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As the question states, if we have the compact embedding of Hilbert spaces $V \subset H$, is $L^2(0,T;V) \subset L^2(0,T;H)$ compact too?

If not true in general, is it true for $V=H^1(\Omega)$ and $H=L^2(\Omega)$?

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    $\begingroup$ What if $V=H=\Bbb R$? $\endgroup$ – David Mitra Jul 11 '13 at 19:56
  • $\begingroup$ What does $L^2(0,T;V)$ mean? $\endgroup$ – Chris Eagle Jul 11 '13 at 20:13
  • $\begingroup$ @ChrisEagle Fairly common notation for the Lebesgue-Bochner space of functions valued in Banach space $V$. $\endgroup$ – 40 votes Jul 11 '13 at 20:14
  • $\begingroup$ @DavidMitra But $\mathbb{R}$ is not compactly embedded in itself. $\endgroup$ – michael_faber Jul 11 '13 at 21:22
  • $\begingroup$ Isn't the identity a compact operator (the image of a bounded sequence has a convergent subsequence)? $\endgroup$ – David Mitra Jul 11 '13 at 21:24
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You want to know whether the unit ball of $L^2(0,T;V)$ is relatively compact (=has compact closure) in $L^2(0,T;H)$. A readable treatment of relative compactness in Lebesgue-Bochner spaces is in Tightness, integral equicontinuity and compactness for evolution problems in Banach spaces by Rossi, and Savaré, see Theorem 1. In a nutshell, you need: boundedness with respect to some Banach space compactly embedded in $H$ (which you have), and integral equicontinuity (which you don't have). Without integral equicontinuity, a counterexample is provided by $f_n(t)=e^{int}\varphi $, where $\varphi$ is any fixed element of $V$.

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