Convergence of subgradients

Question

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 $G$,

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

daw · Accepted Answer · 2018-10-12T13:57:05.180

2

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.

edited Oct 12 '18 at 13:57

answered Oct 12 '18 at 12:56

daw

49,113
2
38
76

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. – Jürgen Sukumaran 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? – Jürgen Sukumaran Oct 12 '18 at 13:33
@copper.hat see edit – daw Oct 12 '18 at 13:57
Sorry, lack of caffeine... – copper.hat Oct 12 '18 at 14:02
@daw perhaps you could take a look at my original question here, https://math.stackexchange.com/questions/2795331/showing-that-the-hausdorff-distance-from-the-subdifferential-to-0-is-bounded the sequence $b_n$ i originally wanted to consider is unbounded by assumption. – Jürgen Sukumaran Oct 12 '18 at 14:33
@daw why do we know that the weak limit is a subgradient? – atul ganju May 15 '22 at 05:08

Convergence of subgradients

1 Answers1