Is the subdifferential always convex and closed set?

Question

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 $x \in \mbox{dom}(f)$. However, the closeness still need to be proved.

All comments would be highly appreciated.

Answer 1

It depends on which subdifferential you are talking about.

• The convex subdifferential is always convex and closed.
• The limiting subdifferential is always closed but not necessarily convex. For instance the limiting subdifferential of $x\mapsto-|x|$ at $0$ is the set $\{-1,1\}$.

Answer 2

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?

Answer 3

For part two , subdifferential can be empty at some points.

Take $f(x) = - \sqrt {1-x^2}$ then at $x=+1 , -1$ subdifferential is empty.