Justify the equality $$A \int_0^\infty e^{-\lambda t} S(t) u \, dt = \int_0^\infty e^{-\lambda t} AS(t) u \, dt$$ used in (16) of §7.4.1. (Hint: Approximate the integral by a Riemann sum and recall $A$ is a closed operator.)
PDE by Evans, 2nd edition: Chapter 7, Exercise 13
The properties of $A$ being a closed operator are outlined in this remark on page 436:
Remark. To say $A$ is closed means that whenever $u_k \in D(A)$ ($k=1,\ldots$) and $u_k \to u$, $Au_k \to v$ as $k \to \infty$, then $$u \in D(A), \quad v = Au.$$
Now, when I tried recalling the Riemann sum definitions for both expressions, I am probably supposed to show that $$A \sum_{k=1}^\infty e^{-\lambda t_k^*} S(t_k^*) u \Delta t = \sum_{k=1}^\infty e^{-\lambda t_k^*} AS(t_k^*) u \Delta t $$ But how can I use the properties of the closed operator $A$ to justify this equality?
Set
$$u_k := \sum_{j=1}^k e^{-\lambda t_j^*} S(t_j^*)u \Delta t.$$
Since $u \in D(A)$ implies $S_t u \in D(A)$ for any $t \geq 0$, we have $u_k \in D(A)$. Moreover, by the linearity of $A$,
$$Au_k = \sum_{j=1}^k e^{-\lambda t_j^*} AS(t_j^*)u \Delta t.$$
Moreover, we know that $u_k$ converges to
$$u := \int_0^{\infty} e^{-\lambda t} S(t) u \, dt.$$
Exactly the same reasoning shows that
$$Au_k \to v:= \int_0^{\infty} e^{-\lambda t} A S(t) u \, dt.$$
As $A$ is a closed operator, it follows that $Au = v$, i.e.
$$A \int_0^{\infty} e^{-\lambda t} S(t) u \, dt = \int_0^{\infty} e^{-\lambda t} A S(t) u \, dt.$$