# Showing that the Hausdorff distance from the subdifferential to $0$ is bounded.

Consider the following problem for a general Hilbert space $$\mathcal{H}$$ over $$\mathbb{R}$$, possibly infinite dimensional. Let $$C$$ be a compact set and $$g:\mathcal{H}\to\mathbb{R}\cup{+\infty}$$ a convex, closed, proper function. Furthermore, assume that $$\partial g(x)\neq\emptyset$$ for all $$x\in C$$. We want to show that

$$\sup\limits_{x\in C}d(0,\partial g(x))<\infty$$

where $$d(x, A)=\inf\limits_{a\in A}\|x-a\|$$ is the Hausdorff distance between the point$$x$$ and the set $$A$$.

Here is my attempt:

Assume that $$\partial g(x)\neq \emptyset$$ for all $$x\in C$$ and for the sake of contradiction, assume that $$\exists \{x_n\}_{n\in\mathbb{N}}\subset C$$ such that $$\lim\limits_{n\to\infty}d(0,\partial g(x_n))=\infty$$.

Since $$C$$ is compact, $$\exists$$ a subsequence $$\{x_{n_k}\}_{k\in\mathbb{N}}$$ such that $$x_{n_k}\to x\in C$$.

For each $$k\in\mathbb{N}$$ let $$b_k = \arg\min\limits_{s\in\partial g(x_{n_k})}\|s\|$$, which exists and is unique since $$\partial g(x_{n_k})$$ is nonempty and closed.

If the sequence $$b_k$$ has a convergent subsequence then we are done since, by upper hemicontinuity of $$\partial g$$, $$b_{k_j}\to b\implies b\in\partial g(x)$$.

So, $$d(0,\partial g(x_{n_{k_j}}))=\|b_{k_j}\|$$ but also we have,

$$\|b\|=\|\lim\limits_{j\to\infty}b_{k_j}\|=\lim\limits_{j\to\infty}\|b_{k_j}\|=\lim\limits_{j\to\infty}d(0,\partial g(x_{n_{k_j}}))=\infty$$

which contradicts that $$b\in\partial g(x)$$.

However, I think showing that $$b_{k}$$ has a convergent subsequence might be impossible since, for any subsequence $$b_{k_j}$$, we have that $$\lim\limits_{j\to\infty}\|b_{k_j}\|=\lim\limits_{k\to\infty}\|b_k\|=\infty$$.

• That's outer semicountinouty . Subdifferential mapping is outer semiconntinuous . And it is indeed easy to prove that. – Red shoes May 26 '18 at 14:32
• @Redshoes do you have a reference where I can read about outer semicontinuity? If it is easy to prove do you mind writing a proof or confirming if what I have is correct? – TSF May 27 '18 at 1:15
• I barely write full answer in this website, that's not helpful for OP if they just see the answer. For reference you can search . – Red shoes May 27 '18 at 3:52
• I am the OP, it would be helpful to me because I have been trying different approaches earnestly and not gotten any closer. – TSF May 27 '18 at 14:52
• For a reference of the proof of upper semicontinuity of the subdifferential check Proposition 4.3.2 in Schirotzek: Nonsmooth analysis. For proving that $b_k$ converges I guess you will need the fact that the subdifferential is locally bounded; check Prop 4.3.1 in the same book for that – John D May 27 '18 at 21:16

The claim is not true. The function in this post Extension of bounded convex function to boundary serves as a counterexample: $$C= \{(x,y): x^2\le y\le 1\}$$ $$f(x,y) = \begin{cases} \frac{x^2}y& \text{ if } y>0\\ 0 & \text{ if } y=0. \end{cases}$$ This function is lower semicontinuous, and also convex: Let $$(x,y)\in C\setminus\{0\}$$, $$\lambda\in(0,1)$$, then $$f(\lambda x, \lambda y) = \lambda \frac{x^2}y = \lambda f(x,y) + (1-\lambda) f(0).$$ In addition, $$\partial f$$ is non-empty everywhere: if $$y>0$$, then $$\partial f(x,y) = \{f'(x,y)\} = \{(\frac{2x}y,-\frac{x^2}{y^2})\}$$, and $$0\in \partial f(0,0)$$, since $$0$$ is a global minimum of this function.
Then $$\partial f(x,2x^2)$$ is unbounded for $$x\to0$$ (note that these are interior points of $$C$$ for $$x$$ small enough).