Timeline for Convergence of Schwartz kernels implies convergence of operators
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 30, 2012 at 17:13 | vote | accept | AlexE | ||
| Dec 25, 2012 at 1:42 | comment | added | fedja | >I have a particular case and want to show that it is a compact operator.< Why don't you just tell us the kernel and the space in which you want the compactness then? That may lead to your goal much faster... | |
| Dec 24, 2012 at 9:41 | comment | added | AlexE | I have a particular case and want to show that it is a compact operator. I managed to do an approximation of the kernel in the sup-norm, s.t. the approximating operator have finite-dimensional range. But as the answer / comments suggest, approximating in the sup-norm is not a good choice. So I have to try something else. But I'm also interested in the general case and Liviu Nicolaescu's answer is a good start for this. | |
| Dec 24, 2012 at 0:10 | answer | added | Liviu Nicolaescu | timeline score: 7 | |
| Dec 23, 2012 at 17:57 | comment | added | Davide Giraudo | You know that $k_j\to k$ uniformly: are you working in a particular case, or you want to deal with the general one? | |
| Dec 23, 2012 at 17:49 | comment | added | AlexE | I know $k_n \to k$ uniformly and tried to deduce convergence of the $K_n$ in the operator norm. But @DavideGiraudo says, that this does not hold in general. So maybe there is some little, small condition (in addition to uniform convergence of the $k_n$), so that I can then deduce $K_n \to K$ in norm? Convergence in $L^2(R^n \times R^n)$ is to strong, since $k_n - k$ is not square-integrable over $R^n \times R^n$. @DeaneYang: I don't see how to apply Hölder here. | |
| Dec 23, 2012 at 16:38 | comment | added | Deane Yang | You can certainly get some results just by writing out the definition of the norms and applying Holder's inequality. Do you need something sharper than that? | |
| Dec 23, 2012 at 16:18 | comment | added | Davide Giraudo | For the first question, we can try to see what happens when $K_j(x,y)=h_j(x)g(y)$, where $h_j$ and $g$ are smooth and square integrable. Even when $g$ is bounded and $h_j\to 0$ uniformly, we don't necessarily have that $K_j(f)\to 0$ for all $f$. However, in the general case, if we have $\lVert K_n-K\rVert_{L^2(\Bbb R^n\times \Bbb R^n)}\to 0$, there is convergence in the operator norm. | |
| Dec 23, 2012 at 15:41 | history | edited | AlexE | CC BY-SA 3.0 |
added 149 characters in body
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| Dec 23, 2012 at 15:06 | comment | added | AlexE | I posted this question some days ago on Math.StackExchange but did not get any answer. math.stackexchange.com/q/262632/7110 | |
| Dec 23, 2012 at 15:05 | history | asked | AlexE | CC BY-SA 3.0 |