TITLE: Does the form domain of the Friedrichs extension of an unbounded operator compactly embed?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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).