Is the gradient map of a convex function convexity preserving?

Question

Suppose $f$ is a convex function, and $\mathcal C$ is a convex set.

1. Is the set $\mathcal A = \nabla f(\mathcal C) = \{\nabla f(x) : x\in \mathcal C\}$ convex?

2. Is the inverse set $\mathcal B = \{x : \nabla f(x) \in \mathcal C\}$ convex?

If not, can you provide a counterexample?

I drew a lot of pictures that suggest it is true, but I can't find a standard proof or a counterexample.

It is possible some of these depend on whether $f$ is strictly or strongly convex. It's ok to add those restrictions, but I'm wondering for the broadest possible answer.

Answer 1

Consider $f(x,y) = \tfrac{1}{2}x^2+\tfrac{1}{4}y^4$ and let $C = \mathbb{R}\cdot(1,1)$.

Then $\nabla f(x,y) = (x,y^3)$. Hence $\nabla f(C)$ is the graph of the cube-function, which is clearly not convex. You can use the same set $C$ to get a counterexample for the second assertion as well.