Consider the set of functions $$\mathcal{B}=\{v\in L^2(0,T;H^1_0(\Omega)): \partial_tv\in L^2(0,T;H^{-1}(\Omega))\},$$ equipped with the norm $$\|v\|_{\mathcal{B}}:=\|v\|_{L^2(0,T;H^1_0(\Omega))}+\|\partial_tv\|_{L^2(0,T;H^{-1}(\Omega))},$$ where $\Omega\subset\Bbb R^n$ is a Lipschitz domain, and $H^{-1}(\Omega)$ is the dual space of $H^1_0(\Omega)$.
Is $(\mathcal{B},\|\cdot\|_{\mathcal{B}})$ a Banach space?
I know that $u\in\mathcal{B}$ implies that $u\in C([0,T];L^2(0,T))$, which is a Banach space in its own right, so I believe one would have to deduce that $\mathcal{B}$ is a closed subspace of $C([0,T];L^2(0,T))$, but I am unsure on how to do this.
I think that a direct proof is possible:
Let $(u_k)$ be a Cauchy sequence in $\mathcal{B}$. Then, $(u_k)$ is Cauchy in $L^2(0,T; H^1_0)$ and $(\partial_tu_k)$ is Cauchy in $L^2(0,T; H^{-1})$. Since these are Banach spaces, we conclude that there exists $u\in L^2(0,T; H^1_0)$ and $w\in L^2(0,T; H^{-1})$ such that
$$\left\{\begin{align}u_k\to u\quad&\mbox{in}\quad L^2(0,T;H^1_0),\\ \partial_t u_k\to w\quad&\mbox{in}\quad L^2(0,T;H^{-1}).\end{align}\right.$$
It follows that $\partial_t u=w$ and thus $\partial_t u\in L^2(0,T;H^{-1})$. Hence $(u_k)$ converges to $u$ in $\mathcal{B}$.