Timeline for Hilbert triples
Current License: CC BY-SA 3.0
19 events
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| Sep 20, 2014 at 23:36 | comment | added | TaQ | I think the question should be closed since it arises from the OP not having "done his homework". The OP should read e.g. Wloka's account (in the PDE book) on Gelfand triples carefully. Then it becomes clear that the definition of OP's space $W$ indeed depends on $H$ . More precisely, it depends on the Hilbert (inner product) space $H$ and the "finer" inner product of $V$ defined on the vector subspace $V$ of $H$ . It is only loose language to express a Gelfand triple as " $V\hookrightarrow H\hookrightarrow V^*$ ". | |
| Sep 20, 2014 at 22:14 | comment | added | Ganesh | @ChristianRemling I see your point, but where precisely are these maps relevant when it comes to defining $W$ is what I'm missing here, I think. | |
| Sep 20, 2014 at 22:10 | comment | added | Christian Remling | I can't answer that because I'm not sufficiently familiar with the definition of your $W$. However, let me give you a silly example that may (or may not) illuminate your difficulty: Let $K=\ell^2(\mathbb Z)$, $L=\ell^2(\mathbb N)$. Then you can say that $K\cong L$, or you could embed $L$ into $K$ as a proper subspace. What I'm trying to say is that we have spaces AND maps, and one and the same space can take various roles. | |
| Sep 20, 2014 at 22:06 | comment | added | Ganesh | @ChristianRemling I'm sorry to insist on this, but I have to be honest here. You say $V^*$ does not depend on $H$, so let's accept that. Does $W$, as constructed above, depend on $H$ or not? If not, then there seems to be a trivial way to strenghten the conclusion of the Sobolev embedding. I don't think this should be so, but I'm struggling to see why it does not follow from what I said in the body of the question. | |
| Sep 20, 2014 at 21:49 | comment | added | Christian Remling | $V^*$ as a Hilbert space does not depend on $H$ (and I don't think that's what Andras and Yemon said). [In fact, if your spaces are separable, there's only one Hilbert space anyway, up to isomorphism.] We don't just consider the spaces in isolation, though, we set up the embeddings you mentioned. The Wikipedia article has a pretty clear summary, I think. | |
| Sep 20, 2014 at 21:48 | comment | added | Ganesh | @ChristianRemling Oh, maybe you meant the dual of the image of $V$, as a subspace of $H$? Is that what $V^*$ is supposed to mean, above? That would depend on $H$, ofc. | |
| Sep 20, 2014 at 21:45 | comment | added | Ganesh | @ChristianRemling If $V^*$ is just the dual of $V$, I don't see how $H$ enters into the definition of $W$.What András and Yemon were saying led me to think that the construction of $W$ does depend on $H$, because $V^*$ itself depends on $H$. So I don't understand which one is right. | |
| Sep 20, 2014 at 21:38 | comment | added | Christian Remling | $V^*$ is the dual of $V$, but we emphasize the relation of these spaces to $H$. | |
| Sep 20, 2014 at 21:32 | comment | added | Ganesh | @AndrásBátkai Ah, so this is my mistake: I assumed $V^*$ was just the dual of $V$. So what you and Yemon are saying is that $W$ does depend on $H$. I see. Thank you. | |
| Sep 20, 2014 at 21:30 | comment | added | Yemon Choi | You might get the stronger statement that W embeds in C([0,1]; V), but then how do you know there are enough interesting things in W? (Also, I second @AndrásBátkai's comment) | |
| Sep 20, 2014 at 21:28 | comment | added | András Bátkai | $V^*$ is built in relation to $H$. In your situation, $V^*=V$, which is not too helpful. | |
| Sep 20, 2014 at 21:28 | comment | added | Christian Remling | Put differently, if you embed $V\mapsto V$ trivially, with the identity, then $V^*$ is just $V$ itself, by the Riesz representation theorem. | |
| Sep 20, 2014 at 21:28 | comment | added | Ganesh | @ChristianRemling I agree, and this is why I asked this question. It would seem that one can strengthen the conclusion of the Sobolev embedding by applying it to the triple $V \hookrightarrow V \hookrightarrow V^*$, where $V$ now plays the role of $H$. This gives the stronger statement that $W$ embeds in $C([0, 1]; V)$, does it not? | |
| Sep 20, 2014 at 21:26 | comment | added | Christian Remling | OK, but then via the embedding, we can think of $V$ as a (non-closed) subspace of $H$, and my comment was that then $V$ has a stronger topology than $H$. | |
| Sep 20, 2014 at 21:15 | comment | added | Ganesh | @ChristianRemling I'm sorry, but I don't understand the relation between my question and what you said. I'm not requiring that $V$ is a subspace of $H$, just that it embeds continuously. Could you elaborate? | |
| Sep 20, 2014 at 21:09 | comment | added | Christian Remling | $V$ and $H$ are not given the same topology, and $V\subset H$ will not be closed with the topology of $H$. | |
| Sep 20, 2014 at 21:05 | history | edited | Ganesh | CC BY-SA 3.0 |
fixed two sentences.
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| Sep 20, 2014 at 20:57 | review | First posts | |||
| Sep 20, 2014 at 21:03 | |||||
| Sep 20, 2014 at 20:57 | history | asked | Ganesh | CC BY-SA 3.0 |