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fixed two sentences.
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Ganesh
Ganesh
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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 $H$$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 $V$$C([0, 1]; V)$, which seems to be a stronger statement. Am I missing something?

Thanks.

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 $H$.

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 $V$, which seems to be a stronger statement. Am I missing something?

Thanks.

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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Ganesh
Ganesh
  • 29
  • 2

Hilbert triples

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 $H$.

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 $V$, which seems to be a stronger statement. Am I missing something?

Thanks.