Points of differentiability of convex functions

Question

Let $U$ be an open neighbourhood of $0 \in \mathbb{R}^2$ and $f\colon U \to \mathbb{R}$ a convex (and bounded) function. Denote by $D \subset U$ the set of points on which $f$ is totally differentiable and the Jacobian of $f$ is continuous on $D$.

It is known that the set $D$ is relatively large; for example, it has full measure in $U$. But I need the following particular statement, which does not seem to follow just from $D$ having full measure: There exists a sequence of points $(x_n)_n \subset \mathbb{R}_{>0}$ with $(x_n,0) \in U$ for all $n \in \mathbb{N}$ and $$(x_n,0) \xrightarrow{n \to \infty} (0,0)$$ such that for each $n \in \mathbb{N}$ exists a sequence $(y_m^{(n)})_m \subset \mathbb{R}_{>0}$ with $(x_n,y_m^{(n)}) \in D$ for all $n,m \in \mathbb{N}$ and $$(x_n,y_m^{(n)}) \xrightarrow{m \to \infty} (x_n,0).$$

Answer 1

Actually, your desired conclusion does "follow just from $D$ having full measure".

Indeed, without loss of generality, $U=(-1,1)^2$. Let $$X:=\{x\in(-1,1)\colon|D_x|=2\},$$ where $D_x:=\{y\in(-1,1)\colon(x,y)\in D\}$ and $|\cdot|$ is the Lebesgue measure. Then, by the Tonelli theorem, $|X|=2$. So, $X$ is dense in $(-1,1)$ and hence there is a sequence $(x_n)$ in $X\cap(0,1)$ such that $x_n\to0$, that is, $(x_n,0)\to(0,0)$. Also, for each $x\in X$, the set $D_x$ is dense in $(-1,1)$ and hence for each $n$ there is a sequence $(y^{(n)}_1,y^{(n)}_2,\dots)$ in $D_{x_n}\cap(0,1)$ such that $y^{(n)}_m\to0$ as $m\to\infty$, so that $D\ni(x_n,y^{(n)}_m)\to(x_n,0)$ as $m\to\infty$. $\quad\Box$