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I apologize if this is a trivial matter for the analysts out there, but I looked this up in a number of books (e.g., H. Brezis, J. Wloka) and could not find a satisfying discussion, one that would settle my question.

Recall that a Hilbert (or Gelfand) triple consists of two Hilbert spaces $V, H$ such that one has continuous, injective and dense inclusions $V \hookrightarrow H \hookrightarrow V^*$.

Given one such triple, one can define the Sobolev space $W \equiv W^{1,2}([0, 1]; V, V^*)$, which is the space of all functions $u : [0, 1] \rightarrow V$ having a weak derivative $u' : [0, 1] \rightarrow V^*$.

Finally, one has the Sobolev embedding theorem, saying that $W$ embeds continuously in $C([0, 1]; H)$.

My question: since $H$ does not play any role in the definition of $W$, why is the Sobolev embedding theorem stated in that way?

It seems to me that $V \hookrightarrow V \hookrightarrow V^*$ is always a Hilbert triple, and then one gets for free that $W$ actually embeds in $C([0, 1]; V)$, which seems to be a stronger statement. Am I missing something?

Thanks.

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    $\begingroup$ $V$ and $H$ are not given the same topology, and $V\subset H$ will not be closed with the topology of $H$. $\endgroup$
    – Christian Remling
    Commented Sep 20, 2014 at 21:09
  • $\begingroup$ @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? $\endgroup$
    – Ganesh
    Commented Sep 20, 2014 at 21:15
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    $\begingroup$ $V^*$ is built in relation to $H$. In your situation, $V^*=V$, which is not too helpful. $\endgroup$
    – András Bátkai
    Commented Sep 20, 2014 at 21:28
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    $\begingroup$ $V^*$ 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. $\endgroup$
    – Christian Remling
    Commented Sep 20, 2014 at 21:49
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    $\begingroup$ 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 " $V\hookrightarrow H\hookrightarrow V^*$ ". $\endgroup$
    – TaQ
    Commented Sep 20, 2014 at 23:36

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