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Say we have a Hilbert space $H$ and a positive symmetric operator $T$ with domain $D$. Define a norm $\|u\|_T = \langle Tu, u\rangle$ for $u\in D$ and take the completion of $D$ with respect to this norm to obtain a new Hilbert space $V$.

Part of the construction of the Friedrichs extension of $T$ is that the inclusion map $D\hookrightarrow H$ extends to an injective bounded map $V\hookrightarrow H$.

If $T$ is unbounded in $H$, is the inclusion $V\hookrightarrow H$ a compact embedding?

Neal
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1 Answers1

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Not in general. For example, with $T=1-\Delta$ with $\Delta=d^2/dx^2$ on the real line, you get the usual Sobolev spaces, and their inclusions are not compact (or else Fourier inversion would be a sum rather than an integral, for example).

paul garrett
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  • Ah, I see, thanks. So part of what makes Rellich-Kondrachov tick is the compactness of the domain. Can that be abstracted into purely functional analytic terms? (e.g., "For all unbounded symmetric operators $T$ satisfying property "$X$," the inclusion...") – Neal Dec 20 '16 at 17:58
  • Not clear to me how to abstract this usefully. But in the example on the real line, if we replace $1-\Delta$ by $-\Delta+x^2$, for example (where $x^2$ functions as a "confining potential") then we do have compactness. – paul garrett Dec 20 '16 at 18:04
  • I'm thinking the statement may be something like "$T$ has only point spectrum iff the embedding is compact" but that's probably trivial. Perhaps there's a useful abstraction by looking at $1$ and $x^2$ as tempered distributions? – Neal Dec 20 '16 at 19:10