Two properties of the subdifferential set are stated as follows:
Given a function $f : \mathbb{R}^n → \mathbb{R}$,
(i) the subdifferential set $\partial f(x)$ is always convex and closed, even if $f$ is nonconvex.
(ii) $\partial f(x)$ can be nonempty set if $f$ is continuous or it could be empty set.
I wonder if there is a proof for such properties.
The way I think of property (i) is that the subdifferential $\partial f(x)$ is the intersection of infinite halfspaces thus it is convex, since $\partial f(x)$ is the set of all subgradients at $x \in dom(f)$. However, the closeness still need to be proved.
All comments would be highly appreciated,,
Best Answer
For a vector $u$ to be an element of the subdifferential, it is necessary and sufficient to have: $$f(y)\geq f(x) + \langle y-x,u\rangle, \forall y$$ Hence the subdifferential can be written as: $$\cap_y \{ u \big| f(y)\geq f(x) + \langle y-x,u\rangle \} $$ This representation is the intersection of closed convex sets. Therefore it is closed and convex.
The second part seems like a tautology to me. The subdifferential could be empty or non-empty?