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Elliptic Partial Differential Equations
Leonardo Abbrescia
Advised by Daniela De Silva, PhD and Ovidiu Savin, PhD.
Contents
1 Introduction and Acknowledgments 2
2 Variational Methods and Sobolev Embedding Theorems 32.1 Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 A non-linear PDE and Method of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2.1 Non-linear variations on Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.2 Regularity Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3 Harnack Inequality for Divergence Equations 233.1 Regularity Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Harnack Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2.1 Structural Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2.2 Moser Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2.3 Applications of the Harnack Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4 Harnack Inequality for Non-Divergence Equations 384.1 ABP Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.2 Measure theory and Harnack Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
5 Curved C1,α Domains 445.1 Estimate for Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.2 General Elliptic Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
1
1 Introduction and Acknowledgments
Second order elliptic partial differential equations are fundamentally modeled by Laplace’s equation ∆u = 0.This thesis begins with trying to prove existence of a solution u that solves ∆u = f using variational methods.In doing so, we introduce the theory of Sobolev spaces and their embeddings into Lp and Ck,α. We thenmove on to applying our techniques to a non-linear elliptic equation on a compact Riemannian manifold. Weintroduce the method of continuity along the way to provide another way of solving the equation. We moveonto proving Schauder estimates for general elliptic equations in divergence form: ∂i(aij∂ju) + c(x)u = fwith various assumptions on a, c, and f . We conclude our study of equations in divergence form by provingthe Harnack Inequality using Moser iteration. Personally I would have liked to have proved the Harnackinequality in my own flavor, but due to lack of time, I had to follow very closely the proof given in [1].
The second half of the thesis revolves around equations in non-divergence form: aijuij = 0. As adisclaimer, I wrote the second half separately from the first and so my notations change heavily. We firststart with the proof of the ABP maximum principle which is used heavily in, not only the proof for theHarnack inequality for non-divergence equations, but for the section on curved C1,α domains. In thatsection I go over a completely new way of getting regularity estimates: approximation by polynomials. Wefirst show how this can be used for ∆u = f and then for general elliptic operators. We conclude the paper byintroducing Krylov’s regularity results for flat domains and generalize it to curved domains whose boundarieslook locally like the graph of a C1,α function.
As one more disclaimer, I was not able to prove every single detail in this book due to lack of time. Ileave tiny bits and pieces as exercises for the reader, but the overwhelming majority is proved in rigorousdetail.
The writing of this paper was a long and arduous process. It grew out of many discussions with ProfessorDaniela De Silva and Professor Ovidiu Savin. I am very grateful for their insights and very very patientguidance. I would not have been able to set my path as a mathematician without them. I’d like to givethanks to Professor Michael Weinstein for being an exceptional guidance counselor and instructor whom I’velearned a lot from about singular integrals, methods of characteristics, and graduate schools. I’d like to givespecial thanks to Professor Duong Phong, whose Analysis II class helped refine the details of this thesis.
2
2 Variational Methods and Sobolev Embedding Theorems
2.1 Laplacian
We begin with the simplest problem. Let Ω ⊂⊂ Rn be a bounded domain, and f a function on Ω. We wishto find a u such that
∆u = f in Ωu = 0 on ∂Ω
The way we will find the solution to this problem is by finding the minimum for a specific functional. Thisis the idea: let g(x) be a function on Ω ⊂ R such that ∃G(x) on Ω with
G′(x) = g(x)
and we want to find an x0 with g(x0) = 0. Then one way to approach this problem is to find an x0 ∈ Ωsuch that G(x) ≥ G(x0) ∀x ∈ Ω =⇒ G′(x0) = 0 =⇒ g(x0) = 0. By comparisons, g(x0) = 0 would be theequivalent of ∆u = 0. So now we try to find the equivalent of G. Define the functional
I(u) =1
2
ˆΩ
|Du|2 +
ˆΩ
fu.
This functional is going to be defined for u ∈W 1,2(Ω). Now lets go over some definitions.
Definition 2.1. Let B be a normed Banach space. Then its completion B :=uk ⊂ B, uk is cauchy
.
Example 2.2. Q = R.
Example 2.3. C∞0 (Ω), ‖u‖p <∞ = Lp(Ω) (modulo the equivalence that f ≡ g if f 6= g on a set ofmeasure zero).
Example 2.4. W 1,20 (Ω) = uk ∈ C∞0 , ‖uk − ul‖2 → 0, ‖Duk − Dul‖2 → 0. Also, W 1,2
0 (Ω) = u ∈L2(Ω), L2(Ω) 3 u = limk→∞ uk, ‖Duk −Dul‖2 → 0, Duk → v ∈ L2..
Now we go back to our question: does there exist a u0 ∈ W 1,20 (Ω) such that I(u) ≥ I(u0) for any
u ∈ W 1,20 (Ω)? We will begin to show this in two steps. Our first is to show that our functional is bounded
below. We will show that ∃C > 0 such that I(u) ≥ −C for any u ∈ W 1,20 (Ω). This will at least give us a
starting point to find a minimum because we no longer have the ambiguity of I(u) exploding to −∞. Next,assuming this, if I(u) did have a minimum, then min I(u) < ∞. Now pick a minimizing sequence ujsuch that I(uj)→ min I(u). The reason why this can be picked will be shown later. We wish to show thatuj → u0 and I(u0) = min I(u).
Claim. We claim that ∃C > 0 such that I(u) ≥ −C ∀u ∈W 1,20 (Ω).
First recall the definition of I:
I(u) =1
2
ˆΩ
|Du|2dx+
ˆΩ
fudx.
Then applying Holder’s inequality and Cauchy’s inequality, we have
I(u) ≥ 1
2
ˆΩ
|Du|2dx− ‖f‖2‖u‖2
≥ 1
2
ˆΩ
|Du|2dx−(ε
2‖u‖22 +
1
2ε‖f‖22
).
Now note that we would be done with our claim if we show that 12‖Du‖
22 ≥ ε
2‖u‖22. This is saying that the
gradient controls our function u. But we are in luck because we choose u to have compact support, so it iszero on the boundary. Then this implies that the gradient not only approximates values near u, but tells uswhat they are.
3
Lemma 2.5. There exists ε = ε(Ω) such that ε2‖u‖
22 ≤ 1
4‖Du‖22.
Proof. First note that this will imply
I(u) ≥ˆ
Ω
(1
4|Du|2 − 1
2εf2
)dx.
Now we move onto a claim:
Claim. Let Ω be convex, and bounded in Rn. Then I will show that ∀u ∈ C∞(Ω), ∀x ∈ Ω, thenˆ
Ω
|u(x)− u(y)| dy ≤ (diam Ω)n
n
ˆΩ
|Du(y)||x− y|n−1
dy.
Proof of Claim. Let x, y ∈ Ω be arbitrary, and denote r = |x− y|. Let ω = (y−x)/|x− y| be the unit vectorin this direction. Then we have that
u(y)− u(x) =
ˆ r
0
d
dtu(x+ tω) dt
=
ˆ r
0
n∑i=1
∂u
∂xi(x+ tω)ωi dt
|u(x)− u(y)| ≤ˆ r
0
|Du(x+ tω)||ω| dtˆ
Ω
|u(x)− u(y)| dy ≤ˆ
Ω
ˆ r
0
|Du(x+ tω)| dt dy.
We transform the RHS of this integral inequality using the polar transformation centered at x: y 7→ (r, ω).Let `ω denote the distance from x to ∂Ω.
ˆΩ
|u(x)− u(y)| dy ≤ˆSn−1
ˆ `ω
0
[ˆ r
0
|Du(x+ tω)| dt]rn−1dr dσ(ω)
=
ˆSn−1
ˆ `ω
0
[ˆ r
t
rn−1 dr
]|Du(x+ tω)| dt dσ(ω)
≤ˆSn−1
ˆ `ω
0
`nωn|Du(x+ tω)|dt dσ(ω)
≤ (diam Ω)n
n
ˆSn−1
ˆ `ω
0
|Du(x+ tω)| dt dσ(ω)
=(diam Ω)n
n
ˆSn−1
ˆ `ω
0
|Du(x+ tω)|tn−1
tn−1 dt dσ(ω)
=(diam Ω)n
n
ˆΩ
|Du(x+ tω)|tn−1
dy
=(diam Ω)n
n
ˆΩ
|Du(y)||x− y|n−1
dy.
With the proof of the claim out of the way, we can prove the following corollary:
Corollary 2.6. Let u ∈ C∞0 (Ω). Then for any x ∈ Rn, we have
|u(x)| ≤ cnˆRn
|Du||x− y|n−1
dy.
4
Proof of Corollary. We begin by introducing a little bit of notation:
uΩ(y) :=
u(y) dy.
Then we have for Ω = BR(0)
|u(x)− uBR(0)| =1
|BR(0)|
∣∣∣∣∣ˆBR(0)
(u(x)− u(y)) dy
∣∣∣∣∣≤ 1
|BR(0)|
ˆBR(0)
|u(x)− u(y)| dy
≤ 1
|BR(0)|(2R)n
n
ˆBR(0)
|Du(y)||x− y|n−1
dy
= cn
ˆBR(0)
|Du(y)||x− y|n−1
dy.
Notice that we will be done with our corollary if we show that uBR(0) → 0 as R→∞. But this is where weuse the fact that u has compact support:
uBR(0) =
BR(0)
|Du(y)||x− y|n−1
dy =
Rn
|Du(y)||x− y|n−1
dy → 0 as R→∞.
Now that we are done with this corollary, we will quote a lemma that will be proved later:
Lemma 2.7 (Estimates for Integral Operator). Assume
|u(x)| ≤ˆK(x, y) · |v(y)| dy.
Then for any 1 ≤ p ≤ ∞, we have ˆ|u(x)|p dx ≤ A
ˆ|v(y)|p dy
where
A = max
supx
ˆK(x, y) dy, sup
y
ˆK(x, y) dx.
.
Applying this lemma to Corollary 2.6 gives us ‖u‖p ≤ C‖Du‖p for u ∈ C∞0 (Ω) (and in fact for u ∈W 1,20 (Ω)
by approximations). We are finally done with our Lemma 2.5.
Now that we are done with the first part of the problem, we go back to the infimum question. Assumewe know that infu∈W 1,2
0 (Ω) I(u) > −∞ and pick a minimizing sequence, i.e. uk ∈ W 1,20 (Ω) such that
I(uk) → inf I(u). By approximations, we may assume uk ∈ C∞0 (Ω). The question now is do the uk’sconverge? To do this we have to go through a few things:
Claim. I claim that uk is a bounded sequence, i.e. ∃C > 0 that is independent of k such that ‖uk‖2 ≤C, ‖Du‖2 ≤ C.
Proof of Claim. The fact that uk is a minimizing sequence of I(u) implies that ∀k,
C1 ≥ I(uk) ≥ 1
4|‖Du‖22 −
1
2ε‖f‖22.
We can bring the f term to the other side and get C2 ≥ ‖Du‖2. Poincare’s inequality then implies thatC3 ≥ ‖u‖22.
5
Ok now we have that uk is a bounded sequence. Recall that in a finite dimensional vector space,boundedness implies pre-compactness. However, our functional space W 1,2
0 (Ω) is infinite dimensional, so weneed to find a weaker substitute called “weak compactness.”
Definition 2.8. Let B be a Banach space and B∗ be its dual space (space of bounded linear functionals),i.e., ` ∈ B∗ is linear and |〈l, u〉| ≤ C‖u‖ for any u ∈ B. Let uk ⊂ B. Then we say that uk u weakly if∀` ∈ B∗,
〈`, uk〉 → 〈`, u〉.
It is easy to see that if uk → u in the usual sense, then uk u weakly. Let ` ∈ B∗ so we have|〈`, uk−u〉| ≤ C‖uk−u‖ → 0. The converse is not true. For an easy example, let uk be an orthonormal basisin an infinite dimensional Hilbert Space. Then ‖uk − ul‖ =
√2, so we obviously do not have convergence.
On the other hand, from Parseval’s formula, we have∑uk
|〈`, uk〉|2 = ‖`‖2 → 0.
Then this implies that 〈l, uk〉 → 0 ∀` =⇒ uk 0 but uk 6→ 0. Now we go to a result from analysis:
Theorem 2.9 (Bamach-Alaoglu). Let B be a reflexive separable Banach space. Then for any boundedsequence uk ⊂ B, there exists a subsequence ukl ⊂ B such that ukl u weakly.
We will apply this to our problem. We have that I(uk)→ inf I(u) and ‖uk‖2 + ‖Du‖2 ≤ C. By passingthrough our subsequence, we have a u∞ ∈ W 1,2
0 (Ω) such that uk u∞ and Duk Du∞. Now thequestion that we have to answer is if u∞ is the minimum that we seek after. However, we can’t say thatuk u∞, Duk Du∞ implies I(uk) → I(u∞). The problem with this is that I(u) is not continuous withrespect to weak convergence. However, it is lower semi-continuous! i.e., uk u weakly in L2 implies that‖u‖2 ≤ lim infk→∞ ‖uk‖2. Here is the proof of this:
Proof.
‖u‖22 =
ˆuu = lim
k→∞
ˆuuk ≤ lim inf
k→∞‖u‖2‖uk‖2.
In particular,
I(u∞) =1
2‖Du∞‖22 + 〈f, u∞〉
≤ lim infk→∞
(1
2‖Duk‖22 + 〈f, uk〉
)= lim inf
k→∞I(uk).
However, since the following inequality is automatic,
inf I(u) ≤ I(u∞) ≤ lim infk→∞
I(uk),
and we have that I(u∞) = inf I(u).
We used the Banach-Alaoglu Theorem: if B is a reflective and separable Banach space and uj is abounded sequence, then there exists a weakly precompact sequence. The missing steps we have are to showthe proof of this, and completely show that ‖u‖2 ≤ C‖Du‖2 for u ∈ C∞0 (Ω). The proof of Theorem 2.9 willbe left as a black box. We move onto the proof of Lemma 2.7.
6
Proof of Lemma 2.7. Our main tool will be Holder’s inequality. Choose p and p∗ such that 1/p+ 1/p∗ = 1.Then we have
|u(x)| ≤ˆ|K(x, y)|
1p∗ |K(x, y)|
1p |v(y)|dy
≤[ˆ|K(x, y)|dy
] 1p∗[ˆ|K(x, y)||v(y)|pdy
] 1p
ˆ|u(x)|p dx ≤
ˆ ([ˆ|K(x, y)|dy
] pp∗[ˆ|K(x, y)||v(y)|pdy
])dx
≤ supx
[ˆ|K(x, y)|dy
] pp∗
ˆ (ˆ|K(x, y)||v(y)|p dy
)dx.
Since´|v(y)|pdy is a constant in terms of x, we can take it out of the integral of the RHS. Then switching
the order of integration and applying the same bounding trick we have
ˆ|u(x)|p dx ≤ sup
x
[ˆ|K(x, y)|dy
] pp∗
supy
[ˆ|K(x, y)|dx
] ˆ|v(y)|pdy
‖u‖pp ≤ App∗+1‖v‖pp
‖u‖p ≤ A‖v‖p.
Notice that this lemma would be pointless if A =∞ because then we learn nothing know with ‖u‖p ≤ ∞.We are in luck because we can actually deduce that A is finite in our case! The reason for this is because|x− y|n−1 yields a singularity of dimension strictly less than n when taking the supremum over the y’s, andso it integrable. Finally since x is over a set of compact support, it is bounded by a constant. Hence foru ∈ C∞0 (Ω) =⇒ ‖u‖p ≤ C‖Du‖p. Now we are finally done with our result from earlier.
Now we propose a question: how do we sharpen our estimates? A better way to visualize this questionis by noticing that the Kernel is integral for any power less than n. We actually used the worst power inour previous proof. The answer to our question comes from the following inequalities:
Theorem 2.10 (Sobolev Inequality). Let Ω ⊂⊂ Rn and u ∈ C∞0 (Ω). Then for any p < n we have
‖u‖ npn−p≤ Cn,p‖Du‖p.
Theorem 2.11 (Trudinger Ineqality). For p = n, there exists constants K,C > 0 such that
ˆΩ
exp
(|u(x)|
K‖Du‖n
) nn−1
≤ C.
Theorem 2.12 (Morrey’s Inequality). For any p > n, we have
‖u‖Cα(Ω) ≤ C‖Du‖p
where α = 1− n/p and
‖u‖Cα(Ω) = supΩ|u|+ sup
x6=y
|u(x)− u(y)||x− y|α
.
Lets first think of why the first inequality is a better estimate than the Poincare inequality that we have.It is better because np/(n− p) > p and we know that the Lp norms grow bigger as p increases. This meansthat we have essentially closed the gap between u and Du that we got from Lemma 2.7. Ok now lets try to
7
prove this. The proof is going to follow the same theme from Lemma 2.7’s proof. From Holder’s inequalitywe have
|u(x)| ≤ˆ|K(x, y)|α
(|K(x, y)|1−α|v(y)|1−β
)|v(y)|βdy
≤(ˆ|K(x, y)|αady
)1/a [ˆ|K(x, y)|(1−α)c|v(y)|(1−β)cdy
]1/c(ˆ|v(y)|βbdy
)1/b
where 1/a+ 1/b+ 1/c = 1. We need to choose our parameters wisely so that we have our desired estimates.One obvious constraint to put is βb = p and (1−β)c = p because we want |v(y)| on the RHS to have powersof p. Raising everything to the power q and integrating gives us
ˆ|u(x)|qdx ≤
ˆ ((ˆ|K(x, y)|αady
)q/a [ˆ|K(x, y)|(1−α)c|v(y)|pdy
]q/c(ˆ|v(y)|pdy
)βq)dx
=
ˆ ((ˆ|K(x, y)|αady
)q/a [ˆ|K(x, y)|(1−α)c|v(y)|pdy
]q/c‖v‖βqp
)dx.
In order to make our calculations a little bit easier take q = c. Then we can do the following:
ˆ|u(x)|qdx ≤
ˆ ((ˆ|K(x, y)|αady
)q/a [ˆ|K(x, y)|(1−α)c|v(y)|pdy
]‖v‖βqp
)dx
≤ supx
(ˆ|K(x, y)|αady
)q/a ˆ [ˆ|K(x, y)|(1−α)c|v(y)|pdy
]‖v‖βqp dx
≤ supx
(ˆ|K(x, y)|αady
)q/asupy
[ˆ|K(x, y)|(1−α)cdx
]‖v‖p+βqp
Recall that so far 0 < α < 1 is arbitrary. Choose it so that αa = (1− α)c. Now lets play around with theseparameters. Recall that we chose q = c and (1 − β)c = p =⇒ β = 1 − p/q. We can then plug this intoβb = p =⇒ b = p/β to get
b =p
β=
p
1− pq
=pq
q − p.
Now that we have parameters b and c, we can plug this into 1/a + 1/b + 1/c = 1 to get the parameter a.After skipping some steps we see that 1/a = 1− 1/p. Finally recall that we have αa = (1−α)c. plugging inour value for a and c and solving for α gives us
α =q(p− 1)
p+ pq − q, αa =
pq
p+ pq − q.
When applied to our gradient estimates, this means we need the integralˆ|K(x, y)|αady
to be finite. Following our explanation after our proof of Lemma 2.7 tells us that we need the integral
ˆ (1
|x− y|n−1
) pqp+pq−q
dy
to be finite. Comparing the powers would require
(n− 1)pq
p+ pq − q< n.
8
After playing around with this inequality we get
q <pn
n− p.
What suffices to show the full proof of Theorem 2.10 is that our coefficient in front of ‖Du‖p must dependonly on n and p. Additionally, setting K(x, y) = |x−y|−n+1 in our generalization of Lemma 2.7 would meanthat A would only be finite when
supx
ˆ (1
|x− y|n−1
) pqpq+p−q
dx <∞.
One can see that our kernel will be integrable ⇐⇒ (n− 1) pqpq+p−q < n ⇐⇒ 1
p −1q <
1n . We now prove the
following general lemma where u no longer has compact support:
Lemma 2.13. Assume 1p −
1q <
1n . If u satisfies
ˆΩ
|u(x)− u(y)|dy ≤ (diam Ω)n
n
ˆΩ
|Du(y)||x− y|n−1
dy
then
‖u− uΩ‖q ≤ cn
(1 + 1
q −1p
1n + 1
q −1p
)1+ 1q−
1p
(diam Ω)n
|Ω|1−1n+ 1
p
‖Du‖p.
Proof. It suffices to show, after dropping some constants,supx
ˆ (1
|x− y|n−1
) pqpq+p−q
dx
1+ 1q−
1p
≤
(1 + 1
q −1p
1n + 1
q −1p
)1+ 1q−
1p
|Ω|1n−
1p+ 1
q . (1)
The way to do this is from this simple fact: ∀0 ≤ µ < 1,ˆ
Ω
dy
|x− y|µn≤ cn|Ω|1−µ
1
1− µ.
The way to prove this fact is to show that it is true for balls and then for general Ω through “rearrangementinequalities.” Then (1) easily follows.
We have finally finished the proof of the Sobolev inequality. Observe now that 1p −
1q <
1n ⇐⇒ q < np
n−p .
We can ask the question if this holds for q ≤ npn−p? We’d think that the answer is no! Before we do that we
give an alternative proof for the Sobolev Inequality:
Theorem 2.14. Let u ∈ C∞0 (Ω) and 1 ≤ p < n. Then
‖u‖ npn−p≤ Cs‖Du‖p.
Proof. Consider first p = 1 and u ≥ 0. Notice that we can write u(x) in the following way:
u(x) =
ˆ ∞0
χu>tdt.
This makes sense because
χu>t =
1 u > t0 u ≤ t
and so the RHS becomes´ u
0dt. Since p = 1 our goal is to show that ‖u‖ n
n−1≤ C‖Du‖1. Then we see
‖u‖ nn−1≤ˆ ∞
0
‖χu>t(·)‖ nn−1
dt.
9
But the inside of this inequality is
‖χu>t(·)‖ nn−1
=
(ˆΩ
(χu>t(·)‖ n
n−1
) nn−1
dx
)n−1n
=
(ˆu>t
dx
) nn−1
= (Volu > t)n−1n
Notice that this has dimension of surface area because we can interpret Vol as having n dimensions, then wetake the nth root of it leaving one spacial dimension, and then we raise it to the n + 1th dimension again.Then we can use the following isoperimetric inequality:
(Volu > t)n−1n ≤ CS Area(∂u > t).
This gives us now the inequality
‖u‖n−1n≤ Cs
ˆ ∞0
Area(∂u > t)dt.
Now we will apply iterated integrals using the coarea formula: let u be a real valued function that isn’tconstant. Then
dx =1
|Du|dσtdt (2)
is the coarea formula. What we are doing is integrating the u = t level set with the dσt measure and thenintegrating with respect to dt. Applying this gives us
‖u‖n−1n≤ Cs
ˆ ∞0
Area(∂u > t)dt
= Cs
ˆ ∞0
(ˆu=t
dσt
)dt
= Cs
ˆΩ
|Du|dx.
We are done with the proof because for p 6= 1 we notice that u ∈ Lp ⇐⇒ up ∈ L1.
With this being said we turn back to the Trudinger inequality: for p = n, there exists constants Kn, Cn >0 such that ˆ
Ω
exp
(|Ω|
Kn(diam Ω)n|u(x)− uΩ|‖Du‖n
) nn−1
dx ≤ Cn|Ω|.
Notice that this is stronger than what the Sobolev inequality says for p < n because exponentials arealways greater than polynomials, and so this is stronger than ‖u − uΩ‖p < ∞. However it is weaker than‖u− uΩ‖∞ ≤ C because this deals with the supremum, which is the best bound we can have.
Proof of the Trudinger Inequality. We will recall the integral inequality
‖u− uΩ‖q ≤ cn
(1 + 1
q −1p
1n + 1
q −1p
)1+ 1q−
1p
‖Du‖p (3)
and drop terms involving Ω because in the end they’re just constants. We expand the exponential as a powerseries to get
ˆΩ
exp
(1
Kn
|u(x)− uΩ|‖Du‖n
) nn−1
dx =
∞∑k=0
1
k!
ˆΩ
(|u(x)− uΩ|Kn‖Du‖n
) knn−1
dx
=
∞∑k=0
1
k!
(‖u− uΩ‖ kn
n−1
Kn‖Du‖n
) knn−1
(4)
10
Now we go back to (3) and plug in p = n to get the following inequality:
‖u− uΩ‖q‖Du‖n
≤ cn(q
(1 +
1
q− 1
n
))1+ 1q−
1n
≤ cnq1− 1n .
This was verified on paper by playing little tricks. Anyway, we plug this into (4) with q = knn−1 :
ˆΩ
exp
(1
Kn
|u(x)− uΩ|‖Du‖n
) nn−1
dx ≤∞∑k=0
1
k!
(1
Kn
(cn
kn
n− 1
)1− 1n
) knn−1
=
∞∑k=0
1
k!
1
Kn
knn−1
(cn
kn
n− 1
)k
=
∞∑k=0
kk
k!
(n
Knn−1n (n− 1)
)k.
Then something called Stirling’s formula makes this converge.
Recall that we have shown this basic inequality before: Consider p, q and assume 1 ≤ p ≤ q ≤ ∞ and1p −
1q <
1n . Then
‖u− uΩ‖q ≤ cn
(1 + 1
q −1p
1n + 1
q −1p
)1+ 1q−
1p
(diam Ω)n
|Ω|1−1n+ 1
p
‖Du‖p
for Ω convex and u satisfying one of our integral inequalities. Then certainly we have
|u− uΩ| ≤ cn
(1 + 1
q −1p
1n + 1
q −1p
)1+ 1q−
1p
(diam Ω)n
|Ω|1−1n+ 1
p
‖Du‖p (5)
We have used this to show the Sobolev Inequality and Trudinger inequality. Now we prove Morrey’s inequal-ity.
Theorem 2.15 (Morrey’s Inequality). Let p > n, α = 1− np ,Ω = Sn−1. Then
‖u‖Cα ≤ C‖Du‖p.
Proof. Recall that
‖u‖Cα = ‖u‖∞ + supx 6=y
u(x)− u(y)
|x− y|α
where 0 < α < 1. It the suffices to show that each term is bounded by ‖Du‖p i.e. ‖u‖∞ ≤ C‖Du‖p and[u]Cα ≤ C‖Du‖p. Clearly the first part follows from (5) with q = ∞. Now we prove the second part. Fix
x, y, x 6= y and let δ = |x − y|. Now define Ω := Bδ(x) ∩ Bδ(y). Clearly this is convex. Then we have thefollowing inequalities
|u(x)− u(y)| ≤ |u(x)− uΩ|+ |uΩ − u(y)|
≤ (diam Ω)n
|Ω|
(ˆΩ
|Du(z)||x− z|n−1
dz +
ˆΩ
|Du(z)||y − z|n−1
dz
)Now notice that diam(Ω) ≤ 2δ and |Ω| ≤ |Bδ(x)|. Finally since all the terms inside the integral are positivethen we can write
|u(x)− u(y)| ≤ C
(ˆBδ(x)
Du(z)
|x− z|n−1dz +
ˆBδ(y)
|Du(z)||y − z|n−1
dz
).
11
Now at this point we don’t have to reinvent the wheel so we see now that our function satisfies the correctrequirements of (5) so we can write
|u(x)− u(y)| ≤ C‖Du‖diam(Bδ(x))n
|Bδ(x)|1−1n+ 1
p
≤ C‖Du‖p(2δ)n · (ωnδn)1n−
1p−1
≤ C‖Du‖pδ1−np−nδn
≤ C‖Du‖pδα
And we are done with the proof.
Theorem 2.16 (Sobolev Embedding Theorem). Let u ∈ W k,p(Ω). Then ‖u‖L∞(Ω) ≤ ‖u‖Wk,p(Ω) for1/p < k/n.
Proof. Lets recall Morrey’s inequality. This says that if 1/p > 1/n, then
‖u‖∞ + supx 6=y
|u(x)− u(y)||x− y|α
= ‖u‖Cα ≤ C‖Du‖p
for u ∈ C∞0 and α = 1− n/p.Let k = 1. Then the theorem follows from Morrey’s inequality because 1/p < 1/n and ‖u‖∞ is obviously
bounded by ‖u‖Cα . Now assume the theorem holds for k and ∀B, consider Dβu where |β| ≤ k. Set v = Dβu.Then by definition we have ‖Dv‖p ≤ ‖u‖Wk+1,p . Since by we are assuming 1
p <k+1n pick some ε > 0 such
that 1p = k+1
n − ε. Now recall the Sobolev inequality: ‖v‖q ≤ ‖Dv‖p for 1p −
1q <
1n ⇐⇒
1p −
1n ≤
1q . We
will choose a q satisfying (for some ε′ < ε)
1
q=
1
p− 1
n+ ε′
=k + 1
n− 1
n+ ε′ − ε
=k
n+ ε′ − ε < k
n.
Now we have ‖v‖q ≤ ‖u‖Wk+1,p for 1q <
kn and by definition ‖u‖Wk,q ≤ ‖u‖Wk+1,p for 1
q <kn . And so our
induction hypothesis tells us‖u‖∞ ≤ ‖u‖Wk,q ≤ ‖u‖Wk+1,p .
As a consequence we have then that if u ∈ W k,p0 (Ω) then u ∈ C0(Ω) if 1/p < k/n. The way to see this
is the following: Let uj ⊂ C∞0 (Ω) and uj → u with respect to ‖ · ‖Wk,p(Ω). This of course exists because
W k,p0 (Ω) = C∞0 (Ω)|‖ · ‖Wk,p(Ω) <∞. Then we apply the Sobolev embedding theorem to uj − um. Then
we have that ‖uj − um‖C0 ≤ C‖uj − um‖Wk,p(Ω) → 0. This implies that uj converges uniformly and solimuj is continuous. In fact, uj → u uniformly in the usual sense because uniform convergence implies thatuj → u in Lp. However, since W k,p(Ω) convergence is stronger, we have uj → u in the usual sense.
Lets summarize what we’ve done. Let Ω ⊂⊂ Rn, f ∈ L2(Ω). Define
I(u) =
ˆΩ
(1
2|Du|2 + fu
)dx
for u ∈ W 1,20 (Ω). Then we showed that ∃u∞ ∈ W 1,2
0 (Ω) with I(u∞) = inf I(u). We will observe that theminimum of a functional I(u) is going to be a generalized solution of the Euler-Lagrange equation for I(u).The basis on which we will set our ground on is that if x∞ = minx∈Ω f(x) =⇒ f ′(x∞) = 0.
12
Let ϕ ∈ C∞0 (Ω) and consider for t << 1 the function R 3 t 7→ A(t) = I(u∞ + tϕ). Then we see thatt = 0 is going to be a minimum for A(t) =⇒
0 =dA
dt
∣∣∣∣t=0
=d
dt
∣∣∣∣t=0
ˆ (1
2|D(u∞ + tϕ)|2 + f · (u∞ + tϕ)
)dx
=d
dt
∣∣∣∣t=0
ˆ (1
2(|Du∞|2 + 2tDu∞Dϕ+ t2|Dϕ|2) + f · (u∞ + tϕ)
)dx
=
ˆ(Du∞Dϕ + fϕ) dx
=
ˆ n∑j=1
∂u∞∂xj
∂ϕ
∂xj+ fϕ
dx.
Now if we assume temporarily that u∞ ∈ C2(Ω) then we are allowed to integrate by parts and get
0 =
ˆ (−∑ ∂2u
∂x2j
+ f
)ϕdx.
Since this is true for any ϕ then we have −∆u∞ + f = 0, which is the Laplace equation that we wanted tosolve from the first page.
2.2 A non-linear PDE and Method of Continuity
2.2.1 Non-linear variations on Manifolds
Before we can extend our current groundwork to general manifolds, we need to expand our theory to moregeneral boundary values. Let Rn+1
+ = x ∈ Rn+1|xn+1 ≥ 0 and let Ω ⊂⊂ Rn+1+ . Assume u ∈ C∞0 (Ω). Then
we will attempt to show
‖u(·, 0)‖Lp(Rn) ≤ CΩ‖∂n+1u‖Lp(Rn+1+ ). (6)
Pick d > diam Ω. Since u has compact support we have
|u(x, 0)| = |u(x, 0)− u(x, d)| =
∣∣∣∣∣−ˆ d
0
∂u
∂xn+1(x, xn+1)dxn+1
∣∣∣∣∣≤
(ˆ d
0
∣∣∣∣ ∂u
∂xn+1(x, xn+1)
∣∣∣∣p dxn+1
)1/p(ˆ d
0
1qdxn+1
)1/q
|u(x, 0)|p ≤ Cˆ d
0
∣∣∣∣ ∂u
∂xn+1(x, xn+1)
∣∣∣∣p dxn+1
ˆ|u(x, 0)|pdx ≤ CΩ
ˆ ˆ d
0
∣∣∣∣ ∂u
∂xn+1(x, xn+1)
∣∣∣∣p dxn+1dx.
Taking the pth root gives us the desired claim.We will now attempt to show that this inequality will allow us to define u|∂Ω for an arbitrary u ∈W 1,p(Ω).
Take uj ∈ C∞(Ω) where uj → u in W 1,p. We know this sequence converges because W 1,p is the completionof C∞ with respect to the p norm. Then by our inequality (6), we can see
‖(uj − uk)|∂Ω‖p ≤ C‖uj − uk‖W 1,p → 0
by definition and so uj is cauchy. But since Lp is complete we can then formally define
u|∂Ω := limuj |∂Ω
13
where the limit is taken over the Lp norm. Recall that we still had Ω being some some simple semi-circle inRn+1
+ .
Lets extend the notion of boundary values in more general Ω with Ω ∈ C∞. Let Ω be a small subset ofΩ such that Ω∩∂Ω 6= ∅. Let v ∈ C∞0 (Ω). We will define (using norms) v|∂Ω. Let y ∈ Ω. Since the boundary
is smooth we can map Ω into an upper half sphere as before with y 7→ x. Of course, we can go backwards.So then we can say v(y) = v(y(x)) =: u(x) and note that via our definitions, v|∂Ω = u(x, 0) and from ourprevious observations we have
‖u(·, 0)‖Lp(Rn) ≤ C∥∥∥∥ ∂u
∂xn+1
∥∥∥∥Lp(Rn+1)
= C
∥∥∥∥ ∂
∂xn+1v(y(x))
∥∥∥∥Lp(Rn+1)
≤ Cn+1∑l=1
∥∥∥∥ ∂v∂yl∥∥∥∥Lp(Rn+1)
(7)
≤ C‖v‖W 1,p(Rn+1).
Noticed that (7) we did something very fishy that I will now justify. We did the change of variables fromintegrating with respect to the x coordinates to integrating with respect to the y coordinates. The problemwith this is that the integrals might not be bounded in the correct way. Recall
dy = det
∣∣∣∣ ∂yl∂xj
∣∣∣∣ dxand c ≤ det
∣∣∣ ∂yl∂xj
∣∣∣ ≤ C. And so we have for a general function f
‖f(·)‖pLpy
=
ˆ|f(y)|pdy
=
ˆ|(f y)(x)|P det
∣∣∣∣ ∂yl∂xj
∣∣∣∣ dxand so we have
c‖f y‖pLpx≤ ‖f‖p
Lpy≤ C‖f y‖p
Lpx
and (7) is valid.
Claim. The following inequality holds for any v (not just those supported in a boundary neighborhood):∥∥v∣∣∂Ω
∥∥Lp(∂Ω)
≤ C‖v‖W 1,p(Ω).
Proof. Note that cdx ≤ dσ ≤ Cdx and we are able to apply the same argument as above (i.e. norms areequivalent under change of variable). So now the problem is to deal with the full Ω. Since Ω is compact wecover it Ω = ∪Nα=1Ωα and pick a partition of unity χα ∈ C∞0 (Ωα) and
∑χ = 1. Then
∥∥v∣∣∂Ω
∥∥Lp(∂Ω)
≤N∑α=1
‖χαv‖LP (∂Ω) ≤N∑α=1
‖χαv‖W 1,p(Ω)
≤N∑α=1
(‖χαv‖Lp(Ω) + ‖D(χαv)‖Lp(Ω)
)≤ C‖v‖W 1,p(Ω)
Where we have bounded ‖χαv‖p ≤ ‖v‖ and expanded the second term using the Leibniz rule.
14
With this done we can now extend our previous work to more general boundary conditions. Say youwanted to solve ∆u = f with u|∂Ω = g. Then if g ∈ Lp(∂Ω) then choose a G ∈W 1,p(Ω) whose restriction isequal to g and then consider v = u−G and the problem ∆v = ∆u−∆G and v|∂Ω.
We will begin by analyzing a Non-Linear PDE. Let M be a compact Riemannian manifold of dimension2. Let
ds2 =
2∑i,j=1
gijdxidxj
be the Riemannian metric. Let v ∈ C∞(M) and let R be a given negative constant. We want to solve
∆u+ λeu+v −R = 0. (8)
Notice that the exponential of u makes this a very non-linear equation. Let√g := det gij . Then we will
define the laplacian as follows
∆u =1√g
2∑j,k=1
∂k(√ggij∂ju
)=
2∑j,k=1
1√g
(√ggjk∂k∂ju
)+ first order terms
=
2∑j,k=1
gjk∂k∂ju+ first order terms.
Example 2.17. In Euclidean space, we have ds2 =∑
(dxi)2 =⇒ gij = gij = δij and so
∆u =∑ ∂2u
∂x2j
+ first order terms.
In order to solve this, we would have to consider the following functional:
I(u) =
ˆM
|Du|2√gdx+R
ˆM
u√gdx
subject to the constraint1
V
ˆM
eu+v√gdx = 1, V =
ˆM
√gdx
and attempt to mimic our work for ∆u = f (show that the function is bounded from below and find aminimum).
Note that Du = ∂ju and |Du|2 = ∂ju∂ku then we need the metric to contract the indices so we introducegjk. Note that we can assume
1
V
ˆv√gdx = 0 (9)
because we can shift everything about constants. We will prove boundedness from below for our functional,but first lets compare with Ω ⊂⊂ Rn and
I(u) =1
2
ˆΩ
|Du|2dx+
ˆΩ
fudx
for any u ∈W 1,20 (Ω) and f ∈ L2(Ω). The way we proved this was by showing that
ˆΩ
|u|2dx ≤ Cˆ
Ω
|Du|2dx,
15
but can we do this on a general compact manifold? No! In general we have the following Poincare inequality
‖u− u‖2L2(M) ≤ C‖Du‖2L2(M), u =
1
V
ˆM
u√gdx.
Now we can write our functional in the following way
I(u) =1
2
ˆM
|Du|2√gdx+R
ˆM
(u− u)√gdx︸ ︷︷ ︸
bounded as before
+Ru
ˆM
√gdx︸ ︷︷ ︸
can blow up
.
In order to deal with this difficulty we exploit our constraint
1 =1
V
ˆM
eu+v√gdx
0 = log
(1
V
ˆM
eu+v√gdx)
and note that 1V
√gdx has total measure 1. Now recall consider Jensen’s Inequality: If
´dµ = 1 then
ˆlog(f)dµ ≤ log
(ˆfdµ
).
We will also use the fact that
log
(a+ b
2
)≥ 1
2log a+
1
2log b
in the following way
0 = log
(1
V
ˆM
eu+v√gdx)≥ 1
V
ˆM
log(eu+v)√gdx
=1
V
ˆM
(u+ v)√gdx =
1
V
ˆM
u√gdx
where we have used (9) in the last step. So now we have that our constraint implies u ≤ 0 and so we doindeed have our bound
I(u) =1
2
ˆ|Du|2√gdx+R
ˆu√gdx ≥ 0.
Now as before lets pick a minimizing sequence uj ∈ C∞(M) such that I(uj)→ min I(u) and each uj satisfiesthe constraint. Clearly ‖Du‖2L2(M) ≤ C and so
‖uj‖2 ≤ ‖uj − uj‖2 + ‖uj‖ ≤ C‖Duj‖2 +1
RI(uj) ≤ C.
This means that our minimizing sequence is bounded and so we can apply the Banach-Alaoglu Theorem toshow that ∃u∞ ∈ W 1,2(M) such that Duj Du weakly and uj u weakly. Now the question is if u∞ isthe minimum or not? We do have
I(u∞) ≤ lim inf1
2
ˆ|Duj |2
√gdx+
ˆRuj√gdx
= lim inf I(uj) = limj→∞
I(uj).
The real question is if u∞ satisfies the constraint i.e. we wish to show
1
V
ˆM
eu∞+v√gdx = 1 (10)
The bottom-line question here is if we can pass through the limit. We need to strengthen our converginghypothesis as much as possible (i.e. from weak convergence to point-wise convergence). In fact, we claimthat for the minimizing sequence we have, Duj Du∞ weakly but uj → u in L2. For this we will need
16
Lemma 2.18 (Rellich’s Lemma). Let M be a compact manifold and uj ⊂W 1,p(M) with ‖uj‖W 1p(M) ≤ Cwith C independent of j, then there exists u∞ ∈ W 1,p(M) and a subsequent ujk such that ujk → u∞ inLp(M).
Note. Note that this also holds for W 1,p(Rn) if suppuj ⊂ k ⊂⊂ Rn ∀j.Now recall a theorem from measure theory:
Theorem 2.19. If uj → u∞ in Lp for 1 ≤ p <∞, then there exists ujk → u∞ point wise a.e.
Note that all together, we can take a subsequence of our subsequence to find a sequence ‖Duj‖ ≤ C, uj →u∞ almost everywhere. We will see that in two dimensions these properties imply that euj+v → eu∞+v inL1. Since these values are always positive, the L1 norm is just convergence of the integral, which is what weneeded. To see this,
euj+v − eu∞+v = −ˆ 1
0
d
dt
[etu∞+(1−t)uj+v
]dt
= −ˆ 1
0
(u∞ − uj)etu∞+(1−t)uj+vdt.
Taking the L1 norm will give usˆM
|euj+v − eu∞+v|√gdx ≤ˆ 1
0
[ˆM
|u∞ − uj |etu∞+(1−t)uj+v√gdx]dt
≤ˆ 1
0
[ˆM
|u∞ − uj |2√gdx
]1/2 [ˆM
e2(tu∞+(1−t)uj+v)√gdx]1/2
dt
= ‖u∞ − uj‖L2
ˆ 1
0
[ˆM
e2(tu∞+(1−t)uj+v)√gdx]1/2
dt
and now note that we need the second integral to be uniformly bounded for the RHS of the inequality to goto zero.
We claim that´ewj√gdx ≤ C (independent of j) if ‖wj‖L2 ≤ C and ‖Duj‖L2 ≤ C. In this case we have
wj = 2(tu∞+(1−t)uj+v) satisfying the condition because ‖Duj‖L2 ≤ C and ‖Du∞‖L2 ≤ lim inf ‖Duj‖L2 ≤C and similarly the L2 norm of uj and u∞ is bounded. Now lets see why this claim is true.
Recall the Trudinger inequality: u ∈ C∞0 (B) implies that
ˆexp
(|u(x)|
K‖Du‖Ln
) nn−1
≤ C
and so for n = 2 we have ˆexp
(|u(x)|
K‖Du‖L2
)2
≤ C.
Now note that we can write
|u(x)| = |u(x)|K‖Du‖L2
K‖Du‖L2 ≤ 1
2
(|u(x)|
K‖Du‖L2
)2
+1
2(K‖Du‖L2)2.
Since the exponential function is increasing we can write
exp |u(x)| ≤ exp
[1
2
(|u(x)|
K‖Du‖L2
)2
+1
2(K‖Du‖L2)2
]ˆ
exp |u(x)| ≤ e 12 (K‖Du‖L2 )2
ˆexp
1
2
(|u(x)|
K‖Du‖L2
)2
≤ e 12 (K‖Du‖L2 )2
C
17
And this concludes the proof. A great exercise is as follows. Let M be a compact Riemannian n-manifoldand show that ˆ
ew√gdx ≤ C exp (C‖Du‖nLn + ‖w‖nLn) .
As a summary, we have shown that for a sequence satisfying I(uj)→ min I(u) and 1V
´euj+v
√gdx = 1 then
uj → u∞ (as explained above) and 1V
´eu∞+v√gdx = 1 so we have
inf I(u) ≤ I(u∞) ≤ inf I(u).
Now we claim that u∞ satisfies our partial differential equation in the Euler-Langrange sense. Fixϕ ∈ C∞(M) and consider u∞ + tϕ + ct. We add the constant ct so that this function still satisfies theconstraint. To see what ct has to be note
1 =1
V
ˆM
eu∞+tϕ+ct+v√gdx
ect =
(1
V
ˆM
eu∞+tϕ+v√gdx)−1
ct = − log
(1
V
ˆM
eu∞+tϕ+v√gdx)
and thus I(u∞ + tϕ+ ct) ≥ I(u∞) for any t and so we leave it as an exercise to show in detail
d
dt
∣∣∣∣t=0
I(u∞ + tϕ+ ct) = 0.
Lets recap our problem. Our functional was
I(u) =1
2
ˆM
|Du|2√gdx+R
ˆM
u√gdx
subject to the constraint1
V
ˆM
eu+v√gdx = 1 (11)
where v is some smooth function and V is the volume over this compact manifold. We obtained a u∞ subjectto the same constraint that satisfies ∆u∞+λeu∞+v−R = 0. Recall that this general laplacian has the form∆u = 1√
g∂i(√g∂ju
). Where does this come from? If we assume ut is smooth then we can check
d
dt
ˆM
|Dut|2√gdx =
d
dt
(ˆgij∂iut∂jut
√gdx
)= − d
dt
(ˆut∂i
(gij∂jut
)√gdx
)= − d
dt
(ˆut∆ut
√gdx
)So we see that this laplacian is the perfect analogue of the one taken on Rn.
We want to derive the generalized Euler-Lagrange equation for our PDE, so let ϕ ∈ C∞0 (M) and ut =u∞ + tϕ + ct, where ct is picked so that ut also satisfies the constraint (11). Lets figure out what λ has tobe in order to for our equation to be solved correctly. Integrate on both sides of our PDE to get
0 =
ˆM
∆u∞√g + λ
ˆM
eu∞+v√g −RˆM
√gdx
=
ˆM
1√g∂i(√ggij∂ju∞
)+ λ
ˆM
eu∞+v√g −RˆM
√gdx
= λV −RV
18
so we see that λ has to equal R. The reason the first term vanishes is because we are integrating an exactform over a compact manifold. So then we have that our PDE is
∆u+Reu+v −R = 0 (12)
and u satisfies (12) in the general sense if ∀ϕ ∈ c∞0 we have
0 =
ˆ (∆u+Reu+v −R
)ϕ√g
=
ˆ (−gij∂i∂jϕ+Reu+v −Rϕ
)√gdx (13)
and (13) is the generalized Euler-Lagrange equation.
2.2.2 Regularity Theory
Lets finally start some regularity theory. Let gij be a smooth Riemannian metric and assume Λ ≥ gij ≥ λ > 0.
If u ∈W 1,2loc satisfies ∆u = f in the weak sense then
1. If f ∈W k,p and 1 < p <∞ the u ∈W k+2,p and for any Ω ⊂⊂ Ω′
‖u‖Wk+2,p(Ω) ≤ CΩ,Ω′(‖f‖Wk,p(Ω′) + ‖u‖Wk,p(Ω′)
)2. If f ∈ Ck,α and 1 < p <∞ the u ∈ Ck+2,α and for any Ω ⊂⊂ Ω′
‖u‖Ck+2,α(Ω) ≤ CΩ,Ω′(‖f‖Ck,p(Ω′) + ‖u‖Wk,p(Ω′)
)Claim. It follows easily that if u satisfies (13) in the generalized sense, then u ∈ C∞ and actually satisfies(12) in the standard sense.
Proof. Indeed we can let −Reu+v +R = f and f ∈ L2 = W 0,2 by the Trudinger’s inequality (it tells us thatthis exponential is bounded by the L2 norm of the weak derivative) then by regularity we will get u ∈W 2,2.Now recall that Morrey’s inequality tells us that if 1
p <kn then W k,p is embedded in a Holder space, but
since p = n = 2 we have u ∈ Cα. But then since ∆u = f this implies f ∈ Cα and we can apply regularityto get u ∈ C2,α. We can iterate this and find of course that u ∈ C∞.
We will prove these regularity statements by the method of continuity and a priori estimates. Consider
∆u+Reu+v −R = 0
on (M, gij(x)) where v is a smooth function. Note that ∆u + Reu − R = 0 admits a solution u ≡ 0 so wesee that the difficulty comes from the v term. Let t ∈ [0, 1] and introduce the family of equations
∆u+Reu+tv −R = 0 (14)
And consider the set I = t ∈ [0, 1]|(14) admits a solution ut. Note that I 6= ∅ because 0 ∈ I is a solution.Then we obviously want to show I = [0, 1], so we need to show that I is open and closed. Lets discuss thisvery briefly (to be made concise later).
Say we want to show I is open. We will do this by an analogue of the implicit function theorem. Recallthat it says given f(x0, y0) = 0 and ∂f
∂y (x0, y0) 6= 0 then ∃ε > 0 such that for |x − x0| < ε, then ∃!y such
that f(x, y) = 0. Now let f(t, u) = ∆u + Reu+tv − R. Then we want to solve f(x, u) = 0 where we knowf(t, u0) = 0. So our goal is going to be an implicit function theorem for Banach spaces and we need to checkthat ∂f
∂u (t0, u0) 6= 0 in a way we will define later.Now lets briefly discuss how we will show that I is closed. Take tj ∈ I such that (14) will admit a solution
called uj , and assume tj → T . Then closeness of I is equivalent to showing that (14) will admit a solutionfor T . It will suffice to have a subsequence converge in C2.
19
Proof. Lets start by showing that I is closed. Let tj → T with tj ∈ I and let uj be the correspondingsolution of (15). Observe that if suffices to show that ∃C independent of j so that ‖uj‖C3 ≤ C where ingeneral
‖u‖C3(Ω) =∑|α|≤3
‖∂αu‖C0(Ω).
The reason why this will help is that this would imply that ∀β ≤ 2, then ∂βuj is an equicontinuous family.Then the Arzela-Ascolati Theorem tells us that if we have an equicontinuous family, then by going througha subsequence ∂βuj we have uniform convergence to DαuT where uT ∈ C2. However we don’t necessarilyhave that T ∈ I because we need uT to be smooth and so far it is only C2. This is fixed however by ourregularity observations.
Before we can apply our regularity conditions we must show that our equation is uniformly elliptic. Recallthat a second order PDE is said to be uniformly elliptic if the leading coefficient satisfies
λ|ξ|2 ≤
∣∣∣∣∣∣∑|α|=2
aαξα
∣∣∣∣∣∣ ≤ Λ|ξ|2.
This we need to show that our Laplacian is elliptic.
∆u =1√g∂i (√g∂ju)
=1√g
√ggij∂i∂ju+ first order terms.
So the symbol (the middle term of the ellipticity inequality requirement) of our Laplacian is going to be
σ∆(x, ξ) = gijξiξj
and since g is positive definite we definitely have the ellipticity requirement. So then we can apply ourregularity theorems by viewing ∆u = f ∈ C2 ⊂ C1,α.
So now we have to prove the a priori estimate ‖uj‖C3 ≤ C. We will use the maximum principle. Sincethere are so many different formulations of maximum principles, it is a good idea to simply examine whathappens near a maximum. Let u ∈ C∞ satisfy ∆u+Rtw+u −R = 0 (by denoting ut = u) and then I claimthat ‖u‖C0 ≤ C where C is independent of t. Let x0 ∈ Ω such that u(x0) is a maximum. Then ∆u(x0) ≤ 0and since R < 0 we have
0 ≤ −∆u = Retw+u −RR ≤ Retw+u
−|R| ≤ −|R|etw+u
1 ≥ etw+u
0 ≥ tw + u
u ≤ −tw≤ ‖w‖C0 .
But since x0 is a maximum we have that ∀x, u(x) ≤ u(x0) ≤ ‖w‖C0and applying the same argument for
the minimum we have ‖u‖C0≤ ‖w‖C0
. In order to get higher derivatives we write ∆u = f ∈ C0 and so forf ∈ Lp ∀1 ≤ p <∞ we have u ∈W 2,p ⊂ Cα for some α when n < kp.
Now lets show that I is open. Let t0 ∈ I i.e. there exists u0 which is a smooth solution of (14). Let(u, t) 7→ F (u, t) = ∆u + Retw+u − R. Then we want to show that ∃δ > 0 so that |t − t0| < δ implies∃ut ∈ C∞|F (ut, t) = 0. The main tool will be the Implicit Function Theorem for Banach Spaces, which goesas follows.
20
Let B1 and B2 be Banach spaces. Let F ∈ C1 and consider B1×R ⊃ Ω 3 (u, t) 7→ F (u, t) ∈ B2. Assumethat F (u0, t0) = 0. Let ∂F
∂u (u0, t0) be the derivative of F at (t0, u0) viewed as a linear operator B1 → B2.Note that if
‖h‖B1 ≤ C∥∥∥∥∂F∂u (u0, t0)h
∥∥∥∥B2
∀h ∈ B1 (15)
then ∂F∂u (u0, t0) is injective and surjective. So assume that (15) holds. Then the Implicit Function Theorem
for Banach Spaces says that ∃δ > 0,∃V that is a neighborhood of u0 such that ∃!ut ∈ V with F (ut, t) = 0.Before we can even apply this theorem, we need to mae sense of derivatives in terms of Banach spaces.
Let B1 3 u 7→ F (u) ∈ B2. Then F is differentiable at u0 if ∃L : B1 → B2 that is a bounded linear operatorsatisfying
F (t, u+ h) = F (t, u) + Lh+ E(t, u, h)
with
limh→0
‖E(u, h)‖B2
‖h‖B1
= 0.
In order to apply our theorem we need so specify our Banach spaces. We will want to have (t, u) ∈ R×C2,α →F (u, t) ∈ C0,α. What we will then need to check are the assumptions of the implicit function theorem. Lett0, u0 ∈ R×C2,α satisfy the conditions of the IFT. Lets determine L. Consider the expression F (t0, u0 +h):
F (t0, u0 + h) = ∆(u0 + h) +Ret0w+u0+h −R= F (t0, u0) + ∆h+Ret0w+u0+h −Ret0w+u
= F (t0, u0) + ∆h+Ret0w+u(eh − 1
)= F (t0, u0) +
(∆ +Ret0+u
)h+Ret0w+u(eh − 1− h).
So we have figured out our L. Now we leave it as an exercise to show that
‖Ret0w+u0(eh − 1− h)‖C0,a ≤ C‖h‖2C2,α .
The main tool for this is the integral form of Taylor’s Remainder Theorem, which starts as follows
h
ˆ 1
0
(1− t) ddt
[f ′(u+ th)] dt =
ˆ 1
0
f ′(u+ th)dt+ h(1− t)f ′(u+ th)
∣∣∣∣10
=
ˆ 1
0
d
dtf(u+ th)dt− hf ′(u)
= f(u+ h)− f(u)− hf ′(u).
However we can write the LHS as
h
ˆ 1
0
(1− t) ddt
[f ′(u+ th)] dt =
ˆ 1
0
(1− t)f ′′(u+ th)dt
and so we have the final part as
f(u+ h) = f(u) + f ′(u)h+ h2
ˆ t
0
(1− t)f ′′(u+ th)dt.
Thus we have that L = ∂F∂u (t0, u0) : C2,α 3 h 7→ ∆h+Ret0w+uh ∈ C0,α, which is clearly a bounded operator.
It is trickier to show that it is injective and bijective.Assume 0 = Lh = ∆h+Ret0w+uh. Let x0 me a maximum (again this means ∆h ≤ 0). Then we have
−Ret0w+uh = ∆h(x0) ≤ 0
21
and since the coefficients are both strictly positive, it means that h(x0) ≤ 0. Since this is a maximum itmeans that ∀x, h(x) ≤ h(x0) ≤ 0. Applying the same process to the minimum we get that h(x) ≥ 0 and soh ≡ 0. This implies that the kernel of L is zero, and so it is injective.
In order to finally complete the proof we will show that L is onto i.e. ∀f ∈ C0,α, we want to show∃h ∈ C2,α so that Lh = f . We will show this by variational methods again. Set
I(h) =
ˆ [√ggij∂ih∂jh−Ret0w+u0h2 + fh
]√gdx
for some h ∈W 1,2(M). We leave it as an exercise to show that I(h) attains its minimum for some h∞. Oneway to do this is by showing that
I(h) ≥ ‖Dh‖22 + ε‖h‖22 −1
ε‖f‖22
and using our tricks. Assuming the exercise, we invoke the black box to make h smooth. Recall that it saysif f ∈ Ck,α then
‖h‖Ck+2,α ≤ C (‖Lh‖Ck,α + ‖h‖Ck,α) .
We now improve the black box by saying that if kerL = 0, then
‖h‖Ck+2,α ≤ C‖Lh‖Ck,α . (16)
Now we will prove this little lemma. We will use weak compactness. If ul ⊂ Ck,α with ‖ul‖Ck,α ≤ Cwith C independent of l, then either ∀k < k, ∀β or k′ = k with 0 < β < α then there exists a convergentsubsequence in Ck
′,β by the routine application of the Arzela-Ascolti Theorem. Now assume (17) does nothold. Then for any N, ∃hN ∈ Ck+2,α such that ‖hN‖Ck+2,α > N‖Lhn‖Ck,α . Set
hN =hN
‖hN‖Ck+2,α
and notice that ‖hN‖Ck+2,α = 1 and so this implies that ‖Lh‖Ck,α < 1N → 0. By the weak compactness,
going through a subsequence, we can assume that hN → h∞ in Ck,α. Applying the black box gives
‖hN − hM‖Ck+2,α ≤ C(‖LhN − hM‖Ck,α + ‖hN − hM‖Ck,α
)which implies that hN → h∞ in Ck+2,α. Thus LhN → Lh∞ in Ck,α and so h∞ ∈ kerL. Since
‖h∞‖Ck+2,α = lim ‖hN‖Ck+2,α = 1
this contradicts the fact that kerL = 0.
22
3 Harnack Inequality for Divergence Equations
3.1 Regularity Estimates
Suppose that u ∈W 1,2(Ω) solves ∂i(aij∂ju) + cu = f in the generalized sense, where aij is uniformly elliptici.e. 0 < λ ≤ aij ≤ Λ. The question we want to answer is: when is u “regular” i.e. u ∈ Cα,W k,2, C∞, etc?
Lets look at the simplest case where aij is constant and c = f = 0. In this case u solving aij∂i∂ju = 0
in the generalized sense. This means that ∀v ∈W 1,20 (Ω),
ˆaij∂iu∂jv = 0. (17)
Then we will show that u ∈ C∞(Ω) and for |α| = k, 0 < r < R, we have
ˆBr(x0)
|Dαu|2 ≤ Cλ,Λ,k(R− r)2k
ˆBR(x0)
|u|2. (18)
Note that this inequality is very powerful because we have that the derivative is being bounded by thefunction, where w usually have it the other way around.
Proof. We apply (17) with v = χ2u with 0 ≤ χ ≤ 1, χ ≡ 1 on Br(x0) and χ ∈ C10 (BR(x0)). Additionally
assume
Dχ(x0) ≤ 2
R− r.
Applying (17) gives us ˆχ2aij∂iu∂ju = −2
ˆaij(χ∂jχ)u∂iu
and note that ellipticity gives us |aijuivj | ≤ (aijuiuj)12 (aijvivj)
12 . Putting absolute values gives us∣∣∣∣ˆ χ2aij∂iu∂ju
∣∣∣∣ ≤ 2
(ˆaij∂iu∂juχ
2
) 12(ˆ
aij∂iχ∂jχ|u|2) 1
2
≤ 1
2
ˆaij∂iu∂juχ
2 + 4
ˆaij∂iχ∂jχ|u|2
1
2
ˆaij∂iu∂juχ
2 ≤ 4
ˆaij∂iχ∂jχ|u|2
where we have used the standard fact that ab ≤ ε2a
2 + 12εb
2 where we chose ε to give us the right coefficients.Then from the ellipticity condition we have
λ
2
ˆBr(x0)
|Du|2 ≤ 1
2
ˆaij∂iu∂juχ
2 ≤ 4
ˆaij∂iχ∂jχ|u|2 ≤ 16
ˆΛ|Dχ|2|u|2 ≤ 8Λ
(R− r)2
ˆBR(x0)
|u|2
which proves the inequality for k = 1. To prove |α| = k ∈ Z, we proceed by induction. Assume u ∈ C∞.Then Dαu ∈ C∞ and satisfies the same equation aij∂i∂j(D
α) = 0. Applying the previous case with k = 1gives (by induction)
ˆBr(x0)
|D(Dαu)|2 ≤ C(r+R
2 − r)2 ˆ
BR−r(x0)
|Dαu|2
≤ C(R+r
2 − r) (R− R+r
2
)2k ˆBR(x0)
|u|2
=C
(R− r)2(k+1)
ˆBR(x0)
|u|2.
23
As one can expect, for the non-smooth case, we use mollifiers. Take η ∈ C∞0 (|x| < 1) with
ˆRnη = 1
and define ηε = 1εn η(xε ). Then define
uε(x) =
ˆu(x− y)ηε(y)dy =
ˆu(y)ηε(x− y)dy
which is well defined for dist(x, ∂Ω) > ε. By an exercise we leave to the reader, note that if u ∈ Lp(Ω) for1 < p < ∞, then for any K ⊂⊂ Ω and ε < εK , then uε → u in Lp(K). Using this, and by a single exercisethat shows ˆ
aij∂iuε∂jv =
ˆaij∂iu∂jvε
we can conclude that if u satisfies aij∂i∂ju = 0 in the generalized sense then so does uε. Thus we have thatuε ∈ C∞ and aij∂i∂juε = 0 and thus we can apply our estimate (18) to find
ˆBr(x0)
|Dαuε| ≤1
(R− r)2k
ˆBR(x0)
|uε|2.
Lets try to generalize this. We will do this with the following theorem.
Theorem 3.1. Let u ∈W 1,2(Ω) be a weak solution to
∂i(aij∂ju) + c(x)u = f (19)
in Ω ⊂ Rn. Assume that
i) 0 < λ ≤ aij ≤ Λ
ii) aij are continuous with modulus of continuity τ i.e. |aij(x)− aij(y)| ≤ τ(|x− y|)
iii) c ∈ Ln, f ∈ Lq where n2 < q < n.
Then for any BR(x) ⊂⊂ Ω, u ∈ Cα(BR(x)) with α = 2− nq and 0 < α < 1 with the estimate
‖u‖Cα(B) ≤ Cn,λ,Λ,τ,‖c‖p(‖f‖Lq(Ω) + ‖u‖W 1,2(Ω)
).
In order to prove this we will need a few lemmas. They are as follows.
Lemma 3.2. Assume u ∈W 1,2(B) and
ˆBr(x0)
|u− ux0|2dx ≤M2rn+2α.
Then u ∈ Cα and ‖u‖Cα(B) ≤ C(M + ‖u‖L1).
Proof. First note that what we are doing makes some sort of sense. In the assumption, we have the differencebetween u and its average at x0 being squared, so that explains the 2α. Then since we are integrating overthe ball, we have the factor of rn.
Let 0 < r < R and look at the difference between averages
|ur(x0)− uR(x0)| ≤ |ur(x0)− u(x)|+ |u(x)− uR||ur(x0)− uR(x0)|2 ≤ 2
(|ur − u(x)|2 + |u(x)− uR|2
).
24
We now integrate over the ball of radius r and using the fact that the LHS is a constant and the assumptionof the lemma we have
rn|ur(x0)− uR(x0)|2 ≤ 2
(ˆBr(x0)
|ur − u(x)|2 +
ˆBr(x0)
|u(x)− uR|2)
≤ 2
(M2rn+2α +
ˆBR(x0)
|u(x)− uR|2)
≤ CM2(rn+2α +Rn+2α
)|ur(x0)− uR(x0)|2 ≤ CM2
(r2α +
(R
r
)nR2α
)≤ CM2
(1 +
(R
r
)n)R2α
|ur(x0)− uR(x0)| ≤ CM
(1 +
(R
r
)n2
)Rα. (20)
Let r = 2−l−1L,R = 2−lL and plug this into (20) to see that
|ur − uR| ≤ CM
(1 +
(2−lL
2−l−1L
)n2
)(2−lL)α
= CM(1 + 2n2 )(2−lL)α
= CM(2−lL)α. (21)
Now use (21) to see that
|u2−m−1L − u2−lL| ≤m∑i=l
|u2−i−1L − u−i|
≤ CMm∑i=l
(2−iL)α
≤ CM(2−lL)α (22)
where we have used the fact that we had a telescoping sequence and that we have a geometric sequence thatwe have bounded by the highest term. This implies that u2−lL is a cauchy sequence and so we can define
u∗ := liml→∞
u2−lL(x0).
We will now show that this is independent of L. Take L < L′. Then (22) implies that
|u2−lL(x0)− u2−lL′ | ≤ CM(
1 +
(L′
L
)n)(2−lL′)α
and taking l→∞ shows that u∗ is independent of L.From all of our hard work, we can say that u∗ = u almost everywhere by the Lebesgue Differentiation
Theorem which says that if u ∈ L1, then limr→0 ur(x0) = u(x0) for almost every x0. Now taking m→∞ in(22) we are able to get
|u2−m−1L(x0)− u2−lL(x0)| ≤ CM(2−lL)α
|u∗(x0)− u2−lL(x0)| ≤ CM(2−lL)α
|u∗(x0)− ur(x0)| ≤ CMrα. (23)
25
Letting r = 1 implies that ‖u∗(x0)‖L∞(B1) ≤ C(M + ‖u‖L1(B1))Now in order to complete this theorem we need to estimate [u∗]Cα . Let x, y ∈ Ω such thatBr(x), Br(y) ⊂⊂
Ω and Br(x)∩Br(y) 6= ∅. Denote δ = |x− y| and let z be the point midway between x and y. By convexitywe can see that Bδ(z) ⊂⊂ Br(x) ∩Br(y). Then we write
|u∗(x)− u∗(y)| ≤ |u∗(x)− uδ(x)|+ |u∗(y)− uδ(y)|+ |uδ(x)− u(z)|+ |uδ(y)− u(z)||u∗(x)− u∗(y)|2 ≤ C
(|u∗(x)− uδ(x)|2 + |u∗(y)− uδ(y)|2 + |uδ(x)− u(z)|2 + |uδ(y)− u(z)|2
)≤ CM2δ2α + C|uδ(x)− u(z)|2 + C|uδ(y)− u(z)|2
where we have used (23) on the first two terms. Integrating the inequality with respect to z and using theassumptions of the lemma yields the required results.
Lemma 3.3. Assume that ˆBr(x0)
|Du|2 ≤M2rn−2+2α.
Then u ∈ Cα.
Proof. The details of the proof will be left as an exercise. Here is a sketch of it. Recall the Poincareinequality: ˆ
S
|u− uS |2 ≤ λSˆS
|Du|2.
Then you need to show that ˆBr(x0)
|u− ur(x0)|2 ≤ Cr2
ˆBr(x0)
|Du|2
where c is independent of r. The way to do this is to apply the Poincare Inequality with r = 1. Then considerthe rescaling u = u(rx). After doing this, apply the assumption of the lemma and you’ll find yourself in theposition of Lemma 3.2.
Lets recall our goals. We were proving Schauder estimates to get regularity. Assume u ∈W 1,2 is a weaksolution of ∂i(aij∂ju) = 0 i.e. ˆ
aij∂ju∂iv = 0 ∀v ∈W 1,20 .
Then we will show that u ∈ Cα where 0 < λ ≤ aij ≤ Λ and |aij(x)− aij(y)| ≤ τ(|x− y|) with τ ↓ 0 as R ↓ 0.We will be using the following key estimate for 0 < r < R
ˆBr(x0)
|Du|2 ≤ C[( rR
)n+ τ(R)
] ˆBR(x0)
|Du|2. (24)
Proof. We first prove the case where we have a constant coefficient i.e. τ ≡ 0. Then we will show that
ˆBr(x0)
|Du|2 ≤ C( rR
)n ˆBR(x0)
|Du|2 (25)
and use a the Lemma of De Giorgi (seen later). By rescaling v(x) = u(Rx) what we need to show becomes
R−2
ˆB rR
(x0)
|Dv|2 ≤ C( rR
)n ˆB1(x)
|Dv|2 ·R−2
mˆBs(x0)
|Dv|2 ≤ CsnˆB1(x0)
|Dv|2.
26
This will follow from previous work. Take s small (s < 12 ). Then By the Sobolev Embedding Theorem and
by the first theorem proved in this section we are able to sayˆBs(x0)
|Dv|2 ≤ supBs
|Dv|2 · sn
≤
∑|α|≤k
ˆB 3
4(x0)
Dα(Dv)2
sn
≤ CsnˆB1(x0)
|Dv|2
and we are done showing (25), which corresponds to the case where aij are constants. Since this case willfollow when we prove the Lemma of De Giorgi, we move on to the non-constant case.
We will prove the non-constant case as a perturbation of the equation aij(x0)∂i∂jw = 0, which we havealready finished. To carry this out, consider the following Dirichlet problem
aij(x0)∂i∂jw = 0 weak sense
w − u ∈W 1,20 (BR(x0)).
Set v = u− w and u = v + w. ThenˆBr(x0)
|Du|2 ≤ 2
[ˆBr(x0)
|Dv|2 +
ˆBr(x0)
|Dw|2]
≤ 2
[ˆBr(x0)
|Dv|2 + c( rR
)n ˆBR(x0)
|Dw|2]
≤ 2
[ˆBr(x0)
|Dv|2 + c( rR
)n ˆBR(x0)
|Du|2 + |Dv|2]
≤ C
[( rR
)n ˆBr(x0)
|Du|2 +
ˆBR(x0)
|Dv|2]
(26)
where we have used the fact that w solves our PDE with constant coefficients, saw that the integral over ris ≤ than the integral over R, and saw max1, r
n
Rn = 1. Now we need to control the integral of |Dv|2, so weuse the fact that v = u−w, where u solves ∂i(aij∂ju) = 0 and w solves aij(x0)∂i∂jw = 0 in the weak sense.
Now since v ∈W 1,20 , we are able to use it as a test function in the definition of weak derivatives to getˆ
aij(x0)DjvDiv =
ˆaij(x0)(Dju+Djw)Div
=
ˆaij(x0)DjuDiv
=
ˆ(aij(x0)− aij(x))DjuDiv +
ˆaij(x)DjuDiv
=
ˆ(aij(x0)− aij(x))DjuDiv.
We now use the fact that aij is elliptic and use its modulus of continuity to see
λ
ˆBR(x0)
|Dv|2 ≤ τ(R)
ˆBR(x0)
|Du||Dv|
≤ τ(R)
ˆBR(x0)
1
2|Du|2 +
1
2|Dv|2
ˆBR(x0)
|Dv|2 ≤ Cτ(R)
ˆBR(x0)
|Du|2
27
which we can plug into (26) to see that we have finally proved (24) for the case of non-constant aij .We still need to relate everything to get that u ∈ Cα. Now we finally state the Lemma of De Girogi:
Lemma 3.4 (Lemma of De Giorgi). Let ϕ(r) ≥ 0 with ϕ(r) ↓ as r ↓, and A,B ≥ 0. Assume that 0 < r < Rand ∃α > β > 0 such that
ϕ(r) ≤ A[( rR
)α+ ε]ϕ(R) +BRβ . (27)
Then ∀0 < β < γ < α,∃ε0 such that ε < ε0 implies
ϕ(r) ≤ C[( rR
)γϕ(R) +Brβ
]. (28)
We will apply the Lemma with
ϕ(r) =
ˆBr(x0)
|Du|2
because (24) implies that ϕ(r) satisfies (27) with B = 0 and α = n. Then by the Lemma of De Giorgi wehave
ϕ(r) ≤ C( rR
)γϕ(R)
and by taking R = 1 we imply that ϕ(r) ≤ Crγ after absorbing ϕ(1) into the constant. We are now in thecase of Lemma 3.3 because we are free to choose any γ < α = n and
ˆ|u− ur(x0)|2dx ≤ Crγ+2.
Hence u ∈ Cα.
Recall that we consider weak solutions of
∂i(aij∂ju) + c(x)u = f
where 0 < λ ≤ aij ≤ Λ and |aij(x) − aij(y)| ≤ τ(|x − y|) where τ ↓ 0 as R ↓ 0, c ∈ Ln and f ∈ Lq forn2 < q < n. We want to show that u ∈ Cα for α = 2− n
q .
Proof. We already got the case for c = f = 0, where for 0 < r < R the main tool was showing
ˆBr(x0)
|Du|2 ≤ C
( rR
)n ˆBR(x0)
|Du|2 + τ(R)
ˆBR(x0)
|Dv|2. (29)
Recall that we defined v = u− w where w solvesaij(x0)∂i∂jw = 0 on BR(x0)
u− w ∈W 1,20 (BR(x0))
in the generalized sense. By the same argument as before, (29) can also be established for non-trivial lowerordered terms. Lets now estimate |Dv| in (29):
ˆaij(x0)DivDjv =
ˆaij(x0)(Diu−Diw)Djv
=
ˆaij(x0)DiuDjv
=
ˆaij(x)DiuDjv +
ˆ(aij(x0)− aij(x))DiuDjv
= −ˆc(x)uv +
ˆfv +
ˆ(aij(x0)− aij(x))DiuDjv. (30)
28
These calculations show two new terms that were not there before in the proof of the previous theorem. Wehandle these terms by bounding their derivatives. Recall the Sobolev inequality:
‖v‖L
2nn−2≤ C‖Dv‖L2
and use the Holder’s and the Sobolev inequality∣∣∣∣∣ˆBR(x0)
c(x)uv
∣∣∣∣∣ ≤ ‖u‖L2
(ˆBR(x0)
|cv|2)1/2
≤ ‖u‖L2
[(ˆ|v|
2nn−2
)?(ˆ|c2|)?]1/2
where we have to figure out which powers to use. We need Lesbegue conjugates 1p + 1
q = 1 and so we let
2p = 2nn−2 =⇒ p = n
n−2 , q = n2 . This implies
ˆ|cv|2 ≤
(ˆ|v|
2nn−2
)n−2n(ˆ
(|c|2)n2
)2/n
= ‖v‖2L
2nn−2‖c‖2Ln
Putting this together yields ∣∣∣∣ˆ c(x)uv
∣∣∣∣ ≤ ‖u‖L2‖v‖L
2nn−2‖c‖Ln
≤ ‖u‖L2‖Dv‖L2‖c‖Ln
≤ ε
2‖Dv‖2L2 +
1
2ε‖u‖2L2‖c‖2Ln
and now the first term gets absorbed in (29). The second term in (30) can be bounded as follows∣∣∣∣ˆBR
fv
∣∣∣∣ ≤ ‖f‖L 2nn+2‖v‖
L2nn−2
≤ ‖f‖L
2nn+2‖Dv‖L2
≤ ε
2‖Dv‖2L2 +
1
2ε‖f‖2
L2nn+2
.
Once again the Dv term gets absorbed in (29). Putting everything together gives us the following estimate
ˆBr(x0)
|Du|2 ≤ C
(( rR
)n+ τ(R)
)ˆBR(x0)
|Du|2
+ C‖u‖2L2(BR)‖c‖2Ln(BR) + C‖f‖2
L2nn+2
. (31)
Since we want to show u ∈ Cα, we look for things of the formˆBr(x0)
|Du|2 ≤Mrn+2−2α.
Now recall the Lemma of De Giorgi: ϕ(r) ↓ 0 as r ↓ 0, ϕ(r) ≥ 0. Assume that ∀0 < r < R, β < α and
ϕ(r) ≤ A(( r
R
)α+ ε)ϕ(R) +BRβ .
Then ∀0 < β < γ < α,∃ε0 > 0 such that ε < ε0 implies
ϕ(r) ≤ C( rR
)γϕ(R) +Brβ .
29
In particular we can let R = 1 above and have
ϕ(r) ≤ Crβ .
The analysis tells us that in order to apply the lemma, we need to estimate the additional terms in (31) byRγ for γ = n− 2 + 2α. Lets begin with the f term
‖f‖2L
2nn+2 (BR(x0))
=
(ˆBR(x0)
|f |2nn+2
)n+2n
≤
((ˆ|f |q
)?(ˆ1
)?)n+2
n
.
In order to figure out the exponents, we need 2nn+2p = q =⇒ p = q(n+2)
2n =⇒ 1m = 1− 1
p = 1− 2n(n+2)q . Thus
‖f‖2L
2nn+2 (BR(x0))
≤
((ˆ|f |q
)1/p(ˆ1
)1/m)n+2
n
=(‖f‖q/pLq R
n/m)n+2
n
= ‖f‖2LqRn+2m = ‖f‖2LqRn−2+2α
where α = 2− nq . Now consider the other term in (31):
‖u‖2L2(BR(x0)) =
ˆBR(x0)
|u|2
≤(ˆ
BR
|u|2nn−2
)n−2n(ˆ
BR
1
)2/n
≤ ‖u‖2L
2nn−2
(Rn)2/n
≤ ‖u‖2W 1,2(BR)R2.
In the end the lower ordered terms are all bounded by
Rminn−2+2α,2 (‖u‖2W 1,2(BR) + ‖f‖2Lq )︸ ︷︷ ︸B
Then (31) and the De Giorgi Lemma tell us
ˆBr(x0)
|Du|2 ≤ C( rR
)γ ˆBR(x0)
|Du|2 +Brminn−2+2α,2
≤ Crminn−2+2α,2
where again we have used the trick of letting R = 1.Now there are two cases. If minn − 2 + 2α, 2 = n − 2 + 2α, we are done by Lemma 3.3. If it is not,
consider the integral of |u|2 over Br(x0). I claim that
ˆBr(x0)
|u|2 ≤( rR
)n ˆBr(x0)
|u|2 + CR2+minn−2+2α,2.
If I manage to show this, observe that we are done by the De Giorgi Lemma. Consider the triangle type
30
inequality |u|2 ≤ |u− uR(x0)|2 + u2R and integrate over the ball of radius r to see
ˆBr(x0)
|u|2 ≤ˆBr(x0)
|u− uR(x0)|2 +
ˆBr(x0)
u2R
≤ˆBR(x0)
|u− uR(x0)|2 +
ˆBr(x0)
u2R (r ≤ R)
≤ R2
ˆBR(x0)
|Du|2 +
ˆBr(x0)
u2R (Poincare)
≤ R2+minn−2+2α,2 +
ˆBr(x0)
u2R
= R2+minn−2+2α,2 +
ˆBr(x0)
(1
Rn
ˆBR(x0)
u
)2
≤ R2+minn−2+2α,2 +
ˆBr(x0)
1
Rn
(ˆBR(x0)
u2
)1/2(ˆBR(x0)
1
)1/22
= R2+minn−2+2α,2 +
ˆBr(x0)
1
Rn/2
(ˆBR(x0)
u2
)1/22
= R2+minn−2+2α,2 +rn
Rn
ˆBR(x0)
u2.
3.2 Harnack Inequality
The Harnack Inequality is essential in proving Regularity estimates in a different way as the previous section.We follow the proof of presented in Gilbarg and Trudinger very closely. We begin by defining our differentialoperator
Lu = Di(aijDju+ biu) + ciDiu+ du
and force boundedness of the coefficients on L:∑|aij(x)|2 ≤ Λ2, λ−2
(∑|bi(x)|2 + |ci(x)|2
)+ λ−1|d(x)| ≤ ν2.
We also make this an elliptic equation by enforcing
aij(x)ξiξj ≥ λ|ξ|2, ∀x ∈ Ω, ξ ∈ Rn.
Then u is a weak solution to Lu = 0(≥ 0,≤ 0) in the domain Ω ifˆ
Ω
(aijDju+ biu
)Div − (ciDiu+ du)v
dx = 0(≤ 0,≥ 0)
for all non-negative functions v ∈ C10 (Ω). Let f i, g be locally integrable functions in Ω. Then u is a weak
solution of the inhomogeneous equationLu = g +Dif
i
in Ω if it satisfies ˆΩ
(aijDju+ biu
)Div − (ciDiu+ du)v
dx =
ˆΩ
(f iDiv − gv
)dx
for all v ∈ C10 (Ω).
31
3.2.1 Structural Inequalities
We rewrite Lu = g +Difi as
DiAi(x, u,Du) +B(x, u,Du) = 0, (32)
where
Ai(x, z, p) = aijpj + biz − f i
B(x, z, p) = cipi + dz − g
for (x, z, p) ∈ Ω × R × Rn. Then we say that u is a weak subsolution (supersolution, solution) of (32) in Ωif Ai(x, u,Du) and B(x, u,Du) are locally integrable and
ˆΩ
(DivA
i(x, u,Du)− vB(x, u,Du))dx ≤ (≥,=)0 (33)
for all non-negative v ∈ C10 (Ω). Writing b = (b1, . . . , bn), c = (c1, . . . , cn), f = (f1, . . . , fn) and using the
Schwarz inequality, we have the inequalities
piAi(x, z, p) ≥ λ
2|p|2 − 1
λ(|bz|2 + |f |2)
|B(x, z, p)| ≤ |c||p|+ |dz|+ |g|.
We can consolidate this further by writing
z = |z|+ k, b = λ−2(|b|2 + |c|2 + k−2|f |2) + λ−1(|d|+ k−1|g|)
for some k > 0. Then for an 0 < ε < 1, we finally have the following inequalities
piAi(x, z, p) ≥ λ
2(|p|2 − 2bz2)
|zB(x, z, p)| ≤ λ
2
(ε|p|2 +
b
εz2
).
If we denote aij by a, then we can write |A(x, z, p)| ≤ |a||p| + |bz| + |f |. Also note that we can divide (32)by 2
λ to finally get the structural inequalities
|A(x, z, p)| ≤ |a||p|+ 2b1/2z
p ·A ≥ |p|2 − 2bz2 (34)
|zB(x, z, p)| ≤ ε|p|2 +1
εbz2
for some ε ∈ (0, 1], z = |z|+ k, b = 14 (|b|2 + |c|2 + k−2|f |) + 1
2 (|d|+ k−1|g|). For the purposes of the followingproof, we will let k = k(R) = 1
2 (Rδ‖f‖q +R2δ‖g‖q/2), for δ = 1− n/q.
3.2.2 Moser Iteration
Before proving The Harnack Inequality, we must prove two theorems first.
Theorem 3.5. Let L be uniformly elliptic with bounded coefficients. f i ∈ Lq(Ω), g ∈ Lq/2(Ω) for q > n. Letu ∈W 1,2(Ω) be a subsolution in Ω. Then for any B2R(y) ⊂ Ω, and p > 1, we have
supBR(y)
u ≤ C(R−n/p‖u+‖Lp(B2R(y)) + k(R)
). (35)
32
Theorem 3.6. Let L be uniformly elliptic with bounded coefficients and suppose that f i ∈ Lq(Ω), g ∈ Lq/2(Ω)for q > n. Let u ∈ W 1,2(Ω) be a supersolution in Ω. If u is non-negative in B4R(y) ⊂ Ω and 1 ≤ p <n/(n− 2), we have
R−n/p‖u‖Lp(B2R(y)) ≤ C(
infBR(y)
u+ k(R)
). (36)
Proof. It is convenient to prove these two theorems conjointly in the case where u is a bounded non-negativesubsolution. We begin by assuming that R = 1, k > 0. The general case will be obtained from transformingx 7→ x/R and letting k → 0. Let β 6= 0, η ∈ C1
0 (B4) be non-negative. We define v := η2uβ . Recall thatu = u+ k. Then we have that
Dv = 2ηDηuβ + βη2uβ−1Du.
Note that v is a valid test function. Then we plug this v into our definition of subsolutions
ˆΩ
(DivA
i(x, u,Du)− vB(x, u,Du))dx ≤ 0
and obtain ˆΩ
(2ηDηuβA(x, u,Du) + βη2uβ−1DuA(x, u,Du)
)dx ≤
ˆΩ
η2uβB(x, u,Du) dx. (37)
We will now attempt to apply our structural inequalities (34) into (37):
η2uβ−1DuA(x, u,Du) ≥ η2uβ−1|Du|2 − 2η2buβ+1
|η2uβB(x, u,Du)| = η2uβ−1|uB(x, u, du)|
≤ η2uβ−1
(ε|Du|2 +
1
εbu2
)= εη2|Du|2uβ−1 +
1
εbη2uβ+1.
The last structural inequality gives us this:
|ηDη ·A(x, u,Du)uβ | ≤ |a|η|Dη|uβ |Du|+ 2b1/2η|Dη|uβ+1
= |a|η|Dη|uβ−1
2 uβ+1
2 |Du|+ 2b1/2η|Dη|uβ+1
≤ ε
2η2uβ−1|Du|2 +
|a|2
2ε|Dη|2uβ+1 + |Dη|2uβ+1 + bη2uβ+1
=ε
2η2uβ−1|Du|2 +
(1 +|a|2
2ε
)|Dη|2uβ+1 + bη2uβ+1.
Applying this to (37) yields
ˆΩ
(βη2uβ−1 − 2bβη2uβ+1
)dx ≤
ˆΩ
εη2|Du|2uβ−1 +
1
εbη2uβ+1 + εη2uβ−1|Du|2
+
(2 +|a|2
ε
)|Dη|2uβ+1 + 2bη2uβ+1
dx
Combining like terms yields
ˆΩ
(β − 2ε)η2|Du|2uβ−1 dx ≤ˆ
Ω
(2β + 2 +
1
ε
)bη2 +
(2 +|a|2
ε
)|Dη|2
uβ+1 dx.
33
Let ε = min
1, β4
. Then we can further consolidate this into
ˆΩ
η2|Du|2uβ−1 dx ≤ C(β)
ˆΩ
(bη2 +
(1 + |a|2
)|Dη|2
)uβ+1 dx. (38)
We introduce the function
w :=
u(β+1)/2 if β 6= −1log u if β = −1
so that
|ηDw| =η β+1
2 u(β−1)/2|Du| if β 6= −1η|Du|u−1 if β = −1
and let γ = β + 1. Then we can rewrite (38) as
ˆΩ
|ηDw|2 dx ≤
C(|β|)γ2
´Ω
(bη2 +
(1 + |a|2
)|Dη|2
)w2 dx if β 6= −1
C´
Ω
(bη2 +
(1 + |a|2
)|Dη|2
)dx if β = 1.
(39)
Before we move any further, we need to introduce a few results from analysis of Sobolev spaces.
Lemma 3.7 (Interpolation Inequality). Let p ≤ q ≤ r. Then for u ∈ Lr(Ω), we have
‖u‖q ≤ ε‖u‖r + ε−µ‖u‖p,
where
µ =
1p −
1q
1q −
1r
.
From the Sobolev Inequality, we have
‖ηw‖22n/(n−2) ≤ Cˆ
Ω
(|Dηw|2 + |ηDw|2
)dx,
for n = n for n > 2, and 2 < 2 < q. Now we apply Holder’s Inequality and the Interpolation Inequality toget
ˆΩ
b(ηw)2 dx ≤ ‖b‖q/2‖ηw‖22q/(q−2)
≤ ‖b‖q/2(ε‖ηw‖2n/(n−2) + ε−σ‖ηw‖2
)2,
where σ = n/(q− n). Now we attempt to plug in these estimates into (39). Let ξ = n/(n− 2). Now we adda factor of
´Ω|wDη|2 dx and carry out some computations to get the following
34
‖ηw‖2χ ≤ Cγ2
ˆΩ
(b(ηw)2 + |wDη|2 + |a|2|wDη|2
)dx
≤ Cγ2‖b‖q/2(ε‖ηw‖2ξ + ε−σ‖ηw‖2
)2+ Cγ2
ˆΩ
(1 + |a|2)|wDη|2 dx
= Cγ2(ε‖ηw‖2χ + ε−σ‖ηw‖2
)2+ Cγ2
ˆΩ
(1 + |a|2)|wDη|2 dx
≤ Cγ2(ε2‖ηw‖22χ + ε1−σ‖ηw‖2χ‖ηw‖2 + ε−2σ‖ηw‖22 + ‖wDη‖22
)‖ηw‖22χ(1− Cγ2ε2) ≤ Cγ2
(ε1−σ‖ηw‖2χ‖ηw‖2 + ε−2σ‖ηw‖22 + ‖wDη‖22
)≤ Cγ2
(ε1−σ
2‖ηw‖22χ +
ε1−σ
2‖ηw‖22 + ε−2σ‖ηw‖22 + ‖wDη‖22
)‖ηw‖22χ
(1− Cγ2
(ε+
ε1−σ
2
))≤ Cγ2
(ε1−σ
2+ ε−2σ
)(‖ηw‖22 + ‖wDη‖22
)‖ηw‖22χ ≤
Cγ2(ε1−σ
2 + ε−2σ)
1− Cγ2(ε1−σ
2 + ε−2σ) (‖ηw‖22 + ‖wDη‖22
)≤ C(1 + |γ|)σ+1‖(η + |Dη|)w‖22
Then we finally get
‖ηw‖2χ ≤ C(1 + |γ|)σ+1‖(η + |Dη|)w‖2, (40)
where C = C(n,Λ, ν, q, |β|) is bounded when |β| is bounded away from zero. We will now get a better cutofffunction η. Let r1, r2 be such that 1 ≤ r1 < r2 ≤ 3, η ≡ 1 in Br1 , β ≡ 0 in Ω\Br2 , with
|Dη| ≤ 2
r2 − r2.
Then we have from (40)
‖w‖L2χ(Br1 ) ≤C(1 + |γ|)σ+1
r2 − r1‖w‖L2(Br2 ). (41)
Before we move on, lets make a quick backtrack to functional spaces. Let Ω ⊂ Rn be a bounded domain.Then I claim that if u is a measurable function on Ω such that |u|p ∈ L1(Ω) for some p ∈ R, and define
φp(u) :=
(1
|Ω|
ˆΩ
|u|p dx)1/p
,
then we have
limp→∞
φp(u) = supΩ|u|. (42)
To start, note that
φp(u) =
(1
|Ω|
ˆΩ
|u|p dx)1/p
≤(
1
|Ω|
ˆΩ
(sup
Ω|u|)p
dx
)1/p
= supΩ|u|.
Taking the lim sup yieldslim supp→∞
φp(u) ≤ supΩ|u|.
35
Now fix ε and define A = x ∈ Ω||u| ≥ supΩ |u| − ε. Then we see that
φp(u) =
(1
|Ω|
ˆΩ
|u|p dx)1/p
≥(
1
|Ω|
ˆA
|u|p dx)1/p
≥(
1
|Ω|
ˆA
(sup
Ω|u| − ε
)pdx
)1/p
=|A||Ω|
(sup
Ω|u| − ε
).
Taking ε→ 0 and then the lim inf yields
lim infp→∞
φp(u) ≥ supΩ|u|.
This and our first inequality yield the claim (42). We go back to our proof. For r < 4, we define the function
φ(p, r) :=
(ˆBr
|u|p dx)1/p
.
From what we just showed, we have that
φ(∞, r) = limp→∞
φ(p, r) = supBr
u.
Now we can rewrite inequality (42) as
φ(χγ, r1) ≤(C(1 + |γ|)σ+1
r2 − r1
)2/|γ|
φ(γ, r2) if γ > 0 (43)
φ(γ, r2) ≤(C(1 + |γ|)σ+1
r2 − r1
)2/|γ|
φ(χγ, r1) if γ < 0 (44)
Now we are in the right position to start our iteration. Recall that when u is a subsolution, we have β > 0and γ > 1. Taking p > 1 and setting γ = γm = χmp and rm = 1 + 2−m. Then plugging this into ourinequality (43) gives us
φ(χmp, 1) ≤(C(1 + |χm−1p|)1+σ
r2 − 1
)2/(χm−1p)
φ(χm−1p, r2)
≤(C(1 + |χm−2p|)1+σ
r3 − r2
)2/(χm−2p)(C(1 + |χm−1p|)1+σ
r2 − 1
)2/(χm−1p)
φ(χm−2, r4) ≤ · · ·
≤ (Cχ)2(1+σ)∑mχ−mφ(p, 2) = Cφ(p, 2).
Letting m→∞ in the above yields
supB1
|u| ≤ C‖u‖Lp(Br2 ), C = C(n,Λ, ν, q, p).
Transforming x 7→ x/R we have the desired estimate (35) if u is a subsolution. Now in the cases when u isa supersolution, we need to approach the problem a bit differently. Recall that when u is a supersolution,β < 0 and γ < 1. Then for any p, p0 such that 0 < p0 < p < χ, we have
φ(p, 2) ≤ Cφ(p0, 3)
φ(−p0, 3) ≤ Cφ(−∞, 1).
Then we will finish proving our theorem if we can show that φ(p0, 3) ≤ C(−p0, 3). This is done in an intricateway and is left as an exercise.
36
Putting our two theorems together give us the full Harnack Inequality:
Theorem 3.8. Let L be uniformly elliptic and have bounded coefficients. Let u ∈ W 1,2(Ω) satisfy u ≥ 0 inΩ and Lu = 0 in Ω. Then for any ball B4R(y) ⊂ Ω, we have
supBR(y)
u ≤ C infBR(y)
u,
where C = C(n,Λ/λ, νR).
3.2.3 Applications of the Harnack Inequality
We can give a new proof of the strong maximum principle (instead of using Hopf’s Lemma) now:
Theorem 3.9. L uniformly elliptic, bounded coeff, u ∈W 1,2(Ω), Lu ≥ 0 in Ω. Then if for some ball B ⊂⊂ Ωwe have
supBu = sup
Ωu ≥ 0,
then the function u must be constant in Ω.
Proof. We apply the Harnack inequality with p = 1 to the function v = M−u and show that M−u = 0.
Finally, we can use the Harnack Inequality to imply Holder continuity.
Theorem 3.10. Let L be uniformly elliptic and have bounded coefficients. Then if u ∈W 1,2(Ω) is a solutionof
Lu = g +Difi,
then u is locally Holder continuous in Ω, and for any ball B0 = BR0(y) ⊂ Ω and R ≤ R0 we have
oscBR(y) u ≤ CRα(R−α0 supB0
|u|+ k).
Proof. Start out by defining M0,M1,M4,m1,m4. Then apply Harnack inequality with p = 1 to M4− u andu−m4 to end up with
ω(R) ≤ γω(4R) + k(R).
Here ω(R) = oscBR u. Before we go on, we have to prove something else first. Suppose ω is non-decreasingon (0, R0], R ≤ R0 satisfies
ω(τR≤γω(R) + σ(R),
where σ is also non-decreasing, 0 < γ, τ,< 1. Then for any µ ∈ (0, 1), we have
ω(R) ≤ C((
R
R0
)αω(R)0 + σ(RµR1−µ
0 )
).
Of course the De Giorgi Lemma concludes the proof.
37
4 Harnack Inequality for Non-Divergence Equations
The formulation of weak solutions to divergence equations relied heavily on the fact that the operator L wasin divergence form. This allowed us to integrate by parts and so a weak solution u needs to be once weaklydifferentiable (W 1,2). A classical solution u must be at least second order continuously differentiable. In thissection we will concern ourselves with the intermediate situation of strong solutions.
Definition 4.1. For operators of the form
Lu = aijuij + biui + c(x)u (45)
with coefficients aij , bi, c(x) defined on a domain Ω ⊂ Rn and a function f on Ω, a strong solution of Lu = fis a function u ∈W 2,p(Ω) that satisfies (45) almost everywhere. See [1][p. 185] for existence and uniquenessof equations of this type.
4.1 ABP Estimate
We begin by proving a maximum principle for strong solutions analogous to the classical one for classicalsolutions. It is called the Alexandrov-Bakelmann-Pucci Maximum Principle. In particular, we will prove thismaximum principle for solutions in the space W 2,n
loc (Ω) as it is the natural environment for such equations.Recall that an operator L in the form (45) is said to be elliptic in the domain Ω ⊂ Rn if the matrix (aij) ispositive definite everywhere in Ω. Let D := det(aij) and D? = D1/n so that
0 < λ ≤ D? ≤ Λ
where λ,Λ are the minimum and maximum eigenvalues of (aij). Assume b = c = 0 so that we haveLu = aijuij . Our condition on aij and f are now
f/D? ∈ Ln(Ω).
Theorem 4.2 (ABP Maximum Principle). Let Lu ≥ f in a bounded domain Ω and u ∈ C0(Ω) ∩W 2,nloc (Ω).
Then
supΩu ≤ sup
∂Ω+
d
nω1/nn
∥∥∥∥ f
D?
∥∥∥∥Ln(Γ+)
where d = diam Ω.
It is important to note that the Morrey’s embedding theorem guarantees that u ∈ W 2,nloc (Ω) will be at
least continuous in Ω because whenever kp > n, W k,p is embedded in Cα for 0 ≤ α < k − np . Before we
begin the proof, we must first go through notions of contact sets and normal mappings.
Definition 4.3. Suppose u is an arbitrary function on Ω. The upper contact set Γ+ or Γ+u is defined to be
the subset of Ω such that the graph of u is below a support hyperplane in Rn+1, i.e.
Γ+ = y ∈ Ω|u(x) ≤ p(x− y) + u(y) ∀x ∈ Ω, for some p ∈ Rn.
From the definition, u will be a concave function on Ω if and only if Γ+ = Ω. It is clear that p = Du(y) ifu ∈ C1(Ω). Finally, if u ∈ C2(Ω), the Hessian D2u ≤ 0 on Γ+. This means that we can essentially think ofΓ+ as the subset of Ω where u is concave down. The upper contact set of u is closed in Ω.
Definition 4.4. Suppose u ∈ C0(Ω) is arbitrary. We define the normal mapping χ(y) = χu(y) for a pointy ∈ Ω to be the set of slopes of supporting hyperplanes at y lying above the graph of u, i.e.
χ(y) = p ∈ Rn|u(x) ≤ p(x− y) + u(y) ∀x ∈ Ω.
Clearly χ(y) is nonempty if and only if y ∈ Γ+. When u ∈ C1(Ω), then χy = Du(y).
38
Proof of Theorem 4.2. Assume that u ∈ C0(Ω)∩C2(Ω). Subtracting sup∂Ω u from u yields the same differ-ential inequality L(u− sup∂Ω) ≥ f and so we can assume u ≤ 0 on the boundary. Note that we can assumesupΩ u ≥ 0 because if it was negative, there would be nothing to prove. This assumption implies that ifu(y) := supΩ u, then y is in the interior of Ω because u ≤ 0 on the boundary. Now I show that
Bu(y)/d ⊂ χ(Γ) = Du(Γ). (46)
This can be seen by sliding hyperplanes onto the graph of u. Consider the cone with vertex u(y) with base∂Ω. Then the slope of the cone is u(y)/d and u ≤ 0 on ∂Ω implies that dropping down hyperplanes of slopeu(y)/d will eventually be tangent to u(x) at a point x ∈ Γ. This can be seen very clearly in pictures:
It is a bit difficult to see, but in the picture, we have u|∂Ω≤0. Now we compute
|Du(Γ)| =ˆDu(Γ)
1
≤ˆ
Γ
|detD2u|. (47)
Lets do a bit of linear algebra. I claim that given two positive matrices A,B, then
detA detB ≤(
TrAB
n
)n.
To see this it suffices to show that
detA ≤(
TrA
n
)n.
Let λ1, . . . , λn be A’s eigenvalues. Then
detA =
n∏i=1
λi, TrA =
n∑i=1
λi.
Thus what we wish to show is equivalent to(∏λi
)1/n
≤ 1
n
∑λi,
39
which is exactly the inequality of arithmetic and geometric means. Taking A = D2u,B = (aij) then on Γ
|detD2u| = det−D2u =1
Ddet(aij) det−D2u
≤ 1
D
(−fn
)n. (48)
Combining (46),(47), and (48) yields
ωn
(u(y)
d
)n≤ 1
nnD‖f−‖nLn(Γ)
u(y) ≤ d
nω1/n
∥∥∥∥ f−D?
∥∥∥∥Ln(Γ)
.
which is precisely the ABP estimate because we had replaced u with u− sup∂Ω u.
The ABP maximum principle can be naturally extended to functions u ∈ C0(Ω) ∩W 2.nloc (Ω) by approxi-
mating with smooth functions. It can also be extended for coefficients satisfying |b|/D? ∈ Ln(Ω) and c ≤ 0in Ω, but these will be taken as a black box. Now we have the correct tools to begin proving the Harnackinequality for non-divergence equations. We say an operator of the form (45) is strictly elliptic when
Λ
λ≤ γ,
(|b|λ
)2
,|c|λ≤ ν.
We will assume throughout the rest of this section that the operator L given by (45) is uniformly elliptic.Note that have the same ABP estimate from the other side: if Lu ≤ f and u|∂B1
≥ 0, then
| infB1
u| ≤ Cn(ˆ
Γ
fn)1/n
where Γ is the convex envelop of u (contrary to the upper contact set Γ+)
4.2 Measure theory and Harnack Inequality
In order to prove the Harnack inequality for non-divergence equations with measurable coefficients, we needa basic measure theoretic estimate that is true at “all scales.” Our first main theorem is as follows
Theorem 4.5. Let u ∈ W 2,n and assume aijuij ≤ 0 for some bounded measurable and uniformly ellipticaij in B2. If u ≥ 0 and u(x) ≤ 1 for some x ∈ ∂B1, then
|u ≤M ∩B1/2| ≥ µ (49)
for some M large and µ small universal.
Heuristically, this is saying that if u goes below 1 at a certain point on the boundary of a ball of radiusone, then it can’t get much bigger in a ball of radius two. That is, the ABP estimate will tell us that thesize of u can be controlled.
Proof. Define α = max1, (n − 1)Λ/λ − 1 and take a barrier function ϕ of the form ϕ(x) = M1 −M2|x|αin B2 − B1/2 with M1 and M2 chosen such that ϕ∂B2 = 0 and ϕ|∂B1 = −2. Note that we can cap off inB1/2 smoothly so that the constants still depend only on n, λ,Λ (we say cap off to smooth out the inner
40
radius with a paraboloid of radius 2; it’ll be made clear with the picture). Suppose that r ≥ 1/2 at the point(r, 0, . . . , 0). Then we compute some derivatives at this point:
Dijϕ = 0 for i 6= j
D11ϕ = −M2α(1 + α)r−α−2r−α−2
Diiϕ = M2αr−α−2.
By rotational symmetry of our function and uniform ellipticity, we have for |x| ≥ 1/4,
aijϕij |x = M2(Λ(n− 1)α|x|−α−2 − λα(1 + α)|x|−α−2)
= M2α|x|−α−2(Λ(n− 1)− λ(1 + α)) ≤ 0 (50)
by our choice of α. However, for |x| ≤ 1/4, aijuij ≤ C = C(n, λ,Λ). This observation along with (50) showsthat aijϕij ≤ Cη for some C universal with η ≡ 1 in B1/4 and η ≡ 0 outside of B1/2, i.e. η ∈ C∞0 (it issmooth because of the smoothness of |x|α in this domain). Finally note that |ϕ| ≤M for some M universalgiven by the distance of the vertex of the parabola that we capped off by and the origin.
By our assumption that u ≤ 1 somewhere on ∂B1 we have that w = u + ϕ ≤ −1 somewhere on ∂B1.Then ABP given gives us
1 ≤ Cˆ
Γw∩B1ηn ≤ C|Γw ∩B1/2|.
Now the crucial part of the proof is to notice that the convex envelop of w is in the set that w ≤ 0 =⇒u + ϕ ≤ 0 =⇒ u ≤ −ϕ ≤ M . Hence our estimate just gave us 1 ≤ C|u ≤ M ∩ B1/2|. This concludes theproof with µ = 1/C.
Corollary 4.6. We can show that this is scaling invariant. That is, if u is defined in B2r and u ≤ αsomewhere on ∂Br, then |u ≤Mα ∩Br/2| ≥ µ.
41
Lemma 4.7. Suppose we have the same u as in Theorem 3.5. Then
|u ≥Mk ∩B1/2| ≤ (1− µ)k
for k = 1, 2, . . . , and M,µ are as in Theorem 3.5.This implies that
|u ≥ t ∩Q1| ≤ dt−ε (51)
for any t > 0 where d, ε are positive universal constants.
Proof. This proof will follow from Calderon-Zygmund decomposition and induction. For k = 1, this is thestatement of Theorem 4.5. Now suppose it holds for k− 1. We introduce the classical decomposition: if Q isa dyadic cube different from Q1, we say Q is the predecessor of Q if Q is one of the 2n cubes obtained fromdividing Q. Then the decomposition states the following: if A ⊂ B ⊂ Q1 are measurable sets and 0 < δ < 1such that
(1) |A| ≤ δ, and
(2) if Q is a dyadic cube such that |A ∩Q| > δ|Q|, then Q ⊂ B,
then |A| ≤ δ|B|.Recall that we assume the Lemma holds for k − 1. For convenience, we use cubes because Theorem 4.5
still holds for them. Let A = u > Mk ∩Q1 and B = u > M ∩Q1. We’ll be done with our lemma if weshow that |A| ≤ (1− µ)|B| by our inductive hypothesis. This is where we’ll use the decomposition. ClearlyA ⊂ B ⊂ Q1 and |a| ≤ |u > M ∩Q1| ≤ 1− µ by Theorem 4.5. Now we need to show (2) also holds.
(2) will follow if we show that if Q = Q1/2i(x0) is a dyadic cube such that |A ∩ Q| > (1 − µ)|B|, then
Q ⊂ B. Suppose it isn’t and take x ∈ Q such that u(x) ≤ Mk−1. Now consider the transformationx = x0 + 1/2iy for y ∈ Q1 and x ∈ Q = Q1/2i(x0) and the function
u(y) = u(x)/Mk−1.
Then I claim that u is under the hypothesis of Theorem 4.5. By (49), it follows that
µ < |u(y) ≤M ∩Q1| = 2in|u(x) ≤Mk ∩Q|.
We see that |Q − A| > µ|Q|, which contradicts our assumption that |A ∩ Q| > (1 − µ)|Q|. Showing that usatisfies the conditions of Theorem 4.5 will be left as an exercise. As a hint, the following property of dyadiccubes: if Q is a dyadic cube Q = Q1/2i(x0) for some i ≥ 0 and x0 ∈ Q1, then
Q4√n/2i(x0) ⊂ Q4
√n,
i ≥ 1 =⇒ Q ⊂ Q3/2i(x0).
(51) follows from the results above by taking d = (1− µ)−1 and ε such that 1− µ = M−ε.
Now we need yet another measure theoretic lemma. Let f be a measurable function on a domain Ω inRn. Define the distribution function µ(t) = |f ≥ t| for t > 0. This measures the “relative” size of f . Notethat µ is a decreasing function on the positive real line.
Lemma 4.8. For any p > 0 and |f |p ∈ L1(Ω),
ˆΩ
|f |p = p
ˆ ∞0
tp−1µ(t)dt.
42
Proof. This is just computations. Suppose f ∈ L1. Then
ˆΩ
|f | =ˆ
Ω
ˆ |f(x)|
0
dtdx
=
ˆ ∞0
µ(t)
and the lemma will hold for general p after change of variables.
Now we are finally able to prove the weak Harnack inequality:
Theorem 4.9 (Weak Harnack Inequality). Suppose u ∈ W 2,n(Q1) satisfies aijuij ≤ 0 in Q1 and u isnon-negative. Then
‖u‖Lp(Q1/4) ≤ C infQ1/2
u
where p > 0 and C universal.
Proof. We already have most of the machinery for the proof. This follows from (51) in Lemma 4.7 becauseif we suppose u is under the assumptions of the emma, |u ≥ t ∩ Q1| ≤ dt−ε and Lemma 4.8 implies forp = ε/2
ˆQ1
up = p
ˆ ∞0
tp−1|u ≥ t ∩Q1|dt
≤ pˆ ∞
0
tp−1 · d · t−εdt
= d · pˆ ∞
1
tp−1t−2pdt (inf u ≥ 1)
= d · pˆ ∞
1
t−p−1dt
= d · pt−p
p
∣∣∣∣∞0
= d.
hence ‖u‖Lp(Q1) ≤ C and we re-scale u away from the assumptions of 4.7 to get the result of the weak HI.
The second piece of HI is called the local maximum principle
Theorem 4.10. Let u ∈W 2,n(Q1) and let aijuij ≥ 0. Then for any 0 < p ≤ n, we have
supQ1/2
u ≤ Cp‖u+‖Lp(Q3/4).
for C universal depending on p.
Proof. See [1][p. 244].
Corollary 4.11. Taking p = 1 in Theorem 4.10, we obtain an extension of the mean value inequality fornon-negative subharmonic functions:
u(y) ≤ C
Rn
ˆBR
u
when Lu ≥ 0, C = C(n, γ, νR2).
43
5 Curved C1,α Domains
5.1 Estimate for Laplacian
We begin with a simple definition.
Definition 5.1. A continuous function u on Rn is said to be C2,α at x0 if there exists a quadratic polynomialP and constants C, ρ such that
‖u− P‖L∞(Br) ≤ Cr2+α, ∀r ≤ ρ. (52)
We say u ∈ C2,α(B1) if u is C2,α at all x ∈ B1.
Now we develop a tool that helps us determine when a function is C2,α.
Proposition 5.2. Let Ω ⊂ Rn be open and bounded. Suppose we can find a sequence of paraboloids Pk =ak + bk · x+ 1
2xT ckx and an r < 1 such that
‖u− Pk‖L∞(Brk
) ≤ Crk(2+α)
where Brk ⊂⊂ Ω. Then u ∈ C2,α(0) where 0 < α < 1.
Proof. Consider the quadratic scaling of a function fr(x) = 1r2 f(rx). By our hypothesis we have
‖Pk+1 − Pk‖L∞(Brk+1 ) ≤ ‖u− Pk+1‖L∞(B
rk+1 ) + ‖u− Pk‖L∞(Brk+1 )
≤ Kr(k+1)(2+α) +Krk(2+α)
≤ 2Krk(2+α).
Quadratic rescaling gives us‖Pk+1,rk − Pk,rk‖L∞(B1) ≤ 2Krkα.
Since the coefficients of polynomials on B1 are controlled by the L∞ norm, we have
|ak+1 − ak| ≤ Krk(2+α), |bk+1 − bk| ≤ Krk(α+1), |ck+1 − ck| ≤ Krkα. (53)
Thus we have that the Pk converges to a polynomial P = a+ bx+ xT cx. Putting these together yield
‖u− P‖L∞(Brk
) ≤ ‖u− Pk‖L∞(Brk
) + |ak − a|+ rk|br − b|+ r2k|ck − c|
≤ CKrk(2+α)
and so we have that u ∈ C2,α.
It is important to realize why we can’t have α = 0, 1. Notice that for α = 0, the proof breaks down in(53) because we’d get |ck+1 − ck| ≤ K, so the sequence of polynomials doesn’t converge. The case for α = 1is left as an exercise. We utilize this proposition to prove the following theorem.
Theorem 5.3. Suppose ∆u = f in B1 where f ∈ Cα(B1). Then u ∈ C2,α(B1/2) and
‖u‖C2,α(B1/2) ≤ Cn,α(‖u‖L∞(B1) + ‖f‖Cα(B1)
).
Proof. By subtracting 12nf(0)|x|2 we assume f(0) = 0. Dividing by ‖u‖L∞(B1) + 1
ε ‖f‖Cα(B1) for some ε > 0to be chosen later, we can assume that ‖f‖Cα(B1) ≤ ε and |u| ≤ 1. Let w be the harmonic function thatagrees with u on ∂B1. Then ∆(u− w) = f and again ‖f‖Cα(B1) ≤ ε. By the maximum principle,
‖u− w‖L∞(B1) ≤ Cε.
44
Let P1 be the harmonic quadratic approximation to w that satisfies
‖w − P1‖L∞(Br) ≤ Cr3.
The constant C that is above is in terms of n because |w| ≤ 1. Putting these together we have
‖u− P1‖L∞(Br) ≤ C(r3 + ε).
Now here we use the fact that 0 < α < 1 so choose r so small that 2Cr3 ≤ r2+α and choose ε so small that2Cε ≤ r2+α. This gives
‖u− P1‖L∞(Br) ≤ r2+α.
Now take the 2 + α rescaling v(x) = 1r2+α (u − P1)(rx). Notice that |v| ≤ 1 and since P1 is harmonic we
have that ∆v = 1rα∆u(rx) = 1
rα f(rx) = g(x). The right side again satisfies |g| ≤ ε so we repeat our processagain to find a harmonic polynomial P2 such that
‖v − P2‖L∞(Br) ≤ r2+α
and so scaling back this will become
‖u(x)− P1(x)− r2+αP2(x/r)‖L∞(Br2 ) ≤ r2(2+α).
Iterating this will give us a sequence of polynomials approximating u. Thus we are in the position ofProposition 1.2, and we are finished.
Lets shift gears a little bit and develop tools for boundary estimates. We say that a Ω is a Schauderdomain, or a Ck,α – domain if it is a domain in Euclidean space with sufficiently regular boundary i.e., ∂Ωcan locally be viewed as the graph of a Ck,α function. For this reason, we will assume that ∂Ω = (x′, g(x′))where g ∈ C2,α, x′ ∈ Rn−1. Recall that the first step in the iteration for Theorem 5.3 required us to obtainthe estimate |u − w| ≤ Cε in B1. Since this sort of iteration process is all we can do at the moment, I willnow attempt to show the first step:
Lemma 5.4. Let B1 be the ball centered at the origin, f ∈ L∞, and suppose that u solves∆u = f in Ω ∩B1
u = 0 on ∂Ω
with |u| ≤ 1 and |f | ≤ δ. Suppose the domain is as flat as
B1 ∩ xn ≥ ε ⊂ B1 ∩ Ω.
Then there is a harmonic function w in B1/4 with w(x′, 0) = 0 so that
|u− w| ≤ C(ε+ δ)
on B 14(0) ∩ Ω.
Proof. We first consider the case where f = 0, so u is harmonic on Ω ∩ B1. Let Γ(x) be the fundamentalsolution to Laplace’s equation, and define the following barrier function as follows
G(x) :=Γ(x)− Γ(r)
Γ(R)− Γ(r)
for 0 < r < R. Notice that 0 ≤ G,G(R) = 1, G(r) = 0, and that G is harmonic in the annulus of littleradius r and big radius R. Consider the lines xn = ±ε and let x0 = (x′0,−ε) ∈ Rn be so small that it is inB1. Lets look at G the barrier function with smaller radius tangent to xn = −ε, and see that it is centeredat (x′0,−ε− r) =: x0.
45
Now I will show that |u| ≤ G on the common domain of influence. First note that G ≥ |u| on ∂Ω becauseu = 0 on ∂Ω. Recall that we assumed |u| ≤ 1 in Ω, and since G = 1 on ∂B(x0, R), we see that |u| ≤ G onΩ ∩ ∂B(x0, R). Putting these two together yields |u| ≤ G on ∂ (Ω ∩B(x0, R)), and the maximum principle(which applies because these two are harmonic on Ω∩B(x0, R)) tells us that |u| ≤ G on the entire commondomain of influence.
Lets restrict ourselves for a moment on rays starting from x0 and moving in the xn direction. Noticethat since G is Lipschitz, we have
|G(0, . . . , 0, t)−G(0, . . . , 0, s)| ≤ C|t− s|.
Let t ∈ R be such that x0 + ten ∈ Ω. We now use the fact that G is zero on the smaller ball of radius r:
|u(x0 + ten)| ≤ G(x0 + ten)
= G(x0 + ten)−G(x0 + ren)
≤ C|t− r| = C|xn + ε|.
Observe that x0 can be moved horizontally as long as the outer ball of radius R is in B1, and the inner ballof r is below −ε, and so,
|u| ≤ C|xn + ε| on all of B 14(0) ∩ Ω (54)
Now that we have this estimate, lets find the corresponding harmonic polynomial that gives us the desiredresult. Let w be the harmonic polynomial such that
w = u on ∂(B+1/4) ∩ Ω ∩ xn ≥ ε
w = 0 on ∂(B+1/4) ∩ 0 ≤ xn < ε
w = 0 on xn = 0.
There is the possibility that w is not continuous on ∂(B+1/4) ∩Ω ∩ xn = ε because on one side, w = u and
on the other w = 0. The way to fix this is the following: move δ > 0 above ε and consider a C∞ cutofffunction vδ that satisfies
0 ≤ vδ ≤ 1 W ⊂⊂ ∂(B+1/4) ∩ Ω ∩ ε ≤ xn ≤ ε+ δ
vδ ≡ 1 V ⊂⊂W
where vδ decreases monotonically to zero. Now define the new harmonic function
wδ :=
w = u on ∂(B+
1/4) ∩ Ω ∩ xn ≥ ε+ δw = v2
δu on ∂(B+1/4) ∩ Ω ∩ ε ≤ xn ≤ ε+ δ
w = 0 on ∂(B+1/4) ∩ 0 ≤ xn < ε
w = 0 on xn = 0.
Then since
46
(i) |wδ| = |u| ≤ G on ∂(B+1/4) ∩ Ω ∩ xn ≥ ε+ δ
(ii) |wδ| = |v2δu| ≤ |u| ≤ G on ∂(B+
1/4) ∩ Ω ∩ ε ≤ xn ≤ ε+ δ
(iii) G ≥ 0 = wδ on ∂(B+1/4) ∩ 0 ≤ xn < ε
(iv) G ≥ 0 = wδ on xn = 0,
putting (i) - (iv) together yields |wδ| ≤ G on ∂(
Ω ∩B+1/4
). Thus the maximum principle then implies
|wδ| ≤ C|xn + ε| on all of B+1/4 ∩ Ω. (55)
Similarly on B−1/4∩Ω, one could extend oddly by wδ(x′, xn) = −wδ(x′,−xn) on B−1/4, and comparing w with
−u, one achieves the same bound as (55) on B−1/4 ∩ Ω. Define the new oddly reflected function
wodd =
wδ on B+
1/4
wδ on B−1/4
which satisfies
|wodd| ≤ C|xn + ε| on all of B1/4 ∩ Ω. (56)
Putting (54) and (56) together finally proves the Lemma for u harmonic:
|u− wodd| ≤ Cε
on B1/4 ∩ Ω. Now lets consider the general case ∆u = f where f ∈ L∞ and |f | ≤ δ. Here, apply the above
to u = u± δ2n |x|
2 and comparison principles allow us to say
|u− wodd| ≤ C(ε+ δ).
Corollary 5.5. Suppose u satisfies ∆u = f in Ωu ≤ ε on ∂Ω
(57)
with again the same flatness assumption on Ω (B1∩xn ≥ ε ⊂ B1∩Ω). Then the same estimate as Lemma5.4 holds.
Proof. In order to see why this is true, we must examine where we used u = 0 on ∂Ω in the proof of Lemma5.4. We were comparing trying to show that |u| ≤ G on the common domain of influence and used the factthat u = 0 ≤ G on ∂Ω. In order to get the same estimates of the Lemma for (57), we simply move the pointx0 = (x′0,−ε − r) to x0 = x′0,−ε − r − 2ε′. The reason for this is that the barrier function G grows at therate
|DG(r)| ≤ Cr1−n,
and so by moving down a sufficiently small amount, G will be bigger than ε, and in turn bigger than u on∂Ω
We are now ready to prove C1,α estimates at the boundary:
47
Theorem 5.6. Suppose u satisfies ∆u = f in Ωu = 0 on ∂Ω
with f(0) = g(0) = ∇g(0) = 0, |u| ≤ 1, |f | ≤ δ. Suppose Ω is a C1,α domain, that is ∂Ω = (x′, g(x′)) forg ∈ C1,α that satisfies |g| ≤ δr1+α. Then u ∈ C1,α(Br/2 ∩ Ω) with the estimate
‖u‖C1,α(Br/2∩Ω) ≤ Cn,α(‖u‖L∞ + ‖f‖L∞ + ‖g‖C1,α(|x′|))
Proof. Consider Br ∩ Ω we want to show that there exists a linear function l such that
‖u− l|L∞(Br∩Ω) ≤ r1+α
and WLOG assume r ≤ 1. Then since u = 0 along the boundary, all the tangential derivatives are then zeroand hence we may reduce this to
‖u− axn‖L∞(Br∩Ω) ≤ r1+α (58)
and we may also assume |a| ≤ 1. In order to achieve this by some sort of iteration used in Theorem 5.3, weneed to conclude
‖u− axn‖L∞(Bρr∩Ω) ≤ (ρr)1+α
for some a. The first thing we do is to re-scale the ball Br into B1 (and hence diluting Ω to Ω) and it isclear that we need to define some u that satisfies
u(x)− axn = r1+αu(x/r)
because u’s domain of influence is B1. Our hypothesis then becomes |u| ≤ 1 and note that we can rewrite itas
u(x) =u(rx)− a(rxn)
r1+α.
Observe that diluting Ω to Ω makes the latter the graph of g(x′) = 1r g(rx′) which satisfies |g| ≤ δrα. Now we
need to figure out what equation u satisfies: ∆u = r2∆u(rx)r1+α = r1−αf(rx) =: f . The bound that f satisfies
is |f | ≤ δr1−α ≤ δ, where we used that r ≤ 1. Now lets figure out the boundary conditions that u satisfies:
∣∣u|∂Ω
∣∣ =
∣∣∣∣u(rx)− a(rxn)
r1+α
∣∣∣∣∂Ω
∣∣∣∣=|axnr|rα+1
=|axn|rα
≤ δ
where we used the fact that u(x) vanishes on ∂Ω =⇒ u(rx) vanishes on ∂Ω. Putting everything togethergives us
|u| ≤ 1
|g| ≤ δrα (59)
∆u = f , |f | ≤ δ|u| ≤ δ on ∂Ω.
Now we are exactly in the position of Corollary 5.5 because the flatness condition is precisely met by (59)(after replacing the ε in the Lemma with δ because ∂Ω is the graph of g, and so its height in B1 is at most δby (59)). Applying the Lemma gives us |u−w| ≤ Cδ in B 1
4∩ Ω. Recall that w is the solution that vanishes
48
on x′ =⇒ |w− bxn|Bρ ≤ Cρ2. Now noticing that the radius of B 14
doesn’t have to be 14 (as long as it is small
enough for the proof of Lemma 5.4 to hold) we reach
|u− bxn|Bρ∩Ω≤ C(δ + ρ2).
By picking δ and ρ small enough, and using the fact that 0 < α < 1 yields
|u− bxn|Bρ∩Ω≤ Cρ1+α.
Rescaling back gives us|u− axn|Bρr∩Ω
≤ C(ρr)1+α
with a = a+ rαb. We can now iterate just like in the last step of Theorem 5.3 to find that u ∈ C1,α.
5.2 General Elliptic Operators
The goal of this section is to generalize Krylov’s estimate for flat domains to general C1,α domains. Firstwe state and prove his estimate
Theorem 5.7. Assume u ∈W 2,n(B+)∩C0(B+) is non-negative and u(en/2) ≥ 1. If u satisfies the equationaijuij = 0 in the half ball B+ = B1 ∩Rn+ with u = 0 on T = B1 ∩ ∂Rn+, and the aij measurable bounded anduniformly elliptic. Then u ∈ C1,α(B+
1/2).
Proof. First we note that if u ≥ 1 at en/2, we can apply the Harnack inequality to find that u ≥ c onB1/10(en/2). Now look at the barrier function φ(x) = C1|x − en
2 |α − C1 with C1 and C2 chosen such that
φ ≡ 0 on ∂B1/2(en/2) and φ = c on ∂B1/10(en/2). Here α = α(n,Λ, λ). By an almost identical computationto the barrier function constructed in the HI for non-divergence equations, one can see that aijφij ≥ 0 onB1/2(en/2) − B1/10(en/2). Now we can use the ABP maximum principle to deduce that u ≥ φ on thecommon domain of influence. Then by Hopf lemma we know that this barrier is going to start at an angle,and so we can actually conclude that u(x) ≥ δ′xn.
49
Note that this was just in this particular domain B1/2(en) but we want to establish the proof in all of
B+1/2. In order to remedy, we have to slide our barrier to the left and right by x0 such that outer ball where φ
vanishes is tangent to B+. The center of these two extreme balls is given by 12 (1, . . . , 1) and 1
2 (−1, . . . ,−1, 1).So how does this help us? We can actually apply the above estimate at each of the translated barriers becausewe can iterate HI to get a thin strip of domain where we know u ≥ C and so we can construct similar barrierson every point along this strip. Then at the end of these iterations take the smallest angle of the planesgenerated by each barrier function and trap all of u ≥ cxn in B1/2. Since u has this flatness associated tois, it is C1,α, as will be seen in the proof of the next theorem.
We now introduce the final theorem of this thesis. We first state it and prove an important lemma thatwill be useful.
Theorem 5.8. Let u ∈ W 2,n(Ω) ∩ C0(Ω) where Ω is a C1,α domain such that 0 ∈ ∂Ω and near zero∂Ω = (x′, g(x′)) with |g(x′)| ≤ δ|x′|1+α. Suppose Lu = aij(x)uij = f(x) strongly on Ω with λI ≤ aij ≤ΛI, |f | ≤ δ in B1 ∩ Ω and u ≤ δ|x′|1+α on ∂Ω. Then u ∈ C1,α(0).
Lemma 5.9. Assume u satisfiesa−0 xn −Mδ ≤ u ≤ a+
0 xn +Mδ
in B1 with |a+0 − a
−0 | = 1 and a+
0 , a−0 ∈ [−10, 10]. Then, in Br0 with r0 ≤ 1, u satisfies
a−1 xn −Mδr1+α0 ≤ u ≤ a+
1 xn +Mδr1+α0
with a+1 ≤ a
+0 + Cδ, a−0 − Cδ ≤ a
−1 , and a+
1 − a−1 ≤ (1− η) for small η.
Proof. First we look at u(1/2en) and notice that
1
2a−0 −Mδ ≤ u(1/2en) ≤ 1
2a+
0 +Mδ.
Now notice that u(1/2en) is either closer to the plane on top or the plane on the bottom: since 1/2|a+0 −a
−0 | =
1/2 and the Mδ is common in both hands of the inequality, we have either
u(1/2en)− 1
2a−0 ≥
1
4or
1
2a+
0 − u(1/2en) ≥ 1
4.
Define u := u− a−0 xn. The hypothesis of the lemma becomes
−Mδ ≤ u ≤ (a+0 − a
−0 )xn +Mδ = xn +Mδ.
Notice that a+0 , a
−0 ∈ [−10, 10] implies
|u| ≤ |u|+ |a−0 xn| ≤ δ|x′|1+δ + 10|xn| = δ|x′|1+α + 10|g(x′)| ≤ 11δ|x′|1+α
on ∂Ω near zero. It is also important to notice that u solves the same equation as u. Since we assumed thatu was closer to the lower plane, we have that u(1/2en) ≥ 1
4 −Mδ. The Harnack Inequality tells us thatu ≥ C0 −Mδ (with C0 independent of δ) on B1/10(1/2en).
Now consider the barrier w(x) = M1|x|−γ −M2 with M1,M2 chosen such that w ≡ 1 on ∂B1/10(1/2en)and w ≡ 0 on ∂B1/2(1/2en). We compute the derivatives of w as
wij(x) =
M1γ(x2
1 + · · ·+ x2n)γ/2−1 +M1γ(γ/2− 1)(x2
1 + · · ·+ x2n)(2xi)(xi) i = j
M1γ(γ/2− 1)(x21 + · · ·+ x2
n)γ/2−2xi(2xj) i 6= j.
Thus, looking in the radial direction (0, . . . , 0, r) we have
D2w =
1rw′(r)
. . .1rw′(r)
w′′(r)
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and so w′′ > 0, w′ < 0 implies that
aijwij ≥ λw′′(r) + Λn− 1
rw′ = γM1r
γ−2(λ(γ − 1) + Λ(n− 1)) ≥ C1
by choosing γ. Now define
W :=C0w
2− 100δ.
We want to do our estimates on Ω ∩ B1 − B1/10(1/2en) and so we examine first ∂B1 ∩ Ω. Here we knowW is very negative from its construction (more so than u ≥ −Mδ) and if it isn’t, we simply move ourcomputations from 1/2en 7→ 1/4en. This gives us u ≥W on Ω∩ ∂B1. From the bounds on g, we know thatthe first points of contact between ∂B1 and ∂Ω will be at most size δ. But from our estimates of u nearzero, we have
u ≥ −11δ ≥ max|xn|≤δ
W ≥ max∂Ω∩|xn|≤δ
W.
Note that W ≡ C0/2− 100δ on ∂B1/10(1/2en) and u ≥ C0 −Mδ on B1/10(1/2) by the harnack inequality.Hence u ≥ W on ∂B1/10(1/2en). If δ is small enough, which can be done by multiplying f by a constantsince it doesn’t change any of our inequalities, we have LW = C0C1 ≥ Lu ∼ δ and so the maximum principledictates that W ≤ u on Ω ∩B1 −B1/10(1/2en).
Observe that in the radial direction x′ = 0, W satisfies
W (0, xn) =C0
2w(0, xn)− 100δ
≥ C0C3xn − 100δ
because of how |x|−γ grows. Also notice that x′ = 0 was not specific and we could have taken any x′ ∈|xn| ≤ δ ∩ ∂Ω as we look for planes in the radial xn direction starting from ∂Ω. Taking the maximum C3
of all such planes we obtain
u− a−0 xn = u ≥W≥ C0C3xn − 100δ
u ≥ (C0C3 + a−0 )− 100δ
and C0C3 + a−0 =: a−1 .
Now assume that u is closer to the upper plane than the lower plane. In this case then u satisfies
1
2a+
0 − u(1/2en) ≥ 1/4.
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We define u = a+0 xn − u and clearly we are looking for a lower bound for u. The hypothesis of the Lemma
becomes
−a−0 xn +Mδ ≥ −u ≥ −a+0 xn −Mδ
(a+0 − a
−0 )xn +Mδ ≥ u ≥ −Mδ.
xn +Mδ ≥ u ≥ −Mδ.
Near zero u satisfies the estimate
|u| ≤ |a+0 xn|+ |u| ≤ 10|xn|+ δ|x′|1+α = 10|g(x′)|+ δ|x′|1+α ≤ 11δ|x′|1+α.
Also since u was closer to the upper plane, u(1/2en) ≥ 1/4 −Mδ. The Harnack Inequality tells us thatu ≥ C0 −Mδ. As one can probably tell, the proof is going to be identical to the one for u.
Now consider the barrier w(x) = M1|x|−γ −M2 with M1,M2 chosen such that w ≡ 1 on ∂B1/10(1/2en)and w ≡ 0 on ∂B1/2(1/2en). In a very similar sense as the lower bound by the barrier in the previous case,we have aijwij ≥ C1 = C1(n, λ,Λ, γ). Define W by
W :=C0w
2− 100δ.
We want to do our estimates on Ω ∩ B1 − B1/10(1/2en) and so we examine first ∂B1 ∩ Ω. Here we knowW is very negative from its construction (more so than u ≥ −Mδ) and if it isn’t, we simply move ourcomputations from 1/2en 7→ 1/4en. This gives us u ≥W on Ω∩ ∂B1. From the bounds on g, we know thatthe first points of contact between ∂B1 and ∂Ω will be at most size δ. But from our estimates of u near zero,we have
u ≥ −11δ ≥ max|xn|≤δ
W ≥ max∂Ω∩|xn|≤δ
W.
Note that W ≡ C0/2− 100δ on ∂B1/10(1/2en) and u ≥ C0 −Mδ on B1/10(1/2) by the harnack inequality.Hence u ≥ W on ∂B1/10(1/2en). If δ is small enough, which can be done by multiplying f by a constantsince it doesn’t change any of our inequalities, we have LW = C0C1 ≥ Lu ∼ δ and so the maximum principledictates that W ≤ u on Ω ∩B1 −B1/10(1/2en).
Observe that in the radial direction x′ = 0, W satisfies
W (0, xn) =C0
2w(0, xn)− 100δ
≥ C0C3xn − 100δ
because of how |x|−γ grows. Also notice that x′ = 0 was not specific and we could have taken any x′ ∈|xn| ≤ δ ∩ ∂Ω as we look for planes in the radial xn direction starting from ∂Ω. Taking the maximum C3
of all such planes we obtain
a+0 xn − u = u ≥W
≥ C0C3xn − 100δ
(a+0 − C0C3)xn + 100δ ≥ u
and a+0 + C0C3 =: a+
1 . Thus we can see that even though the distance from the origin increases a little bit,the slope decreases a lot more.
Proof of Theorem. We rescale u by defining
u(x) =u(r0x)
r0.
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Then by Lemma 5.9, in Br0 we have
a−1 xn −Mδrα0 ≤ u ≤ a+1 xn +Mδrα0
with |a+1 − a
−1 | = 1 − η by direct computations. We also rescale the boundary by g(x′) = g(r0x
′)/r0 =⇒|g| ≤ δrα0 |x′|1+α. We can also compute
Lu = aij(x)uij(x) =aij(r0x)
r0
r20uij(r0x)
r0= f(r0x).
This implies that we can iterate the conclusion of the Lemma by rescaling and so iterating k times withr = rk0 , we have
a−k xn −Mδr1+α ≤ u ≤ a+k xn +Mδr1+α
in Br = Brk0 with |a+k − a
−k | = (1− η)k ≈ rβ for some β. This implies convergence of the a±k to an a∞ in the
sense that |a+k − a∞| ≤ Crβ , |a
−k − a∞| ≤ Crβ . Then
a−k xn −Mδr1+α ≤ u ≤ a+k xn +Mδr1+α
(a−k − a∞)xn −Mδr1+α ≤ u− a∞xn ≤ (a+k − a∞)xn +Mδr1+α.
Our analysis finally gives|u− a∞xn| ≤ Cr1+β +Mδr1+α.
This proves that if β < α, then u ∈ C1,β(0). If β ≥ α then u ∈ C1,α(0).
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References
[1] Gilbarg, D., Trudinger, N.S. Elliptic Partial Differential Equations of Second Order, 2nd ed., Springer-Verlag, NY, 1983.
[2] Caffarelli, L., Cabre, X. Fully Nonlinear Elliptic Equations, AMS,1995
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