Timeline for Embeddings of Weighted Banach Spaces
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2010 at 9:17 | comment | added | Yemon Choi | It's up to you, really. The original question doesn't have to be perfect and complete. | |
| Feb 17, 2010 at 9:07 | comment | added | Yemon Choi | As regards editing the original question: I think you should be able to edit it yourself. Personally I would prefer that you delete as little as possible, and instead add a new part; that way one can see what the original answers of Bill Johnson and myself are referring to. | |
| Feb 17, 2010 at 9:03 | comment | added | Yemon Choi | Does my updated version answer your question? In the case you refer to, the ratio of the weights does not lie in $c_0$, and so you are right that the estimates don't work: indeed, this case, the embedding is NOT compact. | |
| Feb 17, 2010 at 8:46 | comment | added | Leandro | I get stuck about the last remark of your answer. Suppose that $\Omega_p$ now, is given as above but the weights are replaced by a sequence $p(i)\in C_0(\mathbb{Z}^d)$ and for $\Omega_{p'}$, the weights are replaced by $p'(i)\in C_0(\mathbb{Z}^d)$. If $p'(i)\leq p(i)$ then the embedding $\Omega_p\hookrightarrow \Omega_{p'}$ is well defined. Let's suppose also the the sequence decay monotonically, however if $$0<\liminf_{i\in\mathbb{Z}^d} \frac{|p(i)|}{|p'(i)|}$$ Your norm estimates does not works anymore. even in this case the Bill's argument is in hold ? Thanks again | |
| Feb 17, 2010 at 8:41 | history | edited | Yemon Choi | CC BY-SA 2.5 |
clarified last para in response to question in comments
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| Feb 17, 2010 at 8:37 | comment | added | Yemon Choi | I wasn't very clear in my last paragraph: I'll edit it now. Bill is talking about transforming your original question so that instead of looking at a simple operator (inclusion) between two complicated-seeming spaces, one looks at a slightly (but not very) complicated operator on a simple space, namely l^R. | |
| Feb 16, 2010 at 5:53 | vote | accept | Leandro | ||
| Feb 16, 2010 at 5:04 | comment | added | Leandro | Yemon tahnk you very much for the answer, I read Bill's answer also but with my background in functional analysis I took some time to figure out what is happening, but after read your argument everything it was clear for me. Thanks again !! | |
| Feb 16, 2010 at 4:58 | vote | accept | Leandro | ||
| Feb 16, 2010 at 4:58 | |||||
| Feb 16, 2010 at 2:13 | history | answered | Yemon Choi | CC BY-SA 2.5 |