Is $L^2(0,T;H^{-1}(\Omega)) \subset \mathcal{D}^*((0,T)\times \Omega)$?

Question

Let $\Omega \subset \mathbb{R}^n$ be a domain. Consider the space of test functions $\mathcal{D}((0,T)\times \Omega)$ and the space of distributions $\mathcal{D}^*((0,T)\times \Omega).$

Is it true that $L^2(0,T;H^1(\Omega))^* = L^2(0,T;H^{-1}(\Omega))$ is a subset of $\mathcal{D}^*((0,T)\times \Omega)$?

I've not been able to prove this. Given $f \in L^2(0,T;H^{-1}(\Omega))$, we have that $f \in \mathcal{D}^*((0,T)\times \Omega)$ if for $v_n \to v$ in $\mathcal{D}((0,T)\times \Omega)$, it holds that $f(v_n) \to f(v)$ in $\mathbb{R}$ (linearity is easy to check).

How to show this? We know that $D^\alpha v_n \to D^\alpha v$ uniformly for all multiindices $\alpha$, but I can't get $L^p$ convergence from this. Help appreciated.

Answer 1

$D((0,T)\times \Omega)\subset L^2((0,T);H^1(\Omega))\subset L^1_{loc}((0,T);H^1(\Omega))$.

Define $$f(v) = \int_0^Tdt \langle f(t,\cdot),v (t,\cdot)\rangle_{H^{-1};H^1},$$ then $$|f(v)|\le \int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\|v (t,\cdot)\|_{H^1} \le | K |\int_0^Tdt \| f(t,\cdot)\|_{H^{-1}}\sqrt{\|v (t,\cdot)\|_{L^\infty}^2+\|\nabla_x v (t,\cdot)\|_{L^\infty}^2 },$$ where $K$ is a projection of the support of $v$ on $\Omega$. Now we need only to deal with technical details $$|f(v)|\le |K|T \|v\|_{C^1((0,T)\times\Omega)} \int_0^Tdt \| f(t,\cdot)\|_{H^{-1}} \le |K|T^{3/2} \|f\|_{L^2((0,T);H^{-1}(\Omega))}\|v\|_{C^1((0,T)\times\Omega)} . $$ This proves that $f(v)$ is a distribution.