With $V=L^2(0,T;H^1(\Omega))$, let $A:V \to V^*$ with $$\langle Au,v \rangle = \int_0^T \int_{\Omega} \nabla u(t) \cdot \nabla v(t).$$
I want to show that $A$ is a compact operator.
So, one way to do this is, I need to show that if $v_n \in V$ with $\lVert v_n \rVert_V = 1$, then $Av_n$ has a Cauchy subsequence.
Well, $v_n \rightharpoonup v$ in $V$ obviously but this doesn't give me $Av_n \to Av$ which is what I was hoping for. Any other way I can show compactness? Thanks
Given a Hilbert space $H$, we can define the duality map $T:H\to H^*$ by $Tx(y)=\langle y,x\rangle$. This map is a bijection between $H$ and $H^*$; in fact, one sometimes uses it to "identify" $H$ with its dual, thus declaring $H$ to be the identity map. Of course, $T$ is not compact.
Let $H=L^2(0,T,K)$ where $K$ is also a Hilbert space. Then $T$ is defined by $$Tx(y)=\int_0^T \langle y(t),x(t)\rangle_K\,dt\tag1$$ Following Sam's suggestion, specialize the above to $K\subseteq H^1 (\Omega) $ being the set of functions with zero mean, equipped with $\langle u,v\rangle_K=\int_\Omega \nabla u\cdot \nabla v$. Then the duality map $T$ is exactly your operator; as stated in the first paragraph, it is not compact.