Is there a Borel measurable $f:\mathbb{R}^d \to \mathbb{R}^d$ such that $f(x) \in \partial \varphi (x)$ for all $x$?

Question

Let $\varphi: \mathbb{R}^d \to \mathbb{R}$ be a convex function. The subdifferential of $f$ at $x$ is defined as $$ \partial \varphi (x) := \{z \in \mathbb{R}^d : \varphi(y) \geq \varphi(x) + \langle z, y - x\rangle \text{ for all } y \in \mathbb{R}^d\}. $$

It is well-known that

• $\partial \varphi (x)$ is non-empty convex compact for all $x \in \mathbb{R}^d$,
• $\varphi$ is Fréchet differentiable at $x$ IFF $\partial \varphi (x)$ is a singleton,
• $\varphi$ is Fréchet differentiable a.e.,
• the set $D$ on which $\varphi$ is Fréchet differentiable is a Borel set,
• the map $\nabla \varphi : D \to \mathbb R^d$ is Borel measurable.

Because the Lebesgue $\sigma$-algebra is complete, there is a Lebesgue measurable function $f:\mathbb{R}^d \to \mathbb{R}^d$ such that $f(x) \in \partial \varphi (x)$ for all $x \in \mathbb{R}^d$.

Is there a Borel measurable function $f:\mathbb{R}^d \to \mathbb{R}^d$ such that $f(x) \in \partial \varphi (x)$ for all $x \in \mathbb{R}^d$?

Thank you so much for your elaboration!

Answer 1

A Borel measurable function $f:\mathbb{R}^d\to\mathbb{R}^d$ such that $f(x)\in\partial\varphi(x)$ for all $x\in\mathbb{R}^d$ exists.

This follows from the beautiful and surprisingly unknown Federer-Morse Theorem. The following statement is from

V. I. Bogachev, Measure theory. Vol. II. Springer-Verlag, Berlin, 2007.

Here is an explanation how this result implies existence of a Borel selection of $\partial\varphi$.

Let $(I+\partial\varphi)(x)=x+\partial\varphi(x)$. This function has the inverse. Namely, the function $(I+\partial\varphi)^{-1}:\mathbb{R}^d\to\mathbb{R}^d$ is $1$-Lipschitz and surjective (you can read about it in almost any book on convex analysis). Moreover, since $I+\partial\varphi$ maps compact sets to compact sets, it follows that $(I+\partial\varphi)^{-1}(x)\to\infty$ as $x\to\infty$ and hence it extends to a continuous function on the one point compactification of $\mathbb{R}^d$. Denote this compactification by $\mathbb{S}^d$, so $(I+\partial\varphi)^{-1}:\mathbb{S}^d\to\mathbb{S}^d$ and $\infty$ (north pole) is mapped to $\infty$ (north pole).

According to the Federer-Morse Theorem, there is a Borel set $B\subset\mathbb{S}^d$ so that $(I+\partial\varphi)^{-1}|_B:B\to\mathbb{S}^d$ is a bijection and the inverse of this restriction is Borel. Note that this inverse function is a Borel selection of $I+\partial\varphi$. Denote it by $\psi$. Then $f(x):=\psi(x)-x\in\partial\varphi(x)$ is a Borel selection. Here we do not have to worry about $\infty$, because $\infty$ is mapped to $\infty$.