Subdifferential of convex functions is nonempty

Question

I read the counterexample in this post, and I am wondering if it is possible to fix the statement in this way:

Given a convex function $f:\mathbb R^d\rightarrow\mathbb R\cup\{\infty\}$ and $x\in\mathbb R^d$. If $x$ has a neighborhood such that $f<\infty$, is that true that $\partial f(x)\neq\emptyset$?

Answer 1

Yes, this is true. Here is an outline of the proof:

• $f$ is bounded from above by some constant $M \in \mathbb R$ on a neighborhood of $x$
• $f$ is (Lipschitz) continuous on a neighborhood of $x$
• $\partial f(x) \ne \emptyset$

The last step is the most delicate (requires some separation theorem), but should be covered in most texts on convex optimization.

Answer 2

Yes, it is true.

The subgradient set $\partial f(x)$ of a (proper) convex function $f$ is not empty on the relative interior of its domain $x \in \textrm{ri}(\textrm{dom}f)$. A proof is given by Theorem 23.4 in Rockafellar’s Convex Analysis book.

Your neighborhood condition on $x$ implies that $x \in \textrm{ri}(\textrm{dom}f)$.