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

  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.