1
$\begingroup$

I have stumbled upon the following problem during my research:

Let $X$ and $Y$ be Banach spaces, $K\subset X$ nonempty, $F:\overline{K}\rightarrow 2^{Y}$ an upper semicontinuous multivalued map with nonempty, convex, compact values, that is compact on $K$ (in the operatorial sense). Prove that $F$ is also compact on $\overline{K}$.

This seems easy to prove, yet I'm stuck on the details, I think I'm not seeing the woods from the trees. If anybody could please help me out with the proof, it would be much appreciated. I might mention that I only need a weaker version of this statement (but I posted it this way, since it's easier to formulate, and my intuition tells me that this should be true as well), so if it makes the problem easier, we can also suppose that $K$ is a convex cone.

Thanks in advance.

Improve this question
$\endgroup$
5
  • $\begingroup$ What do you mean by compact? If it just the condition that $\overline{T(K)}$ is compact in $Y$ this would be very easy since by upper semicontinuity you have $T(\overline K) \subseteq \overline{T(K)}$. $\endgroup$
    – Jochen Wengenroth
    Commented May 9, 2014 at 11:29
  • $\begingroup$ Hi! Thanks for your answer. By compact I mean that for any bounded set $B\in K$, $\overline{T(B)}$ is compact. So if what you said is truly implied by upper semicontinuity, then the proof would be easy, I actually went down this path, but I couldn't prove that upper semicontinuity implies $T(\overline{B})\subset\overline{T(B)}$. Moreover, I got that lower semicontinuity would imply this. Note that by semicontinuity I mean these: en.wikipedia.org/wiki/Hemicontinuity. I just tried to use the sequential characterizations from that link. So could you please give a detailed proof? $\endgroup$
    – Jozsi
    Commented May 10, 2014 at 14:34
  • $\begingroup$ You are probably right, sorry for the wrong comment. If a rather general statement resists the natural attempts for a proof it might be wrong. I would look for a counterexample. $\endgroup$
    – Jochen Wengenroth
    Commented May 13, 2014 at 15:18
  • $\begingroup$ In the mean time I have found the following proof, which seems correct to me: Let $x_{n}\in\overline{K}$ be a bounded sequence, $y_{n}\in F(x_{n})$. Since $F$ is U.S.C., it has a closed graph. We use this to deduce that for every $(x_{n},y_{n})$ there exists $(u_{n},v_{n})\in Gr(F)$ with $u_{n}\in K$ and $||(x_{n},y_{n})-(u_{n},v_{n})||<1/n$ (here on the product space I mean the sum of the norms). Consequently, $||x_{n}-u_{n}||<1/n$, therefore $(u_{n})$ is also bounded (and in $K$), but since $F$ is compact on $K$, there exists a convergent subsequence $(v_{n_{k}})$. (TBC) $\endgroup$
    – Jozsi
    Commented May 14, 2014 at 18:22
  • $\begingroup$ Now, since $||y_{n}-v_{n}||<1/n$, we get that $(y_{n_{k}})$ is a Cauchy sequence in a Banach space and we are done. I think this proof is correct, but I could use some double-checking. Thank you for your input once again. Have a nice day! :) $\endgroup$
    – Jozsi
    Commented May 14, 2014 at 18:24

0

Reset to default

You must log in to answer this question.