Theorem: A function $f: \textbf{dom}(f) \to \mathbb R$ is convex if and only if $\textbf{dom}(f)$ is convex and $\partial f(\textbf x)$ is not empty for all $\textbf x \in \textbf{dom}(f)$
I know the answer to the question Show that a real-valued function with non-empty subdifferential is convex. What I need to prove is the other way: Given a convex function (on a convex domain, and $\textbf{dom}(f)$ means $f(\textbf x) < +\infty$), then subgradients exist everywhere in the domain.
My attempt so far: I assume that there exists $\textbf x \in \textbf{dom}(f)$ that with any $g$, there always exists $\textbf y'$ that $f(\textbf y') \leq f(\textbf x) + g^\top(\textbf y' - \textbf x)$. From this, I can prove that any point "between" $\textbf x$ and $\textbf y'$ also satisfy the previous inequality. But I don't know how to proceed from here
Can anyone help me if my direction is correct, or should I change the approach? Thanks in advance.
As originally stated, it is not correct.
For instance, if $f(x) = -\sqrt{x}$ for $x\geq 0$ (and $+\infty$ otherwise), then the domain of $f$ is convex, but $\partial f(0)=\varnothing$.
However: It is correct if we assume that $f$ is $L$-Lipschitz and the underlying space is finite-dimensional.
Here is a sketch of the proof.
(1) First, the range of $\partial f$ is bounded; in fact, it lies in the ball centred at $0$ of radius $L$.
(2) Secondly, the graph of $\partial f$ is closed.
(3) Thirdly, the domain of $\partial f$ is dense in the domain of $f$.
(4) Finally, pick $x$ in the domain of $f$, and let $(x_n)$ be a sequence in the domain of $\partial f$ converging to $x$. (This is doable by (3).) Take $x_n^*\in\partial f(x_n)$. Because of (1) and Bolzano-Weierstrass, we may and do assume without loss of generality that $x_n^*\to x^*$. By (2), $(x,x^*)$ belongs to the graph of $\partial f$ and therefore $x$ belongs to the domain of $\partial f$.
(Relevant facts may be found in the book by Bauschke and Combettes.)