The P-Laplace Equation – General Properties and Boundary Behaviour

U.U.D.M. Project Report 2018:36

The p-Laplace equation – general properties and boundary behaviour

Frida Fejne

Examensarbete i matematik, 30 hp Handledare: Kaj Nyström Ämnesgranskare: Wolfgang Staubach Examinator: Denis Gaidashev Augusti 2018

Department of Mathematics Uppsala University

The p-Laplace equation – general properties and boundary behaviour.

Frida Fejne

August 27, 2018 Abstract

In this thesis we investigate the properties of the solutions to the p-Laplace equation, which is the Euler- Lagrange equation of the p-Dirichlet integral, a generalization of the well known Dirichlet integral. It turns out that many of the properties of the harmonic functions also hold for the so called p-harmonic functions. After giving a comprehensive introduction to the subject, where we establish the existence of weak solutions on bounded domains, we discuss general properties such as the Harnack inequality, Hölder continuity, diﬀerentiability, and Perron’s method. In the last part of the thesis we study boundary behaviour and in particular the behaviour of the ratio of two p-harmonic functions near a portion of the boundary where they both vanish.

2 Acknowledgements

I would like to thank Kaj Nyström for introducing me to this fascinating subject and for supervising this thesis. I also thank Lukas Schoug for showing me how to do the pictures and for letting me explain some of the concepts in order to gain deeper understanding of the theory.

3 Contents

1 Introduction 5 1.1 Preliminaries ...... 5 1.1.1 Notation ...... 5 1.1.2 Weighted Sobolev spaces ...... 6 1.2 Variational integrals and the p-Dirichlet integral ...... 8 1.3 Structure of Thesis ...... 10 2 Basic deﬁnitions and existence of weak solutions 10

3 Regularity of weak solutions 17 3.0.1 The case 1 < p < n ...... 19 3.0.2 The case p = n ...... 25 3.0.3 The case n

5 The p-superharmonic functions and their properties 34

6 Perron’s method 41

7 Boundary behaviour 45 7.1 Boundary estimates ...... 50 7.2 Halfspace ...... 51 7.3 The fundamental inequality for the gradient of a p-harmonic function ...... 55 7.4 Estimates for degenerate elliptic equations in weighted Sobolev spaces ...... 62 7.5 Reduction to linear equation and ﬁnal proof ...... 68 A Proofs of basic properties 73

B Some useful inequalities 75

4 1 Introduction

One of the central problems of modern analysis has been the Dirichlet problem, that is, given an open n connected set Ω ⊂ ℝ , and a real-valued function f, continuous on the boundary )Ω, to ﬁnd a function 2 ̄ u ∈ C (Ω) ∩ C(Ω), such that T Δu = 0 in Ω, u = f on )Ω.

The literature on the problem is extensive and goes deep into the realms of many mathematical subjects, such as complex analysis (see e.g. [5]) and probability theory (see e.g. [24]). Another approach to the Dirichlet problem is via the Dirichlet energy integral. The Dirichlet energy integral of a function u ∶ Ω → ℝ is deﬁned as 0 12 0 12 2 )u )u E(u) = ð∇uð dx = + … + dx, (1) ÊΩ ÊΩ )x1 )xn 1 E(u) (sometimes written as multiplied by 2 ). The Euler-Lagrange equation for is easily shown to be the Laplace equation Δu = 0, that is, solving the Dirichlet problem with boundary conditions f is the same as minimizing the Dirichlet integral over functions with boundary data f. In this thesis, we consider a generalized version of (1), namely the p-Dirichlet integral

p I(u) = ð∇uð dx, ÊΩ of which the Euler-Lagrange equation is the p-Laplace equation p−2 Δpu ≡ div(ð∇uð ∇u) = 0.

1.1 Preliminaries

1.1.1 Notation

n In the following we let Ω be a domain in ℝ , which is not necessarily bounded. G and D will always denote n n open sets in ℝ . We let ⟨⋅, ⋅⟩ and dx denote the inner product and the Lebesgue measure on ℝ , respectively. We will express the Euclidean distance between the two points x1 and x2 by d(x1, x2). c will always be a positive constant, depending on at most p and n unless otherwise stated, such that c ≥ 1. The value of c may vary from line to line and between occurrences. Furthermore, we will use the notation c(a1, a2, … , as) when n c also depends on the the additional constants a1, a2, … as. For x ∈ ℝ and r > 0 we define the open ball n B(x, r) = {y ∈ ℝ ∶ ðx − yð < r}. When the center of the ball is arbitrary or clear from the context we will denote the ball by Br, and balls with different radii are always assumed to be concentric unless otherwise stated. The average of a function U over the set E is defined as 1 (u)E = u(x) dx = u(x) dx. ÕE ðEð ÊE

The reader is assumed to be familiar with basic facts of Sobolev spaces, roughly corresponding to the content in chapter 7 in [6] or chapter 5 in [1]. We now, however, repeat the most basic deﬁnitions and properties.

5 1 u u ∈ L (Ω) = ( 1, … , n) Let be a locally summable function, i.e., loc , and let be a multiindex of order 1 = 1 + … + n v ∈ L (Ω) th u ð ð . Then loc is the weak derivative of if the equation

uD ' dx = (−1)ð ð v' dx, ÊΩ ÊΩ where )ð ðu D u = )x 1 , )x n 1 … n ' C∞ D u v holds for all ∈ 0 (Ω), i.e., all smooth functions in Ω with compact support. We use the notation = . k,p The Sobolev space W (Ω) is deﬁned as the normed space that consists of equivalence classes of locally p summable functions u ∶ Ω → ℝ such that D u exists in the weak sense and D u ∈ L (Ω) for all ð ð ≤ k. k,p The norm of u ∈ W (Ω) is deﬁned as

1∕p ⎧ ∑ p ⎪ k ∫Ω ðD uð dx 1 ≤ p < ∞, u k,p ð ð≤ ‖ ‖W (Ω) = ⎨ ∑ D u p = ∞. ⎪ ≤k ð ð ⎩ ð ð ess supΩ ∇u = (u , … , u ) Unless otherwise stated, x1 xn will always denote the distributional gradient. Furthermore, W k,p C∞ W k,p W k,p W k,p 0 (Ω) is the closure of 0 (Ω) in (Ω) and we note that both (Ω) and 0 (Ω) are Banach spaces. W k,p(Ω) Lp (Ω) u ∈ W k,p(Ω) u ∈ W k,p(D) The space loc is deﬁned analogously to loc : loc if and only if for each open set D ⊂⊂ Ω. In the following we will almost exclusively deal with the case k = 1 and 1 < p < ∞. For u W 1,p ∈ 0 (Ω) and Ω bounded we recall the Poincaré inequality 0 11∕n ðΩð ‖u‖Lp(Ω) ≤ ‖∇u‖Lp(Ω), !n n where !n is the volume of a unit ball in ℝ . We will sometimes use the notation A ≈ B which means that the ratio of A and B is bounded from above and below by constants. The dependence of the constants will be speciﬁed at each occurrence.

1.1.2 Weighted Sobolev spaces

We will now give a very brief introduction to weighted Sobolev spaces. These function spaces will only occur in Section 7.4 but it actually turns out that most of the results we discuss in this thesis are also valid for functions that belong to the weighted Sobolev spaces. For a better and deeper introduction to the subject, see ∈ L1 (ℝn) [4] or [9]. We consider a non-negative real-valued function loc and deﬁne the Radon measure by

(E) = (x) dx ÊE n p ̃ p p whenever E ⊂ ℝ . We will denote the L -space corresponding to the measure by L (Ω) or L (Ω; ). We have the following deﬁnition: Deﬁnition 1.1. A weight is called p-admissible if the following conditions hold: n (i) (x) ∈ (0, ∞) a.e. in ℝ and the corresponding measure is a doubling measure, i.e. (B2r) ≤ c(Br) n for all Br ⊂ ℝ .

6 ̃ p ∞ p (ii) If v ∈ L (G) is a vector-valued function and {'i} ⊂ C (G) such that ∫G ð'ið d → 0 and ∫G ð∇'i − p vð d → 0 as i → ∞, then v = 0. n (iii) The weighted Sobolev inequality holds, i.e., there exists > 1 such that for all Br ⊂ ℝ it holds that 0 11∕p 0 11∕p 1 1 ' p d cr ∇' p d , (B ) ð ð ≤ (B ) ð ð r ÊBr r ÊBr ' C∞ B where ∈ 0 ( r). (iv) The weighted Poincaré inequality holds, i.e.,

' − (') p d crp ∇' p d ð Br ð ≤ ð ð ÊBr ÊBr ' ∈ C∞(B ) (') B for all bounded r . Here Br is the average over the ball r using the weighted measure . We note from condition (i) that the Lebesgue measure dx is absolutely continuous with respect to . In the 1,p 1,p following we assume that (x) is a p-admissible weight. The weighted Sobolev space W (Ω) or W (Ω; ) is deﬁned as the closure of smooth functions in Ω, with respect to the weighted Sobolev norm 0 11∕p 0 11∕p ̃ p p ‖'‖ = ð'ð d + ð∇'ð d ÊΩ ÊΩ 1,p p ̃ p Thus, a function u is in W (Ω) if and only if u ∈ L (Ω) and there exists a vector-valued function v ∈ L (Ω) such that for some sequence of smooth functions {'i} such that the following conditions are satisﬁed:

p lim ð'i − uð d = 0 i→∞ ÊΩ p lim ð∇'i − vð d = 0. i→∞ ÊΩ v u W1,p(Ω) v = ∇u W1,p(Ω) W1,p(Ω) We say that is the gradient of in and use the notation . The spaces 0 and loc W1,p(Ω) L1 (Ω) are deﬁned analogous to the unweighted cases. Note that an element in is not necessarily in loc and therefore ∇u does not necessarily have to be the distributional gradient of u.

We now turn to a special class of p-admissible weights, called the Muckenhoupt class Ap, for which the 1,p distributional gradients of the elements in the corresponding Sobolev space W (Ω, ) exist. loc n Deﬁnition 1.2. Assume that 1 < p < ∞. We say that a locally summable function ∶ ℝ → [0, ∞) is an n Ap(ℝ )-weight if 0 1 0 1p−1 1∕(1−p) dx dx ≤ ÕBr ÕBr n for a constant = (p, ) whenever Br ⊂ ℝ . 1,p We end this subsection by proving the existence of the distributional gradient, ∇u, for u ∈ W (Ω). We ﬁrst p 1 show that L (D) ⊂ L (D) for D ⊂⊂ Ω. This can be seen as follows:

1∕p −1∕p ðuð dx = ðuð dx ÊD ÊD 0 11−1∕p 0 11∕p 1∕(1−p) p ≤ dx ðuð d ÊD ÊD

7 1,p where we used the Hölder inequality. Therefore, if 'j → u in W (Ω), it follows that 'j → u and )i'j → )iu 1 in L (D) and thus we see that ó ó ó ó ó ó ó u)i' + ')iu dxó = ó (u − 'j))i' + ()iu − )i'j)' dxó → 0 as j → ∞ óÊΩ ó óÊ ' ó ó ó ó spt ó ' C∞ u whenever ∈ 0 (Ω) which implies that the gradient ∇ exists in distributional sense. We will return to the weighted Sobolev spaces and the Muckenhoupt class Ap in Section 7.4.

1.2 Variational integrals and the p-Dirichlet integral

Let d(x) = (x) dx for a p-admissable weight (x). In the general setting, and using the notation from [9], we consider IF (u, Ω) = F (x, ∇u(x)) dx (2) ÊΩ n n where F (x, ) ∶ ℝ × ℝ → ℝ is a mapping called the variational kernel satisfying the following conditions: n n • x ↦ F (x, ) is measurable for all ∈ ℝ for a.e. x ∈ ℝ , p p n • (x)ðð ≤ F (x, ) ≤ (x)ðð , for 0 < ≤ < ∞ and ∈ ℝ , • the mapping ↦ F (x, ) is strictly convex and continuously differentiable, and p n • F (x, ) = ðð F (x, ), ∈ ℝ, ∈ ℝ . 1 From the second assumption we see that the integral in (2) is finite if and only if F (x, ∇u(x)) ∈ L (Ω). We are interested in minimizing the integral in (2) among a certain class of functions with given boundary values. We have the following definition. 1,p 1,p Definition 1.3. We say that a function u ∈ W (Ω) is an F -extremal in Ω with boundary values g ∈ W (Ω) u g W1,p if − ∈ 0 (Ω) and IF (u, Ω) ≤ IF (v, Ω), v − g ∈ W1,p(Ω) u ∈ W1,p(Ω) F Ω whenever 0 . Furthermore, we say that a function loc is a (free) -extremal in if u is an F -extremal with boundary values u in each open set D ⊂⊂ Ω. The following theorem characterizes all F -extremals in Ω (Theorem 5.18 in [9]). Theorem 1.1. A function u ∈ W1,p(Ω) is an F -extremal in Ω if and only if loc

−div ∇F (x, ∇u) = 0 in Ω, that is,

⟨∇F (x, ∇u), ∇'⟩ dx = 0 ÊΩ for all ' C∞ . ∈ 0 (Ω) u ∈ W 1,p(Ω) 1 < p < ∞ p For loc and we consider the -Dirichlet integral,

p I(u) = ð∇uð dx, (3) ÊΩ

8 p p−2 p−2 i.e., F (x, ) = ðð , so ∇F (x, ) = pðð and thus ∇F (x, ∇u) = pð∇uð ∇u. In order for u to be an F -extremal, it must satisfy p−2 Δpu ≡ div(ð∇uð ∇u) = 0, (4) by the ﬁrst condition of the theorem. This equation is called the p-Laplace equation in Ω. From now on we 2 only consider the p-Dirichlet integral and the p-Laplace equation. If we assume that u ∈ C (Ω) and that ∇u ≠ 0 in Ω we can formally carry out the diﬀerentiation in (4) which yields

n p−2 n 00 1 2 1 0 1 ) É 2 p−4 É p−2 u ux = (p − 2) ∇u ux ux x ux + ∇u ux x , )x xi j ð ð i i j j ð ð j j j i=1 i=1 and thus we obtain p−2 n H n I É ⎛ ) ⎛ É 2 ⎞⎞ ∇ ⋅ ( ∇u p−2∇u) = u2 u ð ð ⎜ ⎜ x xj ⎟⎟ ⎜)xj ⎜ i ⎟⎟ j=1 ⎝ ⎝ i=1 ⎠⎠ n H n I É É = (p − 2) ∇u p−4 u u u + ∇u p−2u ð ð xi xixj xj ð ð xj xj j=1 i=1 n n É É = (p − 2) ∇u p−4 u u u + ∇u p−2 u ð ð xi xj xixj ð ð xj xj i,j=1 j=1 p−4 2 = ð∇uð {ð∇uð Δ2u + (p − 2)Δ∞u}, where n É )u )u )2u Δ u = , ∞ )x )x )x )x i,j=1 i j i j Δ p = 2 Δ u = ∑(u )2 and 2 denotes the Laplace operator. We see that for we obtain the Laplace operator 2 xixi . From the calculations above we see that we can write Δpu as a partial diﬀerential equation in non-divergence form, n É Lu ∶= a (x)u (x), ij xixj i,j=1 with a (x) = (p − 2) ∇u p−4u u + ∇u p−2 . ij ð ð xi xj ð ð ij

We note that aij = aji and it follows that

n n É É a (x) = (p − 2) ∇u p−2u u + ∇u p−2 ij i j ð ð xi j xj i ð ð ij i j i,j=1 i,j=1 p−4 2 p−2 2 = (p − 2)ð∇uð ⟨∇u, ⟩ + ð∇uð ðð .

By the Cauchy-Schwarz inequality we see immediately that for p ≥ 2 n p−2 2 É p−2 2 ð∇uð ðð ≤ aij(x)ij ≤ (p − 1)ð∇uð ðð , i,j=1

9 and for 1 < p < 2 we have that n p−2 2 É p−2 2 (p − 1)ð∇uð ðð ≤ aij(x)ij ≤ ð∇uð ðð . i,j=1

Thus, on compact sets D where ∇u ≠ 0 on D we see that L satisﬁes the uniform ellipticity condition. This can also be seen by using that the Laplace equation is invariant under rotations, which we show in Lemma A.1. Using the rotation invariance it is possible to derive the following fundamental solution for the p-Laplace equation T − log ðxð p = n, Φ(x) = p−n (n − p)ðxð p−1 p < n, which is shown in a calculation following Lemma A.1.

1.3 Structure of Thesis

In the following sections, we first introduce the basic definitions (Section 2), before moving on to regularity of weak solutions (Section 3), differentiability (Section 4), p-superharmonic functions (Section 5) and Perron’s method (Section 6). In these chapters, we closely follow [21], although we try to provide more details and on occasion we also provide theory and results from [9]. In the last section, we proceed with boundary behaviour of weak solutions to the p-Laplace equation (Section 7). There we concern ourselves with the theory and problems from [23], which are based on the results in [13] and [14]-[20]. Many of the statements of the theorems, lemmas, proposition and definitions are taken word for word from sources mentioned above.

2 Basic deﬁnitions and existence of weak solutions

In this section we concern ourselves with basic properties of weak solutions to the p-Laplace equation and discuss some fundamental results that will be used throughout this thesis, such as Caccioppoli’s inequality and the well-known comparison principle. Furthermore, we prove the existence of a p-harmonic function in a bounded domain with boundary values given in Sobolev sense. We end this section by discussing regular points and the Wiener criterion. We begin by deﬁning a weak solution to the p-Laplace equation. Deﬁnition 2.1. u ∈ W 1,p(Ω) p Ω We say that loc is a weak solution of the -Laplace equation in , if

p−2 ⟨ð∇uð ∇u, ∇'⟩ dx = 0, (5) ÊΩ ' C∞ u u p for each ∈ 0 (Ω). If, in addition, is continuous, we say that is a -harmonic function. Furthermore, u ∈ W 1,p(Ω) p Ω we say that loc is a weak supersolution of the -Laplace equation in , if

p−2 ⟨ð∇uð ∇u, ∇'⟩ dx ≥ 0, (6) ÊΩ ' C∞ u u for each non-negative ∈ 0 (Ω). A function is a subsolution if − is a supersolution or equivalently ' C∞ stated that (6) holds for each non-positive ∈ 0 (Ω).

10 If we refer to u as a solution of (4) it will always mean that u is a solution in the weak sense unless otherwise stated. We note that if u is a (super)solution and , ∈ ℝ and ≥ 0 we have that u+ is also a (super)solution. W 1,p W 1,p Since 0 (Ω) is the closure of smooth functions in (Ω) we have the following lemma. 1,p Lemma 2.1. If u ∈ W (Ω) is a solution of (4) in Ω, then

p−2 ⟨ð∇uð ∇u, ∇'⟩ dx = 0 ÊΩ for all ' W 1,p . ∈ 0 (Ω)

Proof. ' W 1,p ' C∞ ' ' W 1,p Since ∈ 0 (Ω) we choose functions i ∈ 0 (Ω) such that i → in (Ω). By Hölder’s inequality we deduce that ó ó ó ∇u p−2∇u, ∇' dx − ∇u p−2∇u, ∇' dxó ó ⟨ð ð ⟩ ⟨ð ð i⟩ ó óÊΩ ÊΩ ó p−1 ≤ ð∇uð ð∇' − ∇'ið dx ÊΩ 0 1(p−1)∕p 0 11∕p p p ≤ ð∇uð dx ð∇' − ∇'ið dx . ÊΩ ÊΩ

The last integral tends to zero as i → ∞ so it follows that

p−2 p−2 ⟨ð∇uð ∇u, ∇'⟩ dx = lim ⟨ð∇uð ∇u, ∇'i⟩ dx = 0. ÊΩ i→∞ ÊΩ

u ∈ W 1,p(Ω) This lemma will be used extensively. From the proof it follows that if loc is a weak solution, then ' W 1,p u W 1,p (5) holds for all ∈ 0 (Ω) with compact support. Moreover, we note that (5) holds if ∈ (Ω) is a p loc weak solution with ∇u ∈ L (Ω). Furthermore, it is also clear from the proof that the analogous version holds for supersolutions as long as ' is non-negative and we pick a non-negative approximating sequence. In the more general setting of Theorem 1.1 we saw that minimizers are the same as weak solutions, which is also established in the following theorem. Theorem 2.2. The following conditions are equivalent for u ∈ W 1,p(Ω): (i) u is minimizing: u p dx v p dx, when v u W 1,p ð∇ ð ≤ ð∇ ð − ∈ 0 (Ω) ÊΩ ÊΩ (ii) The ﬁrst variation vanishes:

u p−2 u, ' dx , when ' W 1,p ⟨ð∇ ð ∇ ∇ ⟩ = 0 ∈ 0 (Ω) ÊΩ

Proof. (i) v x u x "' x " ' W 1,p v u Suppose that holds and take ( ) = ( ) + ( ) for ∈ ℝ and ∈ 0 (Ω) so that − ∈ W 1,p u 0 (Ω). Since is minimizing, the integral

p J(") = ð∇(u + "')ð dx, ÊΩ

11 ¨ attains its minimum for " = 0 so the ﬁrst variation vanishes at " = 0, i.e., J (0) = 0. Furthermore, we have that 0 1 d p p−2 É ) ) (ð∇(u + "')ð ) = pð∇(u + "')ð (u + "'), ' , d" )xi )xi p and, since ð∇(u + "')ð is continuously diﬀerentiable a.e., with respect to " this implies that

¨ p−2 J (0) = ð∇(u)ð ⟨∇u, ∇'⟩ dx, ÊΩ

(i) (ii) (ii) v u W 1,p v W 1,p so ⇒ . Next we assume that holds and that − ∈ 0 (Ω). Note that this implies that ∈ (Ω). Then p−2 ⟨ð∇uð ∇u, ∇(v − u)⟩ dx = 0, ÊΩ and using Cauchy Schwartz and the Hölder inequality we obtain

p p−2 ð∇uð dx = ⟨ð∇uð ∇u, ∇v⟩ dx ÊΩ ÊΩ p−1 ≤ ð∇uð ð∇vð dx ÊΩ 0 11−1∕p 0 11∕p p p ≤ ð∇uð dx ð∇vð dx , ÊΩ ÊΩ from which (i) follows immediately.

(ii) (i) u v u W 1,p v u We note that ⇒ also holds when is a supersolution, provided that − ∈ 0 (Ω) with ≥ . 1,p Next, we define the so called obstacle problem and for this we assume that Ω is bounded and that ∈ W (Ω). Let v W 1,p v , v W 1,p .  ,(Ω) = { ∈ (Ω) ∶ ≥ a.e. in Ω − ∈ 0 (Ω)} Definition 2.2. We say that a function u in  ,(Ω) is a solution to the obstacle problem with obstacle and boundary value if p−2 ⟨ð∇uð ∇u, ∇(v − u)⟩ dx ≥ 0, ÊΩ whenever v ∈  ,(Ω). Then we say that u is a solution to the obstacle problem in  ,(Ω). From the definition, and an application of Hölders theorem, it follows directly that a solution u to the obstacle problem minimizes the p-energy among the functions in  ,(Ω). We note that a solution u to the obstacle problem is always a weak solution to the p-Laplace equation since u+' ∈  ,(Ω) for a non-negative function ∞ p ' ∈ C (Ω) u ∈ L (Ω) u,u(E) 0 . Furthermore, a supersolution loc is a solution to the obstacle problem in  where 1,p 1,p E Ω v − u ∈ W (E) u ∈ W (E) ,(Ω) is a compactly contained set in , since loc is non-negative and . If  is nonempty it can be shown that there exists a unique solution to the obstacle problem (Theorem 3.21 in [9]). Moreover, if the obstacle is continuous the weak solution will also be continuous. We will return to a special case of the obstacle problem in Section 5, when we discuss p-superharmonic functions. The next estimate is known as Caccioppoli’s inequality: Lemma 2.3. If u is a weak solution in Ω, then

p p p p p ð∇uð dx ≤ p ðuð ð∇ð dx, ÊΩ ÊΩ

12 for each C∞ , . In particular, if B ⊂ , then ∈ 0 (Ω) 0 ≤ ≤ 1 2r Ω

p p −p p ð∇uð ≤ p r ðuð dx. ÊBr ÊB2r

Proof. ' pu ' W 1,p ' p u p p−1u u Set = and note that ∈ 0 (Ω) and ∇ = ∇ + ∇ . Since is a weak solution it satisﬁes (5) which yields

p p p−1 p−2 ð∇uð dx = −p u⟨ð∇uð ∇u, ∇⟩ dx ÊΩ ÊΩ p−1 ≤ p ð∇uð ðu∇ð dx ÊΩ 0 11−1∕p 0 11∕p p p p p ≤ p ð∇uð dx ðuð ð∇ð dx , ÊΩ ÊΩ and therefore we conclude that

p p p p p ð∇uð dx ≤ p ðuð ð∇ð dx. ÊΩ ÊΩ

For the second statement we choose as a radial function such that = 1 in Br, ð∇ð ≤ 1∕r and = 0 in Ω ⧵ B2r. The claim now follows immediately from the ﬁrst statement.

We note that a slightly modiﬁed variant of Caccioppoli’s lemma holds for bounded supersolutions. Assume p that u is a bounded supersolution and let L = supΩ u so 0 ≤ (L − u). We choose a test function ' = (L − u) C∞ ' u p p L u p−1 for a non-negative ∈ 0 (Ω) and calculate ∇ = −∇ + ( − ) ∇ . Thus,

p p p−1 p−2 ð∇uð dx ≤ p(L − u) ð∇uð ⟨∇u, ∇⟩ dx ÊΩ ÊΩ p−1 ≤ p ð∇uð ð(L − u)∇ð dx ÊΩ 0 11−1∕p 0 11∕p p p p ≤ p ð∇uð dx ðL − uð ð∇ð dx , ÊΩ ÊΩ so p p p p p ð∇uð dx ≤ p ðL − uð ð∇ð dx. ÊΩ ÊΩ The next lemma also concerns supersolutions. Lemma 2.4. If v > 0 is a weak supersolution in Ω, then 0 1p p p p p ð∇ log vð dx ≤ ð∇ð dx ÊΩ p − 1 ÊΩ whenever C∞ and . ∈ 0 (Ω) ≥ 0

p 1−p Proof. We prove it for u(x) = v(x)+" where " > 0. This is still a supersolution since ∇" = 0. Let ' = u and note that it is well deﬁned since u > 0. It follows that ∇' = p p−1u1−p∇ − (p − 1) pu−p∇u,

13 and since u is a supersolution we have that

p−2 p−1 1−p p −p p 0 ≤ ⟨ð∇uð ∇u, p u ∇⟩ − (p − 1) u ð∇uð dx, ÊΩ so

p −p p p−2 p−1 1−p (p − 1) u ð∇uð dx ≤ p ⟨ð∇uð ∇u, u ∇⟩ ÊΩ ÊΩ p−1 p−1 1−p ≤ p ð∇uð u ð∇ð dx ÊΩ 1− 1 1 0 1 p 0 1 p p p −p p ≤ p ð∇uð u dx ð∇ð . ÊΩ ÊΩ

Thus, it is clear that 0 1p p p −p p p ð∇uð u dx ≤ ð∇ð dx, ÊΩ p − 1 ÊΩ ∇ log(v) = ∇v and since v we obtain

∇v p 0 p 1p p ð ð dx p dx. p ≤ ð∇ ð ÊΩ (v + ") p − 1 ÊΩ

We conclude the proof by letting " tend to 0.

The next theorem is the comparison principle. Theorem 2.5. Suppose that u and v are p-harmonic functions in a bounded domain Ω. If at each ∈ )Ω

lim sup u(x) ≤ lim inf v(x) x→ x→ excluding the situation ∞ ≤ ∞ and −∞ ≤ −∞, then u ≤ v in Ω.

Proof. Take " > 0 and consider the open set

D" = {xðu(x) > v(x) + "}.

Due to the assumptions in the theorem either D" is empty, and then there is nothing to prove, or D" ⊂⊂ Ω. To this end we assume that D" ≠ ç. If we use that u and v are weak solutions we obtain

p−2 p−2 ⟨ð∇vð ∇v − ð∇uð ∇u, ∇⟩ dx = 0, ÊΩ

W 1,p where is any ∈ 0 (Ω) with compact support. By using '(x) = min{v(x) − u(x) + ", 0},

' ⊂ D ' W 1,p and that supp( ) " and ∈ 0 (Ω) it follows that

p−2 p−2 ⟨ð∇vð ∇v − ð∇uð ∇u, ∇v − ∇u⟩ dx = 0. ÊD"

14 For for 1 < p < ∞ we note that p−2 p−2 ⟨ð∇vð ∇v − ð∇uð ∇u, ∇v − ∇u⟩ p−2 p−2 p−2 p−2 = ⟨ð∇vð ∇v − ð∇uð ∇u, ∇v⟩ − ⟨ð∇vð ∇v − ð∇uð ∇u, ∇u⟩ p p−2 p−2 p = ð∇vð − ⟨ð∇uð ∇u, ∇v⟩ − ⟨ð∇vð ∇v, ∇u⟩ + ð∇uð p p p−2 p−2 = ð∇vð + ð∇uð − ⟨∇u, ∇v⟩(ð∇uð + ð∇vð ) p p p−1 p−1 ≥ ð∇vð + ð∇uð − ð∇uð ð∇vð − ð∇vð ð∇uð p−1 p−1 = ð∇vð (ð∇vð − ð∇uð) + ð∇uð (ð∇uð − ð∇vð) p−1 p−1 = (ð∇vð − ð∇uð)(ð∇vð − ð∇uð ) ≥ 0, and therefore the integral is strictly positive if ∇u ≠ ∇v on a set of positive measure. Thus, it follows that ∇u = ∇v a.e. in D" which implies that u = v + c in D". Since u = v + " on )D", we have c = " so u ≤ v + " in Ω. We let " tend to zero to obtain the result.

Studying the proof we see immediately that the comparison principle also holds when u and v are weak sub- and supersolutions, respectively. Furthermore, in Section 5 we show that the comparison principle holds when u is p-subharmonic and v is p-superharmonic. We next establish the existence of a solution to the Dirichlet problem for the p-Laplace operator with boundary values in Sobolev sense. However, we first recall 1,p the definition of a weakly lower semicontinuous function on W (Ω). 1,p Definition 2.3. We say that a function I(⋅) is weakly lower semicontinuous on W (Ω) if

I(u) lim inf I(uk) ≤ k→∞

1,p whenever uk ⇀ u weakly in W (Ω). We are now ready to state and prove the existence theorem. Theorem 2.6. Suppose that g ∈ W 1,p(Ω) where Ω is a bounded domain in ℝn and deﬁne

v W 1,p v g W 1,p .  = { ∈ (Ω) ∶ − ∈ 0 (Ω)}

There exists a unique u ∈  such that

p p ð∇uð dx ≤ ð∇vð dx ÊΩ ÊΩ for all v ∈ . This u is a weak solution to the p-Laplace equation.

Proof. At ﬁrst we show the uniqueness and thus we assume that there exist two diﬀerent minimizers u1 and p u2 s.t. the set {∇u1 ≠ ∇u2} has positive measure. Let v = (u1 + u2)∕2. By the convexity of ðxð we know that p p p ó∇u1 + ∇u2 ó ∇u1 + ∇u2 ó ó ð ð ð ð , ó ó ≤ ó 2 ó 2

15 with strict inequality for {∇u1 ≠ ∇u2}. Since v ∈ , p ó∇u1 + ∇u2 ó ∇u p dx ó ó dx ð 2ð ≤ ó ó ÊΩ ÊΩ ó 2 ó 1 p 1 p < ð∇u1ð dx + ð∇u2ð dx 2 ÊΩ 2 ÊΩ p ≤ ð∇u2ð dx, ÊΩ where the ﬁrst inequality follows from u2 being a minimizer and the third inequality from u1 being a minimizer. We have arrived at a contradiction and hence ∇u1 = ∇u2 a.e. in Ω so u1 = u2 + c a.e. in Ω. Since u u W 1,p c p 2 − 1 ∈ 0 (Ω) it follows that = 0. Next we establish the existence of a -harmonic function with the given boundary values in Ω. Let

p I0 = inf I(v) = inf ð∇vð dx. v∈ v∈ ÊΩ

It is clear that p 0 ≤ I0 ≤ ð∇gð dx < ∞, ÊΩ g v ∞ v j since ∈ . We proceed by choosing a minimizing sequence { j}j=1 such that j ∈  for all and

p 1 ð∇vjð dx < I0 + , j = 1, 2, 3, … ÊΩ j

1,p We next show that {vj} is bounded in W (Ω), i.e.,

p p ðvjð dx + ð∇vjð dx ≤ M. (7) ÊΩ ÊΩ

Since Ω is bounded, we can employ the Poincaré inequality to assert that

p p ðwð dx ≤ c ð∇wð dx, ÊΩ Ê

c c p, n, w W 1,p v g where = ( Ω), holds for all ∈ 0 (Ω). Applying this to j − yields 0 1 p p p ðvj − gð dx ≤ c ð∇vjð dx + ð∇gð dx ÊΩ ÊΩ ÊΩ 0 1 p ≤ c (I0 + 1) + ð∇gð dx , ÊΩ so by using the revered triangle inequality it follows that

p ðvjð dx ≤ M1, ÊΩ j < p < W 1,p v ∞ for all which gives us (7). Since 1 ∞, (Ω) is reﬂexive and therefore the sequence { j}j=1 is Ω u ∈ W 1,p(Ω) {v } weakly precompact in , which implies that there exists a and a subsequence j such that v ⇀ u, ∇v ⇀ ∇u, j j

16 Lp(Ω) v ⇀ u ∈ W 1,p(Ω) v − g ⇀ u − g W 1,p(Ω) weakly in so j . Thus it follows that j in 0 . By deﬁnition, W 1,p W 1,p 0 (Ω) is a closed linear subspace of (Ω) so it is convex and therefore it follows from Mazurs lemma W 1,p u g W 1,p u that 0 (Ω) is closed under weak convergence. Hence − ∈ 0 (Ω), so is an admissible function, i.e., 1,p u ∈ . It is left to show that u is the desired minimizer. I(⋅) is weakly lower semicontinuous on W (Ω) (see e.g. Theorem 8.2.1 in [1]), and thus I(u) lim inf I(v ) = I . ≤ j 0 j →∞

Since u ∈  we conclude that I(u) = I0.

We next consider the Dirichlet problem for the p-Laplace operator. If the boundary function g is continuous, ̄ ̄ i.e., g ∈ C(Ω) and Ω is regular, then u ∈ C(Ω) and uð)Ω = gð)Ω. By regular we mean the following: Definition 2.4. Assume that Ω is a bounded domain. We say that x0 ∈ )Ω is a regular point for the p-Dirichlet g W 1,p C ̄ p u u g W 1,p problem if for each ∈ (Ω) ∩ (Ω), and the unique -harmonic function such that − ∈ 0 (Ω) it holds that lim u(x) = g(x0). x→x0 Ω is regular if each boundary point is regular. Examples of regular sets are balls and polyhedra. Furthermore, every open domain has a so called exhaustion of Ω with regular domains, i.e., there exist domains D1 ⊂ D2 ⊂ … such that Dj ⊂⊂ Ω is regular for each j and Ω = ∪Dj. This can be seen as follows: We first find domains G1 ⊂⊂ G2 ⊂⊂ … ⊂⊂ Ω and then cover each Gj with a finite union  of open cubes and let Dj = int , see [9] and Corollary 6.32 in particular. In 1924 Wiener developed a nifty method to determine if a boundary point w ∈ )Ω is regular, known as the Wiener criterion. Before stating that we need to define the p-capacity of a set. Definition 2.5. Let K ⊂⊂ B(x, r) be a compact set and define ∞ W (K,B(x, r)) = {' ∈ C0 (Ω) ∶ 0 ≤ ' ≤ 1 and ' = 1 in K}.

We deﬁne the p-capacity of K as

p capp(K,B(x, r)) = inf ð∇'ð dx. W (K,B(x,r)) ÊB(x,r)

We now formulate the Wiener criterion. Theorem 2.7. The set Ω is regular at w ∈ )Ω for the p-Dirichlet problem if and only if the following condition holds: 1∕(p−1) 1 H n I capp(B(w, t) ∩ (ℝ ⧵ Ω),B(w, 2t)) dt = ∞. Ê t 0 capp(B(w, t),B(w, 2r))

We will return to the p-Dirichlet problem and regular points in Section 6, where we discuss Perron’s method.

3 Regularity of weak solutions

In this section we will prove that the weak solutions to the p-Laplace equation are locally Hölder continuous. In order to do so one can use the fact that weak solutions to the p-Laplace equation satisfy the Harnack

17 inequality, i.e., that the maximum of the function in a ball is bounded by a constant times the minimum in the same ball, where the constant only depends on p and n. The theorems are stated in the beginning and proved for diﬀerent values of p throughout the section. Theorem 3.1. Suppose that u ∈ W 1,p(Ω) is a weak solution to the p-Laplace equation. Then there exists loc constants > 0 and L such that = (p, n) and L = L(p, n, u p ) such that ‖ ‖L (B2r)

ðu(x) − u(y)ð ≤ Lðx − yð for a.e. x, y ∈ B(x0, r) whenever B(x0, 2r) ⊂⊂ Ω. We next formulate Harnack’s inequality from which the above theorem follows. This inequality will be very useful later in Section 7 when discussing the boundary behaviour of p-harmonic functions. In particular, we will then prove that the ratio of two p-harmonic functions that vanish on a portion of the boundary satisﬁes a Harnack inequality close to that part of the boundary. 1,p Theorem 3.2. Suppose that u ∈ W (Ω) is a weak solution and that u ≥ 0 in B2r ⊂ Ω. We deﬁne the essential minimum and essential supremumloc as follows:

m(r) = ess inf u, M(r) = ess sup u. B r Br

Then there exists c = c(n, p) such that M(r) ≤ cm(r).

Later in this section the Harnack inequality will be proved for n < p < ∞ and for 1 < p < n (see [9] for a proof that holds for 1 < p < ∞). We almost immediately obtain the strong maximum principle. Corollary 3.3. If a p-harmonic function attains its maximum at an interior point, then the function is constant.

Proof. We suppose that u(x0) = maxx∈Ω u(x) for x0 ∈ Ω and apply Harnack’s inequality to the non-negative p-harmonic function v(x) = u(x0)−u(x). Thus, the minimum of v(x) is m(r) = 0 and by applying the Harnack inequality again it follows that M(r) = 0 when 2ðx − x0ð < dist(x0,)Ω). Therefore v(x) is constant and it follows directly that u(x) is also constant. In order to show that u is constant in the whole domain we can either use a chain of intersecting balls or note that the set {x ∈ Ωðu(x) = s} is both open and closed in Ω and thus equals Ω since a domain is connected.

We next show that the Harnack inequality implies Hölder continuity. This type of iterative argument is a standard technique and will be referred to several times throughout this thesis.

Proof of Theorem 3.1. We begin by choosing r suﬃciently small such that B2r ⊂ Ω and applying Harnack’s inequality to the non-negative weak solutions u(x) − m(2r) and M(2r) − u(x) to obtain

M(r) − m(2r) ≤ c(m(r) − m(2r)) M(2r) − m(r) ≤ c(M(2r) − M(r)).

After adding them we see that

M(r) − m(2r) + M(2r) − m(r) ≤ c(m(r) − m(2r) + M(2r) − M(r)) ⇔ M(r) − m(r) + M(2r) − m(2r) ≤ −c(M(r) − m(r)) + c(M(2r) − m(2r)) ⇔ (M(r) − m(r))(1 + c) ≤ (c − 1)(M(2r) − m(2r))

18 so c − 1 !(r) Λ!(2r), Λ ≤ ≡ c + 1 (8) where !(r) = M(r) − m(r) is the essential oscillation of n over B(x0, r). From Harnack’s inequality it follows that c ≥ 1 but if c = 1 the function is constant and in that case there is nothing to prove. For the remaining part of the proof we thus assume that c > 1. We may assume that c ≥ 3 so that = − log(Λ)∕ log(2) ≤ 1. −(m−1) (m−1) m−1 m By iterating (8) it follows that !(2 r) ≤ Λ !(r). Choose m ≥ 1 such that 2 < R∕r ≤ 2 . Then r (2−m) = 2−(2m−1)− = 2−Λm−1, R ≥ and thus r !(r) Λ(m−1)!(2(m−1)r) Λ(m−1)!(2(m−1)R) 2 !(R), ≤ ≤ ≤ R so we see that r (u, B(r)) 2 (u, B(R)), 0 < r < R, osc ≤ R osc u ∈ C (Ω) so loc .

We will prove the Harnack inequality for the 1 < p < ∞, p ≠ n. For p = n we will instead prove the Hölder continuity using Morrey’s lemma. However, we note that Harnack’s inequality also holds for the case where p = n and refer to [9] for a proof.

3.0.1 The case 1 < p < n

We will prove the Harnack inequality for a non-negative weak solution u. In order to do so we will use the interpretations 0 11∕q ess sup = lim uq dx q→∞ Br ÊBr 0 11∕q ess inf = lim uq dx B q→−∞ r ÊBr Furthermore, we will use the notation 1 H I q q ‖u‖q,s = u dx ÊBs

On more than one occasion in this section we will use one of the Sobolev inequalities, which states that for p < n c c n, p u W 1,p there exists a constant = ( ) such that for any ∈ 0 (Ω) the following inequality holds:

‖u‖Lp(Ω) ≤ c‖∇u‖Lp(Ω) (9) where = n(n − p). The proof of the Harnack inequality requires several lemmas and the ﬁrst one concerns subsolutions. Lemma 3.4. Let u W 1,p be a weak subsolution and B ⊂⊂ . Then for > p there exists ∈ 0 (Ω) R Ω − 1 c = c(n, p, ) 0 11∕ 1 ess sup(u ) c u dx + ≤ (R − r)n + Br ÊBR where r < R and u+ = max{u(x), 0}.

19 Proof. v u ' pv −(p−1) C∞ We start by deﬁning = +. Next we set = for some non-negative ∈ 0 (Ω). Since v ∈ W 1,p(Ω) ' ∈ W 1,p(Ω) = − (p − 1) loc we see that 0 . For convenience we set which is positive by assumption. Furthermore, we see that ∇' = p p−1∇v + v −1 p∇v.

Since u is a weak subsolution and ∇v = ∇u+ = ∇u a.e on {u ≥ 0} it follows that

p −1 p p−1 p−2 v ð∇vð dx ≤ −p v ⟨ð∇vð ∇v, ∇⟩ dx. ÊΩ ÊΩ

Note that ( − 1)(p − 1) + p − 1 = + . p p (10) By the Cauchy-Schwarz and Hölder inequalities we obtain

p −1 p p−1 ( −1)(p−1)∕p p−1 ∕p v ð∇vð dx ≤ p v ð∇vð u ð∇ð dx ÊΩ ÊΩ 1− 1 1 0 1 p 0 1 p p −1 p ∕p p ≤ p v ð∇vð dx ðv ∇ð dx ÊΩ ÊΩ and therefore p p −1 p p p v ð∇vð dx ≤ v ð∇ð dx. (11) ÊΩ ÊΩ

Next, we use that p 0 1p ó −p ó ∇(v ∕p) p = ó v p ∇vó = v −1 ∇v p ð ð ó ó ð ðð ð (12) ó p ó p and combined with (11) we obtain 0 1p ∕p p ∕p p ð∇v ð dx ≤ ðv ∇ð dx. ÊΩ − (p − 1) ÊΩ

We note that ∕p ∕p ∕p ð∇(v )ð ≤ ð∇v ð + ðv ∇ð and use this together with Minkowski’s inequality as follows:

1 1 0 1 p 0 1 p ∕p p ∕p ∕p p ð∇(v )ð dx ≤ ð∇v ð + ðv ∇ð ÊΩ ÊΩ 1 1 0 1 p 0 1 p ∕p p ∕p p ≤ ð∇v ð dx + ðv ∇ð dx ÊΩ ÊΩ 1 0 1 0 1 p ∕p p ≤ 1 + ðv ∇ð dx , − p + 1 ÊΩ so 0 1p ∕p p 2 − p + 1 ∕p p ð∇(v )ð dx ≤ ðv ∇ð dx. ÊΩ − p + 1 ÊΩ

20 v ∕p ∈ W 1,p(Ω) Since loc has compact support we can apply the Sobolev inequality which yields:

1 0 1 ∕p p p ∕p p ðv ð dx ≤ c ð∇(v )ð dx, ÊΩ ÊΩ c c n, p n n p C∞ where = ( ) and = ∕( − ). We choose a test function ∈ 0 (Ω) such that 0 ≤ ≤ 1, ð∇ð ≤ 1∕(R − r), = 1 in Br and 0 outside BR so that 0 11∕ 0 11∕ ∕p p ðvð dx ≤ ðv ð dx ÊBr ÊBR 00 1p 1 1 2 − p + 1 1 c v dx . ≤ − p + 1 R − r ÊBR

We ﬁx 0 > p − 1 and note that 2 − p + 1 2 0 − p + 1 ≤ = b, − p + 1 0 − p + 1 when ≥ 0 since 2 − p + 1 p − 1 = 2 + . − p + 1 − (p − 1)

We are now going to iterate the estimate and we start with the radii r0 = R and r1 = r + (R − r)∕2. Since r0 − r1 = (R − r)∕2 the corresponding test function that will be used has ð∇ð ≤ 2∕(R − r) and it follows that

2 p∕ 0 v (cb)p 0 v . ‖ ‖ 0,r1 ≤ R − r ‖ ‖ 0,r0

−2 Then we use r1 and r2 = r + 2 (R − 2) which results in

p 2p p p + + 0 0 2 v 2 cb 0 0 v . ‖ ‖ ,r ≤ ( ) p ‖ ‖ ,r 0 2 + 0 0 (R − r) 0 0

−j Proceeding in this manner, i.e., using radii rj = r + 2 (R − r) yields

p ∑ −l p ∑ l−l+1 cb 0 v j+1 2 0 v . ‖ ‖ 0,rj+1 ≤ R − r ‖ ‖ 0,r0

−1 Since ð ð < 1, the two sums in the exponents are convergent and j É 1 − −(j+1) n −l = ⟶ j → ∞, −1 p as l=0 1 − j+1 É 1 n2 l−l+1 ⟶ = j → ∞. −1 2 2 as l=1 (1 − ) p

21 v j+1 v j+1 Finally, since ‖ ‖ ,r ≤ ‖ ‖ 0,R we obtain

ess sup(v) = lim v j+1 ,r j→∞ ‖ ‖ 0 Br

lim v j+1 ,R ≤ j→∞ ‖ ‖ 0 n n2 0 11∕ 0 cb 0 = 2 0p v 0 dx R − r ð ð ÊBR 0 1 1 1 0 = c v 0 dx . (R − r)n ð ð ÊBR

1,p Lemma 3.5. Suppose that v ∈ W (Ω) is a non-negative supersolution. Then for < 0 and BR ⊂⊂ Ω there loc exists a constant c = c(n, p, ) = c(n, p)−1∕ such that

0 11∕ 1 v dx ≤ c ess inf v. (R − r)n B ÊBR r

Proof. We note that we can assume that v(x) ≥ " > 0 since otherwise we prove inequality for v(x)+" instead p −(p−1) of v(x) and let " tend to zero. In this proof we use the function ' = v and set = − (p − 1) and note that is negative. Since v is a supersolution we obtain:

p −1 p p−1 p−2 v ð∇vð dx ≥ −p v ⟨ð∇vð ∇v, ∇⟩ dx ÊΩ ÊΩ and we use (10) and the Hölder inequality to obtain

p −1 p p−1 p−2 (− ) v ð∇vð dx ≤ p v ⟨ð∇vð ∇v, ∇⟩ dx ÊΩ ÊΩ 1− 1 1∕p 0 1 p 0 1 p −1 p p ≤ p v ð∇vð dx v ð∇ð dx ÊΩ ÊΩ so p ∕p p p p ð∇u ð dx ≤ v ð∇ð dx. ÊΩ − ÊΩ

Using (12) we obtain 0 1p ∕p p ð ð ∕p p ð∇v ð dx ≤ ðv ∇ð dx. (13) ÊΩ − ÊΩ If we now repeat the same procedure using Minkowski’s inequality, as in Lemma 3.4, and note that ð ð = − we deduce that 0 1p ∕p p p ð∇(v )ð dx ≤ 1 − v ð∇ð dx ÊΩ − ÊΩ 0 1p p − 1 − 2 p = v ð∇ð dx. p − 1 − ÊΩ

22 ∕p If we use (9) for v we see that 0 11∕ p p ∕p p v ≤ c ð∇(u )ð dx ÊΩ ÊΩ p p ≤ (2c) v ð∇ð dx ÊΩ since p − 1 − 2 p − 1 − = − 2 p − 1 − p − 1 − p − 1 − ≤ If we choose as in Lemma 3.4 we see that 0 11∕ 2c p v dx v dx, ≤ R − r ÊBr ÊBR and if we continue to use the same technique as in Lemma 3.4 it follows that H I1∕2 H I1∕ p∕ 2c 2 p∕ (v ) dx ≤ (2 ) v dx ÊB R − r ÊB r2 r1 p(1+ 1 ) 2S p(1+ 1 ) ≤ 2 v dx R − r ÊB r0 and eventually we arrive at the estimate 1∕kj H I p ∑ −l k 2c p ∑ l1−l (v ) j dx ≤ 2 v dx. ÊB R − r ÊB rj R

As we let j → ∞ we have that 2c n n2 ess sup v 2 p v dx ≤ R − r Br ÊBR 1 n c v dx, ≤ R − r ÊBR and therefore 0 11∕ 1 1 n ð ð c v dx , ess inf v ≤ R − r Br ÊBR and hence we obtain 0 n 1−1∕ð ð 1 ess inf v ≥ v dx . B R − r r ÊBR

Lemma 3.6. Let v ∈ W 1,p(Ω) be a non-negative weak supersolution. For 0 < " < < (p − 1) = loc n(p − 1)∕(n − p) and BR ⊂⊂ Ω there exists c = c(", , n, p) such that 0 11∕ 0 11∕" 1 1 v dx c v" dx . (R − r)n ≤ (R − r)n ÊBr ÊBR

23 Proof. We start by assuming that 1 < < p − 1. By following the proof of Lemma 3.5 but using that ð ð = in (13) we arrive at the estimate 0 11∕ 0 0 11p∕ 0 11∕ p − 1 1 v c v dx ≤ p − 1 − p∕ ÊBr (R − r) ÊBR

By iterating the estimate as in the previous lemmas we obtain p ∑ −l p cb ∑ l−l+1 v j+1 2 v ‖ ‖ ,rj+1 ≤ R − r ‖ ‖ ,r0 j It is clear that we can iterate the result an appropriate number of steps to obtain an exponent > p − 1. By eventually using Hölder’s inequality to obtain the right exponent and multiply with a suitable constant the conclusion of the lemma follows.

If we combine the the lemmas above we obtain the following bounds for non-negative weak supersolutions: 0 11∕ 1 ess sup u c( , n, p) u dx ≤ (R − r)n Br ÊBr 0 1−1∕ 1 − ess inf u ≥ c( , n, p) u dx B (R − r)n r ÊBr where > 0. In order to prove the Harnack inequality it suﬃces to show that 0 11∕ 0 1−1∕ − u dx ≤ c u dx . (14) ÊBr ÊBr for some value of . In order to do so we will use the John-Nirenberg theorem. Theorem 3.7. Let w ∈ L1 (Ω). Suppose that there is a constant K such that loc

w(x) − w dx K ð Br ð ≤ (15) ÕBr holds whenever B2r ⊂ Ω. Then there exists a constant = (n) > 0 such that

w(x)−w ∕K e ð B−rð dx ≤ 2 ÕBr whenever B2r ⊂ Ω. For a proof of the theorem, see e.g. [7]. We start proving (14) by assuming that u is a positive weak solution 1,p 1,p B2r ⊂⊂ Ω u > 0 u ∈ W (Ω) w = log(u) ∈ W (Ω) and that . Since and loc it follows that loc . We next use the Poincaré inequality for a ball to obtain

w − (w) p dx crp ∇w p dx, ð Br ð ≤ ð ð (16) ÊBr ÊBr where c = c(p, n) and if we combine this with Lemma 2.4 and a suitable test function, e.g., a radial test C∞ , B , r−1 ⧵ B function ∈ 0 (Ω) s.t. 0 ≤ ≤ 1 = 1 in r ð∇ ð ≤ and = 0 in Ω 2r, we see that

p p (n−p) ð∇wð dx ≤ c(p) ð∇ð dx ≤ cr . (17) ÊBr ÊB2r

24 Combining (16) and (17) yields w − (w) p dx K ð Br ð ≤ ÕBr and by applying the Hölder inequality and using that w is locally summable, we see that condition (15) is satisﬁed. Thus, we may apply the John-Nirenberg theorem from which it follows that

±(w−(w) )∕K e Br dx ≤ 2 ÕBr and thus,

(w−(w) )∕K −u(w−(w) )∕K 4 ≥ e Br dx e Br dx ÕBr ÕBr = ew∕K dx e−w∕K dx ÕBr ÕBr = u∕K dx u−∕K dx. ÕBr ÕBr

We next set = ∕K to obtain 0 11∕ 0 1−1∕ − u dx ≤ c u dx (18) ÕBr ÕBr which proves Theorem 3.2 for 1 0. In order to show that the theorem holds for non-negative u we note that if (18) holds for u + " it will continue to hold when we let " tend to 0.

3.0.2 The case p = n

For this case we will not prove Harnack’s inequality but instead that u satisﬁes the conditions for Morrey’s lemma which in turn will imply the Hölder continuity. 1,p Lemma 3.8. Assume that u ∈ W (Ω), 1 ≤ p < ∞. Suppose that

p n−p+p ð∇uð dx ≤ Kr ÊBr whenever B2r ⊂ Ω. Here 0 < 1 and K are independent of the ball Br. Then u ∈ C (Ω). In fact, ≤ loc

1 0 1 p 4 K osc(u) ≤ r ,B2r ⊂ Ω. Br !n

For a proof see [11]. Next we show that a weak solution u satisﬁes these conditions when p = n, using a n method known as the hole ﬁlling technique. We select B2r ⊂⊂ Ω and ' = (u − a) for an arbitrary constant a ∈ ℝ. Thus, we have ∇' = n n−1(u − a)∇ + n∇u.

25 u ' W 1,p Since is a weak solution and ∈ 0 (Ω) with compact support we obtain

n n n−1 n−2 ð∇uð dx = −n (u − a)⟨ð∇uð ∇u, ∇⟩ dx ÊΩ ÊΩ n−1 ≤ n ð∇uð ð(u − a)∇ð dx ÊΩ 0 11−1∕n 0 11∕n n n n n ≤ n ð∇uð dx ðu − að ð∇ð dx ÊΩ ÊΩ so 0 1 n n n n n ð∇uð dx ≤ n ðu − að ð∇ð dx . ÊΩ ÊΩ We intend to use the above calculations in combination with a test function such that 0 ≤ ≤ 1, = 1 in −1 Br, = 0 outside of B2r and ð∇ð < r . Since ∇ = 0 on Br it follows that 0 1 n n n n ð∇uð dx ≤ n ðu − að ð∇ð dx ÊBr ÊB2r 1 nn u a n dx ≤ n ð − ð r ÊH(r) where H(r) = B2r ⧵ Br. We choose the constant a as 1 a = u(x) dx. ðH(r)ð ÊH(r) Using the Poincaré inequality

n n n ðu − að dx ≤ cr ð∇uð dx, ÊH(r) ÊH(r) it follows that n n n ð∇uð dx ≤ cn ð∇uð dx. ÊBr ÊH(r) cnn ∇u n dx We next add ∫Br ð ð to both sides in the above equation in order to ﬁll in the hole in the annulus which results in n n n n (1 + Cn ) ð∇uð dx ≤ cn ð∇uð dx. ÊBr ÊB2r Thus, for the p-energy n D(r) = ð∇uð dxj ÊBr it follows that cnn D(r) ΛD(2r), Λ = < 1. ≤ 1 + cnn −(m−1) (m−1) By iterating the estimate above we see that D(2 ) ≤ Λ D(r), with m ≥ 1. By using the same technique as in the proof of Theorem 3.1 it follows that r D(r) 2 D(R), 0 < r < R, ≤ R with = − log(Λ)∕ log(2) when B2r ⊂ Ω and thus we can invoke Morrey’s lemma.

26 3.0.3 The case n

1,p When p > n the local Hölder continuity is well-known. If u ∈ W (Br) where Br ⊂⊂ Ω it follows that u has u∗ ∈ C0, (B̄ ) = 1 − n a version r for p (see e.g. Theorem 5.6.5 in [1]). Nevertheless we show the Harnack inequality for a positive weak solution u. At ﬁrst we choose r such that B2r ⊂ Ω and apply Theorem 7.17 in [6] which states that for x, y ∈ Br we have the following bound on the oscillation: 1−n∕p u(x) − u(y) cr ∇u p . ð ð ≤ ‖ ‖L (Br)

c c p, n W 1,p where = ( ). Next we want to apply Lemma 2.4 and therefore we choose ∈ 0 (Ω) to be a radial (n−2p)∕p = 1 B ∇ 1∕r = 0 B ∇ log u p function satisfying in r, ð ð ≤ and outside 2r which yields ‖ ‖L (Br) ≤ n−p cr p . We deﬁne v = log u and conclude that ó u(y) ó ólog ó cc ó ó ≤ ó u(x)ó and therefore u(y) −cc log cc ≤ u(x) ≤ so −cc cc e u(x) ≤ u(y) ≤ e u(x) which proves that u satisﬁes the Harnack inequality.

4 Diﬀerentiability

In this section we continue to explore the regularity of the solutions to the p-Laplace equation but we now 1 < p 2 u u ∈ W 2,p(Ω) consider the gradients. For ≤ we prove that has second Sobolev derivatives, i.e., loc , and p 2 ∇u (p−2)∕2∇u ∈ W 1,2(Ω) for ≥ we prove that ð ð loc . However, a much stronger result holds, namely that the gradients are locally Hölder continuous for 1 < p < ∞. This fact is stated, but not proved, in Lemma 4.4 and it will be used later when we consider the boundary behaviour of p-harmonic functions in Section 7. Apart from Lemma 4.4 this section is completely independent from the rest of the thesis. The proofs in this section are based on integrated diﬀerence quotients. We deﬁne óF (x + ℎ) − F (x)ó DF = ó ó ó ó ó ℎ ó where p−2 F (x) = ð∇u(x)ð 2 ∇u(x).

We are now ready to state the ﬁrst theorem. Theorem 4.1. Let p 2. If u is p-harmonic in Ω, then F ∈ W 1,2(Ω). For each subdomain G ⊂⊂ Ω, ≥ loc

cððF ððL2(Ω) ððDF ððL2(G) ≤ (19) dist(G,)Ω) where c = c(p, n).

27 ∞ c(n) Proof. ∈ C (Ω) 0 ≤ ≤ 1 ≡ 1 G ∇ ≤ We let 0 be a cutoﬀ function, satisfying , on and ð ð dist(G,)Ω) (replacing Ω by some Ω1 such that G ⊂⊂ Ω1 ⊂⊂ Ω if necessary). Let ℎ be some constant vector such that ðℎð < dist(supp , )Ω). Furthermore, uℎ(x) = u(x + ℎ) is p-harmonic in {x ∶ x + ℎ ∈ Ω}. We denote by ' the function '(x) = (x)2(u(x + ℎ) − u(x)).

Then, ∇'(x) = (x)2(∇u(x + ℎ) − ∇u(x)) + 2(x)∇(x)(u(x + ℎ) − u(x)), and since u(x + ℎ) and u(x) satisﬁes (5),

p−2 p−2 ð∇u(x + ℎ)ð ∇u(x + ℎ) − ð∇u(x)ð ∇u(x), ∇'(x) dx = 0, ÊΩ so,

2 p−2 p−2 ð∇u(x + ℎ)ð ∇u(x + ℎ) − ð∇u(x)ð ∇u(x), ∇u(x + ℎ) − ∇u(x) dx ÊΩ p−2 p−2 = −2 (x)(u(x + ℎ) − u(x)) ð∇u(x + ℎ)ð ∇u(x + ℎ) − ð∇u(x)ð ∇u(x), ∇ dx ÊΩ ó p−2 p−2 ó ≤ 2 (x)ðu(x + ℎ) − u(x)ð óð∇u(x + ℎ)ð ∇u(x + ℎ) − ð∇u(x)ð ∇u(x)ó ð∇ð dx ÊΩ ó ó

Employing the inequalities (iv) and (v) of Lemma B.1,

p−2 p−2 2 4 ó ó p−2 p−2 ó b 2 b − a 2 aó b b − a a, b − a , 2 óð ð ð ð ó ≤ ð ð ð ð p ó ó p−2 p−2 p−2 p−2 p−2 p−2 ó ó ó b b − a aó (p − 1) a 2 + b 2 ó b 2 b − a 2 aó , óð ð ð ð ó ≤ ð ð ð ð óð ð ð ð ó ó ó ó ó p−2 we get (recall that F (x) = ð∇u(x)ð 2 ∇u(x)) 4 2 F (x + ℎ) − F (x) 2 dx 2 ð ð p ÊΩ 2 p−2 p−2 ≤ ⟨ð∇u(x + ℎ)ð ∇u(x + ℎ) − ð∇u(x)ð ∇u(x), ∇u(x + ℎ) − ∇u(x)⟩ dx ÊΩ ó p−2 p−2 ó ≤ 2 ðu(x + ℎ) − u(x)ð óð∇u(x + ℎ)ð ∇u(x + ℎ) − ð∇u(x)ð ∇u(x)ó ð∇ð dx ÊΩ ó ó p−2 p−2 ≤ 2(p − 1) ðu(x + ℎ) − u(x)ð ð∇u(x + ℎ)ð 2 + ðu(x)ð 2 ðF (x + ℎ) − F (x)ðð∇ð dx ÊΩ 1 1 < = p < = 2 p p 2 2 ≤ 2(p − 1) ðu(x + ℎ) − u(x)ð ð∇ð dx ðF (x + ℎ) − F (x)ð dx ÊΩ ÊΩ p−2 T 2p U 2p p−2 p−2 p−2 × ð∇u(x + ℎ)ð 2 + ð∇u(x)ð 2 dx , (20) Êsupp

28 where we used (iv) in the ﬁrst inequality, (v) in the third and the generalized Hölder inequality with constants p 2 2p∕(p − 2) ∇u ∈ Lp (Ω) , and in the fourth. Note that the last integral is ﬁnite, since loc . We bound it using Minkowski’s inequality:

p−2 T 2p U 2p p−2 p−2 p−2 ð∇u(x + ℎ)ð 2 + ð∇u(x)ð 2 dx Êsupp p−2 p−2 < = p < = p p 2 p 2 ≤ ð∇u(x + ℎ)ð dx + ð∇u(x)ð dx Ê Ê p−2 p−2 < = 2p < = 2p 5 p 5 2 ≤ ð∇u(x)ð dx = ðF ð dx (21) 2 ÊΩ 2 ÊΩ

1 1 2 2 − ℎ ∫ F (x + ℎ) − F (x) dx 2 for ð ð small enough. Multiplying both sides of (20) by ðℎð Ω ð ð , and using (21), we obtain 1 < 2 = 4 óF (x + ℎ) − F (x)ó 2 2 ó ó dx 2 ó ó p ÊΩ ó ℎ ó p−2 1 < = < p = 2p óu(x + ℎ) − u(x)ó p 5(p − 1) F 2 dx ó ó ∇ p dx . ≤ ð ð ó ó ð ð ÊΩ ÊΩ ó ℎ ó By Theorem 5.8.3 in [1],

1 1 < p = < = óu(x + ℎ) − u(x)ó p c(n) p ó ó ∇ p dx ∇u(x) p dx , ó ó ð ð ≤ ð ð ÊΩ ó ℎ ó dist(G,)Ω) ÊΩ 2 p and thus, since ðF (x)ð = ð∇u(x)ð , (21) implies that

1 p−2 1 < 2 = < = + óF (x + ℎ) − F (x)ó 2 c(n, p) 2p p ó ó dx F 2 dx ó ó ≤ ð ð ÊG ó ℎ ó dist(G,)Ω) ÊΩ 1 < = 2 c(n, p) 2 = ðF ð dx . dist(G,)Ω) ÊΩ

1,2 By using Theorem 5.8.3 in [1] again, it follows that that F ∈ W (G).

The following lemma will be used in the proof of the next theorem. Lemma 4.2. Let f ∈ L1 (Ω). Then loc

H 1 I f(x + ℎek) − f(x) )' '(x) dx = − f(x + tℎek)dt dx ÊΩ ℎ ÊΩ )xk Ê0 holds for all ' C∞ . ∈ 0 (Ω)

29 Proof. For a smooth function f we use the Leibniz rule to obtain ) 1 1 ) f(x + tℎek)dt = f(x + tℎek)dt )xk Ê0 Ê0 )xk 1 1 ) = f(x + tℎek) dt Ê0 ℎ )t f(x + ℎe ) − f(x) = k ℎ and by partial integration and since ' has compact support it follows that H 1 I f(x + ℎek) − f(x) ) ' dx = ' f(x + tℎek)dt dx ÊΩ ℎ ÊΩ )xk Ê0 H I )' 1 = − f(x + tℎek)dt dx ÊΩ )xk Ê0

L1 (Ω) Since smooth functions are dense in loc the result follows by approximation. 2,p Theorem 4.3. Let 1 < p ≤ 2. If u is p-harmonic in Ω, then u ∈ W (Ω). Moreover, there exists c = C(D) such that loc p ó 2 ó ó ) u ó p ó ó dx ≤ c ð∇uð dx ÊD ó)x )x ó ÊΩ ó i j ó whenever D ⊂⊂ Ω.

Proof. We use the notation f(x + ℎe ) − f(x) Δℎf(x) = k . ℎ From Lemma 4.2 we have the formula 1 ℎ p−2 ) p−2 Δ (ð∇uð ∇u) = ð∇u(x + tℎek)ð ∇u(x + tℎek)dt )xk Ê0 (in Sobolev sense). Using this (in the second equality) and the calculations from Theorem 7.19 (in the ﬁrst inequality)

2 ℎ p−2 ℎ ℎ ℎ p−2 ⟨Δ (ð∇uð ∇u), Δ (∇u)⟩ dx = −2 Δ u⟨Δ (ð∇uð ∇u), ∇⟩ dx Ê Ê X L 1 M Y ) p−2 ℎ = −2 ð∇u(x + tℎek)ð ∇u(x + tℎek)dt , (Δ u)∇ dx Ê )xk Ê0 X 1 Y p−2 ) ℎ = 2 ð∇u(x + tℎek)ð ∇u(x + tℎek)dt, (Δ u)∇ dx, (22) Ê Ê0 )xk where we used the equality ) ( ) ) ( ) ) f(x), g(x) = f(x), g(x) + f(x), g(x) , )x⟨ ⟩ )x )x and that has compact support.

30 Claim 1. ) (Δℎu)∇ = ∇Δℎu + Δℎu( ∇ + ∇ ) xk xk xk )xk

Proof of claim. 0 1 ) u(x + ℎe ) − u(x) ) u(x + ℎe ) − u(x) ) k , … , k )xk ℎ )x1 ℎ )xn 0 0 1 0 11 ) ) ) ) ) ) = Δℎu + Δℎu , … , Δℎu + Δℎu xk xk )x1 )xk )x1 )xn )xk )xn 0 ) ) )2 ) ) )2 1 = ∇Δℎu + Δℎu + , … , + xk )xk )x1 )xk)x1 )xk )xn )xk)xn 0 ) ) 1 = ∇Δℎu + Δℎu ∇ + ∇ . xk )xk )xk

Thus, by the claim, (22) equals

X 1 Y 2 ∇u(x + tℎe p−2∇u(x + tℎe )dt, ∇Δℎu + (Δℎu)( ∇ + ∇ ) dx. ð kð k xk xk xk (23) Ê Ê0

Next, we choose a ball B3R ⊂⊂ Ω of radius 3R and a smooth such that = 1 in BR, 0 ≤ ≤ 1 in B2R and 0 outside of B2R, and such that −1 2 −2 ð∇ð ≤ R , ðD ð ≤ CR for some positive constant c. We write

p−1 Y (x) = ð∇u(x + tℎek)ð dt, ÊB2R and hence, (23) is smaller than

2 Y ∇ Δℎu dx + 2 Y Δℎu ( ∇ + ∇ ) dx ð ð xk ð ð ð xk ð ð xk ð ÊΩ ÊΩ 2 c Y Δℎu dx + Δℎu Y dx. ≤ R ð xk ð 2 ð ð (24) ÊΩ R ÊB2R

So to summarize we have that 2 C 2 Δℎ( ∇u p−2∇u), Δℎ(∇u) dx Y Δℎu dx + Δℎu Y dx. ⟨ ð ð ⟩ ≤ R ð xk ð 2 ð ð Ê ÊΩ R ÊB2R

W (x)2 = 1 + ∇u(x) 2 + ∇u(x + ℎe ) 2 Δℎ(∇u) Δℎ(u ) Letting ð ð ð k ð and using that ð ð ≥ ð xk ð, together with the inequality (vi) of Lemma B.1,

p−2 p−2 p−2 2 2 2 ⟨ðbð b − ðað a, b − a⟩ ≥ (p − 1)ðb − að (1 + ðað + ðbð ) 2 ,

31 (since 1 < p ≤ 2) we have that

2 p−2 ℎ 2 (p − 1) W ðΔ (∇u)ð dx ÊΩ X p−2 p−2 Y 2 ð∇u(x + ℎek)ð ∇u(x + ℎek) − ð∇u(x)ð ∇u(x) (∇u(x + ℎek) − ∇u(x)) ≤ , dx ÊΩ ℎ ℎ

2 ℎ p−2 ℎ = ⟨Δ (ð∇uð ∇u), Δ (∇u)⟩ dx ÊΩ 2 c Y Δℎ(∇u) dx + Δℎu Y dx. ≤ R ð ð 2 ð ð ÊΩ R ÊB2R

Using Cauchy’s inequality with ", i.e., that for a, b > 0, " > 0 b2 ab "a2 + , ≤ 4"

p−2 2−p ℎ −1 with a = W 2 ðΔ (∇u)ð and b = 2W 2 YR , we get that

p−2 2−p −1 ℎ ℎ −1 2R Y ðΔ (∇u)ð = (W 2 ðΔ (∇u)ð)(2W 2 YR ) 2 p−2 ℎ 2 −1 −2 2−p 2 ≤ " W ðΔ (∇u)ð + " R W Y

Choosing " = (p − 1)∕2, we get (note that the integral inequality is still valid with Ω replaced by BR)

2 p−2 ℎ 2 (p − 1) W ðΔ (∇u)ð dx ÊBR p − 1 2 2W p−2 Δℎ(∇u) 2 dx + R−2W 2−pY 2 dx ≤ 2 ð ð p − 1 ÊBR ÊB2R c + Δℎu Y dx, 2 ð ð R ÊB2R that is, p − 1 W p−2 Δℎ(∇u) 2 dx 2 ð ð ÊBR 2 c R−2W 2−pY 2 dx + Δℎu Y dx, ≤ p − 1 2 ð ð (25) ÊB2R R ÊB2R since = 1 on BR. Next, note that ℎ p p−2 ℎ 2 p ðΔ (∇u)ð ≤ W ðΔ (∇u)ð + W , (26) p 2−p 2 p W Y ≤ W + Y p−1 , (27) p ℎ ℎ p ðΔ uðY ≤ ðΔ uð + Y p−1 . (28)

32 Using ﬁrst (26), then (25) and lastly (27) and (28), we get

ℎ p p−2 ℎ 2 p ðΔ (∇u)ð dx ≤ W ðΔ (∇u)ð dx + W dx ÊBR ÊBR ÊB2R c c W 2−pY 2 dx + Δℎu Y dx + W p dx ≤ 2 ð ð ÊB2R R ÊB2R ÊB2R p p ℎ p ≤ c W dx + c Y p−1 dx + c ðΔ uð dx, ÊB2R ÊB2R ÊB2R where c = c(p, R), and hence we need to bound each of these integrals as ℎ → 0. The ﬁrst one is immediate:

p 2 2 p∕2 W dx = (1 + ð∇u(x)ð + ð∇u(x + ℎek)ð ) dx ÊB2R ÊB2R p ≤ c dx + c ð∇u(x)ð dx. ÊB3R ÊB3R For the second integral, p H I 1 p ⎡ 1 p−1 ⎤ p−1 p−1 Y dx = ⎢ ð∇u(x + tℎek)ð dt ⎥ dx ÊB ÊB ⎢ Ê0 ⎥ 2R 2R ⎣ ⎦ p H I p−1 H Ip p−1 ⎡ 1 p 1 ⎤ p p ≤ ⎢ ð∇u(x + tℎek)ð dt 1 dt ⎥ dx ÊB ⎢ Ê0 Ê0 ⎥ 2R ⎣ ⎦ 1 p p = ð∇u(x + tℎek)ð dt dx ≤ ð∇uð dx ÊB2R Ê0 ÊB3R for small enough ℎ. The integral bound

ℎ p p ðΔ uð dx ≤ c ð∇uð dx ÊB2R ÊB3R is obtained by applying results from Section 5.8 of [1]. Hence,

ℎ p p ðΔ (∇u)ð dx ≤ c ð∇uð dx, ÊBR ÊB3R c = c(p, n, R) ∇u ∈ Lp (Ω) where and the result follows since loc and the fact that any compact set can be covered by a ﬁnite number of balls.

We ﬁnish this section by stating the following lemma, the proof of which can be found in [12]. Lemma 4.4. Let 1 < p < ∞, let u be p-harmonic in B(z, 4r). Then u has a representative in W 1,p(B(z, 4r)) with Hölder continuous partial derivatives in B(z, 4r). In particular, there exist constants c = c(p, n) ∈ [1, ∞) and = (p, n) ∈ (0, 1] such that if x, y ∈ B( ̂z,̂r∕2), B( ̂z, 4̂r) ⊂ B(z, 4r), then 0 1 −1 ðx − yð c ∇u(x) − ∇u(y) ≤ max ∇u . ð ð ̂r B( ̂z,̂r) ð ð Moreover, if u is non-negative, then −1 max ∇u ≤ c ̂r max u. B( ̂z,̂r) ð ð B( ̂z,2̂r)

33 5 The p-superharmonic functions and their properties

In this section we will study the properties of p-superharmonic functions. They will be of great importance when constructing Perron’s solution for the p-Dirichlet problem in Section 6. At first we establish some of their basic properties and later we will explain the connection between p-superharmonic functions and weak supersolutions to the p-Laplace equation. We start by recalling the definition of a lower semi-continuous function. Definition 5.1. We say that v ∶ Ω → (−∞, ∞] is lower semi-continuous (l.s.c) at x ∈ Ω if lim inf v(y) v(x). y→x ≥

We say that v is lower semi-continuous in Ω if v is lower semi-continuous at every point in Ω. We note that it follows directly from the definition that a lower semi-continuous function v attains its infimum on compact sets. Furthermore, following the procedure outlined in [9], one can show that v can be approximated from below with smooth functions, which will be used in the proof of Theorem 5.6. We now turn to the definition of a p-superharmonic function. Definition 5.2. We say that a function v ∶ Ω → (−∞, ∞] is p-superharmonic in Ω if the following conditions are satisfied: (i) v is lower semi-continuous in Ω, (ii) v ≢ ∞ in Ω ̄ (iii) For each compactly contained domain, D ⊂⊂ Ω, the comparison principle holds: if ℎ ∈ C(D) is p-harmonic in D and ℎð)D ≤ vð)D, then ℎ ≤ v in D. We say that a function u ∶ Ω → [−∞, ∞) is p-subharmonic if v = −u is p-superharmonic. The following three results can be found Chapter 7 in [9]. Lemma 5.1. Suppose that vi is p-superharmonic in Ω for i ∈ {1, 2}. Then the minimum, min{v1, v2} is p-superharmonic.

This lemma is not hard to show and can be be naturally generalized, using induction, to hold for

̃v = min{v1, v2, … , vn}, where vi is p-superharmonic for each i. Lemma 5.2. Suppose that {vi} is an increasing sequence of p-superharmonic functions in Ω. Then the limit function v = limi→∞ vi is either p-superharmonic or v ≡ ∞. The next theorem is the comparison principle for p-sub- and p-superharmonic functions. Theorem 5.3. Let Ω be a bounded domain. Furthermore, let v1 and v2 be p-superharmonic and p-subharmonic, respectively. If lim sup v2(y) ≤ lim inf v1(y) (29) y→x y→x for each x ∈ Ω and both sides of (29) are not ∞ or −∞ at the same time, then v2 ≤ v1 in Ω.

Proof. Take " > 0 and let x ∈ Ω. We recall the discussion about regular sets in the end of Section 2, and note that there exists a regular open set D ⊂⊂ Ω such that x ∈ D and v < u + " on )D. Since v is p-subharmonic ̄ there exists a decreasing sequence {'i} of smooth functions in Ω which converges to v on D. Since u + " attains its inﬁmum on compact domains there exists an N such that 'j ≤ u + " on )D for j ≥ N. Next we

34 let ℎ denote the the unique p-harmonic function in D with ℎð)D = 'N . Thus, v ≤ ℎ = 'N ≤ u + " on )D and it now follows from Theorem 2.5 that

v ≤ ℎ ≤ u + " in D.

The conclusion of the lemma follows by letting " tend to zero.

The next proposition states that the set where a p-superharmonic function is ﬁnite, {x ∈ Ω ∶ v(x) < ∞}, is dense in Ω. Proposition 5.4. If v is p-superharmonic in Ω, then the set where v = ∞ does not contain any ball.

Proof. At ﬁrst we show the result for v ≥ 0. Let x0 ∈ Ω and assume that there exists r < R such that v ≡ ∞ in B(x0, r) ⊂ B(x0,R) ⊂⊂ Ω. Next we consider the function

R − n−1 p−1 ∫ x−x t dt ℎ(x) = ð 0ð R − n−1 p−1 ∫r t dt which is p-harmonic for x ≠ x0. Thus ℎ(x) is p-harmonic in the annulus A = {x ∈ Ω ∶ r < ðx − x0ð < R} ℎ = 0 ℎ = 1 p ℎ Ā v 0 and ð)BR and ð)Br . The -harmonicity of and the fact that it is continuous on , ð)BR ≥ and v = ∞ ð)Br allows us to use the comparison principle to conclude that

v(x) ≥ kℎ(x), k = 1, 2, 3, … in A which implies that vðA = ∞. Thus, v ≡ ∞ in BR and we can use a covering argument with a chain of intersecting balls to conclude that v ≡ ∞ in Ω which contradicts the fact that v is p-superharmonic. In order to prove the lemma for an arbitrary p-superharmonic function v we can apply the argument above to ̃v = v − inf v BR .

The next theorem connects the p-superharmonic functions to weak supersolutions of the p-Laplace equation, in the case where the function is continuous. Theorem 5.5. Suppose that v ∈ C(Ω) ∩ W 1,p(Ω). Then the following conditions are equivalent: loc (i) v p dx v ' p dx whenever ' C∞ D , ∫D ð∇ ð ≤ ∫D ð∇( + )ð ∈ 0 ( ) (ii) v p−2 v, ' dx whenever ' C∞ is non-negative, ∫ ⟨ð∇ ð ∇ ∇ ⟩ ≥ 0 ∈ 0 (Ω) (iii) v is p-superharmonic in Ω.

Proof. We assume that (i) holds and consider D ⊂⊂ Ω such that supp ' ⊂ D. Take " ≥ 0. From the assumption it follows that J(0) ≤ J(") where

p J(") = ð∇(v(x) + "'(x)ð dx. ÊD

¨ p−2 From the proof of Theorem 2.2 we know that J (0) = ∫D ð∇vð ⟨∇u, ∇'⟩ dx. By dividing the equation ¨ J(") − J(0) ≥ 0 with " and let " tend zero it follows that J (0) ≥ 0, which is (iii). We next assume that (ii) p p p−2 holds. Then (i) follows directly since ð∇(v+')ð ≥ ð∇vð +p⟨ð∇vð ∇v, ∇'⟩, by Lemma B.1 (viii). We next (ii) (iii) v ∈ C(Ω) ∩ W 1,p(Ω) show that implies . Since loc we need to show that the comparison principle holds.

35 We thus assume that there exists a function ℎ which satisﬁes the conditions in Deﬁnition 5.2 (iii). We use the function ' = max {ℎ − v, 0}. Using (ii) and Hölder’s inequality we obtain for G = {x ∈ Ω ∶ v(x) ≤ ℎ(x)} 0 11−1∕p 0 11∕p p p−2 p p ð∇vð dx ≤ ⟨ð∇vð ∇v, ∇ℎ⟩ dx ≤ ð∇vð dx ð∇ℎð dx ÊG ÊG ÊG ÊG

p p so ∫G ð∇vð dx ≤ ∫G ð∇ℎð dx. Thus, v is the minimizer in G with boundary vales v = ℎ. According to the proof of Theorem 2.6 the minimizer is unique so v = ℎ in G. Thus v satisﬁes the comparison principle and is therefore p-superharmonic. That (iii) implies (ii) follows from Corollary 5.9.

v ∈ C(Ω) ∩ W 1,p(Ω) In the theorem above we assumed that loc , but we can do better. In Corollary 5.9 we will show that a locally bounded p-superharmonic function is always a weak supersolution. In order to prove this we need some more theory. A very nice feature of the p-superharmonic functions is that they can be approximated from below by continuous functions on compactly contained domains as follows: Theorem 5.6. Suppose that v is a p-superharmonic functions in the domain Ω. Given a subdomain D ⊂⊂ Ω ̄ 1,p there exists and increasing sequence of p-superharmonic functions {vj} such that vj ∈ C(D) ∩ W (D) for each vj and v v … v = lim vj 1 ≤ 2 ≤ and j→∞ at each point in D.

In order to prove this theorem we need to return to the obstacle problem that was deﬁned in the introduction n but this time we consider a slightly less general situation. Assume that Ω ⊂ ℝ is bounded and consider the obstacle problem for the class

1,p 1,p (Ω) = {v ∈ C(Ω) ∩ W (Ω) v Ω v − ∈ W (Ω)},  ð ≥ in and loc where is now acting as both the boundary function and the obstacle. Just as in the case for p-harmonic functions there exists a unique minimizer of the p-Dirichlet integral among the functions in  (Ω). 1,p Theorem 5.7. Given ∈ C(Ω) ∩ W (Ω), there exists a unique minimizer v in the class  (Ω) , i.e.,

p p ð∇v ð dx ≤ ð∇vð dx (30) ÊΩ ÊΩ for all similar v. The function v is p-superharmonic in Ω and p-harmonic in the open set S = {x ∈ Ω ∶ ̄ ̄ v (x) > (x)}. If, in addition, Ω is regular enough and ∈ C(Ω), then also v ∈ C(Ω) and v = on )Ω.

Proof. We will not prove the continuity and the boundary value property but instead refer the reader to [21] and the references therein. In order to prove the existence and uniqueness of v we proceed as in Theorem 2.6 with  =  (Ω). However, we must now prove that u ≥ almost everywhere. Since vj ⇀ v in p L (Ω) it is possible to apply Mazurs lemma which states that there exists a sequence of convex combinations v { ̃v } v Lp(Ω) { ̃v } of j , denoted i , which converges strongly to in . By extracting a subsequence of j that converges to v a.e., we conclude that v ≥ a.e. Since  (Ω) is closed under addition with non-negative smooth functions with compact support we can use Theorem 5.5 to assert that v is p-superharmonic. We next assume that v ∈ C(Ω). It remains to show that v is p-harmonic in S. If ̃v = v (x) + "'(x) ≥ (x) ' C∞ ̃v for ∈ 0 (Ω) we have that ∈  (Ω) so (30) and the proof of Theorem 5.5 implies that

p−2 ⟨ð∇v ð ∇v , ∇'⟩ ≥ 0 (31) ÊΩ

36 ' ∈ C∞(Ω) ' ∈ C∞(S) " " ' min (v − ) whenever 0 . We next consider 0 and small enough for ‖ ‖∞ ≤ supp ' to hold which is possible due to the deﬁnition of S. Thus v (x) + "'(x) ≥ (x) which implies that (31) holds with Ω replaced by S.

Proof of Theorem 5.6. We begin by choosing a regular domain D1 such that D ⊂⊂ D1 ⊂⊂ Ω. Since v is lower semi-continuous there exists an increasing sequence of smooth functions { j} such that for all x ∈ Ω it holds that (x) (x) … v(x) lim j(x) = v(x). 1 ≤ 2 ≤ ≤ and j→∞ ∈ C(Ω) ∩ W 1,p(Ω) j (D ) Since j , we now solve the obstacle for each in  j 1 , i.e., we ﬁnd the minimizer v ∶= v D j j in 1 with j as obstacle. We next show that

v1(x) ≤ v2(x) ≤ … , vj(x) ≤ v(x). for each x ∈ Ω. Since j ≤ v we see that vj ≤ v in the set {vj = j} and we therefore consider the open set Aj = {vj > j} and assume that Aj ≠ ç. vj is p-harmonic in Aj according to Theorem 5.7 and since vjð)A = jð)A ≤ vð)A we can use the comparison principle to conclude that vj ≤ v in Aj and thus vj ≤ v in D1. The same argument can be applied to show that vj ≤ vj+1 for j = 1, 2, …. In {vj = j} it holds v = v A v v v directly since j j ≤ j+1 ≤ j+1. Thus we consider j and note that jð)Aj ≤ j+1ð)Aj . Since j+1 is p-harmonic according to Theorem 5.7 we may again use the comparison principle to conclude that vj ≤ vj+1 in D1. Theorem 5.8. If v p-superharmonic and locally bounded from above in Ω, then v ∈ W 1,p(Ω), and the loc approximants vj from Theorem 5.6 can be chosen so that

p lim ð∇(v − vj)ð dx = 0. j→∞ ÊD

Proof. We use the construction in the proof of Theorem 5.6. We note that since v is locally bounded it holds that C = sup v − inf < ∞. D 1 D1 1 We may therefore use the modiﬁed version of Caccioppoli’s inequality, that was discussed after Lemma 2.3, to bound our sequence {vj} such that we obtain

p p p p ð∇vjð dx ≤ p C ð∇ð dx = M, j = 1, 2, … ÊD ÊD1

1,p Thus, since 1 ≤ vj ≤ v, {vj} is a bounded sequence in W (D) and vj → v pointwise a.e. It follows (see p e.g., Theorem 13.44 in [10]) that vj ⇀ v in L (D) and we can use pre-compactness and Mazur’s lemma to 1,p conclude that v ∈ W (D), that ‖∇v‖Lp(D) ≤ M and that there exists a subsequence vj such that ∇vj ⇀ ∇v p in L (D). From now on we let vj denote the subsequence. p Next we turn to the proof of proving the strong L -convergence of the gradients. We note that is suﬃcient to show that

p lim ∇v − ∇vj dx = 0 j→∞ ð ð ÊBr

37 B ⊂ D B ⊂ D C∞ B , whenever r such that 2r 1. Moreover, we consider the test function ∈ 0 ( 2r) 0 ≤ ≤ 1, and = 1 in Br. We next deﬁne 'j = (v − vj),'j ≥ 0 and note that ∇'j = ∇(v − vj) + (∇v − ∇vj). Since vj is a supersolution it holds that

p−2 ⟨ð∇vjð ∇vj, ∇'j⟩ dx ≥ 0 ÊB2r so

p−2 p−2 Jj = ⟨ð∇vð ∇v − ð∇vjð ∇vj, ∇((v − vj))⟩ dx ÊB2r p−2 ≤ ⟨ð∇vð ∇v, ∇((v − vj))⟩ dx. ÊB2r p ∇(vj) ⇀ ∇(v) L lim supj→∞ Jj 0 Using Hölder’s inequality and the fact that in B2r it follows that ≤ for a subsequence. Furthermore,

p−2 p−2 p−2 p−2 Jj = ⟨ð∇vð ∇v − ð∇vjð ∇vj, ∇v − ∇vj⟩ dx + (v − vj)⟨ð∇vð ∇v − ð∇vjð ∇vj, ∇⟩ dx ÊB2r ÊB2r

We use Hölder’s and Minowski’s inequalities to obtain an upper bound of the second integral as follows: 0 11∕p H0 11−1∕p 0 11−1∕pI p p p (v − vj) dx ð∇vð dx + ð∇vjð dx maxð∇ð ÊB2r ÊB2r ÊB2r 0 11∕p 1−1∕p p ≤ 2M maxð∇ð (v − vj) dx ÊB2r which tends to zero as j → ∞, by dominated convergence. Thus,

p−2 p−2 0 lim ∇v ∇v − ∇vj ∇vj, ∇v − ∇vj dx 0 ≤ j→∞ ⟨ð ð ð ð ⟩ ≤ ÊB2r

In order to complete the proof for p ≥ 2 one may use inequality (i) in Lemma B.1. We skip the proof for 1 < p < 2.

We can use this theorem to prove that locally bounded p-superharmonic functions are weak supersolutions. Corollary 5.9. Suppose that v is p-superharmonic and locally bounded in Ω. Then v ∈ W 1,p(Ω) and v is a weak supersolution, i.e., loc p−2 ⟨ð∇vð ∇v, ∇'⟩ dx ≥ 0 ÊΩ for all ' C∞ such that ' . ∈ 0 (Ω) ≥ 0

Proof. By Theorem 5.6 there exists an increasing sequence {vj} of p-superharmonic functions such that v C D̄ W 1,p D v v ' C∞ j ∈ ( ) ∩ ( ) and limj→∞ j = . Take ∈ 0 (Ω). We need to show that

p−2 p−2 ⟨ð∇vð ∇v, ∇'⟩ dx = lim ⟨ð∇vjð ∇vj, ∇'⟩ dx ≥ 0 ÊΩ j→∞ ÊΩ

38 One can easily show that (see Chapter 10 in [21]) p−2 p−2 p−2 p−2 ðð∇vð ∇v − ð∇vjð ∇vjð ≤ (p − 1)ð∇v − ∇vjð(ð∇vð + ð∇vjð ) when p ≥ 2. Thus,

ó p−2 p−2 ó ó lim ∇v ∇v − ∇vj ∇vj, ∇' dxó ój→∞ ⟨ð ð ð ð ⟩ ó ó ÊΩ ó p−2 p−2 ≤ maxð∇'ð lim ðð∇vð ∇v − ð∇vjð ∇vjð dx j→∞ ÊΩ p−2 p−2 ≤ maxð∇'ð(p − 1) lim ð∇v − ∇vjð(ð∇vð + ð∇vjð ) dx j→∞ ÊΩ 0 11∕p 0 11−1∕p p p−2 p−2p∕(p−1) ≤ maxð∇'ð(p − 1) lim ð∇v − ∇vjð dx ð∇vð + ð∇vjð dx j→∞ ÊΩ ÊΩ which tends to zero by the strong convergence of the gradients. We note that the integral in the last parenthesis is ﬁnite since if = (p − 2)∕p − 1 we have that 1∕2 ≤ ≤ 1 so p ≥ p∕2 and 0 11∕2 p 1∕2 2 p ð∇vð dx ≤ ðΩð ð∇vð dx ÊΩ ÊΩ which is ﬁnite since 2 p ≥ p. This also holds for vj, j = 1, 2, … . In the case where 1 < p ≤ 2 we use inequality (iii) in Lemma B.1 and one application of Hölder’s inequality.

p v v ∈ W 1,p(Ω) With a little more work it is actually possible to show that a -superharmonic function such that loc is a supersolution (Corollary 7.21, [9]). The following theorem is known as Harnack’s convergence theorem and states that the limit of an increasing sequence of p-harmonic functions is either inﬁnite or a p-harmonic function. It will be of great importance when we discuss solutions to the p-Dirichlet problem in Section 6. Theorem 5.10. Suppose that uj is p-harmonic and that

0 u u … , u = lim uj ≤ 1 ≤ 2 ≤ j→∞ pointwise in Ω. Then either u ≡ ∞ or u is a p-harmonic function in Ω.

Proof. According to Harnack’s inequality for p-harmonic functions it holds that

uj(x) ≤ cuj(x0), j = 1, 2, 3, … where c = c(p, n), whenever x ∈ B(x0, r) such that B(x0, 2r) ⊂⊂ Ω. From this it follows that if u(x0) < ∞ at some x0 ∈ Ω, then u(x) is locally bounded in Ω. By Cacciopoli’s lemma for p-harmonic functions we obtain

p −p p p p n−p p ð∇ujð dx ≤ cr ðujð dx ≤ ðuð dx ≤ c r u(x0) ÊBr ÊB2r ÊB2r

1,p u ∈ W (Ω) uj p so loc . We can now repeat the argument in Corollary 5.9 but use that each is -harmonic to conclude that p−2 p−2 ⟨ð∇uð ∇u, ∇'⟩ dx = lim ⟨ð∇ujð ∇uj, ∇'⟩ dx = 0 ÊΩ j→∞ ÊΩ ' C∞ whenever ∈ 0 (Ω).

39 Note that it follows from the proof that the theorem is valid even for non-positive functions uj. This can be seen by adding a constant to the sequence since u is locally bounded. We next describe a method known as the Poisson modification which provides a way to locally smooth a p-superharmonic function. Let v be a p-superharmonic function in Ω, D ⊂⊂ Ω is a regular domain and ℎ the p-harmonic function in D such that ℎð)D = v. In the case where v is continuous we define the Poisson modification as the function T v, Ω ⧵ D V = P (v, D) = in ℎ, in D V is clearly p-superharmonic and it follows directly from the comparison principle that V ≤ v. In the case where v is not continuous we can define the Poisson modification as the limit of an increasing sequence of continuous and p-harmonic functions on D as follows: We use the construction explained in Theorem 5.6 to approximate v from below with continuous functions vj and define the Poisson approximation through

V = lim Vj = lim P (vj,D), j→∞ j→∞

̄ 1,p which is possible since vj ∈ C(D) ∩ W (D). We have the following proposition: Proposition 5.11. Suppose that v is p-superharmonic in Ω and that D ⊂⊂ Ω. Then the Poisson modiﬁcation V = P (v, D) is p-superharmonic in Ω, p-harmonic in D and V ≤ v. Moreover, if v is locally bounded, then

p p ð∇V ð dx ≤ ð∇vð dx ÊG ÊG for D ⊂ G ⊂⊂ Ω.

Proof. We assume that v is not continuous and note that from the deﬁnition of the Poisson modiﬁcation we get a corresponding sequence of increasing p-harmonic functions ℎj, j = 1, 2, …. Since ℎj ≤ vj ≤ v ≢ ∞ the limit function is p-harmonic in D according to Theorem 5.10. V is p-superharmonic according to Lemma 5.2. Furthermore, since p-harmonic functions minimize the p-Dirichlet integral it follows that

p p ð∇Vjð dx ≤ ð∇vjð dx ÊD ÊD if v is locally bounded, and by using monotone convergence we obtain

p p ð∇V ð dx ≤ ð∇vð dx. ÊD ÊD

There exists another equivalent definition of the Poisson modification, which highlights the fact that it is a local smoothing procedure. For a regular set D ⊂⊂ Ω we define v = inf{ ̃v ∶ ̃v p , lim inf ̃v(y) v(x)} D is -harmonic y→x ≥ and T v, Ω ⧵ D, V = P (v, D) = in vD, in D. For more details, see [9]. We next use the Poisson modification to show that p-superharmonic functions have the so called essential limit inferior property.

40 Theorem 5.12. If v is p-superharmonic in Ω, then

v(x) = ess inf v(y) y→x at each point x in Ω.

In order to prove the theorem we will need two additional lemmas. Lemma 5.13. Suppose that v is p-superharmonic in Ω. If v(x) ≤ at each x ∈ Ω and if v(x) = for a.e. x ∈ Ω, then v(x) = at each x ∈ Ω.

Proof. If v is continuous there is nothing to prove. Since p-harmonic functions are continuous we choose an arbitrary regular domain, e.g., an open ball, D ⊂⊂ Ω and use the Poisson modiﬁcation V = P (v, D). Since V ≤ v ≤ at each x ∈ Ω it is suﬃcient to show that V (x) = at each x ∈ D. v is locally bounded by assumption, and thus we can use Proposition 5.11 to conclude that

p p p ð∇V ð dx ≤ ð∇vð dx = ð∇ð dx = 0 ÊG ÊG ÊG where D ⊂ G ⊂⊂ Ω. Thus ∇V = 0 so V = almost everywhere in G. However, since V is p-harmonic and thus continuous in D we conclude that V (x) = for each x ∈ D. Lemma 5.14. If v is p-superharmonic in Ω and if v(x) > for a.e. x ∈ Ω, then v(x) ≥ for every x ∈ Ω.

Proof. We note that we may assume that < ∞ because otherwise there is nothing to prove. Next we consider the function ̃v = min{v(x), }. Since v(x) > for a.e. x by assumption it holds that ̃v(x) = for a.e. x ∈ Ω. The conclusion of the lemma follows by applying Lemma 5.13 .

Proof of Theorem 5.12. Since v is lower semi-continuous it follows directly that v(x) ≤ lim inf y→x v(y) ≤ ess lim inf y→xv(y). We next prove that = ess lim infy→xv(y) ≤ v(x). Furthermore, v is p-superharmonic and thus we can, given " > 0, ﬁnd a > 0 s.t.

v(y) > − " for a.e. y ∈ B(x, ).

From Lemma 5.14 it follows that v(x) ≥ − ". Since " is arbitrary the proof of the lemma is complete.

We ﬁnish this section by stating a theorem which provides the ﬁnal connection between supersolutions and p-harmonic functions (Theorem 7.16 in [9]). Theorem 5.15. Let v be a weak supersolution of the p-Laplace equation in Ω. If

v(x) = ess inf v(y) y→x holds for each x ∈ Ω, then v is p-harmonic.

6 Perron’s method

In this section we return to the p-Dirichlet boundary value problem. In particular we consider Perron’s method, which is a technique to solve the Dirichlet problem for the Laplace operator. It is, however, possible to generalize this method for other partial diﬀerential operators including the p-Laplace operator. Throughout n this section we assume that Ω ⊂ ℝ is a bounded domain and g ∶ )Ω → [−∞, ∞].

41 Deﬁnition 6.1. We say that a function v ∶ Ω → (−∞, ∞] is a member of the upper class g if the following conditions are satisﬁes 1. v is p-superharmonic in Ω, 2. v is bounded from below,

3. lim inf x→w v(x) ≥ g(w) when w ∈ )Ω.

We deﬁne the lower class the lower class g symmetrically: a function u is in g if 1. u is p-subharmonic in Ω, 2. u is bounded from above,

3. lim supx→w u(x) ≤ g(w) when w ∈ )Ω. Furthermore, for each x ∈ Ω we deﬁne

the upper Perron solution Hg(x) = inf v(x), v∈g

the lower Perron solution Hg(x) = sup u(x). u∈g

When the boundary function g is clear from context, we sometimes omit the subsript. It follows from the comparison principle that H ≤ H. In Theorem 6.3 we will show that if g is continuous the upper and the lower solution will coincide i.e., H = H =∶ H. Furthermore, under these circumstances H will agree with the unique p-harmonic function in Ω with g as boundary values in Sobolev sense. In order to simplify the calculations, we will from now on assume that g is bounded, i.e., m ≤ g(x) ≤ M for each x ∈ Ω. Since m ∈ g and M ∈ g it follows that m ≤ H ≤ H ≤ M. Due to Lemma 5.1 and a corresponding version for the maximum of p-subharmonic functions we note that we can cut of all the functions and therefore assume that all functions are bounded from below my m and from above by M. There is an obvious symmetry between the upper and the lower Perron solution. Just as in [21] we will focus on proving the statements concerning H. We have the following useful lemma. Lemma 6.1. If g is bounded, H and H are continuous in Ω.

Proof. We will show the lemma for H. Take " > 0. We ﬁx x0 ∈ Ω and consider B(x0,R) ⊂⊂ Ω and 1 2 k 0 < r < R. For x1, x2 ∈ B(x0, r) there exist sequences of functions {vi } and {vi } such that vi ∈ g for 1 2 each i = 1, 2 … and k ∈ {1, 2}. Furthermore, vi (x1) → H(x1) and vi (x2) → H(x2) as i → ∞. We deﬁne 1 2 vi = min{vi , vi } and note that

lim vi(x ) = H(x ), lim vi(x ) = H(x ) i→∞ 1 1 i→∞ 2 2

We next deﬁne the Poisson modiﬁcation Vi = P (vi,B(x0,R)) and note that Vi ∈ g since Vi is p-superharmonic by Proposition 5.11 and lim inf Vi(x) = lim inf vi(x) ≥ g(w). x→w∈)Ω x→w∈)Ω

Thus, it follows that H ≤ Vi ≤ vi in Ω. Moreover, for i suﬃciently large and k ∈ {1, 2}, it holds that vi(xk) < H(xk) + ", so H(xk) > vi(xk) − " > Vi(xk) − ", which implies that

H(x2) − H(x1) ≤ Vi(x2) − Vi(x1) + " ≤ osc Vi + ". B(x0,r)

42 Since Vi is p-harmonic in B(x0,R), and therefore Hölder continuous, we obtain r r osc Vi ≤ L osc Vi ≤ L (M − m). B(x0,r) R B(x0,R) R

Hence we can ﬁnd r > 0 such that ðH(x1) − H(x2)ð ≤ 2" for x1, x2 ∈ B(x0, r), which completes the proof of the lemma.

This lemma will be used in order to prove that Perron’s solutions are either p-harmonic of inﬁnite. Theorem 6.2. The upper solution H satisﬁes one of the following properties: 1. H is p-harmonic in Ω,

2. H ≡ ∞ in Ω, 3. H ≡ −∞ in Ω. A similar result holds for H.

Proof. We need to show that H ≢ ∞ is a solution to the p-Laplace equation. At first we construct a sequence of functions {!i} such that !i ∈ g and the sequence converges to H at the rational points. After that we use Posson modification to smooth the sequence and show that its limit coincides with Perron’s solution on a dense subset. Since H is continuous, this implies that H is p-harmonic. q , q , n H vk ∞ Thus, we let 1 2 … ∈ ℚ ∩ Ω. From the definition of there exists { i }i=1 such that 1 H(q ) vk(q ) H(q ) + , i = 1, 2, 3, … k ≤ i k ≤ k i n for each qk ∈ ℚ ∩ Ω. Next we set 1 1 1 2 2 2 i i i !i = min{v1, v2, … , vi , v1, v2, … , vi , … , v1, v2, … , vi}

k Clearly, !i ∈ g,!1 ≥ !2 ≥ … and H(qk) ≤ !i(qk) ≤ vi (qk) when i ≥ k, so

n lim wi(qk) = H(qk) qk ∈ ℚ ∩ Ω. i→∞ for

Set Wi = P (wi,B) for a ball B ⊂⊂ Ω. From the proof of Lemma 6.1 we know that Wi ∈ g, so H ≤ Wi ≤ n wi. Hence Wi(qk) → H(qk) as i → ∞ for qk ∈ ℚ ∩ Ω. Consider the limit

W = Wi. ilim→∞

Since W1 ≥ W2 ≥ W2 ≥, … we can use Theorem 5.10 to conclude that the W is p-harmonic in B. Since n Wi ≥ H for i = 1, 2, … it follows that W ≥ H and in addition W (qk) = H(qk) for each qk ∈ ℚ ∩ B. Both n W and H are continuous in B and they agree on ℚ ∩ B. Thus they agree everywhere on B and therefore it follows that H = W so H is p-harmonic in B. We conclude the proof by noting that B is arbitrary.

The next theorem is known as Wiener’s resolutivity theorem. Theorem 6.3. Suppose that g ∶ )Ω → ℝ is continuous. Then H = H in Ω.

43 ∞ Proof. We claim that it is suﬃcient to prove the theorem when g ∈ C (Ω). By approximating g with smooth ∞ functions such that, given ", there is a function ' ∈ C (Ω) such that '() − " < g() < '() + ".

Thus, assuming that the theorem holds for smooth functions, i.e., H' = H' it follows that

H' − " = H'−" ≤ Hg ≤ Hg ≤ H'+" = H' + "

∞ n 1,p which imples that Hg = Hg. From now on we thus assume that g ∈ C (ℝ ). Let u ∈ C(Ω)∩W (Ω) be the p u g W 1,p v unique -harmonic function with − ∈ 0 (Ω). Furthermore, let be the solution to the obstacle problem in g(Ω). Since v ≥ g and since v is p-superharmonic, v ∈ g. Since Ω is a domain, we consider the regular sets D1 ⊂ D2 ⊂ … such that Ω = ∪Dj (see the discussion after Deﬁnition 2.4) and the sequence {Vi} such that Vi = P (v, Di). Since Vi+1 = P (v, Di+1) = P (Vi,Di+1) it holds that V1 ≥ V2 ≥, … and Vj ∈ g for each j , , V g W 1,p = 1 2 …. By construction j − ∈ 0 (Ω) and from Theorem 5.7 and Proposition 5.11 it follows that

p p p ð∇Vjð dx ≤ ð∇vð dx ≤ ð∇gð dx (32) ÊΩ ÊΩ ÊΩ

According to Theorem 5.10 the limit function V = limj→∞Vj is p-harmonic in Dj for j = 1, 2, … and we conclude that V is p-harmonic in Ω. By using (32) together with a standard compactness argument, similar to V g W 1,p p the one in the proof of Theorem 3.1, it follows that − ∈ 0 (Ω). Since -harmonic functions in bounded domains with given boundary values are unique, it follows that V = u. In addition, Hg ≤ Vj for each j so we see that Hg ≤ limj→∞ Vj = u. One can also show that Hg ≥ u, which implies that u ≤ Hg ≤ Hg ≤ u.

1,p By studying the proof we see that as long as g ∈ W (Ω) ∩ C(Ω), the solution u obtained in Theorem 3.1 agrees with Perron’s solution Hg. This is so important that it is summarized in the proposition below. Proposition 6.4. If g ∈ W 1,p(Ω) ∩ C(Ω)̄ , then the p-harmonic function with boundary values in Sobolev sense coincides with the Perron solution H.

We recall the discussion about regular points in Section 2 and restate the definition of regular boundary points in terms of Perron’s solution. Definition 6.2. We say that w ∈ )Ω is regular if lim H (x) = g(w) x→w g whenever g ∶ )Ω → ℝ is continuous. A point which is not regular is called irregular. We saw in Section 2 that it is possible to use the Wiener criterion to characterize the regular boundary points of )Ω. We will now present another method, which uses so called barrier functions. Definition 6.3. A point w0 ∈ )Ω has a barrier if there exists a function v ∶ Ω → ℝ such that 1. v is p-superharmonic in Ω,

2. lim inf x→w v(x) > 0 for all w ≠ w0, w ∈ Ω, lim v(x) = 0 3. x→w0 . Theorem 6.5. Let Ω be a bounded domain. The point w0 ∈ )Ω is regular if and only if there exists a barrier at w0.

44 Proof. We ﬁrst show the if direction. We assume that g ∈ C()Ω) and set M = sup ðgð. Using the minimum principle we know that the barrier v is positive inside the domain. Given " > 0 and x ∈ Ω, we can ﬁnd > 0 and > 0 such that ðg(w0) − g(w)ð < " whenever ðw0 − wð < and v(x) ≥ 2M whenever ðx − w0ð ≥ . We consider the function ̃v = g(w0) + " + v(x). ̃v is clearly p-superharmonic and bounded from below. Furthermore, for ∈ )Ω we obtain lim inf ̃v(x) − g() lim inf v(x) = 0 x→ ≥ x→ so ̃v ∈ g. Similarly we see that ̃u(x) = g(w0) − " − v(x) belongs to g. It follows that

g(w0) − " − v(x) ≤ Hg(x) = Hg(x) ≤ g(w0) + " + v(x). so ðHg(x) − g(w0)ð ≤ " + v(x). Thus Hg(x) → g(w0) as x → w0 so w0 is a regular point. We next show p∕(p−1) the existence of a barrier when w0 is a regular boundary point. Consider the function g(x) = ðx − w0ð . p−2 p−1 When x ≠ w0 we see that Δpg = div(ð∇gð ∇g) = (p∕(p − 1)) n. We note that

0 1p−1 p−2 p−2 p ⟨ð∇gð ∇g, ∇'⟩ = − div(ð∇gð ∇g)' = − ' dx ≤ 0 ÊΩ ÊΩ ÊΩ p − 1 ' C∞ g p for all non-negative ∈ 0 (Ω) and by Theorem 5.5 we conclude that is -subharmonic in Ω. It follows from the comparison principle that Hg ≥ g in Ω. By the properties of a regular boundary point it holds that lim H (x) = g(w ) = 0 H x→w0 g 0 , so we can use g as a barrier function.

7 Boundary behaviour

We now discuss the boundary behaviour of p-harmonic functions vanishing in a special type of Lipschitz 1, domains, known as C -domains. The whole section is devoted to proving Theorem 7.4 which is a special case of the theory developed in [13] and [14]-[20]. We let u and v be two non-negative p-harmonic functions vanishing on a portion of the boundary. The conclusion of the theorem is that close to that part of the boundary the ratio of u and v is Hölder continuous and bounded from above and below by constants. Before formulating 1, the theorem we ﬁrst discuss some fundamental properties of Lipschitz- and C -domains. n A Lipschitz domain is a domain in ℝ where locally the boundary is a graph of a Lipschitz continuous function. ¨ ¨ ¨ n−1 That is, after a possible rotation, the boundary can be described by the graph {(w , (w ) ∶ w ∈ ℝ } where n−1 ∶ ℝ → ℝ is a Lipschitz function, i.e., ðð ð∇ið ðð∞ < ∞. More formally, we say that Ω is a Lipschitz n−1 domain if for each xi ∈ )Ω, there exists ri > 0 and a Lipschitz function i ∶ ℝ → ℝ with Lipschitz constant Mi = ðð ð∇ið ðð∞ such that

¨ n ¨ Ω ∩ B(wi, ri) = {x = (x , xn) ∈ ℝ ∶ xn > i(x )} ∩ B(wi, ri) ¨ n ¨ )Ω ∩ B(wi, ri) = {x = (x , xn) ∈ ℝ ∶ xn = i(x )} ∩ B(wi, ri) (33) n hold in an appropriate coordinate system. In the following we assume that Ω ⊂ ℝ is a bounded Lipschitz domain i.e., )Ω is compact and we can find Lipschitz constants M and a r0 that hold for all w ∈ )Ω. This {B(w , ri )} )Ω )Ω can be seen as follows: let i 2 be an open cover of a neighbourhood of . Since is compact we {B(w , ri )}N 2r = min r M = max M can choose a finite subcover i 2 i=1. Next we define 0 i i and i i and note that

45 k, n for these constants the conditions in (33) hold for all x ∈ )Ω. In addition, if i ∈ C (ℝ ) for each i where k, k is a non-negative integer and ∈ (0, 1], we deﬁne Ω to be a bounded C -domain.

We say that a domain Ω satisﬁes the corkscrew condition if there exists constants M > 1 and r0 > 0 such that for any w ∈ )Ω and r ∈ (0, r0), there exists a point ar(w) ∈ Ω, such that r r < a (w) − w r, (a (w),)Ω) . M ð r ð ≤ and dist r ≥ M

Furthermore, we say that Ω satisﬁes the uniform condition if for each w ∈ )Ω, r ∈ (0, r0), and x1, x2 ∈ B(w, r) ∩ Ω, there exists a curve ∶ [0, 1] → Ω with (0) = x1, (1) = x2, such that

L( ) ≤ c(M)ðx1 − x2ð, (34) min{L( ([0, t])),L( ([t, 1]))} ≤ c(M)dist( (t),)Ω), (35) where L( ) denotes the length of . Lemma 7.1. Let Ω be a bounded Lipschitz domain with Lipschitz constant M ≥ 2 and r0 > 0. Then (i) Ω satisfies the corkscrew condition, (ii) ℝn ⧵ Ω̄ satisfies the corkscrew condition, (iii) Ω satisfies the uniform condition.

¨ ¨ Proof. For part (i), if w = (w , (w )) ∈ )Ω is given and 0 < r < r0, then (using the local coordinate system) ¨ ¨ ar(w) = (w , (w ) + r) (36)

−1 does the job. In order to show that dist(ar(w),)Ω) ≥ M r we note that using an appropriate local coordinate n−1 n−1 system we may assume that w = (0 , 0) and that ar(w) = (0 , r). Using the Lipschitz continuity of we see that the distance between ar(w) and the boundary is greater than the distance between the point n−1 n n−2 (0 , r) and the line L = 0 + t(0 , 1,M) for t ∈ ℝ. The orthogonal projection of ar(w) on L is given by n−2 2 Mr(0 , 1,M)∕(1 + M ) and thus it follows that ó n−2 ó ó n−1 Mr(0 , 1,M)ó r 4 2 1∕2 d(ar(w),L) = ó(0 , r) − ó = (M − M + 1) ≥ r∕M ó 1 + M2 ó 1 + M2 ó ó for M ≥ 2. The proof of part (ii) is analogous. In the proof of part (iii) we need to consider diﬀerent cases, ¨ ¨ see Figure 1 for an illustration. For i ∈ {1, 2} we set xi = (xi, (xi) + si) where si > 0 and let di = d(xi,)Ω) denote the distance from xi to the boundary.

1. Here we assume that di ≤ ðx1 − x2ð for i ∈ {1, 2}. Next we deﬁne 0 x − x 1 ̃x = x¨ , (x¨ ) + ð 1 2ð , 1 1 1 2 0 x − x 1 ̃x = x¨ , (x¨ ) + ð 1 2ð , 2 2 2 2

and note that ̃x1, ̃x2 ∈ Ω, since ðx1 − x2ð∕2 < r. The ﬁrst part, 1, is the straight line segment from x ̃x L1 M x x x¨ , x¨ 1 to 1. Clearly ( 1) ≤ ð 1 − 2ð. The second curve 2 follows the path between ( 1 ( 1)) to x¨ , x¨ x x x ̃x ̃x ( 2 ( 2)) but translated ð 1 − 2ð∕2 in the n-direction so it becomes a path between 1 and 2. Thus,

46 t t , t x x t x¨ x¨ 2( ) = ( ( ) ( ( )) + ð 1 − 2ð∕2) where ( ) is the straight line between 1 and 2. Furthermore, we can bound the length of 2 by T n U 1 É L ( 2) ≤ sup ð 2(ti) − 2(ti−1)ð ðti−ti−1ð→0 i=1 T n U É ≤ sup ð(ti) − (ti−1), ((ti)) − ((ti−1))ð ðti−ti−1ð→0 i=1 T n U É ≤ sup ð(ti) − (ti−1)ð + ð((ti)) − ((ti−1))ð ðti−ti−1ð→0 i=1 ¨ ¨ ≤ (1 + M)ðx1 − x2ð ≤ (1 + M)ðx1 − x2ð.

1 The last part, 3, is the straight line between ̃x2 to x2 and L ( 3) ≤ Mðx1 − x2ð. Thus we deﬁne to be the curve = ∪ i and clearly (34) is satisﬁed. We will make a short comment on why (35) holds. We note that we can make sure that the condition holds if we are considering a point on 1, since either we are moving away from the boundary or the distance to the boundary is greater than ðx1 − x2ð/2. On 2 we have control because of the corkscrew condition. By symmetry we have control on 3.

2. Now, d = ðx1 − x2ð ≤ di for i ∈ {1, 2}. For this case we take to be the straight line from x1 to x 2. Obviously (34) is satisﬁed. For the second condition we note that we obtain√ the smallest distance d( (t),)Ω) if d1 = d2 = ðx1 − x2ð. In this case it is clear that d( (t),)Ω) ≥ 3ðx1 − x2ð∕2 so the second condition is also satisﬁed.

3. Assume that d1 ≤ ðx1 − x2ð ≤ d2. Again, we use the notation d = ðx1 − x2ð. We consider to be the x x (t) = x +t(x −x ) B = B(x , d ) B = B(x , d) straight line between 1 and 2, i.e., 1 2 1 . Let d1 1 1 and d 2 . We need to make sure that (35) is satisﬁed. We ﬁrst consider d( (t),)Ω) for 0 ≤ t ≤ min{d1∕d, 1∕2} and d( (t),)Ω) d( (t),)B ∩)B ) = s ̃x (t) s = d( ̃x,)B ∩)B ) note that ≥ d1 d . Let be the point on such that d1 d and y0 = d(x1, ̃x). By solving the following equation system 2 2 2 y0 + s = d1 2 2 2 (d − y0) + s = d

2 2 2 1∕2 2 2 1∕2 we see that y0 = d ∕2d and s = d1(1 − d ∕4d ) so s∕d1 = (1 − d ∕4d ) . Furthermore, we note 1 1 √ 1 that s∕d1 is as small as possible for d1 = d and s∕d1 = 3∕2. Thus, for 0 ≤ t ≤ min{d1∕d, 1∕2} it follows that √ √ d( (t),)Ω) ≥ 3d1∕2 ≥ ( 3∕2) min{L( ([0, t])),L( ([t, 1]))}.

For min{d1∕d, 1∕2} < t ≤ 1 we obtain 2 d( (t),)Ω) ≥ d(x1, (t)) − d1 ∕2d ≥ d(x1, (t)) − d1∕2 ≥ d(x1, (t))∕2 = min{L( ([0, t])),L( ([t, 1]))}∕2

4. The case when d2 ≤ ðx1 − x2ð ≤ d1 follows by symmetry.

47 Figure 1: Illustrations for the proof of Lemma 7.1. Left: case 1. Right: case 3.

From now on, we will denote by ar(w), the point given by (36). We note that it follows from Theorem 6.31 in [9] that a Lipschitz domain is a regular domain. Furthermore, the previous lemma can also be used to show that bounded Lipschitz domains satisfy the so called Harnack chain condition. Lemma 7.2. Let Ω be a Lipschitz domain with Lipschitz constant M ≥ 2, and r0 > 0, and let 1 ", dist(x2,)Ω) > ", and ðx1 − x2ð ≤ A", for some A ≥ 1 and some " > 0. Then there exists a constant c = c(p, n, M, A) such that

u(x1) ≤ cu(x2).

Proof. The existence of a curve ∶ [0, 1] → Ω with (0) = x1, (1) = x2, satisfying (34) and (35) follows from Lemma 7.1. We note that from (34) it follows that L( ) ≤ c(M)A". Clearly, B(xi,"∕8) ⊂ Ω for i ∈ {1, 2}. We note that if x2 ∈ B(x1,"∕16) the conclusion of the lemma follows directly by applying the Harnack inequality for p-harmonic functions, and to this end we therefore assume that x2 ∉ ̃ ̃ B(x1,"∕16). Next we define t1 to be the smallest t s.t. (t) ∈ )B(x1,"∕32) and t2 to be the largest t s.t. ̃ ̃ (t) ∈ )B(x2,"∕32), i.e., t1 denotes the first time the curve leaves B(x1,"∕32) and t2 denotes the last time ̃ ̃ the curve enters B(x2,"∕32). It follows that min{L( ([0, t])),L( ([t, 1]))} ≥ "∕32 whenever t ∈ [t1, t2] and ̃ ̃ from (35) we see that c(M)dist( (t),)Ω) ≥ "∕32 whenever t ∈ [t1, t2]. Since c(M) > 1 it follows that dist( (t),)Ω) ≥ "∕(32c(M)). This will allow us to create a chain of balls with appropriate radius along the curve for which it will be possible to use Harnack’s inequality. We proceed as follows: let ̃r = "∕(4⋅32c(M)), z0 = x1, t0 = 0 and set zj+1 = (tj+1) where tj+1 is defined to be the smallest t > tj such that (t) ∈ )B(zj, ̃r) for all j = 0, … , k − 1. Here k ≥ 1 denotes the greatest k such that zk ∉ B(x2, ̃r), i.e. at time tk the curve has not yet entered B(x2, ̃r). However, by construction zk+1 ∈ B(x2, ̃r). We may now use Harnack’s inequality to conclude that k+1 k+2 u(x1) = u(z0) ≤ cu(z1) ≤ … ≤ c u(zk+1) ≤ c u(x2). In order to conclude the proof we need to make sure that k = k(p, n, M, A). However, using the first property of the curve we have that k̃r ≤ L( ) ≤ c(M)A" which implies that k ≤ ̃c(p, n, M, A) and so the conclusion of the lemma follows.

48 1, Another very useful property of C -domains is that they are relatively ﬂat in the sense that it is possible to locally approximate their boundaries with hyperplanes. In particular, the more we zoom in on the boundary, the ﬂatter it will look. This is summarized in the following lemma. 1, Lemma 7.3. Let Ω be a bounded C -domain for some ∈ (0, 1]. Given << 1, there exists ̄r0 = ̄r0(n, ) << 1 such that for all w ∈ )Ω and 0 < r < ̄r0 there exists a hyperplane Λ = Λ(w) containing w such that

(i) ℎ()Ω ∩ B(w, r), Λ ∩ B(w, r)) ≤ r n (ii) {x ∈ Ω ∩ B(w, r∕2) ∶ d(x, )Ω) ≥ 2r} ⊂ one component of ℝ ⧵ Λ. where ℎ denotes the Hausdorﬀ distance between two sets E, F ⊂ ℝn, i.e., ℎ(E,F ) = max(sup{d(x, E) ∶ x ∈ F }, sup{d(x, F ) ∶ x ∈ E}).

1, Proof. We assume the setting of a local coordinate chart containing w with a C -function , such that ¨ ¨ w = (w , (w )), and we consider the hyperplane ¨ n ¨ ¨ ¨ ¨ Λ = {y = (y , yn) ∈ ℝ ∶ yn = (w ) + ⟨∇(w ), (y − w )⟩}. ¨ ¨ ¨ n−1 Note that this is the tangent plane for Ω at the point (w , (w )). We claim that for y ∈ ℝ it holds that ¨ ¨ ¨ ¨ ¨ ¨ ¨ 1+ ð(y ) − (w ) − ⟨∇(w ), (y − w )⟩ð ≤ cðy − w ð 1, which is just another way to express the C -regularity, i.e., that the magnitude of ð∇ð is not to large. For ̃c ∈ (0, 1) it follows from the mean value theorem that ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ð(y ) − (w ) − ⟨∇(w ), (y − w )⟩ð = ð⟨∇( ̃cw + (1 − ̃c)y ) − ∇(w ), y − w ⟩ð ¨ ¨ ¨ ¨ ¨ ≤ cð ̃cw + (1 − ̃cðw − y ð)ð ðy − w ð ¨ ¨ ¨ ¨ ¨ ¨ ≤ c(ðy − w ð + ̃cðw − y ð) ðy − w ð ¨ ¨ +1 ≤ cðy − w ð . Since Ω is a bounded domain we can apply a covering argument to prove the existence of a uniform constant c that works for each w ∈ )Ω. From the claim it follows that 1+ ℎ()Ω ∩ B(w, r), Λ ∩ B(w, r)) ≤ cr . ̄r , ̄r c < r < ̄r (i) (ii) Next we deﬁne = ( 0 ) = 0 ∕ and note that for 0 0 we obtain . The second claim, , follows directly from the ﬁrst.

We are now ready to state the main theorem of this section. n 1, Theorem 7.4. Let Ω ⊂ ℝ be a C -domain for some > 0. Given p, 1 0 such that the following is true. Let w ∈ )Ω, 0 < r < ̄r0, suppose that u, v are non-negative p-harmonic functions in Ω ∩ B(w, 4r), u, v are continuous in Ω̄ ∩ B(w, 4r), and u = 0 = v on )Ω ∩ B(w, 4r). Then there exists c = c(p, n, ), 1 ≤ c < ∞, and = (p, n, ) ∈ (0, 1), such that if 0 < r < ̄r0, then u(a (w)) u(a (w)) −1 r∕c u(y) r∕c c ≤ ≤ c , v(ar∕c(w)) v(y) v(ar∕c(w)) whenever y ∈ Ω ∩ B(w, r∕c) and u a w 0 1 ó u(y1) u(y2) ó ( r∕c( )) y1 − y2 ó − ó c ð ð ó ó ≤ óv(y1) v(y2)ó v(ar∕c(w)) r whenever y1, y2 ∈ Ω ∩ B(w, r∕c).

49 The proof is divided into a number of steps as follows: • Step 1. In Section 7.2 we begin by proving Theorem 7.4 in the special case of a half space, see Theorem 7.8. One of the key steps of the proof is to make a Schwartz reﬂection and show that the extended function is p-harmonic. This will allow us to apply Lemma 4.4, and in combination with some boundary inequalities and barrier estimates the conclusion of the theorem will follow. 1, • Step 2. In step 2 we will work with estimates for the gradient of a p-harmonic function u in a C - domain vanishing on )(Ω ∩ B(w, 4r)), for some r, and where w ∈ )Ω. In particular we will prove the so called fundamental inequality which states that for some constants c and u(y) u(y) −1 ∇u(y) d(y, )Ω) ≤ ð ð ≤ d(y, )Ω)

whenever y ∈ Ω ∩ B(w, r∕c). This will be proved by approximating the boundary by a hyperplane according to Lemma 7.3 and then apply Theorem 7.8. We do this in Section 7.3. • Step 3. In this step (Section 7.4) we consider degenerate elliptic operators in weighted Sobolev spaces, for weights belonging to the Muckenhoupt class A2. Under certain assumptions it is possible to derive estimates similar to those in Theorem 7.4 for this class of diﬀerential operators. In addition, if u is a p−2 p-harmonic function vanishing on a portion of the boundary containing w we may extend ð∇uð to an A2(B(w, 2̂r))-weight for some small ̂r. We prove this using the estimates obtained in step 2. • Step 4. In this step (Section 7.5) we prove Theorem 7.4. We introduce an elliptic operator L of the type discussed in step 3 such that u − v is a (local) solution to to the equation Lf = 0. Using estimates for the gradients obtained in step 3 and 4, L can be locally reduced to a linear and uniformly elliptic operator. This can be used in order prove Theorem 7.4.

7.1 Boundary estimates

We begin by stating some boundary estimates which will be used throughout the remaining sections. For the statements and references to the proofs, see Lemma 2.1-2.3 in [13]. The ﬁrst lemma is a boundary variant of Caccioppoli’s inequality. n Lemma 7.5. Let Ω ⊂ ℝ be a Lipschitz domain and suppose that 1 < p < ∞. Let x ∈ )Ω, 0 < r < r0, and suppose that u is a non-negative continuous p-harmonic function in Ω̄ ∩ B(w, 2r) and that u = 0 on )Ω ∩ B(w, 2r). Then 0 1p p−n p r ð∇uð dx ≤ c max u . ÊΩ∩B(w,r∕2) Ω∩B(w,r)

The second lemma states that a p-harmonic function that vanished on the boundary is Hölder continuous close to that part of the boundary. n Lemma 7.6. Let Ω ⊂ ℝ be a Lipschitz domain and suppose that 1 < p < ∞. Let x ∈ )Ω, 0 < r < r0, and suppose that u is a non-negative continuous p-harmonic function in Ω̄ ∩ B(w, 2r) and that u = 0 on )Ω ∩ B(w, 2r). Then there exists = (p, n, M) ∈ (0, 1] such that if x, y ∈ Ω ∩ B(w, r), then 0 1 ðx − yð u(x) − u(y) ≤ c max u ð ð r Ω∩B(w,2r)

n Lemma 7.7. Let Ω ⊂ ℝ be a Lipschitz domain and let 1 < p < ∞. Let w ∈ )Ω, 0 < r < r0, and suppose that u is a non-negative continuous p-harmonic function in Ω̄ ∩ B(w, 2r) and that u = 0 on Δ(w, 2r). Then

50 there exists a constant c = c(p, n, M) ∈ [1, ∞) such that if ̃r = r∕c, then

max u ≤ cu(ãr(w)). Ω∩B(w,̃r)

From the last lemma we note that if u is a non-negative continuous p-harmonic function vanishing on a portion of )Ω we are now able to obtain an upper bound of u close to that part of the boundary. This is obviously of great importance since we cannot use the Harnack inequality arbitrarily close to the boundary.

7.2 Halfspace

This section is devoted entirely to proving Theorem 7.4 for the special case where Ω is a halfspace. The key observation for this is that in the case of a completely flat boundary it is possible to make a Schwarz reflection that preserves the p-harmonicity of the p-harmonic functions. This will make it possible to apply Lemma 4.4 which provides valuable estimates for the gradients of the p-harmonic functions. We begin this section by restating Theorem 7.4 tailored to the situation of a halfspace. n n Theorem 7.8. Let ℝ+ = {x ∈ ℝ ∶ xn > 0} and fix 1 0 n such that the following holds. Fix a point w ∈ )ℝ+, i.e., such that wn = 0, let r ∈ (0, ̃r0) and suppose that u n n and v are non-negative p-harmonic functions in ℝ+ ∩ B(w, 4r), continuous in ℝ+ ∩ B(w, 4r) and u = 0 = v n on )ℝ+ ∩ B(w, 4r). Then there exist constants c = c(p, n) ∈ [1, ∞) and = (p, n) ∈ (0, 1), such that if r ∈ (0, ̃r0), then u(a (w)) u(a (w)) −1 r∕c u(y) r∕c c ≤ ≤ c , v(ar∕c(w)) v(y) v(ar∕c(w)) n for y ∈ ℝ+ ∩ B(w, r∕c) and u a w 0 1 ó u(y1) u(y2) ó ( r∕c( )) y1 − y2 ó − ó c ð ð ó ó ≤ óv(y1) v(y2)ó v(ar∕c(w)) r n for y1, y2 ∈ ℝ+ ∩ B(w, r∕c). n Given 1 0 and y = (y1, … , yn) ∈ ℝ , we define the rectangles

Qr(y) = {x ∶ ðxi − yið < r, i ∈ {1, … , n}}, + Qr (y) = {x ∶ ðxi − yið < r, i ∈ {1, … , n − 1}, 0 < ðxn − ynð < r}.

u p Q+ Q+ u )Q+ x Let be a non-negative -harmonic function in 1 (0), continuous on 1 (0) and = 0 on 1 (0)∩{ n = 0}. We extend u to Q1(0) by a Schwarz reﬂection: T u(x¨, x ), x 0, ̃u x¨, x n n ≥ ( n) = ¨ −u(x , −xn), xn < 0.

¨ for x = (x , xn) ∈ Q1(0). Note that ̃u is continuous in Q1(0). Claim 2. ̃u is p-harmonic in Q1(0).

Proof of claim. ̃u W 1,p Q ' W 1,p Q We note that ∈ ( 1(0)). Next, let ∈ 0 ( 1(0)), '(x¨, x ) + '(x¨, −x ) (x¨, x ) = n n n 2

51 and = + , i.e., '(x¨, x ) − '(x¨, −x ) (x¨, x ) = n n . n 2

W 1,p Q+ x Q x Then ∈ 0 ( 1 (0)) and ( ) = 0 on 1(0) ∩ { n = 0} in Sobolev sense. Note that, T (u , … , u , u ), x 0, ∇̃u(x) = x1 xn−1 xn n ≥ (−u , … , −u , u ), x < 0. x1 xn−1 xn n and 1 ∇ = ' (x¨, x ) − ' (x¨, −x ), … ,' (x¨, x ) − ' (x¨, −x ),' (x¨, x ) + ' (x¨, −x ) . 2 x1 n x1 n xn−1 n xn−1 n xn n xn n

Hence, ⎧ 1 ∑n u ' (x¨, x ) − ∑n−1 u ' (x¨, −x ) + u ' (x¨, −x ) , x 0, ⎪ 2 i=1 xi xi n i=1 xi xi n xn xn n n ≥ ⟨∇̃u, ∇ ⟩ = ⎨ 1 − ∑n−1 u ' (x¨, x ) + u ' (x¨, x ) + ∑n u ' (x¨, −x ) , x < 0, ⎪ 2 i=1 xi xi n xn xn n i=1 xi xi n n ⎩ which implies that

p−2 p−2 ð∇̃uð ⟨∇̃u, ∇ ⟩ dx = ð∇̃uð ⟨∇̃u, ∇ ⟩ dx = 0, ÊQ (0)⧵Q+(0) ÊQ+(0) 1 1 1

u p W 1,p Q+ since is -harmonic and ∈ 0 ( 1 (0)). Thus,

p−2 ð∇̃uð ⟨∇̃u, ∇ ⟩ dx = 0. (37) ÊQ1(0) ∇̃u ̃u x What is left is to show is the corresponding equality for . Note that ð ð and xn are even functions of n and ̃u , … , ̃u x , … , that x1 xn−1 are odd functions of n. Furthermore, x1 xn−1 are odd and xn is even, as functions p−2 of xn. Thus, ð∇̃uð ⟨∇u, ∇⟩ is an odd function of xn, and noting that the domain Q1(0) is symmetric around {xn = 0}, we get

p−2 ð∇̃uð ⟨∇̃u, ∇⟩ dx = 0. (38) ÊQ1(0)

Thus, by (37) and (38), ̃u is p-harmonic in Q1(0).

We ﬁrst prove the boundary Harnack inequality in Theorem 7.8. Lemma 7.9. Let < p < and suppose that u and v are non-negative p-harmonic functions in Q+ , 1 ∞ 1 (0) continuous on Q+ and that u v on )Q+ x . Then there exists a constant c c p, n 1 (0) = 0 = 1 (0) ∩ { n = 0} = ( ) ∈ [1, ∞) such that

−1 u(en∕8) u(x) u(en∕8) c ≤ ≤ c v(en∕8) v(x) v(en∕8) for x Q+ , where e is the unit vector in the x -direction. ∈ 1∕8(0) n n

52 Proof. Note that if the lemma holds for a p-harmonic function u(x) and v(x) = xn (note that then v is p- harmoinic as well), then for any p-harmonic functions u and !, we have

−2 u(en∕8) u(x) 2 u(en∕8) c ≤ ≤ c , !(en∕8) !(x) !(en∕8)

u(x) u(x)∕xn = u(x)∕xn !(x)∕xn by writing !(x) !(x)∕xn and using the bounds on and . Thus, we are done if we prove the lemma with v(x) = xn. ̃x x¨, x Q+ x ̂x x¨, ̂u p Given = ( n) ∈ 1∕4(0), such that n ≤ 1∕100, we let = ( 1∕8) and let denote the unique - harmonic function in D = B( ̂x, 1∕8) ⧵ B( ̂x, 1∕100) with boundary values ̂u = u(en∕8) on )B( ̂x, 1∕100) and ̂u = 0 on )B( ̂x, 1∕8). We next prove the following claim. Claim 3. T a x − ̂x (p−n)∕(p−1) + b, p n ̂u(x) = ð ð ≠ a lnðx − ̂xð + b, p = n, for some constants a and b. Furthermore, there exists a constant c = c(n, p) such that

c ̂u(x) ≥ u(en∕8)xn, (39) for x ∈ D, such that xn ≤ 1∕100.

Proof of claim. This is done by comparing with the fundamental solution. We begin with the case p = n. Let a and b be such that T a ln(1∕8) + b = 0, a ln(1∕100) + b = u(en∕8).

a = u(en∕8) b = − ln(1∕8)u(en∕8) Then ln(8∕100) and ln(8∕100) . For the case p ≠ n, we solve p−n ⎧a 1 p−1 + b = u(e ∕8), ⎪ 100 n p−n ⎨ 1 p−1 ⎪a + b = u(en∕8). ⎩ 100

Then, p−n − 1 p−1 u(e ∕8) u(e ∕8) 8 n a = n , b = . p−n p−n p−n p−n 1 p−1 − 1 p−1 1 p−1 − 1 p−1 100 8 100 8

We only do the computation for the constant in the case p < n, the other case is similar.

L p−n M p−n u(en∕8) 1 p−1 ̂u(x) = x − ̂x p−1 − p−n p−n ð ð 8 1 p−1 − 1 p−1 100 8

53 ¨ Letting x = (x1, … , xn) → (x , 0) in D, both ̂u(x) and xn tend to 0, we thus have to check that ̂u does not decrease too quickly (at most to the same order as xn). As ̂u clearly decreases the fastest along the line ¨ t ↦ (x , t), we need only consider the decay there. Let p−n p−n 1 p−1 1 p−1 f(x ) = − x − n 8 n 8 for xn ≤ 1∕100. Then, 1−n n − p 1 p−1 f ¨(x ) = − x c (p, n) n p − 1 8 n ≥ 1 for x ≤ 1∕100. Thus, it is clear that (39) holds.

Using the Harnack inequality and that u is non-negative, we have that ̂u ≤ c1u on )B( ̂x, 1∕100). Hence, ̂u ≤ c1u on )D, so by the comparison principle,

̂u(x) ≤ c1u(x), x ∈ D. Thus, ̂u(x) u(x) u(x) u(en∕8) ≤ c ≤ cc1 = c (40) xn xn xn x D x x Q+ x for ∈ such that n ≤ 1∕100. Since ∈ 1∕4(0) ∩ { n ≤ 1∕100} the estimate u u(en∕8) ≤ c xn x Q+ x in fact holds for ∈ 1∕4(0) such that n ≤ 1∕100. Using the Harnack inequality, it follows that (40) holds Q+ in 1∕8(0). Thus we have proved the lower bound. We now extend u to Q1(0) by Schwarz reﬂection, as above. Then, u is p-harmonic in Q1(0), and by Lemma 7.7, ¨ max u ≤ c u(en∕8). Q1∕4(0)

In combination with Lemma 4.4 and the Harnack inequality we obtain ¨ max ð∇uð ≤ c max u ≤ cc u(en∕8), (41) Q1∕8(0) Q1∕4(0) where the constant only depend on n and p. By an application of the mean value theorem we see that u(x) ≤ cu(en∕8), (42) xn which together with (40) implies the result. Lemma 7.10. Let < p < and suppose that u and v are non-negative p-harmonic functions in Q+ , 1 ∞ 1 (0) continuous on Q+ and that u v on )Q+ x . Then there exist constants c c p, n , 1 (0) = 0 = 1 (0)∩{ n = 0} = ( ) ∈ [1 ∞) and = (p, n) ∈ (0, 1) such that

ó u(x) u(y) ó u(en∕8) ó − ó c x − y ó ó ≤ ð ð óv(x) v(y)ó v(en∕8) whenever x, y Q+ . ∈ 1∕32(0)

54 Proof. As above, we will prove the lemma for v(x) = xn and derive the general result from that. Note that u(x) u(y) 1 − = u (x¨, tx ) − u (y¨, ty )dt, xn n xn n xn yn Ê0 x, y Q+ u Q for ∈ 1∕32(0). We want to apply Lemma 4.4 again, and therefore we extend to 1(0) by Schwarz reﬂection. Thus, 1 óu(x) u(y)ó ó − ó ∇u(x¨, tx ) − ∇u(y¨, ty ) dt ó ó ≤ ð n n ð ó xn yn ó Ê0 1 ¨ ¨ ≤ c max ð∇uð ð(x , txn) − (y , tyn)ð dt B(0,1∕4) Ê0 1 ¨ ¨ 2 2 2 ≤ ̃cu(en∕8) ðx − y ð + t(xn − yn) dt Ê0 u(e ∕8) ̂c n x − y , ≤ 1∕8 ð ð where we used Lemma 4.4 in the second inequality and (41) in the third. Next, we show that this implies the result. ó u(x) u(y) ó ó u(x) u(y) u(y) u(y) ó ó ó ó − ó ó u(x) u(y) ó ó xn yn ó ó xn yn yn yn ó ó − ó = ó − ó = ó + − ó óv(x) v(y)ó ó v(x) v(y) ó ó v(x) v(x) v(y) ó ó ó ó x y ó ó x x y ó ó n n ó ó n n n ó ó ó 1 óu(x) u(y)ó u(y) ó 1 1 ó ó − ó + ó − ó ≤ v(x) ó v(x) v(y) ó ó xn yn ó yn ó ó x ó ó ó x y ó n ó n n ó ó v(y) v(x) ó ó − ó 1 u(e ∕8) u(y) ó yn xn ó c n x − y + ó ó ≤ v(x) ð ð v(x) v(y) 1∕8 yn xn xn yn u(y) u(e ∕8) y v(e ∕8) c n x − y + n ⋅ c n x − y ≤ ð ð v(x) v(y) ð ð v(en∕8) 1∕8 xn yn u(en∕8) u(en∕8) ≤ c ðx − yð + ̃c ðx − yð v(en∕8) v(en∕8) u(en∕8) = C ðx − yð , v(en∕8) where we used Lemma 7.9 for the third and fourth inequality.

Proof of Theorem 7.8. This is an immediate consequence of Lemma 7.9, Lemma 7.10 and Harnack’s inequality.

7.3 The fundamental inequality for the gradient of a p-harmonic function

In this section we will establish an inequality for the gradient of a positive p-harmonic function, u, vanishing 1, on a portion of the boundary. Let Ω be a C -domain for ∈ (0, 1] and u be p-harmonic in Ω ∩ B(w, 4r) and

55 continuous up to and vanishing on )(Ω ∩ B(w, 4r)). Then, for some constants c and it holds that u(y) u(y) −1 ∇u(y) d(y, )Ω) ≤ ð ð ≤ d(y, )Ω) whenever y ∈ Ω ∩ B(w, r∕c). The idea is to ﬁrst prove the inequality for the case where Ω is a half-space and then prove the general case by approximating the boundary with hyperplanes according to Lemma 7.3 and use the translation and rotation invariance of the p-Laplace operator. In combination with the rotation and translation invariance of the p-Laplace operator will result in a situation similar to the half space. These estimates will be used in the next section when we connect the boundary behaviour of p-harmonic functions to linear degenerate elliptic operators in weighted Sobolev spaces. We begin this section by stating and proving a technical lemma (Lemma 3.18 in [13]). n Lemma 7.11. Let G ⊂ ℝ be an open set, suppose that 1 < p < ∞. Also, suppose that u1 and u2 are non-negative p-harmonic functions in G. Let ̃a ≥ 1, y ∈ G and assume that 1 u (y) u (y) 1 ∇u (y) ̃a 1 . ̃a d(y, )G) ≤ ð 1 ð ≤ d(y, )G) (43) Let ̃−1 = (c ̃a)(1+)∕ where is as in Lemma 4.4. Then the following is true for c = c(p, n) suﬃciently large. If u2 1 (1 − ̃")L ≤ (1 + ̃")L in B(y, d(y, )G)) (44) u1 100 for some L such that 0 < L < ∞, then 1 u (y) u (y) 2 ∇u (y) c ̃a 2 . c ̃a d(y, )G) ≤ ð 2 ð ≤ d(y, )G)

̄ −3 Proof. Let z1, z2 ∈ B(y, td(y, )G)) where 0 < t ≤ 10 . By ﬁrst applying Lemma 4.4 and thereafter Harnack’s inequality it follows that ct ∇ui(z1) − ∇ui(z2) ≤ max ui(y)∕d(y, )G) ð ð d(y, )G) B(y,2td(y,)G)) (45) 2 ≤ c t ui(y) (46) for i ∈ {1, 2} and c = c(p, n). From this the upper bound of the gradient of u2 at y follows directly. To establish the lower bound we will argue by contradiction. Thus, we assume that

ð∇u2(y)ð ≤ u2(y)∕d(y, )G) (47) for some > 0 which will be chosen later. We use (45) with z1 = z, z2 = y, (47) and the reversed triangle inequality to obtain 2 ð∇u2(z)ð ≤ ( + c t )u2(y)∕d(y, )G) whenever z ∈ B(y, td(y, )G)) and c = c(p, n). For ̃y ∈ )B(y, td(y, )G)) we deﬁne the line segment from y ¨ to ̃y as r(s) = y + ( ̃y − y)s for s ∈ [0, 1]. Note that ðr (s)ð = ð ̃y − yð = td(y, )G). Integrating along r(s) with t = we see that ó 1 ó ó d ó ðu2( ̃y) − u2(y)ð ≤ ó u2(r(s)) dsó óÊ0 ds ó ó ó 1 ¨ ≤ ð∇u2(r(s))ððr (s)ðds Ê0 ¨ 1+1∕ ≤ c u2(y)

56 ¨ ¨ where c = c (p, n). We let = ∇u1(y)∕ð∇u1(y)ð and use the Cauchy-Schwarz inequality, (45), and (43) to conclude that

⟨∇u1(z), ⟩ = ⟨∇u1(z) − ∇u1(y), ⟩ + ð∇u1(y)ð ≥ ð∇u1(y)ð − (ð∇u1(z) − ∇u1(y)ð) u (y) ∇u (y) − c2 1 ≥ ð 1 ð d(y, )G)

≥ (1 − c ̃a)ð∇u1(y)ð, so ⟨∇u1(z), ⟩ ≥ (1 − c ̃a)ð∇u1(y)ð (48) ̄ 1∕ 1∕ for c = c(p, n) whenever z ∈ B(y, d(y, )G)). Next, we let ̃y = y + d(y, )G) and let 1∕ r(s) = y + d(y, )G)s for s ∈ [0, 1] (49) ¨ 1∕ denote the line segment from y to ̃y and note that r (s) = d(y, )G). By integrating along the line segment and using (48) and (43) we obtain 1 d u1( ̃y) − u1(y) = (u1(r(s)))ds Ê0 ds 1 1∕ = d(y, )G) ⟨∇u1(r(s)), ⟩dt Ê0 1 1∕ ≥ d(y, )G) ⟨∇u1(r(s)), ⟩ds Ê0 1 1∕ ≥ d(y, )G) (1 − c ̃a)ð∇u1(y)ðds Ê0 u (y) 1∕d(y, )G)(1 − c ̃a) 1 . ≥ ̃ad(y, )G) −1 If we let ≤ (2c ̃a) we ﬁnd that 1∕ (u ( ̃y) − u (y)) . 1 1 ≥ 2 ̃a (50) From (48) and (50) we see that ¨ 1+1∕ u2( ̃y) ≤ (1 + c )u2(y) 1∕ u1( ̃y) ≥ (1 + ∕(2 ̃a))u1(y) and in combination with (44) we obtain u ( ̃y) 1 + c¨1+1∕ (1 − ̃")L 2 (1 + ̃")L < (1 − ̃")L ≤ ≤ 1∕ (51) u1( ̃y) 1 + ∕(2 ̃a) 1∕ 1∕ if 1∕( ̃ac) ≥ ≥ ̃ac̃" for c = c(p, n) suﬃciently large. This holds since 1 + c¨1+1∕ ̃" 2 + c¨1+1∕ + 1∕∕(2 ̃a) + c¨1+1∕ − 1∕∕(2 ̃a) (1 + ̃") − (1 − ̃") = 1 + 1∕∕(2 ̃a) 1 + 1∕∕(2 ̃a) 1∕ 1 1 2 + 1+1∕ + 1∕∕(2 ̃a) + c¨ − ̃a c 2 ≤ 1 + 1∕∕(2 ̃a) < 0

57 −1 (1+)∕ −1 for a suﬃciently large value of c. Next we choose ̃" = (c ̃a) and = c ̃a and note that these choices satisfy the earlier claims. Thus, we have reached a contradiction which implies that (47) is false and the conclusion of the lemma follows.

This lemma will be used in order to prove the fundamental inequality in a half space. Lemma 7.12. Let < p < . Suppose that u is a p-harmonic function in Q+ , continuous on the closure 1 ∞ 1 (0) of Q+ and that u on )Q+ y . Then there exists ̃c ̃c p, n and p, n such that 1 (0) = 0 1 (0) ∩ { n = 0} = ( ) = ( ) u(y) u(y) −1 u y y Q+ ≤ ð∇ ( )ð ≤ when ∈ 1∕ ̃c(0) (52) yn yn

Proof. We begin by applying Lemma 7.9 and Lemma 7.10 to the p-harmonic functions u1 = yn and u2 = u. It follows that óu1(y1) u1(y2)ó u1(y2) ó − ó c¨ y − y ó ó ≤ ð 1 2ð (53) óu2(y1) u2(y2)ó u2(y2) y , y Q+ p, n , y , y Q+ for 1 2 ∈ 1∕4(0) and = ( ) ∈ (0 1). We note that the claim in Lemma 7.10 is for 1 2 ∈ 1∕32(0) Q+ but due to scale invariance and the fact that we are looking for an estimate for 1∕ ̃c it will not make any Q+ D y Q+ d y ,)D diﬀerence. For simplicity we denote 1∕4(0) by . If 2 ∈ 1∕8(0) we note that ( 2 ) is always attained for x ∈ )D ∩ {xn = 0}, Therefore it follows that

1 u1(y2) u1(y2) ≤ ð∇u1(y2)ð ≤ ̃a (54) ̃a d(y2,)D) d(y2,)D) where ̃a = ̃a(n) since u1(y2) will go to zero at the same speed as d(y2,)D). We restrict y2 according to ¨ 1∕ d(y2,)D) ≤ 100(̃"∕(2c )) , (55) where ̃" is as in Lemma 7.11. We claim that

u1(y2) u1(y1) u1(y2) (1 − ̃"∕2) ≤ ≤ (1 + ̃"∕2) (56) u2(y2) u2(y1) u2(y2) whenever y1 ∈ B(y2, d(y2,)D)∕100). From (53) it follows that

¨ u1(y2) u1(y1) ¨ u1(y2) 1 − c ðy1 − y2ð ≤ ≤ 1 + c ðy1 − y2ð . (57) u2(y2) u2(y1) u2(y2)

Furthermore, from the restriction in (55) we note that 0d(y ,)D)1 2 ̃"∕(2c¨) 100 ≤

¨ and for ðy1 − y2ð ≤ d(y2,)D)∕100 we see that c ðy1 − y2ð ≤ ̃"∕2. The claim follows by combining this estimate with (57). Finally, we note that by (54) and (56) we can apply Lemma 7.11 and thus we see that (52) holds at y2, so we have indeed proven the existence of the constant ̃c in Lemma 7.12 such that (52) holds for y Q+ ∈ 1∕ ̃c(0).

1, We next prove the fundamental inequality in C -domains (see Lemma 3.35 in [13]).

58 n 1, Lemma 7.13. Let Ω ⊂ ℝ be a C -domain for some ∈ (0, 1]. Given p, 1 0 such that the following is true. Let w ∈ )Ω, and 0 < r < ̄r0. Suppose that u is a positive p-harmonic function in Ω∩B(w, 4r), that u is continuous in Ω∩̄ B(w, 4r), and u = 0 on )Ω∩B(w, 4r). There exists ̃c = ̃c(p, n, ) and ̄ = ̄(p, n, ) such that u(y) u(y) ̄−1 ∇u(y) ̄ d(y, )Ω) ≤ ð ð ≤ d(y, )Ω) whenever y ∈ Ω ∩ B(w, r∕ ̃c).

Proof. We begin by continuously extending u to B(w, 4r) by letting u = 0 in B(w, 4r) ⧵ Ω. The idea of 1, the proof is to use the C - regularity of the domain to obtain a situation similar to when the domain is a halfplane so that Lemma 7.12 is applicable. Therefore we let c1 be a constant which will later be used as ̃c in ¨ Lemma 7.12. Due to Lemma 7.7 and Harnack’s inequality it is possible to ﬁnd c ≥ 100c1 such that for each ¨ ̃y ∈ Ω ∩ B(w, r∕c ) and s = 4c1d( ̃y,)Ω) and z ∈ )Ω with ð ̃y − zð = d( ̃y,)Ω) it holds that

max u ≤ cu( ̃y) B(z,4s) (58) where c = c(p, n). By Lemma 7.3 we have that for << 1, there exists 0 < ̄r0 << 1 such that for 4s ≤ ̄r0 there exists a hyperplane Λ containing z such that

ℎ()Ω ∩ B(z, 4s), Λ ∩ B(z, 4s)) ≤ 4s (59) n {y ∈ Ω ∩ B(z, 4s) ∶ d(y, )) ≥ 8s} ⊂ one component of ℝ ⧵ Λ. (60)

¨ ¨ n−1 Using the invariance properties of the p-Laplace operator we may assume that Λ = {(y , yn) ∶ y ∈ ℝ , yn = 0} and that n {y ∈ Ω ∩ B(z, 4s) ∶ d(y, )Ω) ≥ 8s} ⊂ {y ∈ ℝ ∶ yn > 0} ¨ ¨ We next deﬁne another hyperplane Λ and a domain Ω as follows: ¨ ¨ ¨ n−1 ¨ n Λ = {(y , 0) + 20sen, y ∈ ℝ }, Ω = {y ∈ ℝ ∶ yn > 20s}

¨ and clearly Ω ∩ B(z, 2s) ⊂ Ω ∩ B(z, 2s). For an illustration, see Figure 2. We deﬁne v to be the p-harmonic ¨ function in Ω ∩ B(z, 2s) with continuous boundary values ¨ v(y) = u(y) whenever y ∈ )[Ω ∩ B(z, 2s)] and yn > 40s ¨ v(y) = 0 whenever y ∈ )[Ω ∩ B(z, 2s)] and yn < 30s ¨ ¨ such that v ≤ u on )[Ω ∩ B(z, 2s)]. From the comparison principle it follows that v ≤ u in Ω ∩ B(z, 2s). We ¨ note that the points on the boundary of )[Ω ∩ B(z, 2s)] where u ≠ v lie within 80s from )Ω where u = 0 and therefore we can use Lemma 7.6 and (58) to conclude that ̃ ¨ u ≤ v + c u( ̃y) in )Ω ∩ B(z, 2s). where we have used ̃ to denote the exponent from Lemma 7.6. From the maximum principle for p-harmonic functions it follows that ¨ v ≤ u ≤ v + c u( ̃y) in Ω ∩ B(z, 2s). By applying the Harnack inequality we see that u(y) 1 (1 − c )−1 Ω¨ ∩ B(z, 2s). ≤ v(y) ≤ in

59 ¨ We next apply Lemma 7.12 and use that d( ̃y,)Ω) ≈ d( ̃y,)Ω ) which gives us v( ̃y) v( ̃y) −1 ∇v( ̃y) d( ̃y,)Ω) ≤ ð ð ≤ d( ̃y,)Ω) for = (p, n). We note that if we choose = (̄r0) (i.e. ̄r0) small enough, we can apply Lemma 7.11 with ¨ G = Ω ∩ B(z, 2s) and ̃a = . Using that it follows that u(y) u(y) ̄−1 ∇u(y) ̄ d(y, )Ω) ≤ ð ð ≤ d(y, )Ω) ̄ ̄ for = (p, n, ).

Figure 2: Lemma 7.13.

p−2 We ﬁnish this section by stating and proving a lemma which will be used when we extend ð∇uð to an A2-weight in the next section. Lemma 7.14. Let Ω ⊂ ℝn be a C1, -domain for some ∈ (0, 1]. Given, p, 1 0 such that the following is true. Let w ∈ )Ω, 0 < r < ̄r0, suppose that u is a non-negative p-harmonic function in Ω ∩ B(w, r), u is continuous in Ω̄ ∩ B̄(w, r), and that u = 0 on )[Ω ∩ B(w, r)]. Then there exists c = c(p, n, , "), 1 ≤ c < ∞, such that

1+" 1−" −1 ̃r u(ãr(w)) ̃r c ≤ ≤ c r u(ar(w) r whenever 0 < ̃r< r∕4.

Proof. As a ﬁrst step we extend u continuously to B(w, r) by deﬁning u = 0 in B(w, r) ⊂ Ω. Note that u is not ̃ p-harmonic in B(w, r). By applying Lemma 7.3 we know that given << 1, there exists ̄r0 = ̄r0(n, ) << 1 such that for r < ̄r0, there exists a hyperplane Λ containing w such that

ℎ()Ω ∩ B(w, r), Λ ∩ B(w, r)) ≤ r.̃

60 We note that it is suﬃcient to prove the lemma for r = ̄r0. Using the invariance properties of the p-Laplace equation we, without loss of generality, may assume that u(a4(0)) = 1, r = 4 and w = 0 so that

ℎ(Λ ∩ B(0, 4),)Ω ∩ B(0, 4)) ≤ 4̃ n ̃ where Λ = {y ∈ ℝ ∶ yn = 0}. We let = 4 and note that we may assume that ¨ B(0, 4) ∩ (x , xn) ∶ xn ≥ 2} ⊂ Ω (61) ¨ n B(0, 4) ∩ (x , xn) ∶ xn ≤ −2} ⊂ ℝ ⧵ Ω. (62)

Thus, due to the invariance properties it is enough to prove that −1 1+" 1−" c ̃r ≤ u(ãr(0)) ≤ c̃r when 0 < ̃r< 1. (63)

The proof will be conducted using barrier type estimates. We begin by deﬁning the rectangles + ¨ Q = {x ∶ ðx ð < 1, 2 < xn < 1} − ¨ Q = {x ∶ ðx ð < 1, −2 < xn < 1}

+ − + − + and introduce two p-harmonic functions, u and u on Q and Q , respectively, as follows. We let u be the + p-harmonic function on Q with the (continuous) boundary values

+ + ⎧u (x) = u(x), if x ∈ )Q ∩ {x ∶ xn ≥ 4} ⎪u+(x) = (xn−2) u(x) x ∈ )Q+ ∩ {x ∶ 2 < x < 4} ⎨ 2 if n ⎪u+(x) = 0, x ∈ )Q+ ∩ {x = } ⎩ if n − − − − Furthermore, we let u denote the p-harmonic function in Q with u (x) = u(x) on )Q . It follows directly + − + from the comparison principle that u ≤ u ≤ u . In addition, we also deﬁne the p-harmonic functions v − n + − and v in ℝ such that v (x) = xn − 2 and v (x) = xn + 2. Take 0 < ̂r< 1. We ﬁrst show the upper bound in (63). This will be achieved using induction on the scale of the rectangle. We want to show that k u(ak(0)) ≤ (c) (64) − − for c = c(p, n) and every integer k ≥ 1. Since << 1 we can use Lemma 7.9 on u and v to conclude that − − − u (x) ≤ cu (en∕8)v (x), when x ∈ Ω ∩ B(0, 1∕16).

We see that − − u(x) u (x) max u (x) = max u(x) cu(a4(0)) = c ≤ ≤ )Q− )Q− ≤ (65) − whenever x ∈ Q , where we have used the comparison principle for the ﬁrst inequality and Lemma 7.7 and − the Harnack inequality for the last inequality. Thus, u (en∕8) ≤ c and in combination with (65) we obtain − u(x) ≤ u (x) ≤ c(xn + 2) whenever x ∈ Ω ∩ B(0, 1∕16).

Hence it follows that u(a(0)) ≤ c. Next, we proceed with the induction step and therefore assume that (64) ¨ holds. Using Lemma 7.3 we see that there exists a hyperplane Λ containing w = 0 such that k k k k+1 ℎ(Λ ∩ B(0, 4 ),)Ω ∩ B(0, 4 )) ≤ 4̃ =

61 ¨ k If we now replace Λ and 4 with Λ and 4 , respectively, and repeat the above argument on a smaller scale − − − − − k+1 Qk with uk deﬁned analogous to u but on Qk , and vk = xn + 2 we ﬁnd that − u (a k (0)) k+1 k+1 − k − (xn + 2 ) − (xn + 2 ) u(x) u (x) c v (x) c u (a k (0)) c u(a k (0)) ≤ k ≤ − k ≤ k k+1 k ≤ k k+1 vk (ak (0)) + 2 + 2

k whenever x ∈ Ω ∩ B(0, 1∕ ̃c(k)). Thus, we can conclude that u(ak+1 (0)) ≤ ̂cu(ak (0)) ≤ ( ̂c) , which −" finishes the induction argument. We choose such that ≥ ̂c where ̂c is the same constant as in (64). k Furthermore, let k be the smallest integer so that ≤ ̂r holds. For a sufficiently small value of we can apply the Harnack inequality to âr(0) and ak (0). The length of the Harnack chain depends on which in turn depends on ̄r0 so we conclude that for some c = c(p, n, , ") it holds that 1−" u(âr(0)) ≤ cu(ak (0)) ≤ c̃r and so the right hand inequality i (63) is proved. + + + We begin the proof of the left hand inequality by applying Lemma 7.9 to u , v in Q from which it follows that, for a sufficiently small value of , + + u (a )(0) u (a1∕8(0)) 4M u+ a + ≈ + ≈ ( 1∕8(0)) v (a4M)(0) v (a1∕8(0))

+ where we have used that v (a1∕8(0)) ≈ 1. Assuming that is suﬃciently small it holds that

cu+(a (0) min u+(x) c¨−1 min u(x) ̃c−1u(a (0)) 1∕8) ≥ + ≥ + ≥ 4 )Q ∩{xn≥4}∪{xn=4} )Q ∩{xn≥4}∪{xn=4}

+ −1 so u (a1∕8(0)) ≥ c where c = c(p, n). Using this inequality in combination with Harnack’s inequality and + the fact that v (a4M(0)) ≈ we ﬁnd that −1 −1 + −1 + −1 ̄ u(a8(0)) ≥ c u(a4M(0)) ≥ c u (a4M(0)) ≥ ̂c v (a4M̄(0)) ≥ ̄c for ̄c = ̄c(p, n). By repeating the induction argument for the right hand inequality we see that + u (a k (0)) + −1 k (8) + u(x) ≥ uk (x) ≥ c + vk (x) whenever x ∈ B(0, 1∕ ̃c(k)) vk (a(8)k (0))

−1 −1 k for some small ̃c(k) which implies that u(a(8)k+1 (0)) ≥ ̄c u(a(8)k (0)) ≥ ( ̄c ) . Choose = (p, n, , ") −1 1+" k+1 k small enough for ̄c ≥ (8) to hold and let k be the integer such that ̂r ∈ [(8) , (8) ]. We can now use Harnack’s inequality and the estimate above to conclude that for some c = c(p, n, , ") −1 −1 k(1+") −1 (1+") u(ãr(0)) ≥ c u(a(8)k (0)) ≥ c (8) ≥ c ̃r and so the conclusion of the lemma follows.

7.4 Estimates for degenerate elliptic equations in weighted Sobolev spaces

In this section we will work with a certain class of linear degenerate elliptic diﬀerential operators discussed in [4], [2] and [3]. We consider Lv = div(B(x)∇v) where L is the operator and B(x) is a symmetric matrix whose elements satisfy a degenerate ellipticity conditon involving a non-negative function (x). When the

62 authors to [4] established the Harnack inequality and interior Hölder continuity for weak solutions to the equation Lv = 0, they found that the Poincaré inequality and the Sobolev embedding theorem were needed (cf. Section 1.1.2). Therefore, they imposed the restriction that (x) has to belong to the Muckenhoupt class A2 and showed that these theorems hold for Ap-weights. This section is organized as follows: in the first part we state some general results and properties obtained in [4]- [3]. After that we continue by discussing some boundary Harnack inequalities for the ratio of two L-harmonic functions, i.e., weak solutions to the equation Lv = 0. We end this section by explaining the connection between the theory of linear elliptic differential operators and the theory for p-harmonic functions. In particular we prove that if a p-harmonic function u p−2 satisfies certain assumptions, then ð∇uð can be extended to an A2-weight. The last part of the section is based on the work in [13] and [14]-[20]. Much of the theory needed in order to understand the proofs is outside the scope of this thesis and therefore most of the proofs will either be left out or sketched. Throughout this section we assume that w ∈ )Ω and that Ω is a Lipschitz domain with Lipschitz constants M and r0. For 0 < r < r0, we let (x) be a non-negative Lebesgue measurable function defined a.e. on B(w, 2r) such that ∈ A2(B(w, r)), i.e., 0 1 −2n −1 ̃r sup dy ⋅ dy ≤ ÊB(z,̃r) ÊB(z,̃r) where = (p, ) and the supremum is taken over all balls B(z, ̃r) such that z ∈ B(w, r) and 0 < ̃r ≤ r. Furthermore, let B(x) = {bij(x)} be a symmetric matrix where each bij(x) is a Lebesgue measurable function defined a.e. on B(w, 2r) for i, j ∈ {1, … , n}. In addition, we assume that the coefficients satisfy an ellipticity n condition, namely that for ∈ ℝ n −1 2 É 2 (x)ðð ≤ bij(x)ij ≤ (x)ðð i,j=1 for almost every x ∈ B(w, 2r). We note that since (x) may vanish or be infinite the expression above is a degenerate ellipticity condition. For x ∈ Ω ∩ B(w, 2r) we define the operator n 0 1 É ) ) L = b (x) . )x ij )x (66) i,j=1 i j

We are now ready to state the definition of a weak solution. 1,2 Definition 7.1. Let G be an open set such that G ⊂ B(w, 2r). We say that v ∈ W (G) is a weak solution to Lv = 0 in G, or that v is L-harmonic in G, if the Dirichlet form vanishes, i.e., n É b (x)v ' dx = 0 ij xi xj (67) ÊG i,j=1 ' C∞ G for all ∈ 0 ( ). In a manner similar to the corresponding one for the p-Laplace operator it is possible to show the existence of solutions, the comparison principle, the Harnack inequality, local interior Hölder inequality and the maximum principle, see [4]. Furthermore, in [2] Fabes et al. proved a Wiener criterion for regular boundary points for the L-Dirichlet problem which is defined analogous to the p-Dirichlet problem but with L instead of Δp. Using this it is possible to prove that all boundary points of a Lipschiz domain are regular. The following two lemmas can be found in [13] and summarize some of the work in [4]-[3]. n Lemma 7.15. Let Ω ⊂ ℝ be a bounded Lipschitz domain with constants M, r0, w ∈ )Ω, 0 < r < r0, and let be an A2(B(w, 2r))- weight with constant . Suppose that v is a positive weak solution to L! = 0 in Ω ∩ B(w, 2r). Then there exists a constant c ≥ 1, such that if z ∈ Ω, ̃r> 0,B(z, 2̃r) ⊂ Ω ∩ B(w, r), then

63 (i) 0 1 −1 2 2 2 2 c ̃r ð∇vð d ≤ c d (max v) ≤ c ðvð d ÊB(z,̃r∕2) ÊB(z,̃r) B(z,̃r) ÊB(z,2̃r) where d = dx. (ii) The Harnack inequality holds, i.e., max v ≤ min v. B(z,̃r) B(z,̃r)

(iii) There exists = (n, M, , ) ∈ (0, 1) such that if x, y ∈ B(z, ̃r), then 0 1 ðx − yð v(x) − v(y) ≤ c max v. ð ð r B(z,2̃r)

n Lemma 7.16. Let Ω ⊂ ℝ be a bounded Lipschitz domain with constants M, r0, w ∈ )Ω, 0 < r < r0, and let be an A2(B(w, 2r))-weight with constant . Suppose that v is a positive weak solution to L! = 0 in Ω ∩ B(w, 2r) and that v = 0 on )Ω ∩ B(w, 2r) in the weighted Sobolev sense. Then there exists 1 ≤ c ≤ ∞, such that the following holds with ̃r = r∕c: (i) 2 2 2 r ð∇vð d ≤ ðvð d ÊΩ∩B(w,r∕2) ÊΩ∩B(w,r) where d = dx. (ii) max v ≤ cv(ãr(w)) Ω∩B(w,̃r)

(iii) There exists = (n, M, , ) ∈ (0, 1) such that if x, y ∈ B(w, ̃r), then 0 1 ðx − yð v(x) − v(y) ≤ c max v. ð ð r Ω∩B(w,2̃r)

We note that the properties described for the weak solutions to Lv = 0 are essentially the same as the ones described in Section 7.1. We next state a boundary Harnack inequality for the ratio of two positive L-harmonic functions. n n Lemma 7.17. Let Ω ⊂ ℝ be a Lipschitz domain with constants M, r0, and let be an A2(ℝ )-weight with constant . Suppose that u and v are two non-negative weak solutions to L! = 0 in Ω and that u = v = 0 continuously on )Ω ∩ B(w, 2r). Then there exists c = c(n, M, , ), 1 ≤ c ≤ ∞, such that

−1 u(ar(w)) u(x) u(ar(w)) c ≤ ≤ c v(ar(w)) v(x) v(ar(w)) whenever x ∈ Ω ∩ B(w, r∕4).

The proof of this lemma uses Green functions and an elliptic measures associated to L. The existence of those are established in [2] and [3]. We refer to Lemma 3.13 in [14]. We can use this lemma to obtain local Hölder continuity of the ratio of two L-harmonic functions near a portion of the the boundary where they are both vanishing.

64 n n Lemma 7.18. Let Ω ⊂ ℝ be a Lipschitz domain with constants M, r0, and let be an A2(ℝ )-weight with constant . Suppose that v1 and v2 are two non-negative weak solutions to L! = 0 in Ω and that v1 = v2 = 0 continuously on )Ω ∩ B(w, 2r). Then there exist 1 ≤ c ≤ ∞, and = (n, M, , ), such that 0 1 óv1(x1) v1(x2)ó x1 − x2 v1(ar(w)) ó − ó c ð ð ó ó ≤ (68) óv2(x1) v2(x2)ó r v2(ar(w)) whenever x1, x2 ∈ Ω ∩ B(w, r∕16).

Proof (sketch). We will prove the Hölder continuity using an iterative argument. We consider ̃x ∈ )Ω ∩ B(w, r∕8) and deﬁne the functions v (x) v (x) M(s) = M(s, ̃x) = sup 1 , m(s) = m(s, ̃x) = inf 1 , x∈Ω∩B( ̃x,s) v2(x) x∈Ω∩B( ̃x,s) v2(x) for 0 < s < r∕8. We easily see that

̃u1 = v1 − m(s)v2 ≥ 0 ̃u2 = M(s)v2 − v1 ≥ 0 in Ω ∩ B( ̃x,s). Since these functions are L-harmonic we can apply the Harnack inequality from which it follows that they are either positive or identically zero in B( ̃x,s). First we assume that they are strictly positive and apply Lemma 7.17 two times with u = ̃u1, v = v2 the ﬁrst time and u = ̃u2, v = v2 the second time, which gives 2 M(̃s) − m(s) ≤ ̃c (m(̃s) − m(s)) 2 M(s) − m(̃s) ≤ ̃c (M(s) − M(̃s)) where ̂s = s∕4 and ̃c is the constant in Lemma 7.17. We next set osc(t) = osc(t, ̃x) = M(t) − m(t) and add the above equations to obtain 2 osc(s) + osc(̃s) ≤ ̃c (osc(s) − osc(̃s)) so ̃c2 − 1 osc(̃s) ≤ osc(s). ̃c2 + 1 (69)

If ui(s) ≡ 0 for some i we note that this implies that v1 ≡ v2 and osc(s) = 0 for each s which means that (69) still holds. We need to consider three diﬀerent cases. In all the cases we assume that x1, x2 ∈ B(w, r∕16). At ﬁrst we assume that ðx1 − x2ð > r∕1000. This is in some sense a trivial case and it follows from Lemma 7.17 and (69) that (68) holds. If ðx1 − x2ð ≤ r∕1000 and d(x, )Ω) ≥ 2r∕1000 we are on a safe distance from the boundary and therefore do not need (69). Instead it is possible to obtain (68) by using the Harnack inequality and Lemma 7.15 (iii). For the third case we consider ðx1 − x2ð ≤ r∕1000 and d(x, )Ω) < 2r∕1000. We choose w̃ ∈ )Ω ∩ B(w, r∕8) such that d(x, w̃) = d(x1,)Ω). By iterating (69) in an argument similar to the one described in the proof of Theorem 3.1, where we start with s = r∕16 and end with s ≈ 4ðx1 − x2ð, it is possible to conclude that 0 1 óv1(x1) v1(x2)ó x1 − x2 v1(ar(w)) ó − ó osc(2 x − x , ̃x) c ð ð ó ó ≤ ð 1 2ð ≤ óv2(x1) v2(x2)ó r v2(ar(w)) for some such that 0 < < 1.

65 n In order to get rid of the condition that (x) is a global A2-weight, i.e., that (x) ∈ A2(ℝ ), we state the following theorem, which is a localized version of Lemma 7.17 and Lemma 7.18. n Theorem 7.19. Let Ω ⊂ ℝ be a Lipschitz domain with constants M, r0. Let w ∈ )Ω, 0 < r < r0, and assume that is an A2(B(w, 4r))-weight with constant . Furthermore, suppose that v1 and v2 are two non-negative weak solutions to Lv = 0 in Ω ∩ B(w, 2r) and that v1 = 0 = v2 continuously on )Ω ∩ B(w, 2r). Then there exist 1 ≤ c ≤ ∞ and = (n, M, , ) ∈ (0, 1) such that u(a (w)) u(a (w)) −1 r∕c u(x) r∕c c ≤ ≤ c v(ar∕c(w)) v(x) v(ar∕c(w)) and 0 1 u a w ó u(x1) u(x2) ó x1 − x2 ( r∕c( )) ó − ó c ð ð ó ó ≤ óv(x1) v(x2)ó r v(ar∕c(w)) whenever x1, x2 ∈ Ω ∩ B(w, r∕c). The following lemma relates this section to the previous theory about p-harmonic functions vanishing on a 1, portion of the boundary. We note that the setting is now a C -domain which will allow us to use Lemma 7.14 and Lemma 7.13. (This is Lemma 4.9 in [13].) Lemma 7.20. Let Ω ⊂ ℝn be a bounded C1, -domain for some > 0. Given p, 1 0 such that the following is true. Let w ∈ )Ω, 0 < r < ̄r0, and suppose that u is a positive p-harmonic function in Ω ∩ B(w, 4r), u is continuous in Ω̄ ∩ B(w, 4r), and u = 0 on )Ω ∩ B(w, 4r). Then p−2 there exists ̄c = ̄c(p, n, ) ≥ 1, such that if ̂r = r∕ ̄c, then ð∇uð extends to an A2(B(w, 2̂r))-weight with constant depending on p, n, . p−2 Note that in order to show that ð∇uð extends to an A2(B(w, 2̂r))-weight we ﬁrst need to extend ∇u to n ̄ B(w, 2̂r) ⧵ Ω. In order to do so we use a Whitney decomposition of ℝ ⧵ Ω. The Whitney covering lemma n states that given an open set G ⊂ ℝ there exists a decomposition of G into closed dyadic cubes with disjoint interior. Furthermore, the lengths of the sides of the cubes are proportional to the distance from the boundary of G. For more information see [8].

Proof (sketch). This proof is based on estimates of the gradients obtained from Lemma 7.13 and 7.14. There- fore, we begin by choosing ̄r0 = ̄r0(p, n, ) > 0 suﬃciently small to enable the use of these lemmas. We let the family of open cubes {Q(xj, rj)} where Qj ∶= Q(xj, rj) has center xj and sidelength rj be a Whitney n decomposition of ℝ ⧵ Ω such that ̄ n ̄ ∪jQ(xj, rj) = ℝ ⧵ Ω Q(xj, rj) ∩ Q(xi, ri) = ç when i ≠ j −4n −2n 10 d(Qj,)Ω) ≤ rj ≤ 10 d(Qj,)Ω).

Furthermore, if Qj ∩ B(w, r) ≠ ç there exists a wj ∈ Ω ∩ B(w, ̃cr) such that

d(wj, ̃r) ≈ d(wj, x) ≈ d(xj,)Ω) (70) where ̃c = ̃c(n), 1 ≤ ̃c< ∞ and the proportionality constants only depend on n. The existence of wj and ̃c follows from the fact that the distances from the boundary to the centers of the cubes depends on r and n and the fact that Ω is a Lipschitz domain so an open neighbourhood of the boundary can be covered of balls with uniform radius. We want to apply Lemma 7.13 and therefore let ̂r be so small that if Qj ∩ B(w, 8̂r) ≠ ç it holds that u(x) u(x) c−1 ∇u(x) c ∗ d(x, )Ω) ≤ ð ð ≤ ∗ d(x, )Ω) (71)

66 ̄ for x ∈ B(wj, d(wj,)Ω)∕2) and c∗ = in Lemma 7.13. This can be done by setting ̂r = r∕ max{ ̂c, ̃c} where ̂c = ̂c(p, n) is the constant ̂c from Lemma 7.13. We note that this construction will later allow us to use Harnack’s inequality on u(x). We next deﬁne (x) almost everywhere on B(w, 4̂r) ⧵ )Ω as follows: T ∇u(x) p−2, x ∈ Ω ∩ B(w, 4̂r), x ð ð ( ) = p−2 ð∇u(wj)ð , x ∈ Qj ∩ B(w, 4̂r). Note that (x) is not deﬁned in )Ω ∩ B(w, 4̂r) which has Lebesgue measure zero. By an application of Harnack’s inequality to u(x) and (70), (71) it follows that

(x) = (wj) ≈ (z) when x ∈ Qj and z ∈ B(wj, d(wj,)Ω)∕2). (72) In order to prove the lemma, we need to show that

−2n −1 ̃r dx ⋅ dx ≤ (73) ÊB(z,̃r) ÊB(z,̃r) where z ∈ B(w, ̂r) and ̃r ≤ ̂r. As usual we need to split the proof into diﬀerent cases. If ̃r< d(z, )Ω)∕2 we can use Harnack’s inequality in combination with (71) and (72) to obtain (73). For the case where d(z, )Ω)∕2 ≤ ̃r ≤ ̂r we let ŵ ∈ )Ω denote a point such that d(z, )Ω) = d(z, ŵ ). The we see that B(z, )Ω) ⊂ B(w,̂ 3̂r) ⊂ B(w, 8̂r). ∗ and set c = 3̂r∕̃r. We will again need to consider several cases. We note that when p = 2 the statement in (73) follows trivially. We next assume that p > 2, and note that we can use Hölder’s inequality, Lemma 7.5, Lemma 7.7 and Harnack’s inequality to conclude that

dx ≤ dx ÊB(z,̃r) ÊB(w,̂ 3̃r)

p−2 ≤ c ð∇uð dx ÊΩ∩B(w,ĉ ∗ ̃r) 0 1(p−2)∕p p 2n∕p ≤ c ð∇uð dx ̃r ÊΩ∩B(w,ĉ ∗ ̃r) p−2 n+2−p ≤ cu(ãr(ŵ )) ̃r .

We turn to the task of estimating the second integral in (73). In order to do so we note that we can use Lemma −1 7.14 and Harnack’s inequality to see that given " = min{1, ðp − 2ð }∕20 it follows that 0d(x, )Ω)11+" cu(x) u(a (ŵ )) ≥ ̃r ̃r (74) ∗ when x ∈ Ω ∩ B(w,̂ c ̃r). Using (71) and (74) it follows that

−1 −1 dx ≤ dx ÊB(z,̃r) ÊB(w,̂ 3̃r)

p−2 −1 ≤ c (ð∇uð ) dx ÊΩ∩B(w,ĉ ∗ ̃r) 0d(x, )Ω)1p−2 ≤ c dx ÊΩ∩B(w,ĉ ∗ ̃r) u(x)

(1+")(p−2) 2−p −"(p−2) ≤ c̃r u(ãr(ŵ )) d(x, )Ω) dx. ÊΩ∩B(w,ĉ ∗ ̃r)

67 1, Furthermore, using the C -regularity of Ω it is possible to prove that

−"(p−2) n−"(p−2) d(x, )Ω) dx ≤ c̃r ÊΩ∩B(w,ĉ ∗ ̃r) from which it follows that −1 2−p n+p−2 dx ≤ cu(ãr(ŵ )) ̃r ÊB(z,̃r) −1 −2n and (73) follows by combining the estimates for and and multiplying with ̃r . Due to the symmetry of (73) we can repeat the above argument when p < 2 and instead work with 2 − p < p.

7.5 Reduction to linear equation and ﬁnal proof

n 1, In this section we will prove Theorem 7.4. Throughout this section we let Ω ⊂ ℝ be a bounded C -domain with Lipschitz constants M ≥ 2 and r0 > 0. Let w ∈ )Ω and u and v be two positive p-harmonic functions in Ω ∩ B(w, 2r), where 0 < r < r0 such that u and v are continuous in B(w, 2r) and u = v = 0 on B(w, 2r) ⧵ Ω. 1, Since Ω is a C -domain we can use Lemma 7.13 and Lemma 7.20 to assert the validity of the following two assumptions. Assumption 1. There exists 1 ≤ c0 < ∞ such that for ̂r = r∕c0 and for some ≥ 1 it holds that u(x) u(x) v(x) v(x) −1 ∇u(x) , −1 ∇v(x) d(x, )Ω) ≤ ð ð ≤ d(x, )Ω) d(x, )Ω) ≤ ð ð ≤ d(x, )Ω) whenever x ∈ Ω ∩ B(w, 4̂r). ∗ p−2 ∗ Assumption 2. There exists ̃c0 ≥ 1 such that for r = ̂r∕ ̃c0 we have that (ð∇uð+ð∇vð) is an A2(B(w, 4r ))- weight with constant .

We proceed by deﬁning the partial diﬀerential operator

n 0 1 É ) ) L = b (y) )x ij )x (75) i,j=1 i j in Ω ∩ B(w, 4̂r), where 1 ̃ bij(x) = bij(x, t)dt Ê0 b̃ (x, t) = ∇̃u(x, t) p−4((p − 2)() ̃u)(x, t)() ̃u)(x, t) + ∇̃u(x, t) 2) ij ð ð xi xj ijð ð ̃u(x, t) = tu(x) + (1 − t)v(x) for t ∈ [0, 1]. Furthermore, we deﬁne

e(x) = u(x) − v(x) for x ∈ Ω ∩ B(w, 4̂r).

Next, we will use Assumption 1 to prove two important claims which will be used to prove Theorem 7.21. Claim 4. e is a weak solution to Lv = 0 in Ω ∩ B(w, 4̂r).

Proof of claim. We need to show that e(x) satisﬁes the conditions in Deﬁnition 7.1 in Ω ∩ B(w, 4̂r). Clearly 1,p p−2 e(x) ∈ W (B(w, 4̂r), (ð∇uð + ð∇vð) )

68 ' C∞ B w, ̂r , n ⧵ and thus we only need to show that (67) holds. Take ∈ 0 ( ( 4 )). For ∈ ℝ {0} we deﬁne p−2 p−2 () = ðð so that i() = ðð i. We note that É É É b e ' dx = ( b e , … , b e ), ∇' dx. ij xi xj i1 xi in xi Ê i,j Ê i i

Proceeding with the calculations we see that É 1 1 b e = (u − v ) ∇̃u(x, t) p−2dt + (p − 2) ∇̃u(x, t) p−4 ∇u − ∇v, ∇̃u (u + (1 − t)v )dt ij xi xj xj ð ð ð ð ⟨ ⟩ xj xj i Ê0 Ê0 1 ) = ∇̃u p−2(tu + (1 − t)v ) dt ð ð xj xj Ê0 )t 1 ) = (j(∇̃u(x, t))dt Ê0 )t = j(∇u(x, t)) − j(∇v(x, t)) = ∇u p−2u − ∇v p−2v ð ð xj ð ð xj where we have used the fact that 1 p−2 p−2 d p−2 ðð − ðð = ðt + (1 − t)ð (t + (1 − t)) dt Ê0 dt 1 1 p−2 p−4 = ( − ) ðt + (1 − t)()ð dt + (p − 2) ðt + (1 − t)ð ⟨t + (1 − t), − ⟩dt. Ê0 Ê0 It follows that É b v ' dx = ∇u p−2∇u − ∇v p−2∇v, ∇' dx = 0 ij xi xj ⟨ð ð ð ð ⟩ Ê i,j Ê due to the p-harmonicity of u and v so e is indeed a weak solution to Lv = 0.

By using the Cauchy-Schwarz inequality, we obtain n 2 ̃ É 2 ̃ min{p − 1, 1}ðð (x) ≤ bijij ≤ max{p − 1, 1}ðð (x), i,j=1 where 1 ̃ p−2 (x) = ð∇̃u(x, t)ð dt. Ê0 We are now ready to state the second claim. Claim 5. L can be locally reduced to a linear and uniformly elliptic diﬀerential operator.

Proof.

1 ̃ p−2 (x) = ð∇̃u(x, t)ð dt Ê0 p−2 ≈ (ð∇u(x)ð + ð∇v(x)ð) ≈ (u(x)∕d(x, )Ω) + v(x)∕d(x, )Ω))p−2

69 whenever x ∈ Ω∩B(w, 4̂r). Note that we used Assumption 1 in the calculations above. By using the Harnack ̃ inequality on u(x) and v(x) it follows that (x) is (locally) proportional to a constant. Thus, it follows that locally L is a linear and uniformly elliptic diﬀerential operator.

In the setting described in the beginning of this section, we have the following theorem. ∗ Theorem 7.21. Assume that v ≤ u. Then there exists c ≥ 1, c = c(p, n, M, c0, , ̄c0, ) such that if ̃r = r ∕c, then −1 u(ãr(w)) − v(ãr(w)) u(x) − v(x) u(ãr(w)) − v(ãr(w)) c ≤ ≤ c v(ãr(w)) v(x) v(ãr(w)) whenever x ∈ Ω ∩ B(w, r∗).

Proof. We will only prove the ﬁrst inequality. The proof is divided into two parts. First we show the existence ¨ ∗ of a Λ, 1 ≤ Λ < ∞ and ̂c ≥ 1 s.t. if r = r ∕ ̂c and 0 u(x) − v(x) 1 v(x) e(x) = Λ − u(ar∗(w) − v(a∗ (w))) v(ar∗ (w))

∗ ¨ for x ∈ Ω ∩ B(w, r ), then it holds that e(x) ≥ 0 when x ∈ Ω ∩ B(w, 2r ). From this it follows directly that

u(x) − v(x) −1 u(ar∗ (w)) − v(ar∗ (w)) ≥ Λ . v(x) v(ar∗ (w))

¨ After that we set ̃r = 2r , and show the existence of a c = c(p, n, M, c0, , ̃c0, ) such that Λ ≤ c, which implies that u(x) − v(x) −1 u(ar∗ (w)) − v(ar∗ (w)) ≥ c , v(x) v(ar∗ (w)) and in combination with the corresponding right-hand side inequality, the conclusion of the lemma follows. We ﬁrst allow Λ and ̂c to vary but they will be chosen properly along the proof. Let Λu(x) u¨(x) = , u(ar∗ (w) − v(ar∗ (w)) Λv(x) v(x) v¨(x) = + . u(ar∗ (w) − v(ar∗ (w)) v(ar∗ (w))

¨ ¨ ¨ ¨ We note that u (x) and v (x) are both p-harmonic and that e = u − v and we can therefore deﬁne an operator ̃ ∗ ¨ ¨ L as in (75) for x ∈ Ω ∩ B(w, r ) with u = u and v = v . Next we let e1 and e2 denote the two solutions to ̃ ∗ L! = 0 in Ω ∩ B(w, r ) with the continuous boundary values u(x) − v(x) e1ð)(Ω∩B(w,r∗)) = , u(ar∗ (w)) − v(ar∗ (w)) v(x) e2ð)(Ω∩B(w,r∗)) = . v(ar∗(w))

Recall that the existence of e1 and e2 follows from the Wiener criterion in [3]. Since Assumption 2 holds, we ∗ can apply Theorem 7.19 which ensures the existence of c+ ≥ 1 and r+ = r ∕c+ such that

e1(ar (w)) e (x) e1(ar (w)) c−1 + 1 c + + e (a (w)) ≤ e (x) ≤ + e (a (w)) 2 r+ 2 2 r+

70 ¨ for x ∈ Ω ∩ B(w, r+). We next fix the constants so that ̂c = c+, r = r+, and Λ = ̂ce2(ar¨ (w))∕e1(ar¨ (w)). We ¨ define ̃e = Λe1(x) − e2(x) and note that ̃e ≥ 0 whenever x ∈ Ω ∩ B(w, 2r ). Furthermore, it follows from the ̃ ∗ local linearity of L that both ̃e and e are weak solutions to L! = 0 in Ω ∩ B(w, r ). In addition ̃e and e have the same continuous boundary values by construction which means that we can use the maximum principle ̃ for L to conclude that ̃e = e. Thus, we are finished with the first part of the proof. For the second part of the proof we let L denote the operator constructed in (75) such that u(x) − v(x) is a weak solution in Ω∩B(w, 4̂r). By applying the Harnack inequality from Lemma 7.15 (ii) to the weak solution u − v it follows that u(ar∗ (w)) − v(ar∗ (w)) u(x) − v(x) 1 = ≤ c ≤ ce1(x) u(ar∗ (w)) − v(ar∗ (w)) u(ar∗ (w)) − v(ar∗ (w)) ∗ ∗ ∗ ∗ for x ∈ Ω ∩ )B(w, r ) such that d(x, )Ω) > r ∕c for some c . Thus, we can conclude that there exists a ∗ ∗ −1 ∗ ∈ Ω ∩ )B(w, r ) with d(, )Ω) ≥ r ∕c such that e1(x) > c for x ∈ )B(w, r ) ∩ B(, d(, )Ω)∕4). See Figure 3 for an illustration.

Figure 3: Lemma 7.13.

∗ ∗ We let be a weak solution such that L = 0 in Ω∩B(w, r ) with continuous boundary values on )B(w, r )∩ ∗ ∗ B(, d(, )Ω)∕4) such that = 1 on )B(w, r )∩B(, d(, )Ω)∕16) and = 0 on )B(w, r )∩B(, d(, )Ω)∕4)⧵ ∗ B(, d(, )Ω)∕8). Note that this is possible since the points on )B(w, r )∩B(, d(, )Ω)∕4) are regular since they are part of the boundary of an open ball. Furthermore, by an appropriate choice of boundary values, ∗ an application of the maximum principle and the fact that e1c ≥ 1 on )B(w, r ) ∩ B(, d(, )Ω)∕4) we can ∗ ∗ ensure that ≤ ce1 in Ω ∩ B(w, r ). We set ! = 1 − and note that ! is also a weak solution in Ω ∩ B(w, r ) ∗ with !(w0) = 0 on )B(w, r )∩B(, d(, )Ω)∕16). By applying Lemma 7.16 (iii) to ! we see that there exists s such that 0 y − x 1 (1 − (x)) − (1 − (y)) = (y) − (x) c ð ð ⋅ 1 ð ð ð ð ≤ d(, )Ω)∕32 ∗ ∗ whenever x, y ∈ B(w, r ) ∩ B(, d(, )Ω)∕s). Thus, for ̃c suﬃciently large and letting y → w0 ∈ B(w, r ) ∩ B(, d(, )Ω)∕ ̃c) we note that (x) ≥ 1 − c(1∕ ̃c) , so for ̃c large enough we see that −1 ∗ ≥ ̃c in B(w, r ) ∩ B(, d(, )Ω)∕ ̃c).

71 By combining this with the fact that ≤ ce1 it follows that −1 −1 −1 ∗ e1 ≥ c ≥ c ̃c in B(w, r ) ∩ B(, d(, )Ω)∕ ̃c) −1 and by applying the Harnack inequality to e1 it follows that e1(ar¨ (w)) ≥ ̄c . Furthermore, we can apply the maximum principle, Harnack’s inequality to v, and Lemma 7.7 to see that v(x) sup ∗ x∈)(Ω∩B(w,r )) ∗ e2(x) ≤ ≤ c(p, n, M, r ) v(ar∗ (w)) and thus we conclude that Λ < ̄c holds for some ̄c = ̄c(p, n, M, c0, , ̃c0, ). This concludes the proof of the left-hand inequality.

Theorem 7.22. Assume that v ≤ u. Then there exist c ≥ 1 such that c = c(p, n, M, c0, , ̃c0, ), and = ∗ (p, n, M, c0, , ̃c0, ) where ∈ (0, 1) such that if ̃r = r ∕c, then 0 1 ó u(x1) u(x2) ó u(ãr(w)) x1 − x2 ó − ó c ð ð ó ó ≤ óv(x1) v(x2)ó v(ãr(w)) r whenever x1, x2 ∈ Ω ∩ B(w, ̃r).

Proof (sketch). As usual when we want to prove Hölder continuity from a Harnack type inequality, we use an iterative approach, and thus this proof is similar to the proof of Lemma 7.18. We start by an application of Theorem 7.21 from which it follows that

−1 u(x) −1 u(ãr(w)) u(x) c + 1 − c ≤ ≤ (1 − c) + c v(x) v(ãr(w)) v(x) whenever x ∈ B(w, ̃r), and since c ≥ 1 we see that

u(x1) u(x2) ≤ c whenever x1, x2 ∈ Ω ∩ B(w, ̃r). (76) v(x1) v(x2)

For w̃ ∈ )Ω ∩ B(w, ̃r∕8) and 0 < < ̃r we deﬁne u u M() = sup , m() = inf . Ω∩B(w,̃ ) v Ω∩B(w,̃ ) v

We recall that p-harmonicity is preserved for addition and multiplication by a constant. Thus, while keeping ∗ ﬁxed we can use Theorem 7.21 with u = u and v = m()v to assert the existence of constants c and c∗ such that M( ̃) − m() ≤ c∗(m( ̃) − m()) (77) ∗ where ̃ = ∕c . By another application of the theorem but this time with u = M()v and v = u we obtain

(M()v − u)∕u ≈ constant on Ω ∩ B(w, ̃) where the constant depends on many variables. By combining this estimate with (76) it follows that

(M()v − u)∕v ≈ constant on Ω ∩ B(w, ̃) and thus we can conclude that M() − m( ̃) ≤ ̃c∗(M() − M( ̃)). (78)

72 We next set ̃c = max(c∗, ̃c∗) and by replacing the occurrences of c∗ and ̃c∗ with ̃c and adding (77) and (78), we see that c∗ − 1 osc( ̃) ≤ osc(), (79) c∗ + 1 where osc(t) ∶= M(t) − m(t). By iterating (79) it follows that osc(s) ≤ c(s∕t) osc(t) whenever 0 < s < t ≤ r∕2 (80) for some > 0, c ≥ 1. Similar to the proof of Lemma 7.18 we can use (76), (79), (80), the interior Hölder continuity and Harnack’s inequality for p-harmonic functions to complete the proof.

We are ﬁnally ready to prove Theorem 7.4.

Proof of Theorem 7.4. We let ̂u,̂v be the p-harmonic functions in Ω ∩ B(w, 2r) with

̂u = max{u, v} and ̂v = min{u, v} on )[Ω ∩ B(w, 2r)]

It follows directly from the maximum principle that ̂u ≥ ̂v. Moreover, both ̂u and ̂v satisfy Assumption 1 and Assumption 2, and thus we can apply Theorem 7.21 to obtain a boundary Harnack inequality for the ratio of ̂u and ̂v so −1 ̂u(ãr(w)) ̂u(x) ̂u(ãr(w)) c ≤ ≤ c (81) ̂v(ãr(w)) ̂v(x) ̂v(ãr(w)) for x ∈ Ω ∩ B(w, ̃r). Furthermore, by another application of the maximum principle we conclude that

̂v ≤ v ≤ ̂u, ̂v ≤ u ≤ ̂u and thus it holds that u ̂u ̂v u ̂u = ⋅ ⋅ , v ̂v v ̂u ≤ ̂v ̂u u ̂u v u = ⋅ ⋅ . ̂v v u ̂v ≤ v

By applying (81) we note that −1 ̂u(ãr(w)) u(x) ̂u(ãr(w)) c ≤ ≤ c . ̂v(ãr(w)) v(x) ̂v(ãr(w)) Thus, we have shown that

u(x1) u(x2) ≤ c whenever x1, x2 ∈ Ω ∩ B(w, ̃r). v(x1) v(x2)

We may now ﬁnish the proof of Theorem 7.4 by using the same technique as in Theorem 7.22.

Appendix

A Proofs of basic properties

We ﬁrst prove the rotation invariance of the p-Laplace equation.

73 Lemma A.1. The p-Laplace equation is rotationally invariant.

Proof. Let O = (oij) be an orthogonal n × n-matrix, that is, for k, l ∈ {1, … , n},

n T É 1 k = l, o o = = if ki li kl i=1 0 if k ≠ l.

Let u be a solution to the p-Laplace equation and deﬁne v(x) = u(Ox).

We need to show that Δpv(x) = 0. We use the notation ) D v(x) = v(x) = v i xi )xi and vk = Dku(Ox). Then, n n ) É É v = u(Ox) = D u(Ox)o = v o xi )x k ki k ki i k=1 i=1 and n n )2 É É v = u(Ox) = v o o . xixj )x )x kl ki lj j i l=1 k=1

Clearly, n n n n n n É É É É É É v = v o o = v o o = v = Δ u(Ox), xixi kl ki li kl ki li kl kl 2 i=1 i=1 k,l=1 k,l=1 i=1 k,l=1 and n n H n I É É É v v v = v v v o o xi xj xixj xi xj kl ki lj i,j i,j=1 k,l=1 n H n IH n IH n I É É É É = vmomi vsosj vklokiolj i,j=1 m=1 s=1 k,l=1 n É É = vmomivsvkloki osjolj i,m,s,l,k j=1 n n É É É = vmomivlvkloki = vmvlvkl omioki i,m,k,l m,k,l=1 i=1 n É = vkvlvkl = Δ∞u(Ox). k,l=1

p−2 p−4 2 Hence, ∇ ⋅ (ð∇vð ∇v) = ð∇u(Ox)ð ð∇u(Ox)ð Δ2u(Ox) + (p − 2)Δ∞u(Ox) , and the lemma is proven.

74 p−n Next, we want to show that the fundamental solution of the p-Laplace equation is (n − p)ðxð p−1 if p < n and − log ðxð if p = n. Because of the rotation invariance, we search for a radial solution. Suppose that u(x) = v(r), r = ðxð, is a solution, then )r x x = i , u = v¨(r) i , xi )xi r r and hence p−2 H n I v¨(r)2 É 2 ∇u p−2 = x2 = v¨(r) p−2, ð ð 2 i ð ð r i=1 v¨(r) p−2v¨(r) ∇u p−2∇u = ð ð x ð ð r

Furthermore, 0 ¨ p−2 ¨ 1 ) ðv (r)ð v (r) xi )xi r 4 5 ¨ 4 ¨ 5 ) ¨ p−2 v (r) ¨ p−2 ) v (r) = ðv (r)ð xi + ðv (r)ð xi )xi r )xi r H 2 H 2 II v¨¨(r) x 1 x = (p − 2) v¨(r) p−2 x2 + v¨(r) p−2 v¨¨(r) i + v¨(r) − i , ð ð r2 i ð ð r2 r r3 which gives n − 1 ( ∇u p−2∇u) = (p − 1) v¨(r) p−2v¨¨(r) + v¨(r) p−2v¨(r) = 0. div ð ð ð ð r ð ð ¨ Assuming that ðv (r)ð ≠ 0, we have n − 1 (p − 1)v¨¨(r) + v¨(r) = 0, r i.e., that v¨¨(r) 1 − n = . v¨(r) (p − 1)r (82)

If p = n, then v¨¨(r) 1 = − , v¨(r) r

p−n and thus v(r) = − log r. If p < n, then (82) is solved by v(r) = (n − p)r p−1 .

B Some useful inequalities

Lemma B.1. Let a, b ∈ ℝn.

75 (i) If p ≥ 2, then 1 b p−2b − a p−2a, b − a ( b p−2 + a p−2) b − a 2 22−p b − a , ⟨ð ð ð ð ⟩ ≥ 2 ð ð ð ð ð ð ≥ ð ð

(ii) if p ≤ 2, then 1 b p−2b − a p−2a, b − a ( b p−2 + a p−2) b − a 2, ⟨ð ð ð ð ⟩ ≤ 2 ð ð ð ð ð ð

(iii) if p ≤ 2, then p−2 p−2 p ⟨ðbð b − ðað a, b − a⟩ ≤ (p)ðb − að for some constant (p),

(iv) if p ≥ 2, then

p−2 p−2 2 2 ó ó p p−2 p−2 ó b 2 b − a 2 aó b b − a a, b − a , óð ð ð ð ó ≤ ⟨ð ð ð ð ⟩ ó ó 4

(v) if p ≥ 2, then

p−2 p−2 p−2 p−2 p−2 p−2 ó ó ó b b − a aó (p − 1) b 2 + a 2 ó b 2 b − a 2 aó , óð ð ð ð ó ≤ ð ð ð ð óð ð ð ð ó ó ó ó ó

(vi) if 1 ≤ p ≤ 2, then

p−2 p−2 p−2 2 2 2 ⟨ðbð b − ðað a, b − a⟩ ≥ (p − 1)ðb − að (1 + ðað + ðbð ) 2 ,

(vii) and if p ≥ 2, then p p p−2 p ðbð ≥ ðað + p⟨ðað a, b − a⟩ + C(p)ðb − að ,

(viii) for p ≥ 1 p p p−2 ðbð ≥ ðað + p⟨ðað a, b − a⟩.

p Proof. The proof of this (i)-(vii) is contained in [21]. We only prove (viii). Since ðxð is a convex function n we note that for a, b ∈ ℝ and t ∈ (0, 1) p p p p ða + t(b − a)ð = ð(1 − t)a + tbð ≤ (1 − t)ðað + tðbð p We set c = (b − a) and subtract ðað which yields p p p p ða + tcð − ðað ≤ t(ðbð − ðað ) p−2 and by subtracting tpðað ⟨a, c⟩ and dividing by t we obtain a + tc p − a p ð ð ð ð − p a p−2 a, c b p − a p − p a p−2 a, c t ð ð ⟨ ⟩ ≤ ð ð ð ð ð ð ⟨ ⟩

By letting t tend to 0 we see that the left hand side also tends to 0 from which (viii) follows immediately.

76 References

[1] L. C. Evans, Partial Differential Equations, AMS, 1998. [2] E. Fabes, D. Jerison and C. Kenig, The Wiener test for degenerate elliptic equations, Ann. Inst. Fourier (Grenoble) 32 (1982), 151-182. [3] E. Fabes, D. Jerison and C. Kenig, Boundary behaviour of solutions to degenerate elliptic equations, Conference on harmonic analysis in honor of Antonio Zygmund, Vol I, II Chicago, Ill, 1981, 577-589, Wadsworth Math. Ser, Wadsworth Belmont CA, 1983. [4] E. Fabes, C. Kenig, and R. Serapioni, The local regularity of solutions to degenerate elliptic equations, Comm. Partial Differential Equations 7 (1982), 77-116. [5] J. B. Garnett and D. E. Marshall, Harmonic Measure. New Mathematical Monographs Series (No. 2), Cambridge University Press, New York, 2005. [6] D. Gilbarg and N. S. Trudinger, Elliptic partial differential equations of second order, second ed., Die Grundlehren der Mathematischen Wissenschaften, no. 224, Springer-Verlag, Berlin, 1983. [7] E. Giusto, Direct Methods in the Calculus of Variations, World Scientific Publ. Co., River Edge 2003. [8] L. Grafakos, Classical Fourier Analysis, second edition, Springer-Verlag, 2008. [9] J. Heinonen, T. Kilpeläinen, and O. Martio, Nonlinear potential theory of degenerate elliptic equations, Clarendon Press, Oxford New York Tokyo, 1993. [10] E. Hewitt and K. Stromberg Real and abstract analysis, Springer-Verlag, Berlin, 1965. [11] O. A. Ladyzhenskaya, N. N. Uraltseva. Linear and quasilinear elliptic equations. Academic Press, New-York, 1968. [12] J. Lewis, Note on p-harmonic measure, Comput. Methods Funct. Theory 6 (1) (2006) 109-144. [13] J. Lewis, N.L.P. Lundström and K. Nyström, Boundary Harnack Inequalities for Operators of p-Laplce type in Reifenberg Flat Domains, in Perspectives in PDE, Harmonic analysis and Applications, Proceed- ings of Symposia in Pure Mathematics 79 (2008), 229-266. [14] J. Lewis and K. Nyström, Boundary Behaviour for p-Harmonic Functions in Lipschitz and Starlike Lipschitz Ring Domains, Annales Scientifiques de L’Ecole Normale Superieure, 40 (2007), 765-813. [15] J. Lewis and K. Nyström, Boundary Behaviour and the Martin Boundary Problem for p-harmonic Func- tions in Lipschitz domains, Annals of Mathematics, 172 (2010), 1907-1948. [16] J. Lewis and K. Nyström, Regularity and Free Boundary Regularity for the p-Laplace in Lipschitz and 1 C -domains, Annales Acad. Sci. Fenn. Mathematica, 33 (2008), 523-548. [17] J. Lewis and K. Nyström, Boundary Behaviour of p-harmonic Functions in Domains Beyond Lipschitz Domains, Advances in the Calculus of Variations, 1 (2008), 133-177. [18] J. Lewis and K. Nyström, Regularity of Lipschitz Free boundaries in Two-phase Problems for the p- Laplace Operator, Advances in Mathematics, 225 (2010), 2565-2597. [19] J. Lewis and K. Nyström, Regularity of Flat Free Boundaries in Two-phase Problems for the p-Laplace Operator, Annales de l’Institute Henri Poincare Non Linear Analysis, 29 (2012), 83-108.

77 [20] J. Lewis and K. Nyström, Regularity and Free Boundary Regularity for the p-Laplace Operator in Reifenberg Flat and Ahlfors Regular Domains, Journal of the American Mathematics Society 25 (2012), 827-862. [21] P. Lindqvist, Notes on the p-Laplace equation, University of Jyväskylä, Report 102 (2006). [22] J. Michel and P.Ziemer, Interior regularity for solutions to obstacle problems, Nonlinear analysis 10 (1986), 1427-1448. [23] K. Nyström, Boundary estimates for p-harmonic functions in ﬂat domains - less preliminary notes for Jyväskylä 2012, unpublished lecture notes, 2012. [24] B. Øksendal, Stochastic Diﬀerential Equations: An Introduction with Applications. 6th ed. Berlin- Heidelberg Springer-Verlag, 2003.