Is the preimage of the image of a convex set by the subgradiant of a convex function, also convex?

Question

Let $f : \mathbb{R}^n \to \mathbb{R}$ be a convex function and $\partial f(x) \subseteq \mathbb{R}^n$ its subgradiant at point $x \in \mathbb{R}^n$.

Let $A \subseteq \mathbb{R}^n$ and $B \subseteq \mathbb{R}^n$.

Define $\partial f(A) = \bigcup \left\{\; y \subseteq \mathbb{R}^n \;|\; \exists x \in A, \; y = \partial f(x) \;\right\}$ (image of $A$ by $\partial f$).

Define $(\partial f)^{-1}(B) = \bigcup \left\{\; x \subseteq \mathbb{R}^n \;|\; \exists y \in B, \; y = \partial f(x) \;\right\}$ (preimage of $B$ by $\partial f$).

Let $A \subseteq \text{dom} \;\partial f$ be a convex set. Question: Is $C = (\partial f)^{-1}(\partial f(A))$ also a convex set? I've tried with several convex functions $f$ and it's always verified. I'm also conviced by geometrical arguments. But I cannot prove it.

Thanks

Edit: I believe the result is true because $\partial f$ is maximal monotone. More generaly, I believe if $g$ is maximal monotone, $C = g^{-1}(g(A))$ is a convex set for all convex set $A$.

Answer 1

Attached is a counterexample, of a function G that is convex, but such that for some convex set A, the set C is non-convex. See the image below.

Answer 2

The result is not true. Just take $A=\mathbb{R}^n$. Then the set you claim to be convex is the domain of $\partial f$. Since there are examples of subdifferentials with a non-convex domain even defined in a finite-dimensional space your claim is false.