# Convergence of subgradients

Let $$C$$ be a compact, convex subset of a Hilbert space $$\mathcal{H}$$ and $$g:\mathcal{H}\to\mathbb{R}\cup\infty$$ an extended valued, proper, lower semicontinuous, convex function. Also, assume that $$C\subset dom(\partial g)$$, where $$\partial g$$ is the convex subdifferential of $$g$$ and $$dom(\partial g) = \{x\in\mathcal{H}: \partial g (x)\neq\emptyset\}$$.

I am interested in the following claim:

Given a convergent sequence $$x_n\in C$$ with $$x_n\to x\in C$$ and a sequence of subgradients $$b_n \in \partial g(x_n)$$, we have $$\exists b\in \partial g(x)$$ such that $$b_n\to b$$.

Even a weaker version, in which it is only true that $$\exists b\in \partial g(x)$$ such that $$\exists b_{n_k}$$ with $$b_{n_k}\to b$$ for a subsequence, would be interesting as well.

A related question I am interested in as well is the following: is there a name for this type of 'continuity' for a correspondence/multi-valued function $$G$$,

$$\forall \varepsilon > 0, \exists \delta >0 \mbox{ such that } \|x-y\|<\delta\implies \exists u\in G(x), v\in G(y): \|u-v\|<\varepsilon$$

Without boundedness of $$b_n$$ this will not work.

Here is an example: Take $$H$$ at least two-dimensional, $$x\in H$$, $$y\in H$$ with $$y\ne0$$, $$(x,y)=0$$. Define $$g=I_{\{y\}^\perp}$$, which is the indicator function of the convex set $$\{y\}^\perp$$: $$g(x) = \begin{cases} 0 & \text{ if } (x,y)=0\\ +\infty & \text{ otherwise }\end{cases}$$

Define $$x_n:=x$$, $$b_n:=n\cdot y_n$$.

If the sequence $$(b_n)$$ has a weakly converging subsequence, then the weak limit is a subgradient again.

• Is $g(t) = \infty$ for $t \bot y$? How is $g$ convex? – copper.hat Oct 12 '18 at 13:32
• $g(t)= 0$ for $t\perp y$ and $g(t) = \infty$ otherwise if I understand correctly. – TSF Oct 12 '18 at 13:32
• Is $\bot$ the orthogonal complement? If so, I don't see how $g$ is convex. – copper.hat Oct 12 '18 at 13:33
• I think so yes? – TSF Oct 12 '18 at 13:33
• @copper.hat see edit – daw Oct 12 '18 at 13:57