# Existence of weakly convergent subsequence $(v_{n_k})$ of $(v_n)\subset L^2([0,T],V^*)$

Let $"V\subset H\subset V^*"$ be an evolution triple. Denote $\mathcal{V}=L^{2}([0,T],V)$, $\mathcal{H}=L^2([0,T], H)$, $\mathcal{V}^*=L^2([0,T], V^*)$. Then we have continuous embeddings $$\mathcal{V}\subset \mathcal{H}\subset \mathcal{V^*}.$$ Consider a sequance $(v_{n})\subset\mathcal{V}^*$ which is bounded. Can I extract a weakly convergent subsequance from $(v_{n})$?

One can prove (see Zeidler IIA for example) that $\mathcal{V}$ is reflexive and separable Banach space. This implies that $L^2([0,T], V)^*\cong L^2([0,T],V^*)=\mathcal{V}^*$ is also reflexive (as the dual of relexive space). From the definition of evolution triple, I know that $H$ is Hilbert. Introducing inner product $$\langle v,w\rangle_{\mathcal{H}}=\int_{0}^{T}\langle v(t), w(t)\rangle_{H} \,dt$$ (I didn't check that it is inner product, but I think integration will preserve its properties), I have that $\mathcal{H}$ is Hilbert. So it can be indentified with its dual. Because of continuous embedding $\mathcal{V}\subset \mathcal{H}$ and the indentification - I've just mentioned, I deduce that $\mathcal{H}^*$ is separable. Is it correct? As a Hilbert space it is relexive as well. Now, I want to use Every bounded sequence has a weakly convergent subsequence in a Hilbert space

As a result I can extract a subsequance $(v_{n_k})\subset \mathcal{H}^*$ which is weakly convergent in $\mathcal{H}^*$. Since $\mathcal{H}^*$ is continuously embedded in $\mathcal{V}^*$, $(v_{n_k})\subset\mathcal{V}^*$ is weakly convergent. Do you find it correct?

If you've shown that $\mathcal{V}^*$ is reflexive then that's enough to deduce the existence of a weakly convergent subsequence in $\mathcal{V}^*$. You've identified $H$ with its dual space, so I wouldn't talk about $\mathcal{H}^*$.
I don't see why you would get a weakly convergent subsequence in $\mathcal{H}=\mathcal{H}^*$. Check the definiton of "continuously embedded"; you have the inequality the wrong way around.