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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 $\lim\limits_{j\to\infty}\|b_{k_j}\|=\lim\limits_{k\to\infty}\|b_k\|=\infty$.

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  • $\begingroup$ That's outer semicountinouty . Subdifferential mapping is outer semiconntinuous . And it is indeed easy to prove that. $\endgroup$ – Red shoes May 26 '18 at 14:32
  • $\begingroup$ @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? $\endgroup$ – TSF May 27 '18 at 1:15
  • $\begingroup$ 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 . $\endgroup$ – Red shoes May 27 '18 at 3:52
  • $\begingroup$ I am the OP, it would be helpful to me because I have been trying different approaches earnestly and not gotten any closer. $\endgroup$ – TSF May 27 '18 at 14:52
  • $\begingroup$ 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 $\endgroup$ – John D May 27 '18 at 21:16
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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).

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