I have a function $u \in L^2(0,T;H^1_0)$ with $u' \in L^2(0,T;H^{-1})$, all on some smooth bounded domain $\Omega$.
In addition I have a sequence $u_n$ such that $u_n \to u$ in $L^2(0,T;H^1_0)$ and also the quantity $$\int_0^T \int_{\Omega} |u_n'(t)|^2\varphi(t) \leq C$$ for $C$ independent of $n$, where $\varphi$ is smooth and vanishes at $t=0$, i.e it has compact support near zero, and is non-negative and bounded by one. But otherwise $\varphi$ is arbitrary apart from these conditions.
- From this can we deduce $u' \in L^2(\epsilon, T; L^2(\Omega))$?
- Does it follow that $u' \in L^2(0,T;L^2(\Omega))$?
I have often seen this argument given for a time interval $(0,\infty)$ and in that case the function $\varphi$ is compactly supported near infinity too, and the resulting $u' \in L^2_{loc}(0,\infty,L^2(\Omega))$, but I wonder if we can remove the loc in my specal case.
If the inequality holds in this way: there is $C>0$ such that for all $\phi\in C_0^\infty(0,+\infty)$ with $0\le \phi\le 1$ and all $n$ it holds $$ \int_0^T \int_\Omega |u_n'(t)|^2 \phi(t) dt \le C $$ then $(u_n')$ is uniformly bounded in $L^2(0,T;L^2(\Omega))$, hence $u'$ exists and belongs to that space as well.
To see, this consider smooth approximation $\phi_\epsilon\in C_0^\infty(\Omega)$ of the characteristic function of $[0,T]$: e.g set $\phi_\epsilon = \rho_\epsilon \ast \chi_{[\epsilon,T-\epsilon]}$ with the standard mollifier. Then $0\le\phi_\epsilon\le 1$, and you can pass to the limit $\epsilon\searrow0$ in $$ \int_0^T \int_\Omega |u_n'(t)|^2 \phi_\epsilon(t) dt \le C $$ by Lebesgue dominated convergence theorem, which yields the claim.