Let $e^{At}$ denote an analytic semigroup on Hilbert space $X$ generate by $A:D(A)\to X$. Also, let $f\in L^1(0,\tau;X)$. I want to show that the convolution $$ g(t)=\int_0^t e^{A(ts)}f(s)ds$$ belongs to $W^{1,1}(0,\tau;X)\cap L^1(0,\tau;D(A))$. If I set $f(s)=e^{As}x$ where $x\in X$ then $g(t)=te^{At}x$. Because of analyticity of $e^{At}$ we have $$\sup_{t>0}\tAe^{At}\_{L(X)}<\infty$$ thus $g(t)\in L^{\infty}(0,\tau;D(A))$ and $\dot{g}(t)=e^{At}x+tAe^{At}x\in L^{\infty}(0,\tau;X)$. Does this prove it?

Did you try with $A$ compact and normal – reuns Dec 4 at 0:11