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Mathematische Annalen

, Volume 375, Issue 3–4, pp 1721–1743 | Cite as

Extremal functions for Morrey’s inequality in convex domains

  • Ryan Hynd
  • Erik LindgrenEmail author
Open Access
Article
  • 495 Downloads

Abstract

For a bounded domain \(\Omega \subset \mathbb {R}^n\) and \(p>n\), Morrey’s inequality implies that there is \(c>0\) such that
$$\begin{aligned} c\left\| u \right\| ^p_{\infty }\le \int _\Omega \left| Du\right| ^pdx \end{aligned}$$
for each u belonging to the Sobolev space \(W^{1,p}_0(\Omega )\). We show that the ratio of any two extremal functions is constant provided that \(\Omega \) is convex. We also show with concrete examples why this property fails to hold in general and verify that convexity is not a necessary condition for a domain to have this feature. As a by product, we obtain the uniqueness of an optimization problem involving the Green’s function for the p-Laplacian.

Mathematics Subject Classification

35J60 35J70 35P30 39B62 

1 Introduction

Suppose \(\Omega \subset \mathbb {R}^n\) is a bounded domain and \(p>n\). Morrey’s inequality for \(W^{1,p}_0(\Omega )\) functions u may be expressed as
$$\begin{aligned} c\left\| u\right\| ^p_{C^{1-\frac{n}{p}}(\overline{\Omega })}\le \int _\Omega \left| Du\right| ^pdx, \end{aligned}$$
where \(c>0\) is a constant that is independent of u. In particular,
$$\begin{aligned} c\left\| u\right\| ^p_{\infty }\le \int _\Omega \left| Du\right| ^pdx. \end{aligned}$$
(1.1)
Let us defineObserve that
$$\begin{aligned} \lambda _p\left\| u\right\| ^p_{\infty }\le \int _\Omega \left| Du\right| ^pdx \end{aligned}$$
(1.2)
and that \(c=\lambda _p\) is the largest constant such that (1.1) is valid. Furthermore, if there is a function \(u \in W^{1,p}_0(\Omega ){\setminus }\{0\}\) such that equality holds in (1.1), then \(c=\lambda _p\).

Definition

A function \(u\in W^{1,p}_0(\Omega ){\setminus }\{0\}\) is an extremal if equality holds in (1.2).

It is plain to see that any multiple of an extremal is also an extremal. Using routine compactness arguments, it is not difficult to verify that extremal functions exist. We will argue below that any extremal u satisfies the boundary value problem
$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _pu=\lambda _p \left| u(x_0)\right| ^{p-2}u(x_0)\delta _{x_0}\quad &{}x\in \Omega \\ \qquad u=0 \quad &{}x\in \partial \Omega , \end{array}\right. } \end{aligned}$$
(1.3)
which was derived by Ercole and Pereira in [9]. Here \(\Delta _p\psi :=\text {div}(|D\psi |^{p-2}D\psi )\) is the p-Laplacian, and \(x_0\) is the unique point for which \(\left| u\right| \) is maximized in \(\Omega \). Moreover, using (1.3) we will be able to conclude that any extremal has a definite sign in \(\Omega \). And as the PDE in (1.3) is homogeneous, the optimal constant \(\lambda _p\) can be interpreted as being an eigenvalue.

The primary goal of this work is to address the extent to which extremal functions can be different. In particular, we would like to know if any two extremal functions are necessarily multiples of one another. If they are, we consider the set of extremals to be uniquely determined. For once one extremal is found, all others can be obtained by scaling. We will argue that annuli never have this uniqueness property. We will also exhibit star-shaped domains for which this uniqueness property fails. However, we will see that if a planar domain has certain symmetry, then its extremals are one dimensional.

Our main result is that convex domains always have the aforementioned uniqueness property.

Theorem 1.1

Assume that \(\Omega \subset \mathbb {R}^n\) is open, convex and bounded. If u and v are extremal, then u / v is constant throughout \(\Omega \).

We will also explain how Theorem 1.1 implies the following corollary involving the Green’s function of the p-Laplacian in \(\Omega \).

Corollary 1.2

Assume that \(\Omega \subset \mathbb {R}^n\) is open, convex and bounded. Suppose that \(G(\cdot ,y)\) is the Green’s function of the p-Laplacian in \(\Omega \) with pole \(y\in \Omega \); that is, \(G(\cdot ,y)\) satisfies
$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta _p w=\delta _{y}\quad &{}x\in \Omega \\ \quad w=0\quad &{}x\in \partial \Omega . \end{array}\right. } \end{aligned}$$
(1.4)
Then \(x_0\) given in Eq. (1.3) is the unique point in \(\Omega \) for which
$$\begin{aligned} G\left( x_0,x_0\right) =\max \left\{ G(y,y): y\in \Omega \right\} . \end{aligned}$$
Part of our motivation was to extend a previous result of Talenti. He considered extremal functions for the following inequality, which is also due to Morrey. For each weakly differentiable function \(u:\mathbb {R}^n\rightarrow \mathbb {R}\),
$$\begin{aligned} \left\| u\right\| ^p_{L^\infty (\mathbb {R}^n)} \le c(n,p)\left| \text {supp}(u)\right| ^{\frac{p}{n}-1}\int _{\mathbb {R}^n}\left| Du\right| ^pdx. \end{aligned}$$
(1.5)
Here c(np) is an explicit constant depending only on p and n, and \(|\text {supp}(u)|\) is the Lebesgue measure of the support of u. Employing Schwarz symmetrization, Talenti showed in [22] that if equality holds in (1.5) there are \(a\in \mathbb {R}\), \(r>0\) and \(x_0\in \mathbb {R}^n\) such that
$$\begin{aligned} u(x)= {\left\{ \begin{array}{ll} a\left( r^{\frac{p-n}{p-1}}-\left| x-x_0\right| ^{\frac{p-n}{p-1}}\right) , \quad &{} \left| x-x_0\right| < r\\ 0, &{} \left| x-x_0\right| \ge r. \end{array}\right. } \end{aligned}$$
We remark that a quantitative version of this result has been established by Cianchi [5], and we refer the reader to [4, 8, 21] for work on sharp constants of related inequalities.

Unfortunately, \(\mathbb {R}^n\) and balls are the only known domains for which the extremals have such convenient characterizations. Nevertheless, in this paper, we believe that we have taken significant steps in understanding precisely which domains have a one dimensional collection of extremals. In Sect. 2, we will derive basic properties of solutions of (1.3), and in Sect. 3, we consider the support function of an extremal. In Sect. 5, we will provide examples of domains for which uniqueness fails; these include annuli, bow tie and dumbbell shaped planar domains. In Sect. 4, we verify Theorem 1.1, and in Sect. 6, we use Steiner symmetrization to exhibit some nonconvex planar domains that have unique extremals.

2 Properties of extremals

We now proceed to deriving some properties of extremal functions. These properties will be crucial to our uniqueness study. First, we verify that extremal functions satisfy the boundary value problem (1.3). Then we will study the behavior of solutions of (1.3) near their global maximum or minimum points. We also refer the reader to the recent paper [9] by Ercole and Pereira, where they studied properties of extremal functions in a wide class of inequalities that include (1.1). In particular, they obtained analogous results to Corollaries 2.2 and 2.4 below.

Lemma 2.1

A function \(u\in W_0^{1,p}(\Omega )\) is extremal if and only if
$$\begin{aligned} \int _\Omega \left| Du\right| ^{p-2}Du\cdot D\phi dx=\lambda _p\max \left\{ \left| u(x)\right| ^{p-2}u(x)\phi (x): x\in \overline{\Omega }, \left| u(x)\right| =\left\| u\right\| _\infty \right\} \end{aligned}$$
(2.1)
for all \(\phi \in W_0^{1,p}(\Omega )\).

Proof

  1. 1.
    First let us establish the following identity
    $$\begin{aligned}&\lim _{\epsilon \rightarrow 0^+}\frac{\left\| u+\epsilon \phi \right\| ^p_\infty -\left\| u\right\| ^p_\infty }{\epsilon }\nonumber \\&\quad =p\max \left\{ \left| u(x)\right| ^{p-2}u(x)\phi (x): x\in \overline{\Omega }, \left| u(x)\right| =\left\| u\right\| _\infty \right\} , \end{aligned}$$
    (2.2)
    for \(u,\phi \in C(\overline{\Omega })\). For any \(x_0\) such that \(\left| u(x_0)\right| =\left\| u\right\| _\infty \),
    $$\begin{aligned} \frac{1}{p}\left\| u+\epsilon \phi \right\| _\infty ^p&\ge \frac{1}{p}\left| u(x_0)+\epsilon \phi (x_0)\right| ^p\\&\ge \frac{1}{p}\left| u(x_0)\right| ^p+\epsilon \left| u(x_0)\right| ^{p-2}u(x_0)\phi (x_0)\\&= \frac{1}{p}\left\| u\right\| _\infty ^p+\epsilon \left| u(x_0)\right| ^{p-2}u(x_0)\phi (x_0). \end{aligned}$$
    Therefore,
    $$\begin{aligned} \liminf _{\epsilon \rightarrow 0^+}\frac{\left\| u+\epsilon \phi \right\| ^p_\infty -\left\| u\right\| ^p_\infty }{\epsilon }\ge p\left| u(x_0)\right| ^{p-2}u(x_0)\phi (x_0), \end{aligned}$$
    and so “\(\ge \)” holds in (2.2). Now choose a sequence of positive numbers \((\epsilon _j)_{j\in \mathbb {N}}\) tending to 0 such that
    $$\begin{aligned} \limsup _{\epsilon \rightarrow 0^+}\frac{\left\| u+\epsilon \phi \right\| _\infty ^p-\left\| u\right\| _\infty ^p}{\epsilon }=\limsup _{j\rightarrow \infty }\frac{\left\| u+\epsilon _j\phi \right\| _\infty ^p-\left\| u\right\| _\infty ^p}{\epsilon _j}, \end{aligned}$$
    and select a sequence \((x_{j})_{j\in \mathbb {N}}\) maximizing \(|u+\epsilon _j \phi |\) that converges to a maximizer \(x_0\) of \(\left| u\right| \). Such sequences exist by the continuity of u and \(\phi \), the compactness of \(\overline{\Omega }\), and the inequalities \(|u(x)+\epsilon _j \phi (x)|\le \Vert u+\epsilon _j \phi \Vert _\infty =|u(x_j)+\epsilon _j\phi (x_j)|\). As \(\mathbb {R}\ni z\mapsto \frac{1}{p}|z|^p\) is continuously differentiable,
    $$\begin{aligned} \limsup _{j\rightarrow \infty }\frac{\left\| u+\epsilon _j\phi \right\| _\infty ^p-\left\| u\right\| _\infty ^p}{\epsilon _j}&\le \limsup _{j\rightarrow \infty }\frac{\left| u(x_{j})+\epsilon _j\phi (x_{j})\right| ^p-\left| u(x_{j})\right| ^p}{\epsilon _j}\\&=p\left| u(x_0)\right| ^{p-2}u(x_0)\phi (x_0)\\&\le p\max \{\left| u(x)\right| ^{p-2}u(x)\phi (x) : \left| u(x)\right| =\left\| u\right\| _\infty \}. \end{aligned}$$
    We conclude “\(\le \)” in (2.2).
     
  2. 2.
    Any extremal \(u\in W^{1,p}_0(\Omega ){\setminus }\{0\}\) satisfies
    $$\begin{aligned} \lambda _p=\frac{\displaystyle \int _\Omega \left| Du\right| ^pdx}{\left\| u\right\| _\infty ^p}\le \frac{\displaystyle \int _\Omega \left| Du+\epsilon D\phi \right| ^pdx}{\left\| u+\epsilon \phi \right\| _\infty ^p}, \end{aligned}$$
    for each \(\phi \in W^{1,p}_0(\Omega )\) and \(\epsilon >0\) sufficiently small. Exploiting (2.2)
    $$\begin{aligned} 0&\le \lim _{\epsilon \rightarrow 0^+}\frac{1}{p\epsilon }\left( \frac{\displaystyle \int _\Omega \left| Du+\epsilon D\phi \right| ^pdx}{\left\| u+\epsilon \phi \right\| _\infty ^p}-\frac{\displaystyle \int _\Omega \left| Du\right| ^pdx}{\left\| u\right\| _\infty ^p}\right) \\&= \frac{\displaystyle \int _\Omega \left| Du\right| ^{p-2}Du\cdot D\phi dx}{\left\| u \right\| _\infty ^p}\\&\quad -\frac{\displaystyle \int _\Omega \left| Du\right| ^pdx}{\left\| u \right\| _\infty ^{2p}}\max \{|u(x)|^{p-2}u(x)\phi (x): x\in \overline{\Omega }, |u(x)|=\left\| u \right\| _\infty \}\\&=\frac{1}{\left\| u \right\| _\infty ^p}\left( \int _\Omega \left| Du\right| ^{p-2}Du\cdot D\phi dx\right. \\&\quad \left. -\lambda _p\max \{|u(x)|^{p-2}u(x)\phi (x): x\in \overline{\Omega }, |u(x)|=\left\| u \right\| _\infty \}\right) . \end{aligned}$$
    Canceling the factor \(1/\left\| u \right\| _\infty ^p\) and replacing \(\phi \) with \(-\phi \) gives (2.1).
     
  3. 3.

    Of course if (2.1) holds, we can choose \(\phi =u\) to verify that u is extremal.\(\square \)

     

Corollary 2.2

Each extremal function is everywhere positive or everywhere negative in \(\Omega \).

Proof

Assume that \(u\in W^{1,p}_0(\Omega ){\setminus }\{0\}\) is extremal. Then \(w:=\left| u\right| \) is extremal, as well. Moreover, (2.1) implies
$$\begin{aligned} \int _\Omega \left| Dw\right| ^{p-2}Dw\cdot D\phi dx\ge 0 \end{aligned}$$
for all \(\phi \ge 0\). Therefore, w is p-superharmonic, \(w\ge 0\) and \(w|_{\partial \Omega }=0\). Since w doesn’t vanish identically, \(w=\left| u\right| >0\) (Theorem 11.1 in [18]). Hence, u doesn’t vanish in \(\Omega \) and so u has a definite sign in \(\Omega \). \(\square \)

Observe that the left hand side of (2.1) is linear in \(\phi \), while the right hand side appears to be nonlinear in \(\phi \). We will argue that this forces the set \(\{x\in \overline{\Omega }: |u(x)|=\left\| u \right\| _\infty \}\) to be a singleton for any extremal function.

Proposition 2.3

Assume u is an extremal function. Then \(\{x\in \Omega : |u(x)|=\left\| u \right\| _\infty \}\) is a singleton.

Proof

Without any loss of generality, we may assume \(u>0\) and \(\left\| u \right\| _\infty =1\). In view of (2.1),
$$\begin{aligned} \max _{\left\{ u=1\right\} } \left\{ \phi _1+\phi _2\right\} =\max _{\left\{ u=1\right\} }\phi _1+\max _{\left\{ u=1\right\} }\phi _2, \end{aligned}$$
(2.3)
for any two \(\phi _1,\phi _2\in C^\infty _c(\Omega )\). Suppose that there are distinct points \(x_1\) and \(x_2\) for which \(u(x_1)=u(x_2)=1\). In this case, there are balls \(B_\delta (x_1), B_\delta (x_2)\subset \Omega \) that are disjoint for some \(\delta >0\) small enough. We choose functions \(\phi _1\in C^\infty _c(\Omega )\), \(\phi _2\in C^\infty _c(\Omega )\) that are nonnegative, have maximum value 1, and are supported in \(B_\delta (x_1)\) and \(B_\delta (x_2)\), respectively. It follows that
$$\begin{aligned} \max _{\left\{ u=1\right\} }\left\{ \phi _1+\phi _2\right\} =1, \end{aligned}$$
while \(\max _{\{u=1\}}\phi _1=1\) and \(\max _{\{u=1\}}\phi _2=1\). This contradicts (2.3). \(\square \)

Corollary 2.4

Assume u is an extremal function. Then \(\left| u\right| \) attains its maximum value uniquely at some \(x_0\in \Omega \). Moreover,
$$\begin{aligned} \int _\Omega \left| Du\right| ^{p-2}Du\cdot D\phi dx=\lambda _p |u(x_0)|^{p-2}u(x_0)\phi (x_0) \end{aligned}$$
for each \(\phi \in W^{1,p}_0(\Omega )\). In particular, u is a weak solution of (1.3).

We note that any solution u of (1.3) is differentiable with a locally Hölder continuous gradient in \(\Omega {\setminus }\{x_0\}\), see [10, 15, 23]. However, we show below that u is not differentiable at \(x_0\).

Example 2.5

As we noted above, when \(\Omega =B_r(x_0)\), we have an explicit extremal function
$$\begin{aligned} u(x)=a\left( r^{\frac{p-n}{p-1}}-\left| x-x_0\right| ^{\frac{p-n}{p-1}}\right) , \quad x\in B_r(x_0) \end{aligned}$$
(2.4)
for each \(a\in \mathbb {R}\) (See Fig. 1). Moreover, any extremal is of the form (2.4) for some \(a\in \mathbb {R}\); in particular, any ball has a one dimensional collection of extremal functions. The corresponding optimal constant in (1.2) is
$$\begin{aligned} \lambda _p=\displaystyle \frac{\displaystyle \int _{B_r(x_0)}\left| Du\right| ^pdx}{\displaystyle \left\| u\right\| ^p_\infty } =\left( \frac{p-n}{p-1}\right) ^{p-1}r^{n-p}n\omega _n. \end{aligned}$$
Fig. 1

Example 2.5 with \(a=1\), \(p=4\), \(n=2\), \(r=1\) and \(x_0=0\)

We can use the extremals for balls (2.4) to study the behavior of general extremals near the points which maximize their absolute values. Note in particular, that the family of extremals (2.4) are Hölder continuous with exponent \(\frac{p-n}{p-1}\in (0,1]\), which is a slight improvement of the exponent \(\frac{p-n}{p}\) one has from the Sobolev embedding \(W^{1,p}_0(\Omega )\subset C^{1-\frac{n}{p}}(\overline{\Omega })\). We will first argue that solutions of (1.3), and in particular extremals, have exactly this type of continuity at their maximizing or minimizing points.

Proposition 2.6

Assume u is a solution of (1.3) and that \(B_r(x_0)\subset \Omega \subset B_R(x_0)\). Then
$$\begin{aligned} \frac{\left\| u \right\| _\infty }{R^{\frac{p-n}{p-1}}}\left| x-x_0\right| ^\frac{p-n}{p-1}\le |u(x)-u(x_0)|\le \frac{\left\| u \right\| _\infty }{r^{\frac{p-n}{p-1}}}\left| x-x_0\right| ^\frac{p-n}{p-1} \end{aligned}$$
for each \(x\in \Omega \).

Proof

Without loss of generality we may assume that \(u>0\) and \(u(x_0)=1\). Define
$$\begin{aligned} v(x)&:=\frac{1}{r^{\frac{p-n}{p-1}}}\left( r^{\frac{p-n}{p-1}}-\left| x-x_0\right| ^{\frac{p-n}{p-1}}\right) \\&=1-\frac{1}{r^{\frac{p-n}{p-1}}}\left| x-x_0\right| ^{\frac{p-n}{p-1}} \end{aligned}$$
for \(x\in B_r(x_0)\). Observe that u and v are p-harmonic in \(B_r(x_0){\setminus }\{x_0\}\), \(u(x_0)=v(x_0)\) and \(u\ge v\) on \(\partial B_r(x_0)\). By weak comparison, \(u\ge v\) in \(B_r(x_0)\). That is
$$\begin{aligned} u\left( x\right) \ge 1-\frac{1}{r^{\frac{p-n}{p-1}}}\left| x-x_0\right| ^{\frac{p-n}{p-1}}, \quad x\in B_r(x_0). \end{aligned}$$
Since \(v(x)\le 0\) for \(x\not \in B_r(x_0)\), the above inequality trivially holds for \(x\in \Omega {\setminus } B_r(x_0)\).
Now set
$$\begin{aligned} w(x)&:=\frac{1}{R^{\frac{p-n}{p-1}}}\left( R^{\frac{p-n}{p-1}}-\left| x-x_0\right| ^{\frac{p-n}{p-1}}\right) \\&=1-\frac{1}{R^{\frac{p-n}{p-1}}}\left| x-x_0\right| ^{\frac{p-n}{p-1}}. \end{aligned}$$
Observe that u and w are p-harmonic in \(\Omega {\setminus }\{x_0\}\), \(u(x_0)=w(x_0)\) and \(u\le w\) on \(\partial \Omega \) as \(\Omega \subset B_R(x_0)\). By weak comparison, \(u\le w\) in \(\Omega \). That is
$$\begin{aligned} u(x)\le 1-\frac{1}{R^{\frac{p-n}{p-1}}}|x-x_0|^{\frac{p-n}{p-1}}, \quad x\in \Omega . \end{aligned}$$
\(\square \)

Corollary 2.7

Suppose that u is a non-zero solution of (1.3). Then u is not differentiable at \(x_0\).

Proof

First assume \(\frac{p-n}{p-1}\in (0,1)\). By hypothesis, \(u(x)=u(x_0)+Du(x_0)\cdot (x-x_0)+o(|x-x_0|)\), as \(x\rightarrow x_0\). Choosing R so large that \(\Omega \subset B_R(x_0)\), we have by the previous proposition that
$$\begin{aligned} \frac{\left\| u \right\| _\infty }{R^{\frac{p-n}{p-1}}}\left| x-x_0\right| ^\frac{p-n}{p-1}\le \left| u(x)-u(x_0)\right| \le \left| Du(x_0)\right| \left| x-x_0\right| +o\left( \left| x-x_0\right| \right) , \end{aligned}$$
as \(x\rightarrow x_0\). That is,
$$\begin{aligned} \frac{\left\| u \right\| _\infty }{R^{\frac{p-n}{p-1}}}\frac{1}{\left| x-x_0\right| ^{1-\frac{p-n}{p-1}}}\le \left| Du(x_0)\right| +o(1). \end{aligned}$$
This inequality can not be true since \(\frac{p-n}{p-1}\in (0,1)\). If \(\frac{p-n}{p-1}=1\), then \(n=1\) and the claim trivially holds since u is then of the form (2.4). \(\square \)

We will now refine the above estimates to deduce the exact behavior of a solution u of (1.3) near \(x_0\). The following proposition relies on the results of Kichenassamy and Veron in [12].

Proposition 2.8

Assume that u is a solution of (1.3). Then
$$\begin{aligned} \lim _{x\rightarrow x_0}\frac{\left| u(x)-u(x_0)\right| }{\left| x-x_0\right| ^\frac{p-n}{p-1}} =\left\| u \right\| _\infty \left( \frac{p-1}{p-n}\right) \left( \frac{\lambda _p}{n\omega _n} \right) ^{\frac{1}{p-1}}, \end{aligned}$$
and
$$\begin{aligned} \lim _{x\rightarrow x_0}\frac{\left| Du(x)\right| }{\left| x-x_0\right| ^{\frac{p-n}{p-1}-1}} =\left\| u \right\| _\infty \left( \frac{\lambda _p}{n\omega _n}\right) ^{\frac{1}{p-1}}. \end{aligned}$$
Here \(\omega _n\) is the Lebesgue measure of \(B_1(0)\subset \mathbb {R}^n\).

Proof

Without any loss of generality, we may assume that u is positive in \(\Omega \) and that \(u(x_0)=1\). Recall that u is p-harmonic in \(\Omega {\setminus } \{x_0\}\); and in view of Proposition 2.6, u satisfies \(0\le 1-u(x)\le C|x-x_0|^\frac{p-n}{p-1}\) in \(\Omega \) for some constant C. This permits us to use Theorem 1.1 and Remark 1.6 in [12] to conclude that there is \(\gamma > 0\) such that
$$\begin{aligned}&\lim _{x\rightarrow x_0}\frac{1-u(x)}{\left| x-x_0\right| ^\frac{p-n}{p-1}} =\gamma , \end{aligned}$$
(2.5)
$$\begin{aligned}&\lim _{x\rightarrow x_0}\frac{\left| Du(x)\right| }{\left| x-x_0\right| ^{\frac{p-n}{p-1}-1}} =\left( \frac{p-n}{p-1}\right) \gamma , \end{aligned}$$
(2.6)
and
$$\begin{aligned} \lim _{x\rightarrow x_0}\left( \frac{Du(x)}{\left| Du(x)\right| }+\frac{x-x_0}{\left| x-x_0\right| }\right) =0. \end{aligned}$$
(2.7)
We may integrate by parts and exploit (2.7) to get
$$\begin{aligned} \lambda _p&=\int _{\Omega }\left| Du\right| ^{p}dx\\&=\lim _{r\rightarrow 0^+}\int _{\Omega {\setminus } B_{r}(x_0)}\left| Du\right| ^{p}dx\\&=\lim _{r\rightarrow 0^+}\int _{\Omega {\setminus } B_{r}(x_0)}\text {div}(u\left| Du\right| ^{p-2}Du)dx\\&=\lim _{r\rightarrow 0^+}\int _{\partial B_{r}(x_0)}u\left| Du\right| ^{p-2}Du\cdot \left( - \frac{x-x_0}{\left| x-x_0\right| }\right) d\sigma \\&=\lim _{r\rightarrow 0^+}\int _{\partial B_{r}(x_0)}u\left| Du\right| ^{p-1}d\sigma . \end{aligned}$$
Here \(\sigma \) is \(n-1\) dimensional Hausdorff measure. In view of (2.5), we actually have
$$\begin{aligned} \lambda _p = \lim _{r\rightarrow 0^+}\int _{\partial B_{r}(x_0)}\left| Du\right| ^{p-1}d\sigma . \end{aligned}$$
(2.8)
Now we can apply (2.6). This limit gives
$$\begin{aligned} \left| Du(x)\right| ^{p-1}= \left( \frac{p-n}{p-1}\gamma \right) ^{p-1}|x-x_0|^{-(n-1)}+o(1), \end{aligned}$$
as \(x\rightarrow x_0\). By (2.8),
$$\begin{aligned} \lambda _p = \lim _{r\rightarrow 0^+}\int _{\partial B_{r}(x_0)}\left[ \left( \frac{p-n}{p-1}\gamma \right) ^{p-1}r^{-(n-1)}+o(1)\right] d\sigma = \left( \frac{p-n}{p-1}\gamma \right) ^{p-1}n\omega _n, \end{aligned}$$
which concludes the proof. \(\square \)

Remark 2.9

A function \(w\in W^{1,p}_0(\Omega )\) that satisfies
$$\begin{aligned} \left\{ \begin{array}{rl} -\Delta _pw=0,&{} \quad x\in \Omega {\setminus }\{x_0\}\\ w=1,&{} \quad x=x_0\\ w=0,&{} \quad x\in \partial \Omega \end{array}\right. \end{aligned}$$
(2.9)
weakly is called a potential function. Observe that every extremal is a multiple of a potential function but not vice versa. For instance, if \(\Omega =B_1(0)\), then w is an extremal if and only if \(x_0=0\).
The strong maximum principle for p-harmonic functions implies that \(0<w<1\) in \(\Omega {\setminus }\{x_0\}\). In particular, w is uniquely maximized at \(x_0\). Using similar arguments as in the proof of Proposition 2.8, one can easily show that
$$\begin{aligned} -\Delta _pw=\lambda \delta _{x_0} \end{aligned}$$
in \(\Omega \) where
$$\begin{aligned} \lambda :=\int _\Omega \left| Dw\right| ^pdx. \end{aligned}$$
Therefore, the conclusion of Proposition 2.8 holds for w with \(\lambda \) replacing \(\lambda _p\).

3 Support function of an extremal

Suppose now that \(\Omega \) is convex and that \(u\in W^{1,p}_0(\Omega )\) is a positive extremal which achieves its maximum at \(x_0\in \Omega \). By Corollary 2.4, the results of Lewis [14] imply that
$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( x-x_0\right) \cdot Du\left( x\right)>0\text { for }x\in \Omega {\setminus } \left\{ x_0\right\} , \\ u\text { is locally real analytic in }\Omega {\setminus } \left\{ x_0\right\} ,\\ \left\{ u>t\right\} \text { is convex for each }t\in \mathbb {R}\text {, and}\\ \left\{ u=t\right\} \text { has positive Gaussian curvature for each }t\in (0,\left\| u \right\| _\infty ). \end{array}\right. } \end{aligned}$$
(3.1)
By the implicit function theorem, it also follows that the level sets of u are smooth.
We define the support function of u as
$$\begin{aligned} h(\xi ,t):=\sup \left\{ x\cdot \xi : x\in \overline{\Omega }, u(x)\ge t\right\} . \end{aligned}$$
(3.2)
\(\xi \in \mathbb {R}^n, t\in [0,\left\| u \right\| _\infty ]\). For \(t\in [0,\left\| u \right\| _\infty ]\), \(h(\cdot ,t)\) is the usual support function of the convex set \(\{u\ge t\}\); if \(u(0)\ge t\) and \(|\xi |=1\), \(h(\xi ,t)\) represents the distance from the origin to the hyperplane that supports \(\{u\ge t\}\) with outward normal \(\xi \). It follows from (3.1) and Theorem 4 of [7], \(h\in C^\infty (\mathbb {R}^n{\setminus }\{0\}\times (0,\left\| u \right\| _\infty ))\) and \(h_t<0\).
Suppose \(u(x_0)=\left\| u \right\| _\infty \). Then for
$$\begin{aligned} x\in \Omega {\setminus }\{x_0\},\quad \xi =-\frac{Du(x)}{|Du(x)|},\quad \text {and}\quad t=u(x), \end{aligned}$$
we have
$$\begin{aligned} h(\xi ,t)=x\cdot \xi ,\quad h_t(\xi ,t)=-\frac{1}{|Du(x)|},\quad \text {and}\quad D_\xi h(\xi ,t)=x. \end{aligned}$$
See [16]. In particular, since \(\xi \) is the outward unit normal to the hypersurface \(\{u=t\}\) at the point x, \(D_\xi h(\xi ,t)\) is the inverse image of the Gauss map at x. Moreover, as \(D^2_\xi h(\xi ,t)\xi =0\), the restriction of the linear transformation \(D^2_\xi h(\xi ,t):\mathbb {R}^n\rightarrow \mathbb {R}^n\) to \(\xi ^\perp :=\{z\in \mathbb {R}^n: z\cdot \xi =0\}\) is the inverse of the second fundamental form of \(\{u=t\}\) at the point x (see Section 2.5 of [20] for more on this point). In particular, \(D^2_\xi h(\xi ,t)|_{\xi ^\perp }:\xi ^\perp \rightarrow \xi ^\perp \) is positive definite and its eigenvalues are the reciprocals of the principle curvatures of \(\{u=t\}\) at x.
Recall that \(-\Delta _pu=0\) in \(\Omega {\setminus }\{x_0\}\). Using this equation, Colesanti and Salani proved (in Proposition 1 of [7]) that h satisfies
$$\begin{aligned} h_t^2{\text {tr}}\left[ \left( D^2_\xi h|_{\xi ^\perp }\right) ^{-1}\right] +(p-1)\left( \left( D^2_\xi h|_{\xi ^\perp }\right) ^{-1} \nabla _\xi h_t\cdot \nabla _\xi h_t-h_{tt}\right) =0 \end{aligned}$$
(3.3)
for each \(|\xi |=1\) and \(t\in (0,\left\| u \right\| _\infty )\). Here \(\nabla _\xi h_t:=(I_n-\xi \otimes \xi )D_\xi h_t\) is the projection of the gradient of \(h_t\) onto \(\xi ^\perp \). Equation (3.3) will have an important role in our Proof of Theorem 1.1.

4 Convex domains

Throughout this section, we will assume that \(\Omega \subset \mathbb {R}^n\) is a bounded convex domain. We will also suppose that \(u_0, u_1\in W^{1,p}_0(\Omega )\) are positive extremal functions which satisfy
$$\begin{aligned} {\left\{ \begin{array}{ll} u_0(x_0)=\left\| u_0\right\| _\infty =1\\ u_1(x_1)=\left\| u_1\right\| _\infty =1, \end{array}\right. } \end{aligned}$$
for some \(x_0,x_1\in \Omega \). We aim to show that
$$\begin{aligned} u_0\equiv u_1. \end{aligned}$$
(4.1)
It is easy to see that Theorem 1.1 follows from (4.1).
For each \(\rho \in (0,1)\), we define the Minkowski combination of \(u_0\) and \(u_1\)
$$\begin{aligned} u_\rho (z):=\sup \left\{ \min \left\{ u_0(x),u_1(y)\right\} : z=\left( 1-\rho \right) x+\rho y,\; x,y\in \overline{\Omega }\right\} \end{aligned}$$
\(z\in \overline{\Omega }\). We recall that \(u_0\) and \(u_1\) are quasiconcave. Using the definition above, it is straightforward to verify
$$\begin{aligned} \left\{ u_\rho>t\right\} =\left( 1-\rho \right) \left\{ u_0>t\right\} +\rho \left\{ u_1>t\right\} , \end{aligned}$$
(4.2)
for each \(t\in \mathbb {R}\). Here the addition is the usual Minkowski addition of convex sets. In particular, \(u_\rho \) itself is quasiconcave.

The Minkowski combination was introduced in work of Borell in [1] when he studied capacitary functions; although his work was motivated by the previous papers of Lewis [14] and Gabriel [11]. We also were particularly inspired to utilize the Minkowski combination after we became aware of the work of Colesanti and Salani in [7], who verified a Brunn-Minkowski inequality for p-capacitary functions \((1<p<n)\), and the work of Cardaliaguet and Tahraoui in [3] on the strict concavity of the harmonic radius.

Along the way to proving (4.1), we will need some other useful properties of \(u_\rho \).

Proposition 4.1

Define
$$\begin{aligned} x_\rho :=(1-\rho )x_0+\rho x_1. \end{aligned}$$
Then the following hold:
(i)

\(u_\rho (x_\rho )=\left\| u_\rho \right\| _\infty =1\).

(ii)

\(u_\rho |_{\partial \Omega }=0\).

(iii)

\(u_\rho \in C^\infty (\Omega {\setminus }\left\{ x_\rho \right\} )\cap C(\overline{\Omega })\).

(iv)
For each \(z\in \Omega {\setminus }\left\{ x_\rho \right\} \), there are \(x\in \Omega {\setminus }\left\{ x_0\right\} \) and \(y\in \Omega {\setminus }\left\{ x_1\right\} \) such that
$$\begin{aligned} z&=(1-\rho )x+\rho y,&\\ u_\rho (z)&=u_0(x)=u_1(y),&\\ \frac{Du_\rho (z)}{\left| Du_\rho (z)\right| }&=\frac{Du_0(x)}{\left| Du_0(x)\right| }=\frac{Du_1(y)}{\left| Du_1(y)\right| },&\\ \frac{1}{\left| Du_\rho (z)\right| }&=(1-\rho )\frac{1}{\left| Du_0(x)\right| }+\rho \frac{1}{\left| Du_1(y)\right| },&\\ \frac{D^2u_\rho (z)}{\left| Du_\rho (z)\right| ^3}&\ge (1-\rho )\frac{D^2u_0(x)}{\left| Du_0(x)\right| ^3}+\rho \frac{D^2u_1(y)}{|Du_1(y)|^3}.&\end{aligned}$$

We omit the proof of the above proposition. However, we remark that (i) and (ii) are elementary; Theorem 4 of [7] and Theorem 1 of [14] together imply (iii); and (iv) follows from Sect. 2, [3] or Sect. 7 of [16]. Using these properties we will verify that \(u_\rho \) itself is an extremal for each \(\rho \in (0,1)\).

Lemma 4.2

\(u_\rho \) is extremal.

Proof

We first show that \(u_\rho \) is p-subharmonic and integrate by parts to derive an upper bound on the integral \(\int _\Omega |Du_\rho |^pdz\). Then we show that \(u_\rho \) satisfies the limits in Proposition 2.8 (that are also satisfied by every extremal function). Finally, we combine the upper bound and limits to arrive at the desired conclusion.
  1. 1.
    Let \(z\in \Omega {\setminus }\{x_\rho \}\), and select \(x\in \Omega {\setminus }\{x_0\}\) and \(y\in \Omega {\setminus }\{x_1\}\) such that
    $$\begin{aligned} e:=\frac{Du_\rho (z)}{\left| Du_\rho (z)\right| }=\frac{Du_0(x)}{\left| Du_0(x)\right| }=\frac{Du_1(y)}{\left| Du_1(y)\right| } \end{aligned}$$
    and
    $$\begin{aligned} \frac{D^2u_\rho (z)}{\left| Du_\rho (z)\right| ^3}\ge (1-\rho )\frac{D^2u_0(x)}{\left| Du_0(x)\right| ^3}+\rho \frac{D^2u_1(y)}{\left| Du_1(y)\right| ^3}. \end{aligned}$$
    Recall that such xy exist by Proposition 4.1. We have
    $$\begin{aligned} \left| Du_\rho (z)\right| ^{-(p+1)}\Delta _pu_\rho (z)&=\frac{\Delta u_\rho (z)}{\left| Du_\rho (z)\right| ^3}+(p-2)\frac{D^2u_\rho (z)e\cdot e}{\left| Du_\rho (z)\right| ^3}\\&=\left( I_n+(p-2)e\otimes e\right) \cdot \frac{D^2u_\rho (z)}{|Du_\rho (z)|^3}. \end{aligned}$$
    Note that \(\min \{1,p-1\}>0\) is a lower bound on the eigenvalues of the matrix \(I_n+(p-2)e\otimes e\). Therefore,
    $$\begin{aligned}&\left| Du_\rho (z)\right| ^{-(p+1)}\Delta _pu_\rho (z)\\&\quad \ge \left( I_n+(p-2)e\otimes e\right) \cdot \left( (1-\rho )\frac{D^2u_0(x)}{\left| Du_0(x)\right| ^3}+\rho \frac{D^2u_1(y)}{\left| Du_1(y)\right| ^3}\right) \\&\quad = (1-\rho )\left| Du_0(x)\right| ^{-(p+1)}\Delta _pu_0(x)+ \rho \left| Du_1(y)\right| ^{-(p+1)}\Delta _pu_1(y)\\&\quad = (1-\rho )\cdot 0+ \rho \cdot 0\\&\quad =0. \end{aligned}$$
    Consequently, \(-\Delta _p u_\rho \le 0\) in \(\Omega {\setminus }\{x_\rho \}\).
     
  2. 2.
    The divergence theorem gives
    $$\begin{aligned}&\int _{\Omega {\setminus } B_r(x_\rho )}\text {div}(u_\rho \left| Du_\rho \right| ^{p-2}Du_\rho )dz\\&\quad =\int _{\partial B_r(x_\rho )}u_\rho \left| Du_\rho \right| ^{p-2}Du_\rho \cdot \left( -\frac{z-x_\rho }{|z-x_\rho |}\right) d\sigma . \end{aligned}$$
    On the other hand, since \(u_\rho \) is a positive p-subharmonic function in \(\Omega {\setminus }\{x_\rho \}\)
    $$\begin{aligned} \int _{\Omega {\setminus } B_r(x_\rho )}\text {div}(u_\rho \left| Du_\rho \right| ^{p-2}Du_\rho )dz&=\int _{\Omega {\setminus } B_r(x_\rho )}\left( u_\rho \Delta _p u_\rho +\left| Du_\rho \right| ^{p}\right) dz \\&\ge \int _{\Omega {\setminus } B_r(x_\rho )}\left| Du_\rho \right| ^{p}dz. \end{aligned}$$
    As \(u_\rho \le 1\),
    $$\begin{aligned} \int _{\Omega }\left| Du_\rho \right| ^pdz\le \liminf _{r\rightarrow 0^+}\int _{\partial B_r(x_\rho )}\left| Du_\rho \right| ^{p-1}d\sigma . \end{aligned}$$
    (4.3)
     
  3. 3.
    Let \(w_\rho \) be a solution of the PDE (2.9) with \(x_\rho \) replacing \(x_0\). As \(u_\rho \) is p-subharmonic, \(u_\rho (x_\rho )=1\) and \(u_\rho |_{\partial \Omega }=0\), weak comparison implies \(u_\rho \le w_\rho \). This is a version of Borell’s inequality; see [1, 3]. In particular,
    $$\begin{aligned} w_\rho (z)\ge u_\rho (z)\ge \min \left\{ u_0(x),u_1(y)\right\} , \end{aligned}$$
    (4.4)
    whenever \(z=(1-\rho )x+\rho y\). Now let \(z^k\rightarrow x_\rho \) with \(z^k\ne x_\rho \) for all \(k\in \mathbb {N}\) sufficiently large. Set
    $$\begin{aligned} {\left\{ \begin{array}{ll} x^k:=z^k+\left( x_0-x_\rho \right) \\ y^k:=z^k+\left( x_1-x_\rho \right) \end{array}\right. } \end{aligned}$$
    for each \(k\in \mathbb {N}\). Observe \(z^k=(1-\rho )x^k+\rho y^k\) and
    $$\begin{aligned} \left| z^k-x_\rho \right| =\left| x^k-x_0\right| =\left| y^k-x_1\right| . \end{aligned}$$
    Setting \(\lambda := \int _\Omega |Dw_\rho |^pdz\), we have from Proposition 2.8, Remark 2.9 and (4.4) that
    $$\begin{aligned} \left( \frac{p-1}{p-n}\right) \left( \frac{\lambda }{n\omega _n}\right) ^\frac{1}{p-1}&=\lim _{k\rightarrow \infty } \frac{1-w_\rho (z^k)}{|z^k-x_\rho |^{\frac{p-n}{p-1}}}\\&\le \lim _{k\rightarrow \infty } \max \left\{ \frac{1-u_0(x^k)}{|x^k-x_0|^{\frac{p-n}{p-1}}}, \frac{1-u_1(y^k)}{|y^k-x_1|^{\frac{p-n}{p-1}}}\right\} \\&=\left( \frac{p-1}{p-n}\right) \left( \frac{\lambda _p}{n\omega _n}\right) ^\frac{1}{p-1}. \end{aligned}$$
    It follows that \(\lambda = \lambda _p\). In view of (4.4), and since the sequence \(z^k\) was arbitrary,
    $$\begin{aligned} \lim _{z\rightarrow x_\rho }\frac{1-u_\rho (z)}{\left| z-x_\rho \right| ^{\frac{p-n}{p-1}}}=\left( \frac{p-1}{p-n}\right) \left( \frac{\lambda _p}{n\omega _n}\right) ^\frac{1}{p-1}. \end{aligned}$$
    (4.5)
     
  4. 4.
    Again let \(z^k\rightarrow x_\rho \) with \(z^k\ne x_\rho \) for all \(k\in \mathbb {N}\) sufficiently large. By Proposition 4.1, there are \(x^k\in \Omega {\setminus }\{x_0\}\) and \(y^k\in \Omega {\setminus }\{x_1\}\) such that \(z^k=(1-\rho )x^k+\rho y^k\),
    $$\begin{aligned} u_\rho \left( z^k\right) =u_0\left( x^k\right) =u_1\left( y^k\right) , \end{aligned}$$
    (4.6)
    and
    $$\begin{aligned} \frac{1}{\left| Du_\rho (z^k)\right| }=(1-\rho )\frac{1}{\left| Du_0(x^k)\right| }+\rho \frac{1}{\left| Du_1(y^k)\right| }. \end{aligned}$$
    (4.7)
    Since \(u_\rho (z^k)\rightarrow 1\), (4.6) implies that \(x^k\rightarrow x_0\) and \(y^k\rightarrow x_1\) as \(u_0\) and \(u_1\) are uniquely maximized as these points, respectively. Combining this fact with Proposition 2.8, (4.5) and again with (4.6) also gives
    $$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\left| y^k-x_1\right| }{\left| z^k-x_\rho \right| }=\lim _{k\rightarrow \infty } \frac{\left| x^k-x_0\right| }{\left| z^k-x_\rho \right| }=1. \end{aligned}$$
    (4.8)
    By (4.7),
    $$\begin{aligned} \frac{\left| z^k-x_\rho \right| ^{\frac{p-n}{p-1}-1}}{\left| Du_\rho (z^k)\right| }&=(1-\rho )\frac{ \left| z^k-x_\rho \right| ^{\frac{p-n}{p-1}-1}}{\left| Du_0(x^k)\right| }+\rho \frac{\left| z^k-x_\rho \right| ^{\frac{p-n}{p-1}-1}}{\left| Du_1(y^k)\right| }\\&=\left( \frac{\left| z^k-x_\rho \right| }{\left| x^k-x_0\right| }\right) ^{\frac{p-n}{p-1}-1}(1-\rho )\frac{ \left| x^k-x_0\right| ^{\frac{p-n}{p-1}-1}}{\left| Du_0(x^k)\right| }\\&\quad + \left( \frac{\left| z^k-x_\rho \right| }{\left| y^k-x_1\right| }\right) ^{\frac{p-n}{p-1}-1}\rho \frac{\left| y^k-x_1\right| ^{\frac{p-n}{p-1}-1}}{\left| Du_1(y^k)\right| }. \end{aligned}$$
    We can now employ the second limit in Proposition 2.8 and (4.8) to obtain
    $$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\left| Du_\rho (z^k)\right| }{\left| z^k-x_\rho \right| ^{\frac{p-n}{p-1}-1}}=\left( \frac{\lambda _p}{n\omega _n}\right) ^\frac{1}{p-1}. \end{aligned}$$
    And since \(z^k\) was arbitrary,
    $$\begin{aligned} \lim _{z\rightarrow x_\rho } \frac{\left| Du_\rho (z)\right| }{\left| z-x_\rho \right| ^{\frac{p-n}{p-1}-1}}=\left( \frac{\lambda _p}{n\omega _n}\right) ^\frac{1}{p-1}. \end{aligned}$$
    (4.9)
     
  5. 5.
    Using the upper bound (4.3) and the limits (4.5) and (4.9), we can proceed with the same arguments as in the proof of Proposition 2.8 to conclude
    $$\begin{aligned} \int _{\Omega }\left| Du_\rho \right| ^pdz&\le \liminf _{r\rightarrow 0^+}\int _{\partial B_r(x_\rho )}\left| Du_\rho \right| ^{p-1}d\sigma = \lambda _p. \end{aligned}$$
    \(\square \)
     
In order to verify (4.1), we will employ the respective support functions \(h_0\), \(h_1\), and \(h_\rho \) of \(u_0\), \(u_1\) and \(u_\rho \); recall the support function of an extremal was defined in (3.2). In particular, we note that the identity (4.2) implies
$$\begin{aligned} h_\rho =(1-\rho )h_0+\rho h_1. \end{aligned}$$
Using this identity with the fact that \(h_0, h_1\) and \(h_\rho \) all satisfy Eq. (3.3), Colesanti and Salani showed for each \(t\in (0,1)\) there is \(C(t)>0\) such that
$$\begin{aligned} D^2_\xi h_0(\xi ,t)\left| _{\xi ^\perp }=C(t)D^2_\xi h_1(\xi ,t)\right| _{\xi ^\perp } \end{aligned}$$
(4.10)
and
$$\begin{aligned} \partial _th_0(\xi ,t)=C(t)\partial _th_1(\xi ,t), \end{aligned}$$
(4.11)
for all \(|\xi |=1\) (see the proof of Theorem 1 in [7]).
As \(D^2_\xi h_0(\xi ,t)\xi =D^2_\xi h_0(\xi ,t)\xi =0\in \mathbb {R}^n\), it follows from (4.10) and the homogeneity of \(h_0\) and \(h_1\) that
$$\begin{aligned} D^2_\xi h_0(\xi ,t)=C(t)D^2_\xi h_1(\xi ,t), \end{aligned}$$
for \((\xi ,t)\in (\mathbb {R}^n{\setminus }\{0\})\times (0,1)\). Upon integration we find
$$\begin{aligned} h_0(\xi ,t)=C(t)h_1(\xi ,t)+a(t)\cdot \xi +b(t) \end{aligned}$$
for some \(a(t)\in \mathbb {R}^n\) and \(b(t)\in \mathbb {R}\). Since \(h_0\) and \(h_1\) are homogeneous of degree one, it must be that \(b(t)=0\) for each \(t\in (0,1)\). Thus,
$$\begin{aligned} h_0(\xi ,t)=C(t)h_1(\xi ,t)+a(t)\cdot \xi . \end{aligned}$$
Taking the time derivative of both sides of this equation gives
$$\begin{aligned} \partial _th_0(\xi ,t)=C(t)\partial _th_1(\xi ,t)+C'(t)h_1(\xi ,t)+a'(t)\cdot \xi . \end{aligned}$$
Comparing with (4.11) leads us to
$$\begin{aligned} C'(t)h_1(\xi ,t)+a'(t)\cdot \xi =0. \end{aligned}$$
Suppose that \(C'(t_0)\ne 0\) for some \(t_0\in (0,1)\). Then
$$\begin{aligned} h_1(\xi ,t_0)=\left[ \frac{-a'(t_0)}{C'(t_0)}\right] \cdot \xi . \end{aligned}$$
This would imply the level set \(\{u_1=t_0\}\) is the singleton \(\{-a'(t_0)/C'(t_0)\}\), which is not possible. Therefore, \(C'(t)= 0\) for \(t\in (0,1)\) and thus \(a'(t)\cdot \xi =0\) for all \(\xi \). Consequently, \(a'(t)=0\) for \(t\in (0,1)\). Since \(h_0\) and \(h_1\) coincide at \(t=0\), \(C(t)=1\) and \(a(t)=0\) for all \(t\in [0,1]\). As a result, \(h_0\equiv h_1\) and so \(\{u_0\ge t\}=\{u_1\ge t\}\) for each \(t\in [0,1]\) (Theorem 8.24 in [19]). This verifies (4.1).

Proof of Corollary 1.2

By Eq. (1.4) and inequality (1.2),
$$\begin{aligned} G(y,y)&=\int _\Omega \left| DG(x,y)\right| ^pdx\\&\ge \lambda _p\left\| G(\cdot ,y)\right\| ^p_\infty \\&\ge \lambda _pG(y,y)^p. \end{aligned}$$
Therefore, \(G(y,y)\le \lambda _p^{-1/(p-1)}\) and equality holds if and only if \(G(\cdot ,y)\) is extremal. Theorem 1.1 in turn implies that equality occurs if and only if \(y=x_0\). \(\square \)

5 Nonuniqueness

We will now explain that uniqueness does not hold for general domains by providing a few explicit examples. These instances include planar annuli, bow tie and dumbbell shaped domains. The perceptive reader will also see how to construct other examples from our remarks below.

Example 5.1

Define
$$\begin{aligned} \Omega _{r_1,r_2}:=\left\{ x\in \mathbb {R}^n: r_1<\left| x\right| <r_2\right\} , \end{aligned}$$
for \(r_1,r_2>0\) with \(r_1<r_2\). As mentioned above, there is a positive extremal u that achieves is maximum at a single point \(x_0\in \Omega _{r_1,r_2}\). Notice that for any \(n\times n\) orthogonal matrix O, \(v:=u\circ O\) is a positive extremal and \(\Vert v\Vert _\infty =\left\| u \right\| _\infty \). Consequently, for each \(y_0\in \Omega _{r_1,r_2}\) with \(|y_0|=|x_0|\), there is a distinct positive extremal with supremum norm equal to \(\left\| u \right\| _\infty \). Thus, uniqueness of extremals does not hold for annuli as showed in Fig. 2.
Fig. 2

\(\Omega _{4,6}\) when \(n=2\)

Example 5.2

Consider the “bow tie” domain in the plane
$$\begin{aligned} \Omega _\epsilon :=\left\{ (x_1,x_2)\in \mathbb {R}^2: \left| x_2\right|<\left| x_1\right| +\epsilon , \left| x_1\right| <1\right\} , \end{aligned}$$
for \(\epsilon >0\). Note, in particular, that \(\Omega _\epsilon \) is star-shaped with respect to the origin; see Fig. 3. Let \(u_\epsilon \) be a positive extremal for \(\Omega _\epsilon \) with \(\Vert u_\epsilon \Vert _\infty =1\). If \(u_\epsilon \) is unique, then it must be that
$$\begin{aligned} u_\epsilon (0,0)=1. \end{aligned}$$
(5.1)
This is due to the fact that the \(\Omega _\epsilon \) and the p-Laplacian are invariant with respect to reflection about the \(x_1\) and \(x_2\) axes.
Fig. 3

\(\Omega _{1/10}\)

Let us assume (5.1) holds for each \(\epsilon >0\) and extend \(u_\epsilon \) to be 0 outside of \(\Omega _\epsilon \). Notice that the resulting function, which we also denote as \(u_\epsilon \), belongs to \(W^{1,p}(\mathbb {R}^2)\). Also note that since \(\Omega _0\subset \Omega _\epsilon \)
$$\begin{aligned} \int _{\mathbb {R}^2}\left| Du_\epsilon \right| ^pdx=\int _{\Omega _\epsilon }\left| Du_\epsilon \right| ^pdx =\lambda _p(\Omega _\epsilon )\le \lambda _p(\Omega _0). \end{aligned}$$
Consequently, there is a decreasing sequence of positive numbers \((\epsilon _j)_{j\in \mathbb {N}}\) tending to 0 and a continuous function \(u_0:\mathbb {R}^2\rightarrow [0,1]\) for which \(u_{\epsilon _j} \rightarrow u_0\) locally uniformly on \(\mathbb {R}^2\). In view of (5.1),
$$\begin{aligned} u_0(0,0)=1. \end{aligned}$$
On the other hand, \(u_\epsilon (0,2\epsilon )=0\) for all \(\epsilon >0\). Thus
$$\begin{aligned} u_0(0,0)=\lim _{j\rightarrow \infty }u_{\epsilon _j}(0,2\epsilon _j)=0, \end{aligned}$$
which is a contradiction.

As a result, we conclude that there is some \(\epsilon >0\) such that \(u_\epsilon \) does not achieve its maximum value at (0, 0). For this value of \(\epsilon \), \(\Omega _\epsilon \) will have a least two positive extremals with supremum norm equal to 1.

Example 5.3

The same ideas used in Example 5.2, can be used to show the dumbbell-shaped domain
$$\begin{aligned} B_1(-5,0)\cup \left( [-5,5]\times [-\delta ,\delta ]\right) \cup B_1(5,0)\quad (0<\delta <1), \end{aligned}$$
does not have unique extremals for some \(\delta >0\) chosen small enough. See Fig. 4.
Fig. 4

Dumbbell-shaped domain for \(\delta =1/5\)

6 Steiner symmetric domains

Theorem 1.1 implies that if a convex domain has some reflectional symmetry, then we have additional information on the location of the maximum points of positive extremals. More precisely, we can make the following observation.

Corollary 6.1

Assume \(\Omega \subset \mathbb {R}^n\) is a convex domain that is invariant with respect to reflection across the hyperplanes \(\left\{ x\in \mathbb {R}^n: x_j=0\right\} \) for \(j=1,\dots , n\). Then any positive (negative) extremal achieves its maximum (minimum) value at \(0\in \mathbb {R}^n\).

Proof

Assume \(u\in W^{1,p}_0(\Omega )\) is a positive extremal that achieves it maximum value at z. As \(\Omega \) is invariant with respect to \(\{x\in \mathbb {R}^n: x_1=0\}\), the function
$$\begin{aligned} u_1(x_1,x_2,\ldots , x_n):=u(-x_1,x_2,\ldots , x_n),\quad (x_1,x_2,\ldots , x_n)\in \Omega , \end{aligned}$$
belongs to \(W^{1,p}_0(\Omega )\) and \(\left\| u \right\| _\infty =\Vert u_1\Vert _\infty \). Moreover, it is routine to verify that \(u_1\) is also a positive extremal that achieves it maximum at the reflection of z about the plane \(\{x\in \mathbb {R}^n: x_1=0\}\). By Theorem 1.1, \(u_1=u\) which forces \(z\in \{x\in \mathbb {R}^n: x_1=0\}\). Repeating this argument for \(j=2,\dots , n\), we find \(z\in \{x\in \mathbb {R}^n: x_j=0\}\) for \(j=1,\dots , n\). As a result, \(z=0\). \(\square \)

We now seek to extend this observation. We will show below that certain symmetric two dimensional domains have unique extremals without assuming the domains were convex to begin with. To this end, we employ Steiner symmetrization. In particular, we will make use of the results by Cianchi and Fusco in [6] on the equality condition in the Pólya-Szegö inequality associated with Steiner symmetrization. We also use special properties of the critical points of p-harmonic functions in two dimensions due to Manfredi in [17].

Let us first briefly recall the notion of the Steiner symmetrization of a subset of \(\mathbb {R}^2\). For a given \(A\subset \mathbb {R}^2\) and \(a\in \mathbb {R}\), we will denote \(A\cap \{x_1=a\}\) as the intersection of A with the vertical line \(x_1=a\). We also will write \(\mathcal{L}^m\) for the outer Lebesgue measure defined on all subsets of \(\mathbb {R}^m\)\((m=1,2)\).

Definition 6.2

Assume \(A\subset \mathbb {R}^2\). The Steiner symmetrization of A with respect to the \(x_1\) axis is
$$\begin{aligned} A^{*}_1=\left\{ (a,b)\in \mathbb {R}^2: |b|< \frac{1}{2} \mathcal{L}^1\left( A\cap \{x_1=a\}\right) \right\} . \end{aligned}$$
A is said to be Steiner symmetric with respect to the \(x_1\) axis if \(A^*_1=A\).
Now suppose \(u:\mathbb {R}^2\rightarrow [0,\infty )\) is Lebesgue measurable. We can use the above definition to provide the following rearrangement of u
$$\begin{aligned} u^*_1(x):=\int ^\infty _0\chi _{\{u>t\}^{*}_{1}}(x)dt, \quad x\in \mathbb {R}^2. \end{aligned}$$
This function is called the Steiner rearrangement of u with respect to the \(x_1\) axis. Observe that
$$\begin{aligned} \{u^*_1>t\}=\{u>t\}^*_1 \end{aligned}$$
(6.1)
for each \(t\ge 0\). Note also that \(u^*_1(x_1,\cdot )\) and \(u(x_1,\cdot )\) have the same distribution for \(\mathcal{L}^1\) almost every \(x_1\in \mathbb {R}\).
It is known that if \(p\in [1,\infty )\), \(\Omega \subset \mathbb {R}^2\) is a bounded domain and \(u\in W^{1,p}_0(\Omega )\), then \(u^*_1\in W^{1,p}_0(\Omega )\). Moreover, if \(\Omega ^*_1\) is Lebesgue measurable, the Pólya-Szegö inequality
$$\begin{aligned} \int _{\Omega ^*_1} |D u_1^*|^p dx \le \int _\Omega |D u|^p dx \end{aligned}$$
(6.2)
holds, see [2, 6]. Cianchi and Fusco showed that if \(\Omega ^*_1=\Omega \) and equality holds in (6.2), then \(u^*_1=u\) provided
$$\begin{aligned} \mathcal{L}^2\left( \{u_{x_2}=0\}\right) =0 \end{aligned}$$
(6.3)
(Theorem 2.2 in [6]). All of the above definitions and facts regarding Steiner symmetrization and rearrangements with respect to the \(x_1\) axis have obvious counterparts with respect to the \(x_2\) axis.

Our main assertion regarding the uniqueness of extremals on Steiner symmetric domains is as follows.

Proposition 6.3

Assume \(\Omega \subset \mathbb {R}^2\) is a bounded domain that is equal to its Steiner symmetrization with respect to the \(x_1\) and \(x_2\) axes. Then any positive (negative) extremal achieves its maximum (minimum) value at \(0\in \mathbb {R}^2\).

Proof

Assume \(u\in W^{1,p}_0(\Omega )\) is a positive extremal with \(\left\| u \right\| _\infty =u(z)=1\). In view of (6.1), \(\Vert u^*_1\Vert _\infty =1\), as well. By the Pólya-Szegö inequality (6.2), we easily conclude \(u^*_1\) is extremal and \(\int _\Omega |D u_1^*|^p dx = \int _\Omega |D u|^p dx\). We now claim that u satisfies (6.3). Once we verify this assertion, we would have \(u=u^*_1\) which implies \(u(x_1,x_2)=u(x_1,-x_2)\) for all \((x_1,x_2)\in \Omega \). As a result z belongs to the \(x_1\) axis, and very similarly we would have that z also belongs to the \(x_2\) axis. Therefore, \(z=0\in \mathbb {R}^2\).
Fig. 5

The domain \(\{(x_1,x_2)\in \mathbb {R}^2: |x_1|+2|x_2|<1, 2|x_1|+|x_2|<1 \}\)

Let us now show that any positive extremal \(u\in W^{1,p}_0(\Omega )\) satisfies (6.3). Recall that u is p-harmonic in \(\Omega {\setminus }\{z\}\) and therefore, \(u\in C^1_{\text {loc}}(\Omega {\setminus }\{z\})\). By the results of Manfredi in [17], we know the zeros of Du are isolated in \(\Omega {\setminus }\{z\}\). Consequently, u is locally real analytic in \(S:=\Omega {\setminus }\left( \{z\}\cup \{\left| Du\right| =0\}\right) \), which is an open set of full measure. In particular, \(u_{x_2}\) is also locally real analytic in S. Therefore, if
$$\begin{aligned} \mathcal{L}^2\left( \{x\in S : u_{x_2}(x)=0\}\right) >0, \end{aligned}$$
then it must be that \(u_{x_2}\equiv 0\) in S; see section 3.1 of [13]. Since \(u_{x_2}\) is continuous in \(\Omega {\setminus }\{z\}\), it would then follow that \(u_{x_2}\equiv 0\) in \(\Omega {\setminus }\{z\}\), as well. However, this is clearly not possible as the function
$$\begin{aligned}{}[0,\infty ) \ni t\mapsto u(z+te_2) \end{aligned}$$
is positive at \(t=0\) and vanishes for all \(t>0\) sufficiently large. As a result, (6.3) holds and the assertion follows. \(\square \)
Fig. 6

The domain \(([-3,3]\times [-1,1])\cup ([-1,1]\times [-3,3])\)

Fig. 7

The region bounded by the curve \(r=10+\frac{13}{20}\cos (8\theta )\)

See Figs. 5, 6 and 7 for Steiner symmetric, nonconvex domains \(\Omega \) for which Proposition 6.3 applies to. Figure 5 displays
$$\begin{aligned} \{(x_1,x_2)\in \mathbb {R}^2: \left| x_1\right| +2\left| x_2\right|<1, 2\left| x_1\right| +\left| x_2\right| <1 \}. \end{aligned}$$
Figure 6 shows
$$\begin{aligned} ([-3,3]\times [-1,1])\cup ([-1,1]\times [-3,3]) \end{aligned}$$
and Fig. 7 exhibits the region bounded by the curve
$$\begin{aligned} r=10+\frac{13}{20}\cos (8\theta ) \end{aligned}$$
given in polar coordinates.

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Authors and Affiliations

  1. 1.Department of MathematicsMITCambridgeUSA
  2. 2.Department of MathematicsUppsala UniversityUppsalaSweden

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