Timeline for Norm of differential operator between Sobolev spaces
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2013 at 8:38 | answer | added | Daniel Spector | timeline score: 5 | |
| Jan 18, 2013 at 17:47 | comment | added | Daniel Spector | If you are talking about $U = (0,1)$, $m=1$, and $\alpha=1$ then the answer is yes, as well as more generally (I believe) if $|\alpha|=1$. Consider the following - as you suggest, in general $||f^\prime||_{L^p} \leq ||f||_{W^{1,p}}$, but if you consider a sequence $f_n$ such that $||f_n^\prime||_{L^p}=1$ and $f_n \to 0$ strongly in $L^p$ (consider a variant of $\frac{sin(nx)}{n}$), then the left hand side of the inequality is one while the right hand side tends to one from above. | |
| Jan 17, 2013 at 19:46 | comment | added | Deane Yang | And of course just do everything in $R^1$. | |
| Jan 17, 2013 at 18:49 | comment | added | Mark Meckes | Also, if you really want the exact operator norm, you need to say exactly which "usual" norms you're using on the Sobolev spaces, since there isn't universal agreement among various equivalent options. | |
| Jan 17, 2013 at 17:44 | comment | added | Deane Yang | Any chance you could say a little about why you care about the operator norm? I've never seen anyone need this before. I haven't tried to work out anything at all, but one obvious thing to try is rescaling the support of a compactly supported smooth function down to a single point. | |
| Jan 17, 2013 at 17:31 | history | asked | Sylvester-H | CC BY-SA 3.0 |