Show that if a proper function has a subgradient everywhere in its convex domain, then it’s convex.

Question

Let $f:\mathbb{R}^n\rightarrow (-\infty,\infty]$ be proper. If $\operatorname{dom} f$ is convex and $\operatorname{dom} \partial f=\operatorname{dom} f$, show that $f$ is convex. I wonder what is the angle to prove such a statement?

Answer 1

The idea is straightforward; if we can write $f(x) = \sup_{h \in H} h(x)$ where each $h \in H$ is convex, then $f$ is convex.

Let $H = \{ y \mapsto f(x)+g^T(y-x) | x \in \operatorname{dom}f, g \in \partial f(x) \}$.