In this section we construct examples of solutions to $\det D^{2}u=1$ in $\mathbb{R}^{3}$ such that $\Sigma$ has Hausdorff dimension as close to $2$ as we like. A small modification produces the analagous examples in $\mathbb{R}^{n}$.

For this section, fix $\delta>0$ small. We construct our examples in several steps, which we briefly describe:

1. (i)

   First, we construct functions $w$ with

   $$\det D^{2}w\geq 1$$

   in $\mathbb{R}^{3}$ that degenerate along $\{x_{1}=x_{2}=0\}$ and behave like $x_{1}^{2-\delta}$ along the $x_{1}$ axis.

2. (ii)

   Next, we construct a standard $S\subset[-1,1]$ with Hausdorff dimension close to $1$ and a convex function $v$ on $[-1,1]$ such that for any $x\in S$, there is a tangent line such that $v$ separates from this line faster than $r^{2-\delta}$.

3. (iii)

   Finally, we get our example by solving the Dirichlet problem

   $$\det D^{2}u=1\quad\text{ in }\Omega=\{|x^{\prime}|<1\}\times[-1,1],\quad\quad u|_{\partial\Omega}=C(\delta)(v(x_{1})+|x_{2}|)$$

   and comparing with $w$ at points in $S\times\{0\}\times\{\pm 1\}$.

In the following analysis $c$ and $C$ will denote small and large constants depending on $\delta$.

Construction of $w$: We look for a convex function $w(x_{1},x_{2},x_{3})$ with the homogeneity

$$w(x_{1},x_{2},x_{3})=\frac{1}{\lambda}h(\lambda^{1/\alpha}x_{1},\lambda^{1/\beta}x_{2})(1+x_{3}^{2}),$$

where $\alpha$ and $\beta$ satisfy $1<\alpha,\beta<2$ and

$$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{2}.$$

(It is easy to check that $\geq 3/2$ is necessary for such a function to have $\det D^{2}w$ bounded below). Note that this rescaling preserves the curves $x_{2}=mx_{1}^{\alpha/\beta}$.

Let $f(x)$ denote $1+x^{2}$. An obvious candidate for $w$ is

$$w(x_{1},x_{2},x_{3})=(x_{1}^{\alpha}+x_{2}^{\beta})f(x_{3}).$$

One checks that

	$\displaystyle\det D^{2}w$	$\displaystyle=|x_{1}|^{2\alpha-2}|x_{2}|^{\beta-2}\left(\alpha\beta(\alpha-1)(\beta-1)f^{2}-\alpha^{2}\beta(\beta-1)fx_{3}^{2}\right)$
		$\displaystyle+|x_{1}|^{\alpha-2}|x_{2}|^{2\beta-2}\left(\alpha\beta(\alpha-1)(\beta-1)f^{2}-\alpha\beta^{2}(\alpha-1)fx_{3}^{2}\right).$

Take $\alpha=2-\delta$. Then for $|x_{3}|$ small depending on $\delta$ we have

$$\det D^{2}w\geq c(\delta)(|x_{1}|^{2\alpha-2}|x_{2}|^{\beta-2}+|x_{1}|^{\alpha-2}|x_{2}|^{2\beta-2}).$$

Along the curves $x_{2}=mx_{1}^{\alpha/\beta}$, we compute

$$\det D^{2}w\geq c(\delta)(|m|^{\beta-2}+|m|^{2\beta-2})\geq c(\delta),$$

since $1<\beta<2$.

Thus, up to rescaling the $x_{3}$-axis and multiplying by a constant, we have

$$\det D^{2}w\geq 1\quad\text{ in }\Omega=\{|x^{\prime}|<1\}\times[-1,1].$$

Construction of $S$: Let $\epsilon>0$ be a small constant we will choose shortly depending on $\delta$. Construct a self-similar set in $[-1/2,1/2]$ as follows: First, remove an open interval of length $\gamma=1-2^{-3\epsilon}$ from the center. Proceed inductively by removing intervals a fraction $\gamma$ of each of those that remains. Denote the centers of the intervals removed at stage $k$ by $\{x_{i,k}\}_{i=1}^{2^{k}}$, and the intervals by $I_{i,k}$. Finally, let

$$S=[-1/2,1/2]-\cup_{i,k}I_{i,k}.$$

It is easy to check that $|I_{i,k+1}|=\gamma 2^{-(1+3\epsilon)k}$ and that $S$ has Hausdorff dimension $\frac{1}{1+3\epsilon}$.

Construction of $v$: Let

$$w(x)=\left\{\begin{array}[]{ll}|x|&\quad|x|\leq 1\\ 2|x|-1&\quad|x|>1\end{array}\right.$$

We add rescalings of $w$ together to produce the desired function:

$$v(x)=\sum_{k=1}^{\infty}\sum_{i=1}^{2^{k}}2^{-2(1+2\epsilon)k}w(\gamma^{-1}2^{(1+3\epsilon)k}(x-x_{i,k})).$$

We now check that $v$ satisfies the desired properties:

1. (i)

   v is convex, as the sum of convex functions. Furthermore,

   	$\displaystyle|v(x)|$	$\displaystyle\leq C\sum_{k=1}^{\infty}\sum_{i=1}^{2^{k}}2^{-(1+\epsilon)k}$
   		$\displaystyle\leq C\sum_{k=1}^{\infty}2^{-\epsilon k},$

   so $v$ is bounded.

2. (ii)

   Let $x\in S$. We aim to show that $v$ separates from a tangent line more than $r^{2-\delta}$ a distance $r$ from $x$. By subtracting a line assume that $v(x)=0$ and that $0$ is a subgradient at $x$. Assume further that $x+r<1/2$ and that $2^{-(1+3\epsilon)k}<r\leq 2^{-(1+3\epsilon)(k-1)}$. There are two cases to examine:

   Case 1: There is some $y\in(x+r/2,x+r)\cap S$. Then by the construction of $S$ it is easy to see that there is some interval $I_{i,k}$ such that $I_{i,k}\subset(x,x+r)$. On this interval, $v$ grows by

   $$2^{-2(1+2\epsilon)k}\geq cr^{2\frac{1+2\epsilon}{1+3\epsilon}}=cr^{2-\delta},$$

   where we choose $\epsilon$ so that

   $$\delta=\frac{2\epsilon}{1+3\epsilon}.$$

   Case 2: Otherwise, there is an interval $I_{i,j}$ of length exceeding $r/2$ such that $(x+r/2,x+r)\subset I_{i,j}$. Then at the left point of $I_{i,j}$, the slope of $v$ jumps by at least $2^{-(1+\epsilon)k}$. It follows that at $x+r$, $v$ is at least

   $$\frac{r}{2}2^{-(1+\epsilon)k}\geq cr^{2-\delta}.$$

   Thus, $v$ has the desired properties.

Construction of $u$: We recall the following lemma on the solvability of the Monge-Ampère equation (see [Gut]).

Let $g(x_{1},x_{2},x_{3})=C(v(x_{1})+|x_{2}|)$ for a constant $C$ depending on $\delta$ we will choose shortly, and obtain $u$ by solving the Dirichlet problem

$$\det D^{2}u=1\quad\text{ in }\Omega=\{|x^{\prime}|<1\}\times[-1,1],\quad\quad u|_{\partial\Omega}=g.$$

Take $x\in S\times\{0\}\times\{\pm 1\}$. By translating and subtracting a linear function assume that $x_{1}=0$ and $0$ is a subgradient for $g$ at $x$. Taking $C$ large we guarantee that

$$g(x_{1},x_{2},\pm 1)>C(x_{1}^{2-\delta}+|x_{2}|)>w(x_{1},x_{2},\pm 1)$$

for all $x_{1},x_{2}$, and that that $g>w$ on the sides of $\Omega$. Thus, $u\geq w$ in all of $\Omega$. Since $u=0$ at both $(0,0,\pm 1)$ and $w(0,0,x_{3})=0$ for all $|x_{3}|<1$, we have by convexity that $u=0$ along $(0,0,x_{3})$.

We conclude that $\Sigma$ contains $S\times\{0\}\times(-1,1),$ which has Hausdorff dimension $1+\frac{1}{1+3\epsilon}=2-\frac{3}{2}\delta$.