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I am trying to understand the Hilbert triple $V \subset H \subset V^*$, where $V$ and $H$ are Hilbert spaces and the star denotes the dual space. Eg: $H^1 \subset L^2 \subset H^{-1}.$

If $V \subset H$, then associating it is clear that $H^* \subset V^*$ since every functional in $H^*$ acts on elements of $H$, and since $V$ is in $H$, acts on elements in $V$. Is this understanding right?

Therefore, by assocating $H$ with $H^*$, we can write $V \subset H \equiv H^* \subset V^*$, which is the Hilbert triple.

Is this correct? I feel like I'm missing something deeper.

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I think that you should read Remarque 1 of Chapitre V in the book by Haim Brezis, Analyse Fonctionnelle. It contain a nice discussion on the triple $V \subset H \subset V^*$. –  Siminore May 24 '12 at 16:24
    
Are you sure you meant for both $V$ and $H$ to be Hilbert spaces? The concept I'm familiar with -- en.wikipedia.org/wiki/Rigged_Hilbert_space -- $V$ doesn't have to be a Hilbert space. In fact, you really don't want it to be (although it is allowed). –  Hurkyl May 24 '12 at 16:46
    
@LeonidKovalev Can you please elaborate? Why is $H^1$ not isomorphic to the subspace of $L^2$ that contains $H^1$ functions? As far as I am concerned $H^1$ is a proper ordinary subset of $L^2$. Thanks. –  soup May 24 '12 at 19:09
    
@Hurkyl In the context of PDEs mostly we deal with $H^1$ and $L^2$ which are Hilbert spaces. But I guess greater generality is better. –  soup May 24 '12 at 19:10
    
@Siminore Thanks for the link. –  soup May 24 '12 at 19:10

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