Suppose that $f\in L^\infty(0,1;H_0^1(\Omega))$.
Is it then true that $\int_0^1f(s)\,ds\in H_0^1(\Omega)$?
($\Omega\subset\mathbb R^d$ is a bounded domain, $H_0^1(\Omega)$ is the Sobolev space obtained as the closure of $C_c^\infty(\Omega)$ in the $H^1$-norm).
The only thing I could prove is that $\nabla\int_0^1f(s)\,ds:=\int_0^1\nabla f(s)\,ds\in L^2(\Omega)$, as \begin{align} \int_\Omega \left(\int_0^1\nabla f(s,x)\,ds\right)^2dx&=\int_0^1\int_0^1\int_\Omega\nabla f(s,x)\nabla f(s',x)\,dx\,ds\,ds'\\ &\le\int_0^1\int_0^1\|\nabla f(s)\|_{H^1(\Omega)}\|\nabla f(s')\|_{H^1(\Omega)}\,ds\,ds'\\ &\le \|\nabla f\|_{L^\infty(0,1;H^1(\Omega))}^2. \end{align}