|
|
|
|
|Store|
|
|
|
|
CTK Exchange
Ralph Boles
guest
|
Feb-18-05, 07:19 PM (EST) |
|
"USC correspondence"
|
am working on a proof to present to a reading group I am in, that relies on the following proposition, which I am having trouble proving. Perhaps some of the brilliant minds that frequent this forum may find an opportunity to help. Suppose that M(x) is an USC correspondence from S-->T, S in Rm, T compact in Rn. I want to claim that for every x(n)-->x, there exists a sequence y(n):y(n) is an element of M(x(n)), which converges. Since T is compact, for ANY sequence y(n) st y(n) in M(x(n)), there exists a SUBsequence which converges, but I need an element y(n) in M(x(n)) for EVERY x(n). In case the reader is wondering, I am trying to prove that if f is continuous and M is usc, then the correspondence f(x,M(x)) is also usc, which in turn I am using to prove the Berge "theorem of the maximum." Any ideas would be much appreciated |
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
alexb
Charter Member
1454 posts |
Feb-20-05, 07:47 AM (EST) |
|
1. "RE: USC correspondence"
In response to message #0
|
> am working on a proof to present to a reading group I am >in, that relies on the following proposition, which I am >having trouble proving. Perhaps some of the brilliant minds >that frequent this forum may find an opportunity to help. > >Suppose that M(x) is an USC correspondence from S-->T, S in >Rm, T compact in Rn. > >I want to claim that for every x(n)-->x, there exists a >sequence > >y(n):y(n) is an element of M(x(n)), which converges. > >Since T is compact, for ANY sequence y(n) st y(n) in >M(x(n)), there exists a SUBsequence which converges, but I >need an element y(n) in >M(x(n)) for EVERY x(n). > >In case the reader is wondering, I am trying to prove that >if f is continuous and M is usc, then the correspondence >f(x,M(x)) is also usc, which in turn I am using to prove >the Berge "theorem of the maximum." Any ideas would be much >appreciated I think the Closed Graph theorem is of relevance here. For S and T compact, M is USC iff the graph of M is closed. Take M(x) = sin(pi/x) on some [-N, N], x != 0, and M(0) = [-1,1]. Since the graph of M is closed, M is USC. Now, consider, xn = 1/n. {M(xn)} consists of infinitely many 1's, 0's, and -1's. It clearly contains convergent subsequences, but does not constitute a convergent sequence itself.
|
|
Alert | IP |
Printer-friendly page |
Reply |
Reply With Quote | Top |
|
|
You may be curious to have a look at the old CTK Exchange archive. Please do not post there.
|Front page|
|Contents|
Copyright © 1996-2018 Alexander Bogomolny
|
|