True or False, subgradient proximal mapping of $f = $ gradient $g$

Question

Let $f:{X} \to [-\infty, +\infty]$ be convex, closed and proper. Set $T := \mathrm{Prox}_f$, where $T$ is the proximal mapping of $f$. Is the statement:

"There exists $g: {X} \to \mathbb{R}$ convex and differentiable such that $T = \nabla g"$

True or False? Explain your answer.

I feel this is true. I think that if we show that $g$ needs to be the Moreau envelope of the conjugate of $f$, it'll be true, but I'm not sure how to show it or what to do... Help!

Answer 1

Hint: Denote the Moreau envelope of $f$ by $M_f$. Then $\nabla M_f = \mathrm{Id}-\mathrm{Prox}_f$. Hence $\mathrm{Prox}_f = \mathrm{Id}-(\mathrm{Id}-\mathrm{Prox}_f)$ is the gradient of $\tfrac{1}{2}\|x\|^2-M_f(x) = M_{f^*}(x)$, which is the Moreau envelope of $f^*$ and hence a nice convex function.