I am wondering if it is possible to show $L^{2}((0,T),H^{1}(\Omega))$ is compactly embedded in $L^{2}((0,T),L^{2}(\Omega))$. It's certainly true that $H^{1}(\Omega)$ is compactly embedded in $L^{2}(\Omega)$ however I am unsure if the prior is true.
I have that $\Omega$ is smooth with bounded boundary. So given a bounded subset $\mathcal{F}$ of $L^{2}((0,T),H^{1}(\Omega))$ I can extend trivially in time by zero outside $(0,T)$ and I have the existence of a extension operator $P$ so that $P(\mathcal{F})$ is a bounded subset of $L^{2}(\mathbb{R},H^{1}(\mathbb{R}))\subset L^{2}(\mathbb{R},L^{2}(\mathbb{R}^{N}))$. The aim is to try and show that $\lim_{|h|\to 0}\|\tau_{h}f-f\|_{2}=0$ uniformly in $f\in\mathcal{F}$ (where $\tau_{h}f:=f(x+h)$) which implies compactness of $\mathcal{F}|_{(0,T)\times\Omega}$ in $L^{2}((0,T)\times\Omega)=L^{2}((0,T),L^{2}(\Omega))$ (Theorem 4.26 Brezis).
I believe I have the desired convergence if I translate in space but I am unsure if this will work if I translate in time. Is this a result anyone has come across before? If so, where might I find it?
Set $$\Omega = [0,1]. $$ Define $$ f_n(t,x) = 1_{(2^{-(n+1)},2^{-n})}(t) \cdot 2^{\frac{n+1}{2}}, $$ where $n$ is large enough so the interval is contaiend in $[0,T]$. Then $$ \|f_n\|_{L^2((0,T), H^1([0,1]))}^2 = \int_0^T \int_0^1 f_n(t,x)^2 + \partial_x f_n(t,x)^2 \,dx \,dt $$ $$ = \int_{2^{-(n+1)}}^{2^{-n}} \int_0^1 (2^\frac{n+1}{2})^2 + 0 \,dx \,dt = 2^{-(n+1)} 2^{n+1} = 1. $$ Hence $f_n$ is a bounded sequence in $L^2((0,T), H^1([0,1]))$. But you can directly see that for $k \neq n$ you get $$ \|f_n - f_k\|_{L^2((0,T), L^2([0,1]))}^2 = \|f_n\|_{L^2((0,T), L^2([0,1]))}^2 + \|f_k\|_{L^2((0,T), L^2([0,1]))}^2 = 1 + 1 = 2, $$ because the subsets of $[0,T] \times [0,1]$ where the functions are non-zero are disjoint. Hence the sequence can not have a convergent subsequence in $L^2((0,T), L^2([0,1]))$.