Let $\Omega$ be a open subset of $\mathbb{R}^{d}$. It is well known that $L^2(\Omega\times (0,T))$ can be identified with $L^2(0,T;L^2(\Omega))$. Now, let us consider $$H^{1}(0,T;H_{0}^{1}(\Omega))=\{u: u \in L^2(0,T;H_{0}^{1}(\Omega)) \hbox{ and } u' \in L^2(0,T;H_{0}^{1}(\Omega))\},$$ endowed with the inner product $$(u,v)_{H^1(0,T;H_{0}^{1}(\Omega)}=\int_{0}^{T}(u(t),v(t))_{H_{0}^{1}(\Omega)}dt+\int_{0}^{t}(u'(t),v'(t))_{H_{0}^{1}(\Omega))}dt,$$ where the derivative $u'$ is in the sense of distributions $\mathcal{D}'(0,T;H_{0}^{1}(\Omega))$.
Define $H_{0}^{1}(0,T;H_{0}^{1}(\Omega))=\overline{\mathcal{D}(0,T;H_{0}^{1}(\Omega))}^{H^1(0,T;H_{0}^{1}(\Omega))}$, where $\mathcal{D}(0,T;H_{0}^{1}(\Omega))$ represents the space $C_{0}^{\infty}(0,T;H_{0}^{1}(\Omega))$ endowed with the inductive limit topology.
I would like to know if $H_{0}^{1}(0,T;H_{0}^{1}(\Omega))$ can be identified with $H_{0}^{1}(\Omega\times (0,T))$.
I imagine that I can not make such an identification, since $\mathcal{D}(0,T)\otimes \mathcal{D}(\Omega) \subsetneq \mathcal{D}(\Omega \times (0,T))$, where $\mathcal{D}(0,T)\otimes \mathcal{D}(\Omega)) =\{\eta(t)\varphi(x): \eta \in \mathcal{D}(0,T) \hbox{ and } \varphi \in \mathcal{D}(\Omega) \}$.
No, this is not possible. For $u \in H_0^1(0,T; H_0^1(\Omega))$ you have existence of mixed second-order derivatives $\partial_t \partial x$, whereas $u \in H_0^1((0,T)\times\Omega)$ only possesses first-order derivatives.