Differentiating the variation of constants formula Let


*

*$H$ be a separable $\mathbb R$-Hilbert space

*$S:[0,\infty)\to H$ be an uniformly continuous$^1$ semigroup

*$-A$ be the infinitesimal generator of $S$

*$f:H\to H$ be Lipschitz continuous with sublinear growth

*$t>0$

*$u\in C^0([0,T],H)$


I've read (in An Introduction to Computational Stochastic PDEs on page 111) that $$\frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=-\int_0^tAS(t-s)f(u(s))\:{\rm d}s+f(u(t))\;.\tag 1$$

How can we prove $(1)$?

I've tried to write $$\frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=\lim_{h\to0+}\left(\int_0^t\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))\:{\rm d}s+S(h)\frac1h\int_t^{t+h}S(t-s)f(u(s))\:{\rm d}s\right)\tag 2$$ and it's clear that the first term on the right-hand side of $(2)$ converges to $-\int_0^tAS(t-s)f(u(s))\:{\rm d}s$. However, the second term is not even well-defined, cause $S$ is evaluated at negative points. So, what's the correct approach?

$^1$ i.e. $S\in C^0([0,\infty),\mathfrak L(H))$.
 A: Define $G:\mathbb R\to \mathfrak{L}(H)$ by
$$G(t)=\left\{\begin{align}
S(t),&\quad t\geq 0\\
[S(-t)]^{-1},&\quad t<0
\end{align}\right.$$
As $S$ is uniformly continuous, $S(t)$ is invertible for all $t\geq 0$ and thus $G$ is well defined. In addition, $G$ is a uniformly continuous group.
Using your idea we obtain:
$$\begin{align}
\frac{{\rm d}}{{\rm d}t}\int_0^tS(t-s)f(u(s))\:{\rm d}s=\frac{{\rm d}}{{\rm d}t}\int_0^tG(t-s)f(u(s))\:{\rm d}s\\
=\lim_{h\to0+}\left(\int_0^t\frac{G(h)-\operatorname{id}_H}hG(t-s)f(u(s))\:{\rm d}s+G(h)\frac1h\int_t^{t+h}G(t-s)f(u(s))\:{\rm d}s\right)\\
=\lim_{h\to0+}\left(\int_0^t\frac{S(h)-\operatorname{id}_H}hS(t-s)f(u(s))\:{\rm d}s+S(h)\frac1h\int_t^{t+h}G(t-s)f(u(s))\:{\rm d}s\right)
\end{align}$$
Now the second term is well-defined and converges to
$$S(0)G(t-s)f(u(s))\big|_{s=t}=f(u(t)).$$
A: I will write $f$ instead of $f\circ u$. You have not said what $u$ is, but I will assume it to be continuous.
You have made a mistake, the differential quotient is:
$$\frac1h\left(\int_0^{t+h}S(t+h-s)f(s)\,ds - \int_0^t S(t-s)f(s)\,ds\right) \\= \int_0^t\frac{S(t+h-s)-S(t-s)}hf(s)\,ds+\frac1h\int_t^{t+h}S(t+h-s)f(s)\,ds$$
Although this is only true if $h>0$. If $h<0$ one must reduce to
$$\int_0^{t-|h|}\frac{S(t-|h|-s)-S(t-s)}{-|h|}f(s)\,ds+\frac1{-|h|}\int_{t-|h|}^tS(t-s)f(s)\,ds$$
instead because otherwise the terms are not well defined. I'll only consider the quotient for $h>0$, the other case should be analog.
Now the interior of the first sum is, if $h>0$: $\frac{S(h)-1}hS(t-s)f(s)$ and the $\frac{S(h)-1}h$ term can be pulled out of the integral as you correctly noted.
For the other term:
$$\left\|\frac1h\int_t^{t+h}S(t+h-s)f(s)-f(t)\,ds\right\|≤\frac hh\sup_{s\in[t,t+h]}\|S(t+h-s)f(s)-f(s)+f(s)-f(t)\|\\
≤\sup_{s\in[t,t+h]}\|S(t+h-s)-1\|\cdot\sup_{s\in[t,t+h]}\|f(s)\|+\sup_{s\in[t,t+h]}\|f(s)-f(t)\|$$
From continuity of the functions this goes to $0$ as $h\to0$.
Put everything together and you get the derivative stated.
