We are talking here about the initial value problem on some Hilbert space $H$
$$y'(t)=Ay(t)+f(t), \\ y(0)=y\_0 \in D(A).$$(Problem 1.13 in the reference)


Then $y(t)=e^{At}y\_0 + \int\_0^t e^{A(t-s)}f(s) ds$ (in the reference called the mild solution) is a continuous function 1.13. for $A$ the generator of a $C\_0$ group and $f \in L^1$.


Now, if $A$ is additionally assumed to be self-adjoint and dissipative ($\langle Ax,x \rangle \le 0$ for all $x \in D(A)$) and $f \in L^2,$ then the following reference claims that $y$ is even absolutely continuous. 
[click me to see the statement in Mathematical Methods in Optimization of Differential Systems on p. 174](https://i.stack.imgur.com/fCHo8.jpg) 
But unfortunately, no proof is given. 
Does anybody know how to show this or know a reference where the proof is given? 



Absolutely continuity in variation of constant formula
To build a bit on the comments made, let $P\_t = \exp(A t)$ and write the Duhamel formula as: 
\begin{align\*}
y(t) &= y\_0 + (P\_t - I) y\_0 + \int\_0^t P\_{t-s} f(s) ds \\
&= y\_0 + \int\_0^t g(s) ds
\end{align\*}
where we have introduced $g: \mathbb{R}\_+ \to H$ defined as: 
$$
g(s) = P\_s A y\_0 + P\_{t-s} f(s)
$$
To show that $g(\cdot)$ is integrable, 
\begin{align\*}
\int\_0^t \| g(s) \| ds &\le \int\_0^t \left( \| P\_s A y\_0 \| + \| P\_{t-s} f(s) \| \right) ds \\
&\le \| A y\_0 \| \int\_0^t \| P\_s \|\_{L(H)} ds + \int\_0^t \| P\_{t-s} \|\_{L(H)} \| f(s) \| ds \\
\end{align\*}
where we used the triangle inequality and basic properties of the operator norm. To complete this bound, use the facts that (i) $\| P\_t \|\_{L(H)} \le 1$; (ii) $y\_0 \in D(A)$; and (iii) $f$ is integrable. Afterwards, invoke Khavkine's comment on the equivalent condition for absolute continuity. 


