Analytic content and the isoperimetric inequality in higher dimensions

TL;DR

This paper proves a conjecture linking the analytic content of smooth domains in higher dimensions to the isoperimetric inequality, using innovative methods from partial balayage and optimal transport theory.

Contribution

It introduces a new proof connecting analytic content with geometric inequalities in higher dimensions through advanced mathematical techniques.

Findings

Established the conjecture relating analytic content and isoperimetric inequality in higher dimensions.

Developed a novel proof combining partial balayage and optimal transport theory.

Enhanced understanding of geometric analysis in smooth domains.

Abstract

This paper establishes a conjecture of Gustafsson and Khavinson, which relates the analytic content of a smoothly bounded domain in $\mathbb{R}^{N}$ to the classical isoperimetric inequality. The proof is based on a novel combination of partial balayage with optimal transport theory.

Equations

$$\lambda (\omega )=\inf \{{\Vert \overline{z}-\varphi \Vert }_{\overline{\omega }}:\varphi \in \mathcal{A}(\omega )\}.$$

$$\frac{2A}{P}\le \lambda (\omega )\le \sqrt{\frac{A}{\pi }}.$$

$$\lambda (\Omega )=\inf \{{\Vert x-f\Vert }_{\overline{\Omega }}:f\in A(\Omega )\},$$

$${\Vert f\Vert }_S=\underset{S}{\sup }\Vert f\Vert ,\text{ where }\Vert f\Vert =\sqrt{f_1^2+...+f_N^2}.$$

$$\frac{\partial f_j}{\partial x_k}-\frac{\partial f_k}{\partial x_j}=0\text{ for all }j,k\in \{1,...,N\}\text{ on }\Omega .)$$

$$\mathcal{H}=\{h\in C^1(\overline{\Omega }):\Delta h=0\text{ on }\Omega \},$$

$$\{\nabla h:h\in \mathcal{H}\}\subset A(\Omega ).$$

$$\frac{NV}{P}\le \lambda (\Omega )\le c_N{r}_{\Omega }.$$

$${\lambda }_1(\Omega )$$

$$\Vert \nabla u{\Vert }_{\overline{\Omega }}=\Vert \nabla u{\Vert }_{\partial \Omega }$$

$$\lambda (\Omega )\le {\lambda }_1(\Omega ).$$

$$\frac{NV}{P}\le \lambda (\Omega )\le r_{\Omega }.$$

$$P_{\mu }^{\nu }=G_O\nu +\sup \{s:s\text{ is subharmonic on }O\text{ and }s\le G_O\mu -G_O\nu \}\text{ on }O,$$

$${\mu }_1\ge \mu ⟹\mathcal{B}{\mu }_1\ge \mathcal{B}\mu .$$

$$\mathrm{supp}\mathcal{B}(Nm{|}_{\Omega }+b\tau )\subset {\Omega }_0.$$

$$\epsilon =2^{-1}\mathrm{dist}(\partial {\Omega }^{\prime },\Omega \cup ({\mathbb{R}}^N\backslash {\Omega }_0)).$$

$$\gamma (x)={\int }_{\partial {\Omega }^{\prime }}{\varphi }_{\epsilon }(x-y)d{\tau }^{\ast }(y)(x\in {\mathbb{R}}^N),$$

$$P_{G_O(Nm{|}_{\Omega }+b\gamma m)}^{\nu }=G_O(Nm{|}_{\Omega }+b\gamma m).$$

$$G_O(Nm{|}_{\Omega }+b\gamma m)\le P_{Nm{|}_{\Omega }+b\tau }^{\nu }\le G_O(Nm{|}_{\Omega }+b\tau ),$$

$$\mathrm{supp}\mathcal{B}(Nm{|}_{\Omega }+b\tau )\subset \{x\in O:\mathrm{dist}(x,{\Omega }^{\prime })\le \epsilon \}.$$

$$\frac{\partial G_{\Omega }\gamma }{\partial y_i}(y)={\int }_{\Omega }\frac{\partial G_{\Omega }}{\partial y_i}(x,y)d\gamma (x)(y\in \overline{\Omega };i=1,...,N).$$

$${\psi }_j(y)={\int }_{{\Omega }_j}\frac{\partial G_{\Omega }}{\partial y_i}(x,y)d\gamma (x)(y\in \overline{\Omega };j=1,2),$$

$$\left|{\psi }_2(y)\right|\le C(\Omega )\epsilon {\Vert \frac{d\gamma }{dm}\Vert }_{L^{\infty }(\Omega )}(y\in B(y_0,\epsilon /2)\cap \overline{\Omega }).$$

$$\frac{\partial G_{\Omega }\gamma }{\partial y_i}(y)\to {\int }_{\Omega }\frac{\partial G_{\Omega }}{\partial y_i}(x,y_0)d\gamma (x)(y\to y_0).$$

$$b_N=\frac{m(\{y\in B(0,1):y_N\ge 1/2\})}{m(B(0,1))},$$

$$m(D\backslash R)\le m(D\backslash \Omega )<b_{N}m(\Omega )<b_{N}m(D)$$

$$(\nabla v)(R)\cap \{y\in B(r_D):y\cdot x\ge r_D/2\}\ne \varnothing (x\in \partial B(0,1)).$$

$$w(x)=\sup \left\{v(y)+\nabla v(y)\cdot (x-y):y\in R\right\}(x\in {\mathbb{R}}^N)$$

$$w_{\epsilon }(x)={\int }_{B(\epsilon )}{\varphi }_{\epsilon }(x-y)w(y)dm(y)(x\in {\mathbb{R}}^N),$$

$$O=\{x\in {\mathbb{R}}^N:w_{\epsilon }(x)<C\},$$

Full text

1 Introduction

00footnotetext: 2010 *Mathematics Subject Classification * 31B05. *Keywords: *analytic content, harmonic vector fields, isoperimetric inequality, optimal transport, partial balayage

Let $\omega$ be a bounded domain in the complex plane $\mathbb{C}$ such that $\partial\omega$ is the disjoint union of finitely many simple analytic curves, and let $\mathcal{A}(\omega)$ denote the collection of continuous functions on $\overline{\omega}$ that are analytic on $\omega$. Further, let $\left\|g\right\|_{S}$ denote $\sup_{S}\left|g\right|$ for any bounded function $g:S\rightarrow\mathbb{C}$. The analytic content of $\omega$ is then defined by

[TABLE]

The inequalities for $\lambda(\omega)$ given below, which imply the classical isoperimetric inequality, are due to Alexander [2] and Khavinson [15].

Theorem A**.**

Let $A$ and $P$ denote the area and perimeter of $\omega$, respectively. Then

[TABLE]

An exposition of this circle of ideas may be found in Gamelin and Khavinson [10], and a wider survey of related results is provided by Bénéteau and Khavinson [4]. It was shown in [10] that equality with the upper bound occurs if and only if $\omega$ is a disc. Recently, Abanov et al [1] have shown that equality with the lower bound occurs if and only if $\omega$ is a disc or an annulus.

Rewriting $\lambda(\omega)$ as $\inf\{\left\|z-\overline{\phi}\right\|_{\overline{\omega}}:\phi\in\mathcal{A}(\omega)\}$, it can be seen that a natural generalization of this quantity to smoothly bounded domains $\Omega$ in Euclidean space $\mathbb{R}^{N}$ ($N\geq 2$) is given by

[TABLE]

where $A(\Omega)$ denotes the space of harmonic vector fields $f=(f_{1},...,f_{N})\in C(\overline{\Omega})\cap C^{1}(\Omega)$ and

[TABLE]

(Thus $f$ satisfies $\mathrm{div}f=0$ and $\mathrm{curl}f=0$, where the latter condition means that

[TABLE]

If we write

[TABLE]

then

[TABLE]

Let $r_{\Omega}>0$ be chosen so that a ball of radius $r_{\Omega}$ has the same volume as $\Omega$. Gustafsson and Khavinson [12] established the following inequalities for $\lambda(\Omega)$ in higher dimensions.

Theorem B**.**

Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$ ($N\geq 3$) with volume $V$ such that $\partial\Omega$ is the disjoint union of finitely many smooth components with total surface area $P$. Then there exists a constant $c_{N}>1$ such that

[TABLE]

The lower bound in (2) is sharp, since $\lambda(\Omega)=r$ when $\Omega$ is a ball of radius $r$ (see Theorem 3.1 in [12]). Regarding the upper bound, Gustafsson and Khavinson conjectured that the constant $c_{N}$ may be replaced by $1$, in which case (2) would again contain the classical isoperimetric inequality. However, the methods of [12] do not yield such a conclusion. The purpose of this paper is to verify this long-standing conjecture.

Following [12] we define a related domain constant,

[TABLE]

where we have used the observation that a function $u$ in $C^{1}(\overline{\Omega})$ satisfies $\Delta u=N$ on $\Omega$ if and only if the function $h$ defined by $h(x)=\left\|x\right\|^{2}/2-u(x)$ belongs to $\mathcal{H}$. For future reference we note that, for such $u$, it follows from the harmonicity of the partial derivatives of $u$ that $\|\nabla u\|$ is subharmonic on $\Omega$ (cf. Theorem 3.4.5 of [3]), and so

[TABLE]

by the maximum principle.

It follows from (1) that

[TABLE]

(Equality holds when $\Omega$ is simply connected.) Gustafsson and Khavinson actually showed in [12] that $\lambda_{1}(\Omega)\leq c_{N}r_{\Omega}$ for an explicit constant $c_{N}>1$. We will establish the following estimate.

Theorem 1**.**

Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$ such that $\partial\Omega$ is the disjoint union of finitely many smooth components. Then $\lambda_{1}(\Omega)\leq r_{\Omega}$. Further, equality holds if and only if $\Omega$ is a ball.

In the light of Theorem B we immediately arrive at the following conclusion.

Corollary 2**.**

Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}$ with volume $V$ such that $\partial\Omega$ is the disjoint union of finitely many smooth components with total surface area $P$. Then

[TABLE]

Further, $\lambda(\Omega)=r_{\Omega}$ if and only if $\Omega$ is a ball.

The proof of Theorem 1 combines the technique of partial balayage with results from the theory of optimal transport. Later we will discuss separate necessary and sufficient conditions for a function $u$ in $C^{1}(\overline{\Omega})$ to be a minimizer for $\lambda_{1}(\Omega)$ in (3), whenever such minimizers exist.

The purpose of Corollary 2 is to relate analytic content to the isoperimetric inequality. We do not claim that it offers a novel or shorter proof of the latter, since (apart from other more classical proofs) there are already proofs using optimal transport theory as in McCann and Guillen [16], and Cabré [7] had previously provided a short proof of it based on more geometric methods.

2 Proof of Theorem 1

2.1 Tools for the proof

Let $m$ denote Lebesgue measure on $\mathbb{R}^{N}$ and $O\subset\mathbb{R}^{N}$ be a bounded domain. Further, let $G_{O}(\cdot,\cdot)$ denote the Green function of $O$, and $G_{O}\mu,G_{O}\nu$ denote the potentials of (positive) measures $\mu,\nu$ on $O$, where $\nu\ll m$. The Green function is normalized so that $-\Delta G_{O}\gamma=\gamma$ in the sense of distributions for any potential $G_{O}\gamma$. We define

[TABLE]

whence $P_{\mu}^{\nu}\leq G_{O}\mu$, and recall the following facts (see [14] or [11]).

Theorem C**.**

*(a) $P_{\mu}^{\nu}=G_{O}\eta$ for some measure $\eta$ on $O$ satisfying $\eta\leq\nu$.

(b) $\eta=\nu|_{S}+\mu|_{O\backslash S}$, where $S=\{G_{O}\eta<G_{O}\mu\}$.*

We will refer to the measure $\eta$ in the above theorem as the (partial) balayage of $\mu$ onto $\nu$ in $O$, and denote it by $\mathcal{B}\mu$, where $\nu$ is to be understood from the context. We note that $G_{O}\mu-G_{O}\eta$ is the smallest nonnegative lower semicontinuous function $w$ on $O$ satisfying $-\Delta w\geq\mu-\nu$ in the sense of distributions. Thus, if $\mu_{1}\geq\mu$, then the set $S(\mu)$ associated with $\mu$ is contained in the corresponding set $S(\mu_{1})$. It follows that

[TABLE]

We will also need the following lemma. Let $B(x,r)$ denote the open ball in $\mathbb{R}^{N}$ with centre $x$ and radius $r$.

Lemma 3**.**

Let $\nu=Nm|_{O}$ and $\overline{\Omega}\subset\Omega_{0}\subset O$, where $\Omega_{0}$ is another open set. If $\tau$ is a measure with $\mathrm{supp}\tau\subset\overline{\Omega}$, then there exists $b>0$ such that

[TABLE]

**Proof. **Let $\Omega^{\prime}$ be an open set such that $\overline{\Omega}\subset\Omega^{\prime}$ and $\overline{\Omega^{\prime}}\subset\Omega_{0}$, and let

[TABLE]

Let $\tau^{\ast}$ be the sweeping (classical balayage) of $\tau$ onto $\partial\Omega^{\prime}$, and define

[TABLE]

where $\phi_{\varepsilon}$ is a non-negative rotationally invariant $C^{\infty}$ smoothing kernel on $\mathbb{R}^{N}$ with support $\overline{B}(0,\varepsilon)$ (see, for example, Section 3.3 of [3]). We choose $b$ sufficiently small that $b\gamma\leq N$, whence

[TABLE]

Since $G_{O}(\gamma m)\leq G_{O}\tau^{\ast}\leq G_{O}\tau$, with equality outside $\{x\in O:\mathrm{dist}(x,\Omega^{\prime})\leq\varepsilon\}$, we see that

[TABLE]

again with equality outside $\{x\in O:\mathrm{dist}(x,\Omega^{\prime})\leq\varepsilon\}$, and so

[TABLE]

We recall the following composite result from the theory of optimal transport, in which the existence and smoothness of the function $v$ are due to Brenier [5] and Caffarelli [8], respectively. (See also Chapters 3 and 4 of Villani’s book [18].)

Theorem D**.**

Let $D\subset\mathbb{R}^{N}$ be a bounded open set such that $m(\partial D)=0$. Then there exists a convex function $v:\mathbb{R}^{N}\rightarrow(-\infty,\infty]$ which is $C^{2}$ on $D$, and for which $\nabla v$ maps $D$ into $B(0,r_{D})$ and is measure-preserving, in the sense that $m(A)=m((\nabla v)(A))$ for any Borel set $A\subset D$.

Lemma 4**.**

If a measure $\gamma$ on $\Omega$ has bounded density with respect to $m$, then $G_{\Omega}\gamma\in C^{1}(\overline{\Omega})$, and

[TABLE]

To see this, we note that standard arguments (cf. Theorem 4.5.3 of [3]) show that $G_{\Omega}\gamma\in C^{1}(\Omega)$ and that (7) holds when $y\in\Omega$. We now fix $i$ and note (see Widman [19]) that $\left(\partial G_{\Omega}/\partial y_{i}\right)(x,\cdot)$ has a continuous extension to $\overline{\Omega}$ for each $x\in\Omega$. Let $y_{0}\in\partial\Omega$ and $\varepsilon>0$, and define

[TABLE]

where $\Omega_{1}=\Omega\backslash B(y_{0},\varepsilon)$ and $\Omega_{2}=\Omega\cap B(y_{0},\varepsilon)$. Then $\psi_{1}$ is continuous at $y_{0}$, and (by estimates in [19])

[TABLE]

It follows that

[TABLE]

2.2 Proof of the inequality

Let

[TABLE]

and let $D$ be a bounded open set such that $\overline{\Omega}\subset D$, $m(\partial D)=0$ and $m(D\backslash\Omega)<b_{N}m(\Omega)$. We next choose $v$ as in Theorem D. Since $\nabla v$ is measure-preserving on $D$, the Hessian of $v$, which is positive semi-definite because $v$ is convex, has determinant equal to $1$, and so $\Delta v\geq N$ by the arithmetic-geometric means inequality for the eigenvalues of the Hessian (cf. the argument in Section 1.6 of McCann and Guillen [16]). It will be enough to show that $\lambda_{1}(\Omega)\leq r_{D}$, since $r_{D}$ can be made arbitrarily close to $r_{\Omega}$. This inequality trivially holds if $\Delta v\equiv N$ on $\Omega$, so we assume from now on that $(\Delta v-N)m|_{\Omega}\neq 0$.

Let $R$ be an open set satisfying $\overline{\Omega}\subset R$ and $\overline{R}\subset D$. Since

[TABLE]

and $\nabla v$ is measure preserving on $D$, we see that

[TABLE]

Also, since $v$ is convex, the function

[TABLE]

equals $v$ on $R$. Clearly $w$ is subharmonic (and indeed convex) on $\mathbb{R}^{N}$.

We now define $\varepsilon=2^{-1}\mathrm{dist}(\overline{\Omega},\mathbb{R}^{N}\backslash R)$ and

[TABLE]

where $\phi_{\varepsilon}$ is a smoothing kernel, as before. Then $w_{\varepsilon}$ is also subharmonic (and convex) on $\mathbb{R}^{N}$, and $w_{\varepsilon}\geq w$ (see Theorem 3.3.3 in [3]). We further define

[TABLE]

where

[TABLE]

The set $O$ clearly contains $\overline{R}$. It is also bounded, since for any $x\in\partial B(0,1)$ we see from (8) that there exists $y_{x}\in R$ such that $\nabla v(y_{x})\cdot x\geq r_{D}/2$, and so

[TABLE]

By Sard’s theorem and the smoothness of $w_{\varepsilon}$, we can choose $C$ such that the set $O$ is smoothly bounded and

[TABLE]

Since $w_{\varepsilon}=C$ on $\partial O$, the function $C-w_{\varepsilon}$ is the potential $G_{O}\mu$ of the measure $\mu=(\Delta w_{\varepsilon})m$ on $O$. Further, $\mu\geq Nm|_{\Omega}$, since $\Delta w=\Delta v\geq N$ on $R$, and so

[TABLE]

We next apply Theorem C with $\nu=Nm|_{O}$. The partial balayage $\eta=\mathcal{B}\mu$ satisfies $\eta=Nm|_{S}+\mu|_{O\setminus S},$ where $S=\{G_{O}\eta<G_{O}\mu\}$. Since $G_{O}\mu\geq G_{O}\eta$ and $\mu=(\Delta w_{\varepsilon})m\geq Nm=\eta$ on $\Omega$, the function $G_{O}\mu-G_{O}\eta$ is nonnegative and superharmonic on $\Omega$. In fact, since $\Omega$ is connected and

[TABLE]

it is strictly positive there by the minimum principle, and so $\Omega\subset S$. We note that $G_{O}\mu,G_{O}\eta\in C^{1}(\overline{O})$, by Lemma 4. Since the nonnegative function $G_{O}\mu-G_{O}\eta$ achieves its minimum value [math] on $O\backslash S$, we have $\nabla G_{O}\mu=\nabla G_{O}\eta$ on $\partial S\cap O$. Also, since $G_{O}\eta\leq G_{O}\mu$, we have

[TABLE]

where $n$ denotes the outward unit normal to $\partial O$. Thus

[TABLE]

because $w$ (and hence also $w_{\varepsilon}$) is Lipschitz on $\mathbb{R}^{N}$ with Lipschitz constant at most $r_{D}$. Finally, since $\Omega\subset S$ and $\eta=Nm$ in $S$, it follows (see (4)) that

[TABLE]

as desired.

2.3 The case of equality

The following result strengthens the conclusion of the previous section.

Proposition 5**.**

Let $v$ be as in Theorem D, with $D=\Omega$. If $(\Delta v-N)m|_{\Omega}\neq 0$, then there is a domain $U$ containing $\overline{\Omega}$ and a function $u\in C^{1}(U)$ satisfying $\Delta u=N$ and $\left\|\nabla u\right\|\leq r_{\Omega}$ in $U$.

**Proof. **We may assume that $v(x_{0})=0$ for some $x_{0}\in\Omega$. Let

[TABLE]

where $(\delta_{k})_{k\geq 0}$ is a strictly decreasing sequence of positive numbers with limit [math] and $\delta_{0}<1$ is chosen small enough so that $m(D_{0}\backslash\Omega)<b_{N}m(\Omega)$. For each $k$ we choose $v_{k}$ as in Theorem D, with $D=D_{k}$, and such that $v_{k}(x_{0})=0$. We next choose open sets $R_{k}$ such that $\overline{\Omega}\subset R_{k}$ and $\overline{R_{k}}\subset D_{k}$, and define $\varepsilon_{k}=2^{-1}\mathrm{dist}(\overline{\Omega},\mathbb{R}^{N}\setminus R_{k})$.

For each $k$ we apply the construction of §2.2 with $D=D_{k},R=R_{k},v=v_{k},$ and $\varepsilon=\varepsilon_{k}$. Propositions 3.1 and 3.2 of Brenier [6], applied to the measures

[TABLE]

show that $v_{k}\rightarrow v$ uniformly on $\Omega$.

The functions $w$ and $w_{\varepsilon}$ in §2.2 will now be denoted, by abuse of notation, $w_{k}$ and $w_{k,\varepsilon_{k}}$ respectively. Since $\|\nabla w_{k}\|\leq r_{D_{k}}\leq r_{D_{0}}$ on $\mathbb{R}^{N}$ by construction, we see that $|w_{k}-w_{k,\varepsilon_{k}}|\leq r_{D_{0}}\varepsilon_{k}$ on $\mathbb{R}^{N}$. Hence $\left({w_{k,\varepsilon_{k}}}\right)$ converges uniformly to $v$ on $\Omega$, in view of the fact that $v_{k}=w_{k}$ on $R_{k}$, which contains $\Omega$.

We now choose numbers $C_{k}$ so that the set

[TABLE]

satisfies $D_{0}\subset O_{k}$ for each $k$. Since the sequences $\left(\left\|v_{k}\right\|_{L^{\infty}(D_{k})}\right)$ and

$\left(\left\|\nabla v_{k}\right\|_{L^{\infty}(D_{k})}\right)$ are bounded, we can furthermore arrange that the set

$O=\cup_{k}O_{k}$ is bounded.

Let $\gamma_{k}=(\Delta w_{k,\varepsilon_{k}})m|_{\Omega}$ and $\gamma=(\Delta v)m|_{\Omega}$. These are non-negative measures, and the divergence theorem shows that

[TABLE]

where $\sigma$ denotes surface area measure on $\partial\Omega$. Since $\int_{\Omega}\psi d\gamma_{k}\rightarrow\int_{\Omega}\psi d\gamma$ for any $\psi\in C_{c}^{2}(\Omega)$, we see from (10) and the density of $C_{c}^{2}(\Omega)$ in $C_{0}(\Omega)$ that $(\gamma_{k})$ is weak* convergent to $\gamma$ on $\Omega$. Further, $\gamma_{k}\geq Nm|_{\Omega}$ and $\gamma\geq Nm|_{\Omega}$, as in §2.2.

Since $(\gamma-Nm)(\Omega)>0$ by assumption, we can choose a compact set $K\subset\Omega$ such that $\alpha>0$, where $\alpha=(\gamma-Nm)(K^{\circ})$. It follows that $(\gamma_{k}-Nm)(K^{\circ})\geq\alpha/2$ for all sufficiently large $k$. Let $\gamma_{k}^{\ast}$ denote the sweeping of $(\gamma_{k}-Nm)|_{K^{\circ}}$ onto $\partial\Omega$. Then there exists $b>0$ such that $\gamma_{k}^{\ast}\geq b\sigma$ for all sufficiently large $k$.

If we first consider partial balayage in $O$, then

[TABLE]

for some domain $U$ containing $\overline{\Omega}$. Further, if we choose $b$ sufficiently small, then $\overline{U}\subset D_{0}$, by Lemma 3. By definition

[TABLE]

Since $\overline{U}\subset D_{0}\subset O_{k}\subset O$, this implies that

[TABLE]

Thus (11) remains valid if we henceforth consider partial balayage in $O_{k}$.

Now let $\eta_{k}$ be the measure $\eta$ constructed as in §2.2. Since

[TABLE]

we see from (6) that

[TABLE]

By choosing a suitable subsequence of $(\delta_{k})$ we can arrange that $(G_{O_{k}}\eta_{k})$ converges locally uniformly on $U$ to a function $u$ in $C^{1}(U)$ satisfying $\Delta u=N$ in $U$, and then $(\nabla G_{O_{k}}\eta_{k})$ also converges locally uniformly on $U$ to $\nabla u$. From the final inequality in (9) we see that $\left\|\nabla u\right\|\leq\lim_{k\rightarrow\infty}r_{D_{k}}=r_{\Omega}$ on $U$, so the proposition is proved.

We can now easily address the case of equality in Theorem 1. We first assume that $\lambda_{1}(\Omega)=r_{\Omega}$ and choose $v$ as in Theorem D, with $D=\Omega$. We claim that $\Delta v\equiv N$ on $\Omega$. Indeed, if this were not the case, then there would exist$\ u,U$ as in the above proposition. Since $r_{\Omega}\leq\left\|\nabla u\right\|_{\partial\Omega}$, the subharmonic function $\left\|\nabla u\right\|$ would then achieve its maximum inside $U$, forcing it to be constant. Thus $\left\|\mathrm{Hess}(u)\right\|^{2}=\Delta(\left\|\nabla u\right\|^{2})=0$, contradicting the fact that $\Delta u=N$ on $U$. Hence $\Delta v=N$ on $\Omega$. Since the Hessian of $v$ has determinant equal to $1$, all its eigenvalues are $1$, and so it is the identity matrix. Thus $\nabla v$ is a translation, and $\Omega$ is a ball (of radius $r_{\Omega}$).

Conversely, if $\Omega$ is a ball of radius $r$, then (as we noted in the introduction) $\lambda_{1}(\Omega)=\lambda(\Omega)=r$.

3 Minimizers for $\lambda_{1}(\Omega)$

In general we do not know whether there exist minimizers $u$ for the definition of $\lambda_{1}(\Omega)$ in (3). However, we can give separate necessary and sufficient conditions for a function $u$ in $C^{1}(\overline{\Omega})$ to be a minimizer when such minimizers exist. We begin with a necessary condition.

Proposition 6**.**

Suppose that $u\in C^{1}(\overline{\Omega})$ and $\Delta u=N$ in $\Omega$. If $\|\nabla u\|_{\overline{\Omega}}=\lambda_{1}(\Omega),$ then $\|\nabla u\|$ is constant on $\partial\Omega$ (whence $\left\|\nabla u\right\|=\lambda_{1}(\Omega)$ on $\partial\Omega$, by (4)).

**Proof. **Let $\lambda_{1}=\lambda_{1}(\Omega)$. Suppose that $u\in C^{1}(\overline{\Omega})$ satisfies $\Delta u=N$ and $\|\nabla u\|_{\overline{\Omega}}=\lambda_{1}(\Omega)$, and that there exists $y_{0}\in\partial\Omega$ such that $\left\|\nabla u(y_{0})\right\|<\lambda_{1}$. Then we can choose an open ball $B_{0}$ centred at $y_{0}$ such that

[TABLE]

We choose $\Omega_{0}$ to be a domain with $C^{2}$ boundary such that $\Omega\subset\Omega_{0}\subset\Omega\cup B_{0}$ and $\Omega_{0}\backslash\overline{\Omega}\neq\emptyset$.

Let $f=-G_{\Omega}(Nm)$, let $g$ be a continuous extension of $\left.\dfrac{\partial f}{\partial n}\right|_{\partial\Omega\cap\partial\Omega_{0}}$ to $\partial\Omega_{0}$ and $0<\varepsilon<\lambda_{1}$. Proposition 4 of Sakai [17] tells us that there is a finite sum $\mu$ of point masses (of variable sign) in $\Omega_{0}\backslash\overline{\Omega}$ satisfying

[TABLE]

Since

[TABLE]

there exists $\kappa>0$ such that

[TABLE]

where

[TABLE]

Now let $\delta>0$ and

[TABLE]

Clearly $u_{\delta}\in C^{1}(\overline{\Omega})$, $\Delta u_{\delta}=N$ on $\Omega$ and

[TABLE]

Hence, by (13),

[TABLE]

Also,

[TABLE]

and, in view of (12), the right hand side above can be made less than $\lambda_{1}$ by choosing $\delta$ to be sufficiently small. Combining (14) and (15), we obtain a contradiction to the hypothesis that $u$ achieves the minimum value in (3). Hence $\left\|\nabla u\right\|=\lambda_{1}$ on $\partial\Omega$, as claimed.

The converse to the above proposition is false, as we will now explain. Let us recall that $\lambda_{1}(B(0,1))=1$. We claim that, when $N\geq 3$, there are functions $u\in C^{1}(\overline{B(0,1)})$ satisfying $\Delta u=N$ on $B(0,1)$ and $\left\|\nabla u\right\|=c$ on $\partial B(0,1)$, yet $\lambda_{1}(B(0,1))\neq c$. For example, if we define

[TABLE]

then $\Delta u=N$ on $B(0,1)$ and $\left\|\nabla u\right\|=N/(N-2)$ on $\partial B(0,1)$.

Similarly, if $N=2$ and $E=\{(x,y):4x^{2}+y^{2}<1\}$, then the function $u(x,y)=2x^{2}-y^{2}$ satisfies $\Delta u=2$ on $E$ and $\left\|\nabla u\right\|=2$ on $\partial E$. However, as we will see later, the actual value of $\lambda_{1}(E)$ is $2/3$.

Next we give a sufficient condition for a function $u\in C^{1}(\overline{\Omega})$ to be a minimizer for the definition of $\lambda_{1}(\Omega)$ in (3).

Proposition 7**.**

Let $u\in C^{1}(\overline{\Omega})$, where $\Delta u=N$ on $\Omega$. If $\left\|\nabla u\right\|$ is constant on $\partial\Omega$ and $\nabla u\cdot n\geq 0$ on $\partial\Omega$, then $\lambda_{1}(\Omega)=\left\|\nabla u\right\|_{\overline{\Omega}}$. Further, any two functions satisfying these hypotheses differ only by a constant.

**Proof. **Suppose that $\left\|\nabla u\right\|=c$ and $\nabla u\cdot n\geq 0$ on $\partial\Omega$. Let $v\in C^{1}(\overline{\Omega})$, where $\Delta v=N$ on $\Omega$, and define $w=v-u$. Then $w\in\mathcal{H}$. We choose a point $y\in\partial\Omega$ at which $w$ achieves its maximum value. If $w$ is non-constant, then the Hopf boundary point lemma (see Section 6.4.2 of Evans [9]) tells us that $\nabla w(y)\cdot n_{y}>0$ and so $\nabla w(y)$ is actually a positive multiple of $n_{y}$. Since $\nabla u\cdot n\geq 0$ on $\partial\Omega$, we now see that

[TABLE]

It follows that $c=\lambda_{1}(\Omega)$, as required.

Finally, the preceding argument shows that any two functions satisfying the hypotheses differ only by a constant.

A further useful sufficient condition for a function $u$ to be a minimizer in (3) applies when $u$ is locally convex.

Theorem 8**.**

*Let $u\in C^{1}(\overline{\Omega})$, where $u$ is locally convex on $\Omega$ and satisfies $\Delta u=N$ there. If $\|\nabla u\|=c$ on $\partial\Omega$, then

(i) $\lambda_{1}(\Omega)=c$;

(ii) $\nabla u:\Omega\rightarrow B(0,\lambda_{1}(\Omega))$ is surjective.*

**Proof. **Let $v=\|\nabla u\|^{2}$. Then $\Delta v=2\left\|\mathrm{Hess}(u)\right\|^{2}\geq 0$ on $\Omega$. In fact, $\Delta v>0$ on a dense open subset of $\Omega$, for otherwise $\nabla u$ would be constant on a nonempty open set, contradicting the hypothesis that $\Delta u=N$. Let $\Omega_{\varepsilon}=\{x\in\Omega:\|\nabla u(x)\|<c-\varepsilon\}$, where $\varepsilon>0$ is sufficiently small that $\Omega_{\varepsilon}\neq\emptyset$. Clearly $\overline{\Omega}_{\varepsilon}\subset\Omega$. Further, by the Hopf boundary point lemma, $\partial v/\partial n>0$ on $\partial\Omega_{\varepsilon}$, so $n=\nabla v/\|\nabla v\|$ on $\partial\Omega_{\varepsilon}$. Since $\mathrm{Hess}(u)$ is positive semidefinite on $\Omega$,

[TABLE]

Given any $x\in\partial\Omega$, we choose $B(y,r)\subset\mathbb{R}^{N}\backslash\overline{\Omega}$ such that $\partial B(y,r)\cap\overline{\Omega}=\{x\}$ and then $x_{\varepsilon}\in\partial\Omega_{\varepsilon}$ satisfying $\left\|x_{\varepsilon}-y\right\|=\mathrm{dist}(y,\overline{\Omega}_{\varepsilon})$ to see that

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so part (i) follows from Proposition 7. We note, for future reference, that

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We will now show, further, that $\nabla u\cdot n>0$ on $\partial\Omega_{\varepsilon}$. We write $x=(x^{\prime},x_{N})\in\mathbb{R}^{N-1}\times\mathbb{R}$ and choose our co-ordinate system so that $0\in\partial\Omega_{\varepsilon}$, that the normal $n_{0}$ is in the direction of $(0,...,0,1)$, and that the Hessian of the function $x^{\prime}\mapsto u(x^{\prime},0)$ at $0^{\prime}$ is a diagonal matrix. We know from the proof of part (i) that $(\partial u/\partial x_{N})(0)\geq 0$. Now suppose, for the sake of contradiction, that $(\partial u/\partial x_{N})(0)=0$. Since $v$ is constant on $\partial\Omega_{\varepsilon}$ we see that

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We reorder the first $N-1$ coordinates so that, for some $m\in\{1,...,N\}$,

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and define

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By (17) the Hessian of the function $(x_{m},...,x_{N})\mapsto u(0,...,0,x_{m},...,x_{N})$ at $(0,...,0)$ has the form

[TABLE]

Hence $b_{i}=0$ when $m\leq i\leq N-1$, because this submatrix of $\mathrm{Hess}(u)$ is also positive semidefinite. By (17) and the Hopf boundary point lemma, we now arrive at the contradiction

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Thus

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We will now establish (ii). Let $y\in\partial B(0,1)$ and define

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Then $\nabla u_{t}=\nabla u-ty$ and $\Delta u_{t}=N$ in $\Omega_{\varepsilon}$. Let

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It follows from (18) that $u$ cannot attain the value $\min_{\overline{\Omega}_{\varepsilon}}u$ on $\partial\Omega_{\varepsilon}$, so $0\in A$. We will now show that $A$ is both open and closed relative to $[0,\lambda_{1}(\Omega_{\varepsilon}))$. To see that $A$ is closed, let $(t^{(k)})$ be a sequence in $A$ that converges to some $t\in[0,\lambda_{1}(\Omega_{\varepsilon}))$. There exist points $x^{(k)}$ in $\Omega_{\varepsilon}$ such that $\nabla u_{t^{(k)}}(x^{(k)})=0$ and, by choosing a subsequence, we may arrange that $(x^{(k)})$ converges to some point $x\in\overline{\Omega}_{\varepsilon}$. Clearly $\nabla u_{t}(x)=0$, so $x\notin\partial\Omega_{\varepsilon}$, because

[TABLE]

by (16). Hence $A$ is closed. To see that $A$ is open, let $t\in A$, choose $x\in\Omega$ such that $\nabla u_{t}(x)=0$, and define

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Then $\alpha>0$ and $x\in\Omega^{\prime}$. We can apply the result of the previous paragraph to $u_{t}$ to see that $\nabla u_{t}\cdot n>0$  on  $\partial\Omega^{\prime}$. When $\left|s\right|$ is sufficiently small, the function $u_{t+s}$ thus also has a strictly positive normal derivative on $\partial\Omega^{\prime}$, and so attains the value $\min_{\overline{\Omega^{\prime}}}u_{t+s}$ at some point $x^{(s)}\in\Omega^{\prime}$. Since $\nabla u_{t+s}(x^{(s)})=0$, we see that the set $A$ is also open relative to $[0,\lambda_{1}(\Omega_{\varepsilon}))$.

Hence $A=[0,\lambda_{1}(\Omega_{\varepsilon}))$. It follows that, for any $z\in B(0,\lambda_{1}(\Omega_{\varepsilon}))$, there exists $x\in\Omega_{\varepsilon}$ such that $\nabla u(x)=z$, and so $\nabla u:\Omega_{\varepsilon}\rightarrow B(0,\lambda_{1}(\Omega_{\varepsilon}))$ is surjective. Finally, we let $\varepsilon\rightarrow 0+$ and note from (16) that $\lambda_{1}(\Omega_{\varepsilon})\rightarrow\lambda_{1}(\Omega)$.

**Example. **Consider the ellipsoid

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where $a_{i}>0$ $(i=1,...,N)$. The function

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is clearly convex, satisfies $\Delta u=N$ on $E$ and $\left\|\nabla u(x)\right\|=N/\sum_{i=1}^{N}a_{i}$ on $\partial E$. Thus

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Finally, we remark that minimizers need not be locally convex functions. For example, if $N\geq 3$ and $\Omega=B(0,R)\backslash\overline{B(0,r)}$, where $R>r>0$, then the function

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satisfies $\Delta u=N$ on $\Omega$ and $\nabla u=cn$ on $\partial\Omega$, where

[TABLE]

Thus $\lambda_{1}(\Omega)=c$, by Proposition 7. However, along the $x_{N}$-axis, $\mathrm{Hess}(u)$ is a diagonal matrix in which the first $N-1$ diagonal entries are valued

[TABLE]

so $u$ is not locally convex near the inner boundary. (An analogous example when $N=2$ may be obtained by replacing $\|x\|^{2-N}/(N-2)$ with $\log(1/\|x\|)$ in the formula for $u(x)$ above.)