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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 But unfortunately, no proof is given. Does anybody know how to show this or know a reference where the proof is given?

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    $\begingroup$ Please add to your question the assumptions made on $y_0$. This seems to be an infinite-dimensional analog of en.wikipedia.org/wiki/Carathéodory%27s_existence_theorem $\endgroup$
    – Nawaf Bou-Rabee
    Commented Sep 8, 2016 at 23:15
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    $\begingroup$ Let $H=\mathbb{R}^n$ and consider the ODE $y'(t) = A y(t) + f(t)$ where $f$ is measurable. Then Carathéodory's extension thm implies that this ODE has an absolutely continuous solution, even though $f$ is not necessarily continuous. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Sep 9, 2016 at 1:45
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    $\begingroup$ Also, just see the equivalent definition (3) of absolute continuity on Wikipedia. $\endgroup$
    – Igor Khavkine
    Commented Sep 9, 2016 at 6:35
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    $\begingroup$ @Torpedo, rewriting slightly, you have $y(t) = e^{At} (y(0) + \int_0^t e^{-As} f(s) \, d{s})$. As long as $e^{-As} f(s)$ is in $L^1$ with respect to $s$, you are done. By your hypotheses, $f\in L^2$, but on any compact interval $L^2 \subset L^1$. $\endgroup$
    – Igor Khavkine
    Commented Sep 9, 2016 at 17:00
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    $\begingroup$ @Torpedo, maybe I'm missing something. $e^{At}$ is smooth in $t$, so (smooth) x (absolutely continuous) = (absolutely continuous). No? $\endgroup$
    – Igor Khavkine
    Commented Sep 9, 2016 at 19:03

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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.

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  • $\begingroup$ sorry for asking again, but I looked first at this only on my cell-phone and now I noticed that I do not see why you needed all this: We have $\int_0^t ||P_sA y_0 ||+ ||P_{t-s} f(s)||ds$. What prevents us from estimating this like $||P_sA y_0 || \le ||P_s || \ ||Ay_0|| \le M e^{\omega t} ||Ay_0||$ by the growth condition for semigroups and $||P_{t-s} f(s)|| \le ||P_{t-s}|| \ ||f(s)|| \le M e^{\omega (t-s)} ||f(s)|| \le M e^{\omega t} ||f(s)||.$ Now, it would be sufficient to have $||f( \cdot)|| \in L^1$, right? $\endgroup$
    – Torpedo
    Commented Sep 9, 2016 at 22:44
  • $\begingroup$ I would be very grateful to you if you could help me why you did it that way. $\endgroup$
    – Torpedo
    Commented Sep 9, 2016 at 23:09
  • $\begingroup$ Is $\| f(s) \|$ uniformly bounded by a constant for all $s \in [0,t]$? I don't see that anywhere in the given assumptions. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Sep 10, 2016 at 1:34
  • $\begingroup$ Where did I use this? $\endgroup$
    – Torpedo
    Commented Sep 10, 2016 at 1:37
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    $\begingroup$ Well for evey $t \ge s \ge 0$ we have $P_{t-s} \in L(H)$ and for almost every $s \in [0,t]$ we have $f(s) \in H$. Hence for almost every $s$ and every $t$ satisfying $t \ge s \ge 0$ we have $||P_{t-s} f(s)|| \le ||P_{t-s}||_{L(H)} || f(s)||$. I am sorry, I seem to be completely blind at the moment. What do you think can go wrong here? $\endgroup$
    – Torpedo
    Commented Sep 10, 2016 at 1:46

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