Is this sequence strongly convergent? Consider an analytic semigroup $e^{At}$ on the Hilbert space $H$, and linear operator $G:L^2(0,T;H)\to C(0,T;H)$ as
$$x(t)=Gu(t):=\int_0^te^{A(t-s)}u(s)ds$$
The question is
If the sequence $u_n$ converges weakly to an element $u$ in $L^2(0,T;H)$; does $x_n$ converge strongly to $x$ in $C(0,T;H)$?
Some notes: 


*

*From analyticity of $e^{At}$, it can be shown that $x_n$ belong to $C^{0,1/2}(0,T;H)$.

*If needed, domain of $A$ can be assumed to be a compact subspace of $H$.
 A: Yes, your operator is indeed strongly continuous in this sense. It is seen by the following three steps:
Step 1: The operator $G$ is bounded and linear from $L^2$ to $L^2$.
We have $$\|Gu\|_{L^2}^2 = \int_0^T dt\; \left| \int_0^t ds\; e^{A (t-s)} u(s)\right|^2 \\ \leq \int_0^T dt\; \left( \int_0^t ds\; \|e^{A(t-s)}\|^2 \right) \cdot \left(\int_0^t ds'\; |u(s')|^2 \right) \\
\leq C(\|A\|,T)\; \|u\|^2_{L^2}$$
Step 2: The operator $G$ maps continuously into the Sobolev space $W^{1,2}$.
A simple calculation yields that the ODE
$$ Gu'(t) = A Gu(t) + u(t)$$
holds in a weak/distributional sense. It follows that $\|Gu'\|_{L^2} \leq C \|u\|_{L^2}$. Therefore, $G$ is actually a continuous operator from $L^2$ into $W^{1,2}$.
Step 3: In one dimension, we have the compact Sobolev embedding $W^{1,2} \hookrightarrow C^0$. Since compact maps are completely continuous, that is they map weakly converging sequences onto strongly converging ones, we obtain that $G$ is a strongly continuous operator from $L^2$ into $C^0$.
