Timeline for If an upper semicontinuous multivalued map is compact on a set, is it compact on the boundary as well?
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| May 14, 2014 at 18:24 | comment | added | Jozsi | 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! :) | |
| May 14, 2014 at 18:22 | comment | added | Jozsi | 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) | |
| May 13, 2014 at 15:18 | comment | added | Jochen Wengenroth | 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. | |
| May 10, 2014 at 14:34 | comment | added | Jozsi | 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? | |
| May 9, 2014 at 11:29 | comment | added | Jochen Wengenroth | 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 $T(\overline K) \subseteq \overline{T(K)}$. | |
| May 8, 2014 at 14:25 | review | First posts | |||
| May 8, 2014 at 14:34 | |||||
| May 8, 2014 at 14:09 | history | asked | Jozsi | CC BY-SA 3.0 |