Timeline for Constructing an $\epsilon$-net for a Lipschitz subspace of $L^2$
Current License: CC BY-SA 3.0
8 events
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| Oct 21, 2013 at 21:33 | comment | added | Pietro Majer | If you do not impose a bound on the Lipschitz constant, $X$ may fail to be relatively compact ; its $L^2$ - closure could be the whole closed ball of radius $M$ of $L^\infty$. Possible remedy: Find a smaller domain which is compact but where $F$ has the same infimum. Or use a weaker topology that makes $X$ compact and $F$ lower semicontinuous. | |
| Oct 21, 2013 at 14:08 | comment | added | leo monsaingeon | Ok, then if you can get uniform Lipschitz bounds Ascoli works as suggested by @BillJohnson and you relative compactness of $X$ in the $L^{\infty}$ topology. But this still doesn't help to construct the best posisble $\epsilon$-net, sorry... | |
| Oct 21, 2013 at 13:58 | comment | added | user24451 | @leomonsaingeon It's true that a uniform bound on the Lipschitz constant is needed. I think I can get one. I clarified the assumption $|x| \le M$, I meant $|x(t)| \le M$. Thank you for your help. | |
| Oct 21, 2013 at 13:53 | history | edited | user24451 | CC BY-SA 3.0 |
added 22 characters in body
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| Oct 21, 2013 at 9:07 | comment | added | leo monsaingeon | @ Bill Johnson: the statement is that each $x\in X$ is Lipschitz, but unless I misunderstood something we have no control on the Lipschitz constants uniformly in $x$ so Ascoli doesn't apply. @Nicolas': your assumption $|x|\leq M$ is ambiguous: do you mean mean $|x(t)|\leq M$ uniformly in $t\in (0,1)$ and $x\in X$, or rather $|x|_{L^2}\leq M$ (in which case Ascoli fails)? From what I see I rather recommend Rellich-Kondrachov ($X\subset H^1\subset\subset L^2$) or Riesz-Frechet-Kolmogorov, but in any case you need a uniform bound on the Lipschitz constants! | |
| Oct 21, 2013 at 8:03 | comment | added | Mark Meckes | For your question about (2), see for example en.wikipedia.org/wiki/… | |
| Oct 21, 2013 at 3:03 | comment | added | Bill Johnson | A uniformly bounded set of Lipschitz functions is totally bounded even in $C[0,1]$ by Ascoli-Arzela. | |
| Oct 21, 2013 at 1:25 | history | asked | user24451 | CC BY-SA 3.0 |