# If $V \subset H$ compact, is $L^2(0,T;V) \subset L^2(0,T;H)$ compact too?

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 $H=L^2(\Omega)$?

• What if $V=H=\Bbb R$? – David Mitra Jul 11 '13 at 19:56
• What does $L^2(0,T;V)$ mean? – Chris Eagle Jul 11 '13 at 20:13
• @ChrisEagle Fairly common notation for the Lebesgue-Bochner space of functions valued in Banach space $V$. – 40 votes Jul 11 '13 at 20:14
• @DavidMitra But $\mathbb{R}$ is not compactly embedded in itself. – michael_faber Jul 11 '13 at 21:22
• Isn't the identity a compact operator (the image of a bounded sequence has a convergent subsequence)? – David Mitra Jul 11 '13 at 21:24

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 $V$.