The isodiametric problem on the sphere and in the hypebolic space

TL;DR

This paper establishes the isodiametric inequality in both spherical and hyperbolic geometries, extending classical geometric results to non-Euclidean spaces.

Contribution

The paper provides the first proof of the isodiametric inequality in spherical and hyperbolic spaces, generalizing Euclidean results to curved geometries.

Findings

Proved the isodiametric inequality in spherical space.

Proved the isodiametric inequality in hyperbolic space.

Extended classical geometric inequalities to non-Euclidean geometries.

Abstract

We prove the isodiametric inequality in the spherical and in the hyperbolic space

Equations

$$B_{{\mathcal{M}}^n}(z,r)=\{x\in {\mathcal{M}}^n:d(x,z)\le r\}$$

$$V(X)\le V(B(o,D/2)),$$

$$V(X)\le V(B(z_0,D/2)),$$

$$S^n$$

$$H^n$$

$$\mathcal{B}(x,y)=ts-⟨x_0,y_0⟩.$$

$$\mathcal{B}(x,x)=1\mbox{ for $x\in H^{n}$}.$$

$$T_z$$

$$x=(\mathrm{ch}t)z+(\mathrm{sh}t)u$$

$$x=(\cos t)z+(\sin t)u$$

$$x=z+tu$$

$$\{x\in H^n:\mathcal{B}(x,v)=0\}$$

$$\{x\in S^n:⟨x,v⟩=0\}$$

$$\{x\in {\mathbb{R}}^n:⟨x,v⟩=0\}$$

$$-\mathcal{B}(N_C(x),N_C(x))$$

$$⟨N_C(x),N_C(x)⟩$$

$$y\in {\partial }_{{\mathcal{M}}^n}{B}_{{\mathcal{M}}^n}(z,r)\cap {\partial }_{{\mathcal{M}}^n}C.$$

$${\partial }_{{\mathcal{M}}^n}C\subset \Xi .$$

$${\partial }_{{\mathcal{M}}^n}C=\Xi .$$

$$\pi (x)=\frac{x}{⟨x,e⟩}.$$

$$\pi (L\cap H^n)$$

$$\pi \left(L\cap {\mathrm{int}}_{S^n}{B}_{S^n}\left(e,\frac{\pi }{2}\right)\right)$$

$${\delta }_{{\mathcal{M}}^n}(C_1,C_2)=\max \left\{\underset{x\in C_2}{\max }{d}_{{\mathcal{M}}^n}(x,C_1),\underset{y\in C_1}{\max }{d}_{{\mathcal{M}}^n}(y,C_2)\right\}.$$

$$\underset{m^{\prime }\to \infty }{\lim }{d}_{{\mathcal{M}}^n}(x_{m^{\prime }},C)=\underset{m\to \infty }{\mathrm{limsup}}{d}_{{\mathcal{M}}^n}(x_m,C)\text{mbox}and\underset{m"\to \infty }{\lim }{d}_{{\mathcal{M}}^n}(y_{m"},C_{m"})=\underset{m\to \infty }{\mathrm{limsup}}{d}_{{\mathcal{M}}^n}(y_m,C_m).$$

$$U_r=\bigcup _{x\in C}{\mathrm{int}}_{{\mathcal{M}}^n}{B}_{{\mathcal{M}}^n}(x,r)$$

$$V_{{\mathcal{M}}^n}(C)=\sup \{V_{{\mathcal{M}}^n}(X):X\subset {\mathcal{M}}^n\text{mbox}compactand{\mathrm{diam}}_{{\mathcal{M}}^n}X\le D\}.$$

$$\underset{m\to \infty }{\lim }{V}_{{\mathcal{M}}^n}(C_m)=\sup \{V_{{\mathcal{M}}^n}(X):X\subset {\mathcal{M}}^n\text{mbox}compactand{\mathrm{diam}}_{{\mathcal{M}}^n}X\le D\}.$$

$${\alpha }_1{x}_1+\dots +{\alpha }_{n+2}{x}_{n+2}=z.$$

$$⟨z,z⟩>\left(\sum _{i=1}^{n+2}{\alpha }_i^2\right)-\left(\sum _{i<j}\frac{2{\alpha }_i{\alpha }_j}{n+1}\right)\ge \left(\sum _{i=1}^{n+2}{\alpha }_i^2\right)-\left(\sum _{i<j}\frac{{\alpha }_i^2+{\alpha }_j^2}{n+1}\right)=0$$

$$\pi \left(X\cap B_{S^n}\left(z,\frac{\pi }{2}\right)\right)=\pi (X)\cap H^{+},$$

$$d_{{\mathcal{M}}^n}(x,y)$$

Full text

1. Introduction

Let $\mathcal{M}^{n}$ be either the Euclidean space ${\mathbb{R}}^{n}$, hyperbolic space $H^{n}$ or spherical space $S^{n}$ for $n\geq 2$. We write $V_{\mathcal{M}^{n}}$ to denote the $n$-dimensional volume (Lebesgue measure) on $\mathcal{M}^{n}$, and $d_{\mathcal{M}^{n}}(x,y)$ to denote the geodesic distance between $x,y\in\mathcal{M}^{n}$.

For a bounded set $X\subset\mathcal{M}^{n}$, its diameter ${\rm diam}_{\mathcal{M}^{n}}X$ is the supremum of the geodesic distances $d_{\mathcal{M}^{n}}(x,y)$ for $x,y\in X$. For $D>0$ and $n\geq 1$, our goal is to determine the maximal volume of a subset of $\mathcal{M}^{n}$ of diameter at most $D$. For any $z\in\mathcal{M}^{n}$ and $r>0$, let

[TABLE]

be the $n$-dimensional ball centered at $z$ where it is natural to assume $r<\pi$ if $\mathcal{M}^{n}=S^{n}$. When it is clear from the context what space we consider, we drop the subscript referring to the ambient space. In order to speak about the volume of a ball of radius $r$, we fix a reference point $z_{0}\in\mathcal{M}^{n}$ where $z_{0}=o$ the origin if $\mathcal{M}^{n}={\mathbb{R}}^{n}$.

It is well-known, due to Bieberbach [4] in ${\mathbb{R}}^{2}$ and P. Urysohn [8] in ${\mathbb{R}}^{n}$, that if $X\subset{\mathbb{R}}^{n}$ is measurable and bounded with ${\rm diam}X=D$, then

[TABLE]

and equality holds if and only if the closure of $X$ is a ball of radius $D/2$.

We prove the following hyperbolic- and spherical analogues of (1).

Theorem 1.1**.**

If $D>0$ and $X\subset H^{n}$ is measurable and bounded with ${\rm diam}X\leq D$, then

[TABLE]

and equality holds if and only if the closure of $X$ is a ball of radius $D/2$.

Theorem 1.2**.**

If $D\in(0,\pi)$ and $X\subset S^{n}$ is measurable with ${\rm diam}X\leq D$, then

[TABLE]

and equality holds if and only if the closure of $X$ is a ball of radius $D/2$.

On $S^{2}$, Hernández Cifre, Martínez Fernández [6] proved a stronger version of Theorem 1.2 for centrally symmetric sets of diameter less than $\pi/2$.

The proofs of Theorem 1.1 and Theorem 1.2 build strongly on ideas related to two-point symmetrization in the paper Aubrun, Fradelizi [1]. After reviewing the basic properties of spaces of constant curvature in Secion 2, we consider the extremal sets in Section 3 and convex sets in Section 4 from our point of view. The main tool of this paper, two-point symmetrization, is introduced in Section 5 where we actually prove Theorem 1.2 if $D\leq\frac{\pi}{2}$ and Theorem 1.1. The reason why the argument is reasonably simple (kind of a proof from Erdős’ Book) for Theorem 1.2 if $D\leq\frac{\pi}{2}$ and for Theorem 1.1 is that it is easy to show that the extremal sets are convex in these cases. However, if $D>\frac{\pi}{2}$ in the spherical case, then a priori much less information is availabe on the extremal sets, therefore a more technical argument is provided in Section 6.

2. Spaces of constant curvature

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$. Our focus is on the spherical- and hyperbolic space, and we assume that $S^{n}$ is embedded into ${\mathbb{R}}^{n+1}$ the standard way, and $H^{n}$ is embedded into ${\mathbb{R}}^{n+1}$ using the hyperboloid model. We write $\langle\cdot,\cdot\rangle$ to denote the standard scalar product in ${\mathbb{R}}^{n+1}$, write $z^{\bot}=\{x\in{\mathbb{R}}^{n+1}:\,\langle x,e\rangle=0\}$ for $z\in{\mathbb{R}}^{n+1}\backslash o$ and fix an $e\in{\mathbb{R}}^{n+1}$. In particular, we have

[TABLE]

For $H^{n}$, we also consider the following symmetric bilinear form $\mathcal{B}$ on ${\mathbb{R}}^{n+1}$: If $x=x_{0}+te\in{\mathbb{R}}^{n+1}$ and $y=y_{0}+se\in{\mathbb{R}}^{n+1}$ for $x_{0},y_{0}\in e^{\bot}$ and $t,s\in{\mathbb{R}}$, then

[TABLE]

In particular,

[TABLE]

Again let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$ using the models as above for $H^{n}$ and $S^{n}$. For $z\in\mathcal{M}^{n}$, we define the tangent space $T_{z}$ as

[TABLE]

We observe that $T_{z}$ is an $n$-dimensional real vector space equipped with the scalar product $-\mathcal{B}(\cdot,\cdot)$ if $\mathcal{M}^{n}=H^{n}$, and with the scalar product $\langle\cdot,\cdot\rangle$ if $\mathcal{M}^{n}=S^{n}$ or $\mathcal{M}^{n}={\mathbb{R}}^{n}$.

Let us consider lines and $(n-1)$-dimensional (totally geodesic) subspaces of $\mathcal{M}^{n}$. A line $\ell\subset\mathcal{M}^{n}$ passing through a $z\in\mathcal{M}^{n}$ is given by a unit vector $u\in T_{z}$; namely, $-\mathcal{B}(u,u)=1$ if $\mathcal{M}^{n}=H^{n}$, and $\langle u,u\rangle=1$ if $\mathcal{M}^{n}=S^{n}$ or $\mathcal{M}^{n}={\mathbb{R}}^{n}$, and the line $\ell$ is parameterized by

[TABLE]

where $d_{\mathcal{M}^{n}}(x,z)=|t|$. In addition, the $(n-1)$-dimensional subspace of $\mathcal{M}^{n}$ passing through $z\in\mathcal{M}^{n}$ and having normal vector $v\in T_{z}\backslash\{o\}$ is

[TABLE]

We say that a non-empty compact set $C\subset\mathcal{M}^{n}$ with $C\neq S^{n}$ is solid, if $C$ is the closure of ${\rm int}_{\mathcal{M}^{n}}C$. For such a $C$, we say that $x\in\partial_{\mathcal{M}^{n}}C$ is strongly regular if there exists $r>0$ and $z\in C$ such that $B_{\mathcal{M}^{n}}(z,r)\subset C$ and $x\in\partial_{\mathcal{M}^{n}}B_{\mathcal{M}^{n}}(z,r)$. In this case, we set $N_{C}(x)\in T_{x}$ to be an exterior unit vector at $x$; namely,

[TABLE]

Lemma 2.1**.**

If $C\subset\mathcal{M}^{n}$ is solid, then the strongly regular boundary points are dense in $\partial_{\mathcal{M}^{n}}C$.

Proof.

Let $z\in\partial_{\mathcal{M}^{n}}C$, and let $\varepsilon>0$. As $C$ is solid, we may choose an $x\in{\rm int}_{\mathcal{M}^{n}}C$ such that $d_{\mathcal{M}^{n}}(x,z)<\varepsilon/2$. Let $r>0$ be maximal with the property that $B_{\mathcal{M}^{n}}(x,r)\subset C$, and hence $r\leq d_{\mathcal{M}^{n}}(x,z)$ and there exists

[TABLE]

Therefore $y$ is a strongly regular boundary point, and the triangle inequality yields that $d_{\mathcal{M}^{n}}(y,z)<\varepsilon$. ∎

According to Vinberg [9], the “Standard Hypersurfaces” in either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$ are as follows:

• •

Hyperplanes of the form $\{x\in{\mathbb{R}}^{n}:\,\langle x,p\rangle=t\}$ in ${\mathbb{R}}^{n}$ for $p\in{\mathbb{R}}^{n}\backslash o$ and $t\in{\mathbb{R}}$;

• •

$\partial_{{\mathbb{R}}^{n}}B(z,r)=\{x\in{\mathbb{R}}^{n}:\,\langle x-z,x-z\rangle=r^{2}\}$ in ${\mathbb{R}}^{n}$ for $z\in{\mathbb{R}}^{n}$ and $r>0$;

• •

$\partial_{S^{n}}B(z,r)=\{x\in S^{n}:\,\langle x,z\rangle=\cos r\}$ in $S^{n}$ for $z\in S^{n}$ and $r\in(0,\pi)$, hence for any $p\in{\mathbb{R}}^{n+1}\backslash o$ and $t\in{\mathbb{R}}$, the set $\{x\in S^{n}:\,\langle x,p\rangle=t\}$ is either empty, a point, or the boundary of a spherical ball;

• •

$\partial_{H^{n}}B(z,r)=\{x\in H^{n}:\,\langle x,z\rangle={\rm ch}\,r\}$ in $H^{n}$ for $r>0$ and $z\in H^{n}$, hence for any $t\in{\mathbb{R}}$ and $p\in{\mathbb{R}}^{n+1}\backslash o$ with $\mathcal{B}(p,p)>0$, the set $\{x\in H^{n}:\,\mathcal{B}(x,p)=t\}$ is either empty, a point, or the boundary of a hyperbolic ball;

• •

Hyperplanes of the form $\{x\in H^{n}:\,\mathcal{B}(x,p)=0\}$ in $H^{n}$ for $p\in{\mathbb{R}}^{n}\backslash o$ with $\mathcal{B}(p,p)<0$;

• •

Hyperspheres of the form $\{x\in H^{n}:\,\mathcal{B}(x,p)=t\}$ in $H^{n}$ for $t\in{\mathbb{R}}\backslash 0$ and $p\in{\mathbb{R}}^{n}\backslash o$ with $\mathcal{B}(p,p)<0$;

• •

Horospheres of the form $\{x\in H^{n}:\,\mathcal{B}(x,p)=t\}$ in $H^{n}$ for $t\in{\mathbb{R}}\backslash 0$ and $p\in{\mathbb{R}}^{n}\backslash o$ with $\mathcal{B}(p,p)=0$.

We note the following properties.

Lemma 2.2**.**

If $\Xi$ is a standard hypersurface in $\mathcal{M}^{n}$ where $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, then $\mathcal{M}^{n}\backslash\Xi$ has two connected components, the boundary of both components is $\Xi$, and any of these components is bounded if and only if the component is an open ball.

Corollary 2.3**.**

If $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, and $C\subset\mathcal{M}^{n}$, $C\neq\mathcal{M}^{n}$ is a solid set whose strongly regular boundary points are contained in a fixed standard hypersurface, then $C$ is a ball.

Proof.

Let $\Xi$ be standard hypersurface containing the strongly regular boundary points of $C$. Since $\Xi$ is closed and strongly regular boundary points are dense in $\partial_{\mathcal{M}^{n}}C$ according to Lemma 2.1, we have

[TABLE]

Next we prove

[TABLE]

We suppose that there exists $z\in\Xi\backslash\partial_{\mathcal{M}^{n}}C$, and seek a contradiction. We consider some $y\in({\rm int}_{\mathcal{M}^{n}}C)\backslash\Xi$, thus Lemma 2.2 yields that there exists a continuous curve $\gamma:[0,1]\to\mathcal{M}^{n}$ such that $\gamma(0)=y$, $\gamma(1)=z$ and $\gamma(t)\not\in\Xi$ for $t<1$. As $z\not\in C$, there exists $s=\max\{t:\,\gamma(t)\in C\}<1$. It follows that $\gamma(s)\in(\partial_{\mathcal{M}^{n}}C)\backslash\Xi$, contradicting (3), and proving (4).

We deduce from (4) that ${\rm int}_{\mathcal{M}^{n}}C$ is the union of the components of $\mathcal{M}^{n}\backslash\Xi$, and since $C\neq\mathcal{M}^{n}$, ${\rm int}_{\mathcal{M}^{n}}C$ is one of the components of $\mathcal{M}^{n}\backslash\Xi$ by Lemma 2.2. As $C$ is bounded, we conclude that $C$ is a ball again by Lemma 2.2. ∎

In the final part of this section, we assume that $\mathcal{M}^{n}$ is either $H^{n}$ or $S^{n}$, and use the models in ${\mathbb{R}}^{n+1}$ above. For $k=1,\ldots,n-1$, the $k$-dimensional totally geodesic subspaces are of the form $L\cap\mathcal{M}^{n}$ where $L$ is a linear $(k+1)$-dimensional subspace of ${\mathbb{R}}^{n+1}$ with $L\cap\mathcal{M}^{n}\neq\emptyset$. Next, we define $\pi:{\mathbb{R}}^{n+1}\backslash e^{\bot}\to e^{\bot}+e$ by

[TABLE]

It follows that the restriction of $\pi$ to $H^{n}$ is a diffeomorphism into the “open” $n$-ball $\{x\in e^{\bot}+e:\,d_{{\mathbb{R}}^{n+1}}(x,e)<1\}$, and the restriction of $\pi$ to ${\rm int}_{S^{n}}B_{S^{n}}(e,\frac{\pi}{2})$ is a diffeomorphism into the affine $n$-plane $e^{\bot}+e$ of ${\mathbb{R}}^{n+1}$. In addition, for any $k=1,\ldots,n-1$, $\pi$ induces a natural bijection between certain $k$-dimensional affine subspaces of $e^{\bot}+e$ and certain $k$-dimensional totally geodesic subspaces of $\mathcal{M}^{n}$ not contained in $e^{\bot}$. In particular, if $L$ is a $(k+1)$-dimensional linear subspace of ${\mathbb{R}}^{n+1}$, then

[TABLE]

3. $D$-maximal sets

The main tool to obtain convex bodies with extremal properties is the Blaschke Selection Theorem. First we impose a metric on non-empty compact subsets. Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$. For a non-empty compact set $C\subset\mathcal{M}^{n}$ and $z\in\mathcal{M}^{n}$, we set $d_{\mathcal{M}^{n}}(z,C)=\min_{x\in C}d_{\mathcal{M}^{n}}(z,x)$. For any non-empty compact set $C_{1},C_{2}\subset\mathcal{M}^{n}$, we define their Hausdorff distance

[TABLE]

The Hausdorff distance is a metric on the space of non-empty compact subsets in $\mathcal{M}^{n}$. We say that a sequence $\{C_{m}\}$ of non-empty compact subsets of $\mathcal{M}^{n}$ is bounded if there is a ball containing every $C_{m}$. For non-empty compact sets $C_{m},C\subset\mathcal{M}^{n}$, we write $C_{m}\to C$ to denote if the sequence $\{C_{m}\}$ tends to $C$ in terms of the Hausdorff distance.

The following is well-known and we present the easy argument for the sake of completeness (see Theorem 1.8.8 in R. Schneider [7] for the case when each set is convex).

Lemma 3.1**.**

For non-empty compact sets $C_{m},C\subset\mathcal{M}^{n}$ where $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, we have $C_{m}\to C$ if and only if

**(i): **

assuming $x_{m}\in C_{m}$, the sequence $\{x_{m}\}$ is bounded and any accumulation point of $\{x_{m}\}$ lies in $C$;

**(ii): **

for any $y\in C$, there exist $x_{m}\in C_{m}$ for each $m$ such that $\lim_{m\to\infty}x_{m}=y$.

Proof.

First we assume that $C_{m}\to C$. For (i), if $x_{m}\in C_{m}$, then let $y_{m}\in C$ be a point closest to $x_{m}$. Now $\lim_{m\to\infty}d_{\mathcal{M}^{n}}(x_{m},C_{m})=0$ yields that $\{x_{m}\}$ is bounded and the sequences $\{x_{m}\}$ and $\{y_{m}\}$ have the same set of accumulation points, proving (i). For (ii), let $y\in C$, and let $x_{m}\in C_{m}$ be a point closest to $y$. We have $\lim_{m\to\infty}x_{m}=y$ because $\lim_{m\to\infty}d_{\mathcal{M}^{n}}(y,C_{m})=0$.

Next we assume that (i) and (ii) hold for $\{C_{m}\}$ and $C$. Let $x_{m}\in C_{m}$ be a point farthest from $C$ and let $y_{m}\in C$ be a point farthest to $C_{m}$, and hence both $\{x_{m}\}$ and $\{y_{m}\}$ are bounded. We choose subsequences $\{x_{m^{\prime}}\}\subset\{x_{m}\}$ and $\{y_{m"}\}\subset\{y_{m}\}$ such that

[TABLE]

We may also assume that $\lim_{m^{\prime}\to\infty}x_{m^{\prime}}=x$ and $\lim_{m"\to\infty}y_{m"}=y$ where $x\in C$ by (i) and $y\in C$ by the compactness of $C$. On the one hand, it follows that $\lim_{m\to\infty}d_{\mathcal{M}^{n}}(x_{m},C)=0$. On the other hand, there exists $z_{m}\in C_{m}$ such that $\lim_{m\to\infty}z_{m}=y$ by (ii), therefore $\lim_{m\to\infty}d_{\mathcal{M}^{n}}(y_{m},C_{m})=0$ as well. ∎

The space of non-empty compact subsets of $\mathcal{M}^{n}$ is locally compact according to the Blaschke Selection Theorem (see R. Schneider [7]).

Theorem 3.2** (Blaschke).**

If $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, then any bounded sequence of non-empty compact subsets of $\mathcal{M}^{n}$ has a convergent subsequence.

Let us consider convergent sequences of compact subsets of $\mathcal{M}^{n}$.

Lemma 3.3**.**

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, and let the sequence $\{C_{m}\}$ of non-empty compact subsets of $\mathcal{M}^{n}$ tend to $C$.

**(i): **

${\rm diam}_{\mathcal{M}^{n}}\,C=\lim_{m\to\infty}{\rm diam}_{\mathcal{M}^{n}}\,C_{m}$**

**(ii): **

$V_{\mathcal{M}^{n}}(C)\geq\limsup_{m\to\infty}V_{\mathcal{M}^{n}}(C_{m})$**

Proof.

(i) follows from Lemma 3.1.

For (ii), it is sufficient to prove that for any $\varepsilon>0$, there exists $M$ such that $V_{\mathcal{M}^{n}}(C_{m})\leq V_{\mathcal{M}^{n}}(C)+\varepsilon$ if $m\geq M$. Choose $r>0$ such that the open set

[TABLE]

satisfies $V_{\mathcal{M}^{n}}(U_{r})\leq V_{\mathcal{M}^{n}}(C)+\varepsilon$. Such an $r$ exists as $C$ is compact. We choose $M$ such that $\delta_{\mathcal{M}^{n}}(C_{m},C)<r$ if $m\geq M$, and hence $C_{m}\subset U_{r}$ if $m\geq M$. ∎

For $D>0$ where $D<\pi$ if $\mathcal{M}^{n}=S^{n}$, we say that a compact set $C\subset\mathcal{M}^{n}$ is $D$-maximal if ${\rm diam}_{\mathcal{M}^{n}}\,C\leq D$ and

[TABLE]

Theorem 3.4**.**

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, and let $D>0$ where $D<\pi$ if $\mathcal{M}^{n}=S^{n}$.

**(i): **

There exists a $D$-maximal set in $\mathcal{M}^{n}$.

**(ii): **

For any $D$-maximal set $C$ in $\mathcal{M}^{n}$ and $z\in\partial_{\mathcal{M}^{n}}C$, there exists $y\in\partial_{\mathcal{M}^{n}}C$ such that $d_{\mathcal{M}^{n}}(z,y)=D$.

Proof.

Let $\{C_{m}\}$ be a sequence of compact subsets of $\mathcal{M}^{n}$ with $z_{0}\in C_{m}$, ${\rm diam}_{\mathcal{M}^{n}}\,C_{m}\leq D$ and

[TABLE]

According to the Blaschke Selection Theorem Theorem 3.2, we may asume that the sequence $\{C_{m}\}$ tends to a compact subset $X\subset\mathcal{M}^{n}$. Here $X$ is a $D$-maximal set by Lemma 3.3.

Next let $C$ be any $D$-maximal set in $\mathcal{M}^{n}$, and let $z\in\partial_{\mathcal{M}^{n}}C$. We suppose that

$\Delta=\max_{x\in C}d_{\mathcal{M}^{n}}(z,x)<D$, and seek a contradiction. As $z\in\partial_{\mathcal{M}^{n}}C$, there exists some $y\in B_{\mathcal{M}^{n}}(z,\frac{1}{2}(D-\Delta))\backslash C$, and hence $B_{\mathcal{M}^{n}}(y,r)\cap C=\emptyset$ for some $r\in(0,\frac{1}{2}(D-\Delta))$. Therefore $X=C\cup B_{\mathcal{M}^{n}}(z,r)$ satisfies that $V_{\mathcal{M}^{n}}(X)>V_{\mathcal{M}^{n}}(C)$ and ${\rm diam}_{\mathcal{M}^{n}}\,X\leq D$, which is a contradiction verifying (ii). ∎

4. Convex sets

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$. For $x,y\in\mathcal{M}^{n}$ where $x\neq-y$ if $\mathcal{M}^{n}=S^{n}$, we write $[x,y]_{\mathcal{M}^{n}}$ to denote the geodesic segment between $x$ and $y$ whose length is $d_{\mathcal{M}^{n}}(x,y)$. We call $X\subset\mathcal{M}^{n}$ convex if $[x,y]_{\mathcal{M}^{n}}\subset X$ for any $x,y\in X$, and in addition we assume that $X$ is contained in an open hemisphere if $\mathcal{M}^{n}=S^{n}$. For $Z\subset\mathcal{M}^{n}$ where we assume that $Z$ is contained in an open hemisphere if $\mathcal{M}^{n}=S^{n}$, the convex hull ${\rm conv}_{\mathcal{M}^{n}}Z$ is the intersection of all convex sets containing $Z$.

We observe that for a non-empty compact convex $Z\subset\mathcal{M}^{n}$, the conditions that $V_{\mathcal{M}^{n}}(Z)>0$, ${\rm int}_{\mathcal{M}^{n}}Z\neq\emptyset$ and $Z$ is solid are equivalent.

We deduce from (5) and (6) that if $x,y\in H^{n}$ or $x,y\in{\rm int}_{S^{n}}B_{S^{n}}(e,\frac{\pi}{2})$, and $\mathcal{M}^{n}$ is either $H^{n}$ or $S^{n}$, respectively, then $\pi([x,y]_{\mathcal{M}^{n}})=[\pi(x),\pi(y)]_{{\mathbb{R}}^{n+1}}$; namely, the Euclidean segment in $e^{\bot}+e$. Thus for a subset $Z$ of either $H^{n}$ or ${\rm int}_{S^{n}}B_{S^{n}}(e,\frac{\pi}{2})$, $Z$ is convex on $H^{n}$ or $S^{n}$, respectively, if and only if $\pi(Z)\subset e^{\bot}+e$ is convex.

Since on the sphere, it is an important issue whether a set $Z\subset S^{n}$ is contained in an open hemisphere, we note the following condition:

Lemma 4.1**.**

If ${\rm diam}_{S^{n}}X<\arccos\frac{-1}{n+1}$ for $X\subset S^{n}$, then $X$ is contained in an open hemisphere.

Proof.

We may assume that $X$ is compact. Let $Z={\rm conv}_{{\mathbb{R}}^{n+1}}\,X$, and hence $Z$ is compact as well. Let $z\in Z$ be the unique closest point of $Z$ to $o$. It follows from the Charathéodory theorem applied in ${\mathbb{R}}^{n+1}$ that there exist $x_{1},\ldots,x_{n+2}\in X$ and $\alpha_{1},\ldots,\alpha_{n+2}\in[0,1]$ satisfying $\alpha_{1}+\ldots+\alpha_{n+2}=1$ and

[TABLE]

As ${\rm diam}_{S^{n}}X<\arccos\frac{-1}{n+1}$, we have $\langle x_{i},x_{j}\rangle>\frac{-1}{n+1}$ for any $i\neq j$. We deduce from $2\alpha_{i}\alpha_{j}\leq\alpha_{i}^{2}+\alpha_{j}^{2}$ that

[TABLE]

therefore $X\subset Z\subset\{x\in R^{n+1}:\,\langle x,z\rangle>0\}$. ∎

Lemma 4.2**.**

If either $\mathcal{M}^{n}=H^{n}$ and $r>0$, or $\mathcal{M}^{n}=S^{n}$ and $r\in(0,\frac{\pi}{2})$, then $B_{\mathcal{M}^{n}}(z,r)$ is convex for any $z\in\mathcal{M}^{n}$. In addition, if $X\subset S^{n}$ convex, then $X\cap B_{S^{n}}(z,\frac{\pi}{2})$ is convex.

Proof.

For the case $\mathcal{M}^{n}=H^{n}$ and $r>0$, or $\mathcal{M}^{n}=S^{n}$ and $r\in(0,\frac{\pi}{2})$, we may assume that $z=e$. Thus $\pi(B_{\mathcal{M}^{n}}(z,r))$ is a Euclidean ball in $e^{\bot}+e$, which in turn yields that $B_{\mathcal{M}^{n}}(e,r)$ is convex.

If $X\subset S^{n}$ is convex, then we may assume that $X\subset{\rm int}_{S^{n}}B_{S^{n}}(e,\frac{\pi}{2})$. For $H^{+}=\{x\in{\mathbb{R}}^{n+1}:\,\langle z,x\rangle\geq 0\}$, we have

[TABLE]

which is convex, and hence $X\cap B_{S^{n}}(z,\frac{\pi}{2})$ is convex as well. ∎

We remark that $B_{S^{n}}(z,r)$ is not convex if $r\in[\frac{\pi}{2},\pi)$.

Lemma 4.3**.**

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, and let $X\subset\mathcal{M}^{n}$ be compact, non-empty and satisfy ${\rm diam}\,X\leq\frac{\pi}{2}$ in the case of $\mathcal{M}^{n}=S^{n}$. Then

**(i): **

${\rm diam}_{\mathcal{M}^{n}}{\rm conv}_{\mathcal{M}^{n}}\,X={\rm diam}_{\mathcal{M}^{n}}\,X$;

**(ii): **

$V_{\mathcal{M}^{n}}\left({\rm conv}_{\mathcal{M}^{n}}\,X\right)>V_{\mathcal{M}^{n}}(X)$* if $V_{\mathcal{M}^{n}}\left({\rm conv}_{\mathcal{M}^{n}}\,X\right)>0$ and ${\rm conv}_{\mathcal{M}^{n}}\,X\neq X$.*

Proof.

For (i), let ${\rm diam}\,X=D$ and let $x_{1},x_{2}\in{\rm conv}_{\mathcal{M}^{n}}X$. First, we consider the case when $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$ or $H^{n}$. Since $X\subset B_{\mathcal{M}^{n}}(x_{1},D)$ and $B_{\mathcal{M}^{n}}(x_{1},D)$ is convex by Lemma 4.2, we have $x_{2}\in B_{\mathcal{M}^{n}}(x_{1},D)$. Therefore $d_{\mathcal{M}^{n}}(x_{1},x_{2})\leq D$.

If $\mathcal{M}^{n}=S^{n}$, then Lemma 4.1 yields that $X\subset B_{S^{n}}(z,R)$ for some $z\in S^{n}$ and $R\in(0,\frac{\pi}{2})$. Since $X\subset B_{S^{n}}(x_{1},D)\cap B_{S^{n}}(z,R)$, which is convex by Lemma 4.2, we have $x_{2}\in B_{S^{n}}(x_{1},D)\cap B_{S^{n}}(z,R)$. Therefore $d_{S^{n}}(x_{1},x_{2})\leq D$, finally proving (i).

For (ii), we assume that $V(Z)>0$ for $Z={\rm conv}_{\mathcal{M}^{n}}\,X$ and $Z\neq X$. As $Z$ is convex, it follows that the closure of ${\rm int}Z$ is $Z$ Since $X$ is compact and $X\neq Z$, there exists some $z\in({\rm int}Z)\backslash X$. Therefore $B_{\mathcal{M}^{n}}(z,r)\subset({\rm int}Z)\backslash X$ for some $r>0$, proving that $V(Z)>V(X)$. ∎

Theorem 3.4 guarantees the existence of $D$-maximal sets, and Lemma 4.3 yields

Corollary 4.4**.**

If $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, and $D>0$ where $D\leq\frac{\pi}{2}$ if $\mathcal{M}^{n}=S^{n}$, then any $D$-maximal set in $\mathcal{M}^{n}$ is convex.

Using the map $\pi$, (5) and (6) in the spherical- and hyperbolic case, we deduce from Lemma 2.1

Lemma 4.5**.**

If $\mathcal{M}^{n}$ is either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, $K\subset\mathcal{M}^{n}$ is convex with ${\rm int}_{\mathcal{M}^{n}}K\neq\emptyset$ and $z\in\partial_{\mathcal{M}^{n}}K$, then there exists a supporting hyperplane of $\mathcal{M}^{n}$ containing $z$ and not intersecting ${\rm int}_{\mathcal{M}^{n}}K$. In addition, if $z$ is a strongly regular boundary point, then there exists a unique supporting hyperplane, and its unit exterior normal is $N_{K}(z)$.

In turn, we deduce the following statement, which will be important in identifying boundary points of a two-point symmetrization.

Lemma 4.6**.**

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, let $K\subset\mathcal{M}^{n}$ be convex with ${\rm int}_{\mathcal{M}^{n}}K\neq\emptyset$, and let $x,y\in\partial_{\mathcal{M}^{n}}K$.

**(i): **

If $[x,y]_{\mathcal{M}^{n}}\cap{\rm int}_{\mathcal{M}^{n}}K\neq\emptyset$, then $\ell\cap K=[x,y]_{\mathcal{M}^{n}}$ for the line $\ell$ passing through $x$ and $y$.

**(ii): **

If $[x,y]_{\mathcal{M}^{n}}\subset\partial_{\mathcal{M}^{n}}K$, then there exists a supporting hyperplane $\Pi$ to $K$ containing $[x,y]_{\mathcal{M}^{n}}$.

5. Two-point symmetrization

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, let $H^{+}$ be a closed half space bounded by the $(n-1)$-dimensional subspace $H$ in $\mathcal{M}^{n}$, and let $X\subset\mathcal{M}^{n}$ be compact and non-empty. We write $H^{-}$ to denote the other closed half space of $\mathcal{M}^{n}$ determined by $H$ and $\sigma_{H}(X)$ to denote the reflected image of $X$ through the $(n-1)$-subspace $H$.

The two-point symmetrization $\tau_{H^{+}}(X)$ of $X$ with respect to $H^{+}$ is a rearrangement of $X$ by replacing $(H^{-}\cap X)\backslash\sigma_{H}(X)$ by its reflected image through $H$ where readily this reflected image is disjoint from $X$. Naturally, interchanging the role of $H^{+}$ and $H^{-}$ results in taking the reflected image of $\tau_{H^{+}}(X)$ through $H$. Since this operation does not change any relevant property of the new set, we simply use the notation $\tau_{H}(X)$ (see Figure 1).

Lemma 5.1**.**

Let $\mathcal{M}^{n}$ be either ${\mathbb{R}}^{n}$, $H^{n}$ or $S^{n}$, let $H^{+}$ be a half space, and let $X\subset\mathcal{M}^{n}$ be compact and non-empty such that ${\rm diam}_{\mathcal{M}^{n}}(X)<\pi$ if $\mathcal{M}^{n}=S^{n}$. Then $\tau_{H}(X)=\tau_{H^{+}}(X)$ is compact and satisfies

**(i): **

\tau_{H}(X)\cap H^{+}=\big{(}X\cup\sigma_{H}(X)\big{)}\cap H^{+};

**(ii): **

\tau_{H}(X)\cap H^{-}=\big{(}X\cap\sigma_{H}(X)\big{)}\cap H^{-};

**(iii): **

$V_{\mathcal{M}^{n}}(\tau_{H}(X))=V_{\mathcal{M}^{n}}(X)$;

**(iv): **

${\rm diam}_{\mathcal{M}^{n}}(\tau_{H}(X))\leq{\rm diam}_{\mathcal{M}^{n}}(X)$.

Proof.

Properties (i) and (ii) are just reformulations of the definition of two-point symmetrization, and they directly yield (iii) and the compactness of $\tau_{H}(X)$.

For (iv), let $x,y\in\tau_{H}(X)$. If either $x,y\in X$ or $x,y\in\sigma_{H}(X)$, then readily $d_{\mathcal{M}^{n}}(x,y)\leq{\rm diam}_{\mathcal{M}^{n}}(X)$. Otherwise, we may assume that $x\in X\backslash\sigma_{H}(X)$ and $y\in\sigma_{H}(X)\backslash X$, thus $x\in X\cap(H^{+}\backslash H)$ and $y\in\sigma_{H}(X)\cap(H^{+}\backslash H)$. It follows that $\sigma_{H}(y)\in X\cap(H^{-}\backslash H)$, and hence $[x,\sigma_{H}(y)]_{\mathcal{M}^{n}}$ intersects $H$ in a unique $z$ where $x,\sigma_{H}(y)\in X$ implies that $[x,\sigma_{H}(y)]_{\mathcal{M}^{n}}$ is well-defined even if $\mathcal{M}^{n}=S^{n}$. Applying the triangle inequality to $x,y,z$, we deduce that

[TABLE]

∎

Two-point symmetrization appeared first in V. Wolontis [10]. It is applied to prove the isoperimetric inequality in the spherical space by Y. Benyamini [2], and the spherical analogue of the Blaschke-Santaló inequality by Gao, Hug, Schneider [5].

The following statement was proved by Aubrun, Fradelizi [1] in the Euclidean- and spherical case. Here we provide a somewhat more detailed version of the argument in [1]. Part of the reason for more details is that in the case $\frac{\pi}{2}<D<\pi$ of Theorem 1.2 concerned about $S^{n}$, we use many ideas of the proof of Theorem 5.2 on the one hand, however, many essential ingredients of the argument for Theorem 5.2 do not hold anymore.

Theorem 5.2**.**

If $\mathcal{M}^{n}$ is either $H^{n}$ or $S^{n}$, and the compact, convex $K\subset\mathcal{M}^{n}$ with non-empty interior satisfies that $\tau_{H}(K)$ is convex for any $(n-1)$-dimensional subspace $H$ of $\mathcal{M}^{n}$, then $K$ is a ball.

Proof.

Let the compact and convex $K\subset\mathcal{M}^{n}$ with non-empty interior satisfy that $\tau_{H}(K)$ is convex for any $(n-1)$-dimensional subspace $H$ of $\mathcal{M}^{n}$.

First we prove that for any pair $x,y\in\partial_{\mathcal{M}^{n}}K$, $x\neq y$, of strongly regular points, writing $H$ to denote the perpendicular bisector $(n-1)$-subspace of $[x,y]_{\mathcal{M}^{n}}$, we have

[TABLE]

Let $x\in H^{+}$. To prove (7), we verify that

[TABLE]

by observing $[x,y]_{\mathcal{M}^{n}}\subset\tau_{H}(K)$ and distinguishing two cases.

If $[x,y]_{\mathcal{M}^{n}}\subset\partial_{\mathcal{M}^{n}}K$, then there exists a supporting $(n-1)$-dimensional subspace $\Pi$ to $K$ containing $[x,y]_{\mathcal{M}^{n}}$ according to Lemma 4.6. It follows that $\Pi$ is a supporting $(n-1)$-dimensional subspace to $\sigma_{H}K$, and in turn to $\tau_{H}(K)$, proving (8) in this case.

On the other hand, if $[x,y]_{\mathcal{M}^{n}}\cap{\rm int}_{\mathcal{M}^{n}}K\neq\emptyset$, then let $\ell\subset\mathcal{M}^{n}$ be the one-dimensional subspace of $x$ and $y$, and hence $\ell\cap K=[x,y]_{\mathcal{M}^{n}}$ by Lemma 4.6. As $x\in\partial_{\mathcal{M}^{n}}\sigma_{H}(K)$, we deduce from Lemma 5.1 (i) and (ii) that $\ell\cap\tau_{H}(K)=[x,y]_{\mathcal{M}^{n}}$, and in turn (8) finally follows.

Since $x\in\partial_{\mathcal{M}^{n}}\tau_{H}(K)$, there exists a supporting $(n-1)$-dimensional subspace $\Xi$ to $\tau_{H}(K)$ at $x$, and let $\tau_{H}(K)\subset\Xi^{+}$. We deduce from Lemma 5.1 (i) that $\sigma_{H}(K),K\subset\Xi^{+}$, thus $\sigma_{H}(\Xi)$ is a supporting $(n-1)$-dimensional subspace to $K$ at $y$ with $K\subset\sigma_{H}(\Xi^{+})$. As $x$ and $y$ are strongly regular points, we conclude (7) by Lemma 4.6.

In turn, we deduce from (7) that for any pair $x,y\in\partial_{\mathcal{M}^{n}}K$, $x\neq y$, of strongly regular points, there exists $\lambda(x,y)\in{\mathbb{R}}$ such that

[TABLE]

Obviously, $\lambda(x,y)=\lambda(y,x)$.

We fix a strongly regular point $x_{0}\in\partial_{\mathcal{M}^{n}}K$. We claim that if $x,y\in\partial_{\mathcal{M}^{n}}K$ are strongly regular points different from $x_{0}$, then

[TABLE]

We distiguishing two cases. If $x_{0},x,y$ are not contained in a one-dimensional subspace, then (9) yields that

[TABLE]

Adding up the three relations yields $\lambda(x_{0},x)(x_{0}-x)+\lambda(x,y)(x-y)+\lambda(y,x_{0})(y-x_{0})=o$. Since any two of $x_{0}-x,x-y,y-x_{0}$ are independent in ${\mathbb{R}}^{n+1}$, we have $\lambda(x_{0},x)=\lambda(x,y)=\lambda(y,x_{0})$, and hence (10) holds in this case.

On the other hand, if $x_{0},x,y$ are contained in a one-dimensional subspace, then as strongly regular points are dense in $\partial_{\mathcal{M}^{n}}K$, there exists a strongly regular point $z\in\partial_{\mathcal{M}^{n}}K$ not contained in the one-dimensional subspace passing through $x_{0},x,y$. Applying the previous case first to the triple $x_{0},x,z$, then to the triple $x_{0},y,z$, it follows that $\lambda(x_{0},x)=\lambda(x_{0},z)=\lambda(x_{0},y)$, proving (10).

According to (10), there exists a common value $\lambda$ of $\lambda(x_{0},x)$ for all strongly regular points $x\in\partial_{\mathcal{M}^{n}}K$. Setting $p=N_{K}(x_{0})-\lambda x_{0}\in{\mathbb{R}}^{n+1}$, we deduce from (9) that

[TABLE]

The rest of the argument is split between the hyperbolic- and spherical case.

Case 1 $\mathcal{M}^{n}=H^{n}$

We claim that

[TABLE]

Otherwise $N_{K}(x_{0})=\lambda x_{0}$ by (11), and hence

[TABLE]

what is absurd, verifying (12).

If $x\in\partial_{H^{n}}K$ is a strongly regular point, then $N_{K}(x)\in T_{x}$, thus (11) yields that $0=\mathcal{B}(N_{K}(x),x)=\mathcal{B}(p+\lambda x,x)$, and hence $\mathcal{B}(p,x)=-\lambda$. As $p\neq o$, we deduce that each strongly regular point of $\partial_{H^{n}}K$ is contained in the standard hypersurface $\{x\in H^{n}:\,\mathcal{B}(p,x)=-\lambda\}$ (see the list before Lemma 2.2), therefore Corollary 2.3 yields that $K$ is a ball.

Case 2 $\mathcal{M}^{n}=S^{n}$

We again claim that

[TABLE]

Otherwise $N_{K}(x_{0})=\lambda x_{0}$ by (11), and hence

[TABLE]

what is absurd, verifying (13).

If $x\in\partial_{S^{n}}K$ is a strongly regular point, then $N_{K}(x)\in T_{x}$, thus (11) yields that $0=\langle N_{K}(x),x\rangle=\langle p+\lambda x,x\rangle$, and hence $\langle p,x\rangle=-\lambda$. As $p\neq o$, we deduce that each strongly regular point of $\partial_{S^{n}}K$ is contained in the boundary $\{x\in S^{n}:\,\langle p,x\rangle=-\lambda\}$ of a fixed spherical ball (see the list before Lemma 2.2), therefore Corollary 2.3 yields that $K$ is a spherical ball. ∎

**Remark: ** In ${\mathbb{R}}^{n}$, a similar argument works, but some changes have to be instituted. Instead of (7), we have $N_{K}(x)=\sigma_{\widetilde{H}}(N_{K}(y))$ where $\widetilde{H}$ is the linear $(n-1)$-plane in ${\mathbb{R}}^{n}$ parallel to $H$, therefore (9) still holds. Similarly as above, (9) leads to (11). For the final part of the argument, we have $\lambda\neq 0$, because otherwise (11) yields that $N_{K}(x)=p$ for any strongly regular point $x\in\partial_{R^{n}}K$, contradicting that the strongly regular points are dense in $x\in\partial_{R^{n}}K$. As (11) implies that if $x\in\partial_{R^{n}}K$ is a strongly regular point, then

[TABLE]

and hence each strongly regular point of $\partial_{{\mathbb{R}}^{n}}K$ is contained in the boundary $\{x\in{\mathbb{R}}^{n}:\,\langle x+\lambda^{-1}p,x+\lambda^{-1}p\rangle=\lambda^{-2}\}$ of a fixed ball. Therefore Corollary 2.3 yields that $K$ is a ball.

**Proof of Theorem 1.2 when $D\leq\frac{\pi}{2}$ and of Theorem 1.1 ** Let $\mathcal{M}^{n}=H^{n}$, or let $\mathcal{M}^{n}=S^{n}$ and $D\leq\frac{\pi}{2}$. Theorem 3.4 (i) yields the existence of $D$-maximal sets in $\mathcal{M}^{n}$, and any $D$-maximal set is convex according to Corollary 4.4.

Let $C$ be any $D$-maximal set, and hence $C$ is convex. We deduce from from Lemma 5.1 that for any $(n-1)$-dimensional subspace $H$ of $\mathcal{M}^{n}$, $\tau_{H}(C)$ is $D$-maximal, thus convex, therefore Theorem 5.2 yields that $C$ is a ball. The maximality of $V(C)$ implies that the radius of $C$ is $D/2$, proving Theorem 1.2 when $D\leq\pi/2$ and also Theorem 1.1. $\Box$

**Remark: ** In ${\mathbb{R}}^{n}$, (1) can proved in a similar way.

6. Proof of Theorem 1.2 if $\frac{\pi}{2}<D<\pi$

Let $\frac{\pi}{2}<D<\pi$. We frequently drop the index “$S^{n}$” in the formulas, say we simply write $B(x,r)$, $V(X)$ and $\partial X$ for $x\in S^{n}$, $X\subset S^{n}$ and $r\in(0,\pi)$.

Let us recall from Section 3 that $C\subset S^{n}$ is $D$-maximal if

• •

$C$ is compact;

• •

${\rm diam}_{S^{n}}\,C\leq D$;

• •

$V(C)=\sup\{V(X):\,X\subset S^{n}\mbox{ compact and }{\rm diam}_{S^{n}}X\leq D\}$.

According to Theorem 3.4 (i), there exist $D$-maximal sets in $S^{n}$. We deduce from Lemma 5.1 that if $C\subset S^{n}$ is $D$-maximal, then

[TABLE]

As $D>\frac{\pi}{2}$, it is a priori not clear whether a $D$-maximal set is convex. However, Theorem 3.4 (ii) implies that for any $D$-maximal set $C$ in $S^{n}$ and $z\in\partial C$, there exists $y\in C$ such that

[TABLE]

Lemma 6.1**.**

If $\frac{\pi}{2}<D<\pi$ and $C$ is a $D$-maximal set in $S^{n}$, then there exists a solid $D$-maximal set $C_{0}\subset C$.

Proof.

Let $X\subset C$ be the set of density points; namely,

[TABLE]

thus readily ${\rm int}\,C\subset X$. It follows from Lebesgue’s Density Theorem that

[TABLE]

In addition, if $z\in\partial C$, then (15) yields that $z\not\in X$, and hence $X={\rm int}\,C$. We conclude that $C_{0}$ can be taken as the closure of $X$. ∎

The idea of the proof of Theorem 1.2 when $\frac{\pi}{2}<D<\pi$ is similar to the case $D\leq\frac{\pi}{2}$; more precisely, the idea is to prove that if $\frac{\pi}{2}<D<\pi$ and $C\subset S^{n}$ is a solid $D$-maximal set, then there exist $p\in{\mathbb{R}}^{n+1}$ and $\lambda\in{\mathbb{R}}$ such that

[TABLE]

Thus from now on, the main goal is to understand properties of solid $D$-maximal sets. We use two-point symmetrization again. The difficulty in the $D>\frac{\pi}{2}$ case is that if $x,y\in\partial C$, $x\neq y$, are strongly regular points of a solid $D$-maximal set $C$ in $S^{n}$, and $H$ is the perpendicular bisector of $[x,y]_{S^{n}}$, then a priori $x$ may lie in ${\rm int}\,\tau_{H}C$.

Lemma 6.2**.**

Let $\frac{\pi}{2}<D<\pi$, and let $C$ be a solid $D$-maximal set in $S^{n}$. If $x,y\in\partial C$, $x\neq y$, are strongly regular points, then there exist $\lambda(x,y)\in{\mathbb{R}}$ and $\eta(x,y)\in\{-1,1\}$ such that

[TABLE]

where $\eta(y,x)=\eta(x,y)$ and $\lambda(y,x)=\eta(x,y)\lambda(x,y)$.

Proof.

It is equivalent to prove that

[TABLE]

where $H$ is the perpendicular bisector $(n-1)$-dimensional subspace of $[x,y]_{S^{n}}$. In turn, (18) is equivalent proving that if we assume

[TABLE]

then we have

[TABLE]

Let $x\in H^{+}$. We deduce from (15) and from the fact that $x$ and $y$ are strongly regular boundary points that there exist $r>0$ and $x_{0},x_{1},y_{0},y_{1}\in C$ such that

[TABLE]

Readily, we have $N_{C}(x)=N_{B(x_{0},r)}(x)=N_{B(x_{1},D)}(x)$ and $N_{C}(y)=N_{B(y_{0},r)}(y)=N_{B(y_{1},D)}(y)$.

It follows from (14) that $\tau_{H}C$ is $D$-maximal as well. As $x=\sigma_{H}y\in\tau_{H}C$ and

[TABLE]

by Lemma 5.1, and (19) yields that $x$ is a boundary point of $B(x_{1},D)\cup\sigma_{H}B(y_{1},D)$, we deduce that $x\in\partial\tau_{H}C$. We deduce from (15) that $\tau_{H}C\subset B(z,D)$ and $x\in\partial B(z,D)$ for some $z\in\tau_{H}C$. Now $x\in B(x_{0},r)\cap\sigma_{H}B(y_{0},r)\cap\partial B(z,D)$ and

[TABLE]

by Lemma 5.1, therefore $B(x_{0},r)=\sigma_{H}B(y_{0},r)$. We conclude (20), and in turn (18), proving Lemma 6.2. ∎

As a first step to prove (16), we consider certain specific triples of strongly regular points.

Lemma 6.3**.**

Let $\frac{\pi}{2}<D<\pi$, and let $C$ be a solid $D$-maximal set in $S^{n}$. If $x_{0},x,y\in\partial C$ are strongly regular points not contained in a one-dimensional subspace of $S^{n}$ such that $\eta(x_{0},y)=1$ and $\eta(x_{0},x)=\eta(y,x)$, then for $\lambda=\lambda(y,x_{0})=\lambda(x_{0},y)$, we have $\lambda(x_{0},x)=\lambda$ and

[TABLE]

Proof.

For $\eta(x_{0},x)=\eta(y,x)=\eta$, Lemma 6.2 implies that

[TABLE]

Since $\eta\in\{-1,1\}$, we replace the first equality by $\eta N_{C}(x)-N_{C}(y)=\eta\lambda(x,y)(x-y)$ and add up these three relations, and hence we obtain

[TABLE]

Since any two of $x_{0}-x,x-y,y-x_{0}$ are independent in ${\mathbb{R}}^{n+1}$, we have $\eta\lambda(x,y)=\lambda(x_{0},x)=\lambda(y,x_{0})$ where $\lambda(y,x_{0})=\lambda$. We deduce that $N_{C}(x_{0})-\eta N_{C}(x)=\lambda(x_{0}-x)$, proving Lemma 6.3. ∎

Let us show that we have the setup in Lemma 6.3 if $x$ and $y$ are close enough strongly regular boundary points.

Lemma 6.4**.**

Let $\frac{\pi}{2}<D<\pi$, and let $C$ be a solid $D$-maximal set in $S^{n}$, and let $x_{0}\in\partial C$ be a strongly regular point.

**(i): **

If $\lim_{m\to\infty}y_{m}=x_{0}$ for $y_{m}\in\partial C$ strongly regular boundary points, then

[TABLE]

**(ii): **

For any strongly regular point $z\in\partial C$, $z\neq x_{0}$, there exists $\varrho\in(0,\frac{\pi}{2})$ depending on $x_{0},z$ and $C$ such that $\eta(x_{0},y)=1$ and $\eta(x_{0},z)=\eta(y,z)$ if $y\in B(x_{0},\varrho)\cap\partial C$ is a strongly regular point with $y\neq x_{0}$.

Proof.

There exist some $r\in(0,\frac{\pi}{2})$ and $z_{0},z_{1}\in C$ such that $B(z_{0},r)\subset C\subset B(z_{1},D)$ and $x_{0}\in\partial B(z_{0},r)\cap\partial B(z_{1},D)$ according to Theorem 3.4 (ii).

For (i), there exist some $w_{m}\in C$ with $C\subset B(w_{m},D)$ and $y_{m}\in\partial B(w_{m},D)$, and we may assume that $B(w_{m},D)$ tends to some $B(w,D)$ for $w\in C$. As $B(z_{0},r)\subset B(w,D)$ and $x_{0}\in\partial B(z_{0},r)\cap\partial B(w,D)$, we have $w=z_{1}$. This yields that $\lim_{m\to\infty}N_{C}(y_{m})=N_{C}(x_{0})$. In addition, $\lim_{m\to\infty}\left\langle N_{C}(x_{0}),\frac{y_{m}-x_{0}}{d_{{\mathbb{R}}^{n+1}}(y_{m},x_{0})}\right\rangle=0$ follows from $\lim_{m\to\infty}y_{m}=x_{0}$ and $y_{m}\in B(z_{1},D)\backslash{\rm int}\,B(z_{0},r)$.

We prove (ii) by contradiction, and we assume that there exists a sequence of strongly regular boundary points $y_{m}\in\partial C$ such that $\lim_{m\to\infty}y_{m}=x_{0}$, and

**(a): **

either $N_{C}(x_{0})+N_{C}(y_{m})=\lambda(x_{0},y_{m})(x_{0}-y_{m})$ for each $m$,

**(b): **

or $N_{C}(z)+\eta(z,x_{0})N_{C}(y_{m})=\lambda(z,y_{m})(z-y_{m})$ for each $m$.

If (a) holds, then readily $|\lambda(x_{0},y_{m})|\leq 2/d_{{\mathbb{R}}^{n+1}}(x_{0},y_{m})$, thus (i) yields that

[TABLE]

what is absurd.

If (b) holds, then as $z-x_{0}$ and $N_{C}(x_{0})\in T_{x_{0}}$ are independent, there exists some $u\in S^{n-1}$ such that $\langle u,z-x_{0}\rangle=0$ and $\langle u,N_{C}(x_{0})\rangle>0$. We also observe that if $m$ is large, then $|\lambda(z,y_{m})|\leq 3/d_{{\mathbb{R}}^{n+1}}(z,x_{0})$.

We deduce from $\lim_{m\to\infty}N_{C}(y_{m})=N_{C}(x_{0})$ that

[TABLE]

what is again a contradiction, proving (ii). ∎

Now we choose the right strongly regular “base point” $x_{0}$.

Lemma 6.5**.**

Let $\frac{\pi}{2}<D<\pi$, and let $C$ be a solid $D$-maximal set in $S^{n}$. There exists a strongly regular point $x_{0}\in\partial C$ such that for any one-dimensional subspace $\ell$ of $S^{n}$ passing through $x_{0}$ and any $\varepsilon\in(0,\frac{\pi}{2})$, one finds a strongly regular point $y\in\partial C\cap(B(x_{0},\varepsilon)\backslash\ell)$.

Proof.

First let $n\geq 3$, let $x_{0}\in\partial C$ be any strongly regular point, and let $\varepsilon\in(0,\frac{\pi}{2})$. As $C$ is solid, there exist

[TABLE]

Since ${\rm int}B(x_{0},\varepsilon)\backslash\ell$ is connected (this is the point where we use that $n\geq 3$), connecting $y$ and $z$ by a continuous curve in ${\rm int}B(x_{0},\varepsilon)\backslash\ell$ implies that there exists a $w\in\partial\,C\cap[{\rm int}B(x_{0},\varepsilon)\backslash\ell]$. It follows from the fact that strongly regular points in $\partial C$ are dense that there exists a strongly regular point $\tilde{w}\in\partial\,C\cap[{\rm int}B(x_{0},\varepsilon)\backslash\ell]$.

Next let $n=2$. In this case, the argument is indirect. Since strongly regular points are dense on the boundary, we suppose that for any strongly regular point $w\in\partial C$, there exist a one-dimensional subspace $\ell_{w}$ pasing through $w$ and an $r_{w}\in(0,\frac{\pi}{2})$ such that

[TABLE]

and seek a contradiction. As $C$ is solid, there exist

[TABLE]

For any $x\in B(w,r_{w})\cap\ell_{w}$, the piecewise linear path $[y,x]_{S^{n}}\cup[x,z]_{S^{n}}$ intersects $\partial C$, and the intersection can be only $x$ by (21). Therefore

[TABLE]

It also follows that one component of ${\rm int}B(w,r_{w})\backslash\ell_{w}$ is part of ${\rm int}C$ and the other component is disjoint from $C$, and hence any $x\in\ell_{w}\cap{\rm int}B(z,r_{w})$ is a strongly regular boundary point with

[TABLE]

Let us choose a strongly regular point $w\in\partial C$ and an other strongly regular point $v\in\partial C\backslash\ell_{w}$. It follows that $\ell_{v}\neq\ell_{w}$. We deduce from Lemma 6.4 (ii) and (23) that there exists a $y\in l_{w}\cap\partial C$ with $y\neq w$ and $\eta\in\{-1,1\}$ such that

[TABLE]

On the one hand, we deduce from from $N_{C}(y)=N_{C}(w)$ that $\lambda(w,y)=\lambda(y,w)=0$ in Lemma 6.2, therefore Lemma 6.3 yields that $\eta N_{C}(v)=N_{C}(w)$. However $\ell_{v}\neq\ell_{w}$ and (23) imply $N_{C}(v)\neq\pm N_{C}(w)$ in ${\mathbb{R}}^{n+1}$, thus we have arrived at a contradiction, verifying Lemma 6.5 also if $n=2$. ∎

**Proof of Theorem 1.2 when $\mathbf{\frac{\pi}{2}<D<\pi}$: ** Let $C$ be any $D$-maximal set, and let $C_{0}\subset C$ be the solid $D$-maximal set provided by Lemma 6.1. According to Lemma 6.5, there exists a strongly regular point $x_{0}\in\partial C_{0}$ such that for any one-dimensional subspace $\ell$ passing through $x_{0}$ and any $\varepsilon\in(0,\frac{\pi}{2})$, one finds a strongly regular point $y\in\partial C\cap(B(x_{0},\varepsilon)\backslash\ell)$.

We deduce from Lemma 6.4 (ii) that there exists $\tilde{r}\in(0,\frac{\pi}{2})$ depending on $x_{0}$ and $C$ such that $\eta(x_{0},y)=1$ for any strongly regular point $y\in B(x_{0},\tilde{r})\cap\partial C$ with $y\neq x_{0}$. We claim that there exist $r\in(0,\tilde{r})$ and $\lambda\in{\mathbb{R}}$ such that

[TABLE]

Readily, $\eta(x_{0},y)=1$ by $r<\tilde{r}$. To have the right value of $\lambda(x_{0},y)$, first we fix a strongly regular point $y_{0}\in B(x_{0},\tilde{r})\cap\partial C$ with $y_{0}\neq x_{0}$, and write $\ell_{0}$ to denote the one-dimensional subspace spanned by $x_{0}$ and $y_{0}$. Set

[TABLE]

Lemma 6.3 and Lemma 6.4 (ii) applied to the triple $x_{0},y,y_{0}$ imply the existence of an $r_{0}\in(0,\tilde{r})$ such that $\lambda(x_{0},y)=\lambda$ holds for any strongly regular point $y\in\partial C\cap(B(x_{0},r_{0})\backslash\ell_{0})$. We fix such a strongly regular point $y_{1}\in\partial C\cap(B(x_{0},r_{0})\backslash\ell_{0})$, and write $\ell_{1}$ to denote the one-dimensional subspace spanned by $x_{0}$ and $y_{1}$. In particular, $\ell_{0}\neq\ell_{1}$.

Finally, applying Lemma 6.3 and Lemma 6.4 (ii) to the triple $x_{0},y,y_{1}$, there exists $r\in(0,r_{0})$ such that $\lambda(x_{0},y)=\lambda$ holds for any strongly regular point $y\in\partial C\cap(B(x_{0},r)\backslash\ell_{1})$. Since either $y\not\in\ell_{0}$ or $y\not\in\ell_{1}$ hold for any point $y\in\partial C\cap(B(x_{0},r)\backslash x_{0})$, we conlcude (24).

Our next goal is to verify that $p=N_{C}(x_{0})-\lambda x_{0}$ satisfies that if $x\in\partial C$ is a strongly regular point, then

[TABLE]

where we set $\eta(x_{0},x_{0})=1$. Writing $\ell$ to denote the one-dimensional subspace passing through $x_{0}$ and $x$, the choice of $x_{0}$ and Lemma 6.4 (ii) yield the existence of a strongly regular point $y\in\partial C\cap(B(x_{0},r)\backslash\ell)$ such that $\eta(x_{0},y)=1$ and $\eta(x_{0},x)=\eta(y,x)$. Therefore we conclude (25) by Lemma 6.3.

We again claim that

[TABLE]

Otherwise $N_{K}(x_{0})=\lambda x_{0}$ by (25), and hence

[TABLE]

what is absurd, verifying (26).

If $x\in\partial C$ is a strongly regular point, then $N_{K}(x)\in T_{x}$, thus (25) yields that $0=\langle\eta(x_{0},x)N_{C_{0}}(x),x\rangle=\langle p+\lambda x,x\rangle$, and hence $\langle p,x\rangle=-\lambda$. As $p\neq o$, we deduce that each strongly regular point of $\partial C_{0}$ is contained in the boundary $\{x\in S^{n}:\,\langle p,x\rangle=-\lambda\}$ of a fixed spherical ball, therefore Corollary 2.3 yields that $C_{0}$ is a spherical ball. As $C_{0}$ has maximal volume among sets of diameter at most $D$, it follows that the radius of $C_{0}$ is $D/2$, say $C_{0}=B(z,D/2)$.

Finally, we show that $C=B(z,D/2)$. To prove this, let $x\in S^{n}\backslash B(z,D/2)$, and let $\ell$ be the (or a) one-dimensional subspace of $S^{n}$ passing through $x$ and $x_{0}$. As $\ell$ intersects $B(z,D/2)$ in an arc of length $D$, this arc contains a point $y$ with $d_{S^{n}}(x,y)>D$. Therefore $x\not\in C$, completing the proof of Theorem 1.2 when $D>\pi/2$. $\Box$

**Acknowledgement: ** We are grateful for the careful referee whose remarks have substantially improved the paper. Károly J. Böröczky is supported by NKFIH projects ANN 121649, K 109789, K 116451 and KH 129630.