Proof Step 1. We first claim that for $u\in {C^\mathrm{con}_0}(\bar \Omega )$ and any $a,b\in \Omega $ , there exists $\bar a\in \Omega $ such that

(2.4) $$ \begin{align} d_{\Omega}(\bar a)\le |a-b|, \qquad |u(\bar a)|\ge |u(a)-u(b)|. \end{align} $$

We recall the proof, which is standard. Consider $a,b\in \Omega $ such that $u(a) \le u(b)\le 0$ . Let $\bar b$ be the point in $\partial \Omega $ on the ray that starts at a and passes through b. Then there exists some $\theta \in (0,1)$ such that $ b = (1-\theta )a + \theta \bar b$ . We next define $\bar a = \theta a + (1-\theta ) \bar b$ . These definitions imply that

$$\begin{align*}\bar a - \bar b = \theta(a - \bar b) = a - b. \end{align*}$$

Thus, $d_{\Omega }(\bar a)\le |\bar a -\bar b| = |a-b|$ . Moreover, by convexity,

$$ \begin{align*} u( b) - u(a) &= u((1-\theta)a + \theta \bar b) - u(a) \ \le \ \theta (u(\bar b) - u(a))\\ &= u(\bar b) - [ (1-\theta)u(\bar b) + \theta u(a)]\\ &\le u(\bar b) - u((1-\theta) \bar b + \theta a) \ = \ u(\bar b) - u(\bar a) = |u(\bar a)|. \end{align*} $$

Since $d_{\Omega }(\bar a)\le |a - b|$ , this proves (2.4).

Step 2. Given $u\in {C^\mathrm{con}_0}(\bar \Omega )$ , $\delta>0$ and $x,y\in \Omega $ such that $|x-y|\le \delta $ , fix $a\in \Omega $ such that $d_{\Omega }(a)\le \delta $ and $|u(x)-u(y)|\le |u(a)|$ , and define $w(x) = u(a) u_a(x)$ . Then $u\le w\le 0$ in $\Omega $ and $u=w=0$ on $\partial \Omega $ , so standard arguments (see, for example, [Reference Gutiérrez4, Lemma 1.4.1]) imply that

(2.5) $$ \begin{align} |\partial u(\Omega)| \ge |\partial w(\Omega)| = |u(a)|^n f_\Omega(a), \end{align} $$

with equality if and only if $u = w$ . The definition (1.3) of $\omega _\Omega $ then implies that

(2.6) $$ \begin{align} |u(x)-u(y)|\le |u(a)| \overset{({2.5})}\le f_\Omega(a)^{-1/n} |\partial u(\Omega)|^{1/n}. \end{align} $$

Thus, for nonzero $u\in {C^\mathrm{con}_0}(\bar \Omega )$ ,

$$\begin{align*}\text{ if }|x-y|\le \delta, \text{ then }\ \quad \ \frac{|u(x)-u(y)|}{ |\partial u(\Omega)|^{1/n} } \le \sup_{d_{\Omega}(a)\le \delta} f_\Omega(a)^{-1/n}. \end{align*}$$

It follows from this and the definition of $\omega _\Omega $ that

$$\begin{align*}\omega_\Omega(\delta)\le\sup_{d_{\Omega}(a)\le \delta} f_\Omega(a)^{-1/n}. \end{align*}$$

However, given any $a\in \Omega $ such that $d_{\Omega }(a)\le \delta $ , consider $u = u_a$ , and fix $b\in \partial \Omega $ such that $d_{\Omega }(a) = |a-b|$ . Then

$$\begin{align*}|u_a(a)-u_a(b)| = |u_a(a)| = 1 = \frac {|\partial u_a(\Omega)|^{1/n}}{f_\Omega(a)^{1/n}}, \end{align*}$$

and thus,

$$\begin{align*}\omega_\Omega(\delta) \ge \sup_{|x-y|\le \delta} \frac{|u_a(x)-u_a(y)|}{|\partial u_a(\Omega)|^{1/n}} \ge f_\Omega(a)^{-1/n} \qquad\text{ whenever }\ \ d_{\Omega}(a)\le \delta.\\[-43pt] \end{align*}$$

Proof It is clear that $\partial u_a(a) \subset \partial u_a(\Omega )$ . To prove the other inclusion, assume that $p\in \partial u(x_0)$ for some $x_0\in \Omega $ . We must show that $p\in \partial u(a)$ . We may assume that $x_0\ne a$ , so we can write $x_0 = \theta a + (1-\theta )y$ for some $y\in \partial \Omega $ and $\theta \in (0,1)$ .

For $x\in \Omega $ and $p\in \mathbb {R}^n$ , we will write $\ell _{x,p}(z) := u_a(x) + p\cdot (z-x)$ , so that $p\in \partial u_a(x)$ if and only if $\ell _{x,p}\le u_a$ in $\Omega $ . Since $u_a$ and $\ell _{x_0, p}$ are both linear when restricted to the segment $\{ sa+(1-s)y : s\in (0,1]\}$ , and because $x_0$ belongs to the interior of this segment and $u_a\ge \ell _{x_0,p}$ on this segment, we see that $u_a = \ell _{x_0,p}$ on this segment, and in particular at $x = a$ . Thus, $\ell _{x_0,p}$ is a supporting hyperplane at a; in fact, $\ell _{x_0,p} = \ell _{a,p}$ . It follows that $p\in \partial u_a(a)$ .

Proof If $p\in \partial u_{\Omega ',a}(\Omega ')$ , then $p\in \partial u_{\Omega ', a}(a)$ , which implies that $\ell _{a,p}\le u_{\Omega ',a}$ in $\Omega '$ . But it is easy to check that $u_{\Omega ',a} \le u_{\Omega ,a}$ in $\Omega $ , and it follows that $\ell _{a,p}\le u_{\Omega ,a}$ in $\Omega $ , which implies that $p\in \partial u_{\Omega , a}(a)$ .

Thus, $\partial u_{\Omega ',a}(\Omega ') \subset \partial u_{\Omega ,a}(\Omega )$ , from which we deduce that $f_{\Omega '}(a)\le f_\Omega (a).$

Proof We first prove the lemma for $a=0$ . We know from Lemma 2.1 that $\partial u_0(\Omega ) = \partial u_0(0)$ . Then

$$ \begin{align*} p\in \partial u_0(0) &\iff u_0(x) \ge u_0(0) + p\cdot x\text{ for all }x\in \Omega\\ &\iff u_0(x) \ge -1 + p\cdot x\text{ for all }x\in\bar \Omega. \end{align*} $$

Since $u_0(x)\le 0$ in $\Omega $ , it follows that

$$\begin{align*}p\in \partial u_0(0) \ \ \Longrightarrow \ \ -1+p\cdot x\le 0\text{ for all }x\in \Omega \ \ \Longrightarrow \ \ p\in \Omega^\circ. \end{align*}$$

However, if $p\in \Omega ^\circ $ , then $\ell _{0,p}(x):= -1+x\cdot p$ is an affine function such that $\ell _{0,p} \le 0 = u_0$ on $\partial \Omega $ and $\ell _{0,p}(0)=-1 = u_0(0)$ . It follows from this and the definition of $u_0$ that $\ell _{0,p} \le u_0$ in $\Omega $ , and hence that $p\in \partial u_0(0)$ .

It follows that $\partial u_0(\Omega ) = \Omega ^\circ $ , and hence that $f_\Omega (0) = |\Omega ^\circ |$ .

For general $a\in \Omega $ , the definitions imply that for every $x\in \Omega $ ,

$$\begin{align*}u_{\Omega, a}(x) = u_{\Omega-a,0}(x-a), \qquad\text{ and thus, }\partial u_{\Omega,a}(x) = \partial u_{\Omega-a,0}(x-a). \end{align*}$$

Thus, $f_\Omega (a) = |\partial u_{\Omega , a}( a) | = |\partial u_{\Omega -a,0}(0)| = |(\Omega -a)^\circ | $ .

Proof The definitions imply that for every $x\in \Omega $ ,

$$\begin{align*}u_{M\Omega, Ma}(Mx) = u_{\Omega,a}(x). \end{align*}$$

Thus, for every $x\in \Omega $ ,

$$ \begin{align*} &p\in \partial u_{M\Omega, Ma}(Mx) \\ &\qquad \ \ \Longleftrightarrow \ \ u_{M\Omega, Ma}(Mz) \ge u_{M\Omega, Ma}(Mx) + p \cdot (Mz-Mx)\qquad\text{ for all }Mz\in M\Omega\\ &\qquad \ \ \Longleftrightarrow \ \ u_{\Omega, a}(z) \ge u_{\Omega, a}(x) + M^Tp \cdot (z-x) \qquad\text{ for all }z\in \Omega. \\ &\qquad \ \ \Longleftrightarrow \ \ M^Tp\in \partial u_{\Omega,a}(x). \end{align*} $$

We deduce that $\partial u_{M\Omega , Ma}(M\Omega ) = \{ M^{-T}p : p\in \partial u_{\Omega ,a}(\Omega )\}$ . Now the conclusion follows from basic properties of Lebesgue measure.

Proof We will show that $(\pi _P(\Omega ^\circ ))^\circ = (\Omega _P^\circ )^\circ = \bar \Omega _P$ , where our convention is that if $A\subset P$ is convex, then $A^\circ $ denotes the polar within P, whereas if A is a convex set not contained in P, then $A^\circ $ denotes its polar in $\mathbb {R}^n$ . Then

$$ \begin{align*} y\in (\pi_P(\Omega^\circ))^\circ &\ \ \Longleftrightarrow \ \ y\in P\text{ and }y\cdot x\le 1\quad\text{ for all }x\in \pi_P(\Omega^\circ)\\ &\ \ \Longleftrightarrow \ \ y\in P\text{ and } y\cdot \pi_P x\le 1\quad\text{ for all }x\in \Omega^\circ\\ &\ \ \Longleftrightarrow \ \ y\in P\text{ and } y\cdot x\le 1\quad\text{ for all }x\in \Omega^\circ\\ &\ \ \Longleftrightarrow \ \ y\in P \cap (\Omega^\circ)^\circ = P\cap \bar \Omega = \bar \Omega_P, \end{align*} $$

completing the proof.

Proof Step 1. After a translation and a rotation, we may assume that $a=0$ and that $x = (0,\ldots , 0,-\delta )$ , where $\delta = d_{\Omega }(a)$ . Then $-e_n$ is the outer unit normal at x, and hence, $\Omega \subset \{ y\in \mathbb {R}^n : y_n> -\delta \}$ . One can then quickly check that

(2.8) $$ \begin{align} \{ -s e_n : 0 \le s \le \frac 1 \delta\} \subset \Omega^\circ. \end{align} $$

Let $P := \mathbb {R}^{n-1}\times \{0\}$ , so that $\Omega _P = S(a,\nu )$ . It then follows from Lemma 2.5 that

(2.9) $$ \begin{align} S^\circ(a,\nu) = \pi_P (\Omega^\circ). \end{align} $$

Now let T denote the Steiner symmetrization of $\Omega ^\circ $ with respect to the hyperplane $x_n=0$ . Well-known properties of Steiner symmetrization imply that $|T| = |\Omega ^\circ |$ , that T inherits the convexity of $\Omega ^\circ $ , and, owing to (2.8), (2.9), that

$$\begin{align*}S^\circ(a,\nu)\subset T, \qquad \{ \pm \frac 1{2\delta} e_n \} \subset T. \end{align*}$$

By convexity, T contains the cones in $\mathbb {R}^n$ with base $S^\circ (a,\nu )\subset \mathbb {R}^{n-1}\times \{0\}$ with vertices at $\pm \frac 1 {2\delta }e_n$ . Each of these cones has measure $\frac 1{2 \delta n} |S^\circ (a,\nu )|$ . We conclude that

$$\begin{align*}|\Omega^\circ| = |T| \ge \frac 1{ \delta n} |S^\circ(a,\nu)| = \frac 1 n d_{\Omega}(a)^{-1}|S^\circ(a,\nu)|. \end{align*}$$

Step 2. Let $P^\perp = \{0^{n-1}\}\times \mathbb {R}$ , the orthogonal complement of P. Since $a=0$ and $d_{\Omega }(a) \ge \delta $ , it is clear that $\Omega \cap P^\perp = \Omega _{P^\perp } \supset \{0^{n-1}\}\times (-\delta ,\delta )$ . It easily follows that $\Omega _{P^\perp }^\circ \subset \{0^{n-1}\}\times (-\frac 1\delta , \frac 1\delta )$ . In addition, Lemma 2.6 implies that

$$\begin{align*}\pi_{P^\perp}(\Omega^\circ) \subset \Omega_{P^\perp}^\circ. \end{align*}$$

It follows from these facts and (2.9) that

$$\begin{align*}\Omega^\circ\subset S^\circ(a,\nu) \times (-\frac 1\delta, \frac 1\delta), \end{align*}$$

(writing $S^0(a,\nu )$ as a subset of $\mathbb {R}^{n-1}$ rather than of $\mathbb {R}^{n-1}\times \{0\}$ ). Thus,

$$\begin{align*}|\Omega^\circ| \le |S^\circ(a,\nu)|\times \frac 2\delta = 2d_{\Omega}(a)^{-1}|S^\circ(a,\nu)|.\\[-37pt] \end{align*}$$

Proof of Theorem 1.3

Estimate (1.11) follows directly from Proposition 2.1 and Lemma 2.6.

Proof Let $\delta>0$ and $a\in \Omega $ with $d_{\Omega }(a)=|a-b|=\delta $ for some $b\in \partial \Omega $ . After a translation and rotation, we may assume that (3.1) holds. We necessarily have that $a=(0, \ldots , 0, \delta )$ . Indeed, suppose $a_i\ne 0$ for some $1\le i \le n-1$ . Then from the supporting hyperplane $\{x_{n}=0\}$ , we obtain

$$\begin{align*}d_\Omega(a)\le \operatorname{dist}(a,\{x_{n}=0\})=|a_{n}|<|a|=d_\Omega(a), \end{align*}$$

a contradiction, verifying the claim.

Now, relabeling coordinates, we write the unit outer normal at b as $\nu =-e_{n}$ and have that

$$ \begin{align*} S(a,\nu) &\subseteq \left\{x\in \mathbb{R}^{n-1}\times \{0\}: \delta> \eta \sum_{i=1}^k |x_i|^{p_i}, \ \ |(x_{k+1}, \ldots x_{n-1})|< \text{diam}(\Omega)\right\}. \end{align*} $$

Thus, $|S(a,\nu )|$ is bounded by the volume of the set on the right, which is

$$\begin{align*}C\delta^{\frac 1{p_1}+\ldots + \frac 1{p_k} } \end{align*}$$

for a constant C dependingFootnote ¹ on $\eta , p_1,\ldots , p_k, k, n-1, \text {diam}(\Omega )$ . Since a was arbitrary, (1.13) and Theorem 1.3 (or see (1.14)) imply that

$$\begin{align*}\omega_{\Omega}(\delta) \le C\,\delta^{\alpha}. \end{align*}$$

Hence, the result on the Hölder estimate.

Proof Assume that $0<\delta <\delta _0$ , to be fixed below.

Step 1. It is clear from (3.2) and properties of h that if $\delta _0$ is sufficiently small (in fact, here, $\delta _0<D/2$ is sufficient), then the origin is the unique closest boundary point to $(0,\delta )$ , and hence that $N(a)$ as defined in (1.10) consists of $\{ -e_2\}$ . Then the definitions imply that

$$\begin{align*}S((0,\delta), -e_2) = ( - h^{-1}(\delta) , h^{-1}(\delta)) \times \{0\}. \end{align*}$$

Recall our convention that $S^\circ (a,\nu )$ denotes the polar within the subspace $\nu ^\perp $ . If a and b are positive numbers, then $(a,b)^\circ = [\frac 1a, \frac 1b]$ , so it follows that

$$\begin{align*}|S^\circ((0,\delta), -e_2) | = \frac 2 { h^{-1}(\delta)}. \end{align*}$$

This and Theorem 1.3 imply the lower bound for $\omega _\Omega (\delta )$ in (3.3).

Step 2. To complete the proof of the Lemma, again by Theorem 1.3, it suffices to show that if $d_{\Omega }(a)=\delta $ and $\nu \in N(a)$ , then

(3.4) $$ \begin{align} |S^\circ(a,\nu)| \ge \frac 1{h^{-1}(\delta)} , \end{align} $$

if $\delta _0$ is small enough. Fix any $a\in \Omega $ such that $d_{\Omega }(a)=\delta $ and any $b\in \partial \Omega $ such that $d_{\Omega }(a)=|a-b|$ , and let $\nu = \frac {b-a}{|b-a|}$ . Noting from (3.2) that $\Omega $ is symmetric about the $x_2$ axis (since h is even) and about the line $x_2 = D/2$ , we can assume that $b\in \{(x_1, x_2)\in \partial \Omega : 0\le x_1\le R, x_2 = h(x_1)\}$ .

Then we define $\widetilde \Omega $ to be the set obtained by translating b to the origin and rotating so that $\widetilde \Omega \subset \{(x_1, x_2): x_2>0\}$ . This operation moves a to the point $(0,\delta )$ . Next, we let $\tilde h_1$ be the function whose graph parametrizes the lower part of $\partial \widetilde \Omega $ , defined by $\tilde h(x_1) := \inf \{ x_2\in \mathbb {R} : (x_1, x_2)\in \widetilde \Omega \}$ . By our assumption about the monotonicity of the boundary curvature along the short arc connecting $(0,0)$ to $(R,D/2)$ , we see that if $\delta _0$ is small enough, then

$$\begin{align*}\text{curvature of }\partial \Omega\text{ at }(x_1, h(x_1)) \le \text{curvature of }\partial \widetilde \Omega\text{ at }(x_1, \tilde h(x_1)) \end{align*}$$

for $0< x_1 < h^{-1}(\delta _0)$ . Since $\tilde h(0) = \tilde h'(0) = h(0) = h'(0)$ , and because $0=h"(0)\le \tilde h"(0)$ , this implies that $\tilde h(x_1) \ge h(x_1)$ for $0<x_1< h^{-1}(\delta _0)$ .

Computing $S^\circ (a,\nu )$ in the coordinate system of $\widetilde \Omega $ , we find that $S(a,\nu ) = (-\alpha , \beta )\times \{0\}$ , where $-\alpha ,\beta $ are the negative and positive solutions, respectively, of the equation $\tilde h(x)=\delta $ , and thus,

$$\begin{align*}|S^\circ(a, \nu)| = \frac 1\beta+ \frac 1 \alpha \ge \frac 1\beta. \end{align*}$$

But the fact that $\tilde h \ge h$ for $0<x_1< h^{-1}(\delta _0)$ implies that $\beta \le h^{-1}(\delta )$ , proving (3.4).

Proof We first claim that for $R, c>0$ and $S\subset \mathbb {R}^k$ ,

(3.5) $$ \begin{align} \text{ if }S\subset B_R \text{ and }|S^\circ|<c, \qquad\text{ then } B_r\subset S \text{ for }r= \frac {|B^{k-1}_1|}{2c(2R)^{k-1}}. \end{align} $$

Indeed, for $r<R$ , suppose $S\subset B_R$ does not contain $B_r$ . By a rotation, we may assume that there is a point of the form $b = (0,\ldots , 0, r_1)$ with $0<r_1<r$ such that $d_S(0) = |0 - b| = r_1$ . Then the plane $\{ x : x_k = r_1\}$ is a supporting hyperplane at b, so $S\subset B_R \cap \{ x : x_k < r_1\}$ . We claim that

(3.6) $$ \begin{align} \{ (y', y_k)\in \mathbb{R}^{k-1}\times \mathbb{R} : |y'|< \frac 1{2R}, \ 0<y_k < \frac 1{2r_1}\} \subset S^\circ. \end{align} $$

This is clear, since if $y = (y', y_n)$ belongs to the set on the left, then one readily checks that $x\cdot y\le 1$ for all $x\in S \subset B_R \cap \{ x : x_k<r_1\}$ , proving (3.6). It follows that

$$\begin{align*}c> |S^\circ| \ge \frac {|B^{k-1}_1|}{2r_1(2R)^{k-1} } \ge \frac {|B^{k-1}_1|}{2r(2R)^{k-1}}. \end{align*}$$

This cannot happen if $r \le \frac {|B^{k-1}_1|}{2c(2R)^{k-1}}$ . So for such r, it must be the case that $B_r\subset S$ , proving (3.5).

Now assume that there exists $A>0$ such that $\omega _\Omega (\delta )\ge A \delta ^{1/n}$ , and fix a sequence $a_j\in \Omega $ and $\nu _j\in N(a_j)$ such that $d_{\Omega }(a_j) := \delta _j\to 0$ , and $|S^\circ (a_j, \nu _j)| < n A^{-n}$ . The existence of such sequences follows directly from Theorem 1.3. Upon passing to subsequences (still labeled $a_j, \nu _j, \delta _j$ ) we may assume that $a_j\to b\in \partial \Omega $ . After a translation and a rotation, we may assume that $b=0$ and $\Omega \subset \{ x\in \mathbb {R}^n : x_n>0\}$ .

Appealing to (3.5) with $k=n-1$ , we find that $B_r\subset S(a_j, \nu _j)$ with $r = \frac {|B^{n-2}_1| A^n }{2^{n-1}n R^{n-2}}$ . Then the definition of $S(a_j,\nu _j)$ implies that

$$\begin{align*}\{ a_j + x : x\cdot \nu_j=0, |x|< r\}\subset \Omega \subset \{ x : x_n>0\} \quad\text{for every }j. \end{align*}$$

This implies that $\nu _j\to - e_n $ as $j\to \infty $ and $a_j\to b=0$ . Then

$$\begin{align*}\{ a_j + x : x\cdot \nu_j=0, |x|< r\} \ \ \longrightarrow \ \ \{ x : x\cdot(- e_n)=0, |x|<r\} = B^{n-1}_r\times \{0\} \end{align*}$$

as $j\to \infty $ , in the Hausdorff distance. It follows that $B^{n-1}_r\times \{0\}\subset \bar \Omega $ , and hence, since $\Omega \subset \{x : x_n>0\}$ , we conclude that $B^{n-1}_r\times \{0\}\subset \partial \Omega $ . Thus, we have found a flat spot.

We omit the proof that if $\partial \Omega $ has a flat spot, then $\omega _\Omega (\delta ) \ge c\delta ^{1/n}$ for some c, which is a very direct consequence of Theorem 1.3.

Proof Choosing coordinates so that $x=0$ and $\nu = - e_n$ , we find that locally near $0$ , $\Omega $ has the form $\{ x = (x', x_n)\in \mathbb {R}^{n-1}\times \mathbb {R}: x_n> h(x')\}$ for h such that $h(x') = \frac 12 x'\cdot Q x'(1+o(1))$ as $x'\to 0$ , with $\det Q = \kappa $ . From there, the definitions imply that

$$\begin{align*}S(x-\delta\nu, \nu) = \{ x'\in \mathbb{R}^{n-1} : h(x') < \delta \} \times \{0\}. \end{align*}$$

The expansion of h for small $x'$ implies that for any $\varepsilon>0$ , there exists $\delta _0>0$ such that if $0<\delta <\delta _0$ , then

$$ \begin{align*} \{x'\in \mathbb{R}^{n-1} : x'\cdot Q x' < 2\delta(1-\varepsilon) \} &\subset \{ x'\in \mathbb{R}^{n-1} : h(x') < \delta \} \\ &\subset \{x'\in \mathbb{R}^{n-1} : x'\cdot Q x' < 2\delta(1+\varepsilon) \}. \end{align*} $$

Since the ellipse $\{ x' : x'\cdot Q x' < r^2\}$ has volume $r^{n-1}|B^{n-1}_1|/\sqrt {\det Q}$ , we deduce (3.7) from the standard fact that $|E| \, |E^\circ | = |B^{n-1}_1|^2$ for any ellipse E in $\mathbb {R}^{n-1}$ , a consequence of affine invariance.

Proof We recall that if $S\subset \mathbb {R}^n$ is a convex set, the support function $\sigma _S$ is defined by

$$\begin{align*}\sigma_S(y) = \sup_{x\in S}x\cdot y. \end{align*}$$

It is rather clear from the definitions that

$$\begin{align*}S^\circ = \{ y\in \mathbb{R}^n : \sigma_S(y)\le 1\}\qquad\quad \sigma_{S-a}(y) = \sigma_S(y) - a\cdot y,\qquad\quad \sigma_{B^n_1}(y) = |y|. \end{align*}$$

Step 1. Let B denote the unit ball $B^n_1$ . Then using the properties of the support function noted above,

$$\begin{align*}f_B(a) = |(B-a)^\circ| = |\{ y\in \mathbb{R}^n : \sigma_{B-a}(y) \le 1\}| = |\{ y\in \mathbb{R}^n : |y|-a\cdot y \le 1\}|. \end{align*}$$

For a as above, by writing $y = (y', y_n)\in \mathbb {R}^{n-1}\times \mathbb {R}$ , by squaring both sides, completing a square, and rearranging, we find that

$$\begin{align*}|y| \le 1 + a\cdot y = 1-\alpha y_n \quad \iff \quad (1-\alpha^2) |y'|^2 + (1-\alpha^2)^2(y_n + \frac{\alpha}{1-\alpha^2})^2 \le 1. \end{align*}$$

The inequality on the right defines an ellipsoid whose volume is easily found, yielding

$$\begin{align*}f_B(a) = |(B-a)^\circ| = (1-\alpha^2)^{-(n+1)/2}|B^n_1|. \end{align*}$$

Since $d_{\partial B}(a) = 1-\alpha $ , we can rewrite this as

$$\begin{align*}f_B(a) = \frac{|B^n_1|}{ [d_{ B}(a)(2 - d_{ B}(a)]^{\frac{n+1}2}}. \end{align*}$$

Step 2. Now let E denote a general ellipsoid as in the statement of the theorem. Noting that

$$\begin{align*}E = MB \quad\text{ for }M = \text{diag}(\ell_1,\ldots, \ell_n), \end{align*}$$

we find from Lemma 2.4 that

$$\begin{align*}f_E(a) = f_{MB}(a) = (\ell_1 \cdots \ell_n)^{-1} f_B(M^{-1}a). \end{align*}$$

Also, since $M^{-1}a = (0,\ldots , 0, -\alpha /\ell _n)$ , we see that $d_B(M^{-1}a) = 1-\frac \alpha {\ell _n} = \frac {\ell _n-\alpha }{\ell _n}$ . Substituting into the above formula, we obtain

$$\begin{align*}f_E(a) = \frac {\ell_n^{\frac{n-1}2}}{\ell_1\cdots\ell_{n-1}} \frac{|B^n_1|}{ [ ( \ell_n-\alpha) (2 - \frac{\ell_n-\alpha}{\ell_n})]^{\frac{n+1}2}}. \end{align*}$$

It is clear that if $\alpha $ is close enough to $\ell _n$ , then $\ell _n-\alpha = d_{ E}(a)$ . So to complete the proof, we must show that $\kappa (p) = \ell _n^{n-1}/(\ell _1^2\cdots \ell _{n-1}^2)$ . This is an easy computation. Near p, we write $\partial E$ as the graph

$$\begin{align*}x_n = g(x'), \qquad g(x') = -\ell_n\left(1 - ( \frac {x_1^2}{\ell_1^2} +\cdots + \frac {x_{n-1}^2}{\ell_{n-1}^2})\right)^{1/2}. \end{align*}$$

We compute that $Dg(0)=0$ and $D^2 g(0) = \ell _n \text {diag}(\ell _1^{-2}, \ldots , \ell _{n-1}^{-2})$ , and it follows that $\kappa (p) = \det D^2 g(0) = \ell _n^{n-1}/(\ell _1^{2} \cdots \ell _{n-1}^{2})$ , as claimed.

Proof Given $\varepsilon>0$ , choose a point $p_\varepsilon \in \partial \Omega $ at which $\partial \Omega $ is twice differentiable and $\kappa (p_\varepsilon ) < (1+\varepsilon )\kappa _0$ . We may assume after a translation and a rotation that $p_\varepsilon =0$ and that there exists $r>0$ such that in a neighborhood of $p_\varepsilon $ ,

(4.2) $$ \begin{align} \Omega\cap B^n_r = \{ (x', x_n)\in B^n_r : x_n> g(x') \} \end{align} $$

for a convex g such that

(4.3) $$ \begin{align} g(x') = \frac 12 \sum_{j=1}^{n-1}\lambda_j x_j^2 + o(|x'|^2) \quad\text{ as } |x'|\to 0, \end{align} $$

with

$$\begin{align*}\prod_{j=1}^{n-1} \lambda_j = \kappa(p_\varepsilon) = \kappa(0), \qquad \lambda_j>0\text{ for all }j. \end{align*}$$

Now let $E_\varepsilon $ be the ellipse

$$\begin{align*}E_\varepsilon = \{ x\in \mathbb{R}^n : \frac {x_1^2}{\ell_1^2} + \cdots+ \frac {x_{n-1}^2}{\ell_{n-1}^2} + \frac{(x_n - \ell_n)^2}{\ell_n^2} < 1\} \end{align*}$$

for

$$\begin{align*}\ell_n = \eta\text{ to be chosen}, \quad \ell_j = \left( \frac{\ell_n}{\lambda_j(1+\varepsilon)}\right)^{1/2}. \end{align*}$$

We claim that

(4.4) $$ \begin{align} \text{if }\eta>0\text{ is taken to be small enough, then } E_\varepsilon \subset \Omega. \end{align} $$

In view of (4.2), it suffices to show that if $\eta $ is small enough and $x = (x',x_n)\in E_\varepsilon $ , then $x\in B_r$ and $x_n> g(x')$ .

It is clear that there exists $\eta _0>0$ such that $E_\varepsilon \subset B^n_r$ whenever $0<\eta <\eta _0$ .

Second, we use the concavity of the square root to see that

$$ \begin{align*} x\in E_\varepsilon &\quad\Longrightarrow \quad \frac{(\ell_n-x_n)^2}{\ell_n^2}\le 1 - \sum_{j=1}^{n-1} \frac {x_j^2}{\ell_j^2} \\ &\quad\Longrightarrow \quad \frac{(\ell_n-x_n)}{\ell_n}\le \left(1 - \sum_{j=1}^{n-1} \frac {x_j^2}{\ell_j^2} \right)^{1/2} \le 1 - \frac 12\sum_{j=1}^{n-1} \frac {x_j^2}{\ell_j^2} \\ &\quad\Longrightarrow \quad x_n \ge \frac 12 \sum_{j=1}^{n-1} \frac{\ell_n}{\ell_j^2} x_j^2 = \frac {(1+\varepsilon)}2 \sum_{j=1}^{n-1}\lambda_j x_j^2. \end{align*} $$

Note also that the definitions imply that $|x'| < C \sqrt \eta $ for $(x',x_n)\in E_\varepsilon $ , so the above inequality and (4.3) imply that there exists $\eta _1\in (0,\eta _0)$ such that if $0<\eta <\eta _1$ , then $x_n> g(x')$ , completing the proof of (4.4).

We henceforth fix $\eta <\eta _1$ . Let $a_\delta = (0,\ldots , 0,\delta )$ for $\delta < \ell _n$ , and note that $a_\delta \in E_\varepsilon $ when $\delta <2\ell _n$ . Note also that $d_{\Omega }(a_\delta ) = d_{ E_\varepsilon }(a_\delta )= \delta $ for all sufficiently small $\delta>0$ .

It follows from Lemma 2.2 that $f_\Omega (a_\delta )\le f_{E_\varepsilon }(a_\delta ) (a)$ , so we use (an easy modification of) Lemma 4.1 to conclude

$$\begin{align*}\lim_{\delta\searrow 0} \delta^{\frac{n+1}2}f_\Omega(a_\delta) \le \lim_{\delta\searrow 0} \delta^{\frac{n+1}2}f_{E_\varepsilon}(a_\delta) = \lim_{\delta\searrow 0} \frac {\sqrt{\kappa_{E_\varepsilon}(0)}|B^n_1|}{(2- \frac \delta {\ell_n})^{(n+1)/2}} = \frac {\sqrt{\kappa_{E_\varepsilon}(0)}|B^n_1|}{2^{(n+1)/2}}, \end{align*}$$

where $\kappa _{E_\varepsilon }(0)$ denotes the curvature of $\partial E_\varepsilon $ at $0$ , which is

$$\begin{align*}\kappa_{E_\varepsilon}(0) = (1+\varepsilon)^{n-1}\kappa(0) < (1+\varepsilon)^n \kappa_0. \end{align*}$$

Applying Proposition 2.1, we find that

$$\begin{align*}\liminf_{\delta\searrow 0} \delta^{-\frac{n+1}{2n}} \omega_\Omega(\delta)> \frac 1{\sqrt{1+\varepsilon}} \left( \frac{2^{(n+1)/2}}{ {\sqrt \kappa_0}|B^n_1|} \right)^{1/n}. \end{align*}$$

Since $\varepsilon>0$ was arbitrary, conclusion (4.1) follows.

Proof We will show that for any $a\in \Omega $ ,

(4.5) $$ \begin{align} f_\Omega(a) \ge \frac 1{d_{\Omega}(a)^{ \frac 32} } \frac{ \sqrt{\kappa_0} \pi}{2^{3/2} }. \end{align} $$

In view of Proposition 2.1, this implies that

$$\begin{align*}\sup_{\delta>0} \frac{\omega_\Omega(\delta)}{\delta^{ \frac 34 } } \le \left( \frac {2^{3/2}} {\sqrt{\kappa_0} \pi}\right)^{1/2}. \end{align*}$$

This will complete the proof of the Proposition, as the opposite inequality follows from Proposition 4.1.

To prove (4.5), fix any $a\in \Omega $ , and let $b\in \partial \Omega $ be a point such that $d_{\Omega }(a) = |a-b| =: \delta $ , not necessarily small. After a rotation and a translation, we may assume that $b = (0,0)$ and $a = (0,\delta )$ . Clearly, $B_\delta (a)\subset \Omega $ , and $0\in \partial B_\delta (a)\cap \partial \Omega $ . From these facts and the convexity of $\Omega $ , one easily sees that $\Omega \subset \{ (x_1,x_2) : x_2>0\}$ and

$$\begin{align*}\partial \Omega\cap [(-\delta,\delta)\times (0,\delta)] \ \ \subset \ \ \left\{( x_1,x_2) : 0< x_2 \le \delta - \sqrt{ \delta^2 - x_1^2}\right\}. \end{align*}$$

Let $I = \{ x_1 : (x_1,x_2)\in \Omega \text { for some }x_2\}$ be the projection of $\Omega $ onto the $x_1$ -axis. Then writing the lower part of $\partial \Omega $ as the graph of a function $g:I\to \mathbb {R}$ , we have

$$\begin{align*}g\text{ is convex}, \qquad 0\le g(x_1) \le \delta - \sqrt{\delta^2-x_1^2},\qquad \Omega\subset \{(x_1,x_2) : x_1\in I, x_2> g(x_1)\}. \end{align*}$$

Note that g is differentiable at $x_1=0$ , with $g'(0)=0$ .

We now claim that

(4.6) $$ \begin{align} g(x_1) \ge \frac {\kappa_0}2 x_1^2 \quad\text{ for }x_1\in I, \quad \text{ and thus, }\Omega \subset D := \{ (x_1,x_2) : x_2> \kappa_0 x_1^2/2 \}. \end{align} $$

We will prove this for $x_1>0$ ; the argument for $x_1<0$ is basically identical. Let us write $S := \{ x_1\in I : g\text { is twice differentiable at }x_1\}$ . To prove (4.6), we recall assumption (1.6), which implies that

$$\begin{align*}\frac{g"(x_1)}{(1+(g'(x_1)^2)^{3/2}} \ge \kappa_0\qquad\text{ for every }x_1\in S. \end{align*}$$

This clearly implies that $g"(x_1)\ge \kappa _0$ in S. Since g is convex, $g'$ is increasing function, so for positive $x_1\in I$ , elementary real analysis yields

$$\begin{align*}g'(x_1) = g'(x_1) - g'(0) \ge \int_{\{t\in S : 0<t< x_1\}} g"(t) dt \ge \kappa_0 x_1. \end{align*}$$

Since $g'$ is locally Lipschitz, we obtain (4.6) by integrating again.

In view of (4.6) and Lemma 2.2, in order to prove (4.5), it suffices to show that

(4.7) $$ \begin{align} f_D(a) = \frac {\sqrt{\kappa_0}\pi}{(2\delta)^{3/2}}. \end{align} $$

This is a straightforward computation. First, let

$$\begin{align*}M := \left(\begin{array}{cc} \sqrt{\frac {\kappa_0}{2\delta} }&0\\ 0&\frac 1 \delta\end{array}\right). \end{align*}$$

Then $\widetilde D := MD = \{ (x_1, x_2) : x_2> x_1^2 \}$ and $Ma = (0,1) = e_2$ . By Lemma 2.4,

$$\begin{align*}f_D(a) = \sqrt{\frac{ \kappa_0}{2\delta^{3/2} } }\ f_{\widetilde D}(e_2) \end{align*}$$

and

$$\begin{align*}f_{\widetilde D}(e_2) = |(\widetilde D - e_2)^\circ| = |\{ y\in \mathbb{R}^2 : \sigma_{\widetilde D - e_2}(y) \le 1\}|. \end{align*}$$

Given $y\in \mathbb {R}^2$ , we compute $\sigma _{\widetilde D - e_2}(y)$ by attempting to find x that maximizes $x\mapsto y\cdot (x-e_2)$ subject to the constraint that $x\in \widetilde D$ . It is clear that a maximum can only occur for $x\in \partial \widetilde D$ , so one can use Lagrange multipliers to find that

$$\begin{align*}\sigma_{\widetilde D - e_2}(y_1,y_2) = \begin{cases} +\infty&\text{ if }y_2\ge 0\\ -\frac {y_1^2}{4y_2} - y_2 &\text{ if }y_2<0. \end{cases} \end{align*}$$

It easily follows that

$$\begin{align*}(\widetilde D-e_2)^\circ = \left\{ (y_1, y_2) : y_1^2 + 4\left(y_2-\frac 12\right)^2 \le 1\right\}, \qquad\text{ and hence, }\quad f_{\widetilde D}(e_2) = \frac \pi 2, \end{align*}$$

concluding the proof of (4.7).

Proof In view of Proposition 4.1 and Proposition 2.1, we only need to prove that

(4.9) $$ \begin{align} \liminf_{\delta\searrow 0} \inf_{d_{\Omega}(a)=\delta} \delta^{(n+1)/2} f_\Omega(a) \ge \frac{|B^n_1| \sqrt{\kappa_0}} {2^{(n+1)/2}}. \end{align} $$

Step 1. We first claim that for any $\varepsilon _1>0$ , there exists $r_0>0$ such that for any $b \in \partial \Omega $ , there exists a rigid motion S (that is, the composition of a rotation and a translation) and a convex $C^2$ function $g:B^{n-1}_{r_0}\to [0,\infty )$ such that

$$\begin{align*}S(0) = b, \qquad S(\{ (x', g(x')): |x'|<r_0 \}) \subset \partial \Omega, \end{align*}$$

and $g(0)=0$ , with

(4.10) $$ \begin{align} \|D^2 g(x') - D^2 g(0)\| < \varepsilon_1 \qquad\text{ for }x'\in B_{r_0}^{n-1}(0), \end{align} $$

where $\| \cdot \|$ denotes the operator norm. Informally, this states that $\partial \Omega $ is uniformly $C^2$ . Since $\partial \Omega $ is $C^2$ and compact, on some level this is clear, but we provide some details nonetheless. Our proof of (4.10) will also show that there exist positive $\Lambda _{min}\le \Lambda _{max}$ , independent of $b\in \partial \Omega $ , such that

(4.11) $$ \begin{align} \Lambda_{min} \le \lambda_1\le \cdots \le \lambda_{n-1}\le\Lambda_{max}, \qquad \{ \lambda_j \} = \text{ eigenvalues of }D^2g(0). \end{align} $$

First, the compactness of $\partial \Omega $ implies that there exists $R>0$ , $J\ge 2$ and

• • maps $S_j: \mathbb {R}^n\to \mathbb {R}^n$ for $j=1,\ldots , J$ , each one a rigid motion (the composition of a translation and a rotation),

• • $C^2$ functions $h_j:B_{4R}^{n-1}\to [0,\infty )$ for $j=1,\ldots , J$ ,

such that $|\nabla h_j| \le \frac 14$ in $B_{4R}^{n-1}$ and

$$\begin{align*}\partial \Omega = \cup_{j=1}^J U_j, \qquad U_j :=S_j \left( \{ (x', h_j(x')) : x'\in B^{n-1}_R \} \right). \end{align*}$$

For every j, clearly, $h_j, D h_j$ , and $D^2 h_j$ are uniformly continuous on $B_{3R}^{n-1}$ , and there are only finitely many of these functions, so there exists a common $C^2$ modulus of continuity for $\{ h_j\}_{j=1}^J$ on $B_{3R}^{n-1}$ , by which we mean a continuous, increasing function $\mu :[0,\infty )\to [0,\infty )$ such that $\mu (0)=0$ and

(4.12) $$ \begin{align} |h_j(x)-h_j(y)| + |Dh_j(x)-Dh_j(y)| + |D^2h_j(x)-D^2 h_j(y)| \le \mu(|x-y|) \end{align} $$

for all x and y in $B_{3R}^{n-1}$ and $j=1,\ldots , J$ .

Now consider any $b\in \partial \Omega $ . Fix some $\bar x'\in B^{n-1}_R(0)$ and $j\in \{1,\ldots , J\}$ such that $b = S_j(\bar x', h_j(\bar x'))$ . By a further translation, we may send $(\bar x', h_j(\bar x'))$ to the origin in $\mathbb {R}^n$ . Then by a rotation in $\mathbb {R}^n$ , we can arrange that part of the (translated and rotated) graph of $h_j$ over $B_{2R}(\bar x')$ can be written as the graph over some ball $B_{r}^{n-1}$ of a convex $C^2$ function $g:B^{n-1}_r\to [0,\infty )$ such that $g(0)=0$ and $\nabla g(0)=0$ . By applying Lemma 4.2 below to $h(x') = h_j(x' - \bar x') - h_j(\bar x')$ , we find that independent of the choice of b, the domain of the resulting function g can be taken to be $B^{n-1}_R$ , and g has a $C^2$ modulus of continuity in $B_R^{n-1}$ that can be estimated solely in terms of $\mu $ from (4.12). This proves (4.10). Similarly, (4.11) follows from (4.25) in Lemma 4.2, which we prove below, together with the fact that for every $j\in \{1,\ldots ,J\}$ , the eigenvalues of $D^2 h_j$ are bounded away from $0$ in $B^{n-1}_R$ , being positive on $B^{n-1}_{2R}$ .

Step 2. We now prove (4.9).

Step 2.1: Normalization and approximation by a quadratic. Because $\partial \Omega $ is $C^2$ and compact, there exists $\delta _0>0$ such that if $\delta := d_{\Omega }(a) <\delta _0$ , then there is a unique $b\in \partial \Omega $ such that $d_{\Omega }(a)=|a-b|$ . Fix some such a and b. In view of Step 1, we may assume after a rigid motion that $a = \delta e_n = (0,\ldots , 0,\delta )$ and $b=0$ , and that there is a nonnegative convex function g, vanishing at $x'=0$ , such that (4.10) holds for some $\varepsilon _1$ and $r_0(\varepsilon _1)$ to be specified in a moment, and with $\{(x', g(x')) : |x'|<r_0\}$ contained in the (rotated and translated) $\partial \Omega $ .

Fix $\varepsilon>0$ . Using (4.10) for a suitably small choice of $\varepsilon _1$ and calculus, we can guarantee that

(4.13) $$ \begin{align} \left. \begin{aligned} g(x')> \frac {(1-\varepsilon)}2 x'\cdot Qx' &\\ |Dg(x') - Q x'| < \varepsilon |x'| & \end{aligned}\right\} \ \text{ in }B_{r_0}^{n-1}, \quad\text{ for }Q = D^2g(0). \end{align} $$

Let $M:= Q^{1/2}$ , the positive definite symmetric square root of Q, and define

$$\begin{align*}\widetilde \Omega := \left\{ \left(\frac {M x'}{\sqrt \delta}, \frac{x_n}\delta \right) : (x', x_n)\in \Omega\right\}, \qquad \tilde g(y') := \frac 1 \delta g(\sqrt \delta M^{-1} y'). \end{align*}$$

Due to (4.11), the eigenvalues of M are bounded between $ \Lambda _{min}^{1/2}$ and $\Lambda _{max}^{1/2}$ . Thus,

(4.14) $$ \begin{align} |y'|< r_\delta := \frac{r_0}{\sqrt{\delta\Lambda_{min}}} \qquad\Longrightarrow \quad |\sqrt \delta M^{-1} y'| < r_0. \end{align} $$

The definition of $\widetilde g$ is chosen so that

$$\begin{align*}\{ (y', \tilde g(y')) : |y'| < r_\delta \} \ \subset \ \partial \widetilde \Omega. \end{align*}$$

Hypothesis (1.6) implies that $\det Q = (\text {curvature of }\partial \Omega \text { at }b=0) \ge \kappa _0$ , so we deduce from Lemma 2.4 that

(4.15) $$ \begin{align} \delta^{(n+1)/2}f_\Omega(a) = |\! \det M| \, f_{\widetilde \Omega}(\tilde a) \ge \sqrt \kappa_0\ f_{\widetilde \Omega}(\tilde a) \qquad\text{ for } \ \ \tilde a = \frac a \delta = e_n. \end{align} $$

In addition, using (4.14) and requiring that $\varepsilon < \frac 12 \Lambda _{min}$ , we can translate properties (4.13) into statements about $\tilde g$ . It follows that $(y',\tilde g(y'))\in \partial \widetilde \Omega $ and

(4.16) $$ \begin{align} \left. \begin{aligned} \tilde g(y')> \frac {(1-\varepsilon)}2 |y'|^2 & \\ |D\tilde g(y') - y'| < \varepsilon \Lambda_{min}^{-1} |y'| <\frac 12 |y'| & \end{aligned}\right\} \text{ in }B_{r_\delta}^{n-1}. \end{align} $$

Fix $z'$ such that $|z'|< r_\delta /2$ and let $\Phi (y') := z'+y' - D\tilde g(y')$ . Then the second inequality in (4.16) implies that $\Phi $ maps $\bar B_{r_2}^{n-1}$ to itself, for $r_2 = 2|z'| < r_\delta $ . Thus, the Brouwer Fixed Point Theorem implies that $\Phi $ has a fixed point $y'$ in $B_{r_2}^{n-1}$ . But $\Phi (y') =y'$ exactly when $D\tilde g(y') = z'$ . It follows that

(4.17) $$ \begin{align} B^{n-1}_{r_\delta/2} \subset \{ D\tilde g(y') : |y'|< |B^{n-1}_{r_\delta}| \}. \end{align} $$

Step 2.2: finding a large subset of $\partial u_{\widetilde \Omega , a}(\tilde \Omega )$ .

We will write $\tilde u_{\tilde a} := u_{\widetilde \Omega , \tilde a}$ . We next will show that

(4.18) $$ \begin{align} E_{\varepsilon,\delta} := \left\{ s( z', -1) :|z'|< \frac {r_\delta}2, \ \ 0\le s \le \left( 1 + \frac 1{2(1-\varepsilon)} |z'|^2 \right)^{-1} \right \} \subset \partial \tilde u_{\tilde a}(\widetilde \Omega). \end{align} $$

To see this, fix $z'\in B^{n-1}_{r_\delta /2}$ , and using (4.17), find $y'\in B^{n-1}_{r_\delta }$ such that $\nabla \tilde g(y') = z'$ . Let

$$\begin{align*}\ell_{y'}(x) := \frac{ (\nabla \tilde g(y'), -1) \cdot(x - (y', \tilde g(y')))} {1 + y'\cdot \nabla \tilde g(y')- \tilde g(y')}. \end{align*}$$

We claim that $\ell _{y'}$ is a supporting hyperplane to the graph of $\tilde u_{\tilde a}$ at $\tilde a =e_n$ . We must show that $\ell _{y'}(\tilde a) = \tilde u_{\tilde a}(\tilde a) = -1$ , which follows directly from the definition, and that $\ell _a \le \tilde u_{\tilde a}$ in $\widetilde \Omega $ . Since both $\ell _a$ and $\tilde u_{\tilde a}$ are linear on line segments connecting $\partial \widetilde \Omega $ to $\tilde a$ , it suffices to check that $\ell _a \le \tilde u_{\tilde a} = 0$ on $\partial \widetilde \Omega $ . This follows from noting that $\ell _{y'}$ vanishes exactly on the hyperplane $\{ x\in \mathbb {R}^n : \nu (y) \cdot (x - y) = 0\}$ , where $y = (y', \tilde g(y'))\in \partial \widetilde \Omega $ and $\nu (y)$ is the outer unit normal to $\partial \widetilde \Omega $ at y. This is a supporting hyperplane to $\partial \widetilde \Omega $ , so $\ell _{y'}$ does not change sign in $\widetilde \Omega $ . Since $\ell _{y'}(\tilde a)<0$ , the claim follows.

Thus, $\nabla \ell _{y'}(\tilde a)\in \partial \tilde u_{\tilde a}(\tilde a)$ . Since it is clear that $0\in \partial \tilde u_{\tilde a}(\tilde a)$ and $\partial \tilde u_{\tilde a}(\tilde a)$ is convex, it follows that the segment $\{ s\nabla \ell _{y'}(\tilde a) : 0\le s \le 1\} \subset \partial \tilde u_{\tilde a}(\tilde a)$ ; that is,

(4.19) $$ \begin{align} \left\{ s(\nabla\tilde g(y') , -1) \ : \ 0 \le s \le \frac 1{1+y'\cdot \nabla\tilde g(y')-\tilde g(y')} \right \} \subset \partial \tilde u_{\tilde a}(a). \end{align} $$

However, recalling that $\nabla \tilde g(y') = z'$ , and using (4.16) and elementary inequalities,

$$\begin{align*}y'\cdot \nabla \tilde g(y') - \tilde g(y') \le y'\cdot z' - \frac {(1-\varepsilon)}2|y'|^2 \le \frac 1{2(1-\varepsilon)}|z'|^2. \end{align*}$$

We deduce (4.18) by combining this and (4.19).

Step 2.3: conclusion of proof. Since a was an arbitrary point such that $d_{\Omega }(a)=\delta $ , it follows from (4.18) and (4.15) that for all sufficiently small $\delta $ ,

(4.20) $$ \begin{align} \inf_{d_{\Omega}(a)=\delta} \delta^{(n+1)/2}f_\Omega(a) \ge \sqrt{\kappa_0} |E_{\varepsilon,\delta}|. \end{align} $$

We will show that

(4.21) $$ \begin{align} \lim_{\delta\to 0}|E_{\varepsilon,\delta}| = (1-\varepsilon)^{(n-1)/2}\frac {|B^n_1|}{2^{ (n+1)/2}}. \end{align} $$

Since $\varepsilon>0$ is arbitrary, this and (4.20) imply (4.9) and thus complete the proof of the Proposition.

To establish (4.21), note first that $E_{\varepsilon ,\delta }$ forms an increasing family of sets as $\delta \searrow 0$ . Thus, the Monotone Convergence Theorem implies that $\lim _{\delta \to 0}|E_{\varepsilon ,\delta }| = |E_{\varepsilon ,0}|$ , for

$$\begin{align*}E_{\varepsilon,0} := \cup_{\delta>0} E_{\varepsilon,\delta} = \left\{ s(p', -1) \,:\ p'\in \mathbb{R}^{n-1}, \ 0 \le s \le \left(1 + \frac 1{2(1-\varepsilon)} |p'|^2 \right)^{-1} \right\}. \end{align*}$$

We claim that, in fact,

(4.22) $$ \begin{align} E_{\varepsilon,0} = \left\{ (q', q_n) \in \mathbb{R}^n : \frac 2{(1-\varepsilon)}|q'|^2 + 4(q_n +\frac 12)^2 \le 1 \right\} =: \mathcal E_{\varepsilon,0}. \end{align} $$

Indeed, both $\mathcal E_{\varepsilon ,0}$ and $E_{\varepsilon ,0}$ are contained in the set $\{ 0\} \cup \{(q', q_n) : q_n < 0\}$ . It is clear that the origin belongs to both sets. Any point with $(q', q_n)$ with $q_n<0$ can be written

(4.23) $$ \begin{align} (q', q_n) = \frac{ t(p',-1)}{1 + \frac 1{2(1-\varepsilon)} |p'|^2 } \qquad\text{ for some }p'\in \mathbb{R}^{n-1}\text{ and }t>0. \end{align} $$

Then

$$\begin{align*}|q'|^2 = t^2 \frac{|p'|^2}{(1 + \frac 1{2(1-\varepsilon)} |p'|^2)^2} \end{align*}$$

and

$$\begin{align*}4(q_n+\frac t2)^2 = (2q_n+t)^2 = t^2 \frac{\left( -1+ \frac 1{2(1-\varepsilon)} |p'|^2\right)^2}{(1 + \frac 1{2(1-\varepsilon)} |p'|^2)^2}, \end{align*}$$

from which we see that

$$\begin{align*}\frac 2{(1-\varepsilon)}|q'|^2 + 4(q_n +\frac t2)^2 = t^2. \end{align*}$$

One then checks that

$$\begin{align*}\frac 2{(1-\varepsilon)}|q'|^2 + 4(q_n +\frac 12)^2 = 1 +4(1-t)q_n. \end{align*}$$

For $q_n<0$ , the right-hand side is larger than $1$ if and only if $t>1$ . Thus,

$$\begin{align*}0 <t\le 1\text{ in }({4.23}) \quad\Longleftrightarrow\quad (q',q_n)\in \mathcal E_{\varepsilon,0}. \end{align*}$$

Since $(q', q_n)\in E_{\varepsilon ,0}$ if and only if $0<t\le 1$ in (4.23), this implies (4.22).

And it is clear that $|\mathcal E_{\varepsilon ,0}|= \frac {(1-\varepsilon )^{(n-1)/2}|B^n_1|}{2^{(n+1)/2}}$ , proving (4.21).

Proof We may assume by a suitable choice of coordinates that $\nabla h (0) = m e_{n-1}$ for some $m\in [-\frac 14, \frac 14]$ , where $e_{n-1}$ is the standard basis vector along the $x_{n-1}$ axis.

We then define (temporarily adopting column vector notation for ease of reading)

$$\begin{align*}S\left(\begin{array}{c}x_1\\ \vdots \\ x_{n-2} \\ x_{n-1} \\ x_n\end{array}\right) = \left(\begin{array}{c}x_1\\ \vdots \\ x_{n-2} \\ \alpha x_{n-1}+\beta x_n \\ -\beta x_{n-1} + \alpha x_n\end{array}\right), \qquad \alpha = \frac 1{\sqrt {1+m^2}}, \quad \beta = \frac m{\sqrt {1+m^2}}. \end{align*}$$

Clearly, $S\in SO(n)$ . Note that $\alpha \ge 4/\sqrt {17}> 4/5$ and $|\beta |\le 1/\sqrt {17} < 1/4$ . We next define $\Phi :B_{2R}^{n-1} \to \mathbb {R}^n$

$$\begin{align*}S \binom{x'}{h(x')} =: \Phi(x') = \binom {\phi(x')}{\phi^n(x')}, \end{align*}$$

where we will write the components of $\phi $ as $(\phi ^1,\ldots , \phi ^{n-1})$ , for $\phi $ and $\phi ^n$ taking values in $\mathbb {R}^{n-1}$ and $\mathbb {R}$ , respectively. We now define

$$\begin{align*}\psi = \phi^{-1}, \qquad g = \phi^n\circ \psi. \end{align*}$$

We now verify that these functions have the stated properties.

Proof that g is well-defined on $B_R^{n-1}$ : From the definitions, we see that

(4.26) $$ \begin{align} \partial_i\phi^j = \delta^j_i \ \ \text{ if }j\le n-2, \qquad D \phi^{n-1} = \alpha e^{n-1} + \beta D h. \end{align} $$

Thus, writing $\| A\|$ to denote the operator norm of a matrix A and $I_k$ the $k\times k$ identity matrix, we check that

(4.27) $$ \begin{align} \| D\phi - I_{n-1} \| \le |1-\alpha| + |\beta| < \frac 9{20} \qquad\text{ everywhere in }B_{2R}^{n-1}. \end{align} $$

Now fix any $y'\in B_R^{n-1}$ , and define $\Phi (x') = y'+x' -\phi (x')$ . Then for any $x', z'\in B_{2R}^{n-1}$ ,

(4.28) $$ \begin{align} \left|\Phi(x') - \Phi(z')\right| =\left| \int_0^1 \frac d{ds}\Phi(s x'+(1-s )z')\,ds\right| &\le \sup_{B^{n-1}_{2R}} \| D\phi - I_{n-1} \| \, |x'-z'| \nonumber \\ & \le \frac 9{20}|x'-z'|. \end{align} $$

Thus, $\Phi $ is a contraction mapping. Note also, that when $y=0$ , we find from (4.28) that $|\phi (x')-x'| \le \frac 9{20}|x'|$ . So if $|x'| \le 20R/11$ , then

$$\begin{align*}|\Phi(x')| \le |y'| + |\phi(x')-x'| \le \frac {20R}{11}. \end{align*}$$

So $\Phi $ maps $B^{n-1}_{20R/11}$ to itself, and the Contraction Mapping Principle thus implies that there is a unique $z'\in B^{n-1}_{20R/11}$ such that $\Phi (z') = z'$ , which says exactly that $\phi (z') = y'$ .

These facts imply that $\phi ^{-1} =\psi $ is well-defined in $B^{n-1}_R$ , taking values in $B^{n-1}_{20R/11}$ , and hence that g is well-defined in $B^{n-1}_R$ as well.

Proof that (4.24) holds: The definitions imply that if $y' = \phi (x')$ , then $g(y') = \phi ^n(x')$ , and thus,

$$\begin{align*}\binom {y'}{g(y')} = \binom{\phi(x')}{\phi^n(x')} = \Phi(x') = S\binom {x'}{h(x')}. \end{align*}$$

We deduce (4.24) from this and remarks above about the range of $\psi $ .

Proof that $\nabla g(0)=0$ : We compute

(4.29) $$ \begin{align} g_{y_i} = (\phi^n_{x_k}\circ \psi )\, \psi^k_{y_i}. \end{align} $$

Since $\phi (0)=0$ , it is clear that $\psi (0)=0$ . It thus suffices to check that $\nabla \phi ^n (0)=0$ . This follows from the choice of $\alpha $ and $\beta $ , which guarantees that

$$\begin{align*}\phi^n_{x_k}(0) = \begin{cases} \alpha h_{x_k}(0) = 0&\text{ if }k\le n-2\\ -\beta + \alpha h_{x_{n-1}}(0) = 0&\text{ if }k=n-1. \end{cases}\end{align*}$$

$C^2$ modulus of continuity of g: By differentiating (4.29), we obtain

(4.30) $$ \begin{align} g_{y_i y_j} = (\phi^n_{x_k x_l}\circ \psi)\psi^k_{y_i}\psi^l_{y_j} + (\phi^n_{x_k}\circ \psi )\psi^k_{y_i y_j}. \end{align} $$

Moreover,

$$\begin{align*}\psi^i_{y_j} = [(D\phi)^{-1}\circ \psi]^i_j, \qquad \psi^i_{y_j y_k} = - (\phi^m_{x_a x_b}\circ \psi)\, \psi^i_{y_m}\ \psi^a_{y_j}\ \psi^b_{y_k}. \end{align*}$$

For every $x'\in B_{2R}^{n-1}$ , it follows from (4.27) that $D\phi (x')$ belongs to the compact set $\{ A\in M^{n-1} : \| A - I\| \le 9/20 \}$ on which the map $A\mapsto A^{-1}$ is smooth and hence bounded and Lipschitz. Hence, $D\psi $ is bounded in $B_R^{n-1}$ , and $\psi $ is Lipschitz continuous. Then elementary estimates show that first $D\psi $ and then $D^2\psi $ have moduli of continuity estimated only in terms of the $C^2$ modulus of continuity of $\phi $ , which in turn is controlled by the $C^2$ modulus of continuity of h. Then similar arguments show that the $C^2$ modulus of continuity of g can be estimated only in terms of that of h.

Formula for $D^2 g(0)$ . Computing as in our verification that $\nabla g(0)=\nabla \phi ^n(0)=0$ , and recalling that $\psi (0)=0$ , one easily checks that $D^2 \phi ^n(0) = \alpha D^2 h(0)$ . Similarly, nothing from the definitions that $\partial _{n-1}\phi ^{n-1}(0) = \alpha +\beta m = 1/\alpha $ , one checks that

$$\begin{align*}D\phi(0) = \text{diag}(1,\ldots,1,\frac 1 \alpha), \quad\text{ and so }\quad D\psi(0) = \text{diag}(1,\ldots,1, \alpha). \end{align*}$$

We deduce (4.25) from these facts and (4.30).

Finally, it is clear that if h is convex, then g is convex, as then the graph of g is part of the lower boundary of a convex set.