3. Proof of Theorem 1.1 [04TW]

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3. Proof of Theorem 1.1

In this section assume that

\det D^{2}u\geq 1

in $B_{1}\subset\mathbb{R}^{n}$ . Fix $x\in\Sigma$ and a subgradient $p$ at $x$ . By translation and subtracting a linear function assume that $x=p=0$ . Then $\{u=0\}$ contains a line segment of some length $l$ . By Lemma 2.2,

S_{h,0}(0)\leq C(n)h^{n/2}

for all $h>0$ .

Letting $v=u+\frac{1}{2}|x|^{2}$ and denoting the sections of $v$ by $S_{h,p}^{v}$ , it follows that

|S_{h,0}^{v}(0)|\leq\frac{C(n)}{l}h^{\frac{n+1}{2}}

for all $h$ small.

Theorem 1.1 thus follows from the following more general result:

Theorem 3.1.

Let $v$ be any convex function on $B_{1}\subset\mathbb{R}^{n}$ with sections $S_{h,p}^{v}$ , and let $\Sigma_{v}$ denote the set of points $x$ such that for all supporting slopes $p$ at $x$ , there is some $C_{p}$ such that

|S_{h,p}^{v}(x)|<C_{p}h^{\frac{n+1}{2}}

for all $h$ small. Then

\mathcal{H}^{n-1}(\Sigma_{v})=0.

Proof of Theorem 1.1:.

Let $v=u+\frac{1}{2}|x|^{2}$ . By the discussion preceding the statement of Theorem 3.1, $\Sigma\subset\Sigma_{v}$ . The conclusion follows from Theorem 3.1. ∎

We briefly discuss the main ideas of the proof. Fix $x\in\Sigma_{v}$ and a subgradient $p$ at $x$ . In the following analysis $c,\,C$ will denote small and large constants depending on $n$ and the $C_{p}$ . If $S_{h,p}^{v}(x)\subset\subset B_{1}$ then the definition of $\Sigma_{v}$ and Lemma 2.4 give

Mv(S_{h}^{v}(x))\geq ch^{\frac{n-1}{2}}=c(h^{1/2})^{n-1}

for all $h$ small.

An important technique of the proof is to replace $v$ by $v+\frac{1}{2}|x|^{2}$ . Then all of the sections are compactly contained in $B_{1}$ for $h$ small, and the diameter of sections is at most $h^{1/2}$ . By replacing the sections $S_{h,p}^{v}(x)$ by $B_{\sqrt{h}}(x)$ and using a covering argument, we easily obtain that $\Sigma_{v}$ has Hausdorff dimension at most $n-1$ .

Lemmas 3.2 and 3.3 improve this result as follows. We aim to rule out behavior like

|x|^{2}+|x_{n}|,

which has a singular hyperplane. For this example, the sections at $\{x_{n}=0\}$ have the correct growth when we take supporting slopes with no $x_{n}$ -component, but the sections are too large when we take supporting slopes with $x_{n}$ -component $1$ .

In the first lemma we use that the sections are small for all supporting planes at $x\in\Sigma_{v}$ to show that $v$ must grow much faster than quadratically in at least two directions, unlike the example above. In the second lemma we use the above observation about the Monge-Ampère mass of $v$ in the directions where $v$ grows much faster than quadratically from $x$ . Since we replaced $v$ by $v+\frac{1}{2}|x|^{2}$ we also know that $v$ grows at least quadratically in the remaining directions. This allows us to cover $\Sigma_{v}$ with balls in which the Monge-Ampère mass of $v$ is much larger than the radius to the $n-1$ , giving the desired improvement.

Lemma 3.2.

Fix $x\in\Sigma_{v}$ . For a supporting slope $p$ at $x$ , let

d_{1}(h)\geq d_{2}(h)\geq...\geq d_{n}(h)

denote the axis lengths of the John ellipsoid of the section $S_{h,p}^{v}(x)$ . Then

\frac{d_{n-1}(h)}{h^{1/2}}\rightarrow 0\text{ as }h\rightarrow 0.

Proof.

By translating and subtracting a linear function assume that $x=p=0$ . Assume by way of contradiction that we can find $h_{k}\rightarrow 0$ and some $\delta>0$ such that

d_{n-1}(h_{k})>\delta h_{k}^{1/2}

for all $k$ . We first show that $v$ is trapped by two tangent planes at $0$ .

Let $x_{1,k}$ and $x_{2,k}$ be the points on $\partial S_{h_{k},0}^{v}(0)$ where the hyperplanes perpendicular to the shortest axis of the John ellipsoid become tangent to $\partial S_{h_{k},0}^{v}(0)$ , and let $p_{1,k}$ and $p_{2,k}$ denote subgradients at these points. Since

d_{1}(h_{k})d_{2}(h_{k})...d_{n}(h_{k})<Ch_{k}^{\frac{n+1}{2}},

we have that $d_{n}(h_{k})<\frac{C}{\delta^{n-1}}h_{k}$ for all $k$ . By this observation and convexity we can rotate and pass to a subsequence such that

p_{1,k}\rightarrow c_{1}(\delta)e_{n},\quad p_{2}(k)\rightarrow-c_{2}(\delta)e_{n}.

Then $v$ is trapped by the planes $\pm c(\delta)x_{n}$ . We conclude that

S_{h_{k},0}^{v}(0)\subset\{|x_{n}|<C(\delta)h_{k}\}.

To complete the proof, we show that the volumes of sections obtained with tilted supporting planes are too large. Take the largest $a$ such that $v\geq ax_{n}$ and consider the sections

S_{k}=S_{(1+aC(\delta))h_{k},a}^{v}(0).

Then $S_{k}$ engulf $S_{h_{k},0}^{v}(0)$ . Furthermore,

\sup\{|x_{n}|:x\in S_{k}\}=R_{k}h_{k},

where $R_{k}\rightarrow\infty$ as $k\rightarrow\infty$ . Indeed, if not, then for some small $\epsilon$ and a sequence $b_{i}\rightarrow 0$ we would have $v(x^{\prime},b_{i})>(a+\epsilon)b_{i}$ for all $x^{\prime}$ . Convexity and $v(0)=0$ imply that $v>(a+\epsilon)x_{n}$ for all $x_{n}>b_{i}$ , which in turn implies that

v>(a+\epsilon)x_{n},

contradicting the definition of $a$ .

Finally, let $(x^{\prime}_{k},R_{k}h_{k})\in S_{k}$ be the point in $S_{k}$ furthest in the $e_{n}$ direction. Since $v$ grows at least quadratically, we have

|x^{\prime}_{k}|<C(\delta,a)h_{k}^{1/2}.

Recall that $S_{h_{k},0}^{v}(0)\subset\{|x_{n}|<C(\delta)h_{k}\}$ . Since $d_{i}(h_{k})>\delta h_{k}^{1/2}$ for all $i\leq n-1$ , $S_{k}$ contains the cone with vertex $(x^{\prime}_{k},R_{k}h_{k})$ and base given by a ball of radius $h_{k}^{1/2}(\delta-C(a,\delta)/R_{k})$ on the hyperplane $\{x_{n}=C(\delta)h_{k}\}$ . We conclude that

|S_{k}|\geq c(\delta,a)R_{k}h_{k}^{\frac{n+1}{2}},

contradicting our definition of $\Sigma_{v}$ for $k$ large. ∎

Lemma 3.3.

Fix $x\in\Sigma_{v}$ . For any $\epsilon>0$ , there is a sequence $r_{k}\rightarrow 0$ such that

Mv(B_{r_{k}}(x))>\frac{1}{\epsilon}r_{k}^{n-1}.

Proof.

Fix a subgradient $p$ at $x$ and let $d_{1}(h),...,d_{n}(h)$ be defined as in the statement of Lemma 3.2. Let

I=\min\left\{i:\frac{d_{i}(h)}{h^{1/2}}\rightarrow 0\text{ as }h\rightarrow 0\right\}.

Fix $\delta$ small. Then we can find a sequence $h_{k}\rightarrow 0$ and $\eta$ depending only on $p$ such that

d_{I}(h_{k})<\delta h_{k}^{1/2},

and $d_{i}(h_{k})>\eta h_{k}^{1/2}$ for all $i<I$ . Rotate the axes so that the $e_{i}$ are the axes for the John ellipsoid of $S_{h_{k},p}^{v}(x)$ and assume by translation that $x=0$ .

Take the restriction of $v$ to the subspace spanned by $e_{I},...,e_{n}$ , and call this restriction $w$ . Let

S_{k}^{w}=S_{h_{k},p}^{v}(x)\cap\{x_{1}=...=x_{I-1}=0\},

the slice of the section $S_{h_{k},p}^{v}(x)$ in this subspace. Then since

d_{1}(h_{k})d_{2}(h_{k})...d_{n}(h_{k})\leq Ch_{k}^{\frac{n+1}{2}}

and $v$ grows at most quadratically in the first $I-1$ directions, we have

|S_{k}^{w}|_{\mathcal{H}^{n-I+1}}\leq\frac{C}{\eta^{(I-1)/2}}h_{k}^{\frac{n+2-I}{2}}.

Using this and Lemma 2.4,

Mw(S_{k}^{w})\geq c\eta^{(I-1)/2}h_{k}^{\frac{n-I}{2}}.

Finally, let $r_{k}=C(n)d_{I}(h_{k})$ , with $C(n)$ taken large enough that

S_{k}^{w}\subset B_{r_{k}/2}(x).

By strict quadratic growth, $\nabla v(B_{r_{k}}(x))$ contains a ball of radius $r_{k}/2$ around every point in $\nabla v(S_{k}^{w})$ . It follows that

	$\displaystyle Mv(B_{r_{k}}(x))$	$\displaystyle\geq c(n)Mw(S_{k}^{w})r_{k}^{I-1}$
		$\displaystyle\geq ch_{k}^{\frac{n-I}{2}}r_{k}^{I-1}$
		$\displaystyle\geq\frac{c}{\delta^{n-I}}r_{k}^{n-1}.$

By Lemma 3.2 we have $I\leq n-1$ , so the conclusion follows. ∎

We can complete the proof of theorem 3.1 with a covering argument.

Proof of Theorem 3.1:.

Fix $\epsilon$ small. By Lemma 3.3, for each $x\in\Sigma_{v}$ we can choose an arbitrarily small $r$ such that

Mv(B_{r}(x))>\frac{1}{\epsilon}r^{n-1}.

Cover $\Sigma_{v}\cap B_{1/2}$ with such balls, and choose a Vitali subcover $\{B_{r_{i}}(x_{i})\}_{i=1}^{N}$ , i.e. a disjoint subcollection such that $B_{3r_{i}}(x_{i})$ cover $\Sigma\cap B_{1/2}$ . Then

	$\displaystyle\sum_{i=1}^{N}(3r_{i})^{n-1}$	$\displaystyle\leq C\epsilon\sum_{i=1}^{N}Mv(B_{r_{i}}(x_{i}))$
		$\displaystyle\leq C\epsilon,$

since $v$ is locally Lipschitz and the $B_{r_{i}}$ are disjoint. This means exactly that

\mathcal{H}^{n-1}(\Sigma_{v}\cap B_{1/2})=0.

∎

Remark 3.4.

Replacing $\Sigma_{v}$ with

\Sigma_{v}^{k}=\{|S_{h,p}^{v}|<C_{p}h^{\frac{n+k}{2}}\},\quad 1\leq k\leq n-1

and replacing $1$ with $k$ in the preceding, one obtains that $\mathcal{H}^{n-k}(\Sigma_{v}^{k})=0$ . If $\det D^{2}u\geq 1$ , such growth happens for $v=u+\frac{1}{2}|x|^{2}$ at points where $u$ agrees with a linear function on a $k$ -dimensional subspace. This shows that the Hausdorff dimension of the $k$ -dimensional singularities is at most $n-k$ . In particular, we recover Lemma 2.3 since for $k\geq\frac{n}{2}$ we would have a $k$ -dimensional singularity with Hausdorff $k$ -dimensional measure $0$ .