princeton University Press, Princeton, NJ. Withthe assistance of Timothy S. Murphy, Monographs in Harmonic Analysis, III.

[Talenti, 1976] Talenti, G. (1976). Best constant in Sobolev inequality. Ann. Mat. Pura Appl. (4), 110:353–372.

[Tao, 2016] Tao, T. (2016). Finite time blowup for an averaged three-dimensional Navier-Stokes equation.J. Amer. Math. Soc., 29(3):601–674.

[Tartar, 2007] Tartar, L. (2007). An introduction to Sobolev spaces and interpolation spaces, volume 3 ofLecture Notes of the Unione Matematica Italiana. Springer, Berlin; UMI, Bologna.

[Ziemer, 1989] Ziemer, W. P. (1989). Weakly differentiable functions, volume 120 of Graduate Texts inMathematics. Springer-Verlag, New York. Sobolev spaces and functions of bounded variation.

.1. INTRODUCTION AND SOME BASIC NOTATION 8

.1. Introduction and some basic notation

When studying Partial Differential Equations (PDEs) the first question that arises is: whatare partial differential equations.

Let Ω ⊂ Rn be an open set and u : Ω → R be differentiable. The partial derivatives ∂1 isthe directional derivative

∂1u(x) ≡ ∂x1u(x) = d

dx1u(x) = d

dt

��t=0

u(x + te1),

where e1 = (1, 0, . . . , 0) is the first unit vector. The partial derivatives ∂2, . . . ∂n are definedlikewise.

Sometimes it is convenient to use multiinidces: an n-multiindex γ is a vector γ = (γ1, γ2, . . . , γn)where γ1, . . . , γn ∈ {0, 1, 2, . . . , }. The order of a multiindex is |γ| defined as

|γ| =n�

i=1γi.

For a suitable often differentiable function u : Ω → R and a multiindex γ we denote with∂γu the partial derivatives

∂γu(x) = ∂γ1x1∂γ2

x2 . . . ∂γnxn

u(x).For example, for γ = (1, 0, 0, . . . , 0) we have

∂γu(x) = ∂x1u,

i.e. a partial derivative of first order; and for γ = (1, 2, 0, . . . , 0) we have

∂γu = ∂122u ≡ ∂1∂2∂2u,

i.e. a partial derivative of 3rd order.

The collection of all partial derivatives of k-th order of u is usually denoted by Dku(x) ∈ Rnk

or (the “gradient”) ∇ku. Usually these are written in matrix form, namely

Du(x) = (∂1u(x), ∂2u(x), ∂3u(x), . . . , ∂nu(x))

and

D2u(x) = (∂iju)i,j=1,...n ≡

∂11u(x) ∂12u(x) ∂13u(x) . . . ∂1nu(x)∂21u(x) ∂22u(x) ∂23u(x) . . . ∂2nu(x)

... ...∂n1u(x) ∂n2u(x) ∂n3u(x) . . . ∂nnu(x)

Definition .1.1. Let Ω ⊂ Rn an open set and k ∈ N∪ {0}. A partial differential equation(PDE) of k-th order is an expression of the form

(.1.1) F (Dku(x), Dk−1u(x), Dk−2u(x), . . . , Du(x), u(x), x) = 0 x ∈ Ω,


where u : Ω → R is the unknown (also the “solution” to the PDE) and F is a givenstructure (i.e. map)

F : Rnk × Rnk−1 × . . . × Rn × R × Ω → R

• (.1.1) is called linear if F is linear in u: meaning if we can find for every n-multiindex γ with |γ| ≤ k a function aγ : Ω → R (independent of u) such thatF (Dku(x), Dk−1u(x), Dk−2u(x), . . . , Du(x), u(x), x) =

�

|γ|≤k

aγ(x)∂γu(x)

• (.1.1) is called semilinear if F is linear with respect to the highest order k, namelyif

F (Dku(x), Dk−1u(x), Dk−2u(x), . . . , Du(x), u(x), x) =�

|γ|=k

aγ(x)∂γu(x)+G(Dk−1u(x), Dk−2u(x), . . . , Du(x

• (.1.1) is called quasilinear if F is linear with respect to the highest order k butthe coefficient for the highest order may depend on the lower order derivatives ofu. Namely if we have a representation of the form

F (Dku(x), Dk−1u(x), Dk−2u(x), . . . , Du(x), u(x), x) =�

|γ|=k

aγ(Dk−1u(x), Dk−2u(x), . . . , Du(x), u(x), x)∂γu

• If all the above do not apply then we call F fully nonlinear.

We have a system of partial differential equations of order k, if u : Ω → Rm is a vectorand/or the structure function F is also a vector

F : Rm nk × Rm nk−1 × . . . × Rm n × Rm × Ω → R�

for m, � ≥ 1.

The goal in PDE is usually (besides modeling what PDE describes what situation) to solvePDEs, possibly subject to sidecondition (such as prescribed boundary data on ∂Ω). Thisis rarely possible explicitely (even in the linear case) mostly the best one can hope for isaddress the following main questions for PDEs are

• Is there a solution to a problem (and if so: in what sense? – we will learn thedistributional/weak sense and strong sense)

• Are solutions unique?• What are properties of the solutions (e.g. does the solution depend continuously

on the data of the problem)?

It is important to accept that there are PDEs without (classical) solutions and there is nogeneral theory of PDEs. There is theory for several types of PDES.Example .1.2 (Some basic linear equations). • Laplace equation

Δu :=n�

i=1uxixi

= 0.


• Eigenvalue equation (aka Helmholtz equation)Δu = λu.

• Transport equation∂tu −

n�

i=1biuxi

= 0

• Heat equation∂tu − Δu = 0

• Schrodinger equationi∂tu + Δu = 0

• Wave equationutt − Δu = 0

Example .1.3 (Some basic nonlinear equations). • Eikonal equation|Du| = 1

• p-Laplace equation

div (|Du|p−2Du) ≡n�

i=1∂i(|Du|p−2∂iu) = 0

• Minimal surface equation

div Du�

1 + |Du|2

= 0.

• Monge-Amperedet(D2u) = 0.

• Hamilton-Jacobi∂tu + H(Du, x) = 0

In this course we will focus on the linear theory (the nonlinear theory is always based onideas on the linear theory). Almost each of the linear and nonlinear equations warrants itsown course, so we will focus on the basics (namely: mainly elliptic equations).

CHAPTER 1

Model equations and special solutions

I.1. Transport equation

We consider solutions u : Rn × (0,∞) → R of

(I.1.1) ∂tu + b · Du = 0 in Rn × (0,∞).

The variables in Rn we denote by x (space) and the variable in (0,∞) by t (time).

Here b = (b1, . . . , bn) is a constant vector, and Du is the gradient of u, so that

b · Du(x) =n�

i=1bi∂iu

If we were to assume that u is sufficiently differentiable, thend

dsu(x + sb, t + s) = b · Du(x + sb, t + s) + ∂tu(x + sb, t + s)

(I.1.1)≡ 0

That is, u is constant in the direction (b, 1) in Rn+1. But this means we can solve explicitelyan equation of the form

∂tu + b · Du = 0 on Rn × (0,∞)u = g on Rn × {0}

Namely, since (assuming enough differentiability!) u is constant on lines with slope (b, 1)we have

u(x, t) = u(x + λb, t + λ) ∀λ ≥ −t.

Taking λ = −t we then find

(I.1.2) u(x, t) = u(x − tb, 0) = g(x − tb)

as the solution.

That is, if u ∈ C1(Rn × (0,∞))∩C0(Rn × [0,∞)) solves (I.1.1) then u is of the form (I.1.2).Also if g ∈ C1(Rn) the u of the form (I.1.2) solves (I.1.1).

On the other hand if g �∈ C1 then there cannot be C2-solutions to the transport equation!In that case one reverts to weak solutions.

11

I.2. LAPLACE EQUATION 12

(I.1.1) is called the homogeneous transport equation, since the right-hand side is zero. Ifwe consider the inhomogeneous problem

(I.1.3)

∂tu + b · Du = f in Rn × (0,∞)u = g in Rn × {0}

for a given f we can try to do the same spiel as above:

This time we have for a sufficiently smooth solutiond

dsu(x + sb, t + s) = b · Du(x + sb, t + s) + ∂tu(x + sb, t + s)

(I.1.3)≡ f(x + sb, t + s).

That is, we do not know that u(x, t) − g(x − tb) is constantly zero, but we have by thefundamental theorem of calculus,

u(x, t) − g(x − tb)=u(x, t) − u(x − tb, 0)

=� 0

−t

d

dsu(x + sb, t + s) ds

=� 0

−tf(x + sb, t + s) ds

σ:=t+s=� t

0f(x + (σ − t)b, σ) dσ.

That is, as in the homogeneous case we can conclude that (under enough differentiabilityassumptions),

u(x, t) = g(x − tb) +� t

0f(x + (σ − t)b, σ) dσ.

is the unique solution to the inhomogeneous (linear) transport problem (I.1.3).

I.2. Laplace equation

Let Ω ⊂ Rn be an open set (this will always be the case from now on). We consider thehomogeneous Laplace equation(I.2.1) Δu = 0 in Ωwhere we recall that Δu = tr(D2u) = �n

i=1 ∂iiu.

The inhomogenous equation (sometimes: Poisson equation) is, for a given function f :Ω → R,(I.2.2) Δu = f in Ω

Definition I.2.1. A function u ∈ C2(Ω) is called harmonic if u pointwise solvesΔu(x) = 0 in Ω

We also say, u is a solution to the homogeneous Laplace equation.


We say that u is a subsolution1 or subharmonic ifΔu(x) ≥ 0 in Ω.

IfΔu(x) ≤ 0 in Ω

we say that u is a supersolution or superharmonic.

I.2.1. Fundamental Solution, Newton- and Riesz Potential. There are manytrivial solutions (polynomials of order 1) of Laplace equation. But these are not veryinteresting. There is a special type of solution which is called fundamental solution (which,funny enough, is actually not a solution).

It appears when we want to compute the solution to an equation on the whole space(I.2.3) Δu(x) = f(x).For this we make a brief (formal) introduction to Fourier transform:

The Fourier transform takes a map f : Rn → R and transforms it into Fu ≡ f : Rn → Ras follows

f(ξ) := 1(2π)n

2

�

Rne−i�ξ,x� f(x) dx.

The inverse Fouriertransform f∨ is defined as

f∨(ξ) := 1(2π)n

2

�

Rne+i�ξ,x� f(x) dx.

It has the nice property that (f ∧)∨ = f .

One of the important properties (which we will check in exercises) is that derivativesbecome polynomial factors after Fourier transform:

(∂xig)∧ (ξ) = −iξig(ξ).

For the Laplace operator Δ this implies(Δu)∧(ξ) = −|ξ|2u(ξ).

This means that if we look at the equation (I.2.3) and apply Fourier transform on bothsides we have

−|ξ|2u(ξ) = f(ξ),that is

u(ξ) = −|ξ|−2f(ξ),Inverting the Fourier transform we get an explicit formula for u in terms of the data f .

u(x) = −�|ξ|−2f(ξ)

�∨(x).

1yes that notion is confusing


This is not a very nice formula, so let us simplify it. Another nice property of Fouriertransform (and its inverse) is that products become convolutions. Namely

(g(ξ)f(ξ))∨ (x) =�

Rng∨(x − z)f∨(z) dz.

In our case, for g(ξ) = −|ξ|−2 we get that

u(x) =�

Rng∨(x − z) f(z) dz.

Now we need to compute g∨(x − z), and for this we restrict our attention to the situationwhere the dimension is n ≥ 3. In that case, just by the definition of the (inverse) Fouriertransform we can compute that since g has homogeneity of order 2 (i.e. g(tξ) = t−2g(ξ),then g∨ is homogeneous of order 2 − n. In particular

g∨(x) = |x|2−ng∨(x/|x|).Now an argument that radial functions stay radial under Fourier transforms leads us toconclude that

g∨(x) = c1|x|2−n.

That is, we have arrived that (by formal computations) a solution of (I.2.3) should satisfy

(I.2.4) u(x) = c1

�

Rn|x − z|2−n f(z) dz.

The constant c1 can be computed explicitely, and we will check below that this potentialrepresentation of u really is true. This potential is called the Newton potential (which isa special case of so-called Riesz potentials). The kernel of the Newton potential is calledthe fundamental solution of the Laplace equation (which, again, is not a solution)

Definition I.2.2. The fundamental solution Φ(x) of the Laplace equation for x �= 0 isgiven as

Φ(x) =− 1

2πlog |x| for n = 2

− 1n(n−2)ωn

|x|2−n for n ≥ 2Here ωn is the Lebesgue measure of the unit ball ωn = |B(0, 1)|.

One can explicitely check that ΔΦ(x) = 0 for x �= 0 (indeed, ΔΦ(x) = δ0 where δ0 is theDirac measure at the point zero, cf. remark I.2.4).

The following statement justifies (somewhat) the notion of fundamental solution: thefundamental solution Φ(x) can be used to construct all solutions to the imhomogeneousLaplace equation:

Theorem I.2.3. Let u be the Newton-potential of f ∈ C2c (Rn), that is

u(x) :=�

RnΦ(x − y) f(y) dy.

Here C2c (Rn) are all those functions in C2(Rn) such that f is constantly zero outside of

some compact set.


We have

• u ∈ C2(Rn)• −Δu = f in Rn.

Proof. First we show differentiability of u. By a substitution we may write

u(x) :=�

RnΦ(x − y) f(y) dy =

�

RnΦ(z) f(x − z) dz.

Now if we denote the difference quotient

Δeih u(x) := u(x + hei) − u(x)

h

where ei is the i-th unit vector, then we obtain readily

Δeih u(x) :=

�


�

RnΦ(z) (Δei

h f)(x − z) dz.

One checks that Φ is locally integrable (it is not globally integrable!), that is for everybounded set Ω ⊂ Rn,

(I.2.5)�

Ω|Φ| < ∞.

Indeed, (we show this for n ≥ 3, the case n = 2 is an exercise), if Ω ⊂ Rn is a boundedset, then it is contained in some large ball B(0, R).

(I.2.6)�

Ω|Φ| ≤ C

�

B(0,R)|x|2−n dx

Using Fubini’s theorem,�

B(0,R)|x|2−n dx

=� R

0

�

∂B(0,r)|θ|2−n dHn−1(θ) dr

=� R

0r2−n

�

∂B(0,r)dHn−1(θ) dr

=cn

� R

0r2−nrn−1dr

=cn

� R

0r1dr

=cn12R2 < ∞.

This establishes (I.2.5)


On the other hand (Δeih f) has still compact support for every h. In particular, by dominated

convergence we can conclude that

limh→0

Δeih u(x) :=

�


�

RnΦ(z) lim

h→0(Δei

h f)(x − z) dz.

that is∂iu(x) :=

�


�

RnΦ(z) (∂if)(x − z) dz.

In the same way

∂iju(x) :=�


�

RnΦ(z) (∂ijf)(x − z) dz.

Now the right-hand side of this equation is continuous (again using the compact supportof f). This means that u ∈ C2(Rn).

To obtain that Δu = f we first use the above argument to get

Δu(x) =�


�

RnΦ(z) (Δf)(x − z) dz.

Observe that(Δf)(x − z) = Δx(f(x − z)) = Δz(f(x − z)).

Now we fix a small ε > 0 (that we later send to zero) and split the integral, we have

Δu(x) =�

RnΦ(x−y) f(y) dy =

�

B(0,ε)Φ(z) (Δf)(x−z) dz+

�

Rn\B(0,ε)Φ(z) (Δf)(x−z) dz =: Iε+IIε.

The term Iε contains the singularity of Φ, but we observe thatIε

ε→0−−→ 0.

Indeed, this follows from the absolute continuity of the integral and since Φ is integrableon B(0, 1):

|Iε| ≤ supRn

|Δf |�

B(x,ε)|Φ(z)| ε→0−−→ 0.

The term IIε does not contain any singularity of Φ which is smooth on Rn\Bε(0), so wecan perform an integration by parts2

IIε =�

Rn\B(0,ε)Φ(z) (Δf)(x−z) dz =

�

∂B(0,ε)Φ(z) ∂νf(x−z) dHn−1(z)−

�

Rn\B(0,ε)∇Φ(z)·∇f(x−z) dz.

Here ν is the unit normal to the ball ∂B(0, ε), i.e. ν = −zε

.

By the definition of Φ one computes that (using (I.2.5))��

�

∂B(0,ε)Φ(z) ∂νf(x − z) dHn−1(z)

�� ≤ supRn

|∇f |�

∂B(0,ε)|Φ(z)| ε→0−−→ 0.

2 �

Ωf ∂ig =

�

∂Ωf g νi −

�

Ω∂if g,

where ν is the normal of ∂Ω pointing outwards (from the point of view of Ω). ν i is the i-th component ofν. Fun exercise: Check this rule in 1D, to see the relation what we all learned in Calc 1.


So we perform another integration by parts and have

IIε =o(1) −�

∂B(0,ε)∂νΦ(z) f(x − z) dz +

�

Rn\B(0,ε)ΔΦ(z)� ��

=0

f(x − z) dz

= o(1) −�

∂B(0,ε)∂νΦ(z) f(x − z) dz

Here in the last step we used that ΔΦ = 0 away from the origin.

Now we observe that the unit normal on ∂B(0, ε) is ν(z) = − zε

and

DΦ(z) =− 1

2π1

|z|z

|z| n = 2,

− 1n(n−2)ωn

(2 − n)|z|1−n z|z| n ≥ 3.

Thus, for |z| = ε,

∂νΦ(z) = ν · DΦ(z) = 1nωn

ε1−n

Thus we arrive at

IIε =o(1) −�

∂B(0,ε)

1nωnεn−1 f(x − z) dHn−1(z)

=o(1) −�

∂B(0,ε)f(x − z) dHn−1(z)

=o(1) − f(x) +�

∂B(0,ε)(f(x) − f(x − z)) dHn−1(z)

Here we use the mean value notation�

∂B(0,ε)= 1

Hn−1(∂B(0, ε)

�

∂B(0,ε).

Now one shows (exercise!) that for continuous f

limε→0

�

∂B(0,ε)(f(x) − f(x − z)) dHn−1(z) = 0.

(Indeed this is essentially Lebesgue’s theorem). That isIIε = o(1) − f(x) as ε → 0

and thusΔu(x) = −f(x) + o(ε),

and letting ε → 0 we haveΔu(x) = −f(x),

as claimed. �Remark I.2.4. One can argue (in a distributional sense, which we learn towards the endof the semester)

−ΔΦ = δ0,


where δ0 denotes the Dirac measure at 0, namely the measure such that�

Rnf(x) dδ0 = f(0) for all f ∈ C0(Rn).

Observe that δ0 is not a function, only a measure. In this sense one can justify that

−Δu(x) =Δ�

RnΦ(x − z)f(z)

=�

RnΔΦ(x − z)f(z) dz

=�

Rnf(z) dδx(z)

=f(x)

I.2.2. Mean Value Property for harmonic functions. An important property(but very special to the “base Operator Δ”, i.e. not that easily generalizable to moregeneral PDEs) is the mean value property

Theorem I.2.5 (Harmonic functions satisfy Mean Value Property). Let u ∈ C2(Ω) suchthat Δu = 0, then

(I.2.7) u(x) =�

∂B(x,r)u(z) dHn−1(z) =

�

B(x,r)u(y) dy

holds for all balls B(x, r) ⊂ Ω.

If Δu ≤ 0 then we have “≥”in (I.2.7). If Δu ≥ 0 then we have “≤” in (I.2.7).

Proof. Setϕ(r) :=

�

∂B(x,r)u(y) dHn−1(y).

Observe that by substitution z := y−xr

we have

ϕ(r) :=�

∂B(0,1)u(x + rz) dHn−1(z).

Taking the derivative in r we have

ϕ�(r) =�

∂B(0,1)Du(x + rz) · z dHn−1(z).

Transforming back we get

ϕ�(r) =�

∂B(x,r)Du(y) · y − x

rdHn−1(y).

Observe that y−xr

is the outer unit normal of ∂B(x, r). That is

ϕ�(r) = |∂B(x, r)|−1�

∂B(x,r)∂νu(y)dHn−1(y).


By Stokes or Green’s theorem (aka, integration by parts)

ϕ�(r) = |∂B(x, r)|−1�

B(x,r)Δu(y)dy

(I.2.7)= 0.

That is,ϕ�(r) = 0 ∀r if B(x, r) ⊂ Ω.

which implies that ϕ is constant, and in particularϕ(r) = lim

ρ→0ϕ(ρ).

But (exercise!) for continuous functions u,

limρ→0

ϕ(ρ) = limρ→0

�

∂B(x,ρ)u(y) dHn−1(y) = u(x),

we have shown that

(I.2.8) u(x) =�

∂B(x,r)u(y) dHn−1(y)

holds whenever B(x, r) ⊂ Ω.

Moreover, by Fubini’s theorem�

B(x,r)u(y) dy = 1

|B(x, r)|� r

0

�

∂B(x,ρ)u(θ) dHn−1(θ) dρ

= 1|B(x, r)|

� r

0|B(x, ρ)|

�

∂B(x,ρ)u(θ) dHn−1(θ) dρ

(I.2.8)= 1|B(x, r)|

� r

0|∂B(x, ρ)| u(x)dρ

=u(x) 1|B(x, r)|

� r

0

�

∂B(x,ρ)1 dHn−1(θ)dρ

=u(x) |B(x, r)||B(x, r)|

=u(x).Together with (I.2.8) we have shown the claim for Δu = 0. The inequality arguments areleft as an exercise. �

The converse holds as well (and there is actually a whole literature on “how many balls”one has to assume the mean value property to get harmonicity, cf. [Llorente, 2015,Kuznetsov, 2019])

Theorem I.2.6 (Mean Value property implies harmonicity). Let u ∈ C2(Ω). If for allballs B(x, r) ⊂ Ω,

(I.2.9) u(x) =�

∂B(x,r)u(θ) dHn−1(θ)


thenΔu = 0 in Ω

Proof. Assume the claim is false.

Then there exists some x0 ∈ Ω such that Δu(x0) �= 0, so (by continuity of Δu) w.l.o.g.Δu > 0 in a small neighbourhood B(x0, R) of x0.

On the other hand, setting as above

ϕ(r) :=�

∂B(x0,r)u(θ)

(I.2.9)≡ u(x0)

we have ϕ�(r) = 0 for all r > 0 such that B(x0, r) ⊂ Ω. But as computed before, for r < R,

ϕ�(r) = C(r)�

B(x0,r)Δu dy > 0.

This (0 = ϕ�(r) > 0) is a contradiction, so the claim is established. �

I.2.3. Maximum and Comparison Principles. The mean value property as aboveis very rigid in the sense that it holds only for very special operators such as the Laplacian.A much more general property (which for the Laplacian Δ is a direct consequence ofthe mean value property) are maximum principles, which should be seen as a “forcedconvexity/concavity property” for sub-/supersolutions of a large class of PDEs (2nd orderelliptic, see Chapter 2 later.

In one-dimension a subsolution of Laplace’s equation satisfiesu�� ≥ 0

that is, subsolutions are exactly the convex C2-functions. Convexity means that on anyinterval (a, b) the maximum of u is obtained at a or at b – and if the maximum is obtainedin a point c ∈ (a, b) then u is constant. The curious fact is that these properties stillhold in arbitrary dimension for solutions of the Laplace equation (and later a large class ofelliptic 2nd order equations), they are the so-called weak maximum principle and strongmaximum principle.

Corollary I.2.7 (Strong Maximum-principle). Let u ∈ C2(Ω) be subharmonic, i.e. Δu ≥ 0in Ω. If there exists x0 ∈ Ω at which u attains a global maximum then u is constant in theconnected component of Ω containing x0.

Proof. By taking a possibly smaller Ω we can assume w.l.o.g. Ω is connected and ustill attains its global maximum in x0 ∈ Ω.

LetA := {y ∈ Ω : u(y) = u(x0)}.

We will show that A = Ω (and thus u is constant) by showing that the following threeproperties hold


• A is nonempty• A is relatively closed (in Ω).• A is open

Then A is an open and closed set in Ω, and since A is not the empty set it is all of Ω.

Clearly A is nonempty since x0 ∈ A.

Also A is relatively closed by continuity of u: If Ω � ykk→∞−−−→ y0 ∈ Ω then

u(y0) = limk→∞

u(yk) = u(x0)

and thus y0 ∈ A.

To show that A is open let y0 ∈ A. Since Ω is open we can find a small ball B(y0, ρ) ⊂ Ω.

Observe that x0 is a global maximum of u in B(y0, ρ).

The mean value property, Theorem I.2.5, and then the fact that u(x0) ≥ u(y) for all y inB(y0, ρ), imply

u(x0) = u(y0) ≤�

B(y0,ρ)u(y) dy ≤

�

B(y0,ρ)u(x0) dy = u(x0).

Since left-hand side and right-hand side coincide the inequality is actually an equality.

That is, we haveu(x0) =

�

B(y0,ρ)u(y) dy,

in other words �

B(y0,ρ)u(y) − u(x0) dy = 0.

Since u(y) − u(x0) by assumption ≤ 0 the above integral becomes

−�

B(y0,ρ)|u(y) − u(x0)| dy = 0.

that isu(y) ≡ u(x0) in B(y0, ρ),

that is B(y0, ρ) ⊂ A. That is, A is open. �Remark I.2.8. The statement of Corollary I.2.7 is false if one replaces global with localmaximum (even though local maxima are locally global maxima). A counterexample is forexample

u(x) :=

0 x ≤ 0x3 x > 0

Then u ∈ C2(R) andΔu = u�� ≥ 0 in (−1, 1)


Clearly u attains several local maxima, namely in (−1, 0) we have u ≡ 0, but also clearlyu is not constant. The argument above in the proof of Corollary I.2.7 fails, since the point0 is not a local maximum, and thus the set

A := {x ∈ (−1, 1) : u(x) = 0}is not open.

For the next statement we use the notation A ⊂⊂ B (“A is compactly contained in B)which means that A is bonded and its closure A ⊂ B. I.e. for two open sets A, B thecondition A ⊂⊂ B means in particular that ∂A has positive distance from ∂B.

Corollary I.2.9 (Weak maximum principle). Let Ω ⊂⊂ Rn and u ∈ C2(Ω) ∩ C0(Ω) besubharmonic, i.e. Δu ≥ 0 in Ω. Then

supΩ

u = sup∂Ω

u,

i.e. “the maximal value is attained at the boundary”3.Remark I.2.10. This statement also holds on unbounded sets Ω, one just has to define themeaning of sup∂Ω in a suitable sense (i.e. “sup∂Rn” should be interpreted as lim sup|x|→∞).

Proof of Corollary I.2.9. Clearly by continuitysup

Ωu ≥ sup

∂Ωu.

To prove the converse let us argue by contradiction and assume that(I.2.10) sup

Ωu > sup

∂Ωu.

Since u is continuous and Ω bounded this must mean that there exists a local maximumpoint x0 ∈ Ω such that(I.2.11) u(x0) = sup

Ωu > sup

∂Ωu.

But in view of Corollary I.2.7 (strong maximum principle) u is then constant on theconnected component of Ω containing x0. But this implies that on the boundary of thisconnected component the value of u is still u(x0), which implies

sup∂Ω

u ≥ u(x0).

But this contradicts the assumption (I.2.11). �Remark I.2.11. A particular consequence of the strong maximum principle is the follow-ing. If for Ω ⊂⊂ Rn we have u ∈ C2(Ω) ∩ C0(Ω) satisfying

Δu ≥ 0 in Ωu = g on ∂Ω

3again: think of convex functions which do have this property


for some g ∈ C0(∂Ω). Then the following (equivalent) statements are true:

• If g ≤ 0 but g �≡ 0 on ∂Ω then we have that u < 0 in all of Ω.• If g ≤ 0 then either u ≡ 0 or u < 0 everywhere in Ω.

Such a behaviour is special to the PDEs of order two. Even forΔ2u = Δ(Δu) = 0 in Ω

the above statement may not hold (see e.g. [Gazzola et al., 2010]).Corollary I.2.12 (Strong Comparison Principle). Let Ω ⊂⊂ Rn open and connected.Assume that u1, u2 ∈ C2(Ω) ∩ C0(Ω) satisfy

Δu1 ≥ Δu2 in Ω.

If u1 ≤ u2 on ∂Ω, then exactly one of the following statements is true

(1) either u1 ≡ u2(2) or u1(x) < u2(x) for all x ∈ Ω.

Proof. Let w := u1 − u2, then we have

Δw ≥ 0 in Ωw ≤ 0 in ∂Ω

The claim now follows from Remark I.2.11. �

The maximum principle is a great tool to get uniqueness for linear equations!Theorem I.2.13 (Uniqueness for the Dirichlet problem). Let Ω ⊂⊂ Rn, f ∈ C0(Ω) andg ∈ C0(∂Ω) be given. Then there is at most(!) one solution u ∈ C2(Ω) ∩ C0(Ω) of

Δu = f in Ωu = g on ∂Ω

Proof. Assume there are two solutions, u, v solving this equation. If we set w := u−vthen w is a solution to the equation

Δw = 0 in Ωw = 0 on ∂Ω

In view of Corollary I.2.9 we then havesup

Ωw ≤ sup

∂Ωw = 0.

That is, w ≤ 0 in Ω. But observe that −w solves the same equation, which implies thatsup

Ω(−w) ≤ sup

∂Ω(−w) = 0,

that is −w ≤ 0 in Ω. But this readily implies that w ≡ 0 in Ω, that is v ≡ w. �


So comparison principles are a fantastic tool for obtaining uniqueness for PDEs. Let usalso note that via the so-called Perron’s method, see e.g. [Koike, 2004], one can alsoobtain existence from such comparison principles, but we shall not investigate this furtherhere.

I.2.4. Weak Solutions, Regularity Theory. Now we look at our first encounterwith distributional solutions. Let u ∈ L1

loc(Ω), that is u is a measurable function on Ωwhich is integrable on every compactly contained set K ⊂ Ω, i.e.

�

K|u| < ∞.

u certainly has no reason to be differentiable, it might not even be continuous. How onearth are we going to define

Δu = 0 in Ω?The idea is that if u ∈ C2(Ω) then(I.2.12) Δu = 0 in Ωis equivalent to saying that

(I.2.13)�

Ωu Δϕ = 0 for all ϕ ∈ C∞

c (Ω).

(Recall that C∞c (Ω) are those smooth functions that have compact support supp ϕ ⊂⊂ Ω).

Indeed, for ϕ ∈ C∞c (Ω) and u ∈ C2(Ω) we have by integration by parts

�

Ωu Δϕ =

�

ΩΔu ϕ.

So for u ∈ C2(Ω) we clearly have that (I.2.13) is equivalent to

(I.2.14)�

ΩΔu ϕ = 0 for all ϕ ∈ C∞

c (Ω).

Now if (I.2.12) holds then clearly (I.2.14) holds.

On the other hand assume that (I.2.14) holds, but (I.2.12) is false. That is assume thereis x0 ∈ Ω such that (w.l.o.g.)

Δu(x0) > 0.

Since u ∈ C2(Ω) we have Δu ∈ C0(Ω) and thus there exists a ball B(x0, r) ⊂⊂ Ω suchthat(I.2.15) Δu > 0 on B(x0, r)

Now let ϕ ∈ C∞c (Ω) a bump function (or cutoff function), namely a function ϕ such that

ϕ ≥ 1 in B(x0, r/2) and ϕ ≡ 0 in Ω\B(x0, r), and ϕ ≥ 0 everywhere. These bump functionsreally exist: they can be build by essentially scaled and glued versions of

η(x) :=

e− 1

1−|x|2 for |x| < 10 for |x| > 1


(JoshDif [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], fromWikimedia Commons)

Figure I.2.1. A bump function

See Figure I.2.4.

For this bump function ϕ we have from (I.2.15)�

ΩϕΔu > 0

which contradicts (I.2.14). This proves the equivalence of (I.2.13) and (I.2.12) for C2-functions u.

However, we notice that while (I.2.12) only makes sense for functions u that are twice dif-ferentiable, the statement (I.2.13) makes sense for all functions u ∈ L1

loc(Ω). This warrantsthe following definition:Definition I.2.14 (Weak solutions of the Laplace equation). For a function u ∈ L1

loc(Ω)we say that (I.2.12) is satisfied in the weak sense (or in the distributional sense) if

(I.2.13)�


c (Ω).

holds. The functions ϕ used to “test” the equation are for this very reason called test-functions.

To distinguish the notion of solution we used before, we say that if Δu = 0 in a differentiablefunction sense tjem u is a strong solution or classical solution.

We also have shown above the following statementProposition I.2.15. Let u ∈ C2(Ω). Then the following two statements are equivalent:

(1) u is a weak solution to the Laplace equation Δu = 0 in Ω(2) u is a classical solution of Δu = 0 in Ω.

Weyl proved that this equivalence holds for u ∈ L1loc (i.e. with no a priori differentiablity

at all) – this is our first result of regularity theory: showing that weak solutions which area priori only integrable are actually differentiable. Observe: the reason this works here isthat we have a homogeneous equation Δu = 0, and that Δ is a constant-coefficient linearelliptic operator (and one can spend much more time for proving similar results for more


general linear elliptic operators). Having said that, in some sense, the regularity theoryfor elliptic equations is always somewhat based on the following Theorem, Theorem I.2.16(albeit in a hidden way).

Theorem I.2.16 (Weyl’s Lemma). Let u ∈ L1loc(Ω) for Ω ⊂ Rn open. If u is a weak

solution of Laplace equation, i.e.

(I.2.13)�


c (Ω).

then u ∈ C∞(Ω) and Δu in the classical sense.

Observe that this theorem (rightfully) does not say anything about u on ∂Ω, this is apurely interior result!

The proof of Theorem I.2.16 exhibits the structure that many proofs in PDE have. First onobtains some a priori estimates (namely under the assumption that everything is smoothwe find good estimates). Then we show that these estimates hold also for rough solutionsby an approximation argument.

The a priori estimates for the Laplace equations are called the Cauchy estimates. Theseare truly amazing: They say that if we solve the Laplace equation we can estimate allderivatives, in pretty much any norm simply by the L1-norm of the function.

Lemma I.2.17 (Cauchy estimates). Let u ∈ C∞(Ω) be harmonic, Δu = 0 in Ω. Then wehave for any ball B(x0, r) ⊂ Ω and for any multiindex γ of order k,

|∂γu(x0)| ≤Ck

rn+k�u�L1(B(x0,r)).

In particular we have for any Ω2 ⊂⊂ Ω that

supΩ2

|Dku| ≤ C(dist (Ω2, Ω), k)�u�L1(Ω)

Proof of the Cauchy estimates, Lemma I.2.17. For k = 0 we argue with themean value property for harmonic functions, Theorem I.2.6. We have for any ρ such thatB(x0, ρ) ⊂ Ω and any x ∈ B(x0, ρ/2),

|u(x)| =��

�

B(x,ρ/2)u(z) dz

�� ≤C

ρn

�

B(x,ρ/2)|u(z)| dz ≤ C

ρn

�

B(x0,ρ)|u(z)| dz.

That is, we have obtained that for if Δu = 0 on B(x0, ρ) then

(I.2.16) supB(x0,ρ/2)

|u| ≤ C

ρn�u�L1(B(x0,ρ)).

This proves in particular the case k = 0 (taking ρ =: r).


For the case k = 1 we use a technique called “differentiating the equation” (and in moregeneral situations where this is used in a discretized version we will study later is due toNirenberg, cf. Section III.3.2). Observe that Δu = 0 in Ω implies

Δ∂iu = ∂iΔu = 0 in Ω

So if we set v := ∂iu we have that Δv = 0 in Ω. For x ∈ B(x0, ρ/4), again from themean value property for harmonic functions, Theorem I.2.6, we get with an additionalintegration by parts

|∂iu(x)| =��

�

B(x,ρ/4)∂iu(z) dz

�� = C

ρn

��

�

∂B(x,ρ/4)u(θ) νidHn−1(θ)

��

≤ C

ρnρn−1 sup

B(x,ρ/4)|u|

≤ C

ρnρn−1 sup

B(x0,ρ/2)|u|

Now in view of the estimates in the step k = 0, namely (I.2.16), we arrive at

supB(x0,ρ/4)

|∇u(x)| ≤ C

ρn+1 �u�L1(B(x0,ρ)).

Differentiating the equation again, we find by induction that (the constant changes in eachappearance!)

|∇ku(x0)| ≤ supB(x0,4−kρ)

|∇ku(x)| ≤ C

ρn+1 �∇k−1u�L1(B(x0,41−kρ)) ≤ . . . ≤ C

ρn+k�u�L1(B(x0,ρ).

If we want to show the estimate on Ω2 ⊂⊂ Ω we now pick ρ < dist (Ω2, ∂Ω) and obtainthe claim. �

Proof of Weyl’s Lemma: Theorem I.2.16. We use a mollification argument, i.e.we approximate u with smooth functions uε that also solve (in the classical sense) theLaplace equation.

Let η ∈ C∞c (B(0, 1)) be another bump function, this time with the condition η(x) = η(−x),

i.e. η is even, η ≥ 0 everywhere, and normalized such that�

Rnη = 1.

We rescale η by a factor ε > 0 and set

ηε(x) := ε−nη(x/ε).


Then the convolution4 is defined asuε(x) := ηε ∗ u(x) :=

�

Rnηε(y − x) u(y) dy

Clearly this is not well-defined for all x, if u ∈ L1loc(Ω) only. But it is defined for all x ∈ Ω

such that dist (x, ∂Ω) > ε, since supp ηε(· − x) ⊂ B(x, ε).

But observe that derivatives on uε hit only the kernel ηε (which is smooth) (there is adominated convergence to be used to show that, and for this we need L1

loc!)

∂γuε(x) := ηε ∗ u(x) :=�

Rn∂γηε(y − x) u(y) dy

That is uε ∈ C∞(Ω−ε) whereΩ−ε = {x ∈ Ω, dist (x, ∂Ω) > ε}

The fun part (which we used above already) is that convolutions behave well with differ-ential operators, namely we will show now that Δuε = 0 in Ω−ε:

For this let ψ ∈ C∞c (Ω−ε) a testfunction, then we have

�

Ω−ε

uε(x) Δψ(x) dx =�

Rn

�

Rnu(y)ηε(x−y) Δψ(x) dy dx =

�

Rnu(y)

�

Rnηε(x−y) Δψ(x) dx dy

Now, by integration by parts (for any fixed y ∈ Rn)�

Rnη(x−y) Δψ(x) dx =

�

RnΔxηε(x−y) ψ(x) dx =

�

RnΔyηε(x−y) ψ(x) dx = Δy

�

Rnηε(x−y) ψ(x) dx

So if we setϕ(y) := ηε ∗ ψ(y) ≡

�

Rnηε(x − y) ψ(x) dx

then we have by the support condition on ψ that ϕ ∈ C∞c (Ω), and thus

�

Ω−ε

uε(x) Δψ(x) dx =�

Rnu(y) Δϕ(y) dy

(I.2.13)= 0.

This argument works for any ψ ∈ C∞c (Ω−ε), that is uε is weakly harmonic in Ω−ε. But

since uε ∈ C∞(Ω−ε) this implies in view of Proposition I.2.15 that in the strong senseΔuε = 0 in Ω−ε.

So now uε is a smooth solution to Laplace’s equation, so we use the a priori estimates ofLemma I.2.17.

Fix Ω2 ⊂⊂ Ω. Between Ω2 and Ω we can squeeze two more set Ω3, and Ω4,Ω2 ⊂⊂ Ω3 ⊂⊂ Ω4 ⊂⊂ Ω.

For any ε small enough, namelyε < dist (Ω3, ∂Ω4) and ε < dist (Ω3, ∂Ω4)

4we have seen this operation for the Fourier Transform argument above after (I.2.3), there we used anonsmooth kernel | · |2−n for the convolution


we have that Δuε = 0 in Ω3, so by the Cauchy estimates, Lemma I.2.17, we have for anyk ∈ N

supΩ2

|∇kuε| ≤ C(k, Ω2, Ω3) �uε�L1(Ω3).

Now we estimate, by Fubini,

�uε�L1(Ω3) ≤�

Ω3

�

Rn|ηε(x − y)| |u(y)| dy dx =

�

Rn|u(y)|

�

Ω3|ηε(x − y)|dx dy

Since ε is small enough we have that

supp��

Ω3|ηε(x − ·)|dx

�⊂ Ω4.

So we get

�uε�L1(Ω3) ≤ �u�L1(Ω4) supy∈Rn

�

Ω3|ηε(x − y)|dx ≤ �u�L1(Ω4)

�

Rn|ηε(z)|dz.

Now we use the definition of ηε to compute via substitution5�

Rn|ηε(z)|dz = ε−n

�

Rn|η(z/ε)|dz = ε−n

�

Rn|η(z/ε)|dz =

�

Rn|η(z)|dz = 1.

The last equality is due to the normalization of η,�

η = 1.

That is, we have shown that for any k ∈ N ∪ {0}supΩ2

|∇kuε| ≤ C(k, Ω2, Ω3) �u�L1(Ω4),

and the right-hand side is finite since u ∈ L1loc(Ω) and Ω4 ⊂⊂ Ω.

This estimate holds for any ε > 0, so uε and all its derivative are uniformly equicontinuous(in ε). By Arzela-Ascoli (and a diagonal argument in k) we find a converging subsequenceε → 0 and a function u0 ∈ C∞(Ω2) such that for any k ∈ N ∪ {0}.

|∇kuε(x) −∇ku0(x)| ε→0−−→ 0 locally uniformly in Ω2.

We claim that u = u0 in almost every point (since u is an L1loc-function it is actually a the

class of maps equal up to almost every point, u0 is a continuous representative of the classu). Indeed, by the normalization

�η = 1 which implies

�ηε = 1 we have

|uε(x) − u(x)| =��

ηε(y − x) (u(y) − u(x)) dy

�� ≤ C(η)�

B(x,ε)|u(y) − u(x)| dy.

So, by the Lebesgue differentiation theorem, we have for almost every x ∈ Ω2,limε→0

|uε(x) − u(x)| = 0,

that isu0 = u a.e. in Ω2.

Thus u ∈ C∞(Ω2), and Δu = 0 in classical sense in Ω2.5observe for z = z/ε we have in n space dimensions dz = ε−ndz


Since this holds for any Ω2 ⊂ Ω we have shown

u ∈ C∞(Ω), and Δu = 0 in classical sense in Ω. �

Corollary I.2.18 (Liouville). Let u ∈ C2(Rn) and Δu = 0 in all of Rn. If u is a boundedfunction then u ≡ const.

Proof. Fix x0 ∈ Rn. In view of Lemma I.2.17 we have for such a function u, for anyradius r > 0,

|Du(x0)| ≤C

rn+1 �u�L1(B(x0,r))

If u is bounded,�u�L1(B(x0,r)) ≤ C rn sup

Rn|u| < ∞

and thus|Du(x0)| ≤ Cr−1 sup

Rn|u|.

This holds for any r > 0, so if we let r → ∞, we get

|Du(x0)| = 0,

which holds for any x0 ∈ Rn. That is, Du ≡ 0, and by the fundamental theorem of calculusthis means u is a constant. �

I.2.5. Harnack Principle. Above we learned, e.g. in Corollary I.2.7 of the strongmaximum principle. Another type of maximum principle is the Harnack inequality.

Theorem I.2.19. Let Ω ⊂ Rn open. For any open, connected, and bounded U ⊂⊂ Ω thereexists a constant C = C(U, Ω) such that for any solution u ∈ C2(Ω) with u ≥ 0 and suchthat

Δu = 0 in Ωwe have

supU

u ≤ C infU

u

Proof. The proof is based on the mean value formula, Theorem I.2.5, namely for anyx ∈ U and any r < dist (U, ∂Ω) we have

u(x) =�

B(x,r)u(z) dz

Let now R := 14dist (U, ∂Ω). For any x0 ∈ U and any x ∈ B(x0, R) we have (here we use

u ≥ 0 and that B(x, R) ⊂ B(y, 2R) for x, y ∈ B(x0, R))

u(x) =�

B(x,R)u(z) dz ≤ 2n

�

B(y,2R)u(z) dz = 2n u(y).


Again, this holds for any x, y ∈ B(x0, R). Taking the supremum for x ∈ B(x0, R) and theinfimimum on y ∈ B(x0, R) we get(I.2.17) sup

B(x0,R)u ≤ 2n inf

B(x0,R)u.

That is we have the Harnack principle on any Ball B(x0, R). Since U is bounded, openand compactly contained in Ω we can now cover all of U by finitely many balls (B�)N

�=1which lie inside Ω centered at points in U and of radius R. The supremum of u on Ω canbe located in some ball Bi1 , in the sense that(I.2.18) sup

Uu ≤ sup

Bi0

u,

and the infimum is attained in some ball Bi2 in the sense of(I.2.19) inf

Uu ≥ inf

Bi1u.

(Observe that Bi0 , Bi1 may contain points in Ω\U . Since U is connected, there is a chainof balls (B�i

)Ki=1, K ≤ N , such that B�i

∩ B�i+1 �= ∅ and such that �1 = i1 and �K = i2.Observe that this implies in particular,(I.2.20) inf

B�i

u ≤ supB�i+1

u.

Then we can conclude via the following chain of estimates

supU

u(I.2.18)≤ sup

Bi0

u = supB�1

u(I.2.17)≤ 2n inf

B�1u

(I.2.20)≤ 2n sup

B�2

u ≤ . . . ≤ K2n infB�K

u(I.2.19)≤ K2n inf

Uu.

�

I.2.6. Green Functions. Our next goal are Green’s functions. In some way Greenfunctions are a restriction of the fundamental solution to domains Ω ⊂ Rn factoring in alsoboundary data.

Recall that for the fundamental solution Φ we showed in Theorem I.2.3 that for the Newtonpotential

(I.2.21) u(x) :=�

RnΦ(x − y)f(y) dy

we have Δu = f . It is an interesting observation that (for reasonable f) we havelim

|x|→∞u(x) = 0.

That is the Newton potential approach solves an equation of

Δu = f in Rn

u = 0 on the boundary, i.e. for |x| → ∞.


The Greens function is a way to restrict this construction to domains Ω. Instead of theFundamental solution Φ(x − y) we get the Green kernel G(x, y) . Instead of the Newtonpotential we consider

u(x) =�

ΩG(x, y) f(y) dy

and hope that this object solves

Δu = f in Ωu = 0 on ∂Ω.

The Greens function G (which depends on Ω) can be computed explicitely only for veryspecific Ω (balls, half-spaces) – which is somewhat related to the fact that there is notnecessarily a reasonable Fourier transform for generic sets Ω.

But one can abstractly show that the Green functions exists for reasonable sets Ω. The ideais as follows: We know that the Newton potential as in (I.2.21) solves the right equationΔu = f , but it does not satisfy u = 0 on ∂Ω. So let us try to correct the Newton potentialand choose the Ansatz

u(x) :=�

ΩΦ(x − y) f(y) dy −

�

ΩH(x, y) f(y) dy

By our computatins for Theorem I.2.3 we have that then for x ∈ Ω

Δu(x) := f(x) −�

ΩΔxH(x, y) f(y) dy,

so it would be nice ifΔxH(x, y) = 0 ∀ x, y ∈ Ω.

Moreover we would like that u(x) = 0 on ∂Ω, which would be satisfied ifΦ(x − y) = H(x, y) ∀x ∈ ∂Ω, y ∈ Ω.

That is, for each fixed y ∈ Ω we should try to find a function H(·, y) that solves

(I.2.22)

ΔxH(·, y) = 0 in Ω,

H(·, y) = Φ(· − y) on ∂Ω.

Observe that for fixed y ∈ Ω the boundary condition Φ(· − y) ∈ C∞(∂Ω) is a smoothfunction, since for y ∈ Ω we clearly have

infx∈∂Ω

|x − y| > 0.

That is, there is a good chance to solve this equation (I.2.22) (and from Theorem I.2.13we know that there is at most one solution).

Definition I.2.20 (Green function). For given Ω, if there exists H as in (I.2.22) then wecall

G(x, y) := Φ(x − y) − H(x, y)the Green function on Ω.


One can show that G is symmetric, i.e. that

(I.2.23) G(x, y) = G(y, x) ∀ x �= y ∈ Ω

While the Green function are usually not explicit, some properties and estimates can beshown, and there is an extensive research literature on the subject, e.g. see [Littman et al., 1963].The Green function is also specially important from the point of view of stochastic pro-cesses, see e.g. [Chen, 1999].

We will only investigate the most basic property:

Theorem I.2.21. Let Ω ⊂⊂ Rn, ∂Ω ∈ C1 f ∈ C0(Ω) and g ∈ C0(∂Ω). Assume thatu ∈ C2(Ω) ∩ C0(Ω) is a solution to

(I.2.24)−Δu = f in Ωu = g on ∂Ω

Then if G is the Green function for Ω from Definition I.2.20 we have for any x ∈ Ω,

u(x) =�

ΩG(x, y) f(y) dy −

�

∂Ωg(θ)∂ν(θ)G(x, θ) dHn−1(θ).

Proof. Recall the Gauss-Green formula6 on (smooth enough) domains A,

(I.2.25)�

Au(y)Δv(y) − Δu(y) v(y) dy =

�

∂Au(θ)∂νv(θ) − ∂νu(θ) v(θ) dHn−1(θ).

We apply this to formula to A = Ω\B(x, ε) and v(y) := G(x, y). Observe that by symmetryof G, (I.2.23),

ΔyG(x, y) = ΔxG(x, y) = 0 x �= y,

so, also in view of (I.2.24), (I.2.25) becomes

(I.2.26) −�

AG(x, y) f(y) dy =

�

∂Au(θ)∂νG(x, θ) − ∂νu(θ) G(x, θ) dHn−1(θ).

Now we argue as in the proof of Theorem I.2.3. Observe that H is a smooth function.

We have�

∂Au(θ)∂νG(x, θ)dHn−1(θ)

=�

∂Ωg(θ)∂νG(x, θ) −

�

∂B(x,ε)u(θ)∂νΦ(x − θ) dHn−1(θ) +

�

∂B(x,ε)u(θ)∂νH(x − θ) dHn−1(θ)

ε→0−−→�

∂Ωg(θ)∂νG(x, θ) − u(x) + 0.

6this is a special case of the integration by parts formula


and �

∂A∂νu(θ)G(x, θ)dHn−1(θ)

=�

∂Ω∂νu(θ)G(x, θ) −

�

∂B(x,ε)∂νu(θ)G(x, θ) dHn−1(θ)

=0 −�

∂B(x,ε)∂νu(θ)G(x, θ) dHn−1(θ)

ε→0−−→0.

This proves the claim. �

In special situations one can actually construct explicit Green’s function. Let us firstlyconsider the Half-space

Rn+ = {x = (x1, . . . , xn) ∈ Rn : xn > 0} .

So we need to find a solution to the equation

ΔxH(·, y) = 0 in Rn+,

H(·, y) = Φ(· − y) on Rn−1 × {0} ≡ ∂Rn+.

Since H at the boundary has to coincide with Φ it is likely that H should be somewhat ofthe form of Φ – only the singularity has to be getten rid of – the idea is a reflection:

H(x, y) := Φ(x − y∗)where

y∗ = (y1, . . . , yn)∗ = (y1, . . . , yn−1,−yn).It is a good exercise to check that

(1) H is symmetric, H(x, y) = H(y, x)(2) H is smooth in Rn

+ × Rn+ (since x∗ = y implies xn = −yn, so x and y cannot both

lie in the upper half-space if this happens)(3) The reflection does not change the PDE, namely ΔxH = 0 for x, y ∈ Rn

+.(4) Indeed H(x, y) = Φ(x − y) for x ∈ Rn−1 × {0} and y ∈ Rn

+.

So we setG(x, y) := Φ(x − y) − Φ(x − y∗) = Φ(x − y) − Φ(x∗ − y)

When we now use the representation formula, as in Theorem I.2.21, then we need tocompute ∂ν(y)G(x, y) for y ∈ Rn−1 × {0}. Observe that the outwards unit normal ν(y) =(0, . . . , 0,−1), so we compute

∂ν(y)G(x, y) = −∂ynΦ(x − y) + ∂ynΦ(x∗ − y) = cnxn − yn

|x − y|n − cnxn + yn

|x − y|n = cnxn

|x − y|n .


If we write the variables in Rn+ as x = (x�, xn), x� ∈ Rn−1 and xn > 0, then as in Theo-

rem I.2.21 we indeed obtain, e.g., if

u(x) := cn

�

Rn−1

xn

(|x� − y�|2 + |xn − yn|2)n2

g(y�) dy�

then u satisfies indeed (for “reasonable” g)

Δu = 0 in Rn+

limxn→0 u(x) = g(x�)limxn→∞ u(x) = 0.

The formula for u is called the Poisson formula on the Half-space Rn+, also the harmonic

extension of g from Rn−1 to Rn+.

The other situation where we can compute the Green’s function is the ball. For simplicitylet us consider Ω = B(0, 1), the unit ball centered at zero. Again the first goal is to findH(x, y) that corrects the fundamental solution. In the case of the half-space Rn

+ we setH(x, y) = Φ(x − y), i.e. we reflected the y-variable in a way that did not interfere withthe PDE but removed the singularity (and coincided with Φ(x − y) on the boundary.

So lets do the same for the ball. The canonical operation that reflects points from the ballB(0, 1) outside and vice versa is called the inversion at a sphere,

y∗ := y

|y|2 : B(0, 1) → B(0, 1)c.

(Although it is not explicitely used here, it is good to know: the inversion at the sphere isa conformal transform, i.e. it preserves angles). So a first attempt would be to set

H(x, y) := Φ��x − y

|y|2��

�,

which takes care of the singularity of Φ (for y, x ∈ B(0, 1) we have |x − y|y|2 | �= 0, and

does not disturb the PDE for G(x, y). However such a G(x, y) is not equal to Φ(x − y) for|x| = 1. So we need to adapt G to the boundary data. Observe that for |x| = 1,

|y|2��x − y

|y|2��

2

=|y|2�

|x|2 + 1|y|2 − 2�x,

y

|y|2 ��

=�|y|2|x|2 + 1 − 2�x, y�

�

|x|=1=�|y|2 + |x|2 − 2�x, y�

�

=|x − y|2.


That is why we set

GB(0,1)(x, y) := Φ�

|y|��x − y

|y|2��

�,

which satisfies all the requested properties.

From this we obtain (without proof)

Theorem I.2.22 (Poisson’s formula for the ball). Assume g ∈ C0(∂B(0, r)). Define

u(x) := cn

�

∂B(0,r)

1r

r2 − |x|2|x − θ|n g(θ) dHn−1(θ)

Then

(1) u ∈ C∞(B(0, r))(2) Δu = 0 in B(0, r)(3)

limB(0,r)�x→x0

u(x) = g(x0) for any x0 ∈ ∂B(0, r)

I.2.7. Methods from Calculus of Variations – Energy Methods. As we haveseen, comparison principles is a strong tool for uniqueness (and also existence). Thesearguments also work in some situations of nonlinear pdes, where the theory of distribu-tional solutions does not work, but the theory of Viscosity solutions can be applied, see[Koike, 2004].

On the other hand, the comparison methods are (currently) restricted to first or second-order equations, and to scalar equations. For systems or higher-order PDEs they seem notto be that helpful.

In this section we have a short look on energy methods, which is a basic tool of distributionaltheory. They do not rely on any comparison principle, and they are often used for higher-order differential equations and systems. On the other hand for some fully nonlinearequations (“non-variational” equations, equtions “not in divergence form”) they cannot bewell applied.

The ideas should be reminiscent of the arguments we employed for the weak solutions inTheorem I.2.16.

Assume that we have

(I.2.27)

Δu = f in Ωu = 0 on ∂Ω

We have seen before Theorem I.2.16 that this equation is related to the integral equation�

ΩDu · Dϕ + fϕ = 0 ∀ϕ ∈ C∞

c (Ω).


The interesting point is that this expression is a Frechet-Derivative of a function acting onthe map u in direction ϕ.

Indeed one can characterize solutions as minimizers of an energy functional. This is some-times called the Dirichlet principle.

Theorem I.2.23 (Energy Minimizers are solutions and vice versa). Assume f ∈ C0(Ω).

Denote the class of permissible functions

X :=�u ∈ C2(Ω), u = 0 on ∂Ω

�

and define the energyE(u) :=

�

Ω

12 |Du|2 + fu = 0.

Let u ∈ X be a minimizer of E in X, i.e.E(u) ≤ E(v) ∀v ∈ X.

Then u solves (I.2.27).

Conversely, if u ∈ X solves (I.2.27), then u is a minimizer of E in the set X.

Proof. We compute what is called the Euler-Lagrange-equations of E : Let ϕ ∈ C∞c (Ω),

then certainly u + tϕ ∈ X for all t ∈ R. That is the minimizing property says that thefunction

E(t) := E(u + tϕ)has a minimum in t = 0. By Fermats theorem (one checks easily that E is differentiablein t)

d

dt

��t=0

E(t) ≡ E �(0) = 0.

Now observe thatd

dt

��t=0

|D(u + tϕ)|2 = 2�Du, Dϕ�and

d

dt

��t=0

f (u + tϕ) = f ϕ.

Thus, we arrive at0 = d

dt

��t=0

E(t) =�

ΩDu · Dϕ + fϕ = 0.

That is, u is a weak solution of (I.2.27). But u ∈ C2(Ω), so we argue similar to the proofof Proposition I.2.15:

By an integration by parts (for ϕ ∈ C∞c (Ω) there are no boundary terms), we thus have

0 =�

ΩDu · Dϕ + fϕ = 0 = −

�

Ω(Δu − f)ϕ.


Since Δu − f is continuous, and the last estimate holds for any smooth ϕ ∈ C∞c (Ω) we

get that (as for Proposition I.2.15, or otherwise by the fundamental lemma of calculus ofvariations, Lemma I.2.24,

Δu − f = 0.

That is the first claim is proven: minimizers are solutions.

For the converse assume u solves (I.2.27). Let w be any other map in X. Then we have�

Ω(Δu − f)(u − w) = 0.

Observe that u and w have the same boundary value 0 on ∂Ω. Thus, when we performthe following integration by parts we do not find boundary terms,

(I.2.28) 0 = −�

Ω∇u · ∇(u − w) + f(u − w) = 0.

Now we compute (using Young’s inequality or Cauchy-Schwarz 2ab ≤ a2 + b2)�

Ω|∇u|2 + fu

(I.2.28)=�

Ω∇u · ∇w + fw

≤�

Ω

12 |∇u|2 + 1

2 |∇w|2 + fw

=12

�

Ω|∇u|2 + E(w)

Subtracting 12�

Ω |∇u|2 from both sides in the estimate above we obtainE(u) ≤ E(w).

That is, we have shown: if u solves the equation, then u is a minimizer. �

Above we have used the following statement.

Lemma I.2.24 (Fundamental Lemma of the Calculus of Variations). Let Ω ⊂ Rn be anyopen set and assume f ∈ L1

loc(Ω), i.e. for any Ω� ⊂⊂ Ω we have�

Ω�|f | < ∞.

(1) If�

Ωf(x) ϕ(x) ≥ 0 for all ϕ ∈ C∞

c (Ω) that are nonnegative, ϕ ≥ 0,

thenf ≥ 0 almost everywhere in Ω.

(2) If�

Ωf(x) ϕ(x) = 0 for all ϕ ∈ C∞

c (Ω) that are nonnegative, ϕ ≥ 0,

thenf ≡ 0 almost everywhere in Ω.


The proof is left as an exercise, it is a combination of convolution arguments as in Theo-rem I.2.16 and the argument used for Proposition I.2.15.Theorem I.2.25 (Uniqueness). Assume f ∈ C0(Ω) ∩ L1(Ω)

Denote the class of permissible functionsX :=

�u ∈ C2(Ω), u = 0 on ∂Ω

�

Then there is at most one solution u ∈ X to (I.2.27)

Proof. Assume u, w ∈ X are two solutions, thenΔ(u − w) = 0.

Multiplying by u − w and integrating by parts (observe that there are no boundary termssince u = w on ∂Ω, we obtain �

Ω|∇(u − w)|2 = 0.

But this implies that ∇u − w ≡ 0, so u − w ≡ const. Since u = w on the boundary thatconstant is zero, and u ≡ w. �

These methods can be extended, e.g. for higher order differential equations (where nomaximum principle holds), e.g.

Δu = f in Ωu = 0 on ∂Ω∂νu = 0 on ∂Ω

See exercises.

CHAPTER 2

Second order linear elliptic equations, and maximum principles

From now on, we will use the Einstein summation convention, which is summarized as “wesum over repeated indices”.

For example, for vectors a = (a1, . . . , an), b = (b1, . . . , bn), we write

�a, b� = aibi :=n�

i=1aibi.

This notation is used a lot in coordinate and tensor computation in physics (e.g. relativity,hence the name). In contrast to Physics (and Geometry) we will not care about “raised”and “lowered” indices. I.e.

aibi = aibi = aibi = aibi ≡

n�

i=1aibi

We will also use this for matrices, namely if A = (aij)ni,j=1 and b = (b1, . . . , bn) is a vector

aijbj =

n�

j=1aijb

j ≡ (Ab)i.

Observe that in particular we implicitely always assume we know what dimension n wething about. This notation needs some time to get accostumed to, but it is very beneficialfor computations.

II.1. Linear Elliptic equations

Second order elliptic equationd are a class of equations that in some sense are governed bythe Laplacian operator.

Definition II.1.1 (Linear elliptic equations). (1) (“non-divergence form”) linear sec-ond order operators are defined to be operators of the form

L := aij∂ij + bi∂i + c

for coefficents aij, bi, c : Ω → R. They act as follows on functions u ∈ C2(Ω)Lu(x) := aij(x)∂iju(x) + bi(x)∂iu(x) + c(x) u(x).

L is called a constant coefficient operator, if the coefficients aij, bi and c are allconstant.

40

II.2. MAXIMUM PRINCIPLES 41

(2) (“divergence form”) linear second order operators are defined to be operators ofthe form

L := ∂i (aij∂j) + bi∂i + c

for coefficents aij, bi, c : Ω → R. They act as follows on functions u ∈ C2(Ω)

Lu(x) := ∂i (aij(x)∂ju(x)) + bi(x)∂iu(x) + c(x) u(x).

(3) Clearly, divergence on non-divergence form are very similar if aij is smooth enough,but they are not if that is not the case (and in general

(4) (divergence-form or non-divergence form) operators L are called elliptic (also oftencalled uniformly elliptic and bounded) if there exists an ellipticity constants Λ > 0such that

ξT Aξ ≡ ξiaijξj ≥ 1

Λand

supΩ

|aij|, |bi|, |c| < ∞.

For simplicity, although this is not strictly necessary we will below always assume A issymmetric.

Example II.1.2. • The operator Δ is clearly elliptic in the above sense, with

aij = δij :=

1 if i = j

0 else

• Operators like div (|∇u|p−2∇u) are not (uniformly elliptic), since |∇u| = 0 cannotbe excluded. These operators are called degenerate elliptic.

Definition II.1.3. u ∈ C2(Ω) is called a subsolution of −Lu = f for an elliptic operatorL, if

−Lu ≤ 0 in Ωand a supersolution if

−Lu ≥ 0 in Ω.

u ∈ C2(Ω) is called a solution if it is both sub- and supersolution.

In the following we will restrict ourselves to elliptic non-divergence operators!

II.2. Maximum principles

The first result is a generalization of the weak maximum principle for Δ, Corollary I.2.9.


Theorem II.2.1 (Weak maximum principle for c = 0). Let Ω ⊂⊂ Rn, u ∈ C2(Ω) ∩ C0(Ω)be an L-subsolution, i.e.(II.2.1) − Lu ≤ 0 in ΩIf L is (non-divergence form) linear elliptic operator with c ≡ 0, then

supΩ

u = sup∂Ω

u.

If instead of (II.2.1) we have−Lu ≥ 0 in Ω

theninfΩ

u = inf∂Ω

u.

Proof. First we assume instead of (II.2.1)(II.2.2) − Lu > 0 in Ω

Clearly, by continuity of u in Ω,sup

Ωu ≥ sup

∂Ωu

If we hadsup

Ωu > sup

∂Ωu,

then we would find the global (and thus a local) maximum x0 ∈ Ω, at which we haveDu(x0) = 0 and D2u(x0) ≤ 0. But this implies (recall c ≡ 0)

Lu(x0) = aij(x0)∂iju(x0) + bi(x0) ∂iu(x0)� �� =0

Since aij(x0) is elliptic, and ∂iju(x0) ≥ 0 we haveaij(x0)∂iju(x0) ≥ 0.

(This is a general Linear Algebra fact, if A, B are symmetric, nonnegative matrices, thentheir Hilbert-Schmidt Scalar product A : B := aijbij ≥ 0.) That is, we have

Lu(x0) ≥ 0which is a contradiction to (II.2.2).

We conclude that under the assumption (II.2.2) we havesup

Ωu = sup

∂Ωu.

In order to weak the assumption to (II.2.2) we consider, for some γ > 0, vγ(x) := eγx1 ,where x1 is the first component of x = (x1, . . . , xn). Observe that

Lvγ(x) =�a11(x)γ2 + b1(x)γ

�eγx1


Since L is elliptic we have a11 ≥ 1Λ and b1 ≥ −Λ, so

Lvγ(x) = a11(x)γ2 + b1(x)γ ≥ eγx1γ� 1

Λγ − Λ�

.

If we choose γ = 3Λ we thus findLvγ(x) > 0 in Ω.

Consequently, under the assumption (II.2.1) we have for any ε > 0, for wε := u + εvγ,Lwε(x) > 0 in Ω.

and thus by the first stepsup

Ωwε = sup

∂Ωwε

Since wε = u + εvγ and vγ is continuous (and Ω is bounded) we have��sup

Ωu − sup

∂Ωu

�� ≤ C(Ω)ε.

Letting ε → 0 we obtain the claim.

The inf claim follows by taking −u instead of u. �

Also in the case c �≡ 0 a type of weak maximum principle holds (essentially mimmickingthe above argument):

Theorem II.2.2 (Weak maximum principle for c ≤ 0). Let Ω ⊂⊂ Rn, u ∈ C2(Ω) ∩ C0(Ω)solve

−Lu ≤ 0.

If c ≤ 0 in Ω we havesup

Ωu ≤ sup

∂Ωu+,

where u+, u− denotes the positive part of u, namelyu+ = max{0, u}

If on the other hand−Lu ≥ 0.

and c ≤ 0 in Ω we haveinfΩ

u ≥ inf∂Ω

(−u−),

where u+, u− denotes the positive part of u, namelyu+ = max{0, u}, u− = −min{0, u}

In particular, if Lu = 0 thensup

Ω|u| ≤ sup

∂Ω|u|


Proof. Let us assume −Lu ≤ 0. First we observe that if

supΩ

u ≤ 0

then there is nothing to show, since we have u+ ≥ 0 by definition and thus

supΩ

u ≤ 0 ≤ sup∂Ω

u+.

So w.l.o.g. we may assume that supΩ u > 0. Set

Ω+ := {x ∈ Ω : u(x) > 0} .

Since u is continuous Ω+ = u−1((0,∞)) is an nonempty, open set.

Define the elliptic operator L0 by

L0u := Lu − cu = aij∂iju + bi∂iu.

Since −Lu ≤ 0 we have −L0u ≤ cu ≤ 0 in Ω+ — since by assumption c ≤ 0. So we have,using also Theorem II.2.1,

supΩ

uu≤0: Ω\Ω+

≤ supΩ+

uII.2.1= sup

∂Ω+u = sup

∂Ω+u+ ≤ sup

∂Ωu+.

In the last step we used that ∂Ω+ ⊂ Ω can be split into two parts: the part ∂Ω+ ⊂ Ω (onthis part we have u = u+ = 0), and the part ∂Ω+ ⊂ ∂Ω where u+ ≥ 0.

This settles the claim for −Lu ≤ 0, the claim for −Lu ≥ 0 is a similar argument.

For the last case assume that −Lu = 0. By the arguments before we have then (observethat |u| = u+ + u−).

supΩ

u ≤ sup∂Ω

u+ ≤ sup∂Ω

|u|.

andinfΩ

u ≥ inf∂Ω

(−u−),

which can be rewritten as

− infΩ

u ≤ − inf∂Ω

(−u−) = sup∂Ω

(u−) ≤ sup∂Ω

|u|.

Now at least one of the following cases holds:

supΩ

|u| = supΩ

u, or supΩ

|u| = − infΩ

u

but in both cases the estimates above imply

supΩ

|u| ≤ sup∂Ω

|u|

�


Example II.2.3 (Counterexample for c ≥ 0). ConsiderLu = Δu + 5u

for Ω = (−1, 1) × (−1, 1). Thenu = (1 − x2) + (1 − y2) + 1

satisfies−Lu = −1 ≤ 0.

However,sup

Ωu ≥ u(0) = 3,

andsup∂Ω

u = 1.

As it was the case for the Δ-operator, Theorem I.2.13, the weak maximum principle impliesuniqueness results.

Corollary II.2.4 (Uniqueness for the Dirichlet problem). Let L be as above a non-divergence form linear elliptic operator, Ω ⊂⊂ Rn with smooth boundary, c ≤ 0, f ∈ C0(Ω),g ∈ C0(∂Ω). Then there exists at most one solution u ∈ C2(Ω) ∩ C0(Ω) of the Dirichletboundary problem

Lu = f in Ωu = g on ∂Ω

Proof. exercise. �Corollary II.2.5 (Comparison principle). Let L be a linear elliptic differential operator(non-divergence form), and assume that c ≤ 0 in Ω ⊂⊂ Rn. Let u, v ∈ C2(Ω) ∩ C0(Ω)satisfy −Lu ≤ −Lv in Ω. Then u ≤ v on ∂Ω implies u ≤ v in Ω.

Proof. exercise �Corollary II.2.6 (Continuous dependence on data). Let L be a linear elliptic differentialoperator (non-divergence form), and assume that c ≤ 0 in Ω ⊂⊂ Rn.

Let u ∈ C2(Ω) ∩ C0(Ω) satisfy−Lu = f in Ωu = g in Ω

where f ∈ C0(Ω) and g ∈ C0(∂Ω).

Then for some constant C = C(b, Λ) we havesup

Ω|u| ≤ sup

∂Ω|g| + C(b, Λ) sup

Ω|f |.


Proof. exercise �

Our next goal is the the strong maximum principle, for this we use the following result byHopf:

Lemma II.2.7 (Hopf Boundary point Lemma). Let B ⊂ Rn be a ball, and let L be asabove. Let u ∈ C2(B) ∩ C0(B) and assume that for x0 ∈ ∂B we have

• u(x) < u(x0) for all x ∈ B• −Lu ≤ 0 in B.• One of the following

(1) c ≡ 0(2) c ≤ 0 and u(x0) ≥ 0(3) u(x0) = 0Then for ν the outwards facing normal of B at x0 (i.e. if B = B(y0, ρ) then forν = x0−y0

ρ

∂νu(x0) > 0,

if that derivative exists.

Proof. W.l.o.g. we may assume(II.2.3) B = B(0, R), c ≤ 0, u(x0) = 0, u < 0 in B(0, R) :Indeed, the condition B = B(0, R) can be assumed simply by shifting. As for the otherconditions set (recall that c+ = max{c, 0})

L := L − c+.

andu := u − u(x0).

Then in B,−Lu = −(L − c+)(u − u(x0)) = −Lu + c+u + cu(x0) − c+u(x0) ≤ c+ (u − u(x0)) + c u(x0)

If c ≡ 0 then we readily have −Lu ≤ 0.

If c ≤ 0 we have c+ ≡ 0, and again obtain −Lu ≤ 0.

If u(x0) = 0 then c+u ≤ 0, since u ≤ u(x0) = 0 by assumption.

Since c − c+ ≤ 0 we observe that L is an operator that satisfies the missing conditions in(II.2.3). Thus, indeed, (II.2.3) can be assumed w.l.o.g.

So assume (II.2.3) from now on.

Set for some α > 0vα(x) := e−α|x|2 − e−αR2

.


Clearly 0 ≤ vα ≤ 1 in B = B(0, R). Moreovervα ≡ 0 on ∂B(0, R).

For ρ ∈ (0, R) denote by A(ρ, R) the annulus B(0, R)\B(0, ρ). We will show next(II.2.4) For any ρ ∈ (0, R) there exists α > 0 such that − Lvα < 0 in A(ρ, R)For this we first compute(II.2.5) ∂ivα(x) = −2αxie

−α|x|2 .

Next we compute∂ijvα(x) =

�−2αδij + 4α2xixj

�e−α|x|2

so (using the ellipticity conditions, aijxixj ≥ |x|2, and |a|, |b|, |c| ≤ Λ,−Lv(x) = − aij∂ijv − bi∂iv − cv

= − aij

�−2αδij + 4α2xixj

�e−α|x|2 − bi

�−2αxie

−α|x|2�− ce−α|x|2 + ce−αR2

� �� ≤0

≤�2αΛ − 4α2Λ|x|2 + 2αΛ|x| + Λ

�e−α|x|2 .

That is, for x ∈ A(ρ, R),−Lv(x) ≤

�−4α2Λρ2 + 2αΛ + 2αΛR + Λ

�

� �� ≤0 for α � 1

e−α|x|2

If we take α large, the (negative) α2-term dominates, that is for α � 1 (depending onρ > 0), Λ and R we have (II.2.4).

Next, we consider the equation for u + εv, which in view of (II.2.4) becomes−L(u + εv) < 0 in A(ρ, R).

The weak maximum principle, Theorem II.2.2, implies(II.2.6) sup

A(ρ,R)u + εv ≤ sup

∂A(ρ,R)(u + εv)+.

The boundary ∂A(ρ, R) is the union of ∂B(0, R) where v ≡ 0 and since u is continuousand u < 0 in B(0, R) we have u ≤ 0 on ∂B(0, R). That is (u + εv)+ = 0 on ∂B(0, R).

On ∂B(0, ρ), since u < 0 on B(0, R) we have sup∂B(0,ρ) u < 0, and consequently, sincev ≤ 1 we have for all 0 < ε < ε0 := − sup∂B(0,ρ) u

u + εv < 0 on ∂B(0, ρ)That is (II.2.6) implies(II.2.7) u + εv ≤ 0 in A(ρ, R).Now fix ρ ∈ (0, R), choose ε, α so that the above is true.

Denote ν := x0|x0| the outwards unit normal to ∂B at x0 ∈ ∂B. Observe that for all small

0 < t � 1 (depending on ρ) we have x0 − tν ∈ A(ρ, R).


Recall that by assumption u(x0) = 0, then (II.2.7) implies for any small t > 0,

u(x0 − tν) + εv(x0 − tν)(II.2.7)≤ 0 = u(x0) + εv(x0).

This leads to (again: for all 0 < t � 1)u(x0 − tν) − u(x0)

t≤ −ε

v(x0 − tν) − v(x0)t

Letting t → 0+ on both sides we obtain(II.2.8) − ∂νu(x0) ≤ ε∂νv(x0).Observe that (II.2.5) implies

∂νv(x0) = ∂iv(x0)(x0)i

R= −2α

|x0|2R

e−αR2< 0

That is (II.2.8) implies−∂νu(x0) < 0

which implies the claim. �

The Hopf Lemma, Lemma II.2.7 implies the strong maximum principle.

Corollary II.2.8 (Strong maximum principle). Let Ω ⊂ Rn be an open and connected set,(but Ω may be unbounded). Let u ∈ C2(Ω) ∩ C0(Ω) satisfy

−Lu ≤ 0 in Ω.

If c ≡ 0 or if c ≤ 0 and supΩ u ≥ 0 then we have the following.

If there exists x0 ∈ Ω such thatu(x0) = sup

Ωu

then u ≡ u(x0) in Ω.

Proof. Assume the claim is false. Via the modification as in the proof of Lemma II.2.7,we may assume w.l.o.g. u ≤ 0 in Ω and u(x0) = 0 for some x0 ∈ Ω, but u �≡ 0.

LetΩ− := {x ∈ Ω : u(x) < 0}.

Observe that Ω− is open (u is continous) and Ω− �= ∅ (because u ≤ 0 and u �≡ 0).

Observe that the boundary of Ω− does cannot be contained in ∂Ω, i.e.∂Ω− ∩ Ω �= ∅.

Indeed this follows from connectedness: Let γ ⊂ Ω be a continuous path from x0 to a pointin Ω−. Then there has to be a point on γ where γ leaves Ω−. This point lies in ∂Ω− andin Ω.


This means we can find a point x1 ∈ Ω− which is close to ∂Ω− but not close to ∂Ω, i.e.x1 ∈ Ω−, ρ := dist (x1, ∂Ω−) < 10dist (x1, ∂Ω).

By definition of the distanceB(x1, ρ) ⊂ Ω−, B(x1, ρ)\Ω− �= ∅.

Let x2 ∈ ∂B(x1, ρ)\Ω−. Since by construction x2 ∈ ∂Ω− ∩ Ω we have u(x2) = 0 bycontinuity. Moreover u < 0 in B(x1, ρ) ⊂ Ω−.

Since everything takes place well within Ω, the conditions of the Hopf Lemma, Lemma II.2.7,are satisfied and thus for ν the outwards facing normal at x2 to ∂B(x1, ρ)

∂νu(x2) > 0.

But on the other hand x2 ∈ Ω is a local maximum for u, so Du(x2) = 0, which is acontradiction. The claim is then proven. �

A consequence of the Hopf Lemma, Lemma II.2.7, and the strong maximum principle,Corollary II.2.8, is the uniqueness for the Neumann problem.

Corollary II.2.9 (Uniqueness for Neumann-boundary problem). Let Ω ⊂⊂ Rn be openand connected. Moreover we assume a boundary regularity of ∂Ω, the interior spherecondition1:

Assume that for any x0 ∈ ∂Ω there exists a ball B ⊂ Ω such that x0 ∈ B.

Then the following holds for any elliptic operator as above with c ≡ 0: For any givenf ∈ C0(Ω) and any g ∈ C0(∂Ω) there is at most one solution u ∈ C2(Ω) ∩ C1(Ω) of theNeumann boundary problem

−Lu = f in Ω∂νu = g on ∂Ω,

up to constant functions. That means, the difference of two solutions u, v is constant,u − v ≡ c.

Proof. The difference of two solutions u, v, w := u − v satisfies2−Lw ≤ 0 in Ω∂νw = 0 on ∂Ω,

Firstly, assume that there exists x0 ∈ Ω such that supΩ w = w(x0). Then, by the strongmaximum principle, Corollary II.2.8, we have w ≡ w(x0) and the claim is proven. If thisis not the case, then there must be x0 ∈ ∂Ω with w(x0) > w(x) for all x ∈ Ω. If we take aball from the interior sphere condition of ∂Ω at x0 then on this ball B we can apply Hopf

1This condition does not allow for outwards facing cusps. One can show that every set Ω whoseboundary ∂Ω is a sufficiently smooth manifold satisfies the interior sphere condition

2actually we have = in the equation below, but the argument works for ≤ as well


Lemma, Lemma II.2.7, which leads to ∂νw(x0) > 0, which is ruled out by the Neumannboundary assumption ∂νw = 0. �

CHAPTER 3

Sobolev Spaces

III.1. Basic concepts from Functional Analysis

We will always consider real vector spaces with a norm (X, � · �), where the norm needs tosatisfy

• for all x ∈ X: �x�X = 0 ⇔ x = 0• �λx�X = |λ| �x�X for all x ∈ X and λ ∈ R.• �x + y�X ≤ �x�X + �y�X for all x, y ∈ X.

A normed vector space is a metric space via the metric dX(x, y) := �x − y�.

A normed vector space (X, � · �) is complete if every Cauchy sequence has a limit in X.We then say X is a Banach space (with the norm � · �).

Sometimes normed spaces have an interior product (aka scalar product), �·, ·� : X ×X → Rwhich satisfy

• �x, y� = �y, x� for all x, y ∈ X.• �λx + µy, z� = λ�x, z� + µ�y, z� for all x, y, z ∈ X and λ, µ ∈ R.

If X has such an interior product, then then it is called pre-Hilbert. If X is moreovercomplete (i.e. a Banach space) then it is called a Hilbert space.

Between Banach spaces X, Y we define

L(X, Y ) := {T : X → Y : T continuous and linear} ,

which is a Banach space with the norm

�T�L(X,Y ) := sup�x�X≤1

�Ty�Y .

The dual space X∗ = L(X,R) is the space of continuous linear functionals on X (calledthe dual space of X).

Example III.1.1 (Example 1: Holder spaces). The space Ck,γ(Ω) for k ∈ N0 and γ ∈ [0, 1]is defined as the set of all functions f ∈ C0(Ω) such that f ∈ Ck(Ω) and the k-the

51

III.1. BASIC CONCEPTS FROM FUNCTIONAL ANALYSIS 52

derivatives of f , Dkf are Holder continuous

[Dkf ]Cγ(Ω) := supx,y∈Ω

|Dkf(x) − Dkf(y)||x − y|γ < ∞

For γ = 1 we say this is Lipschitz continuity.

One norm is�f�Ck,γ(Ω) := �f�L∞(Ω) + [Dkf ]Cγ(Ω).

where�f�L∞(Ω) ≡ �f�C0(Ω) := sup

Ω|f |.

Example III.1.2 (Example 2: Lebesgue spaces). For p ∈ [1,∞) the class f ∈ Lp(Ω) isgiven as

Lp(Ω) =�f : Ω → R measurable s.t.�f�Lp(Ω) < ∞

�/ ∼,

where�f�Lp(Ω) =

��

Ω|f |p

� 1p

and we factor module the equivalence relation ∼ wheref ∼ g if f − g = 0 almost everywhere.

For p = ∞ we set�f�L∞(Ω) = ess sup

Ω|f | = inf {Λ > 0 : Ln(|f | > Λ) = 0} .

For p = 2 Lp has a scalar product

�f, g�L2(Ω) :=�

Ωfg.

Observe that the set of functions C∞c (Ω) has no proper norm. Often one resorts to using

the Schwartz class of functions which is at least nicely metrizable.

We will work a lot with Lp-spaces, so let us state some of the basic properties:

Lemma III.1.3 (Holder inequality). For 1 ≤ p, q, r ≤ ∞, if 1p

+ 1q

= 1r

we have�fg�Lr(Ω) ≤ �f�Lp(Ω) �g�Lq(Ω)

whenever f ∈ Lp(Ω), g ∈ Lq(Ω).

In particular we have �

Ωfg ≤ �f�Lp(Ω) �g�Lp� (Ω)

where we denote (from now on) p� = pp−1 the Holder dual.

Moreover, f ∈ Lp(Ω) implies f ∈ Lqloc(Ω) for any 1 ≤ q ≤ p. (Set g := χΩ ∈ Lq�

loc and useHolder)

III.2. PHILOSOPHY OF DISTRIBUTIONS AND SOBOLEV SPACES 53

Dual spaces of Lp are characterized, namely:

Theorem III.1.4 (Riesz Representation theorem). Let 1 < p < ∞. Assume thatT ∈ (Lp(Ω))∗,

i.e. T : Lp(Ω) → R is linear, and bounded:|Tf | ≤ �f�Lp(Ω) for all f ∈ Lp(Ω)

Then for q = p� = pp−1 we for some g ∈ Lq(Ω)

Tf =�

Ωfg.

Theorem III.1.5 (Density of smooth functions in Lp(Ω)). For 1 ≤ p < ∞ smooth, com-pactly supported functions are dense in Lp(Ω), that is for every f ∈ Lp(Ω) there existsfk ∈ C∞

c (Ω) such that�f − fk�Lp(Ω)

k→∞−−−→ 0.

In particular, in view of Hahn-Banach theorem, we can weaken a little bit the condition ofTheorem III.1.4.

Corollary III.1.6 (Riesz Representation theorem). Let 1 < p < ∞. Assume that T :C∞

c (Ω) → R is linear, and bounded in Lp(Ω):|Tf | ≤ �f�Lp(Ω) for all f ∈ C∞

c (Ω).Then for q = p� = p

p−1 we for some g ∈ Lq(Ω)

Tf =�

Ωfg for all f ∈ C∞

c (Ω.

In particular, Tf can be extended to a bounded, linear operator on all of Lp(Ω).

A nice consequence for p ∈ (1,∞):

�f�Lp(Ω) = supϕ∈C∞

c (Ω), �ϕ�Lp� (Ω)≤1

�

Ωf ϕ

III.2. Philosophy of Distributions and Sobolev spaces

Let f ∈ C0(Ω). From the first time we learned about functions we thought about them asa collection of points x ∈ Ω with there respective value f(x). I.e.

f ∈ C0(Ω) ↔ {(x, f(x)) x ∈ Ω}In some sense we test/sample f at every test-point x to obtain f(x) and a representationfor f .


The idea of distributions1is that we sample functions not at test-points x ∈ Ω but attest-functions ϕ ∈ C∞

c (Ω), i.e. we identify(III.2.1) f ∈ C0(Ω) ↔ {(ϕ, f [ϕ]) ϕ ∈ C∞

c (Ω)}where f [ϕ] measures the action of f on ϕ (from the functional analytic perspective weidentify f with an operator of a dual space of a function space containing C∞

c , and this iscalled the distributional interpretation of functions),

f [ϕ] :=�

Ωfϕ.

For continuous functions we know that the relation (III.2.1) is one-to-one on continuousfunctions, this was used for Weyl’s lemma, Proposition I.2.15.

Indeed we can substantially weaken (III.2.1) and have that the following is a one-to-onecorrespondence, for any p ∈ [1,∞).(III.2.2) f ∈ Lp(Ω) ↔ {(ϕ, f [ϕ]) ϕ ∈ C∞

c (Ω)}One-to-one means in particular that f [ϕ] = g[ϕ] implies f = g in L1, which is the funda-mental theorem of calculus of variations, Lemma I.2.24.

Clearly we have that the mapϕ �→ f [ϕ]

is linear in ϕ. Moreover, by Holder’s inequality, we have boundedness of f [ϕ] as a linearoperator acting on Lp�(Ω).

|f [ϕ]| ≤ �f�Lp(Ω) �ϕ�Lp� (Ω) ∀ϕ ∈ C∞c (Ω).

In view of Riesz Representation Theorem, Corollary III.1.6 linearity and boundedness off [ϕ] are enough to find f as a map in Lp(Ω).

Let us now talk about derivatives. For smooth functions f ∈ C1(Ω) we have, by integrationby parts,

(∂if)[ϕ] =�

Ω∂if ϕ = −

�

Ωf ∂iϕ = f [−∂iϕ] ∀ϕ ∈ C∞

c (Ω).

Why not define this for everything. If T : C∞c (Ω) → R is a distribution (i.e. linear in ϕ)

then we can just define the distributional derivative ∂iT as(∂iT )[ϕ] := T [−∂iϕ].

In this sense every distribution has infinitely many derivatives (in distributional sense).But not every function has a derivative.Example III.2.1. Take the Heaviside function:

f(x) :=

1 x > 00 x < 0.

1Observe that the Fourier Transform is a somewhat similar idea: For f ∈ L2(Rn) it is equivalent toconsider (x, f(x)) for almost all x ∈ Rn, or (ξ, f(ξ))


This function is measurable, locally integrable, but certainly not continuous in x = 0.

The pointwise derivative of f is almost everywhere zero (since f is almost everywhereconstant). But this is not true for the distributional derivative: For any ϕ ∈ C∞

c (R),

f �[ϕ] = −�

Rf ϕ� = −

�

Rf ϕ�

= −� 0

−∞f ϕ� −

� ∞

0f ϕ�

= −� 0

−∞0 ϕ� −

� ∞

01 ϕ�

= −� ∞

0ϕ�

=ϕ(0).

So, in the distributional sense,f � = δ0

where δ0 is the dirac measure. �

Rϕdδ0 = ϕ(0).

This is an important lesson: pointwise derivative do not necessarily predict distributionalderivatives.

Also we see, the distributional derivative of a function may not be again a function (thereis no function representing the dirac-measure).

But sometimes distributional derivatives of a function is again a function.

For example for any C1-function its distributional derivative ∂if [] coincides with its usualderivative ∂if .

Sobolev spaces W 1,p are the function spaces of Lp-functions whose distributional derivativeis also an Lp-function. That means, f ∈ W 1,p(Ω) if f ∈ Lp(Ω) and if there exist gi ∈ Lp(Ω),1 ≤ i ≤ n, such that ∂if = gi in the sense of distributions, i.e.

�

Ωf ∂iϕ = −

�

Ωg ϕ ∀ϕ ∈ C∞

c (Ω).

When we have this condition satisfied we will not bother with saying g is ∂if in the senseof distribution, but simply denote g by ∂if (it is unique by the fundamental theorem ofcalculus of variations, Lemma I.2.24).

Again observe: pointwise derivative are not the same as the distributional derivative, e.g.the Heaviside-function above does not belong to the Sobolev space W 1,p, even though itspointwise derivative is almost everywhere zero.

III.3. SOBOLEV SPACES 56

III.3. Sobolev Spaces

A remark on literature: A standard reference for Sobolev spaces is [Adams and Fournier, 2003].Very readable is also [Evans and Gariepy, 2015]. The introduction here takes a lot fromthe introduction to Sobolev spaces in [Evans, 2010]. A classical reference Sobolev spacesin PDEs is [Gilbarg and Trudinger, 2001]. Also [Ziemer, 1989]. For very delicateproblems one might also consult [Maz’ya, 2011].

Definition III.3.1. (1) Let 1 ≤ p ≤ ∞, k ∈ N and Ω ⊂ Rn open, nonempty. TheSobolev space W k,p(Ω) is the set of functions

u ∈ Lp(Ω)such that for any multiinidex γ, |γ| ≤ k we find a function (the distributionalγ-derivative or weak γ-derivative) “∂γu”∈ Lp(Ω) such that

�

Ωu ∂γϕ = (−1)|γ|

�

Ω“∂γu” ϕ ∀ϕ ∈ C∞

c (Ω).

Such u are also sometimes called Sobolev-functions.(2) For simplicity we write W 0,p = Lp.(3) The norm of the Sobolev space W k,p(Ω) is given as

�u�W k,p(Ω) =�

|γ|≤k

�∂γu�Lp(Rn)

or equivalently (exercise!)

�u�W k,p(Ω) =�

|γ|≤k

�∂γu�pLp(Rn)

1p

.

(4) We define another Sobolev space Hk,p(Ω) as follows

Hk,p(Ω) = C∞(Ω)�·�

W k,p(Ω) .

that is the (metric) closure or completion of the space (C∞(Ω), � ·�W k,p(Ω)). In yetother words, Hk,p(Ω) consists of such functions u ∈ Lp(Ω) such that there existapproximations uk ∈ C∞(Ω) with

�uk − u�W k,p(Ω)k→∞−−−→ 0.

We will later see that Hk,p is the same as W k,p locally, or for nice enough domains;and use the notation H or W interchangeably. For k = 0 this fact follows fromTheorem III.1.5 for any open set Ω.

(5) Now we introduce the Sobolev space Hk,p0 (Ω)

Hk,p0 (Ω) = C∞

c (Ω)�·�

W k,p(Ω) .

We will later see that this space consists of all maps u ∈ Hk,p(Ω) that satisfyu,∇u, . . .∇k−1u ≡ 0 on ∂Ω in a suitable sense (the trace sense, for a precise


formulation see Theorem III.3.21). – Again, later we see that H = W and thus,W k,p

0 (Ω) = Hk,p0 (Ω) for nice sets Ω.

Observe that in view of Theorem III.1.5, Lp(Ω) = W 0,p(Ω) = W 0,p0 (Ω).

(6) The local space W k,ploc (Ω) is similarly defined as Lp

loc(Ω). A map belongs to u ∈W k,p

loc (Ω) if for any Ω� ⊂⊂ Ω we have u ∈ W k,p(Ω�).

Remark III.3.2. Some people write Hk,p(Ω) instead of W k,p(Ω). Other people use Hk(Ω)for Hk,2 – notation is inconsistent...

Some people claim that W stand for Weyl, and H for Hardy or Hilbert.

Example III.3.3. For s > 0 letf(x) := |x|−s.

Observe that f is only defined for x �= 0, but since measurable functions need only bedefined outside of a null-set this is still a reasonable function.

We have already seen, when working with fundamental solutions, that f ∈ Lploc(Rn) for

any 1 ≤ p < ns.

We can compute for x �= 0 that(III.3.1) ∂if(x) = −s |x|−s−2xi

and by the same argument as above we could conjecture that ∂if ∈ Lqloc(Rn) for any

1 ≤ q < ns+1 .

It is an exercise to show that

(1) (III.3.1) holds in the distributional sense, i.e. that if n ≥ 2 and 0 < s < n − 1then for any ϕ ∈ C∞

c (Rn),�

Rnf(x) ∂iϕ(x) dx =

�

Rns |x|−s−2xiϕ(x) dx.

(2) to conclude that f ∈ W 1,qloc (Rn) for any 1 ≤ q < n

s+1 .

Example III.3.4. Letf(x) := log |x|.

One can show that f ∈ Lploc(Rn) for any 1 ≤ p < ∞, and f ∈ W 1,p

loc (Rn) for all p ∈ [1, n), ifn ≥ 2.

Example III.3.5. Letf(x) := log log 2

|x| in B(0, 1)

One can show that for n ≥ 2, f ∈ W 1,n(B(0, 1)).

Moreover, for n = 2, in distributional senseΔf = |Df |2


Observe that this serves as an example for solutions to nice differential equations that arenot continuous!

Proposition III.3.6 (Basic properties of weak derivatives). Let u, v ∈ W k,p(Ω) and |γ| ≤k. Then

(1) ∂γu ∈ W k−|γ|,p(Ω).(2) Moreover ∂α∂βu = ∂β∂αu = ∂α+βu if |α| + |β| ≤ k.(3) For each λ, µ ∈ R we have λu + µv ∈ W k,p(Ω) and

∂α(λu + µv) = λ∂αu + µ∂αv

(4) If Ω� ⊂ Ω is open then u ∈ W k,p(Ω�)(5) For any η ∈ C∞

c (Ω), ηu ∈ W k,p and (if k ≥ 1), and we have the Leibniz formula(aka product rule)

∂i(ηu) = ∂iη u + η∂iu.

Proof. (1) We show that ∂iu ∈ W k−1,p(Ω), only. The general statement thenfollows accordingly. By definition of the distributional derivative we have that∂iu ∈ Lp(Ω). For any |β| ≤ k − 1 and ϕ ∈ C∞

c (Ω) we have�

Ω∂iu ∂βϕ = −

�

Ωu ∂i∂

βϕ = −(−1)|β|+1�

Ω∂i∂

βu ϕ = (−1)|β|�

Ω∂i∂

βu ϕ.

The first inequality comes from the fact that ∂βϕ ∈ C∞c (Ω) and from the definition

of the weak derivative ∂i. The second equation comes from the definition of theweak derivative of ∂i∂

β for W k,p-functions.(2) We show ∂i∂ju = ∂j∂iu, again the general case follows. And as above this is proven

by deducing respective properties from the properties in the space of test-funtions:For ϕ ∈ C∞

c (Ω) we have ∂i∂jϕ = ∂j∂iϕ, and thus�

Ω∂i∂ju ϕ =

�

Ωu ∂i∂jϕ =

�

Ωu ∂j∂iϕ =

�

Ω∂j∂iu ϕ.

(3) Follows from the linearity of the definition of weak derivative and the equivalentstatements for smooth functions ϕ ∈ C∞

c (Ω)(4) If Ω� ⊂ Ω then any ϕ ∈ C∞

c (Ω�) belongs also to C∞c (Ω). That is any property true

for test functions ϕ ∈ C∞c (Ω) holds also for testfunctions in ϕ ∈ C∞

c (Ω�).(5) For ϕ ∈ C∞

c (Ω) we have by the usual Leibniz rule�

Ωηu∂iϕ =

�

Ωu ∂i(ηϕ) −

�

Ωu ∂iη ϕ

= −�

Ω∂iu ηϕ −

�

Ωu ∂iη ϕ

= −�

Ω(∂iu η + u ∂iη) ϕ

The second equation is the definition of weak derivative ∂iu (since ηϕ ∈ C∞c (Ω) is

a permissible testfunction).


That is we have shown for all ϕ ∈ C∞c (Ω),

�

Ωηu∂iϕ =

�

Ωu ∂i(ηϕ) −

�

Ωu ∂iη ϕ.

This means that in distributional sense ∂i(ηu) = ∂iη u + η∂iu. Now observe thatηu ∈ Lp(Ω) and ∂iη u + η∂iu ∈ Lp(Ω), so ηu ∈ W 1,p(Ω).

�

Proposition III.3.7. (W k,p(Ω), � · �W k,p(Ω)), (Hk,p(Ω), � · �W k,p(Ω)), (Hk,p0 (Ω), � · �W k,p(Ω))

are all Banach spaces.

For p = 2 they are Hilbert spaces, with inner product

�u, v� =�

|γ|≤k

�∂γu ∂γv.

Proof. � ·�W k,p(Ω) is a norm. By definition (Hk,p(Ω), � ·�W k,p(Ω)), (Hk,p0 (Ω), � ·�W k,p(Ω))

are complete and thus Banach spaces.

As for the completeness of W k,p(Ω), it essentially follows from the completeness of Lp(Ω).

Let (ui)i∈N ⊂ W k,p(Ω) be a Cauchy sequence of W k,p-functions, i.e.∀ε > 0∃N = N(ε) ∈ N s.t. ∀i, j ≥ N : �ui − uj�W k,p(Ω) < ε.

We have to show that ui converges to some u ∈ W k,p(Ω) in the W k,p(Ω)-norm.

Observe that by the definition of the W k,p-norm, if ui is a Cauchy sequence for W k,p, thenfor any |γ| ≤ k, (∂γui)i∈N are Cauchy sequences of Lp(Ω).

Since Lp(Ω) is a Banach space, i.e. complete, each ∂γui converges in Lp(Ω) to some objectwhich we call ∂γu,

�∂γui − ∂γu�Lp(Ω)i→∞−−−→ 0 ∀|γ| ≤ k.

Observe that as of now we do not know that ∂γu is actually the weak derivative of u! Butwe can check this is the case.

Since ∂γui is the weak derivative of ui, we have�

Ω∂γui ϕ = (−1)|γ|

�

Ωui ∂γϕ ∀ϕ ∈ C∞

c (Ω).

But on both sides we have strong convergence in Lp(Ω). For any (fixed) ϕ ∈ C∞c (Ω),

�

Ω∂γu ϕ

i→∞←−−−�

Ω∂γui ϕ = (−1)|γ|

�

Ωui ∂γϕ

i→∞−−−→ (−1)|γ|�

Ωu ∂γϕ

and thus for any ϕ ∈ C∞c (Ω),

�

Ω∂γu ϕ = (−1)|γ|

�

Ωu ∂γϕ.


That is, ∂γu is indeed the weak derivative of u, thus u ∈ W k,p(Ω) and by the definition ofthe W k,p-norm

�ui − u�W k,p(Ω)i→∞−−−→ 0.

�

III.3.1. Approximation by smooth functions. We mentioned above the H = Wproblem, i.e. we would like to approximate Sobolev functions by smooth functions. Why?Because then we don’t have to deal that many times with the weak definition of derivatives,but show desired results for smooth functions, then pass to the limit and hopefully obtainthe result for Sobolev maps. Observe that since W k,p(Ω) is a Banach space, and C∞(Ω) ⊂W k,p(Ω) (exercise!) we clearly have Hk,p(Ω) ⊂ W k,p(Ω). for the other direction we nowobtain the first result:

Proposition III.3.8 (Local approximation by smooth functions). Let u ∈ W k,p(Ω), 1 ≤p < ∞. Set

uε(x) := ηε ∗ u(x) =�

Rnηε(y − x) u(y) dy.

Here ηε(z) = ε−nη(z/ε) for the usual bump function η ∈ C∞c (B(0, 1), [0, 1]),

�B(0,1) η = 1.

Then

(1) uε ∈ C∞(Ω−ε), where as beforeΩ−ε := {x ∈ Ω : dist (x, ∂Ω) > ε}

for each ε > 0 such that Ω−ε �= ∅.(2) Moreoever for any Ω� ⊂⊂ Ω,

�uε − u�W k,p(Ω�)ε→0−−→ 0.

Proof. (1) As in the proof of Theorem I.2.16 we have uε ∈ C∞(Ω−ε) – we do notneed that u is a Sobolev function, but merely that u ∈ Lp(Ω).

(2) Next we claim that ∂γuε(x) = (∂γu)ε(x) for x ∈ Ω−. Indeed, for x ∈ Ω−ε,

∂γuε(x) =�

Ω∂γ

x (ηε(x − z)) u(z) dz = (−1)|γ|�

Ω∂γ

z (ηε(x − z)) u(z) dz.

Now we observe that ηε(x − ·) ∈ C∞c (Ω) if x ∈ Ω−ε: observing size of the support

of ηε, supp ηε ⊂ B(0, ε).Thus by the definition of weak derivative,

(−1)|γ|�

Ω∂γ

z (ηε(x − z)) u(z) dz =�

Ωηε(x − z)∂γu(z) dz = (∂γu)ε(x).

Now similar to the argument in the proof of Theorem I.2.16, for any Ω� ⊂⊂ Ω andε < dist (Ω�, ∂Ω), for any 1 ≤ p < ∞2

�(∂γu)ε − ∂γu�Lp(Ω�)ε→0−−→ 0.

2but not for p = ∞!


This holds for any γ such that ∂γu ∈ Lp, i.e. for all |γ| ≤ k. We conclude that

�uε − u�W k,p(Ω�)ε→0−−→ 0.

�

Even though Proposition III.3.8 is only about local approximation, it is very useful toprove properties of Sobolev function.

Lemma III.3.9. For 1 ≤ p < ∞3, if v ∈ W 1,p(Ω) and f ∈ C1(R,R) with [f ]Lip (R) ≡�f ��L∞(Rn) < ∞ then f(v) ∈ W 1,p(Ω), and we have in distributional sense

(III.3.2) ∂α(f(v)) = f �(v) ∂αv.

Proof. Let vε be the (local) approximation of v in W 1,ploc (Ω) from Proposition III.3.8.

First we observe that (III.3.2) is true if v was a differentiable function, in particular,

∂α(f(vε)) = f �(vε) ∂αvε in Ω−ε.

Now let ϕ ∈ C∞c (Ω), and take ε0 so small such that Ω� := supp ϕ ⊂ Ω−ε for all ε ∈ (0, ε0).

Then we have for all ε < ε0,

(III.3.3)�

Ωf(vε) ∂αϕ = −

�

Ωf �(vε) ∂αvε ϕ

Now we observe that f(vε) ε→0−−→ f(v) with respect to the Lp(Ω�)-norm. Indeed, observethat Ω� ⊂⊂ Ω, so by Proposition III.3.8,

�f(vε) − f(v)�Lp(Ω�) ≤ �f ��L∞(Ω�) �vε − v�Lp(Ω�)ε→0−−→ 0.

That is, the left-hand side of (III.3.3) converges (recall supp ∂αϕ ⊂ supp ϕ = Ω�)�

Ωf(v) ∂αϕ ≡

�

Ω�f(v) ∂αϕ = lim

ε→0

�

Ω�f(vε) ∂αϕ ≡ lim

ε→0

�

Ωf(vε) ∂αϕ.

As for the right-hand side of (III.3.3) we have that ∂αvεε→0−−→ v in Lp(Ω�), and f �(vε) ε→0−−→

f �(v) almost everywhere in Ω (up to taking a subsequence ε → 0)4. By dominated conver-gence this implies �

Ωf �(vε) ∂αvε ϕ

ε→0−−→�

Ωf �(v) ∂αv ϕ

Then from (III.3.3) we get the claim, observing that f �(v)∂αv ∈ Lp(Ω), since f � ∈ L∞. �

3we can later conclude, using Theorem III.3.17, that this also holds for p = ∞, since then Sobolevmaps are simply Lipschitz maps

4these are results from measure theory:, since f � is continuous, and since L1-convergence implies almosteverywhere convergence, see [Evans and Gariepy, 2015, Theorem 1.21]


Actually, a stronger statement is true: if u ∈ W 1,p(Ω) and f : R → R is Lipschitzcontinuous, f ∈ C0,1 then f ◦ u ∈ W 1,p(Ω). Again, formally this is no problem since∇(f ◦ u) = Df(u)∇u– since f is almost everywhere differentiable. But the precise proofis tedious. We just sketch the proof of a special case:

Lemma III.3.10. Let u ∈ W 1,1(Ω), then |u| ∈ W 1,1(Ω).

Moreover we have Du = 0 almost everywhere in {u(x) = 0}5.

Also we haveD|u| = u

|u|Du.

Proof. We only sketch the proof.

The difficulty lies in the fact that | · | is merely Lipschitz continuous, so we mollify it:

fε,θ(t) :=�

(t + θε)2 + ε2 −�

(θε)2 + ε2.

fε,θ is a smooth function.

One approximates |u| by uε := fε,θ(u) for some θ ∈ R

Since fε,θ is smooth we have in distributional sense, by Lemma III.3.9,

Duε = u + εθ�(u + εθ)2 + ε2

Duε→0−−→ Du ·

1 in {u > 1}θ√

θ2+1 in {u = 0}−1 in {u < 1}

Now uε → |u| in L1(Ω), and Duε converges also in L1(Ω). Using test functions and theconvergence as ε → 0 we get that

L1(Ω) � D|u| = Du ·

1 in {u > 1}θ√

θ2+1 in {u = 0}−1 in {u < 1}

But weak derivatives are unique as L1-functions. The nonunique looks independent in θ.This means either Du = 0 almost everywhere in {u = 0} or {u = 0} is a zeroset (whichstill means that Du = 0 almost everywhere in {u = 0}). �

In general, crazy sets, it might be difficult to extend Proposition III.3.8 to the boundary(think of an open set whose boundary is the Koch-curve). To rule this out we make thefollowing definition of Ck-boundary data

5Check this for smooth functions: Either {u(x) = 0} is a zeroset. On the other hand, on the “sub-stantial” parts of {u(x) = 0} we should think of u as constant


Definition III.3.11 (Regularity of boundary of sets). Let Ω ⊂ Rn be an open set. We saythat ∂Ω ∈ Ck (more generally in Ck,α) if ∂Ω ⊂ Rn is a Ck (or Ck,α, respectively) manifold,that is if

for any x ∈ ∂Ω there exists a radius r > 0 and a Ck-diffeomorphism Φ : B(x, r) → B(0, r)(i.e. the map Φ is a bijection between B(x, r) and B(0, r) and Φ and Φ−1 are both of classCk) such that

• Φ(x) = 0• Φ(Ω ∩ B(x, r)) = B(0, r) ∩ Rn

+• Φ(B(x, r)\Ω) = B(0, r) ∩ Rn

−.Theorem III.3.12 (Smooth approximation for Sobolev functions). Let Ω ⊂ Rn be openand bounded, and ∂Ω ∈ C1. For any u ∈ W k,p(Ω) there exist a smooth approximatingsequence ui ∈ C∞(Ω) such that

�ui − u�W k,p(Ω)i→∞−−−→ 0.

Proof. First we consider the situation close to the boundary.

Let x0 ∈ ∂Ω.

Observe first the following: If x ∈ B(0, r)+ and |z − x| < ε (for ε � r) then x + εen ⊂B(x, 2r)+. Since ∂Ω belongs to C1 one can show that the same holds (on sufficiently smallballs B(x0, r)) as well: For some λ = λ(x0), a unit vector ν = ν(x0), if z ∈ B(x, ε) andx ∈ Ω ∩ B(x0, r) then

z + λεν ∈ Ω.

One should think of ν the inwards facing unit normal at x0 (which can be computed fromthe derivatives of Φ and is continuous around x0).

That is for x ∈ Ω� := B(x0, r/2) ∩ Ω we may set

uε(x) :=�

Rnηε(z − x) u(z + λνε)dz =

�

Rnηε(z − λνε − x) u(z)dz.

Clearly, uε is still smooth, but now in all of of Ω�. Moreover observe that if we setvε(x) := u(z + λνε).

we have�vε − u�W k,p(Ω�)

ε→0−−→ 0since vε is merely a translation. Moreover, uε = ηε ∗ vε, and thus as before

�uε − vε�W k,p(Ω�) = ε→0−−→ 0.

We conclude that uε → u in W k,p(Ω�).

Now we cover all of ∂Ω by (finitely many, by compactness) balls B(xi, ri) and choose theapproximation uε,i on Ωi := B(xi, ri) ∩ Ω as above. In Ω0 := Ω\�B(xi, ri) ⊂⊂ Ω we canfind another approximation uε,0.


Now we pick a smooth decomposition of unity ηi with support in Ωi ∩ ∂Ω such that�

i∈Nηi ≡ 1 in Ω.

Settinguε :=

�

i

ηiuε,i ∈ C∞(Ω).

We then use the Leibniz rule to conclude that�uε − u�W k,p(Ω)

ε→0−−→ 0.

�

On Rn approximation is much easier, indeed we can approximate with respect to the W k,p-norm any u ∈ W k,p(Rn) by functions uk ∈ C∞

c (Rn). That is, W k,p(Rn) = W k,p0 (Rn). We

could describe this as “u ∈ W k,p(Rn) implies that u and k − 1-derivatives of u all vanishat infinity”.

Proposition III.3.13. (1) Let u ∈ W k,p(Ω), p ∈ [1,∞). If supp u ⊂⊂ Rn then thereexists uk ∈ C∞

c (Ω) such that

�u − uk�W k,p(Ω)k→∞−−−→ 0.

(2) Let u ∈ W k,p(Rn), p ∈ [1,∞). Then there exists uk ∈ C∞c (Rn) such that

�u − uk�W k,p(Rn)k→∞−−−→ 0.

(3) Let u ∈ W k,p(Rn+) = Rn−1 × (0,∞)). Then there exists u ∈ C∞

c (Rn−1 × [0,∞)(i.e.e u may not be zero on (x�, 0) for small x�) such that

�u − uk�W k,p(Rn+)

k→∞−−−→ 0.

Proof. (1) follows from the proof of Proposition III.3.8: Observe that supp u ⊂⊂ Ωimplies that ηε ∗ u ∈ C∞

c (Ω) if ε is only small enough.

(3) is an exercise, a combination of the proof of (2) and Theorem III.3.12.

So let us discuss (2). Let η ∈ C∞c (B(0, 1)) again be the typical mollifier bump function,

ηε(x) = ε−nη(x/ε). We have already seen thatηε ∗ u ∈ C∞(Rn).

But there is no reason that ηε ∗ u ∈ C∞c (Rn). Set (without rescaling by Rn!)

ϕR(x) := η(x/R) ∈ C∞c (B(0, R)).

Then we setuε,R := ηε ∗ (ϕR u)

Now before uε,R ∈ C∞c (B(0, R + ε)) ⊂ C∞

c (Rn).


Moreover we have for any � = 0, . . . , k

�∇�(u − uϕR)�Lp(Rn) =�∇�(1 − ϕR)u�Lp(Rn)

≤C(�)��

i=0�∇i(1 − ϕR)∇�−iu�Lp(Rn)

≤C(�) �(1 − ϕR)∇�u�Lp(Rn) + C(�)��

i=0�∇i(1 − ϕR)�∞�∇�−iu�Lp(Rn)

≤C(�, η) �(1 − ϕR)∇�u�Lp(Rn\B(0,R)) + C(�, η)��

i=0R−i�u�W k,p(Rn)

R→∞−−−→0by Lebesgue dominated convergence theorem.

On the other hand, as already seen for R fixed,�ηε ∗ (ϕR u) − ϕRu�W k,p(Rn)

ε→0−−→ 0.

Now we show that for any � > 0 there exists ε�, R� such that for u� := uε�,R�we have

(III.3.4) �u� − u�W k,p(Rn) <1�

�→∞−−−→ 0,

that is C∞c (Rn) � u� → u in W k,p(Rn).

First, by the arguments above we can choose R� large enough such that

�u − uϕR��W k,p(Rn) ≤

12�

.

Next we can choose ε� small enough such that

�u� − uϕR��W k,p(Rn) �ηε�

∗ (uϕR�) − uϕR�

�W k,p(Rn) ≤12�

.

Thus, by triangular inequality,

�u� − u�W k,p(Rn) ≤ �u� − uϕR��W k,p(Rn) + �uϕR�

− u�W k,p(Rn) ≤ 2 12�

= 1�.

This proves (III.3.4), and thus (2) is established. �

III.3.2. Difference Quotients. We used above, e.g. for the Cauchy estimates, Proofof Lemma I.2.17 the method of differentiating the equation (e.g. that if Δu = 0 then alsofor v := ∂iu we have Δv = 0 – so we can easier estimates for ∂iu). In the Sobolev spacecategory this is also a useful technique. Sometimes, the “first assume that everything issmooth, then use mollification”-type argument as for Lemma I.2.17 is difficult to put intopractice. In this case, a technique developed by Nirenberg, is discretely differentiating theequation (which does not require the function to be a priori differentiable):

Δu = 0 ⇒ v(x) := (Δeih u)(x) := u(x + hei) − u(x)

h: Δv = 0


For this to work, we need some good estimates. Recall that (by the fundamental theoremof calculus), for C1-functions u,

�Δe�h u�L∞ ≤ �∂�u�L∞ .

This also holds in Lp for W 1,p-functions u, which is a result attributed to Nirenberg, seeProposition III.3.15.

One important ingredient is that the fundamental theorem of calculus holds for Sobolevfunctions:Lemma III.3.14. Let u ∈ W 1,1

loc (Ω). Fix v ∈ Rn. Then for almost every x ∈ Ω such thatthe path [x, x + v] ⊂ Ω we have

u(x + v) − u(x) =� 1

0∂αu(x + tv)vα dt.

Proof. Let Ω� ⊂ Ω. In view of Proposition III.3.8 we can approximate u by uk ∈C∞(Ω�) such that

�uk − u�W 1,1(Ω�)k→∞−−−→ 0.

The claim holds for the smooth functions uk, namely we have that whenever [x, x+v] ⊂ Ω�,

(III.3.5) uk(x + v) − uk(x) =� 1

0∂αuk(x + tv)vα dt.

Now we have�uk(· + v) − uk(·) − (u(· + v) − u(·)) �Lp(Ω�)

k→∞−−−→ 0,

in particular (up to taking a subsequence),

uk(x+v)−uk(x) k→∞−−−→ u(x+v)−u(x) for almost every x ∈ Ω� such that [x, x + v] ⊂ Ω�.

Also the right-hand side converges. Observing that6

��

Ω

�� 1

0|f(x, t)| dt

�p� 1

p

≤�� 1

0

�

Ω|f(x, t)|p dt

� 1p

we have

�� 1

0∂αuk(· + tv)vα dt −

� 1

0∂αu(· + tv)vα dt�Lp(Ω�)

≤�� 1

0|v|�(Duk − Du�p

Lp(Ω�) dt� 1

p

= �Duk − Du�Lp(Ω�)k→∞−−−→ 0.

So again, up to possibly a subsubsquence,� 1

0∂αuk(x + tv)vα dt

k→∞−−−→� 1

0∂αu(x + tv)vα dt

6This can be seen by Jensens inequalty: For any p ∈ [1,∞),

(�

A

|f |)p ≤�

A

|f |p,

this can also be shown by Holder’s inequality. Then Fubini gets to the claim.


for all x such that [x, x + v] ⊂ Ω.

Taking the limit in (III.3.5) we conclude. �Proposition III.3.15. (1) Let k ∈ N, (i.e. k �= 0), and 1 ≤ p < ∞. Assume that

Ω� ⊂⊂ Ω are two open (nonempty) sets, and let 0 < |h| < dist (Ω�, ∂Ω). Foru ∈ W k,p(Ω) we have

�Δe�h u�W k−1,p(Ω�) ≤ �∂�u�W k−1,p(Ω).

Moreover we have�Δe�

h u − ∂�u�W k−1,p(Ω�)h→0−−→ 0.

(2) Let u ∈ W k−1,p(Ω), 1 < p ≤ ∞. Assume that for any Ω� ⊂⊂ Ω and any � =1, . . . , n there exists a constant C(Ω�) such that

sup|h|<dist (Ω�,∂Ω)

�Δe�h u�W k−1,p(Ω�) ≤ C(Ω�, �)

Then we u ∈ W k,ploc (Ω), and for any Ω� ⊂ Ω we have

(III.3.6) �∂�u�W k−1,p(Ω�) ≤ sup|h|<dist (Ω�,∂Ω)

�Δe�h u�W k−1,p(Ω�).

If p = ∞ we even have u ∈ W k,∞(Ω) with the estimate(III.3.7) �∂�u�W k−1,∞(Ω) ≤ sup

Ω�⊂⊂Ωsup

|h|<dist (Ω�,∂Ω)�Δe�

h u�W k−1,∞(Ω�).

Proof of Proposition III.3.15(1). The proof of (1) is essentially the same as fordifferentiable function, we use the fundamental theorem of calculus.

By the fundamental theorem of calculus, Lemma III.3.14,

Δe�h u(x) = 1

h

� 1

0

d

dt(u(x + the�)) dt = 1

hh� 1

0∂�u(x + the�) dt =

� 1

0∂�u(x + the�) dt,

Similarly, for any |γ| ≤ k − 1, � = 1, . . . , n

|Δe�h ∂γu(x)| ≤

� 1

0|∂�∂

γu(x + the�)|.

Taking the Lp-norm, observing that7

��

Ω

�� 1

0|f(x, t)| dt

�p� 1

p

≤�� 1

0

�

Ω|f(x, t)|p dt

� 1p

7This can be seen by Jensens inequalty: For any p ∈ [1,∞),

(�

A

|f |)p ≤�

A

|f |p,

this can also be shown by Holder’s inequality. Then Fubini gets to the claim.


we have

�Δe�h ∂γu�Lp(Ω�) ≤

�� 1

0�∂�∂

γu(· + the�)�pLp(Ω�) dt

� 1p

.

Now observe that by substitution and |h| < dist (Ω�, ∂Ω),�∂�∂

γu(· + the�)�Lp(Ω�) = �∂�∂γu(·)�Lp(Ω�+he�) ≤ �∂�∂

γu(·)�Lp(Ω)

Consequently, for any |h| < dist (Ω�, ∂Ω).

�Δe�h ∂γu�Lp(Ω�) ≤

�� 1

0�u�p

W k,p(Ω) dt� 1

p

= �u�W k,p(Ω).

This shows (1), under the assumption that the fundamental theorem of calculus works. �

Before we proof Proposition III.3.15(2), we need results from Functional Analysis:

Let X be a Banach space, then X∗ = L(X) denotes the dual space (namely the space ofall linear, continuous functionals X → R. The bi-dual space X∗∗ = L(X∗) has X as acanonical subset, meaning there is a canonical isometric embedding JX : X → X∗∗ givenby (for x ∈ X and x∗ ∈ X∗,

JX(x)[x∗] := x∗[x].If JX is surjective, then we say that X is a reflexive space. The important fact for us isthat Lp-spaces are reflexive if p ∈ (1,∞) (L1 and L∞ are not reflexive).

Then we have the important theorem (essentialy a consequence of Banach-Alaoglu the-orem) that if X is a reflexive Banach space then every bounded set es weakly (sequen-tially) pre-compact, which means that whenever (xk) ⊂ X is a bounded sequence, i.e.supk �xk�X < ∞ then there exists a subsequence xki

such that xkiis weakly convergent to

some x ∈ X. Weak convergence means that for any dual element y∗ ∈ X∗ we have

y∗(xki) i→∞−−−→ y∗(x) as numbers in R

Lucky for us, the dual space of Lp-spaces, 1 < p < ∞, is characterized by the Rieszrepresentation theorem, Theorem III.1.4. This leads to the following theorem

Theorem III.3.16 (Weak compactness). Let 1 < p < ∞, Ω ⊂ Rn open, and assume that(fk)k∈N is a bounded sequence in Lp(Ω), that is

supk∈N

�fk�Lp(Ω) < ∞.

Then there exists a function f ∈ Lp(Ω) and a subsequence fkisuch that fki

weakly Lp-converges to f , that is for any g ∈ Lp�(Ω), where p� = p

p−1 is the Holder dual of p, wehave �

Ωfki

gi→∞−−−→

�

Ωf g.

In particular we have�f�Lp(Ω) ≤ sup

k�fk�Lp(Ω).


Proof of Proposition III.3.15(2). First let us assume that p < ∞.

Assume that for all � ∈ {1, . . . , n} we havesup


h u�W k−1,p(Ω�) < ∞.

In view of Theorem III.3.16 we can choose a sequence hii→∞−−−→ 0 such that for any |γ| ≤

k − 1,Δe�

hi∂γu

i→∞−−−→ f�,γ weakly in Lp(Ω�).Since we are optimists, we call ∂�∂

γu := f�,γ ∈ Lp(Ω�). We still need to show that ∂�∂γu

is actually the distributional derivative of u! Also, for simplicity of notation we drop thei in hi and write h → 0 (meaning always this subsequence). Weak convergence means inparticular, that for any ϕ ∈ C∞

c (Ω�) ⊂ Lp(Ω�),

(III.3.8)�

Ω�Δe�

h ∂γu ϕh→0−−→

�

Ω�∂�∂

γu ϕ

Since supp ϕ ⊂⊂ Ω� for |h| small enough we have that Δe�−hϕ ∈ C∞

c (Ω�). Now we performa discrete integration by parts, namely by substitution,

�

Ω�Δe�

h ∂γu ϕ = −�

Ω�∂γu Δe�

−hϕ

Now since Δe�−hϕ ∈ C∞

c (Ω�) is a testfunction and u ∈ W k−1,p,�

Ω�Δe�

h ∂γu ϕ = −�

Ω�∂γu Δe�

−hϕ = (−1)|γ|+1 −�

Ω�u Δe�

−h∂γϕh→0−−→ (−1)|γ|+1

�

Ω�u ∂�∂

γϕ,

in the last step we used dominated convergence and the smoothness of ϕ.

But then in (III.3.8) we obtain

(−1)|γ|+1�

Ω�u ∂�∂

γϕ =�

Ω�∂�∂

γu ϕ

This holds for any � ∈ {1, . . . , n} and so we have shown that ∂�∂γu is indeed the weak

derivative of u which belongs to Lp, and thus u ∈ W k,p(Ω�). Since this holds for anyΩ� ⊂ Ω we conclude that u ∈ W k,p

loc (Ω). The estimate (III.3.6) follows from the estimate ofTheorem III.3.16.

As for the case p = ∞, we observe first that for Ω� ⊂⊂ Ω the estimate(III.3.9) sup


h u�W k−1,∞(Ω�) ≤ C(Ω�, �)

implies (by Holders inequality) alsosup


h u�W k−1,2(Ω�) ≤ C(Ω�, �)

Thus (III.3.9) implies u ∈ W k,2loc (Ω) and in view of Proposition III.3.15(1) we have that

Δe�h u → ∂�u in W k−1,2

loc (Ω).

In particular, we already have the existence of the distributional derivative ∂�u ∈ W k−1,2loc (Ω).


SetΛ := sup

Ω�⊂⊂Ωsup


h u�W k−1,∞(Ω�).

For simplicity of notation in the following we shall assume k = 1.

We now claim that the above observations, together with (III.3.7), for any ϕ ∈ C∞c (Ω),

(III.3.10)�

Ω∂�u ϕ ≤ Λ �ϕ�L1(Ω).

Indeed, since for ϕ ∈ C∞c (Ω) let supp ϕ ⊂ Ω� ⊂⊂ Ω, then we have�

Ω∂�u ϕ = lim

|h|→0

�

ΩΔe�

h u� ��

≤Λ

ϕ ≤ Λ�ϕ�L1(Ω).

which is exactly (III.3.10).

Let x ∈ Ω be a Lebesgue point of ∂�u in Ω, i.e.

∂�u(x) = limr→0

�

B(x,r)∂�u.

Observe that almost all points in Ω are Lebesgue points (since ∂�u ∈ L2loc(Ω)).

SetΩ� = {z ∈ Ω : dist (x, ∂Ω) <

12dist (x, ∂Ω)} ⊂⊂ Ω.

Then for all r < 14dist (x, ∂Ω) we can set ϕ := |B(x, r)|−1χB(x,r) ∈ L2(Ω) which can be

approximated by smooth C∞c (Ω�) functions ϕi → ϕ in L2(Ω). Since Ω� ⊂⊂ Ω we also have

ϕi → ϕ in L1(Ω) (observe �ϕ�L1(Ω) = 1 by construction of ϕ). Then�

B(x,r)∂�u = lim

i→∞

�

B(x,r)∂�uϕi,

which leads to|�

B(x,r)∂�u| ≤ Λ lim

i→∞�ϕi�L1(Ω�) ≤ Λ�ϕ�L1(Ω�) = Λ.

Since x was chosen to be a Lebesgue point of ∂�u, and since the last estimate holds for anyr > 0, we find

|∂�u(x)| = limr→0

|�

B(x,r)∂�u| ≤ Λ.

This again holds for any Lebesgue point x ∈ Ω, and since almost all points in Ω areLebesgue points,

|∂�u(x)| ≤ Λ a.e. x ∈ Ω,

which implies�∂�u�L∞(Ω) = ess sup

Ω|∂�u| ≤ Λ,

which was the claim. �


Theorem III.3.17 (Ck−1,1 ≈ W k,∞). (1) Let Ω ⊂ Rn be open and nonempty, k ∈ Nthen

Ck−1,1(Ω) ⊂ W k,∞(Ω),and

�Dku�L∞(Ω) ≤ C [Dk−1u]Lip (Ω)

(2) Let Ω ⊂⊂ Rn connected, ∂Ω ∈ C0,1. Then for k ∈ N

W k,∞(Ω) ⊂ Ck−1,1(Ω),and

[Dk−1u]Lip (Ω) ≤ C(k, Ω) �Dku�L∞(Ω).

The above holds in the following sense: recall that functions in Lp (and thusin particular in W k,∞ are classes (namely: two functions f, g ∈ Lp(Ω) are thesame if they coincide almost everywhere). So what we mean above is: For everyf ∈ W k,∞(Ω) there exists a representative g ∈ Ck−1,1(Ω) that coinides with f a.e.

Proof. We restrict our attention to k = 1 and leave the other cases as exercise.

For (1): Let u ∈ C0,1(Ω). Since u is Lipschitz,sup

Ω�⊂⊂Ωsup


h u�L∞(Ω�) ≤ [u]Lip (Ω).

From Proposition III.3.15(2) we then obtain u ∈ W 1,∞(Ω) with the claimed estimate.

For (2):

First we assume that Ω = B(0, 1) is a ball, u ∈ W 1,∞(B(0, 1)). We argue by mollification(what else can we do): Let uε be the usual mollification uε = ηε ∗ u which, as we alreadyknow, converges in W 1,2

loc (B(0, 1)) to u. Moreover (also as seen before), for any δ ∈ (0, 1),x ∈ B(0, δ), if ε < δ then

∂�uε(x) =�

∂�u(y) ηε(y − x) dy,

and thus whenever x ∈ B(0, δ), if ε < δ

|∂�uε(x)| ≤ �∂�u(y)�L∞(B(0,1))�ηε�L1(B(0,1)) = �∂�u(y)�L∞(B(0,1)).

In particular, by the fundamental theorem of calculus (recall: uε is differentiable), wheneverε < δ

[uε]Lip ,B(0,1−δ) ≤ �Du�L∞(B(0,1)).

Observe there is no constant on the right-hand side. Since moreover �uε�L∞(B(0,δ0) ≤�u�L∞(B(0,1)) we have that uε is equicontinuous and bounded, and thus by Arzela-Ascoli(up to a subsequence ε → 0) we have uε → u in C0(B(0, δ)) (Here is where we find the“continuous representative of u”, the limit of uε coinicides a.e. with u). In particular, u iscontinuous in B(0, 1 − δ). Also observe that for any x �= y ∈ B(0, 1 − δ), for any ε < δ,|u(x)−u(y)| ≤ 2�u−uε�L∞(B(0,1−δ))+|uε(x)−uε(y)| ≤ 2�u−uε�L∞(B(0,1−δ))+|x−y|�Du�L∞(B(0,1)).


This holds for any ε < δ, so letting ε → 0 we obtain by the uniform convergence uε → uin B(0, 1 − δ).

|u(x) − u(y)| ≤ |x − y|�Du�L∞(B(0,1)) for all x, y ∈ B(0, 1 − δ)

This again holds for any δ > 0 so that

|u(x) − u(y)| ≤ |x − y|�Du�L∞(B(0,1)) for all x, y ∈ B(0, 1).

That is, u is Lipschitz continuous and we have

[u]Lip (B(0,1)) ≤ �Du�L∞(B(0,1)).

So Theorem III.3.17(2) is established for Ω = B(0, 1).

Next assume that Ω ⊂⊂ Rn and ∂Ω ∈ C0,1. Moreover we assume Ω is path-connected.

The regularity of the boundary is used in the following way: For any two points x, y ∈ Ωthere exists a continuous path γ connecting x and y inside Ω such that the lenght of γ,L(γ) ≤ C(Ω)|x − y| (essentially take the straight line connecting x and y, when it hits ∂Ωfollow ∂Ω, then regularize and shift it away from ∂Ω).

Since Ω is open, and by the argument above in every open ball we can replace u by itscontinuous representative we may assume that u w.l.o.g. is continous, and we just want toshow that u is Lipschitz continuous.

Let x, y ∈ Ω and let γ be such a path connecting x and y. Set δ := 12dist (γ, ∂Ω) > 0.

Setting L := �L(γ)δ

� + 2 points (xi)Li=1 in γ, such that

L�

i=1B(xi, δ) ⊃ γ.

and such that B(xi, δ) ∩ B(xi+1, δ) �= ∅, x ∈ B(x1, δ) and y ∈ B(xL, δ). In every B(xi, δ)we use the argument from above, and have

[u]Lip (B(xi,δ)) ≤ �Du�L∞(B(xi,δ)) ≤ �Du�L∞(Ω).

Now, by triangular inequality

|u(x) − u(y)| ≤|u(x) − u(x0)| + |u(y) − u(xL)| +L−1�

i=1|u(xi) − u(xi+1)|

≤�Du�L∞(Ω) (L + 1)2δ ≤ CL(γ) ≤ C(Ω)�Du�L∞(Ω) |x − y|.

This implies that u is Lipschitz continuous with

[u]Lip ≤ C(Ω)�Du�L∞(Ω).

which establishes the theorem. �


III.3.3. Weak compactness in W k,p. In the proof of of Proposition III.3.15(2) wederived and used the following consequence of Theorem III.3.16, which we want to record(so we don’t have to argue always with Theorem III.3.16.

Theorem III.3.18 (Weak compactness). Let 1 < p < ∞, k ∈ N, Ω ⊂ Rn open. Assumethat (fi)i∈N is a bounded sequence in W k,p(Ω), that is

supi∈N

�fi�W k,p(Ω) < ∞.

Then there exists a function f ∈ W k,p(Ω) and a subsequence fijsuch that fij

weakly W k,p-converges to f , that is for any |γ| ≤ k and any g ∈ Lp�(Ω), where p� = p

p−1 is the Holderdual of p, we have �

Ω∂γfij

gi→∞−−−→

�

Ω∂γf g.

In particular we have�f�W k,p(Ω) ≤ sup

i�fi�W k,p(Ω).

III.3.4. Extension Theorems. If f is a Lipschitz function on a set Ω ⊂ Rn, then f

can be thought of as a restriction of a map f : Rn → R, f = f��Ω. This is (a special case of)

the so-called Kirszbraun theorem. This is in general not true for Sobolev functions, evenif Ω is open.

Definition III.3.19. Let Ω ⊂ Rn be open. Ω is called a W k,p-extension domain, if thereexists a linear operator E : W 1,p(Ω) → W 1,p(Rn) such that

Eu(x) = u(x) for all x ∈ Rn, u ∈ W k,p(Ω)and E is bounded, i.e.

sup�u�

W k,p(Ω)≤1�Eu�W k,p(Rn) < ∞.

Theorem III.3.20. Any open set Ω ⊂⊂ Rn with boundary ∂Ω ∈ Ck is a W k,p(Ω) extensiondomain for k ∈ N, 1 ≤ p < ∞.

More precisely, for any Ω ⊃⊃ Ω there exists an operator E : W k,p(Ω) → W k,p0 (Ω) with

Eu = u in Ω and�Eu�W k,p(Ω) ≤ C(Ω, Ω, n, k) �u�W k,p(Ω).

Proof. We will first show how to extend W k,p-functions from Rn+ to all of Rn. Then

by “flattening the boundary” (for this we need the regularity ∂Ω) we extend this argumentto general Ω as claimed.

From Rn+ to Rn:

Denote the variables in Rn by (x�, xn) where x� ∈ Rn−1 and xn ∈ R.

We can explicitely define E0 : W k,p(Rn+) → W k,p(Rn) by a type of reflection.


The main point is that we know (from the heaviside function example) that W 1,k-functionscannot have a jump, so at least for smooth functions u, if we hope for E0u ∈ W 1,p we needthat

limyn→0−

E0u(y�, yn) != limyn→0+

E0u(y�, yn) = limyn→0+

u(y�, yn)

So, for k = 1 we could simply use the even reflection,

E0u(y�, yn) := u(y�, |yn|) =

u(y�, yn) if yn > 0u(y,−yn) if yn < 0.

which indeed takes C∞(Rn+)-functions into Lipschitz-functions (i.e. W 1,∞

loc (Rn)-functions,hence W 1,p

loc (Rn)).

More generally, for W k,p-functions, k ≥ 1 we then need that for any � = 1, . . . , k the(� − 1)-th derivatives in yn-direction coincide:

(III.3.11) limyn→0−

(∂n)�−1E0u(y�, yn) != limyn→0+

(∂n)�−1E0u(y�, yn) = limyn→0+

(∂n)�−1u(y�, yn).

So again, we use a reflection, but a more complicated one,

E0u(y�, yn) :=

u(y�, yn) if yn > 0�k

i=1 σi u(y,−iyn) if yn < 0.

Here, (σi)ki=1 are constants to be chosen, such that (III.3.11) is true for smooth functions:

For all � = 1, . . . , k

k�

i=1σi(−i)�−1(∂n)�u(x�, 0) = (∂n)�u(x�, 0) ⇐

k�

i=1σi(−i)�−1 = 1.

Such a σ exists by linear algebra: Defining a matrix A by Ai� := (−i)�−1, and interpretingσ as a vector in Rk we want to solve

Aσ =

1...1

,

which is possible if A is invertible (to check this is the case is left as an exercise).

Now we argue as follows: Let u ∈ W k,p(Rn+). By Proposition III.3.13 there exists uj ∈

C∞c (Rn−1 × [0,∞)) that approximate u in W k,p(Rn

+).

One now checks that E0uj ∈ Ck−1,1(Rn), moreover we have for almost any x ∈ Rn (namelywhenever x = (x�, xn), with xn �= 0), for any |γ| ≤ k,

|∂γ(E0uj)(x�, xn)| ≤ C(σ, k)

|∂γuj(x�, xn)| xn > 0�k

i=1 |∂γu(y,−iyn)| xn < 0.


Let us illustrate this fact for k = 1, for k > 1 it is an exercise.∂xα(E0uj)(x�, xn) = ∂xαuj(x�, |xn|) = (∂xαuj)(x�, |xn|) α = 1, . . . , n − 1.

∂xn(E0uj)(x�, xn) = ∂xαuj(x�, |xn|) = (∂xnuj)(x�, |xn|) xn

|xn| .

In particular we get that u ∈ W k,p(Rn) and�E0uj�W k,p(Rn) ≤ C(k) �uj�W k,p(Rn

+)

In particular we getlim sup

j→∞�E0uj�W k,p(Rn) ≤ C(k) lim sup

j→∞�uj�W k,p(Rn

+) = C(k) �u�W k,p(Rn+).

Thus, in view of Theorem III.3.18 we find g ∈ W k,p which is the weak W k,p-limit of E0uj.By strong Lp-convergence of uj to u we see that indeed E0g = u, and thus we get

�E0u�W k,p(Rn) ≤ C(k) �u�W k,p(Rn+).

as claimed.

From Ω to Rn We only sketch the remaining arguments. If ∂Ω ∈ Ck then from small ballsB centered at boundary points there exists Ck-charts φ : B → Rn such that φ(B∩Ω) ⊂ Rn

+φ(B ∩ Ωc) ⊂ Rn

−. By a decomposition of unity, we set u = �i ηiu such that ηi are

supported only in one of these balls Bi ∩ Ω. Then (ηiu) ◦ φi ∈ W k,p(Rn+) (since it is

locally in W k,p and then it is constantly zero). Here we use that (we haven’t shown it) theTransformation rule still holds for Sobolev functions. Then we extend (ηi ◦ u) ◦ φi to all ofRn, i.e. considerE0((ηiu) ◦ φi). Finally we set

E1u :=�

i

(E0((ηiu) ◦ φi)) ◦ φ−1i .

The transformation rule shows that Eu ∈ W k,p(Rn).

From Ω to Ω� To get E2u ∈ W k,p0 (Ω�) we simply take a cuttoff function η ∈ C∞

c (Ω�), η ≡ 1in Ω, and set

E2u := ηE1u.

�

III.3.5. Traces. Let Ω ⊂ Rn and ∂Ω ∈ C∞.

If u ∈ Cα(Ω), α ∈ (0, 1] with�u�Cα(Ω) < ∞

then we find a unique map u��∂Ω

∈ Cα(∂Ω). Indeed, for any x ∈ ∂Ω there exists exactlyone value u(x) such that u(x) = limΩ�x→x u(x), because �u�L∞ < ∞ and since we haveuniform continuity

|u(x) − u(y)| ≤ �u�Cα(Ω)|x − y|α |x−y|→0−−−−−→ 0.


Moreover, for any x, y ∈ ∂Ω and x, y ∈ Ω we have|u(x) − u(y)| ≤|u(x) − u(x)| + |u(x) − u(y)| + |u(x) − u(y)|

≤|u(x) − u(x)| + �u�Cα |x − y|α + |u(x) − u(y)|Taking x → x and y → y we thus find

|u(x) − u(y)| ≤ �u�Cα |x − y|α

that is�u�Cα(∂Ω) ≤ �u�Cα(Ω).

The map that computes from u the trace map u we may call T , u = Tu. Then we have alinear operator

T : Ck,α(Ω) → Ck,α(∂Ω),By the computations above, T is linear and bounded

�Tu�Ck,α(∂Ω) ≤ �u�Ck,α(Ω).

On the other hand, when u ∈ Lp(Ω) there is absolutely no reasonable (unique) sense of atrace u

��∂Ω

.

One interesting and important fact of Sobolev spaces is that there is such a trace operatorT if k− 1

p> 0, that associates to a Sobolev function u ∈ W k,p(Ω) a map Tu ∈ W k− 1

p,p(∂Ω).

Observe that formally, if p = ∞ (i.e. in the Lipschitz case, the trace map is of the sameclass as the interior map, but for p < ∞ the trace map has less differentiability than theinterior map. We do not want to deal with fractional Sobolev spaces here, so instead ofproving the sharp trace estimate

T : W 1,p(Ω) → W 1− 1p

,p(∂Ω)we will only show the following:

Theorem III.3.21. Let Ω ⊂⊂ Rn, ∂Ω ∈ C1, 1 ≤ p < ∞. There exists a (unique) boundedand linear Trace operator T

T : W 1,p(Ω) → Lp(∂Ω)such that

(1) Tu = u

��∂Ω

whenever u ∈ C0(Ω) ∩ W 1,p(Ω).(2) for each u ∈ W 1,p(Ω) we have

�Tu�Lp(∂Ω) ≤ C(Ω, p) �u�W 1,p(Ω).

Proof. For u ∈ C1(Ω) ∩ W 1,p(Ω) we define

Tu := u��∂Ω

.


It now suffices to show that for all u ∈ C1(Ω) ∩ W 1,p(Ω) we have

(III.3.12) �Tu�Lp(∂Ω) ≤ C(Ω, p) �u�W 1,p(Ω).

Then, by density of smooth functions C∞(Ω) in W 1,p(Ω), Theorem III.3.12, linearity andboundedness of the trace operator, there exists a (unique) extension of T to all of W 1,p(Ω).

To see (III.3.12) we argue again first on a flat boundary Ω = Rn+. A flattening the boundary

argument as above, then leads to the claim.

Observe the following, which holds by the integration-by-parts formula:

�u�pLp(Rn−1) =

�

Rn−1×{0}|u(x�)|pdHn−1(x) =

�

Rn+

∂n (|u(x)|p) dx =�

Rn+

p|u(x)|p−2u(x) ∂nu(x) dx

Then by Young’s inequality, ab ≤ C(ap + bp�) (where p� = pp−1 is the Holder dual of p),

�

Rn+

p|u(x)|p−2u(x) ∂nu(x) dx ≤ C�

Rn+

(|u|p + |∂nu|p) ≤ C�u�pW 1,p(Rn

+).

This establishes (III.3.12) for Ω = Rn+. For general Ω we use a decomposition of unity and

flattening the boundary argument as in the theorems above. �

Theorem III.3.22 (Zero-boundary data and traces). Let Ω ⊂⊂ Rn and ∂Ω ∈ C1. Letu ∈ W 1,p(Ω).

Then u ∈ H1,p0 (Ω) is equivalent to u ∈ W 1,p

0 (Ω), where

W 1,p0 (Ω) = {u ∈ W 1,p(Ω) : Tu = 0}

for the trace operator T from Theorem III.3.21.

Remark III.3.23. By induction one obtains that if ∂Ω ∈ C∞ then Hk,p0 (Ω) are exactly

those functions where T (∂γu) = 0 for any |γ| ≤ k − 1.

For time reasons we will not give the proof here. For a proof see [Evans, 2010, §5.5,Theorem 2].

III.3.6. Embedding theorems. Let X, Y be two Banach spaces. T : X → Y is a(we assume always: linear) embedding if T is injective. We say that the embedding X ⊂ Yis continuous under the operator T , if T is a linear embedding and T is continuous (i.e. abounded operator). If (as it often happens) T is (in a reasonable sense) the identity map,then we say that X embedds into Y continuously, and write X �→ Y . E.g., clearly (bydefinition)

W 1,p(Ω) �→ Lp(Ω)since, by definition of the norm

�u�Lp(Ω) ≤ �u�W 1,p(Ω).


We say than an embedding X �→ Y is compact if the operator T : X → Y is compact, i.e.if for any bounded sequence (xn)n∈N ⊂ X, supn �xn�X < ∞, we have that (T (xn))n∈N ⊂ Yhas a convergent subsequence in Y .

From functional analysis we also have: If T : X → Y is compact, then if (xn)n∈N is weaklyconvergent in X then Txn is strongly convergent in Y .

By Arzela-Ascoli, it is easy to check that Ck,α(Ω) embedds compactly into C�,β(Ω) if k ≥ �and k + α > � + β.

The first implortant theorem is that for bounded sets Ω with smooth boundary we haveW 1,p(Ω) embedds compactly into Lp(Ω). (By induction: W k,p(Ω) embedds compactly intoW �,p(Ω) whenever k ≥ p).

Observe that by Theorem III.3.18 we have weak compactness in W 1,p(ω) for bounded set,but strong convergence in Lp(Ω).

Theorem III.3.24 (Rellich-Kondrachov). Let Ω ⊂⊂ Rn, ∂Ω ⊂ C0,1, 1 ≤ p ≤ ∞. Assumethat (uk)k∈N ∈ W 1,p(Ω) is bounded, i.e.

supk∈N

�uk�W 1,p(Ω) < ∞.

Then there exists a subsequence ki → ∞ and u ∈ Lp(Ω) such that ukiis (strongly) conver-

gent in Lp(Ω), moreover the convergence is pointwise a.e..

Proof. If p = ∞, from Theorem III.3.17 we have W 1,∞(Ω) = C0,1(Ω). By Arzela-Ascoli it is clear that C0,1 is compactly embedded in C0(Ω), so in particular in L∞ (whichhas the same norm as C0(Ω).

Now let p ∈ [1,∞). By the extension theorem, Theorem III.3.20 we may assume thatuk ∈ W 1,p(Rn) with supp uk ⊂ B(0, R) for some (fixed) large R > 0.

The main idea is to use Arzela-Ascoli for mollified versions of uk. Denote by η ∈ C∞c (B(0, 1))

the usual bump function,�

η = 1, and ηε = ε−nη(·/ε). Set

uk,ε := ηε ∗ uk ∈ C∞c (B(0, 2R)).

Observe that|uk,ε(x)| ≤ C(R)ε−n�uk�Lp(B(0,R))

|Duk,ε(x)| ≤ C(R)ε−n−1�uk�Lp(B(0,R)),

so since uk is bounded (even Lp-boundedness is enough for now) we have

supk∈N

�uk,ε�Lip (Rn) ≤ C(ε).

That is, for any εj := 1j

there exists a subsequence uki(εj ),εjthat is convergent in L∞(Rn).


By a diagonalizing this subsequences we obtain only one subsequence uki,εjso that for any

fixed εj we have convergence in L∞(Rn), i.e. for any j ∈ N and any δ > 0 there existsNj,δ ∈ N such that

�uki1 ,εj− uki2 ,εj

�L∞(Rn) ≤ δ ∀i1, i2 > Nj,δ.

Next we observe, by the fundamental theorem of Calculus,��uki1

(x) − uki1 ,εj(x)

�� =�

Rn|ηε(z)|

��uki1(x − z) − uki1

(x)�� ≤

� 1

0

�

B(0,ε)|ηε(z)||Duki1

(x − tz))| |z|dz dt

≤ε1−n� 1

0

��

B(0,ε)|Duki1

(x − tz))|p� 1

p

εn− np dz dt

Thus, by Fubini

�uki1− uki1 ,εj

�Lp(Rn) =�uki1− uki1 ,εj

�Lp(B(0,2R)) ≤ ε1− np

�� 1

0

�

B(0,R)

�

B(0,ε)|Duki1

(x − tz)dxdzdt

� 1p

≤ε�Duki1�Lp(B(0,2R)).

(III.3.13)

Now we claim that this leads to a Cauchy-sequence for the (non-mollified) uki: Let δ > 0.

�uki1− uki2

�Lp(Rn) ≤�uki1− uki1 ,εj

�Lp(Rn) + �uki1,εj− uki2,εj

�Lp(Rn) + �uki2 ,εj− uki2

�Lp(Rn)

(III.3.13)≤ 2 C εj sup

k�uk�W 1,p(Rn) + C(R)�uki1,εj

− uki2,εj�L∞(B(2R))

Choosing now first εj small enough so that

2 C εj supk

�uk�W 1,p(Rn) <δ

2and then choosing for this εj the N(εj, δ) large enough so that for any i1, i2 > N(εj, δ)

C(R)�uki1,εj− uki2,εj

�L∞(B(2R)) <δ

2we see that

�uki1− uki2

�Lp(Rn) ≤ δ for any i1, i2 > N(εj, δ).

That is, ukiis a Cauchy sequence in Lp(Rn) and thus converges. �

One important consequence of Rellich’s theorem, Theorem III.3.24 is Poincare’s inequality.In 1D it is called sometimes Wirtinger’s inequality – and it is quite easy to prove. LetI = (a, b) ⊂ R, then for any u ∈ W 1,p(I),(III.3.14) �u − (u)I�Lp(I) ≤ C(I, p)�u��Lp(I).

Here(u)I :=

�

Iu

denotes the mean value of u on I.


The proof of (III.3.14) is done by the fundamental theorem of calculus, Lemma III.3.14.We have (using Holder’s inequality and Fubini many times)

�u− (u)I�pLp(I) ≤ |I|−1

�

I

�

I|u(x)−u(y)|p ≤ |I|−1

� 1

0

�

I

�

I|u�(tx+(1− t)y)|p |x−y|p dx dydt

Now observe that by substituting y := tx + (1 − t)y, we have

|I|−1� 1

2

0

�

I

�

I|u�(tx + (1 − t)y)|p |x − y|p dx dydt

≤C(I, p)� 1

2

0

�

I

�

I

11 − t

|u�(y)|p dy dxdt

=C(I, p)� 1

2

0

11 − t

dt |I| �u��pLp(I)

=C(I, p)�u��pLp(I).

In the same way, substituting x := tx + (1 − t)y

|I|−1� 1

12

�

I

�

I|u�(tx + (1 − t)y)|p |x − y|p dx dydt

≤C(I, p)�u��pLp(I).

So (III.3.14) is established.

The Poincare inequality says that (III.3.14) holds also in higher dimensions,

(III.3.15) �u − (u)Ω�Lp(Ω) ≤ C(Ω, p)�∇u�Lp(Ω).

If Ω is convex, the above proof works almost verbatim, in general open sets Ω this is moretricky.

Clearly, (III.3.15) does not hold if we remove (u)Ω from the left-hand side. Indeed, justtake u ≡ const to find a counterexample. And indeed, a W 1,p-Poincare-type inequalityholds whenever constants are excluded in a reasonable sense.

Theorem III.3.25 (Poincare). Let Ω ⊂⊂ Rn be open and connected, ∂Ω ∈ C0,1, 1 ≤ p ≤∞.

Let K ⊂ W 1,p(Ω) be a closed (with respect to the W 1,p-norm) cone with only one constantfunction u ≡ 0. That is, let K ⊂ W 1,p(Ω) be a closed set such that

(1) u ∈ K implies λu ∈ K for any λ ≥ 0.(2) if u ∈ K and u ≡ const then u ≡ 0.

Then there exists a constant C = C(K, Ω) such that

(III.3.16) �u�Lp(Ω) ≤ C�∇u�Lp(Ω) ∀u ∈ K.


The proof is a standard method from analysis, called a blow-up proof. One assumes thatthe claim is false, and then tries to compute/construct the “most extreme” counterexample– which one then hopes to see cannot exist. Before we begin, we need the following smallLemma.

Lemma III.3.26. For Ω ⊂ Rn open assume that u ∈ W 1,1loc (Ω). If ∇u ≡ 0 then u is

constant in every connected component of Ω.

Proof. This follows from (local) approximation by smooth functions. If ∇u ≡ 0 then∇uε ≡ 0 in Ω−ε, where uε = ηε ∗ u. This implies that uε ≡ const in every connectedcomponent of Ω−ε. Pointwise a.e. convergence of uε to u gives the claim. �

Proof of Theorem III.3.25. Assume the claim is false for a given K as above. Thatmeans however we choose the constant C there will be some countexample u that dails theclaimed inequality (III.3.16).

That is, for any m ∈ N there exists um ∈ K such that (III.3.16) is false for C = m, i.e.

(III.3.17) �um�Lp(Ω) > m �∇um�Lp(Ω).

Now we construct the “extreme/blown up” counterexample (that, as we shall see, does notexist – leading to a contradicition).

Firstly, we can assume w.l.o.g.

(III.3.18) �um�Lp(Ω) = 1, �∇um�Lp(Ω) ≤1m

.

Indeed otherwise we can just take um := um

�um�Lp(Ω)which satisfies (III.3.18).

(III.3.18) implies in particular,

supm

�um�W 1,p(Ω) < ∞.

In view of Rellich’s theorem, Theorem III.3.24, we can thus assume w.l.o.g. (otherwisetaking a subsequence) that um is convergent in Lp(Ω). In particular um is a Cauchysequence in Lp(Ω). Observe that also ∇um is a cauchy sequence in Lp(Ω), indeed by(III.3.18) ∇um

m→∞−−−→ 0 in Lp(Ω). In particular, u is a Cauchy sequence in W 1,p(Ω). SinceW 1,p(Ω) is a Banach space we find a limit map u ∈ W 1,p(Ω) such that

(III.3.19) �um − u�W 1,p(Ω)m→∞−−−→ 0.

In view of (III.3.18) this implies that ∇u ≡ 0. From Lemma III.3.26 and since Ω isconnected, u is a constant map. But since K is closed we have that u ∈ K, and sincethe only constant map in K is the constant zero map, we find u ≡ 0 in Ω. But then by(III.3.19)

�um�W 1,p(Ω)m→∞−−−→ 0.


which contradicts the conditions in (III.3.18), namely

�um�W 1,p(Ω) ≥ �um�Lp(Ω)(III.3.18)= 1.

We have found a contradiction, and thus the assumption above (that for any m there existsum that contradicts the claimed equation) is false. So there must be some number m suchthat for C := m the equation (III.3.16) holds. �Corollary III.3.27 (Poincare type lemma). Let Ω ⊂⊂ Rn be open, connected, and ∂Ω ∈C0,1.

(1) There exists C = C(Ω) such that for all u ∈ W 1,p(Ω) we have�u − (u)Ω�Lp(Ω) ≤ C(Ω)�∇u�Lp(Ω)

(2) For any Ω� ⊂⊂ Ω open and nonempty there exists C = C(Ω, Ω�) such that for allu ∈ W 1,p(Ω) we have

�u − (u)Ω��Lp(Ω) ≤ C(Ω, Ω�)�∇u�Lp(Ω)

(3) There exists C = C(Ω) such that for all u ∈ W 1,p0 (Ω)

�u�Lp(Ω) ≤ C(Ω)�∇u�Lp(Ω)

If Ω = B(x, r) (and in the second claim Ω�B(x, λr)) then C(Ω) = C(B(0, 1)) r (and forthe second claim: C(Ω, Ω�) = C(B(0, 1), B(0, λ)) r).

Proof. The last claim can be proven by a scaling argument, and it is given as anexercise.

Regarding the first claim, we simply letK := {u ∈ W 1,p(Ω), (u)Ω = 0}.

By Rellich’s theorem, Theorem III.3.24 this is a closed cone in W 1,p. Observe that if u ∈ Kis constant, u ≡ C then (u)Ω = C = 0 by assumption, so C = 0. That is, the only constantfunction in K is the zero-function. Clearly u − (u)Ω belongs to K, so we get the claim.

Regarding the second claim, we argue similarly settingK := {u ∈ W 1,p(Ω), (u)Ω� = 0}.

Regarding the third claim, observe that W 1,p0 (Ω) is (by definition) a closed set, and since

it is a linear space it is in particular a cone. Now if u ∈ W 1,p0 (Ω) is constant, u ≡ c then

u is in particular continuous, but then by the zero trace theorem, Theorem III.3.22 c ≡ 0.Again, the only constant function in K is the zero-function. �

We have seen in Theorem III.3.24 and used in the Poincare inequality that W 1,ploc (Ω) em-

bedds compactly into Lploc(Ω). There is a meta-theorem/feeling that “above” any compact


embedding there is a merely continuous embedding, for more precise versions of this effectsee [Haj�lasz and Liu, 2010].

In our case it is that W 1,p embedds into Lp∗ where p∗ follows the following rule

(III.3.20) 1 − n

p= 0 − n

p∗

(we will see this numerology appear later again for Morrey and Sobolev-Poincare embed-ding, Corollary III.3.31 and Theorem III.3.35). Observe that p∗ = np

n−p∈ (1,∞) for p < n.

We set p∗ := ∞ for p ≥ n. p∗ is called the Sobolev exponent. What happens if p∗ > n (whichshould be interpreted from this numerological point of view as p∗ > ∞? Theorem III.3.35will tell us: u is Holder continuous.

Theorem III.3.28 (Sobolev inequality). Let p ∈ [1,∞) such that p∗ := npn−p

∈ (1,∞)(equivalently: p ∈ [1, n)). Then W 1,p(Rn) embedds into Lp∗(Rn). That is, if u ∈ W 1,p(Rn)then u ∈ Lp∗(Rn) and we have8

�u�Lp∗ (Rn) ≤ C(p, n) �Du�Lp(Rn).

Proof of Theorem III.3.28. There are more than one way to prove Sobolev’s in-equality. One (the “Harmonic Analysis” one) is by convolution, using the Riesz potentialrepresentation and boundedness of Riesz transform on Lp-spaces. It is very strong andgeneral but beyond the scope of these lectures.

The one we present here is an elegant trick due to Nirenberg (here we are again!). It ismuch less stable, relies heavily on the structure of Rn, etc., but it obtains the case p = 1(that in general is much more difficult to obtain), see e.g. [Schikorra et al., 2017].

By approximation it suffices to assume that u ∈ C∞c (Rn).

Let x = (x1, x2, . . . , xn) ∈ Rn. Then we have by the fundamental theorem of calculus

u(x1, x2, . . . , xn) = u(y1, x2, . . . , xn) +� y1

x1∂1u(z1, x2, . . . , xn) dz1.

Taking y1 large enough we have u(y1, x2, . . . , xn) = 0, since supp u ⊂⊂ Rn. Thus we obtainthe estimate

|u(x1, x2, . . . , xn)| ≤�

R|Du(z1, x2, . . . , xn)| dz1.

The same way we obtain, for any � = 1, . . . , n,

|u(x1, x2, . . . , xn)| ≤�

R|Du(x1, x2, . . . , z�, . . . xn)| dz�.

Multiplying these estimates for � = 1, . . . , n we obtain

|u(x1, x2, . . . , xn)|n ≤ Πn�=1

�

R|Du(x1, x2, . . . , z�, . . . xn)| dz�.

8The optimal constant C(p, n) has actually a geometric meaning, and is related to the isoperimetricinequality, cf. [Talenti, 1976]


Now we prove the case p = 1, when p∗ = nn−1 . We have

�

R|u(x1, x2, . . . , xn)| n

n−1 dx1 ≤�

RΠn

�=1

��

R|Du(x1, x2, . . . , z�, . . . xn)| dz�

� 1n−1

dx1

≤��

z1∈R|Du(z1, x2, . . . , xn)| dz1

� 1n−1

�

RΠn

�=2

��

R|Du(x1, x2, . . . , z�, . . . xn)| dz�

� 1n−1

dx1

Now by Holder’s inequality9

�

RΠn

�=2

��

R|Du(x1, x2, . . . , z�, . . . xn)| dz�

� 1n−1

dx1 ≤�

Πn�=2

�

R

�

R|Du(x1, x2, . . . , z�, . . . xn)| dz� dx1

� 1n−1

and thus�

R|u(x)| n

n−1 dx1 ≤��

z1∈R|Du(z1, x2, . . . , xn)| dz1

� 1n−1

�Πn

�=2

�

R

�

R|Du(x1, x2, . . . , z�, . . . xn)| dz� dx1

� 1n−1

Now we integrate this with respect to x2, and again by Holder’s inequality,�

R

�

R|u(x)| n

n−1 dx1dx2

≤�

x2∈R

��

z1∈R|Du(z1, x2, . . . , xn)| dz1

� 1n−1

�Πn

�=2

�

R

�

R|Du(x1, x2, . . . , z�, . . . xn)| dz� dx1

� 1n−1

dx2

≤��

R

�

R|Du(x1, z2, x3, . . . xn)| dz2 dx1

� 1n−1

·�

x2∈R

��

z1∈R|Du(z1, x2, . . . , xn)| dz1

� 1n−1

�Πn

�=3

�

R

�

R|Du(x1, x2, . . . , z�, . . . xn)| dz� dx1

� 1n−1

dx2

≤��

R

�

R|Du(x1, z2, x3, . . . xn)| dz2 dx1

� 1n−1

·��

x2∈R

�

z1∈R|Du(z1, x2, . . . , xn)| dz1 dx2

� 1n−1

�Πn

�=3

�

x2∈R

�

R

�

R|Du(x1, x2, . . . , z�, . . . xn)| dz� dx1, dx2

�

If n = 2 we are done (the Πn�=3-term is one). If n ≥ 3 we see a pattern, continuing to

integrate in x3, . . . , xn we obtain�

R

�

R|u(x)| n

n−1 dx1dx2 ≤Πn�=1

��

Rn|Du(x1, . . . , x�−1, z�, x�+1 . . . xn)|dx1, . . . , x�−1, z�, x�+1 . . . xn)

� 1n−1

=��

Rn|Du|

� nn−1

.

Taking the exponent n−1n

on both sides we obtain(III.3.21) �u�

Ln

n−1 (Rn) ≤ �Du�L1(Rn)

9the generalized version for k := n−1 and all pi := n−1: whenever p1, . . . , pk ∈ [1,∞] and�

i1pi

= 1,�

Rd

Πki=1|fi| ≤ Πk

i=1

��

Rd

|fi|pi

� 1pi


This is the claim for p = 1 (i.e. p∗ = nn−1).

The general claim follows when we apply the p = 1 Sobolev inequality to v := |u|γ forsome γ > 1 that we choose later. We have

|Dv| = |D|u|γ| ≤ γ|u|γ−1|Du|,thus (III.3.21) applied to v

(III.3.22) �|u|γ�L

nn−1 (Rn) ≤ C(γ)�|u|γ−1 |Du|�L1(Rn)

Now observe that�|u|γ�

Ln

n−1 (Rn) = �u�γ

Lγ n

n−1 (Rn)

Moreover, by Holder’s inequality, p� = pp−1 ,

�|u|γ−1 |Du|�L1(Rn) ≤ �|u|γ−1�Lp� (Rn)�Du�Lp(Rn) = �u�γ−1Lp�(γ−1)(Rn)�Du�Lp(Rn)

So (III.3.22) becomes�u�γ

Lγ n

n−1 (Rn)�u�1−γ

Lp�(γ−1)(Rn) ≤ C(γ)�Du�Lp(Rn)

Choosing γ := p(n−1)n−p

> 1 we have γ nn−1 = p�(γ − 1) = p∗, and then

�u�Lp∗ (Rn) ≤ C(γ)�Du�Lp(Rn).

�

The relation between p and p∗ in Theorem III.3.28 is sharp in the following sense

Lemma III.3.29. Assume that p, q ∈ (1,∞) are such that for all u ∈ C∞c (Rn)

(III.3.23) �u�Lq(Rn) ≤ C(p, n) �Du�Lp(Rn).

Then p = q∗.

Proof. This is proven by a scaling argument. Assume (III.3.23) holds. Take anarbitrary u ∈ C∞

c (Rn) such that �Du�Lp(Rn) ≥ 1, �u�Lp(Rn) ≥ 1.

We rescale u and set for λ > 0,uλ(x) := u(λx).

We apply (III.3.23) to uλ. Observe that by substitution

�uλ�Lq(Rn) = λ− nq �u�Lq(Rn),

and since ∇uλ = λ(∇u)λ we have

�∇uλ�Lp(Rn) = λ1− np �∇u�Lp(Rn).

From (III.3.23) applied to uλ we then obtain for any λ > 0,

λ− nq �u�Lq(Rn) ≤ λ1− n

p �∇u�Lp(Rn).


Equivalently, setting Λ := �∇u�Lq(Rn)/�u�Lq(Rn) > 0 we obtain

λ0− nq

−(1− np

) ≤ Λ ∀λ > 0The exponent above the λ is exactly the numerology of (III.3.20)! In particular, if q �= p∗

then σ := 0 − nq− (1 − n

p) �= 0, and we have

λσ ≤ Λ ∀λ > 0If σ > 0 we let λ → ∞, if σ < 0 we let λ → 0+ to get a contradiction. Thus, necessarilyσ = 0, that is q = p∗. �Corollary III.3.30 (Sobolev-Poincare embedding). Let u ∈ W 1,p(Rn), 1 ≤ p < n. Forany q ∈ [p, p∗] we have u ∈ Lq(Rn) with the estimate

�f�qLq(Rn) ≤ C(q, n)

��f�p

Lp(RN ) + �Df�p∗

Lp(Rn)

�.

Proof. Clearly the claim holds for q = p and, by Theorem III.3.28, for q = p∗.

Now observe that for q ∈ [p, p∗] we can estimate the Lq-norm by the Lp-norm and theLp∗-norm (this technique is called interpolation).

�

Rn|f |q =

�

Rn|f |qχ|f |>1 +

�

Rn|f |qχ|f |≤1 ≤

�

Rn|f |p∗ +

�

Rn|f |p.

That is,�f�q

Lq(Rn) ≤ �f�pLp(Rn) + �f�p∗

Lp∗ (Rn) � �f�pLp(Rn) + �Df�p∗

Lp(Rn).

�Corollary III.3.31 (Sobolev-Poincare embedding on domains). Let Ω ⊂ Rn and ∂Ω beC1. For 1 ≤ p < n we have for any u ∈ W 1,p(Ω),

�u�Lp∗ (Ω) ≤ C(Ω)��u�Lp(Ω) + �Du�Lp(Ω)

�

Also, for any q ∈ [p, p∗] 10

�u�Lq(Ω) ≤ C(Ω, q, �u�W 1,p(Ω)).

If moreover Ω ⊂⊂ Rn and u ∈ W 1,p0 (Ω) then

�u�Lp∗ (Ω) ≤ C(Ω) �Du�Lp(Ω).

Lastly, if 1 ≤ p < ∞ and Ω ⊂⊂ Rn, u ∈ W 1,p(Ω) then for any q ∈ [1, p∗] (if p < n) or forany q ∈ [1,∞) (if p ≥ n)

�u�Lq(Ω) ≤ C(Ω, q, p, n) �u�W 1,p(Ω).

10This means the following: For any Λ > 0 there exists a constant C(Ω, q, Λ) such that�u�Lq(Ω) ≤ C(Ω, q, Λ) ∀u : �u�W 1,p(Ω) ≤ Λ.


Proof. By the extension theorem, Theorem III.3.20, we can extend u to Eu ∈ W 1,p(Rn).Then from Sobolev inequality, Theorem III.3.28, we get�u�Lp∗ (Ω) ≤ �Eu�Lp∗ (Rn) � �DEu�Lp(Rn) � C(Ω)�u�W 1,p(Ω) ≤ C(Ω)

��u�Lp(Ω) + �Du�Lp(Ω)

�.

The second claim follows from the same argument using Corollary III.3.30. The third claimfollows from Poincare inequality, Corollary III.3.27, since for u ∈ W 1,p

0 (Ω), Ω ⊂⊂ Rn wehave

�u�Lp(Ω) ≤ C(Ω)�Du�Lp(Ω).

The last claim follows by additionall using Holder’s inequality: if p < n, q ∈ [1, p∗],�u�Lq(Ω) ≤ C(|Ω|, q, p)�u�Lp∗ (Ω) ≤ C(|Ω|, q, p)�u�W 1,p(Ω).

If p ≥ n and q ∈ [p,∞) we can find r ≤ p such that r∗ > q. Thus, from first Holder’sinequality, then Sobolev inequality, and then again Holder’s inequality,

�u�Lq(Ω) ≤ C(|Ω|, q, p)�u�Lr∗ (Ω) ≤ C(|Ω|, q, p)�u�W 1,r(Ω) ≤ C(|Ω|, q, p)�u�W 1,p(Ω).

�Theorem III.3.32 (Sobolev Embedding). Let Ω ⊂⊂ Rn be open, ∂Ω ∈ C0,1, k ≥ � fork, � ∈ N ∪ {0}, and 1 ≤ p, q < ∞ such that (compare with (III.3.20))

(III.3.24) k − n

p≥ � − n

q.

Then the identity is a continuous embedding W k,p(Ω) �→ W �,q(Ω). That is,(III.3.25) �u�W �,q(Ω) ≤ C(�u�W k,p(Ω))If k > � and we have the strict inequality

(III.3.26) k − n

p> � − n

q,

then the embedding above is compact. That is, whenever (ui)i∈N ⊂ W k,p(Ω) such thatsup

i�ui�W k,p(Ω) < ∞

then there exists a subsequence (uij)j∈N such that (uij

)j∈N is convergent in W �,q(Ω).

Proof. If k = �, then (III.3.24) implies p ≥ q. Thus, in that case (III.3.25) followsfrom the Holder’s inequality:

�u�W �,q(Ω) ≤ C(|Ω|, n) �u�W �,p(Ω) = C(|Ω|, n) �u�W k,p(Ω)

Next we us assume k = � + 1. Then (III.3.24) implies that q ≤ p∗ (if p < n) or q < ∞(for p > n), wher we recall the Sobolev exponent p∗ := np

n−p. Then by Sobolev inequality,

Corollary III.3.31,�f�Lq(Ω) ≤ C(q, Ω)�f�W 1,p(Ω).

Applying this inequality to f := ∂γu for |γ| ≤ � we obtain (III.3.25) for k = � + 1, namelyfor q ≤ p∗,

�u�W �,q(Ω) ≤ C(p, q, Ω)�u�W �+1,p(Ω)


More generally If k = �+N for some N ∈ N, set ri := (ri−1)∗ for i = 1, . . . , N with r0 := p.This works well if all of the r∗

i �= ∞ (otherwise we choose ri ≤ (ri1)∗ and r0 < p, butlarge enough such that rN > q). Then (III.3.24) implies that q ≤ rN , and we get first byHolder’s inequality then by the argument above iterated

�u�W �,q(Ω) � �u�W �,rN (Ω) � �u�W �+1,rN−1 (Ω) � . . . � �u�W k,r0 (Ω) � �u�W k,p(Ω).

This proves the continuous embedding, (III.3.26) in full generality.

As for the compact embedding, it suffices to assume k = �+1. This is because combinationsof continuous and compact embeddings are compact, so if we show the compactness of theembedding satisfying (III.3.26) for k = � + 1 then we can build a chain of embeddings asabove to get a compact embedding for all k > �.

Moreover, we can assume w.l.o.g. k = 1, � = 0. The general case then follows by considering∂γu for |γ| ≤ �.

So let 1 ≤ q < p∗ (i.e. (III.3.26) and assume that we have a sequence (ui) such thatsup

i�ui�W 1,p(Ω) < ∞.

By Sobolev’s inequality, Corollary III.3.31 this implies also for some r ∈ (q, p∗),(III.3.27) Λ := sup

i�ui�Lr(Ω) < ∞.

By Rellichs theorem, Theorem III.3.24, we can find a subsequence uijthat is strongly

convergent in Lp(Ω) and in particular we can choose the subsequence such that uijconverges

pointwise a.e. to some u ∈ Lq(Ω) (that u belongs to Lr, and thus to Lq follows from theweak compactness, Theorem III.3.16).

Now we use Vitali’s convergence theorem11. To show the uniform absolute continuity of theintegral let ε > 0 and for some δ to be chosen (independent of j) let E ⊂ Ω be measurablewith |E| < δ. Then we have by Holder’s inequality (recall Λ from (III.3.27))

supj

�uij�Lq(E) ≤ |E| 1

q− 1

r supj

�uij�Lr(E) ≤ δ

1q

− 1r Λ.

So if we choose δ = δ(ε, Λ) > 0 small, so that

δ1q

− 1p Λ < ε,

thensup

j�uij

�Lq(E) < ε whenever E ⊂ Ω measurable and |E| < δ.

This is uniform absolute continuity, and by Vitali’s theorem uijis convergent in Lq(Ω).

This shows compactness, and Sobolev’s embedding theorem is proven. �11See wikipedia. It essentially shows that almost everywhere convergent sequences fi of functions

converge also in Lq(Ω) if (and only if) we have uniform absolute continuity of the integrals


Our next goal is Morrey’s embedding theorem, Theorem III.3.35. For this we use a char-acterization of Holder functions by so-called Campanato spaces.

Theorem III.3.33 (Campanato’s theorem). Let u ∈ L1(Rn) and assume that for someλ > 0

(III.3.28) Λ := supB(x,r)⊂Rn

r−λ�

B(x,r)|u − (u)B(x,r)| < ∞,

where(u)B(x,r) =

�

B(x,r)u.

Then u ∈ Cλloc(Rn) and we have for some uniform constant C = C(n, λ) > 0

supx�=y

|u(x) − u(y)||x − y|λ ≤ C Λ.

Remark III.3.34. The converse also holds, if u ∈ Cα than Λ < [u]Cα , which is an easyexercise to check.

Proof of Theorem III.3.33. First we claim that for any R > 0 and almost anyx ∈ Rn we have for some uniform constant C > 0(III.3.29) |u(x) − (u)B(x,R)| ≤ C RλΛ.

To see this, observe that for almost every x ∈ Rn, by Lebesgue’s theorem, limk→∞(u)B(x,2−kR) =u(x). Thus, by a telescoping sum

(III.3.30) |u(x) − (u)B(x,R)| ≤∞�

k=0

��(u)B(x,2−kR) − (u)B(x,2−(k+1)R)

��

Now,��(u)B(x,2−kR) − (u)B(x,2−(k+1)R)

��

≤�

B(x,2−(k+1)R)

��u(x) − (u)B(x,2−kR)

��

≤ |B(x, 2−kR)||B(x, 2−k+1R)|� ��

=C(n)

�

B(x,2−kR)

��u(x) − (u)B(x,2−kR)

��

(III.3.28)≤ C(n) Λ (2−kR)λ.

Plugging this into (III.3.30) we get

|u(x) − (u)B(x,R)| ≤ C(n)ΛRλ∞�

k=02−kλ λ>0= C(λ, n)Λ Rλ,

i.e. (III.3.29) is established.


Now let x, y ∈ Rn. Set R := |x − y|. Then

|u(x) − u(y)| ≤|u(x) − (u)B(x,R)| + |u(x) − (u)B(y,R)| + |(u)B(y,R) − (u)B(x,R)|(III.3.29)

≤ C(n, λ)|x − y|λ + |(u)B(y,R) − (u)B(x,R)|.(III.3.31)

We have to estimate the last term, which we do as above: Observe that B(x, 2R) ⊃B(y, R) ∪ B(x, R),

|(u)B(y,R) − (u)B(x,R)|

≤�

B(x,R)

�

B(y,R)|u(z1) − u(z2)|dz1dz2

≤ |B(x, 2R)||B(y, R)|

|B(x, 2R)||B(x, R)|

� �� =C(n)

�

B(x,2R)

�

B(x,2R)|u(z1) − u(z2)|dz1dz2

=C(n)�

B(x,2R)

�

B(x,2R)|u(z1) − (u)B(x,2R)|dz1dz2 + C(n)

�

B(x,2R)

�

B(x,2R)|(u)B(x,2R) − u(z2)|dz1dz2

=C(n)�

B(x,2R)

�

B(x,2R)|u(z1) − (u)B(x,2R)|dz1dz2 + C(n)

�

B(x,2R)

�

B(x,2R)|(u)B(x,2R) − u(z2)|dz1dz2

=2C(n)�

B(x,2R)|u(z) − (u)B(x,2R)|dz

(III.3.28)≤ 2C(n, λ)Rλ.

Since R = |x − y|, together with (III.3.31) we have shown

|u(x) − u(y)| ≤ C(n, λ)|x − y|λ,

and can conclude. �

Theorem III.3.35 (Morrey Embedding). Let Ω ⊂⊂ Rn with ∂Ω ∈ Ck, k ∈ N. Assumethat for p ∈ (1,∞), α ∈ (0, 1) and � < k we have

k − n

p≥ � + α.

Then the embedding W k,p(Ω) �→ C�,α(Ω) is continuous.

If k − np

> � + α then the embedding is compact.

Proof. Let u ∈ W k,p(Ω). By Extension Theorem, Theorem III.3.20, we can assumeu ∈ W k,p(Rn) and supp u ⊂⊂ B(0, R) for some large R > 0.

As in the Sobolev theorem it suffices to assume � = k − 1, and indeed we can reduce to thecase k = 1 and � = 0.


We use Campanato’s Theorem, Theorem III.3.33. For B(x, r) ⊂ Rn, we have by Poincare’sinequality, Corollary III.3.27, and then Holder’s inequality,�

B(x,r)|u− (u)B(x,r)| ≤ r1−λ

�

B(x,r)|Du| = Cr1−n

�

B(x,r)|Du| ≤ Cr1−nrn− n

p

��

B(x,r)|Du|p

� 1p

That is,sup

B(x,r)⊂Rn

r−(1− np )�

B(x,r)|u − (u)B(x,r)| ≤ C�Du�Lp(Rn)

Thus, by Campanato’s theorem, if 1 − np

= 0 + α ∈ (0, 1), then (using also the extensiontheorem estimate),

[u]Cα(Ω) ≤ [u]Cα(Rn) ≤ C�u�W 1,p(Ω),

which is the continuity of the embedding of W 1,p(Ω) in C0,α if 1 − np

= 0 + α.

If on the other hand 1− np

> 0 + α, then we use Arzela-Ascoli to show that the embeddingL∞ ∩ Cβ(Rn) �→ L∞ ∩ Cα(Rn) is compact if β > α, and from this we conclude thecompactness of the embedding W 1,p(Ω) �→ Cα(Ω) if 1 − n

p> α. �

CHAPTER 4

Existence and Regularity for linear elliptic PDE

Our main goal in this section is the following.

Theorem IV.0.1. Let Ω ⊂⊂ Rn with ∂Ω ∈ C∞. For any f ∈ L2(Ω) there exists a uniquesolution u ∈ W 1,2

0 (Ω) −Δu = f in Ωu = 0 on ∂Ω

I.e. �

Ω∇u · ∇ϕ =

�

Ωfϕ ∀ϕ ∈ C∞

c (Ω).

Moreover we have interior regularity: if for any k ≥ 0 we have additionally f ∈ W k,2loc (Ω)

then u ∈ W k+2,2loc (Ω) and the equation above holds almost everywhere in Ω.

In particular if f ∈ C∞(Ω) then u ∈ C∞(Ω) and the equation holds pointwise1

We split the proof of Theorem IV.0.1 into two parts. First we prove existence and unique-ness for f in dual space, f ∈ (W 1,2(Ω))∗. That is assume that f is a linear, boundedfunctional on W 1,2(Ω) with

�f�(W 1,2(Ω))∗ := sup�ϕ�W 1,2(Ω)≤1

f [ϕ] < ∞.

One should think of two standard objects: For some g ∈ L2(Ω)

f [ϕ] :=�

Ωgϕ

andf [ϕ] :=

�

Ωg∂iϕ

Oberserve that in both cases (by Poincare inequality),

�f�(W 1,2(Ω))∗ ≤ �f�L2(Ω).

Theorem IV.0.2. Let Ω ⊂⊂ Rn with ∂Ω ∈ C∞.

1this last part follows from the Morrey embedding, Theorem III.3.35

92

4. EXISTENCE AND REGULARITY FOR LINEAR ELLIPTIC PDE 93

Assume that f ∈ (W 1,2(Ω))∗, i.e. assume that f is a linear, bounded functional on W 1,2(Ω)with

�f�(W 1,2(Ω))∗ := sup�ϕ�W 1,2(Ω)≤1

f [ϕ] < ∞.

Then there exists a unique (weak) solution to

(IV.0.1)


and it satisfies�u�W 1,2(Ω) ≤ C �f�(W 1,2(Ω))∗ .

Here the equation

Proof. We use what is called the direct method of Calculus of Variations2: Set

E(u) := 12

�

Ω|Du|2 + f [u].

As in Section I.2.7 one can check that there is at most one minimizer in W 1,20 (Ω) of this

functional, that any minimizer is a solution to (IV.0.1) and that any solution is a minimizer.

So all that is needed to show is the existence of a minimizer with the claimed estimate.

Let uk ∈ W 1,20 (Ω) be a sequence that approximates inf E (this exists by the very definition

of inf),lim

k→∞E(uk) = inf

W 1,20 (Ω)

E .

In particular, we can assume that E(uk) ≤ E(0) = 0 for all k ∈ N. Now observe that12�Duk�2

L2(Ω) = E(uk) + f [uk] ≤ 0 + �f�(W 1,2(Ω))∗ �uk�W 1,2(Ω).

That is, by Poincare inequality, Corollary III.3.27,�uk�2

W 1,2(Ω) ≤ C�f�(W 1,2(Ω))∗ �uk�W 1,2(Ω).

Dividing both sides by �uk�W 1,2(Ω) we getsup

k�uk�W 1,2(Ω) ≤ C�f�W 1,2(Ω).

That is uk is uniformly bounded in W 1,2(Ω). By the weak compactness theorem, Theo-rem III.3.18, we can thus (up to taking a subsequence) assume uk weakly converging tou ∈ W 1,2

0 (Ω), which in particular implies

f [uk] k→∞−−−→ f [u].Moroever,

�Du�L2(Ω) ≤ lim infk→∞

�Duk�L2(Ω) ≤ �f�(W 1,2(Ω))∗ .

2we did something very similar in the variational methods section, Section I.2.7, but we did not havethe tools to show existence of a minimizer


Thus, we concludeE(u) ≤ lim inf

k→∞E(uk) = inf

W 1,20 (Ω)

E.

But since u ∈ W 1,20 (Ω) we also have

E(u) ≥ infW 1,2

0 (Ω)E,

and thusE(u) = inf

W 1,20 (Ω)

E.

That is we have found a minimizer of E. �Theorem IV.0.3. Let u ∈ W 1,2

0 (Ω) solve


for some f ∈ W k,2loc (Ω), k ≥ 0. Then u ∈ W k+2,0

loc (Ω).

Idea of the proof. Let k = 0.

Essentially we differentiate the equation,Δ∂αu = ∂αf.

Observe that if f ∈ L2loc then ∂αf ∈ W −1,2

loc , where W −1,2 denotes the dual (W 1,2)∗.

By Theorem IV.0.2 we could hope that since there is only one solution to the aboveequation, we get

�∂αu�W 1,2 ≤ C �∂αf�(W 1,2)∗ ≤ �f�L2 .

For k ≥ 1 we would then argue by an induction.

The above is a great idea, but there are different problems: ∂αu does not a priori belongto W 1,2, but even if we knew that: ∂αu has no reason to have the zero-boundary dataW 1,2

0 (Ω).

To overcome the first obstacle we use difference quotients, Proposition III.3.15, for thesecond one we cut off. �

Proof. We show the claim by induction, for any k = 0, 1 . . . , and any Ω1 ⊂⊂ Ω2 ⊂⊂ Ω�u�W k+2,2(Ω1) ≤ �f�W k,2(Ω2) + �u�W k+1,2(Ω2).

For k = −1 this already holds by Theorem IV.0.2. Let η ∈ C∞c (Ω2) with η ≡ 1 in Ω1:

observe thatΔ(ηu) = ηΔu + 2∇η · ∇u + uΔη = ηf + 2∇η · ∇u + uΔη.

Set g := Δ(ηu). If f ∈ W k,2 and u ∈ W k+1,2 then g ∈ W k,2(Ω). So we can work withv := ηu and g instead of f and u to obtain our theorem (away from the boundary ∂Ω).


The advantage of supp v ⊂⊂ Ω2 is that now, for |h| � 1, that if u ∈ W k+1,2loc (Ω) then

Δhe�1

. . . Δhe�k

v ∈ W 1,20 (Ω).

Moreover,Δ(Δh

e�1. . . Δh

e�kv) = Δh

e�1. . . Δh

e�kΔv = Δh

e�1. . . Δh

e�kg.

In order to apply Theorem IV.0.2, we interpret Δhe�1

. . . Δhe�k

g ∈ (W 1,2(Ω))∗, and��

ΩΔh

e�1. . . Δh

e�kg ϕ

�� =��

ΩΔh

e�2. . . Δh

e�kg Δh

e�1ϕ�� g�W k−1,2(Ω) �ϕ�W 1,2(Ω).

Then Theorem IV.0.2 implies for any |h| � 1�Δh

e�1. . . Δh

e�kv�W 1,2(Ω1) ≤ C �g�W k−1,2(Ω2)

By Proposition III.3.15 we get v ∈ W k+1,2(Ω1)�v�W k+1,2(Ω1) � C �g�W k−1,2(Ω2),

since v ≡ u in Ω1 we get for any k ≥ 1,�u�W k+1,2(Ω1) � C �f�W k−1,2(Ω2) + �u�W k,2(Ω1).

�

CHAPTER 5

Fractional Sobolev spaces as trace spaces

V.1. The Fractional Sobolev space W s,p

There is no fractional Sobolev space, there are many. Some of them are called Besov spaces.We restrict our attention on the so-called Gagliardo-Slobodeckij-Sobolev space that appearsas trace operator. We call it W s,p but be aware that some authors denote by W s,p anotherspace (and technically: W s,p for s = 1 is a different space that W 1,p – it is very messy;The most general version are so-called Triebel-Lizorkin spaces F s

p,q and Besov-spaces Bsp,q

– each of them is one fractional Sobolev space. Our choice is F sp,p = Bs

p,p).

As for references, many people like the Hitchhiker’s guide [Di Nezza et al., 2012], seealso [Bucur and Valdinoci, 2016]. More related to Interpolation theory is [Tartar, 2007].Overviews on Besov- and Triebel spaces can be found beginner-friendly in Grafakos’ Har-monic Analysis books [Grafakos, 2014a, Grafakos, 2014b]. The book [Runst and Sickel, 1996]gives in particular a good recollection with references on Triebels’ books. A book very fo-cussed on these Sobolev spaces is also [Samko, 2002].

Let Γ be a (smooth) n-dimensional set (since we will interpret W s,p as trace-spaces later,one can think of Γ as the boundary of a nice set ∂Ω, but it is acceptable to think of Γ asa subset of Rn.

For s ∈ (0, 1) and p ∈ (1,∞) we set

[f ]W s,p(Γ) :=��

Γ

�

Γ

|f(x) − f(y)|p|x − y|n+sp

dx dy

� 1p

.

This object (depending usually on your geographic location) is called the Gagliardo-normor Slobodeckij-norm.

It is only a seminorm, for f = const we have [f ]W s,p = 0.

It is a fun exercise to check that [f ]W s,p(Γ) < ∞ for s ≥ 1 means that f is constant, hencethe restriction to s ∈ (0, 1).

In some sense one can interpret (and should do so) as [f ]W s,p(Γ) measuring some sort ofLp-norm of some sort of an s-derivative. Namely

|f(x) − f(y)||x − y|s

96

V.2. ITS A TRACE SPACE! 97

measures some sort of the s-Holder constant (at a point x and y), indeed if we were to takesupx,y then we would get the Holder norm of order s.

So|f(x) − f(y)|p

|x − y|sp=�

|f(x) − f(y)||x − y|s

�p

,

is some sort of s-Derivative to the power p.

Nowdx dy

|x − y|nis a slightly singular measure (it is singular on the axis x = y), but not very. Almost likea mean value or (for the harmonic analysis enthusiast: maximal function).

So

[f ]W s,p(Γ) =��

Γ

�

Γ

|f(x) − f(y)|p|x − y|n+sp

dx dy

� 1p

.

is somehow the s-derivative of f in Lp (ish).

It is easy (and fun) to check for any ε > 0 then if Γ is compact (or f has compact support)(V.1.1) [f ]W s,p(Γ) ≤ �f�L∞(Γ) + [f ]Cs+ε(Γ).

The Sobolev space W s,p(Γ) is then defined as all f ∈ Lp(Γ) such that [f ]W s,p(Γ) < ∞, withnorm

�f�W s,p(Γ) = �f�Lp(Ω) + [f ]W s,p(Γ)

This is very similar to the W 1,p-norm which is Lp plus Gradient in Lp.

V.2. Its a trace space!

Why do we care about W s,p? First and foremost they are trace spaces.

In Theorem III.3.21 we learned that if u ∈ W 1,p(Ω) then u

��∂Ω

∈ Lp(∂Ω). Indeed the precise

rule is u ∈ W 1,p(Ω) then u��∂Ω

∈ W 1− 1p

,p(∂Ω). And this is a sharp embedding, in the sense

that for any u ∈ W 1− 1p

,p(∂Ω) there exists an extension u ∈ W 1,p(Ω) with u as its trace.

Let us investigate this statement for p = 2 and Ω = Rn+1+ the upper halfplane. For

simplicity we will now start to call functions on the boundary ∂Ω = Rn × {0} with smallletters u, v, w etc. and functions in the set U, V, W .

Theorem V.2.1. For any u ∈ C∞c (Rn), p ∈ (1,∞), we have

[u]W

12 ,2(Rn)

≈ infU |Rn =u

�DU�L2(Rn+1+ ).


where A ≈ B means that there exist a constant C > 0 such that C−1A ≤ B ≤ CA.

In particular,

• If U ∈ C∞c (Rn × [0,∞)) then its trace belongs to W

12 ,2(Rn) and we have the

estimate[u]

W12 ,2(Rn)

≤ C �DU�L2(Rn+1+ ).

• For any u ∈ C∞c (Rn) there exists U ∈ W 1,2(Rn) (The dot in W 1,2 means that U

does not need to belong to L2 but only DU ∈ L2), such that�DU�L2(Rn+1

+ ) ≤ C[u]W

12 ,2(Rn)

Indeed, the proof will show that such a U can be chosen as the variation minimizerof �DU�L2(Rn+1

+ ), i.e. the Harmonic extension of u

ΔRn+1U = 0 in Rn+1+

U = u on Rn × {0}Here we assume (often implicitely) that lim|x|→∞ U(x) = 0.

Proof. It is helpful to consider the variables in Rn+1 by (x, t), x ∈ Rn, t ∈ R.

We need to show that a solution U to

ΔRn+1U = ΔxU + ∂ttU = 0 in Rn+1+

U = u on Rn × {0}with lim|(x,t)|→∞ U = 0 (which thus minimizes the inf in the statement) satisfies

[u]W

12 ,2(Rn)

≈ �DU�L2(Rn+1+ ).

Computing by an integration by parts, (observe that ∂t = −∂ν on ∂Rn+1+ = Rn × {0}),

�DU�2L2(Rn+1

+ ) =�

Rn+1+

DU · DU = −�

Rn×{0}∂tU U −

�

Rn+1+

Δx,tU� �� =0

U

So,

(V.2.1) �DU�2L2(Rn+1

+ ) = −�

Rn×{0}∂tU U.

So we really would like to know the Neumann-derivative ∂tU of a solution to the Dirichletproblem ΔU = 0 and U = u on Rn – this is why we talk about a Dirichlet-to-Neumannproperty.

Lucky for us, we have the Fourier transform.


Taking the Fouriertransform in x-direction, Δx,tU = 0, recalling that Δx becomes −c|ξ|2after Fourier transform (Lets for simplicity pretend that c = 1),

∂ttU(ξ, t) − |ξ|2U(ξ, t) = 0.

The only solution to the ordinary differential equations with limt→∞ U = 0, isU(ξ, t) = e−t|ξ|U(ξ, 0) = e−t|ξ|u(ξ)

Observe that |ξ|2 is the Fourier symbol of the Laplacian (−Δ), we could think of |ξ| as theFourier symbol of an operatoron we shall call (−Δ) 1

2 , the half-laplace or fractional laplace.

Thus, reverting the Fourier transform, we could write ΔU = 0 from our assumption as

U(x, t) = e−t(−Δ)12 u(x).

However we want to interpret this, we have

∂tU(x, 0) = −(−Δ) 12 u(x).

From (V.2.1) we get�DU�2

L2(Rn+1+ ) =

�

Rn(−Δ) 1

2 u u.

That does not look much better, but we can do an integration by parts,�

Rn(−Δ) 1

2 u u =�

Rn(−Δ) 1

4 u (−Δ) 14 u = �(−Δ) 1

4 u�2L2(Rn).

Indeed, using twice Plancherel theorem (Fouriertransform: yay! – a nice exercise: try toshow integration by parts formula via Fourier transform for the classical derivative ∂i)

�

Rn(−Δ) 1

2 u u =�

Rn|ξ|u u =

�

Rn|ξ| 1

2 u |ξ| 12 u =

�

Rn|(−Δ) 1

4 u|2

So we have shown that for our U -u combination�DU�2

L2(Rn+1+ ) = C�(−Δ) 1

4 u�2L2(Rn).

Now counting derivatives: The Laplacian has two derivatives, so (−Δ) 14 has “1/2”deriva-

tives. So one would hope that

�(−Δ) 14 u�L2(Rn) ≈ [u]

W12 ,2(Rn)

This is indeed true, but this is because we have L2 (and thus: Plancherell). In general

(WARNING) �(−Δ) s2 u�Lp(Rn) �≈ [u]W s,p(Rn).

unless p = 2. I.e. for s ∈ (0, 1),

�(−Δ) s2 u�L2(Rn) ≈ [u]W s,2(Rn).

For this let us compute a new formula for (−Δ) s2 u. Recall that after Fourier transform

((−Δ) s2 u)∧(ξ) = |ξ|su(ξ)


. We computed in the lines after (I.2.3) how products and homogeneous functions behaveunder fourier transform, so we could (see the Newton potential which was derived fromthis for s < 0)

(−Δ) s2 u(x) = | · |−n−s ∗ u(x).

This is almost true, (but in a much more precise sense it is false, for any function u �≡ 0.),one needs to take into account the (now hypersingular singularity of |z|−n−s for z = 0).Namely

(−Δ) s2 u(x) = c| · |−n−s ∗ (u − u(x)) (x) = c

�

Rn

u(y) − u(x)|x − y|n+s

dy.

This is good for s ∈ (0, 1), for s ∈ [1, 2) one should to write a principal value symbol. Withthis formula one checks that�

Rn(−Δ) s

2 u(x) v(x) =�

Rn

�

Rn

u(y) − u(x)|x − y|n+s

v(x) dy dx = −12

�

Rn

�

Rn

(u(y) − u(x)) (v(y) − v(x))|x − y|n+s

dy dx

where the last step follows from symmetry: using fubini with interchanging x and y we seethat

�

Rn

�

Rn

(u(y) − u(x)) (v(y) + v(x))|x − y|n+s

dy dx = 0

and noting that v(x) = 12(v(x) − v(y)) + 1

2(v(x) + v(y)).

In particular,�

Rn(−Δ) 1

2 u(x) u(x) = −[u]2W

12 ,2(Rn)

,

and the claim is shown. �

An adaptation (using some delicate harmonic analysis, like square function estimates) leadsto the general statement of Theorem V.2.1:

Theorem V.2.2. For any u ∈ C∞c (Ω), s ∈ (0, 1), p ∈ (1,∞), for some set Ω with nice

boundary. Then[u]W s,p(∂Ω) ≈ inf

U |∂Ω=u�t1− 1

p−sDU�Lp(Ω).

Where U is a solution of

ΔU = 0 in ΩU = u on ∂Ω

This is somewhat generally known fact in the Harmonic Analysis community and intrinsi-cally contained in Harmonic Analysi classics like [Stein, 1993]. See also [Bui and Candy, 2015]and [Lenzmann and Schikorra, 2019].

V.3. COOL THINGS ONE CAN DO WITH THIS: INTEGRATION BY PARTS REVISITED 101

V.3. Cool things one can do with this: Integration by parts revisited

Let us do an example (which surprisingly is not known to a wider community, but knownfor a long time to some experts):

Let f, g ∈ C∞c (R,R), and consider

A :=�

Rf(x) g�(x) dx = −

�

Rf �(x) g(x) dx

By Holder’s inequality, we easily get the estimates|A| ≤ �f�L2(R) �g��L2(R)

or|B| ≤ �f ��L2(R) �g�L2(R).

But we can get an “intermediate” estimate where for f and g only half a derivative counts:Proposition V.3.1. Let f, g ∈ C∞

c (R,R), then��

Rf(x) g�(x) dx

�� ≤ [f ]W

12 ,2(R)

[g]W

12 ,2(R)

One application of this estimate (which in its statement does not use the notion of fractionalSobolev spaces) is (cf. (V.1.1)):Corollary V.3.2. Let f, g ∈ C∞

c (R,R). For any s ∈ (1/2, 1]��

Rf(x) g�(x) dx

�� ≤ C(s, supp f, supp g) �f�Cs�g�Cs

Also,

(V.3.1)�� 1

0f(x) g�(x) dx

�� ≤ C(s) (|f(0)| + [f ]Cs) (|g(0)| + [g]Cs)

Proof of (V.3.1), proving also the Proposition and the Corollary. First weassume that u(0) = v(0) = u(1) = v(1) = 0. Then we can extend u, v by zero to a Lipschitzfunction on all of R (obviously with compact support). Denote by U, V : R2

+ → R exten-sions of u, v : R → R with decay at infinity of U , V , and ∂tU , ∂tV , i.e. lim|(s,t)|→∞ U(s, t) =0, lim|(s,t)|→∞ |DU(s, t)| = 0, and likewise for V .

Then, following the ideas in [Lenzmann and Schikorra, 2019], see also [Brezis and Nguyen, 2011],� 1

0u(t) v�(t)dt

=�

Ru(t) v�(t)dt

=�

R

� ∞

0∂s (U(t, s)∂tV (t, s)) ds dt

=�

R

� ∞

0(∂sU(t, s)∂tV (t, s)) ds dt +

�

R

� ∞

0U(t, s)∂s,tV (t, s) ds dt

V.3. COOL THINGS ONE CAN DO WITH THIS: INTEGRATION BY PARTS REVISITED 102

For the second term observe�

R

� ∞

0U(t, s)∂s,tV (t, s) ds dt = −

�

R

� ∞

0∂tU(t, s)∂sV (t, s) ds dt

That is, we obtain�� 1

0u(t) v�(t)dt

��

≤2�

R2+

|DU | |DV |

≤2�DU�L2(R2+) �DV �L2(R2

+)

Choosing U and V the harmonic extension we obtain�� 1

0u(t) v�(t)dt

�� ≤ C

��

R

�

R

|u(x) − u(y)|2|x − y|2 dx dy

� 12��

R

�

R

|v(x) − v(y)|2|x − y|2 dx dy

� 12

From this one easily obtains for any s > 12 ,

�� 1

0u(t) v�(t)dt

�� ≤ C (�u�L∞ + [u]Cs) (�v�L∞ + [v]Cs) .

Recall that u and v have support in [0, 1]. Thus we arrive at�� 1

0u(t) v�(t)dt

�� ≤ C�[u]Cs((0,1))

� �[v]Cs((0,1))

�.

This shows the claim of the proposition under the assumption that u(0) = v(0) = u(1) =v(1) = 0.

If this assumption is not satisfied, we set u(t) := (1 − t)u(0) + tu(1) and likewise v(t) :=(1 − t)v(0) + tv(1). Then�� 1

0u(t)v�(t)

�� ≤�� 1

0(u − u) (t) (v − v)� (t)

��+�� 1

0u(t)v�(t) dt

��+�� 1

0u(t)v�(t)

��+�� 1

0u(t)v�(t)

��

The first term we can estimate by the first part of the proof. Clearly

[u]Cs((0,1)) ≤ |u(0)| + |u(1)| ≤ |u(0)| + [u]Cs((0,1)),

and likewise for v. For the second term, observe�� 1

0u(t)v�(t)

�� ≤ |v(1) − v(0)|�u�L∞(0,1) ≤ [v]Cs((0,1))�|u(0)| + [u]Cs((0,1))

�

We argue similarly for the third term. For the fourth term, we do an integration by parts,before arguing as above.�� 1

0u(t)v�(t)

�� ≤ |v(1)||u(1)|+|v(0)||u(0)|+�� 1

0u�(t)v(t)

�� ≤�|u(0)| + [u]Cs((0,1))

� �|v(0)| + [v]Cs((0,1))

�.

�

V.6. “THE” OTHER FRACTIONAL SOBOLEV SPACE HS,P 103

These kind of estimates can be generalized to several situations that involve “commutator”-type structures (it falls under the realm of “compensated compactness”). See, e.g., [Brezis and Nguyen, 2011for Jacobians, [Lenzmann and Schikorra, 2019] for general commutator estimates.

V.4. W s,p is not a gradient to the power p

Here we have a word of warning. We explained W s,p as “essentially” the s-derivativef(x)−f(y)

|x−y|s to the power p.

This leads to a good intuition for many things, but also one important false analogy:

Recall: If f ∈ W 1,p(Ω) then for any q < p we have W 1,qloc (Ω). This is simply because the

same holds for Lp(Ω) ⊂ Lqloc(Ω) if q < p.

But this is false for W s,p. This can be easily be seen by a characterization of W s,p as so-called Triebel-spaces. But we do not want to go there, but refer to [Mironescu and Sickel, 2015]which used this relation to construct a counterexample for W s,p(Ω) �⊂ W s,q

loc (Ω) even thoughtq < p.

That is one odd result, and it should serve as a warning that W s,p behave sometimes alittle bit different from W 1,p.

V.5. W s,p becomes W 1,p as s goes to one

The notion of

[u]W s,p(Ω) =��

Ω

�

Ω

|u(x) − u(y)|p|x − y|n+sp

dx dy

� 1p

does not make sense as s → 1. Indeed,��

Ω

�

Ω

|u(x) − u(y)|p|x − y|n+p

dx dy

� 1p

< ∞ ⇔ u is constant.

However, one can show, see [Bourgain et al., 2001], that

(1 − s)1p [u]W s,p

s→1−−→ �∇u�Lp .

V.6. “The” other fractional Sobolev space H s,p

We found above the operator (−Δ) s2 . Another fractional Sobolev space, often denoted by

Hs,p(Rn) is given as all functions such that[u]Hs,p(Rn) := �(−Δ) s

2 u�Lp(Rn) < ∞.

For p = 2 there is nothing new: Hs,2 is W s,2, for p �= 2 these two spaces are different.

V.7. EMBEDDING THEOREMS, TRACE-THEOREMS ETC. HOLD TRUE 104

The space Hs,p is a more natural fractional Sobolev space in some features (for exam-ple u ∈ H1,p

loc implies u ∈ H1,qloc for q < p) but it also has some disadvantages: it is

a bit more complicated to define on subsets Ω ⊂ Rn (since (−Δ) s2 is a global opera-

tor), it is not a trace space (but there are some trace-space-type characterizations, see[Lenzmann and Schikorra, 2019]).

But for example it satisfies that Hs,p s→1−−→ W 1,p (and this is almost trivial to show).

Warning: The notation Hs,p for one type, W s,p for the other type of Sobolev spaces is notuniform in the literature. Check carefully what space is meant if you find it in a researchpaper!

V.7. Embedding theorems, Trace-theorems etc. hold true

The good news is: All the embedding theorems we learned before have an analogue. More-over, up to the Sobolev-inequality (where we used the very elegant proof by Nirenberg inTheorem III.3.28) even the proofs stay the same (up to minor adjustments).

Recall the numerology for Sobolev spaces: W 1,p was locally embedded into Lq if

1 − n

p≥ 0 − n

q

The 1 is the derivative order of W 1,p and 0 is the derivative order of Lq. Here are the mainresults:

• W s,p(Ω) embedds (locally) continuously into W t,q(Ω) if s ≥ t and

s − n

p≥ t − n

q.

• If the inequality above is strict, the embedding is compact.• In particular we have Rellich: W s,p compactly embedds into Lp if s > 0 (locally)• In particular we have Poincare inequalty

�u − (u)�Lp(Ω) ≤ C(Ω) [u]W s,p(Ω)

whenever s > 0 (and Ω is sufficiently nice).• If s − n

p≥ k + α and s > k then W s,p embedds into Ck,α (compactness if there is

a strict inequality)

We mentioned before that Hs,p and W s,p are different spaces. However they are in somesense “very close by”. This can, e.g., be seen with Sobolev embedding: If if s ≥ t and

s − n

p≥ t − n

q.

then W s,p(Ω) embedds into H t,q(Ω) (locally) and Hs,p(Ω) embedds into W t,q(Ω) (locally).

V.7. EMBEDDING THEOREMS, TRACE-THEOREMS ETC. HOLD TRUE 105

As for traces the numerology before was W 1,p(Ω) embedds into W 1− 1p

,p(∂Ω). This staysthe same, as long as s − 1

p> 0 then W s,p(Ω) embedds into W s− 1

p,p(∂Ω).

When s − 1p

= 0 things get difficult. For an illustration: In the class W12 ,2 there are “two

spaces” with 0 on the boundary. One, W12 ,2

0 (Ω) where the trace is zero, and the Lions-Magenes-space W

12 ,2

00 (Ω) which are maps which can be extended by zero to a W12 ,2(Rn)-map.

See [Tartar, 2007].

CHAPTER 6

Parabolic PDEs

VI.1. The heat equation: Fundamental solution and Representation

We consider∂tu − Δu = f in Rn+1

+

u(0, ·) = g on Rn.(VI.1.1)

First assume f = 0. Then (VI.1.1) is called homogeneous heat equation. For f �= 0 it iscalled inhomogeneous.

Trivial solutions of the homogeneous equation constant maps u(x, t) ≡ c, or (not completelytrivial) time-independent harmonic functions u(x, t) := v(x) with Δv = 0.

For elliptic equations we had the notion of a fundamental solution, Section I.2.1; Thereexists a similar concept for the heat equation, the heat kernel, which we will (formally)derive now.

If we fix x ∈ Rn and look at (VI.1.1) as an equation in time t then it looks like an ODE,and naıvely the solution should be

u(x, t) = etΔu(x, 0).Of course, etΔ does not make any sense for now, but we will define this later via semi-grouptheory, Chapter 8.

To make (still formally, but more precise sense) of the “ODE argument”, we use theFourier-transformation (with respect to the variables x ∈ Rn):

Let u be a solution of ∂tu = Δu. Taking the Fourier transform (in x) on both sides we findd

dtu(ξ, t) = �∂tu(ξ, t) = �Δu(ξ, t)

= −|ξ|2u(ξ, t).Let ξ be fixed and let

v(t) = u(ξ, t).Then the above reads as

d

dtv(t) = −|ξ|2v(t).

106

VI.1. THE HEAT EQUATION: FUNDAMENTAL SOLUTION AND REPRESENTATION 107

There is one solution to this ODE (starting from a given value v(0)):v(t) = e−t|ξ|2v(0).

Observe that in particular v(∞) = 0 (i.e. “decay at infinity”).

Ansatz: v(0) = 1, resp. u(0) = δ0. This meansu(ξ, t) = e−t|ξ|2 .

In this case we haveu(x, t) = 1

(4πt)n2

e− |x|24t ,

which seems to be a special solution.

Definition VI.1.1.

Φ(x, t) =

1(4πt)

n2

e− |x|24t , t > 0, x ∈ Rn

0, t < 0, x ∈ Rn.

is called fundamental solution or heat kernel.

One has∂tΦ − ΔΦ = 0, for t > 0

and

limt→0

Φ(x0, t) =

0, x0 �= 0∞, x0 = 0.

Lemma VI.1.2.∀t > 0 :

�

RnΦ(x, t) dx = 1.

Proof. �

RnΦ(x, t) dx = Φ(0, t) = 1.

�

Analogously to the fundamental solution for the Laplace equation, the heat kernel Φ gen-erates solutions to the heat equation. Indeed, if we set

u(x, t) := Φ(·, t) ∗ g(x)

=�

RnΦ(x − y, t)g(y) dy

Thenu(ξ, t) = �(Φ(·, t) ∗ g)(ξ) = Φ(ξ, t)g(ξ).

That is,u(ξ, 0) = g(ξ), ( d

dt+ |ξ|2)u(ξ, t) = 0.

VI.1. THE HEAT EQUATION: FUNDAMENTAL SOLUTION AND REPRESENTATION 108

Revert the Fourier-transformation to obtain

( d

dt− Δ)u = 0 in Rn+1

+ , u(x, 0) = g(x), x ∈ Rn.

Motivated by this calculation we set

u(x, t) =�

RnΦ(x − y, t)g(y) dy.

Theorem VI.1.3 (Potential representation). Let g ∈ C0(Rn)∩L∞(Rn). Let u as in (VI.1).Then u is defined in Rn and there holds:

(i) u ∈ C∞(Rn+1+ ),

(ii) ∂tu − Δu = 0 in Rn+1+ und

(iii)∀x0 ∈ Rn : lim

(x,t)→(x0,0)u(x, t) = g(x0).

Next we search a potential representation for

( d

dt− Δ)u = f in Rn+1

+

u(·, 0) = 0 on Rn.(VI.1.2)

Again we use the Fourier transformation for intuition. Set v(t) = u(ξ, t). Then∂tv(t) + |ξ|2v(t) = f(ξ, t) =: h(t).

We use what is called Duhamel’s formula.

For s > 0 solve∂tws(t) + |ξ|2ws(t) = 0, t > s

ws(s) = h(s).Again, from ODE theory we know what ws is,(VI.1.3) ws(t) = e−(t−s)|ξ|2h(s).Now we try to solve (VI.1.2) by the Ansatz (which we may recognize from ODE calculus)

v(t) =� t

0ws(t) ds.

Then v(0) = 0, and moreover

∂tv(t) = wt(t) +� t

0∂tws(t) ds = h(t) − |ξ|2

� t

0ws(t) ds = h(t) − |ξ|2v(t).

Thus, using (VI.1.3) we know that

v(t) =� t

0e−(t−s)|ξ|2h(s) ds

VI.3. MAXIMUM PRINCIPLE AND UNIQUENESS 109

solves (VI.1.2).

Using the inverse Fourier transform,

u(x, t) =� t

0

�

RnΦ(x − y, t − s)f(y, s) dyds.

Theorem VI.1.4. Let f ∈ C21(Rn × [0,∞)) with compact support and let u as in (VI.1).

Then

(i) u ∈ C21(Rn × (0,∞)),

(ii) ( ddt

− Δ)u = f in Rn × (0,∞)

(iii) ∀x0 ∈ Rn : lim(x,t)→(x0,0) u(x, t) = 0.

VI.2. Mean-value formula

(cf. [Evans, 2010, Chapter 2.3])

Use the fundamental solution to construct a parabolic ball, or heat ballE(x, t; r) ⊂ Rn+1.(VI.2.1)

Definition VI.2.1 (Heat ball). Let (x, t) ∈ Rn+1. Set

E(x, t; r) =�

(y, s) ∈ Rn+1 : s ≤ t, Φ(x − y, t − s) ≥ 1rn

�.(VI.2.2)

Theorem VI.2.2 (mean value). Let X ⊂ Rn+1 be open and u ∈ C21(X) solve (∂t−Δ)u = 0

in X. Then there holds

u(x, t) = 14rn

�

E(x,t;r)u(y, s) |x − y|2

(t − s)2 dyds(VI.2.3)

for all E(x, t; r) ⊂ X.

VI.3. Maximum principle and Uniqueness

Definition VI.3.1. Let Ω ⊂ Rn be an open set and denote with ΩT := Ω× (0, T ] for sometime T > 0. It is important to note that the top Ω × {T} belongs to ΩT . The parabolicboundary ΓT of ΩT is the boundary of ΩT without the top,

ΓT = ΩT \ΩT = ∂Ω × [0, T ) ∪ Ω × {0}.

Theorem VI.3.2. Let U be bounded and u ∈ C21(UT ) ∩ C0(UT ) be a solution of ut = Δu

in UT . Then there holds the weak maximum principle

(i)max

UT

u = maxΓT

u(VI.3.1)


and the strong maximum principle:

(ii) If U is connected and if there is (x0, t0) ∈ UT withu(x0, t0) = max

UT

u,(VI.3.2)

thenu(x, t) = u(x0, t0) ∀(x, t) ∈ Ut0 .(VI.3.3)

Proof. (ii)⇒(i), since ifmax

UT

u > maxΓT

u(VI.3.4)

then by (ii) u is constant at all prior times, which contradicts (VI.3.4).

Now we prove (ii). Suppose there is (x0, t0) ∈ UT withu(x0, t0) = M = max

UT

u.(VI.3.5)

Since t0 > 0, there exists a small heat ball E(x0, t0, r0) ⊂ UT and we have by Theo-rem VI.2.2

M = u(x0, t0) = 14r0n

�

E(x0,t0,r0)u(y, s) |y − x|2

(t − s)2 dsdy ≤ M.(VI.3.6)

Hence u ≡ M in E(x0, t0; r0).

Now we need to show u = M in all of Ut0 . It suffices to show u ≡ M in any Ut1 for anyt1 < t0, by continuity u ≡ M in all of Ut0 . So let (x1, t1) ∈ Ut0 ., t1 < t0. Then there existsa continuous path γ : [0, 1] → U connecting x0 and x1. In the spacetime set

Γ(r) = (γ(r), rt1 + (1 − r)t0).(VI.3.7)

Letρ = max{r ∈ [0, 1] : u(Γ(r)) = M}.(VI.3.8)

Show that ρ = 1. Suppose ρ < 1. Then we use the proof above to find a heat ballE = E(Γ(ρ), r�),(VI.3.9)

where u = M. Since Γ crosses E (time parameter is decreasing along Γ), we obtain acontradiction to the maximality of ρ. �Remark VI.3.3. The same holds for −u and hence we have a minimum principle. Hence,if in particular

ut − Δu = 0 in UT

u = 0 on ∂U × [0, T ]u = g in U × {0}

(VI.3.10)


with g(x) > 0 for some x ∈ U then u > 0 in UT (infinite speed of propagation, non-relativistic).

Remark VI.3.4. For general X ⊂ Rn+1 open we have a similar result, see exercises.

Theorem VI.3.5 (Uniqueness on bounded domains). Let U � Rn bounded and g ∈C0(ΓT ), f ∈ C0(UT ). Then there is at most one solution C2

1(UT ) ∩ C0(UT ) tout − Δu = f in UT

u = g on ΓT .(VI.3.11)

Proof. Apply the maximum (and minimum) principle to show that the difference oftwo solutions is zero. �Theorem VI.3.6. Let u ∈ C2

1(Rn × (0, T ]) ∩ C0(Rn × [0, T ]) be a solution of(∂t − Δ)u = 0 in Rn × (0, T )

u = g on Rn × {t = 0}(VI.3.12)

with the growth condition

u(x, t) ≤ Aea|x|2(VI.3.13)

for some a, A > 0. Then there holdssup

Rn×[0,T ]u ≤ sup

Rng.(VI.3.14)

Proof. It suffices to show this estimate for small times, by splitting up the timeinterval into many small time steps. For this reason we assume first:

4aT < 1.(VI.3.15)

For ε > 0 and µ chosen below, let

v(x, t) = u(x, t) − µ

(T + ε − t) n2

e|x|2

4(T +ε−t)(VI.3.16)

for some µ > 0. Then vt − Δv = 0. (observe that t appears in the negative above).Theorem VI.3.2 implies

∀U � Rn : maxUT

v ≤ maxΓT

v ≤ max(max v(·, 0), max∂U×[0,T ]

v(x, t)).(VI.3.17)

We have

v(x, 0) = g(x) − µ

(T + ε)n2

e|x|2

4(T +ε) ≤ supRn

g.(VI.3.18)

Let U = BR(0), then

maxBR(0)×[0,T ]

v ≤ max�

supRn

g, max|x|=R,t∈[0,T ]

v(x, t)�

.(VI.3.19)


For |x| = R and t ∈ (0, T )

v(x, t) = u(x, t) − µ

(T + ε − t) n2

eR2

4(T +ε−t)

≤ Aea|x|2 − µ

(T + ε − t) n2

eR2

4(T +ε−t)

≤ Aea|x|2 − µ

(T + ε)n2

eR2

4(T +ε)

Since 4aT < 1, there exist ε > 0, γ > 0, such that

a + γ = 14(T + ε)(VI.3.20)

and hence

v(x, t) ≤ AeaR2 − µ

(T + ε)n2

eaR2+γR2.(VI.3.21)

In particular, the right term dominates for R >> 0: in particular for all large R > 0 wehave v(x, t) ≤ g(0). So for large R and |x| = R we have for all t ∈ (0, T ],

v(x, t) ≤ g(0) ≤ supRn

g(VI.3.22)

and somax

(x,t)∈BR(0)×(0,T ]v(x, t) ≤ sup

Rng ∀R >> 1.(VI.3.23)

Letting R → ∞ we find thatsup

Rn×[0,T ]v(x, t) ≤ sup

Rng,(VI.3.24)

i.e.

supRn×[0,T ]

�u(x, t) − µ

(T + ε − t) n2

e|x|2

4(T +ε−t)

�≤ sup

Rng(VI.3.25)

This holds for any any µ > 0.

Letting µ → 0 for fixed x gives the claim under the assumption that 4aT < 1.

If 4aT ≥ 1, we can slice the time interval (0, T ] into parts (0, T1] ∪ (T1, T2] ∪ . . . ∪ (TK , T ]with 4a(Ti+1 − Ti) < 1 for all i. Using the estimate in each of these time intervals weconclude. �

Theorem VI.3.7. Let g ∈ C0(Rn), f ∈ C0(Rn × [0, T ]). Then there is at most one solutionu ∈ C2

1(Rn × (0, T ]) ∩ C0(Rn × [0, T ]) of(∂t − Δ)u = f in Rn × (0, T )

u = g on Rn × {0}(VI.3.26)


with

|u(x, t)| ≤ Aea|x|2 ∀(x, t) ∈ Rn × (0, T ).(VI.3.27)

Proof. Exercise VI.3.9 �

Exercise VI.3.8. In Theorem VI.3.7 we learned of the strong maximum principle in par-abolic Cylinders. Use this to obtain the strong maximum principle in general open setsX:

let X ⊂ Rn+1 be a bounded, open set. Assume that u ∈ C∞(X) and

∂tu − Δu in X.

Assume moreover that for some (x0, t0) ∈ X we have

M := u(x0, t0) = sup(x,t)∈X

u(x, t).

(1) Describe (in words) in which set C the function is necessarily constant

C := {(x, t) ∈ X : u(x, t) = M} .

(2) Assume the set X (grey) and the point (x0, t0) are given in the picture. Draw (inorange) the set C from the question above.

Exercise VI.3.9. Show Theorem VI.3.7: let g ∈ C0(Rn), f ∈ C0(Rn × [0, T ]) for someT > 0.

VI.4. HARNACK’S PRINCIPLE 114

Assume there are two solutions u1 and u2 ∈ C21(Rn ×(0, T ))∩C0(Rn × [0, T ]) of the problem

(∂t − Δ)u = 0 in Rn × (0, T ),u(x, 0) = g(x) fur x ∈ Rn.

If moreover we know that there are a1, a2 and A1, A2 > 0 with|u1(x, t)| ≤ A1 ea1 |x|2 , |u2(x, t)| ≤ A2 ea2 |x|2 ∀(x, t) ∈ Rn × [0, T ],

show that thenu1 ≡ u2 auf Rn × [0, T ].

Hint: Use Theorem VI.3.6.

Exercise VI.3.10. (cf. [John, 1991]) Define the following Tychonoff-function,

u(x, t) :=∞�

k=0

g(k)(t)(2k)! x2k.

Here g(k) denotes the k-th derivative of g, given as

g(t) :=

e(−t−α) t > 00 t ≤ 0.

(1) Show that u ∈ C21(R2

+) ∩ C0(R × [0,∞)).(2) Show moreover that

(VI.3.28)

(∂t − Δ)u = 0 in Rn × (0, T ),u(x, 0) = 0 fur x ∈ Rn.

(3) Find a different solution v �≡ u of (VI.3.28).(4) Why (without proof) does this not contradict Question VI.3.9?

VI.4. Harnack’s Principle

In the parabolic setting an “immediate” Harnack principle is not true in general, to comparesup and inf of a function one needs to wait for an (arbitrary short) amount of time.

Theorem VI.4.1 (Parabolic Harnack inequality). Assume u ∈ C21(Rn × (0, T ])∩L∞(Rn ×

[0, T ]) and solves∂tu − Δu = 0 in Rn × (0, T )

andu ≥ 0 in Rn × (0, T )

Then for any compactum K ⊂ Rn and any 0 < t1 < t2 < T there exists a constant C, sothat

supx∈K

u(x, t1) ≤ C infy∈K

u(y, t2)

VI.4. HARNACK’S PRINCIPLE 115

Proof. By the representation formula, Section VI.1, and uniqueness of the Cauchyproblem

u(x2, t2) =�

Rn

1(4πt2)

n2

e− |x2−y|2

4t2 u0(y) dy.

Now, for t1 < t2 whenever |x1|, |x2| ≤ Λ < ∞, there exists a constant C = C(|t1 − t2|, Λ)so that

− |x2 − y|24t2

≥ − |x1 − y|24t1

− C. ∀y ∈ Rn

See Exercise VI.4.2.

Consequently,

u(x2, t2) ≥�

t1

t2

�n2

e−C�

Rn

1(t1)

n2

e− |x1−y|2

4t1 u0(y) dy =�

t1

t2

�n2

e−Cu(x1, t1).

�Exercise VI.4.2. Show the following estimate, which we used for Harnack-principle, The-orem VI.4.1:

If K ⊂ Rn is compact and 0 < t1 < t2 < ∞, then there exists a constant C > 0 dependingon K and (t2 − t1), such that

|x1 − y|2t2

≤ |x2 − y|2t1

+ C ∀x1, x2 ∈ K, y ∈ Rn.

Exercise VI.4.3 (Counterexample Harnack). (1) Sei u0 : Rn → [0,∞) eine glatteFunktion mit kompaktem support mit u0(0) = 1. Setze

u(x, t) :=�

RnΦ(x − y, t) u0(y) t > 0

Zeigen Sie,inf

x∈Rnu(x, t) = 0 fur alle t > 0.

Abersupx∈Rn

u(x, t) > 0 fur alle t > 0.

Warum ist dies kein Widerspruch zum Harnack-Prinzip, Theorem VI.4.1?(2) Zeigen Sie, dass das folgende Sei ξ ∈ Rn gegeben, und u definiert als

uξ(x, t) := (t + 1)− 12 e− |x+ξ|2

4(t+1) .

Zeigen Sie dass u eine Losung von (∂t − Δ)u = 0 auf Rn × (0,∞) ist. Zeigen Sieaber auch, dass es jedes feste t > 0 keine Konstante C = C(t) > 0 gibt fur die gilt

supx∈[−1,1]

uξ(x, t) ≤ C infy∈[−1,1]

uξ(y, t) ∀ξ ∈ Rn.

Warum ist dies kein Widerspruch zum Harnack-Prinzip, Theorem VI.4.1?Hinweis: Wahlen Sie x = − ξ

|ξ| und y = 0. Was passiert, wenn |ξ| → ∞?

VI.5. REGULARITY AND CAUCHY-ESTIMATES 116

VI.5. Regularity and Cauchy-estimates

Theorem VI.5.1 (Smoothness). Let u ∈ C21(UT ) satisfy

ut = Δu in UT .(VI.5.1)

Then u ∈ C∞(int(UT )).

Proof. This is a standard technique to transfer local questions to global situations,using a cut-off function. Let

C(x, t; r) = {(y, s) : |x − y| ≤ r, t − r2 ≤ s ≤ t}(VI.5.2)

and

C1 = C(x0, t0; r), C2 = C�

x0, t0;34r�

, C3 = C�

x0, t0;r

2

�(VI.5.3)

for some r such that C1 ⊂ UT . Choose a cut-off functionη ∈ C∞(Rn × [0, t0])(VI.5.4)

with 0 ≤ η ≤ 1, η|C2 ≡ 1, η ≡ 0 around Rn × [0, t0]\C1. Suppose first that u is smooth. Setv(x, t) = η(x, t)u(x, t) ∀(x, t) ∈ Rn × (0, t0],(VI.5.5)

extended by 0. Then∂tv − Δv = utη + ηtu − ηΔu − uΔη − 2 �∇u,∇η�

= ηtu − uΔη − 2 �∇u,∇η�=: f(x, t)

(VI.5.6)

with bounded v and f ∈ C21 by smoothness of u. Let (x, t) ∈ C3. Then by Theorem VI.1.4

v(x, t) =� t

0

�

RnΦ(x − y, t − s)f(y, s) dyds

=� t

0

�

RnΦ(x − y, t − s)

�u(y, s)ηt(y, s) − u(y, s)Δη(y, s)

− 2 �∇u(y, s),∇η(y, s)��

dyds

(VI.5.7)

We note: The singularity y = x and s = t is cut off due to (x, t) ∈ C3. Hence (η ≡ 1around C1)

v(x, t) =�

Cc1

Φ(x − y, t − s)�(∂t − Δ)η(y, s)u(y, s)

�dyds

+�

Cc1

2DΦ(x − y, t − s)Dη(y, s)u(y, s).(VI.5.8)

By convolution: If u ∈ C21(UT ), we have a representation

v(x, t) =�

CK(x, y, s, t)u(y, s) dyds(VI.5.9)

VI.5. REGULARITY AND CAUCHY-ESTIMATES 117

with no singularities in the kernel. Thus v is smooth and so is u around (x0, t0). �Theorem VI.5.2 (Cauchy estimates). For all k, l ∈ N there exists C > 0 such that for allu ∈ C2,1(UT ) (u ∈ L1

loc will be sufficient), solving(∂t − Δ) u = 0,(VI.5.10)

there holds

maxC(x0,t0; r

2 )|Dk

x∂ltu| ≤ C

rk+2l+n+2�u�L1(C(x0,t0;r))(VI.5.11)

for all C(x0, t0; r) ⊂ UT .

Proof. Suppose first (x0, t0) = (0, 0) and r = 1. SetC(1) = C(0, 0; 1).(VI.5.12)

Then as in the proof of Theorem VI.5.1 we have

u(x, t) =�

C(1)K(x, t, y, s)u(y, s) dyds ∀(x, t) ∈ C

�12

�.(VI.5.13)

ThenDk

x∂ltu(x, t) =

�

C(1)

�Dk

x∂ltK(x, t, y, s)

�u(y, s) dyds(VI.5.14)

and hence

|Dkx∂l

tu(x, t)| ≤ Ck,l�u�L1(C(1)) ∀(x, t) ∈ C�1

2

�.(VI.5.15)

Thus the claim is proven for r = 1. For r > 0 and (x0, t0) ∈ Rn+1 setv(x, t) = u(x0 + rx, t0 + r2t).(VI.5.16)

ThenmaxC( 1

2)|Dk

x∂ltv| ≤ Ck,l�v�L1(C(1)).(VI.5.17)

Hencemax

C(x0,r0; r2 )

|Dkx∂l

tu|rk+2l ≤ Ck,lr−(n+2)�u�L1(C(1)).(VI.5.18)

�

CHAPTER 7

linear parabolic equations

VII.1. Definitions

The heat equation is the simplest or most pure parabolic equation. In general we want tostudy equations of the form

∂tu − Lu,

where L is a uniformly elliptic differential operator (for each time t). More precisely, westudy L which for given coefficient functions aij(x, t), bi(x, t) and c(x, t) has the form

Lu(x, t) = aij(x, t) ∂iju(x, t) + bi(x, t) ∂iu(x, t) + c(x, t) u(x, t).

Recall that we use Einstein’s summation convention,

=n�

i,j=1aij(x, t) ∂iju(x, t) +

n�

i=1bi(x, t) ∂iu(x, t) + c(x, t) u(x, t).

We want L to be elliptic (and equivalently ∂t − L to be parabolic), which simply meansthat the leading order coefficients form a non-degenerate, positive matrix.

Definition VII.1.1 (Parabolic). We say that an operator ∂t − L is uniformly parabolic,if there exists a constant λ > 0 so that

aij(x, t) ξi ξj ≥ λ|ξ|2 ∀(x, t) ∈ ΩT , ξ ∈ Rn.

Equivalently, the matrix A(x, t) = (aij(x, t))1≤i,j≤n satisfies

�A(x, t)ξ, ξ�Rn ≥ λ ∀(x, t) ∈ ΩT , ξ ∈ Rn, |ξ| = 1.

We also say that L is uniformly elliptic.

The simplest example of a parabolic operator is the heat operator. Indeed take

aij := δij =

1 i = j

0 i �= j

and b ≡ c ≡ 0. Then L = +Δ. Indeed, parabolic operators have many features similar to∂t − Δ.

118

VII.2. MAXIMUM PRINCIPLES 119

Definition VII.1.2. Let X ⊂ Rn+1 be an n + 1-dimensional domain. The parabolicboundary PX of X is defined as follows. For ρ > 0, (x0, t0) ∈ Rn+1 define the (backwards-in-time) cylinder Qρ(x0, t0) as

Qρ(x0, t0) =�(x, t) ∈ Rn+1 : |x − x0| < ρ, t ∈ (t0 − ρ2, t0),

�.

Then the parabolic boundary PX of X is defined asPX := {(x0, t0) ∈ ∂X so that Qρ(x0, t0) ∩ Xc �= ∅ ∀ρ > 0}

Exercise VII.1.3. Let Ω ⊂ Rn be a domain and ΩT = Ω × (0, T ]. Show that PΩT = ΓT .

VII.2. Maximum principles

VII.2.1. Weak maximum principle. We will always assume that the operators∂t + L are uniformly parabolic and the coefficients aij, bi, c are continuous. Moreover weassume symmetry,

aij = aji 1 ≤ i, j ≤ n.

Also X ⊂ Rn+1 bounded.

Theorem VII.2.1 (Weak maximum principle, c ≡ 0). Let X ⊂ Rn+1 be open and boundedand let L be an elliptic operator with(VII.2.1) c = 0.

Let u ∈ C21(X) ∩ C0(X).

(1) If u is a subsolution of ∂t − L, i.e.(∂t − L)u ≤ 0,(VII.2.2)

thensup

X

u = sup∂P X

u.(VII.2.3)

(2) If u is a supersolution of ∂t − L, i.e.(∂t − L)u ≥ 0,(VII.2.4)

theninfX

u = inf∂P X

u.(VII.2.5)

Proof. We only prove the first claim, the second one follows by replacing u with −u.Also we will assume that X = ΩT

For now assume that we have a strict subsolution. That is,(VII.2.6) (∂t − L)u < 0 in ΩT .


Assume that there exists a point (x0, t0) ∈ ΩT with u(x0, t0) = maxΩTu. Then x0 ∈ Ω and

t0 ∈ (0, T ], so the maximality condition tells us∂tu(x0, t0) ≥ 0, Du(x0, t0) = 0, D2u(x0, t0) ≤ 0.

In particular, observing (VII.2.1),∂tu(x0, t0) − Lu(x0, t0) ≥ aij(x0, t0) ∂iju(x0, t0).

In view of Exercise VII.2.2 this implies∂tu(x0, t0) − Lu(x0, t0) ≥ 0,

a contradiction to (VII.2.6). So what do we do if we had only (VII.2.2)? We consider asubsolution slightly below u. Let uε(x, t) := u(x, t) − εt. Then, again with (VII.2.1),

∂tuε − Luε = ∂tu − Lu − ε < 0 in ΩT .

The above argument implies thatmax

ΩT

uε = maxΓT

uε ∀ε > 0.

In particular we havemax

ΩT

u ≤ εT + maxΩT

uε ≤ εT + maxΓT

uε ≤ εT + maxΓT

u.

Letting ε → 0 we havemax

ΩT

u ≤ maxΓT

u.

The inverse estimate is always true, so the claim is proven. �

Exercise VII.2.2. A matrix A ∈ Rn×n is nonnegative, A ≥ 0, if�Av, v� ≥ 0 ∀v ∈ Rn.

A matrix A is symmetric, if AT = A.

Show that

(1) A ≥ 0 implies P T AP ≥ 0 for any matrix P ∈ Rn×n.(2) A ≥ 0 implies that the diagonal entries Aii ≥ 0 for any i ∈ {1, . . . , n}.(3) A ≥ 0 and B ≥ 0 and B is symmetric then

A : B :=n�

i,j=1AijBij ≥ 0.

If c ≥ 0, then we have to adapt the claim. For a function f let f+ := max{f, 0} andf− := max{−f, 0}.

Exercise VII.2.3. Complete the above proof for general domain X.


Theorem VII.2.4 (Weak maximum principle, c ≤ 0). Let u and X as in Theorem VII.2.1and ∂t − L parabolic with c ≤ 0. Then if ut − Lu ≤ 0 then

supX

u ≤ sup∂P X

u+.(VII.2.7)

For ut − Lu ≥ 0, theninfX

u ≥ − sup∂P X

u−,(VII.2.8)

where u+ = max(0, u) and u− = −min(u, 0). If ut = Lu, thensup

X

|u| = sup∂P X

|u|(VII.2.9)

Proof. We just prove the first claim, the second and third are simple corollaries.

Again, we assume ΩT , general X is an exercise. we first simplify the equation, and assumethat

(∂t − L)u < 0 in ΩT .

The only situation we have to exclude is that there exists (x0, t0) ∈ ΩT at which there is apositive maximum value u(x0, t0) > 0. With the arguments as in Theorem VII.2.1,

∂tu(x0, t0) − Lu(x0, t0) ≥ −c(x0, t0) u(x0, t0) ≥ 0,

and we have our contradiction. The full claim is obtained if we consider again uε(x, t) :=u(x, t) − εt. Then

maxΩT

uε ≤ maxΓT

(uε)+ ≤ maxΓT

(u)+.

We let ε → 0 to conclude. �

A consequence of the weak maximum principle is uniqueness of solutions and the compar-ison principle.

Corollary VII.2.5 (Uniqueness). Let X ⊂ Rn+1 and L as above with c ≤ 0. Let u, v ∈C2

1(X) ∩ C0(X) satisfyut − Lu = vt − Lv.(VII.2.10)

Then if u = v on ∂P X, we have u = v in X.Corollary VII.2.6 (Comparison Principle). Let X and L as above and u, v ∈ C2

1(X) ∩C0(X) with

ut − Lu ≤ vt − Lv(VII.2.11)

in X with u ≤ v on ∂P X, then we have u ≤ v in X.

We leave the proofs as exercises, Exercise VII.2.7.

Exercise VII.2.7. Prove Corollaries VII.2.5 and VII.2.6. Hint: What equation does u−vsatisfy?


VII.2.2. Strong Maximum principle. Let∂tu − Lu = 0 in ΩT

We want to understand better the relation between u at different times. We have thefollowing very important “propagation of positivity” property. See [Lieberman, 1996, II,Lemma 2.6]

Lemma VII.2.8. [Propagation of positivity] For R > 0 and α > 0 let BR(0) ⊂ Rn.Let Q(R) = BR × (0, αR2). Let 0 ≤ u ∈ C2

1(Q(R)) satisfyut − Lu ≥ 0,(VII.2.12)

where L is elliptic with b = c = 0. Ifu(x, 0) ≥ h ∀|x| < εR(VII.2.13)

for some h > 0 and 0 < ε < 1, then

u(x, αR2) ≥ c(ε, λ, R, �aij�∞)h ∀|x| ≤ R

2(VII.2.14)

for some positive c.

Proof. Let Q ⊂ Rn+1 be a cone so that at time t = 0, Q ∩ (Rn × {t = 0}) is theball {|x| < εR} and at time t = αR2, Q ∩ (Rn × {t = αR2}) is the ball {|x| < R}. SeeFigure VII.2.1. In formulas, Q can be written

Q =�(x, t) ∈ Rn+1 : |x|2 < ψ(t), 0 < t < αR2

�

forψ(t) := (1 − ε2)

αt + ε2R2.

On Q we will construct a comparison (“barrier”) function v with the following properties:

Figure VII.2.1. Q and its parabolic boundary PQ (green)

(VII.2.15)

vt − Lv ≤ 0 in Q

v ≤ u on PQ

and moreover

(VII.2.16) v(x, αR2) ≥ c h whenever |x| ≤ R

2


If we have such a v, then by Corollary VII.2.6 (the general domain version)

u(x, αR2) ≥ v(x, αR2) ≥ ch whenever |x| ≤ R

2

So how do we construct such a v? We essentially rescale (in time) the map (1 − |x|2)2.Choose the Ansatz

v(x, t) := µ(t) (ν(t) − |x|2)2.

For µ, ν nonnegative functions. In general, away from t = 0, we only know that u ≥ 0, soto make v as large as possible, it seems reasonable to set v(x, t) ≡ 0 on the positive partof the parabolic boundary PQ ∩ {t > 0}. That is,

ν(t) := ψ(t).Now we compute the equation. Firstly

∂xixj v(x, t) = 8µ(t) xj xi − 4µ(t) (ψ(t) − |x|2)δij

Consequently, by ellipticity−aij(x, t) ∂xixj v(x, t) ≤ µ(t)

�−8 ψ(t) λ + 8 (ψ(t) − |x|2) λ + 4(ψ(t) − |x|2) tr(A)

�.

Also,vt(x, t) = µ�(t) (ψ(t) − |x|2)2 + 2µ(t) (ψ(t) − |x|2)ψ�(t).

This vt has to be the positive guy, so we would like to be able to compare µ�(t) and ν �(t).We thus choose (note that ψ(t) > 0) for some constant η > 0,

µ(t) := ηψ(t)−q.

Then

−aij(x, t) ∂xixj v(x, t) ≤ ηψ1−q(t)�−8 λ + 8

�(ψ(t) − |x|2)

ψ(t)

�λ + 4

�(ψ(t) − |x|2)

ψ(t)

�tr(A)

�.

and (observe that ψ�(t) = 1−ε2

αR,

vt(x, t) =η�−qψ−q−1(t) (ψ(t) − |x|2)2 + 2ψ(t)−q (ψ(t) − |x|2)

� 1 − ε2

αR

=ηψ(t)1−q

−q

�(ψ(t) − |x|2)

ψ(t)

�2

+ 2ψ(t)�

(ψ(t) − |x|2)ψ(t)

� 1 − ε2

αR.

We see a quadratic structure in

ξ(t) :=�

(ψ(t) − |x|2)ψ(t)

�,

namelyvt(x, t) − aij(x, t)∂xixj v(x, t)

≤ηψ1−q(t)�−�

q1 − ε2

αR

�ξ(t)2 +

�21 − ε2

αR ψ(t)2 + 8λ + 4 tr(A)

�ξ(t) − 8 λ

�.


Observe that the leading order term and the zero-order term are negative, hence (seeExercise VII.2.9) there exists a large q > 0 so that

vt(x, t) − aij(x, t) ∂xixj v(x, t) ≤ 0 in Q.

On the other hand, for t = 0, in view of (VII.2.13),

v(x, 0) = ηε−2qR−2q (ε2R2 − |x|2)2 ≤ η (εR)4−2q ≤ 1h

η (εR)4−2q u(x, 0).

So we chooseη := h (εR)2q−4.

Then v satisfies (VII.2.15). It remains to check (VII.2.16). For |x| ≤ R2 ,

v(x, αR) = h (εR)2q−4 R−2q (R2 − |x|2)2 ≥ hε2q−4 916 .

This finishes the proof of Lemma VII.2.8. It is worth noting that we actually get anestimate of the form εκ, where κ is a uniform constant depending on R, λ, etc. For thisassume w.l.o.g. that ε < 1

2 , for any ε > 12 the claim follows from the ε < 1

2 case since thepositivity set is larger than required. �

Exercise VII.2.9. Assume that a, b, c ∈ R be fixed. To any λ ∈ R we associate thepolynomial

pλ(x) := λax2 + bx + c x ∈ R.

Show that if a < 0 and c < 0 then there exists a λ > 0 so that

supx∈R

pλ(x) < 0.

Hint: p-q formula

Theorem VII.2.10 (Strong Maximum Principle). Let b, c = 0, L elliptic, X ⊂ Rn+1 openand bounded, u ∈ C2

1(X) ∩ C0(X) and assume in X :

(∂t − L)u ≤ 0.(VII.2.17)

Assume there is (x0, t0) ∈ X, such thatu(x0, t0) = sup

Xu,(VII.2.18)

then

u(x, t) = u(x0, t0) ∀(x, t) ∈ S(x0, t0),(VII.2.19)

whereS(x0, t0) = {(x, t) : ∃ g ∈C0 ([0, 1], X\∂pX) , g(0) = (x0, t0),

g(1) = (x, t), g decreasing in t}.(VII.2.20)


Proof. SetM := max

Xu.(VII.2.21)

Claim: Assume a maximal point (y0, t0) ∈ X, r > 0, such thatQ(y0, t0, 3r) ⊂ X(VII.2.22)

and such that there is (y1, t1) ∈ Q(y0, t0, r) withu(y1, t1) < M.(VII.2.23)

Set v = M − u and

R = 2|y1 − y0| < 2r, α := t0 − t1

R2 .(VII.2.24)

By continuity there exists ε > 0 and h > 0 such thatv(x, t1) > h, |y| < εR.(VII.2.25)

By Lemma VII.2.8 there exists c > 0, such that v(y, t0) > ch > 0 for all |y − y1| < R/2,a contradiction. Hence if u(x0, t0) = M, then u(y, t) = M for all (y, t) ∈ Q(x0, t0; r),whenenver Q(x0, t0; 3r) ⊂ X. Hence {u = M}∩S(x0, t0) is (parabolically) open and closedand hence all of S(x0, t0). �

Without proof, now we state:Theorem VII.2.11 (Parabolic Harnack inequality). Assume u ∈ C2

1(UT ) and solves(∂t − L)u = 0 in UT

andu ≥ 0 in UT

Assume moreover that b ≡ 0 and c ≡ 0 and a is smooth.

If V � U is connected, then for each time 0 < t1 < t2 ≤ T there is a constant C such thatsupx∈V

u(x, t1) ≤ C infx∈V

u(x, t2).

Proof. See [Evans, 2010, Theorem 10, p.391]. �

CHAPTER 8

Principles of Semi-group theory

As references we refer to [Evans, 2010, §7.4] and [Cazenave and Haraux, 1998].

In Section VI.1 we looked at (∂t − Δ) u = 0 and naıvely we should have

u = etΔu(0).(VIII.0.1)

We made this precise with the help of the Fourier Transform.

Is there a similar relation if we look at an elliptic operator L instead of Δ?

Generally: Let X be a real Banach space and a linear map A,

A : D(A) ⊂ X → X,(VIII.0.2)

where D(A) is the domain of A, a linear (usually dense) subset of X. We are looking forsolutions u ∈ C1((0, T ), X) of

u = Au, t ∈ (0, T ),u(0) = ϕ.

(VIII.0.3)

A is in general not bounded, but closed. Assume there exists a solution to (VIII.0.3), then

T (t)ϕ := u(t)(VIII.0.4)

defines an operator. Resonable properties of T : are

• T (t) : X → X is linear.

• T : [0, T ) → L(X). (hopefully)

• T (0) = id ,

• T (t + s) = T (t) ◦ T (s), (from uniqueness hopefully)

• t �→ T (t)ϕ is continuous.

The latter three properties are characteristic for a semigroup.

Assume now that we have a semigroup

T : [0,∞) × X → X.(VIII.0.5)126

VIII.1. M-DISSIPATIVE OPERATORS 127

Then we find some A such that T is the semigroup of A. A will then be called the generatorof T.

u(t) = lims→0

u(t + s) − u(t)s

= lims→0

T (t + s)ϕ − T (t)ϕs

= lims→0

T (s) − T (0)s

u(t)

≡ Au(t).

(VIII.0.6)

Hence let

Au = lims→0

T (s) − T (0)s

u,(VIII.0.7)

whenever the limit exists. Call D(A) the set of u ∈ X where this limit exists.

One might conjecture there is some sort of equivalence between generators A and semi-groups T .

Questions: Which generators A allow semigroups? Which generators are implies by semi-groups?

The main theorem which gives us an answer to this question is the Hille-Yoshida Theorem,Theorem VIII.2.6 and VIII.2.8. For Schrodinger equations this is (a generalization of)Stones’ theorem, it also appears under the name Lumer–Phillips theorem.

VIII.1. m-dissipative operators

We want to solveu�(t) = Au, t > 0u(0) = ϕ

(VIII.1.1)

with some operator

D(A) ⊂ X → X,(VIII.1.2)

where X is a Banach space and D(A) a linear subspace, e.g. X = L2 and D(A) = H2 andA = Δ. The norm of our space X is the L2-norm, and then A is not a bounded operator.On the other hand, since C∞

c is dense in L2, D(A) is dense in L2 (everything with respectto the L2-norm).

VIII.1.1. linear bounded operators. (i) Let X = Rn or Cn, A : X → X linear (andthus bounded), then

u(t) = etAϕ(VIII.1.3)


is the unique solution to (VIII.1.1), where

etA =∞�

k=0

1k!t

kAk.(VIII.1.4)

(ii) Let X be a general Banach space and A ∈ L(X), where L(X) is the space of boundedlinear operators. Here etA also makes sense.

Lemma VIII.1.1. Let A, B ∈ L(X). Then

(i) eA converges absolutely,

(ii) e0 = id ,

(iii) AB = BA ⇒ eA+B = eAeB,

(iv) e−A =�eA�−1

.

Theorem VIII.1.2. Let A ∈ L(X), ϕ ∈ X, T > 0. Then there exists a unique solutionu ∈ C1((0, T ), X) of

u�(t) = Au(t)u(0) = ϕ.

(VIII.1.5)

Proof. Put

u(t) = etAϕ.(VIII.1.6)

Then

u�(t) = etAAϕ = Au(t).(VIII.1.7)

For a second solution v set

w(t) = e−tAv(t),(VIII.1.8)

then w�(t) = 0 and hence w(t) = w(0) = ϕ. �

VIII.1.2. unbounded operators. Let X be a real or complex Banach space. Anoperator

A : D(A) ⊂ X → X(VIII.1.9)

is called linear, if and only if D(A) is a linear subspace and A ist linear on D(A). We sayA is densely defined, if

D(A) = X.(VIII.1.10)


A is bounded, if and only if1

�A� := sup�x�≤1

�Ax� < ∞.(VIII.1.11)

Otherwise it is called unbounded.

examples

(1) X = L2(Rn), A = Δ, D(A) = H2(Rn) or D(A) = C∞.(2) X = C0([0, 1]), D(A) = X, K ∈ C0([0, 1] × [0, 1])

Au(x) =� 1

0K(x, y)u(y) dy(VIII.1.12)

is bounded.

We use the following notation.G(A) = {(u, Au) ⊂ X × X : u ∈ D(A)}(VIII.1.13)

is the graph of A,R(A) = {Au : u ∈ D(A)}(VIII.1.14)

the range of A. An extension of A is

A : D(A) ⊂ X → X,(VIII.1.15)

such that

D(A) ⊂ D(A) and Au = Au ∀u ∈ D(A).(VIII.1.16)

A is called closed, if G(A) is closed in X × X. That is, whenever (uk)k∈N ⊂ D(A) andu, g ∈ X with uk

k→∞−−−→ u in X, Aukk→∞−−−→ g ∈ X we have u ∈ D(A) and g = Au.

A is called closable, if there exists a closed extension A.

Theorem VIII.1.3 (Closed Graph Theorem). Let A : X → X be linear. Then A is con-tinuous (i.e. bounded) if and only if A is closed.

Observe that this assumes that A is defined on all of X (i.e. D(A) = X).

VIII.1.3. Notion of m-dissipative operators. (X, � · �) Banach space, D(A) ⊂ Xdense subspace. Let A : D(A) → X linear (not necessarily bounded, closed).

Definition VIII.1.4. A is dissipative, if�u − λAu� ≥ �u� ∀u ∈ D(A), λ > 0.(VIII.1.17)

A is called accretive, if −A is dissipative.1this notion is equivalent with A being continuous, see any functional analysis book


Lemma VIII.1.5. Let X be a Hilbert2 space,

A : D(A) ⊂ X → X(VIII.1.18)

linear, then A is dissipative if and only if

Re �u, Au� ≤ 0 ∀u ∈ D(A).(VIII.1.19)

Proof of Lemma VIII.1.5. Assume A dissipative, then:

�u�2 + λ2�Au�2 − 2λRe �u, Au� − �u�2 = �u − λAu�2 − �u�2 ≥ 0.(VIII.1.20)

Dividing by λ and letting λ → 0 gives

Re �u, Au� ≤ 0.(VIII.1.21)

For the converse, assume that

Re �Au, u� ≤ 0,(VIII.1.22)

then

�u − λAu�2 = �u�2 + λ2�Au�2 − 2λRe �u, Au� ≥ �u�2.(VIII.1.23)

�

Example VIII.1.6. • Heat equation (∂t −Δ)u = 0: A = Δ, X = L2(Rn), D(A) =H2(Rn), then

�u, Δu� = −�

Rn|∇u|2 ≤ 0.

• Heat equation (∂t − Δ)u = 0: A = Δ, X = L2(Ω), D(A) = H2 ∩ H10 (Ω), then

�u, Δu� = −�

Rn|∇u|2 ≤ 0.

• Schrodinger equation (∂t − iΔ)u = 0. A = iΔ, D(A) = H2(Rn),

�u, ±iΔu� = ∓i�

Rn|∇u|2

and hence the real part is 0 and both iΔ and −iΔ are dissipative.

Definition VIII.1.7 (m-dissipative). A linear operator A : D(A) ⊂ X → X is calledm-dissipative, if A is dissipative and I − λA is surjective for all λ > 0. (hence I − λA iscontinuously invertible as a map from D(A) to X.)

2i.e. there exists a scalar product �, � : X × X → C such that �v�2 = �v, v�. For L2(Ω) the scalarproduct �f, g� :=

�Ω fg (real) or �f, g� :=

�Ω fg (complex)


Observe that for the notion of m-dissipative the right choice of D(A) becomes relevant.I.e. if we choose D(Δ) = C∞

c (Ω) then Δ is dissipative, but not m-dissipative; but if wechoose D(Δ) = H2(Ω) then Δ is

Our aim is to show that for any m-dissipative A we can define (some sort of) eA. We alsocall A m-accretive, if −A is m-dissipative. Set

Jλ = (I − λA)−1 : X → D(A).(VIII.1.24)

Then (VIII.1.17) implies for m-dissipative A

�Jλv� ≤ �v� ∀v ∈ X.(VIII.1.25)

Lemma VIII.1.8. Let A be dissipative, then A is m-dissipative if and only if there existsλ0 > 0 such that I − λ0A is surjective.

Proof. Let λ ∈ (0,∞) and v ∈ X. Our goal is to find u ∈ D(A) such that u−λAu = v.Observe that it is equivalent to find u such that

u − λ0Au = λ0

λv +

�1 − λ0

λ

�u,(VIII.1.26)

or equivalently (recall Jλ0 = (I − λ0A)−1 : X → D(A))

u = Jλ0

�λ0

λv +

�1 − λ0

λ

�u

�=: F (u).(VIII.1.27)

That is, u is a fixed point of u = F (u). We show that F : D(A) → D(A) is a contraction,to apply Banach Fixed Point theorem.

Observe, by (VIII.1.25),

�F (u) − F (w)� =��Jλ0

��1 − λ0

λ

�(u − w)

�� ≤��1 − λ0

λ

�� u − w�.(VIII.1.28)

Hence F is a contraction, whenever λ > λ0/2.

Then we apply Banach Fixed Point theorem , on D(A) (which is not closed in general)! Theargument of Banach Fixed point theorem shows there uk := F k(0) converges in X to someu ∈ X. To see that u ∈ D(A), observe that uk → u in X implies Jλ0(uk) → Jλ0u ∈ D(A)in X by continuity. By affine linearity of F we conclude that F (uk) k→∞−−−→ F (u) ∈ D(A).Since uk+1 = F (uk) we see that u = F (u) ∈ D(A).

That is, there is a unique u ∈ D(A) with F (u) = u, and I − λA is surjective, wheneverλ > λ0

2 .

Iterating this argument, e.g setting λi+1 := 23λi, we find that for any λ > λi, i ∈ N, I −λA

is surjective, and letting i → ∞ we see that for any λ > 0 we have I − λA is surjective.�


Proposition VIII.1.9. All m-dissipative operators (A, D(A)) are closed.

Proof. Let uk → u in X and Auk → g in X. We need to show that u ∈ D(A) andAu = g.

Observe that (I − A)uk → u + g. Since A is m-dissipative, J1 = (I − A)−1 exists and is acontinuous map X → D(A). Thus

uk→∞←−−− uk = J1((I − A)uk) k→∞−−−→ J1(u − g) = J1u − J1g ∈ D(A)

That is, u = J1u − J1g ∈ D(A), and applying I − A we obtain (I − A)u = u − g, i.e.Au = g. �Example VIII.1.10. X = L2(Rn), A = Δ, D(A) = H2(Rn). Then A is m-dissipative.

We already know that A is dissipative (Example VIII.1.6), so by Lemma VIII.1.8 we onlyneed to show that

∀v ∈ L2 ∃u ∈ H2 : u − Δu = v.(VIII.1.29)

Here we see that the choice of D(A) is important (the above will not work for D(A) = C∞.)We solve this by Fourier-transform (on domains: exercise!)

u(ξ) + |ξ|2u(ξ) = v(ξ)(VIII.1.30)

and hence we conjecture

u(ξ) := 11 + |ξ|2 v(ξ).(VIII.1.31)

Hence u ∈ L2 andξ1ξ2

1 + |ξ|2 v(ξ) ∈ L2(VIII.1.32)

implies that u,∇2u ∈ L2.

Proposition VIII.1.11. Let A be m-dissipative, then

∀u ∈ D(A) : �Jλu − u� λ→0−−→ 0.(VIII.1.33)

Proof. Observe that by (VIII.1.25), Jλ − I : X → X is bounded linear operator, i.e.�Jλ − I� ≤ �Jλ� + �I� ≤ 2.(VIII.1.34)

By density D(A) ⊂ D(A), it thus suffices to prove the result for u ∈ D(A). Since Jλ =(I − λA)−1 using again (VIII.1.25)

�Jλu − u� = �Jλ (u − (I − λA)u) � ≤ λ�Au� → 0, λ → 0.(VIII.1.35)

�

VIII.2. SEMIGROUP THEORY 133

Set

Aλ := AJλ = 1λ

(Jλ − I).(VIII.1.36)

This Aλ ∈ L(X) will serve as an “approximation” for A, so that we can make (certain)sense of an operator etA in terms of limλ→0 etAλ . This is justified by the following

Proposition VIII.1.12. Let A be m-dissipative and D(A) = X. Then

Aλuλ→0−−→ Au, ∀u ∈ D(A).(VIII.1.37)

Proof. We have(I − λA)A = A(I − λA).(VIII.1.38)

Thus, multiplying both sides with Jλ from the left and also from the right, we have Aλ =AJλ = JλA.

Now observe that by Proposition VIII.1.11,

JλAuλ→0−−→ Au,(VIII.1.39)

since D(A) is dense in X �

VIII.2. Semigroup Theory

Let X be a Banach space. A semigroup is an operatorT : [0,∞) → L(X),(VIII.2.1)

such that

(i) T (0) = I,(ii) T (t + s) = T (t)T (s).

T is called C0-semigroup (strongly continuous semigroup), if

(iii) limt→0 �T (t)u − u� = 0 ∀u ∈ X.

Note, that by (ii) we necessarily have T (s)T (t) = T (t)T (s).

Example VIII.2.1. (1) A ∈ L(X), T (t) = etA.(2) X = Lp(R), p ∈ [1,∞].

T (t)u(x) = u(t + x).(VIII.2.2)

If p < ∞, then T is a continuous semigroup, since C∞c is dense and hence for

u ∈ Lp and ε > 0 there exists f ∈ C∞c with

�f − u�p < ε/3.(VIII.2.3)


We have for all small t,

supx

|f(x − t) − f(x)| < t�∇f�∞ < ε/3,(VIII.2.4)and thus for t � 1

��

R|T (t)f − f |p

� 1p

<ε

3 (diam(supp f) + 1) .(VIII.2.5)

Moreover, by the definition of T (t)

�T (t)(u − f)�p,Rn = �u − f�p,Rn ≤ ε

3 .

Thus,�T (t)u − u�p ≤ �T (t)f − f�p + �T (t)(u − f)�p + �u − f�p

≤ ε

The situation is different for p = ∞: let u = χ[0,1], then�u − T (t)u�∞ = sup

x|u(x) − u(x + t)| ≥ 1 ∀t > 0.(VIII.2.6)

Thus T is no C0-semigroup for p = ∞.

Proposition VIII.2.2. Let T (t) be a C0-semigroup. Then ∃M ≥ 1 and ω ∈ R such that�T (t)� ≤ Meωt.(VIII.2.7)

Proof. We show that there exists δ > 0 such thatM := sup

0<t<δ�T (t)� < ∞.(VIII.2.8)

If this was not the case, then there exists a sequence tn → 0 with(VIII.2.9) �T (tn)� → ∞.

Recall Banach-Steinhaus Theorem: If for a sequence An ∈ L(X) we have∀u ∈ X : sup

n�Anu� < ∞,(VIII.2.10)

then supn �An� < ∞.

Hence, if (VIII.2.9) holds, then there must be u ∈ X such that �T (tn)u� → ∞. But thisis in contradiction to the C0-property. Hence (VIII.2.9) cannot hold, i.e. (VIII.2.8) mustbe true.

Now let t > 0, then there exists n ∈ N and s ∈ (0, δ), such thatt = nδ + s.(VIII.2.11)

Then by the semigroup propertyT (t) = T (δ) ◦ · · · ◦ T (δ)

� �� n times

◦T (s).(VIII.2.12)


That is, with (VIII.2.8),

�T (t)� ≤ �T (δ)�n�T (s)� ≤ Mn+1 ≤ MMtδ = Met log M

δ .(VIII.2.13)

�Proposition VIII.2.3. Let T (t) be a C0-semigroup. Then the map

(t, u) �→ T (t)u(VIII.2.14)

is continuous.

Proof. Exercise. �Definition VIII.2.4. Let T (t) be a C0-semigroup. Then

ω0 = inf{w ∈ R : ∃M ≥ 1, �T (t)� ≤ Meωt}(VIII.2.15)

ist called the growth bound of the semigroup.

Definition VIII.2.5. A C0-semigroup is called contraction semigroup, if∀t > 0: �T (t)� ≤ 1.(VIII.2.16)

Recall that

�Jλ� ≤ 1, �Aλ� ≤ 2λ

,(VIII.2.17)

where

Aλ := AJλ = 1λ

(Jλ − I).(VIII.2.18)

We define

Tλ(t) = etAλ .(VIII.2.19)

For any λ > 0, Tλ is a C0-semigroup (because Aλ ∈ L(X)). Moreover, for any λ > 0 wehave that Tλ it is a contraction semigroup:

(VIII.2.20) �Tλ(t)�L(X) = �etJλ1λ e− t

λI� = e− t

λ�etλ

Jλ� ≤ e− tλ e

tλ = 1.

Observe that in contrast, et�Aλ� is in generally not uniformly bounded w.r.t λ → 0.

The semigroup for an m-dissipative operator A will be constructed out of Tλλ→0−−→ T (t).

The following is the (first part) of the main theorem of the semigroup theory:

Theorem VIII.2.6 (Hille Yoshida (Part I)). Let A : D(A) ⊂ X → X m-dissipative anddensely defined. Then for all u ∈ X the limit

T (t)u = limλ→0

Tλ(t)u(VIII.2.21)

exists and the convergence is uniform (w.r.t. t) on time-intervals of the form [0, T ].


Furthermore (T (t))t≥0 is a contraction semigroup, T (t)(D(A)) ⊂ D(A), and for all u ∈D(A),

u(t) := T (t)u(VIII.2.22)

is the unique solution u ∈ C0([0,∞), D(A)) ∩ C1((0,∞), X) to

(VIII.2.23)

u(t) = Au(t) t > 0u(0) = u

Proof. Step (1): On the contraction semigroup property

First we show that for µ, λ, t > 0(VIII.2.24) �Tλ(t)u − Tµ(t)u�X ≤ t �Aµu − Aλu�X .

Indeed, observe that (Aλ and Aµ commute!)

Tλ(t) − Tµ(t) =etAλ − esAλ =� t

0

d

ds

�e(t−s)AµesAλ

�ds

=� t

0e(t−s)AµesAλ ds (Aλ − Aµ)

Consequently, (recall (VIII.2.20))

�Tλ(t)u − Tµ(t)u�X ≤� t

0�e(t−s)Aµ�L(X)� ��

≤1

�esAλ�L(X)� �� ≤1

ds �(Aλ − Aµ)u�X

≤t�(Aλ − Aµ)u�X .

Thus (VIII.2.24) is established.

From (VIII.2.24) we conclude for any fixed u ∈ D(A), using also Proposition VIII.1.12,that (Tλ(t)u)λ>0 is a Cauchy-sequence as λ → 0 in X w.r.t λ, and this Cauchy sequence isuniform in t ∈ [0, T ] for any fixed T < ∞.

Hence the proposed limit T (t)u := limλ→0 Tλ(t)u exists for any u ∈ D(A), and Precisely,�Tλ(t)u − T (t)u�X ≤ t�(Aλ − A)u�X ∀u ∈ D(A).

Clearly, T (t) a linear operator, and from the above estimate together with (VIII.2.20) wefind

�T (t)u�X ≤�Tλ(t)u − T (t)u�X + �Tλ(t)u�X

≤�Tλ(t)u − T (t)u�X + �u�X

≤t�(Aλ − A)u�X + �u�X .

Letting λ → 0, we find�T (t)u�X ≤ �u�X ∀u ∈ D(A).

That is T (t) : D(A) → X is a linear bounded operator; since D(A) is a dense set in X,T (t) can be extended to a linear bounded operator on all of X.


Now let u ∈ X with approximating sequence un ∈ D(A).�Tλ(t)u − T (t)u� ≤ �Tλ(t)u − Tλ(t)un� + �Tλ(t)un − T (t)un�

+ �T (t)(un − u)�≤ 2�un − u� + �Tλ(t)un − T (t)un�≤ 2�un − u� + t�(Aλ − A)un�X

(VIII.2.25)

Hence by (VIII.2.24) we find Tλ(t)u → T (t)u for all u ∈ X, and indeed the above estimateimplies even uniformity in t: namely, for any t0 > 0,

(VIII.2.26) supt∈[0,t0)

�Tλ(t)u − T (t)u� λ→0−−→ 0 ∀u ∈ X.

Furthermore we have the semigroup property�T (t)T (s)u − T (t + s)u� ≤ �T (t)T (s)u − T (t)Tλ(s)u�

+ �T (t)Tλ(s)u − Tλ(t)Tλ(s)u�+ �Tλ(t + s)u − T (t + s)u�

→ 0.

(VIII.2.27)

As for the C0-continuity, we have�T (t)u − u�X ≤ �Tλ(t)u − u�X + �Tλ(t)u − T (t)u�X .

By (VIII.2.26), for any ε > 0 and any t0 > 0 there exists λ > 0 such that

supt∈t0

�Tλ(t)u − T (t)u�X <12ε.

On the other hand for this fixed λ > 0 there exists t1 ∈ (0, t0) such that

supt<t1

�Tλ(t)u − u�X <ε

2 .

In particular,supt<t1

�T (t)u − u�X ≤ ε.

That is T (t) is a contractive C0-semigroup.

Next we show that T (t) maps D(A) to D(A). First we observe that Tλ maps D(A) toD(A). Indeed, since we can write eJλ = I + JλB (with B convergent since Jλ ∈ L(X))and since Jλ maps X into D(A) we have that eJλ maps D(A) into D(A). Moreover,by (VIII.2.18) we have Tλ(t) = etAλ = et 1

λ(Jλ−I) = e− t

λ eJλ , so Tλ(t) maps D(A) intoD(A). Now let u ∈ D(A), then Tλ(t)u λ→0−−→ T (t)u, and ATλ(t)u = Tλ(t)Au → T (t)Au.By Proposition VIII.1.9, A is a closed operator, which implies that T (t)u ∈ D(A) andlimλ→0 ATλ(t)u = AT (t)u, and in particular,(VIII.2.28) AT (t)u = T (t)Au ∀u ∈ D(A).


Step (2): On the equation (VIII.2.23)

Let u ∈ D(A) and set

uλ(t) = etAλu.(VIII.2.29)

Thend

dtuλ = etAλAλu = Tλ(t)Aλu.(VIII.2.30)

Equivalently, for u ∈ D(A) using Aλu → Au and Tλ → T ,

u(t) ← uλ(t) = u +� t

0Tλ(s)Aλu ds → u +

� t

0T (s) Au ds.(VIII.2.31)

Thus u(·) ∈ C1 and (cf. (VIII.2.28))

u(t) = T (t)Au = Au(t).(VIII.2.32)

Uniqueness proceeds as in Theorem VIII.1.2. �

VIII.2.1. Generators of semigroups. Let T (t) be a contraction semigroup. Define

D(A) :=�

u ∈ X : limh→0+

T (h)u − u

hexists

�.(VIII.2.33)

For u ∈ D(A) set

Au = limh→0+

T (h)u − u

h.(VIII.2.34)

Example VIII.2.7. X = Cub(R) be the set of uniformly continuous, bounded functionswith the A∞-norm.

T (t)u(x) := u(x + t).(VIII.2.35)

Then T (t) is a contraction semigroup. Then

Au = u�, D(A) = {u, u� ∈ Cub(R)}.(VIII.2.36)

Proof. It is clear that u, u� ∈ Cub(R) implies��

u(x + h) − u(x)h

− u�(x)��

∞→ 0.(VIII.2.37)

Now let u ∈ D(A), then u�+ ∈ Cub(R) and hence u�

+ = u� ∈ Cub(R). �

Theorem VIII.2.8 (Hille Yoshida Part II). Let T (t) be a contraction semigroup with gen-erator A. Then A is m-dissipative and densely defined.


Proof. (i) A is dissipative, i.e. for all λ > 0, �u − λAu� ≥ 0. Indeed,��u − λ

T (h)u − u

h

�� ≥��

�1 + λ

h

�u

��−��

λ

hT (h)u

��

=�

1 + λ

h

��u� − λ

h�T (h)u�

≥�

1 + λ

h�u� − λ

h�u�

�= �u�.

(VIII.2.38)

In the last step we used that T (h) is contracting, i.e. �T (h)u�X ≤ �u�X . Letting h → 0on the left hand side shows A is dissipative.

(ii) A is m-dissipative. It suffices to show that (I − A) is surjective. Thus we want to findJu, such that

(I − A)Ju = u.(VIII.2.39)

Ansatz:

Ju =� ∞

0e−tT (t) dt.(VIII.2.40)

Why? Because (formally!) we know(I − A)T (t)u = T (t)u − ∂t(T (t)u)

This is equivalent to(I − A)e−tT (t)u = −∂t(e−tT (t)u)

Integrating on both sides on t ∈ (0,∞) we then should get(I − A)Ju = T (0)u = u.

To make this more precise, first observe

�Ju� ≤� ∞

0e−t�T (t)u� dt ≤ �u�(VIII.2.41)

and hence �J� ≤ 1. Now let us compute (with the semigroup property T (h)T (t) = T (h+t)for h, t > 0),

(T (h) − I) Ju =� ∞

0e−tT (t + h)u dt −

� ∞

0e−tT (t)u dt

=� ∞

he−t+hT (t)u dt −

� ∞

0e−tT (t)u dt

=� ∞

0

�e−t+h − e−t

�T (t)u −

� h

0e−t+hT (t)u dt

= (eh − 1)� ∞

0e−tT (t)u dt − eh

� h

0e−tT (t)u dt

= (eh − 1)Ju − eh� h

0e−tT (t)u dt.

(VIII.2.42)

VIII.3. AN EXAMPLE APPLICATION OF HILLE-YOSHIDA 140

HenceT (h) − I

hJu = eh − 1

hJu − eh

h

� h

0e−tT (t)u dt.(VIII.2.43)

The right-hand side converges as h → 0 (we use that T is continuous at 0), consequentlyso does the left-hand side. Thus Ju ∈ D(A) and

AJu = Ju − u,(VIII.2.44)

which is the claim.

(iii) D(A) is dense. Let u ∈ X, then we claim that the following uh ∈ D(A) and uhh→0−−→ u

in X. Namely,

uh := 1h

� h

0T (s)u ds.(VIII.2.45)

First we show that uhh→0−−→ u holds. Indeed, we use again that T is continuous at 0,

�uh − u� =��

1h

� h

0(T (s) − I) u ds

��

≤ 1h

� h

0� (T (s) − I) u� h→0−−→ 0.

(VIII.2.46)

Now show that uh ∈ D(A) for all h > 0. Let t << h. We calculateT (t) − I

tuh = 1

ht

� h

0T (t) ◦ T (s)u ds − 1

ht

� h

0T (s)u ds

= 1ht

� t+h

tT (s)u ds − 1

ht

� h

0T (s)u ds

= 1ht

� t+h

hT (s)u ds + 1

ht

� h

tT (s)u ds

− 1ht

� t

0T (s)u ds − 1

ht

� h

tT (s)u ds

t→0−−→ 1h

T (h)u − 1h

T (0)u ∈ X.

(VIII.2.47)

Hence the left hand side converges in X, that is uh ∈ D(A). �

VIII.3. An example application of Hille-Yoshida

Example VIII.3.1. Let Ω ⊂ Rn be a bounded domain with smooth boundary ∂Ω. Wewant to find solutions to the equation

(VIII.3.1)

(∂t − L)u = 0 in Ω × (0,∞)u = 0 in (∂Ω) × (0,∞)u = u0 in Ω × {0}


where the operator L given asLu(x) := ∂i(aij(x)∂ju) + bi(x)∂iu(x) + c(x)u(x)

has smooth (time-independent) coefficients a, b, c ∈ C∞(Ω), and is uniformly elliptic, i.e.for some λ > 0

aij(x)ξi ξj ≥ λ|ξ|2 ∀x ∈ Ω, ξ ∈ Rn.

We claim that there exists a solution to (VIII.3.1) (in the sense of a semigroup) such that�u(t)�L2(Ω) ≤ ect�u0�L2(Ω) ∀t ∈ (1,∞).

Proof. To see this let X = L2(Ω), D(A) := H2(Ω)∩H10 (Ω), and A := L−ω for an ω ≥

0 yet to be chosen. Observe that if we show that A is m-dissipative then by Hille-Yoshida,Theorem VIII.2.6, we find a contraction semigroup T (t) such that for u(t) := T (t)u0

∂t(u(t)) = Au(t)with the contractive property

�u(t)�L2(Ω) ≤ �u0�L2(Ω).

Setting u(t) := eωtu(t) then u(0) = u(0) = u0 and the requested equation is satisfied∂t(u(t)) = ωeωtu(t) + Aeωtu(t) = Lu(t).

Moreover, we obtain�u(t)�L2(Ω) ≤ eωt�u0�L2(Ω).

That is, we need to show that A is m-dissipative (clearly D(A) is dense in L2). �

First we would like to show that A is dissipative.

Lemma VIII.3.2. There exist ω0 > 0 such that for any ω ≥ ω0, A as above is dissipative.

Proof. In view of Lemma VIII.1.5 we need to show that�Au, u�L2(Ω) ≤ 0 ∀u ∈ D(A).

Here is where we need to make use of ω:�Au, u�L2(Ω) ≤ −

�

Rnaij∂iu∂ju + C(b)

��∇u�L2�u�L2 + �u�2

L2

�− ω�u�L2(Ω)

By ellipticity,−�

Rnaij∂iu∂ju ≤ −λ�∇u�L2(Ω)

Moroever by Young’s inequality, for any ε > 0,

�∇u�L2�u�L2 ≤ ε�∇u�2L2(Ω) + C

1ε�u�2

L2(Ω),

so that for small enough ε > 0 we arrive at

�Au, u�L2(Ω) ≤ −λ

2�∇u�2L2 + C�u�2

L2 − ω�u�L2(Ω)


Choosing ω := C we conclude

(VIII.3.2) �Au, u�L2(Ω) ≤ −λ

2�∇u�2L2 ≤ 0.

that is, by Lemma VIII.1.5, A is dissipative. �

The next step is to prove that A is actually m-dissipative.

In view of Lemma VIII.1.8 it suffices to show that I − λ0A is surjective for some λ0 > 0.

That is, for some fixed λ0 > 0 and ω > ω0 we need to show that for any f ∈ L2(Ω) thereexists u ∈ H1

0 (Ω) ∩ H2(Ω) with (I − λ0A)u = f , that is we need to find u such that

(VIII.3.3) ∂i(aij∂ju) = − 1λ0

f + ( 1λ0

+ ω)u − bj∂ju − cu

The ∂i(aij∂ju) is the leading order term, the remaining terms are lower order terms thatcan be dealt with by a fixed point argument. First we treat the leading order term:

Lemma VIII.3.3. Let g ∈ L2(Ω)3 then there exists exactly one u ∈ H10 (Ω) with

(VIII.3.4) ∂i(aij∂ju) = g

in distributional sense. Moreover,(VIII.3.5) �u�H1(Ω) ≤ C �g�L2(Ω)

Proof. This can be done, e.g. by a variational argument, the direct method (cf.Theorem IV.0.2)

Let E : H10 (Ω) → R be an energy defined as

E(u) := 12

�

Ωaij ∂iu ∂ju +

�

Ωg u.

• Assume that E(·) has a minimizer u ∈ H10 (Ω), i.e. E(u) ≤ E(v) for all v ∈ H1

0 (Ω).Then E(u) ≤ E(u + tϕ) for any ϕ ∈ C∞

c (Ω), t ∈ R. Thus t �→ E(u + tϕ) as aminimum at t = 0, and consequently, by Fermat’s theorem (aij is symmetric!)

0 = d

dt

��t=0

E(u + tϕ) =�

Ωaij ∂iu ∂jϕ +

�

Ωg ϕ.

This holds for any ϕ ∈ C∞c (Ω); That is, (VIII.3.4) is satisfied (in distributional

sense). Indeed, we say that (VIII.3.4) is the Euler-Lagrange equation of the energyE .

• E(·) is coercive (it controls �∇u�L2(Ω).

E(u) ≥ λ�∇u�2L2(Ω) − �g�L2(Ω) �u�L2(Ω).

3g ∈ H−1(Ω) suffices


By Poincare inequality, and Youngs inequality (C changes from line to line)

E(u) ≥ λ�∇u�2L2(Ω) − C�g�L2(Ω) �∇u�L2(Ω) ≥ λ�∇u�2

L2(Ω) − C�g�2L2(Ω) − λ

2�∇u�2L2(Ω).

That is,λ

2�∇u�2L2(Ω) ≤ E(u) + C�g�2

L2 .

Again, by Poincare inequality, this implies(VIII.3.6) �u�2

H1(Ω) � E(u) + C�g�2L2 .

• Now let uk ∈ H10 (Ω) be a sequence such that E(uk) k→∞−−−→ infH1

0 (Ω) E . Since 0 ∈H1

0 (Ω) we may assume, w.l.o.g., that E(uk) ≤ E(0) = 0. By coercivity, (VIII.3.6),�uk�2

H1(Ω) � E(uk) + C�g�2L2 ≤ 0 + C�g�2

L2 ,

that issupk∈N

�uk�2H1(Ω) < ∞.

By weak compactness, Rellich, Theorem III.3.18 and Theorem III.3.24 we mayassume w.l.o.g. (otherwise taking a subsequence) that there is u ∈ H1

0 (Ω) and– uk

k→∞−−−→ u strongly in L2(Ω) and a.e.– ∇uk weakly converges to ∇u in L2(Ω), that is for any G ∈ L2(Ω,Rn),

�

Ω∇ukG

k→∞−−−→�

Ω∇uG.

Consequently, �

Ωguk

k→∞−−−→�

Ωgu.

Moreover,�

Ωaij∂iuk∂juk =

�

Ωaij∂iuk∂ju +

�

Ωaij∂iuk∂j(uk − u)

=�

Ωaij∂iuk∂ju +

�

Ωaij∂iu∂j(uk − u)

� �� k→∞−−−→0

+�

Ωaij∂i(uk − u)∂j(uk − u)

� �� ≥0

so

(VIII.3.7) lim infk→∞

�

Ωaij∂iuk∂juk ≥ lim inf

k→∞

�

Ωaij∂iuk∂ju =

�

Ωaij∂iu∂ju

Together we have shownlim inf

k→∞E(uk) ≥ E(u).

But since u ∈ H10 (Ω) we have that

E(u) ≥ infH1

0 (Ω)E ,


and by constructionlim

k→∞E(uk) = inf

H10 (Ω)

E

So (VIII.3.7) implies thatE(u) = inf

H10 (Ω)

E ,

that is u minimizes E in H10 (Ω).

• we have found a minimizer of E and this minimizer satisfies the equation.

We still need to show uniqueness and the estimate (VIII.3.5) of the solution. Observethat uniqueness follows once we show that (VIII.3.5) holds for any solution u ∈ H1

0 (Ω) of(VIII.3.4). Assume this is true, then if we have two solutions u and v for the same g, thenw := u − v ∈ H1

0 (Ω) satisfies∂i(aij∂jw) = 0.

By (VIII.3.5) we have �w�H1 = 0 that is w ≡ 0.

It remains to show (VIII.3.5) for an arbitrary solution u ∈ H10 (Ω) of (VIII.3.4). This

follwos from multiplying (VIII.3.4) by u and integrating. By ellipticity we then find,

�∇u�2L2(Ω) �

�

Ωaij∂iu ∂ju = |

�

Ωgu| � �g�L2(Ω) �u�L2(Ω).

By Poincare on the left-hand side we find�u�2

H1(Ω) ≤ |g�L2(Ω) �u�H1(Ω).

Dividing both sides by �u�H1(Ω) we have established (VIII.3.4). �

Now we need to take care of the lower order guys in (VIII.3.3), and for this we use a fixedpoint argument. Let f be fixed and for u ∈ H1

0 (Ω) set Tu ∈ H10 (Ω) be the solution of

(VIII.3.8) ∂i(aij∂j(Tfu)) = − 1λ0

f + ( 1λ0

+ ω)u − bj∂ju − cu.

By Lemma VIII.3.3 is well-defined, linear, and bounded (observe if u ∈ H1 then the right-hand side of the equation above belongs to L2(Ω)). But more is true: for any f ∈ L2(Ω),Tf is compact, which will be a consequence of the following estimate

Lemma VIII.3.4. For f ∈ L2(Ω) let Tf : H10 (Ω) → H1

0 (Ω) be defined as above. Then

�Tf (u)�H1(Ω) ≤ C��u�L2(Ω) + �f�L2(Ω)

�

and�Tf (u) − Tf (v)�H1(Ω) ≤ C �u − v�L2(Ω)

Proof. Testing (VIII.3.8) we have by ellipticity and Poincare inequality

�Tf (u)�2H1(Ω) � C

��u�L2(Ω) + �f�L2(Ω)

��Tf (u)�L2(Ω) +

��

Ωbj∂ju Tf (u)

��


By another integration by parts, (recall: b ∈ C∞)��

Ωbj∂ju Tf (u)

�� ≤ C(b,∇b) �u�L2(Ω)�Tf (u)�H1(Ω).

Consequently, we find�Tf (u)�2

H1(Ω) � C��u�L2(Ω) + �f�L2(Ω)

��Tf (u)�H1(Ω),

and dividing both sides by �Tf (u)�H1(Ω) we find the first estimate.

The second estimate follows by the same argument: observe that Tf (u)−Tf (v) = T0(u−v).Then, by the above argument,

�T0(u − v)�2H1(Ω) � C �u − v�L2(Ω)

�

From Lemma VIII.3.4 we readily conclude that Tf is actually compact for any f ∈ L2(Ω)

Corollary VIII.3.5. For f ∈ L2(Ω), let Tf be defined as above. Then Tf : H10 (Ω) →

H10 (Ω) is a continuous, compact operator. That is, for any bounded sequence (uk)k ∈

H10 (Ω),

supk

�uk�H1(Ω) < ∞

there exists a subsequence ukisuch that (Tfuki

)i is convergent in H10 (Ω).

Proof. Clearly from Lemma VIII.3.4 we obtain that Tf is Lipschitz continuous.

Now let uk be a bounded H10 -sequence. Up to a subsequence we can assume that uk weakly

converge to some u ∈ H10 (Ω) and by Rellich, Theorem III.3.24, the convergence is strong

in L2(Ω). By Lemma VIII.3.4,

�Tfuk − Tfu�H1(Ω) � �uk − u�L2(Ω)k→0−−→ 0.

Thus, Tf is compact. �

Observe that this does not mean that Tf is contracting, so we cannot apply e.g. BanachFixed Point argument. Instead we use the Schauder fixed point theorem in the form ofSchaefer’s fixed point theorem also known as the Leray-Schauder theorem. For a proof see[Gilbarg and Trudinger, 2001, Theorem 11.3].

Theorem VIII.3.6 (Leray-Schauder). Let T : X → X be a continuous and compactmapping of a Banach space X into itself, such that the set

{x ∈ X : x = µTx for some 0 ≤ µ ≤ 1}is bounded, i.e. there exists M > 0 such that whenever for some µ ∈ [0, 1] and some x ∈ Xwe have x = µTx then

�x�X ≤ M.

Then T has a fixed point.


We can apply this theorem to the compact operator T essentially since we dissipativity:

Lemma VIII.3.7. There exist ω0 > 0 such that for any ω > ω0 and any λ0 > 0 we havethe following:

Let µ ∈ [0, 1] and assume that u ∈ H10 (Ω) satisfies

u = µTu

then with a constant independent of µ,�u�H1(Ω) ≤ C�f�L2(Ω)

Lemma VIII.3.8. From u = µTu we conclude in view of the definition of T , (VIII.3.8),

−∂i(aij∂ju) = −µ�− 1

λ0f + ( 1

λ0+ ω)u − bj∂ju − cu

�.

Similar to the proof of A being dissipativ, we test this equation with u, and obtain

�∇u�2L2(Ω) ≤ C�f�L2(Ω) +

�

Ω

�c − ω − 1

λ0

�|u|2 +

�

Ωbj∂ju u.

Observe that ∂juu = 12∂j|u|2 and integrating by parts we find

�∇u�2L2(Ω) ≤ C�f�L2(Ω) −

�ω + 1

λ0

� �

Ω|u|2 + C(c, b) �u�2

L2(Ω)

If ω is large enough, namely ω > C(c, b) this implies�∇u�2

L2(Ω) ≤ C�f�L2(Ω).

By Poincare inequality we obtain the claim.

No we can conclude:

A is m-dissipative. We argued above that A is m-dissipative if for any f ∈ L2(Ω)there exists u ∈ H2(Ω)∩H1

0 (Ω) such that Tu = u. Since T is compact by Corollary VIII.3.5and in view of Lemma VIII.3.7, Leray-Schauder in form of Theorem VIII.3.6 is applicable,and implies that there exists a fixed point u ∈ H1

0 (Ω) such that Tu = u.

Observe that this implies that u solves an equation of the form

∂i(aij∂ju) = g in Ωu = 0 in ∂Ω

for g ∈ L2(Ω).

Now we argue as in Theorem IV.0.1, to obtain interior regularity, indeed we find�u�W 2,2

loc(Ω) ≤ C(Ω) �g�L2(Ω).

Close to the boundary one needs to do a reflection argument to obtain the same estimateup to the boundary.

VIII.4. FORMAL REGULARITY THEORY (A PRIORI ESTIMATES) 147

Up to doing this, we have shown that there is u ∈ D(A) and (I − λ0A)u = f ;

That is, A is m-dissipative, and thus generates a contractive semigroup as claimed. �Exercise VIII.3.9. Find a solution to the inhomogeneous equation

∂tu(x, t) − Lu(x, t) = f(x, t) in Ω × (0,∞)u(x, 0) = u0(x).

Here L is the usual elliptic operator with smooth coefficients.

Find an estimate of �u(t)�L2(Ω) in terms of �f�L2 , �u0�L2 and t.

hint: Use the Duhamel Ansatz from the heat equation in Rn, i.e. set

u(t) =� t

0vs(t) ds

where vs(t) is the solution associated to

∂tv(x, t) − Lv(x, t) = 0 in Ω × (s,∞)v(x, s) = f(x, s).

Observe that vs(t) can be written as semigroup!

VIII.4. Formal regularity theory (a priori estimates)

Here we show estimates that hold for the heat equation (and which can be easily extendedto linear parabolic equations).

For simplicity we prove only a priori estimates, i.e. estimates under the assumption thatu is already sufficiently regular. It is further work to show that these estimates holdalso without this a priori regularity assumption (we did something similar for the Laplaceequation, cf. Lemma I.2.17).

Cf. [Evans, 2010, 7.1.3].

Assume that Ω ⊂⊂ Rn with smooth boundary, and we have

∂tu − Δu = f in Ω × (0,∞)u = 0 on ∂Ω × (0,∞)u(0, x) = u0(x) on Ω × {0}

We first test the equation with u, (i.e. we multiply with u and integrate by parts).

(VIII.4.1)�

Ω∂tu · u −

�

ΩΔu · u =

�

Ωfu

Observe that∂tu · u = 1

2∂t|u|2,

VIII.4. FORMAL REGULARITY THEORY (A PRIORI ESTIMATES) 148�

ΩΔu · u = −

�

Ω|∇u|2

and by Holder, then Poincare inequality and then Young’s inequality,�

Ωfu ≤ �f�L2(Ω) �u�L2(Ω) ≤ C ≤ �f�L2(Ω) �∇u�L2(Ω) ≤ C(ε) �f�2

L2(Ω) + ε�Du�2L2(Ω)

Consequently (VIII.4.1) becomes (for ε = 12)

∂t

�

Ω|u|2 + 1

2

�

Ω|Du|2 ≤ C(ε)

�

Ω|f |2

We integrate in t from 0 to some s > 0 and find�

Ω|u(s)|2 −

�

Ω|u0|2 + 1

2

�

Ω×(0,s)|Du|2 ≤ C(ε)

�

Ω×(0,s)|f |2

From which we obtain�

Ω×(0,s)|Du|2 +

�

Ω|u(s)|2 ≤ C

�

Ω×(0,s)|f |2 +

�

Ω|u0|2.

From this we conclude

Lemma VIII.4.1.�

Ω×(0,T )|Du|2 + sup

t∈(0,T )

�

Ω|u(t)|2 ≤ C

�

Ω×(0,T )|f |2 +

�

Ω|u0|2.

That is, we get an L∞t (L2

x)-bound on u and an L2t L

2x-bound on u. in terms of the L2-

bound of u0 and the L2t L

2x-bound of f . (These are the very simple versions of what is for

Schrodinger equations referred as Strichartz estimates).

We obtain another estimate from testing with f , i.e. integrating(VIII.4.2) (∂tu − Δu)2 = f 2

Observe that(∂tu − Δu)2 = |∂tu|2 + |Δu|2 − 2∂tuΔu,

and (using that ∂tu = 0 on ∂Ω since u = 0 on ∂Ω for all times)�

Ω∂tuΔu = −

�

Ω(∂t∇u) · ∇u = −1

2∂t

�

Ω|∇u|2

Thus, after integrating in x and t, (VIII.4.2) becomes�

Ω×(0,s)

�|∂tu|2 + |Δu|2

�+�

Ω|∇u(s)|2 −

�

Ω|∇u(0)| =

�

Ω×(0,s)f 2

which leads to

Lemma VIII.4.2.�

Ω×(0,T )

�|∂tu|2 + |Δu|2

�+ sup

t∈(0,T )

�

Ω|∇u(t)|2 ≤

�

Ω×(0,s)f 2 +

�

Ω|∇u(0)|2

VIII.4. FORMAL REGULARITY THEORY (A PRIORI ESTIMATES) 149

As in the elliptic case, cf. Chapter 4: higher order estimates in t are obtained by differen-tiating the equation in t, i.e. considering that u := ∂tu is a solution to

∂tu − Δu = ∂tf in Ω × (0,∞)u = 0 on ∂Ω × (0,∞)u(0, x) = ∂tu(0, x) = Δu0 + f on Ω × {0}

In this way we obtain from Lemma VIII.4.1Lemma VIII.4.3.�

Ω×(0,T )|D∂tu|2 + sup

t∈(0,T )

�

Ω|∂tu(t)|2 ≤ C

�

Ω×(0,T )|∂tf |2 +

�

Ω|Δu0|2 + |f(0)|2.

Remark VIII.4.4. Observe that these estimate do not work if the sign in front of Δchanges! Parabolic equations (in contrast to, e.g., Schrodinger equations) have controlonly in one time direction!

∂tu + Δu = f in Ω × (0,∞)u = 0 on ∂Ω × (0,∞)u(0, x) = u0(x) on Ω × {0}

CHAPTER 9

Linear Hyperbolic equations: waves

IX.1. Halfwave-decomposition

The model case for hyperbolic equations is the wave equation

∂ttu − Δu = 0 in Rn × (0,∞)u = g in Rn × {0}∂tu = h on Rn × {0}

For the elliptic (Laplace equation) and parabolic (heat equation) it was useful to use theFourier transform to get a feeling of the equation.

Taking the Fourier transform in x, we find that∂ttu(ξ, t) + c|ξ|2u(ξ, t) = 0.

Solutions of this second order ODE are of the formu(ξ, t) = eit|ξ|a(ξ) + e−it|ξ|b(ξ).

where i is the complex unit! a and b have to satisfy

a(ξ) + b(ξ) = g(ξ)i|ξ|a(ξ) − i|ξ|b(ξ) = h(ξ)

that is

a(ξ) + b(ξ) = g(ξ)−a(ξ) + b(ξ) = i|ξ|−1h(ξ)

which leads to

a(ξ) = 12

�g(ξ) − i|ξ|−1h(ξ)

�

b(ξ) = 12

�g(ξ) + i|ξ|−1h(ξ)

�

In the semigroup approach we interpreted e−t|ξ|2 as the Fourier symbol of e−tΔ (since |ξ|2is the Fourier symbol of Δ). Since |ξ| =

�|ξ|2 we denote the operator with Fourier symbol

|ξ| as half-Laplacian (−Δ) 12 – and the solution u as

u = eit(−Δ)12 1

2�g − i(−Δ)− 1

2 h�

+ e−it(−Δ)12 1

2�g + i(−Δ)− 1

2 h�

150

IX.2. D’ALAMBERT’S FORMULA IN ONE DIMENSION 151

If we call u− := eit(−Δ)12 1

2

�g − i(−Δ)− 1

2 h�

and u+ := e−it(−Δ)12 1

2

�g + i(−Δ)− 1

2 h�

then weobserve that u± satisfies

∂tu± − i(−Δ) 12 u± = 0 in Rn × (0,∞)

u± = 12

�g ± i(−Δ)− 1

2 h�

in Rn × {0}

Observe that the differential operator ∂t−i(−Δ) 12 looks formally similar to the Schrodinger

equation ∂t+iΔ. This decomposition of the wave-equation is called half-wave decompositionof the wave equation.

IX.2. D’Alambert’s formula in one dimension

In one dimension we consider

∂ttu − ∂xxu = 0 in R × (0,∞)u = g in R × {0}∂tu = h on R × {0}

This equation can be rewritten as(∂t + ∂x) (∂t − ∂x) u = 0.

So if we set(IX.2.1) v(x, t) := (∂t − ∂x) u,

then v satisfiesvt + ∂xv = 0.

This is a transport equation, cf. (I.1.1), which has the solutionv(x, t) = v(x − t, 0)

Observe thatv(z, 0) = ∂tu(z, 0) − ∂xu(z, 0) = h(z) − g�(z).

Plugging this into (IX.2.1)h(x − t) − g�(x − t) = (∂t − ∂x) u,

i.e. an inhomogeneous transport equation. This, in turn, has the solution

u(x, t) =� t

0v(x + (t − s) − s, 0)ds + u(x + t, 0)

=12

� x+t

x−th(s) − g�(s)ds + g(x + t)

=12

� x+t

x−th(s)ds + 1

2(g(x + t) − g(x − t))

This formula is called the D’Alambert formula.

IX.3. SECOND-ORDER HYPERBOLIC EQUATIONS 152

As we have seen for the transport equation, this solution is not smoothing: If g ∈ Ck,h ∈ Ck−1 then u might as well only be in Ck as well for all t > 0 (this is contrast to theheat equation which is immediately smooth).

Another observation is finite speed of propagation: u(x, t) takes into account h and g onlyin a neighborhood of x – for the heat equation if we change the boundary data in one pointit changes everywhere. This is finite speed of propagation vs infinite speed of propagation.In particular no such thing as a maximum principle will hold.

In higher dimensions, a solution representation can be computed with a method calledspherical means, see [Evans, 2010, 2.4.1].

IX.3. Second-order hyperbolic equations

Let L be the usual elliptic operator, then the operator ∂tt − L is called hyperbolic (themodel case being the wave equation).

We need to define weak solutions of (∂tt − L)u.

As usual we would like to have u(·, t) ∈ H1(Ω) for any fixed time t. Since ∂ttu = Lu isin general not a function, but a distribution, which we have to test with a H1-Sobolevfunction. This is denoted as H−1(Ω).

Definition IX.3.1. A distribution f belongs to H−1(Ω) if for all ϕ ∈ H1(Ω),|f [ϕ]| ≤ Λ�ϕ�H1(Ω)

The minimal value of Λ is denoted by �f�H−1(Ω).

We have H1(Ω) ⊂ L2(Ω) ⊂ H−1(Ω).

Definition IX.3.2. A functionu ∈ L2(0, T ; H1

0 (Ω))u� ∈ L2(0, T ; L2(Ω))

u�� ∈ L2(0, T ; H−1(Ω))1 is a weak solution to

(IX.3.1)

(∂tt − L)u = f in Ω × (0, T )u = 0 on ∂Ω × (0, T )u = g, ∂tu = h on Ω × {0}

if for almost every t ∈ (0, T ) and any ϕ ∈ C∞c (Ω)

∂ttu(t)[ϕ] −�

ΩLu(t) ϕ =

�

Ωf(t)ϕ

1in particular u ∈ H2(0, T ; H−1(Ω)) and as such u and u� are continuous in t as mappings into L2 orH−1, respectively.

IX.4. EXISTENCE VIA GALERKIN APPROXIMATION 153

and if the boundary data is attained in the trace sense.

While there is a method based on semigroups to show existence, here we use anotherargument (that could be used also for parabolic equations), that is less algebraic and morenumerical:

IX.4. Existence via Galerkin approximation

The basic idea of Galerkin method is to approximate the infinite dimensional problem(∂tt−L)u = f (plus boundary data) by finite dimensional problems: we project the probleminto finite dimensional subspaces (of H1

0 (Ω)), solve it there, and pray for convergence asthe dimension of the subspaces goes to infinity.

So assume2 that we have (wk)∞k=1 ⊂ H1

0 (Ω) that is an basis of H10 (Ω) and of L2(Ω) such

that its orthogonal in H10 (Ω) and orthonormal in L2(Ω), i.e.�

∇wk · ∇w� =�

wk · w� = 0 k �= �,

and �wk�L2 = 1.

For m ∈ N we “project the equation” onto span{w1, . . . , wm}, i.e. we set the “approximat-ing solution”

um(t) :=m�

k=1dk

m(t) wk

2One can take an eigenvalue basis for the Laplacian. One can show, [Evans, 2010, 6.5.1], that thereexist an infinite sequence of eigenvalues 0 < λ1 ≤ λ2 ≤ . . . with λk

k→∞−−−−→ ∞ such the solutions wk ∈ H10 (Ω)

�−Δwk = λkwk in Ωwk = 0 on ∂Ω

form an orthonormal basis of L2(Ω), i.e. �wk�L2(Ω) = 1,�

Ω wkwj = 0 whenever k �= j, and for anyf ∈ L2(Ω) there is a unique sequence µk ∈ �2 such that

f =∞�

k=1µkwk

(with convergence in L2(Ω) if f ∈ L2(Ω) and convergence in H10 (Ω) if f ∈ H1

0 (Ω)).Observe that then also (integrating by parts)

�

Ω∇wk∇wj =

�0 k �= j

µk k = j

in particular wk forms an orthonormal basis of H10 (Ω) with respect to the scalar product

�f, g�H1(Ω) =�

∇f · ∇g.


where the coefficients dkm(t) have to be chosen such that for all t ∈ (0, T ) and all k =

1, . . . , m,

(IX.4.1)�

Ω∂ttumwk − Lumwk =

�

Ωfwk.

with initial value for k = 1, . . . , m

dkm(0) =

�gwk, ∂td

km(0) =

�hwk.

Observe that by orthonormality,�

Ω∂ttumwk = ∂ttd

km(t).

Since L is linear and the derivatives are spacial only, we can write�

ΩLumwk =

m�

�=1ek�(t) d�

m(t).

for some coefficients ek�(t). That is, setting fk :=�

Ω fwk (IX.4.1) becomes a linear secondorder initial value ODE,

(IX.4.2) ∂ttdkm(t) +

m�

�=1ek�(t) d�

m(t) = fk(t).

ODE theory tells us there is a unique C2-solution to this equation.

The hope is to find a weak solution u to (IX.3.1) as limit m → ∞ of um, and for this weneed uniform estimates. First we use

Proposition IX.4.1.

supm

�sup

t∈(0,T )

��um(t)�H1

0 (Ω) + �∂tum(t)�H10 (Ω)

�+ �∂ttum�L2(0,T ;H−1(Ω)

�

≤C��f�L2(0,T ;L2(Ω) + �g�H1

0 (Ω) + �h�L2(Ω)�

.

Here C = C(T, Ω, L).

An important ingredient is

Lemma IX.4.2 (Gronwall’s inequality). Let f : [a, b] → R be differentiable on (a, b) andcontinuous on [a, b]. If f satisfies the differential inequality

f �(t) ≤ α(t) f(t) ∀t ∈ (a, b),for some integrable function α, then for any t ∈ (a, b),

f(t) ≤ f(a) exp�� t

aα(s)ds

�.

whenever the right-hand side makes sense.


Proof of Lemma IX.4.2. Set g(t) := exp(� t

a α(s)ds).

Theng�(t) = α(t) g(t).

Observe moreover that g(0) = 1 and g ≥ 0. Then for

h(t) := f(t)g(t)

we haveh�(t) = f �(t) g(t) − f(t)g�(t)

(g(t))2 ≤ α(t) f(t)g(t) − f(t)α(t)g(t)(g(t))2 = 0.

That is, h is monotonically decreasing, soh(t) ≤ h(a) ∀t ∈ (a, t),

that isf(t)g(t) ≤ f(a),

which gives the claim. �

Proof of Proposition IX.4.1. Multiply (IX.4.1) with ∂tdmk and sum over k. Then

(IX.4.3)�

Ω∂ttum∂tum − Lum∂tum =

�

Ωf∂tum.

We observe �

Ω∂ttum∂tum = 1

2∂t�∂tum�2L2(Ω).

Also �

Ωaij(x, t) ∂xi

um ∂xj∂tum

=12∂t

��

Ωaij(x, t) ∂xi

um ∂xjum

�− 1

2

�

Ω∂taij(x, t) ∂xi

um ∂xjum

So from (IX.4.3) we obtain

∂t

��∂tum�2

L2(Ω) + 12

��

Ωaij(x, t) ∂xi

um ∂xjum

��

≤C��∂tum�2

L2(Ω) + �um�2H1

0 (Ω) + �f�2L2(Ω)

�.

Since by ellipticity�um�2

H10 (Ω) �

�

Ωaij(x, t) ∂xi

um ∂xjum

we can setη(t) := �∂tum�2

L2(Ω) + 12

��

Ωaij(x, t) ∂xi

um ∂xjum

�

and have for all t ∈ (0, T )

η�(t) ≤ C�η(t) + �f(t)�2

L2(Ω)

�


This is an ODE inequality, and Gronwall’s inequality, Lemma IX.4.2,3 implies

supt∈(0,T )

�∂tum�2L2(Ω) + �um�2

H10 (Ω) ≤ C eCt

�η(0) +

� T

0�f(t)�2

L2(Ω).

�

By the initial values for um we find the first part of the claim, namelysup

msup

t∈(0,T )

��um(t)�H1

0 (Ω) + �∂tum(t)�H10 (Ω)

�≤ C

��f�L2(0,T ;L2(Ω) + �g�H1

0 (Ω) + �h�L2(Ω)�

.

To obtain the H−1(Ω) estimate of ∂ttum, we need to estimate

�∂ttum�H−1(Ω) ≡ sup�v�

H10 (Ω)≤1

�

Ω∂ttumv.

Fix v ∈ H10 (Ω). Since wk is an L2-orthonormal basis, there exists a unique λ ∈ �2 with

v =∞�

k=1λkwk.

Setv1 :=

m�

k=1λkwk,

and v2 := v − v1. Since wk are still orthogonal as H10 (Ω)-maps, we have

�v1�2H1

0 (Ω) + �v2�2H1

0 (Ω) = �v�2H1

0 (Ω) ≤ 1.

Moreover, orthogonality and the definition of um also implies�

Ω∂ttumv =

m�

k=1λk

�

Ω∂ttumωk.

Thus, (IX.4.1) gives �

Ω∂ttumv =

�

ΩLumv1 +

�

Ωfv1,

that is�∂ttum�H−1(Ω) � �um�H1

0 (Ω) + �f�L2(Ω).

Squaring this estimate and integrating in t we obtain�∂ttum�2

L2(0,T )H−1(Ω) ≤ T supt∈(0,T )

�um�H10 (Ω) + �f�2

L2(0,T ;L2(Ω),

which by the estimates we already proved leads to the claim. �3Here in the following version: let η be nonnegative, absolutely continuous function on [0, T ] such that

for almost every t,η�(t) ≤ φ(t)η(t) + ψ(t),

for φ(t) and ψ(t) nonnegative, integrable functions on [0, T ]. Then

η(t) ≤ exp(� t

0φ(s) ds)

�η(0) +

� t

0ψ(s) ds

�.

See [Evans, 2010, §B.2]. (exercise!)


Theorem IX.4.3. The Galerkin method above converges to a weak solution of (IX.3.1)which satisfies the same estimates as Proposition IX.4.1.

Proof. By Proposition IX.4.1 the sequence um is bounded in L2(0, 1; H10 (Ω)), ∂tum is

bounded in L2(0, 1; L2(Ω)), and ∂ttum is bounded in L2(0, 1; H−1(Ω)). These are all Hilbertspaces, and by weak compactness we find

u ∈ L2(0, 1; H10 (Ω)), with ∂tu in L2(0, 1; L2(Ω)), and ∂ttu in L2(0, 1; H−1(Ω)) such that (up

to taking a subsequence of um which by an abuse of notation we call again um)um �→ u weakly in L2(0, 1; H1

0 (Ω))∂tum �→ ∂tu weakly in L2(0, 1; L2(Ω))

∂ttum �→ ∂ttu weakly in L2(0, 1; H−1(Ω))In particular we have for any v ∈ C1(0, 1; H1

0 (Ω))� T

0

�

Ω∂ttum v

m→∞−−−→� T

0

�

Ω∂ttuv.

as well as � T

0

�

ΩLum v

m→∞−−−→� T

0

�

ΩLuv.

Now let for K ≤ m the projection onto span(w1, . . . , wK) of w be vK , i.e.

vK :=K�

k=1wk

�

Ωwkv.

Then from (since K ≤ m) the equation (IX.4.2)� T

0

�

Ω∂ttum vK −

� T

0

�

ΩLum vK =

� T

0

�

ΩfvK

Letting m → ∞ we obtain for any K� T

0

�

Ω∂ttu vK −

� T

0

�

ΩLu vK =

� T

0

�

ΩfvK

Since (wk)k∈N is a basis of H10 (Ω) we can take the limit as K → ∞ to obtain

� T

0

�

Ω∂ttu v −

� T

0

�

ΩLu v =

� T

0

�

Ωfv.

We still need to verify u = g and ∂tu = h – this follows along the arguments above bychoosing testfunctions v that vanish at time T but not at time 0. We refer to [Evans, 2010,7.2.2., Theorem 3] for the details.

The estimates of Proposition IX.4.1 survive the convergence. �

Let us remark that the solution is actually unique, see [Evans, 2010, 7.2.2, Theorem 4].Regularity estimates are also possible depending on the data, see [Evans, 2010, 7.2.2,Theorem 5,6].

IX.5. PROPAGATION OF DISTURBANCES 158

IX.5. Propagation of Disturbances

Parabolic equations have infinite speed of propagation. For example, if we consider asolution to �

∂tu − Δu = 0 in Rn × (0,∞)then if u(x, 0) ≥ 0 everywhere but we know that u(x0, 0) > 0 for some x0 ∈ Rn then (bymaximum principle) u(x, t) > 0 for all x ∈ Rn and t > 0. I.e., the positivity informationhas travelled from the point x0 with infinite speed to every point x ∈ Rn in space. Thisis closely related to the maximum principles. For hyperbolic equations this is not trueanymore, indeed we have the opposite, which is called propagation of disturbances.

(IX.5.1)�∂ttu − Δu = 0 in Rn × (0,∞)

In this case we have

Theorem IX.5.1 (Finite propagation of speed). Assume that u solves (IX.5.1) and u(x, 0) ≡0 and ∂tu(x, 0) ≡ 0 in B(x, r) then u(x, t) = 0 in the cone C(x0, r),

C(x0, r) := {(x, t) ∈ Rn × (0,∞) : 0 ≤ t ≤ r, |x − x0| < r − t}

Proof. W.l.o.g. x0 = 0. For 0 < t < r we set

e(t) := 12

�

B(r−t)|∂tu(x, t)|2 + |Du(x, t)|2 dx

We take the derivative

e�(t) = −12

�

∂B(r−t)|∂tu|2dHn−1−1

2

�

∂B(r−t)|Du(x, t)|2dHn−1 dx+

�

B(r−t)∂tu∂ttu+Du·∂tDu dx

Integrating by parts we obtain�

B(r−t)Du · ∂tDu dx =

�

∂B(r−t)∂νu ∂tu dHn−1 −

�

B(r−t)Δu ∂tu dx

=�

∂B(r−t)∂νu ∂tu dHn−1 −

�

B(r−t)∂ttu ∂tu dx

that is

e�(t) = −12

�

∂B(r−t)|∂tu|2dHn−1 − 1

2

�

∂B(r−t)|Du|2dHn−1 dx +

�

∂B(r−t)∂νu ∂tu dx

Now by Young’s inequality (in this case simply because 2ab ≤ a2 + b2)�

∂B(r−t)∂νu ∂tu dx ≤ 1

2

�

∂B(r−t)|Du|2 + |∂tu|2 dx.

That is for all t ∈ (0, r),e�(t) ≤ 0.

This implies for all t ∈ (0, r)0 ≤ e(t) ≤ e(0).

IX.5. PROPAGATION OF DISTURBANCES 159

Since u(x, 0) ≡ const in B(r), |Du| ≡ 0 in B(r). Since moreover ∂tu(x, 0) = 0 in B(r)we have e(0) = 0, and consequently e(t) ≡ 0 for all t ∈ (0, r). This in turn means that|Du| = |∂tu| = 0 in C(r), i.e u is constant in the cone C(0, r). Thus u ≡ 0. �

CHAPTER 10

Introduction to Navier–Stokes

For a thorough exposition we refer to [Galdi, 2011, Robinson et al., 2016].

X.1. The Navier–Stokes equations

The incompressible Navier-Stokes system is supposed to represent the evolution of thedynamics of an incompressible viscous fluid.

For u = (u1, u2, u3) : R3 → R (represents the velocity field of a fluid) and p : R3 → R (thepressure) the Navier-Stokes equation is

∂tu − Δu + (u · ∇)u + ∇p = 0 in R3 × (0,∞)∇ · u = 0 in R3 × (0,∞)u(x, 0) = u0(x) for x ∈ R3.

In more physical version, there are factors of viscosity and density of the fluid, whichfor analytical simplicity we set here equal to 1. We also assume that no external forcesare acting on the fluid. Also let us remark that there is (often: implicitely) assumed aboundedness for |x| → ∞ of u.

We do not have a good understanding of this equation, indeed it is one of the MilleniumProblems formulated by Charles Fefferman for the Clay Math Institute.

Conjecture X.1.1. Let ν > 0 and let u0 : R3 → R3 be a divergence-free vector field in theSchwartz class. Then there exist a smooth vector field u : [0,∞] × R3 → R3 (the velocityfield) and a smooth function p : R3 → R (the pressure field) obeying the equations

∂tu + (u · ∇)u = νΔu −∇p∀(0,∞) × R3

∇ · u(t, x) = 0 ∀t ∈ (0,∞), x ∈ R3

u(0, ·) = u0(·)as well as the finite energy condition u ∈ L∞

t (L2x([0, T ] × R3)) for every 0 < T < ∞.

We will look at two-dimensional and three-dimensional versions of this equation (in twodimensions things are a bit easier), i.e. for (nice!) domains Ω ⊆ Rn, n = 2, 3 we consider

160

X.1. THE NAVIER–STOKES EQUATIONS 161

solutions u : Ω → Rn, p : Ω → R of

(X.1.1)

∂tu − Δu + (u · ∇)u + ∇p = 0 in Ω × (0, T )∇ · u = 0 in Ω × (0, T )

We need to define some operators that appear here. The operator ∇· denotes the diver-gence, in two dimensions:

∇ · (u1, u2) ≡ div ((u1, u2)) = ∂1u1 + ∂2u2.

and in three dimensions

∇ · (u1, u2, u3) ≡ div ((u1, u2, u3)) = ∂1u1 + ∂2u2 + ∂3u3.

The notion u · ∇ is to be understood as a vector product of u (a vector field) and theGradient, i.e. in twod dimensions,

u · ∇ := u1∂1 + u2∂2

and thus

u · ∇u := u1∂1

�u1

u2

�+ u2∂2

�u1

u2

�.

In three dimensions,u · ∇ := u1∂1 + u2∂2 + u3∂3

and thus

u · ∇u := u1∂1

u1

u2

u3

+ u2∂2

u1

u2

u3

The pressure term is never explicitely specified because it follows intrinsically from u: Sincediv u = 0 we can apply divergence to the equation (X.1.1) and find

div ((u · ∇)u) + div (∇p) = 0.

Since div (∇p) = Δp we have that p must solve the following equation,

Δp = −div ((u · ∇)u) in Ω.

Observe that this equation is uniquely solvable if boundary data on ∂Ω is given (and u issufficiently regular in the right Sobolev space).

In the following we will always assume that Ω is a smoothly bounded set or the whole space,in particular we assume that C∞(Ω) is dense in H1(Ω).

X.2. WEAK FORMULATION 162

X.2. Weak formulation

For n = 2, 3 and Ω ⊂ Rn open denote by

C∞c,σ(Ω) :=

�ϕ ∈ [C∞

c (Ω)]3 : div ϕ = 0�

SetL2

σ,n ≡ L2σ,n(Ω) := C∞

c,σ(Ω)�·�L2

equipped with the L2-norm.

Lemma X.2.1. Let Ω ⊂ Rn be a domain with (if existing) smooth boundary. For anyf ∈ L2

σ,n(Ω) and any g ∈ H1(Ω) (without any boundary assumption) we have�

Ωf∇g = 0

Proof. Since f ∈ L2σ,n(Ω) there are C∞

c (Ω,R3) � fk → f ∈ L2(Ω), with ∇ · fk = 0.Thus, �

Ωf∇g = lim

k→∞

�

Ωfk∇g = lim

k→∞

�

Ω∇ · fkg = 0

�Remark X.2.2. Observe that formally (if f was nicer; ν denotes the outer unit normalof ∂Ω ) �

Ωf∇g =

�

∂Ωf · νg −

�

Ω∇ · f g.

So if ∇ · f = 0 this implies that�

∂Ωf · νg = 0 ∀g ∈ H1(Ω).

Since the trace space of H1(Ω) is H12 (∂Ω), this implies

�

∂Ωf · νg = 0 ∀g ∈ H

12 (∂Ω).

Of course, if f ∈ L2(Ω), f is not well-defined on ∂Ω – but this argument shows that f · νhas a distributional meaning.

Indeed, one can show [Galdi, 2011, Theorem III.2.3] that whenever Ω is a smooth boundeddomain, then

(X.2.1) L2σ,n(Ω) =

�u ∈ L2(Ω,R3) : ∇ · u = 0, u

��∂Ω

· ν = 0�

Here, ∇ · u = 0 is to be understood in the distributional sense and u · ν has again adistributional meaning (in the dual space of trace spaces).

An important theorem is the Helmoltz-Weyl decomposition (in more geometric contextsknown as Hodge decomposition).


Theorem X.2.3. Let Ω ⊂ R3 be an open smoothly bounded set in R3. Then for anyf ∈ L2(Ω) there are h ∈ L2

σ,n(Ω) and g ∈ H1(Ω) such thatf = h + ∇g,

with�h�L2(Ω) + �∇g�L2(Ω) ≤ �f�L2(Ω)

h in the decomposition above is sometimes called Helmholtz-Leray projection of f .

Sketch of the proof. First we solve for g1 ∈ H10 (Ω),

Δg1 = ∇ · f in Ω,

g1 = 0 on ∂Ω.

This can be done with the estimate�∇g1�L2(Ω) � �f�L2(Ω).

We set h1 := f −∇g1, which satisfies�h1�L2(Ω) � �f�L2(Ω).

Observe that div h1 = 0. However this does not mean that h1 ∈ L2σ,n(Ω) because h1 · ν is

not necessarily zero (however, as in Remark X.2.2, the notion of h1 · ν��∂Ω

can be definedin a distributional sense). So we solve

Δg2 = ∇ · h1 = 0 in Ω,

∂νg2 = h1 · ν on ∂Ω.

Again (this needs to be proven),�∇g2�L2(Ω) � �h1�L2(Ω) � �f�L2(Ω).

Now,h := f −∇g1 −∇g2 ≡ h1 −∇g2

satisfies div (h) = 0 and h · ν = 0 on ∂Ω, so by (X.2.1) we have that h ∈ L2σ,n(Ω). �

Now we setV := L2

σ,n(Ω,Rn) ∩ H10 (Ω,Rn).

In particular,V ⊂

�f ∈ H1

0 (Ω,Rn) : div f = 0�

but these two spaces are not the same. Also observe that H10 (Rn,Rn) = H1(Rn,Rn).

Lemma X.2.4. For n = 1, 2, 3, 4 and Ω ⊂ Rn open with smooth boundary the map

(u, v, w) �→�

Ωuα∂αvβwβ

is a trilinear continuous map on H1(Ω) × H1(Ω) × H1(Ω).


Proof. It is clearly tri-linear, all that is needed to show is boundedness. By Holder’sinequality

�

Ωuα∂αvβwβ ≤ �v�H1(Ω) �u�L4(Ω) �w�L4(Ω)

By Sobolev embedding, Corollary III.3.31, H1(Ω) ⊂ Lp(Ω) for any p ∈ [2, p∗], p < ∞ forp∗ = 2n

n−2 (for n ≥ 3), p ∈ [2,∞) for n ≤ 2. As long as n ≤ 4 this implies that

�u�L4(Ω) ≤ �u�H1(Ω),

�w�L4(Ω) ≤ �w�H1(Ω).

�

What makes Navier-Stokes equation a “nonlinear” equation (this notion is debatable) isthe term (u ·∇)u. One important observation is that it vanishes when testing the equationwith u. For this we introduce the notion of the L2-scalar product,

�f, g� =�

Ωfg.

and the L2-norm�f� :=

��f, f�.

Lemma X.2.5. Let u ∈ V and v, w ∈ H1(Ω,Rn) then

�u · ∇v, w� = −�u · ∇w, v�

and in particular�u · ∇v, v� = 0.

Proof. By Lemma X.2.4,

(u, v, w) �→ �u · ∇v, w�

is a bounded trilinear form on the spaces involved (in particular it makes sense as anintegral). By approximation (here we use that Ω is a nice set) we can assume that v, w ∈C∞(Ω,R3) ∩ W 1,∞(Ω).

Observe now that

�u · ∇v, w� + �u · ∇w, v�=�uα · ∂αvβ, wβ� + �uα · ∂αwβ, vβ�


Observe that ∂αvβwβ ∈ L∞(Ω) and since uα ∈ L2σ,n(Ω) we assume w.l.o.g. (by approxima-

tion) that uα ∈ C∞c with div (u) = 0. Then

=�

Ωuα · ∂α(vβwβ)

= −�

Ω∂αuα · vβwβ

= −�

Ωdiv (u) · (vβwβ)

=0.

In the third to fourth line we used integration by parts and that uα is zero in trace senseon ∂Ω. In the last line we use that div u = 0.

This proves the claims. �

Observe that we also have

Lemma X.2.6. Let u ∈ V , and p ∈ H1(Ω) then�u,∇p� = 0

Proof. Again by approximation we may assume that p ∈ C∞(Ω) ∩ W 1,∞(Ω).

So, integration by parts again, since u = 0 on ∂Ω and div u = 0,�u,∇p� = −�div u, p� = 0.

�

So if we have a smooth solution u(t) ∈ V of the Navier-Stokes equation, (X.1.1), then we(formally) should have

�∂tu − Δu + (u · ∇)u + ∇p, u� = 0.

By Lemma X.2.4, Lemma X.2.5, Lemma X.2.6 this implies�∂tu(t), u(t)� − �Δu(t), u(t)� = 0.

Integration by parts (u(t) ∈ V so no boundary terms)

�Δu(t), u(t)� = −�∇u(t),∇u(t)� = −�

Ω|∇u(t)|2dx.

Also (formally!) interchanging derivative ∂t and the integral we have

�∂tu(t), u(t)� = 12∂t

�

Ω|u(t)|2.

Integrating formally in t = 0 to some s we then find that�

Ω|u(s)|2 +

� s

0

�

Ω|∇u(t)|2dx ds =

�

Ω|u0|2,


that is we find that any (nice enough) solution on (0, T ) of the Navier stokes equation withu(0) = u0 satisfies immediately u ∈ L∞(0, T ; L2

σ,n)∩L2(0, T ; V ). We like this so much thatwe put this into the definition of solution.Definition X.2.7. We say that u : [0, T ) → V is a weak solution of the Navier-Stokesequation (X.1.1) with initial value u0 ∈ H1

0 (Ω,R3) if

• u ∈ L∞(0, T ; L2σ,n) ∩ L2(0, T ; V ) for all T > 0 and

• u satisfies the equation� s

0−�u, ∂tϕ� +

� s

0�∇u,∇ϕ� ds +

� s

0�(u · ∇)u, ϕ� ds = �u0, ϕ(0)� − �u(s), ϕ(s)�

for almost all s ∈ (0, T ) and all test functions ϕ ∈ Dσ, whereDσ = {ϕ ∈ C∞

c (Ω × [0,∞)) : div (ϕ(t)) = 0 ∀t ∈ [0,∞)} .

Remark X.2.8 (A problem in 3D). Observe that in Definition X.2.7 one can not just testwith ϕ = u (even with a density argument). It is true that by Lemma X.2.5 we can makesense (by approximation) of

�(u(t) · ∇)u(t), u(t)�,if u(t) ∈ V ; and this expression is zero a.e.; however

� s

0�(u(t) · ∇)v(t), w(t)� dt

is not well defined if u, v, w ∈ L∞(0, T ; L2σ,n) ∩ L2(0, T ; V ) if Ω ⊂ R3. This holds, however,

for Ω ⊂ R2, which is why 2D Navier-Stokes is easier than 3D. See Proposition X.2.9 below.Proposition X.2.9. For n = 2 and Ω ⊂ R2 open with smooth boundary the map

(u, v, w) �→� T

0

�

Ωuα∂αvβwβ

is a trilinear continuous map onu ∈ L∞(0, T ; L2

σ,n).v, w ∈ L2(0, T ; V ).

The proof is based on a technique called compensated compactness or commutator estimates.The version we use here is due to [Coifman et al., 1993].Theorem X.2.10 (Coifman-Lions-Meyer-Semmes). Let A ∈ L2

σ,n(Rn,Rn), b ∈ H1(Rn).Then for any ϕ ∈ C∞

c (Rn) we have�

RnA · ∇b ϕ ≤ C�A�L2(Rn) �∇b�L2(Rn) [ϕ]BMO,

where BMO denotes the space of bounded mean oszillation, defined via the seminorm

[ϕ]BMO := supB

�

B|ϕ − (ϕ)B|,

the supremum being over all balls with finite radius in Rn.


Proof of Proposition X.2.9. We assume Ω = Rn, the other cases follow by ap-proximation (observe that u has zero boundary data). By Theorem X.2.10,

�

R2uα∂αvβwβ ≤ �u�L2(R2) �Dv�L2(R2) [w]BMO

From Poincare inequality it is elementary to observe (here we use the dimension n = 2)

[w]BMO � �∇w�L2(Rn) ≤ �w�H1(Rn).

So we have (by approximation) the estimate for almost all s,�

R2uα∂αvβwβ ≤ �u�L2(R2) �∇v�L2(R2) �∇w�L2(R2)

Thus,� T

0

�

R2uα∂αvβwβ ≤ �u�L∞((0,T );L2(R2)) �∇v�L2(0,T ;L2(R2) �∇w�L2(0,T ;L2(R2)).

�

Sketch of the proof of Theorem X.2.10. Theorem X.2.10 is a deep theorem ofharmonic analysis.

We present here a sketch of the proof of the following (weaker, but for our purposessufficient) estimate (same conditions on A, b, ϕ):

�

R2A · ∇b ϕ ≤ C�A�L2(R2) �∇b�L2(R2) �∇ϕ�L2(R2).

We follow essentially a proof by Brezis-Nguyen [Brezis and Nguyen, 2011], sharp esti-mates can also be obtained with these arguments [Lenzmann and Schikorra, 2019].

• A can be written as ∇⊥a :=�

−∂ya∂xa

�. This is called Hodge decomposition,

Helmholtz decomposition, or in this case the Poincare Lemma. The idea is thatwe can solve

Δa = ∂2A1 − ∂1A

2

then B := A − ∇⊥a is divergence free, div B = 0 but also curl-free, ∂1B2 −

∂2B1 = 0. One compute that this implies in particular that B is harmonic,

ΔB1 = ΔB2 = 0. If we choose the right boundary data for a above then we canensure that B = 0 on its boundary (i.e. decay at infinity), but then ΔB = 0implies B ≡ 0 by uniqueness of solutions. That is, A = ∇⊥a.

• Observe that ∇⊥a · ∇b = det(∇a,∇b).• Use the harmonic extension to R3

+, solve

Δα = 0 in R3+

α = a in R2 × {0}


Δβ = 0 in R3+

β = b in R2 × {0}

ΔΦ = 0 in R3+

Φ = ϕ in R2 × {0}

By Stokes’ theorem we then have (and this is a cancellation property)�

R2det(∇a,∇b)ϕ =

�

R3+

det(∇R3α,∇R3β,∇R3Φ).

Thus, by Holder’s inequality, we find the estimate�

R2A · ∇b ϕ ≤ C�∇α�L3(R3

+) �∇β�L3(R3+) �∇Φ�L3(R3

+).

Now harmonic extensions attain the trace inequality, i.e. one can show�∇α�L3(R3

+) ≈ [α]W 1,3(R3+)[a]

W 1− 13 ,3(R2)

,

and by Sobolev inequality for fractional Sobolev spaces[a]

W 1− 13 ,3(R2)

� �∇a�L2(R2) = �A�L2(R2).

We do this for α, β, ϕ and obtain the claim. �

The following is the time-independent version:

Corollary X.2.11. Let Ω ⊆ R2. Let u ∈ L2σ,n(Ω), v, w ∈ H1

0 (Ω), then

�(u · ∇)v, w� � �u�L2 �∇v�L2 �∇w�L2 .

Remark X.2.12. The div − curl argument is actually not necessary, one can base thison Gagliardo-Nirenberg inequality (in the context of Navier-Stokes equation also calledLadyzhenskaya inequality).

The following is essentially the equation of Definition X.2.7 for ϕ independent of time.

Lemma X.2.13. Let u be a weak solution of the Navier-Stokes equation as in Defini-tion X.2.7, then for all ϕ ∈ C∞

c,σ(Ω) (i.e. constant in time) and almost every t2 ≥ t1, foralmost every t1 ≥ 0 we have

(X.2.2)� t2

t1�∇u,∇ϕ� +

� t2

t1�u · ∇u, ϕ� = �u(t1), ϕ� − �u(t2), ϕ�.

Secondly, setX :=

�C∞

c,σ(Ω)�·�H1

0 (Ω)�∗

.


Then for almost every time t > 0 there exists an operator ∂tu ∈ X∗ such that for everyϕ ∈ C∞

c,σ(Ω)

�∂tu(t), ϕ� = lims→0

�u(t + s) − u(t)s

, ϕ�,and moreover(X.2.3) �∂tu, ϕ� + �∇u,∇ϕ� = �u · ∇u, ϕ�

Proof. Recall from Definition X.2.7, u is a weak solution if� s

0−�u, ∂tϕ� +

� s

0�∇u,∇ϕ� ds +

� s

0�(u · ∇)u, ϕ� ds = �u0, ϕ(0)� − �u(s), ϕ(s)�

for almost all s ∈ (0, T ) and all test functions ϕ ∈ Dσ, whereDσ = {ϕ ∈ C∞

c (Ω × [0,∞)) : div (ϕ(t)) = 0 ∀t ∈ [0,∞)} .

Applying this definition for s = t1 and s = t2 and subtracting we obtain� t2

t1−�u, ∂tϕ� +

� t2

t1�∇u,∇ϕ� ds +

� t2

t1�(u · ∇)u, ϕ� ds = �u(t1), ϕ(t1)� − �u(t2), ϕ(t2)�.

This holds in particular if we take ϕ(x, t) := η(t)ψ(x) for some η ≡ 1 in (t1, t2). In thatcase, ∂tϕ ≡ 0 in (t1, t2), consequently we have

� t2

t1�∇u,∇ψ� ds +

� t2

t1�(u · ∇)u, ψ� ds = �u(t1), ψ� − �u(t2), ψ�.

This establishes (X.2.2).

For the (X.2.3) we argue as follows: let now ϕ ∈ C∞c (Ω), div ϕ = 0. We set

G(s)[ϕ] := �∇u(s),∇ϕ� + �(u(s) · ∇)u(s), ϕ�then from (X.2.2) we have for almost every t and s,

�u(t + s) − u(t)

s, ϕ

�= 1

s

� t+s

tG(σ)[ϕ]dσ.

Observe that by Lemma X.2.4,

G(σ)[ϕ] ��∇u(σ)�L2(Ω) + �∇u(σ)�2

L2(Ω)

��∇ϕ�L2(Ω)

�1 + �u(σ)�2

H1(Ω)

��ϕ�H1(Ω)

So we have ��

�u(t + s) − u(t)

s, ϕ

�� 1

s

� s

t

�1 + �u(σ)�2

H1(Ω)

�dσ�

�ϕ�H1(Ω).

Recall that this holds whenever ϕ ∈ C∞c (Ω,R3) with div (ϕ) = 0. So if we set

X :=�

C∞c,σ(Ω)�·�

H10 (Ω)

�∗

X.4. UNIQUENESS OF WEAK SOLUTIONS IN DIMENSION TWO 170

then

Tt,sϕ := ϕ �→�

u(t + s) − u(t)s

, ϕ

�

belongs to X∗.

On the other hand, u ∈ L2((0, T ), H10 (Ω)), and thus by Lebesgue’s Theorem for almost any

t > 0 we have

lims→0

1s

� s

t

�1 + �u(σ)�2

H1(Ω)

�dσ =

�1 + �u(t)�2

H1(Ω)

�< ∞.

That is, for any such t the sequence (Tt,s)s∈(0,T −t) is a bounded sequence of elements in X∗,which is a reflexive Banach space. By reflexivity, X∗ = X∗∗ and X = X∗∗, there exists aweak limit operator ∂tu := Tt,0 ∈ X∗ such that for any ϕ ∈ X,

�∂tu, ϕ� := Tt,0[ϕ] = lims→0

�u(t + s) − u(t)

s, ϕ

�.

and on the other hand (by Lebesgue’s theorem)

�∂tu, ϕ� = lims→0

1s

� t+s

tG(σ)[ϕ]dσ = G(t)[ϕ].

This establishes the second part of the claim. �

X.3. Existence of weak solutions

See [Robinson et al., 2016, Theorem 4.4]

Theorem X.3.1. For each u ∈ L2σ,n(Ω), for Ω ⊆ Rn, n = 2, 3 either a smoothly bounded

domain or all of Rn, there exists a weak, global in time, solution u which for any T > 0belongs toL∞(0, T ; L2

σ,n(Ω)) ∩ L2(0,∞; V ) in the sense of Definition X.2.7.

X.4. Uniqueness of weak solutions in dimension two

Cf. [Robinson et al., 2016, Theorem 3.15 ]

Theorem X.4.1. Let Ω ⊂ R2 be a smooth bounded domain and assume that u0 ∈ L2σ,n(Ω).

If u and v both solve (X.1.1) in Ω × (0, T ) in the weak sense defined in Definition X.2.7,then u(t) ≡ v(t) for almost every t ∈ (0, T ).

Proof. Set w := u − v ∈ L∞(0, T ; L2σ,n(Ω)) ∩ L2(0, T ; V ).

By Lemma X.2.13 we have

−�∂tw, ϕ� = �∇w,∇ϕ� + �u · ∇u, ϕ� − �v · ∇v, ϕ�.


Plugging ϕ = w into Lemma X.2.13 (here we need a suitable approximation argumentwhich we drop for now) we obtain

12∂t�w�2

L2 = −�∇w�2L2 + �u · ∇u, w� − �v · ∇v, w�.

Observe that�u · ∇u, w� − �v · ∇v, w�

=�u · ∇w, w� + �w · ∇v, w�.By Lemma X.2.5, �u · ∇w, w� = 0. So, by Corollary X.2.11,

|�u · ∇u, w� − �v · ∇v, w�| � �w�L2 �∇u�L2 �∇w�L2 .

By Young’s inequality ab ≤ εa2 + C(ε)b2 we find that

∂t�w�2L2 + �∇w�2

L2 � 12 �∇w�2

L2 + �∇u�2L2 �w�2

L2 .

Absorbing to the left-hand side∂t�w�2

L2 � �∇u�2L2 �w�2

L2 .

By Gronwall’s inequality, Lemma IX.4.2,

(X.4.1) �w(t)�2L2 � �w(0)�2

L2 exp�� t

0�∇u(s)�2

L2ds�

Observe that (here is where we use 2D, so that �∇u�L2 has the right exponent!),

exp�� t

0�∇u(s)�2

L2ds�

≤ exp �∇u�L2(0,T ;L2(Ω) < ∞.

Moreover �w(0)�L2 = 0, so (X.4.1) implies w ≡ 0, i.e. u ≡ v. �

Instead of using above the Theorem X.2.10 one can also use the Gagliardo-Nirenberg-inequality (also referred to as Ladyzhenskaya-inequality in special cases):

Proposition X.4.2. For n < 4,

�w�L4(Rn) ≤ C �w�4−n

4L2(Rn) �∇w�

n4L2(Rn).

Proof. By Sobolev inequality, for all w ∈ W 1,2(Rn),

�w�L4(Rn) ≤ C��w�L2(Rn) + �∇w�L2(Rn)

�.

The claim now follows from scaling: Fix w ∈ W 1,2(Rn) and set for r > 0wr(x) := w(rx).

Apply the Sobolev inequality to this wr

�wr�L4(Rn) ≤ �wr�L2(Rn) + �∇wr�L2(Rn).

Now�wr�L4(Rn) = r− n

4 �w�L4(Rn)


�wr�L2(Rn) = r− n2 �w�L4(Rn)

and since (∇wr)(x) = r(∇w)(rx),�∇wr�L2(Rn) = r r− n

2 �∇w�L2(Rn),

so thatr− n

4 �w�L4(Rn) ≤ r− n2 �w�L2(Rn) + r1− n

2 �∇w�L2(Rn).

i.e.�w�L4(Rn) ≤ C

�r− n

4 �w�L2(Rn) + r1− n4 �∇w�L2(Rn)

�.

This holds for any r > 0, so let us choose

r := �w�L2(Rn)

�∇w�L2(Rn)

then�w�L4(Rn) ≤ 2C�w�

4−n4

L2(Rn) �∇w�n4L2(Rn).

�Remark X.4.3. If we apply Proposition X.4.2 in the proof of Theorem X.4.1 for estimatingthe crucial term

�w · ∇v, w�we find for n = 2, 3

|�w · ∇v, w�| � �w�2L4 �∇v�L2

��w�

4−n4

L2(Rn) �∇w�n4L2(Rn)

�2�∇v�L2

=�w�4−n

2L2(Rn) �∇w�

n2L2(Rn) �∇v�L2

Since we can absorb an �∇w�2L2(Rn)-term to the left-hand side, Young’s inequality ab ≤

εa4n + C(ε)b

44−n

|�w · ∇v, w�| � �w�2L4 �∇v�2

L2

≤ 12�∇w�2

L2(Rn) + C��w�

4−n2

L2(Rn) �∇v�L2

� 44−n

= 12�∇w�2

L2(Rn) + C �w�2L2(Rn) �∇v�

44−n

L2 .

The argument in the proof of Proposition X.4.2 thus will lead to an estimate

∂t�w�2L2 + c�∇w�2

L2 � C �w�2L2(Rn) �∇v�

44−n

L2

And Gronwall’s inequality, Lemma IX.4.2 will lead to the analogue of (X.4.1),

�w(t)�2L2 � �w(0)�2

L2� �� =0

exp�� t

0�∇u(s)�

44−n

L2 ds�

X.5. SOME REGULARITY-TYPE THEORY FOR N = 2 (GLOBAL), N = 3 (DOMAINS) 173

What is the problem? The problem is that in order to conclude that w ≡ 0 we need toensure that � t

0�∇u(s)�

44−n

L2 ds < ∞

For n = 2 this follows from u ∈ L2(0, T ; H1). For n = 3 we would need u ∈ L4(0, T ; H1)(which we don’t have).

We see that for n = 2 the Navier-Stokes equation is critical because the “nonlinearity” iscontrolled by L2(0, T ; H1) which is what the ∂t −Δ-leading order term controls. For n = 3the energy is supercritical because the nonlineary needs to be controlled L4(0, T ; H1) whilethe parabolic operator only provides L4(0, T ; H1).

Notice that essentially all our argument relied on scaling (scaling invariant Sobolev in-equalities, Gagliardo-Nirenberg) etc. – One can flip through the spaces, i.e. replace H1

with something else, but the scaling argument will always lead to a problem for n = 3.

[Tao, 2016] calls this the “super-criticality” barrier for the the Navier-Stokes problem in3D, Conjecture X.1.1, – and he argued that in order to solve the Navier-Stokes problemin the positive one should need more than just sharp embedding theorems to rule outsingularities – and if one wants to solve the Navier-Stokes problem in the negative, onecan make use of that supercriticality.

We see in the above argument the difference between 2D and 3D. This might become moreclearer via the (instead of the div-curl estimate):

X.5. Some Regularity-type theory for n = 2 (global), n = 3 (domains)

To illustrate the differences in two dimensions and three dimensions we show some improvedregularity arguments for the following two cases

Definition X.5.1. For n = 2, 3 we treat the case

• Ω � Rn smoothly bounded set• Ω = R2.

First we observe (we essentially did this before in the motivation for weak solution) thatLemma X.2.13 implies some initial estimates.

Proposition X.5.2. Let n = 2, 3, Ω as in Definition X.5.1.

Assume for almost every t ∈ (0, T ) the map u ∈ L∞((0, T ), L2σ,n) ∩ L2((0, T ), H1

0 (Ω))satisfies

−�u(t), ∂tϕ� + �∇u(t),∇ϕ� = �u(t) · ∇u(t), ϕ�for all ϕ ∈ C∞

c (Ω,Rn) with div (ϕ) = 0.


Thensup

t�u(t)�L2 +

� t

0�∇u(s)�2

L2 ds � �u(0)�L2(Ω)

Proof. If Ω � Rn is bounded (with smooth boundary), we observe that (in a distribu-tional sense) ∂tu exists and belongs to H1

0 (Ω)∗, because thats what is true for �∇u(t),∇ϕ�and �u(t) · ∇u(t), ϕ�. If Ω = R2 this is still true by Corollary X.2.11. See Lemma X.2.13.

Since moreover u ∈ H10 (Ω) we can approximate it by uk ∈ C∞

c (Ω) with respect to theH1-norm. By Helmholtz decomposition, Theorem X.2.3

uk = ϕk + ∇gk

with ϕk ∈ L2σ,n(Ω) and gk ∈ H1(Ω).

By density we can use ϕk as a testfunction,−�∂tu(t), ϕk� + �∇u(t),∇ϕk� = �u(t) · ∇u(t), ϕk�

Observe that gk satisfies

Δgk = div (uk) in Ω,

∂νgk = 0 on ∂Ω.

Since div (uk) → 0 in L2(Ω), by a reflection argument ∇gk → 0 in W 1,2(Ω). By density,thus �

Ωu · ∇Δgk = 0,

moreover �

Ω∂tu · ∇gk = 0.

and either by div-curl (n = 2, global) or boundedness (Poincare),

�u · ∇u, gk� k→∞−−−→ 0.

Thus, we can test with uk, and in the limit with u, and find∂t�u(t)�2

L2 + �∇u(t)�2L2 = 0.

Integrating this we have

�u(t)�2L2 +

� t

0�∇u(s)�2

L2ds ≤ �u(0)�L2 .

�

When passing to higher order estimates the pressure comes into play. Observe that theequation

∂tu − Δu + u · ∇u = ∇p

implies (together with div (u) = 0) thatΔp = div (u · ∇u) = ∂αβ(uαuβ).


The problem is that we have no boundary data given for p.

In R2 we can argue as follows: by CLMS�

R2∇p · ∇ϕ =

�

R2∂αuβ∂βuαϕ � �∇u�2

L2 �∇ϕ�L2 .

By duality we obtain (using implicitely that p decays at infinity which can be deducedessentially from the fact that u ∈ L2)(X.5.1) �∇p�L2(R2) � �∇u�2

L2(R2).

For n = 3, in a bounded domain, it is more complicated (see [Robinson et al., 2016,Theorem 5.7]), in this case (cf. [Robinson et al., 2016, Theorem 6.12]) we assume apriori that the following formal estimate holds:(X.5.2) �∇p�L2(Ω) � �u�L4Ω �∇u�L4(Ω) � �∇u�L4(Ω).

Proposition X.5.3. Let n = 2, 3, Ω as in Definition X.5.1.

Assume for almost every t ∈ (0, T ) the map u ∈ L∞((0, T ), L2σ,n) ∩ L2((0, T ), H1

0 (Ω))satisfies

−�u(t), ∂tϕ� + �∇u(t),∇ϕ� = �u(t) · ∇u(t), ϕ� + �∇p, ϕ�for all ϕ ∈ C∞

c (Ω,Rn) (without the assumption div (ϕ) = 0) for p satisfying (X.5.1) (forn = 2) or (X.5.2) for n = 3.

If moreover�u(0)�L2 + �∇u(0)�L2 < ε

thensup

t�∇u(t)�2

L2 +� t

0�∇2u(s)�2

L2 ds � C(�u�H2)

Proof. Test with Δu (here we loose that Δu is divergence free as in the previousarguments, thats why we deal with ∇p), then

��∂t�∇u�2L2 + �Δu�2

L2

�� ≤ |�(u · ∇)u, Δu�| + |�∇u, Δu�|In n = 3 but a bounded domain we get

|�(u · ∇)u, Δu�| ≤�u�L∞�∇u�L2 �∇2u�L2

Now we use the so-called Agmon’s inequality (this clearly holds only in a bounded set), cf.Proposition X.4.2,

�u�L∞(Ω) ≤C �∇u�12L2(Ω) �u�

12H2(Ω) � �u�L2 �∇u�

12L2 + �∇u�

12L2 �∇2u�

12L2

��∇u�32L2 + �∇u�

12L2 �∇2u�

12L2

From this we find with the help of Youngs inequality etc. for p > 4(X.5.3) ∂t�∇u�2

L2 ≤ C�∇u�pL2 − c�Δu�2

L2 .


For n = 3 and on bounded domains we can estimate by Sobolev (not Poincare)�Δu�2

L2 � �∇u�L2 ,

so we have∂t�∇u�2

L2 ≤ C �∇u�2L2 (�∇u�p−2

L2 − c)Since the initial data is small, ∂t�∇u�L2 ≤ 0, thus

supt>0

�∇u�2L2(t) ≤ �∇u(0)�L2 .

Using this estimate we also obtain by integrating (X.5.3) again using Proposition X.5.2� T

0�Δu�2

L2 ≤C(u(0)).

For n = 2 but on R2 we use CLMS and integration by parts to obtain|�(u · ∇)u, Δu�| ≤ |�(u · ∇)∂αu, ∂αu�|+|�(∂αu · ∇)u, ∂αu�| � �∇u�2

L2 �∇2u�L2+�u�L2 �∇2u�2L2 .

In view of (X.5.1) we have the same estimate for|�∇p, Δu�| � �∇u�2

L2 �∇2u�L2 .

By Proposition X.5.2 and the smallness condition we have�u�L2 ≤ ε,

so we can absorb as in the 3D-case and find(X.5.4)

��∂t�∇u�2L2 + c�Δu�2

L2

�� ≤ C�∇u�4L2 .

Now we can argue as in the 3D case to get the claim. �

CHAPTER 11

Short introduction to Calderon-Zygmund Theory

Calderon-Zygmund theory is the Lp-regularity theory for elliptic equations. For exampleassume that u solves

Δu = ∂βf in Rn

and that f ∈ Lp(Rn), p ∈ (1,∞). We would like to conclude that ∇u ∈ Lp(Rn) with theestimates

�∇u�Lp � �f�Lp .

Example: Assume that u ∈ W 1,2 solves

Δu = ∂αu

then we would like to conclude that u is smooth and a classical solution. Observe thatu ∈ L2∗ by Sobolev embedding so we would like to conclude that ∇u ∈ L2∗ and thenbootstrap our way to smoothness of u.

This theory is closely connected with harmonic analysis and Calderon-Zygmund operators.Denote by I2 = (−Δ)−1 the Riesz potential (we called this Newton potential before) (weassume for simplicity that n ≥ 3), we have the formula (I.2.4)

I2g(x) = c�

Rn|x − y|2−n g(y) dy.

Then,∂αu = ∂αΔ−1Δu = ∂αΔ−1∂βf.

Computing the derivative we find that

(XI.0.1) ∂αΔ−1∂βf(x) = c�

Rn

(x−y)α

|x−y|(x−y)β

|x−y||x − y|n f(y) dy.

We will see below that this operator is an Calderon-Zygmund operator which as such is abounded linear operator from Lp to Lp, namely

�∂αΔ−1∂βf�Lp(Rn) � �f�Lp(Rn) ∀f ∈ Lp(Rn), p ∈ (1,∞).

From this we obtain immmediately that

�∇u�Lp(Rn) � �f�Lp(Rn).

In the following we make these statements precise.177

XI.1. CALDERON-ZYGMUND OPERATORS 178

XI.1. Calderon-Zygmund operators

The typical Calderon-Zygmund operator is the Riesz transform

Rαf(x) := c�

Rn

(x−y)α

|x−y||x − y|n f(y)dy.

One can compute that the Fourier symbol of Rα is i ξα

|ξ| , α = 1, . . . , n, i.e.

(Rαf)∧(ξ) = iξα

|ξ| f∧(ξ).

In particular we have that ∂α∂βf = cRαRβΔf , which is what we used in (XI.0.1).

Observe that the symbol of Rα belongs to L∞(Rn),��i

ξα

|ξ|

��L∞

≤ 1.

It is easy to show that such an operator is bounded on L2:Lemma XI.1.1. Let m ∈ L∞(Rn) and define

Tf := (m(ξ)f∧(ξ))∨.

Then T is a linear bounded operator on L2(Rn) with�Tf�L2(Rn) � �m�L∞�f�L2(Rn).

Such a T is usually called a multiplier operator, and m is the symbol.

Proof. By Plancherel identity, �g�L2(Rn) = �g∧�L2(Rn). Thus,�Tf�L2(Rn) = �(Tf)∧�L2(Rn) = �m f∧�L2(Rn) ≤ �m�L∞ �f∧�L2 = �m�L∞ �f�L2 .

�

Observe that we cannot simply replace L2 with Lp in Lemma XI.1.1, since there is noPlancherel identity on Lp for p �= 2.Theorem XI.1.2 (Boundedness of Calderon-Zygmund-Operators). Let T : L2(Rn) →L2(Rn) be a bounded linear operator, which for f ∈ C∞

c (Rn) can be written as

Tf(x) =�

Rn

Ω(x − y)|x − y|n f(y) dy

(in a principle value sense). If moreover, the kernel Ω satisfies

• Ω : Rn\{0} → R is bounded, �Ω�L∞ < ∞

• Ω is homogeneous of order 0, i.e. Ω(rz) = Ω(z) for all r > 0, z ∈ Rn\{0}.• Ω : Rn\{0} is Lipschitz with the bound supz∈Rn\{0} |z||∇Ω(z)| < ∞


then T is1 a bounded linear operator from Lp(Rn) → Lp(Rn).

Lemma XI.1.3. Let f ∈ L2(B(r)) then for all c ∈ R,�f − (f)B(r)�L2(B(r)) ≤ �f − c�L2(B(r)).

Proof. Exercise! �

We are not proving Theorem XI.1.2 in its full generality, but only for the case we need (arelatively easy adaptation of the following does the job).

Proposition XI.1.4. . For a monomial p of degree k let

Tf :=�

Rn

p(x − y)|x − y|n+k

f(y) dy

Then T is a linear bounded operator from L2(Rn) to L2(Rn), and for f ∈ L∞(Rn)∩L2(Rn),[Tf ]BMO � �f�L∞(Rn).

Here

[g]BMO = supB(x,ρ)

��

B(x,ρ)|g − (g)B(x,ρ)|2

� 12

.

Proof. The L2-boundedness follows from Lemma XI.1.1.

For x0 ∈ Rn and r > 0 let fr,x0(x) := f(x0 + rx). Observe that by the structure of T ,T (fr,x0)(x) = (Tf)(x0 + rx).

This implies that �

B(0,1)T (fr,x0) =

�

B(x0,r)Tf,

and �

B(0,1)

��T (fr,x0) − (T (fr,x0))B(0,1)��2

=�

B(x0,r)

��T (f) − (T (f)B(x0,r)��2

,

and�fr,x0�L∞(Rn) = �f�L∞(Rn).

Thus, if we can show that for any f ∈ L2 ∩ L∞(Rn)�

B(0,1)|Tf − (Tf)B(0,1)|2 � �f�2

L∞(Rn),

then the full claim follows via scaling and translation.

Now letf := f1 + f2,

1more precisely: extends to


with f1 = χB(0,2)f and f2 = χRn\B(0,2)f . Then,�

B(0,1)|Tf − (Tf)B(0,1)|2 �

�

B(0,1)|Tf1 − (Tf1)B(0,1)|2 +

�

B(0,1)|Tf2 − (Tf2)B(0,1)|2

�2�

B(0,1)|Tf1|2 +

�

B(0,1)|Tf2 − (Tf2)B(0,1)|2.

Observe that by the L2-boundedness, Lemma XI.1.1,�

B(0,1)|Tf1|2 � �Tf1�2

L2(Rn) � �f1�2L2(B(0,2)) = �f�2

L2(B(0,2)) � �f�2L∞(Rn).

Now in view of Lemma XI.1.3,

(XI.1.1)�

B(0,1)|Tf2 − (Tf2)B(0,1)|2 �

�

B(0,1)|Tf2 − Tf2(0)|2

Now,

Tf2(x)−Tf2(0) =�

Rn

�p(x − y)

|x − y|n+k− p(−y)

|y|n+k

�f2(y) dy =

�

Rn\B(0,2)

�p(x − y)

|x − y|n+k− p(y)

|y|n+k

�f(y) dy

If x ∈ B(0, 1) and y �∈ B(0, 2) then |x − y| ≈ |y| � 1. In this case we obtain from thefundamental theorem of calculus that

�p(x − y)

|x − y|n+k− p(−y)

|y|n+k

�� |x|

|x − y|n+1 .

Consequently,

|Tf2(x) − Tf2(0)| � |x|�

Rn\B(0,2)|x − y|−n−1 |f(y)| dy � �f�L∞(Rn)

�

|x−y|�1|x − y|−n−1 dy.

Observe thatsup

x

�

|x−y|�1|x − y|−n−1 dy < ∞.

So we have shown thatsup

x∈B(0,1)|Tf2(x) − Tf2(0)| � �f�L∞(Rn).

which together with (XI.1.1) implies�

B(0,1)|Tf2 − (Tf2)B(0,1)|2 � �f�L∞(Rn).

Thus we have shown �

B(0,1)|Tf − (Tf)B(0,1)|2 � �f�L∞(Rn),

which by the scaling argument leads to the claim. �

Why are we happy about Proposition XI.1.4? Because BMO represents “almost L∞”, andwe have

XI.2. W 1,P -THEORY FOR THE LAPLACE EQUATION 181

Theorem XI.1.5. Let 1 ≤ p < ∞ and T be a linear operator of “strong (p,p)-type”,meaning that

�Tf�Lp(Rn) � �f�Lp(Rn) ∀f ∈ Lp(Rn),and bounded from L∞ to BMO, i.e

[Tf ]BMO � �f�L∞(Rn) ∀f ∈ L∞(Rn),Then for any q ∈ [p,∞), T maps Lq(Rn) into Lq(Rn) with

�Tf�Lq(Rn) � �f�Lq(Rn) ∀f ∈ Lq(Rn).

Proof. [Giaquinta and Martinazzi, 2012, Theorem 6.29] �

Proof of Theorem XI.1.2. Observe that T is bounded from L2 to L2, Lemma XI.1.1,and from L∞ to BMO, Proposition XI.1.4, and thus by Theorem XI.1.5 for any p ∈ [2,∞)we have

�Tf�Lp(Rn) ≤ Cp �f�Lp(Rn).

For p < 2 we argue by duality. Observe that by Riesz Representation Theorem

�Tf�Lp(Rn) = sup�g�

Lp� (Rn)

�

RnTf g = sup

�g�Lp� (Rn)

�

Rnf T ∗g � �f�Lp(Rn) �T ∗g�Lp� (Rn).

Now observe that T ∗ is of the same type of operator, so we have for any q ∈ [2,∞) we have�T ∗g�Lq(Rn) ≤ Cp �g�Lq(Rn).

Since for p < 2 we have that q := p� > 2 this concludes the proof. �Remark XI.1.6. There is another, older, way, using the Calderon-Zygmund decomposi-tion and an L1-L1-weak type estimate to obtain Theorem XI.1.2.

XI.2. W 1,p-theory for the Laplace equation

Theorem XI.2.1. Let Ω1 ⊂⊂ Ω ⊂⊂ Rn be two smoothly bounded domains, and let p ≥ 2.

Assume that for some f ∈ Lp(Ω) there is u ∈ W 1,2(Ω) that satisfies in distributional senseΔu = ∂αf in Ω

Then�∇u�Lp(Ω1) ≤ C(Ω1, Ω, p)

��f�Lp(Ω) + �u�L2(Ω)

�.

Remark XI.2.2. • The L2-norm for u on the right-hand side is necessary, sinceotherwise f = 0 would imply that u is constant (which is false without the as-sumption of appropriate boundary values).

• This statement holds for more general equations, e.g. ∂i(Aij∂ju) = ∂αf , if A issmooth enough (the sharp assumption being V MO, [Iwaniec and Sbordone, 1998])


• This is an interior statement, but it holds up to the boundary: for example if

Δu = ∂αf in Ω2

u = 0 on ∂Ω2

then �∇u�Lp(Ω2) ≤ �f�Lp(Ω2); Cf. [Giaquinta and Martinazzi, 2012, Chapter7].

The proof of Theorem XI.2.1 follows a sequence of cutoff arguments, such as the following

Lemma XI.2.3. For p, q ∈ [2,∞) assume that u ∈ W 1,p(Ω) satisfies for some f ∈ Lq(Ω)

Δu = ∂αf in Ω

Let η ∈ C∞c (Ω). Then for v := ηu we have

Δv = g + ∂αf in Rn

with�f�Lq(Rn) � �f�Lq(Ω),

and for 1 ≤ r ≤ min{p, q} we have

�g�Lr(Rn) � �f�Lq(Ω) + �u�W 1,p(Ω)

Moreover f and g have compact support, so does v and we have

�v�W 1,p(Rn) � �u�W 1,p(Ω).

All the constants depend on η.

Proof.Δv = (Δη)u

� �� Lp(Rn)

+ 2∇η · ∇u� �� Lp(Rn)

− (∂αη) f� ��

Lq(Rn)

+∂α( ηf��Lq(Rn)

)

�

Moreover we use the following global result:

Proposition XI.2.4. Assume that v ∈ W 1,p(Rn),

Δv = g + ∂αf in Rn

with f ∈ Lq(Rn), g ∈ Lr(Rn) all with compact support.

Then for 1 < σ ≤ q and if r < n additionally σ ≤ nrn−r

�∇v�W 1,σ(Rn) � C(f, g).


Proof. By the compact support of v we only need to estimate ∇v (the rest followsfrom Poincare). By the boundedness of the Riesz transform,

�∇v�Lσ(Rn) � �(−Δ) 12 v�Lσ(Rn).

Now we estimate for any Φ ∈ C∞c (supp (v),Rn) with �Φ�Lσ� (Rn) ≤ 1 the expression

�

Rn∇v · Φ

Setϕ := I1RβΦβ.

By the compact support of Φ one can show that ϕ ∈ Lσ�(Rn), by the boundedness of theRiesz transforms ∇ϕ ∈ Lσ�(Rn). Moreover ∇ϕ ∈ Lσ(Rn) for any σ ∈ (1, σ�) by bound-edness of Riesz transform and the compact support of Φ. In particular ϕ ∈ W 1,σ�(Rn) ≡W 1,σ�

0 (Rn), and �

Rn∇v · Φ =

�

Rn(−Δ) 1

2 v (−Δ) 12 ϕ = c

�

Rn∇v ∇ϕ

the last equality can be seen by Fourier transform. From the equation of v we get�

Rn∇v ∇ϕ =

�

Rnf∂αϕ + gϕ

��f�Lσ(Rn)�∂αϕ�Lσ� (Rn) + �g�Lr(Rn) �ϕ�Lr� (Rn).

Since σ ≤ q, using compact support of f ,

�f�Lσ(Rn)�∂αϕ�Lσ� (Rn) � �f�Lq(Rn) �ϕ�W 1,σ� (Rn)� �� ≤1

.

As for the second term, if r < n then since σ < nrn−r

we have that nσ�n−σ� > r� and by Sobolev

embedding (since also ∇ϕ ∈ W 1,σ(Rn) for any σ ∈ (1, σ�))

�g�Lr(Rn) �ϕ�Lr� (Rn) � �g�Lr(Rn) �ϕ�W 1,σ� (Rn)� �� 1

.

We have shown that for any Φ ∈ C∞c (supp v,Rn), �Φ�Lσ� (Rn) ≤ 1 we have

�

Rn∇v · Φ �

��g�Lr(Rn) + �f�Lq(Rn)

�.

By duality/Riesz representation theorem this means that ∇v ∈ Lσ(Rn) and

�∇v�Lσ(Rn) ��g�Lr(Rn) + �f�Lq(Rn)

�.

Since v has compact support, by Poincare inequality, we finally obtain

�v�W 1,σ(Rn) ��g�Lr(Rn) + �f�Lq(Rn)

�.

�


Proof of Theorem XI.2.1. Let Ω ⊃⊃ Ω1 ⊃⊃ Ω2 ⊃⊃ . . . and take ηi ∈ C∞c (Ωi)

with η ≡ 1 in Ωi+1.

Following Lemma XI.2.3 and Proposition XI.2.4 we obtain thatη1u ∈ W 1,σ1(Ω1)

where we take σ1 ≤ p and σ1 ≤ 2nn−2 (if n ≥ 3). If we can take σ1 = p we are done,

otherwise we observe that σ1 = 2nn−2 > 2. We then repeat the argument for η2η1u: from

Lemma XI.2.3 and Proposition XI.2.4 we then obtainη2η1uW 1,σ2(Rn)

for σ2 ≤ p and σ2 ≤ σ1nn−σ1

(if σ1 < n). Again, either we can choose σ2 = p or σ2 = σ1nn−σ1

.

In this way we obtain a sequencevk := ηkηk−1 . . . η1u ∈ W 1,σk(Rn)

where

σk =

p if σk−1 < n or p ≤ σknn−σk

σk−1nn−σk−1

else.

This sequence terminates after finitely many steps. Indeed, let

σk :=

σk−1nn−σk−1

if σk−1 < n

∞ otherwise.

The sequence σk is increasing, σk ≥ σk−1 and strictly increasing unless σk = n. The onlypossibility that σk is not ∞ after finitely many steps k, is that σk < n for all k – then wehave a monotone, bounded sequence which has a limit σ ≤ n, which has to satisfy

σ = σn

n − σ

There is no positive, finite solution to this equation. Contradiction. So σk is infinite fork ≥ K for some K, which means that σk = p for k ≥ K.

That is, we have shown that vK ∈ W 1,p(Rn), and since vK ≡ u in ΩK we have u ∈ W 1,p(Rn)in ΩK .

The number K is independent of the equation, it just depends on the dimension and p, soif we choose ΩK well, then we get the claim. �

CHAPTER 12

Short introduction to Viscosity solutions

Cf. Koike

• There is no C1-solution to the equation

|u�(t)|2 = 1 t ∈ (−1, 1)u(−1) = u(1) = 0.

• If we consider a.e. solutions u ∈ C0,1 there are many.

u1(t) =

1 + t t ∈ (−1, 0)1 − t t ∈ [0, 1).

oru2(t) = −u1(t)

but also really any saw-tooth-shaped function is a solution.• In particular there are sequences of solutions approximating the function u ≡ 0

(which in no way is a solution)

u1 and u2 are special solutions, they are in some sense maximal. Why do we choose u1over u2? well this is a choice of convexity we make (and for our purposes we will want tomake things concave).

In order to define notion a solution one can use the vanishing viscosity method: one addsa viscosity term −εu�� to the equation, i.e.

−εu��

ε(t) + |u�ε(t)|2 = 0 t ∈ (−1, 1)

uε(−1) = uε(1) = 0.

There is one unique solution to this equation, and it is

uε(x) = −ε log�

cosh(x/ε)cosh(1/ε)

�.

One can show that uεε→0−−→ 1 − |x| uniformly in [−1, 1].

We repeat this argument for general second-order PDEs(XII.0.1) F (x, u, Du, D2u) = 0 in Ω.

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Partial Diﬀerential Equations