Is this operator invertible? Let $T(t)$ be a strongly continuous semi-group on a Banach space $X$, and let $A(\cdot)\in C(0,\tau; \mathcal{L}(X))$ for some $\tau>0$. The operator $G:C(0,\tau;X)\to C(0,\tau;X)$ maps every $h\in C(0,\tau;X)$ to
$$(Gh)(t)=h(t)-\int_0^tT(t-s)A(s)h(s)ds.$$
Is $G$ invertible? What are the conditions on $T(t)$ and $A(s)$ to ensure that $G$ is invertible?
 A: The operator $G$ is always invertible - and this is true for general strongly continuous mappings $T: [0,\tau] \to \mathcal{L}(X)$, no matter whether they fulfil the semigroup law.
Proof. As indicated in Nik Weaver's comment, we write $G$ as $G = \operatorname{id} - B$, where
\begin{align*}
  (Bh)(t) = \int_0^t T(t-s) A(s) h(s) \, ds.
\end{align*}
(One might call $B$ a "vector-valued Volterra integral operator".) We show that the spectral radius of $B$ is $0$.
To this end, let $C := \sup_{t \in [0,\tau]} \|T(t)\|_{\mathcal{L}(X)}$ and $D = \sup_{t \in [0,\tau]} \|A(t)\|_{\mathcal{L}(X)}$ and note that $C,D < \infty$ by the uniform boundedness theorem.
For every function $h \in C(0,\tau;X)$ one easily checks by induction that
\begin{align*}
  \|(B^n h)(t)\|_X \le (CD)^n \frac{t^n}{n!} \|h\|_{C(0,\tau;X)}
\end{align*}
for each integer $n \ge 0$ and each $t \in [0,\tau]$. Hence,
\begin{align*}
  \|B^n\|_{\mathcal{L}(C(0,\tau;X))} \le (CD)^n \frac{\tau^n}{n!}
\end{align*}
for each $n$ and consequently, the spectral radius $r(B)$ fulfils
\begin{align*}
 r(B) = \lim_{n \to \infty} \|B^n\|_{\mathcal{L}(C(0,\tau;X))}^{1/n} \le \lim_{n \to \infty} \frac{CD\tau}{(n!)^{1/n}} = 0.
\end{align*}
This proves the assertion.
Remark. The above proof is very close to the standard proof for the Picard–Lindelöf theorem for ODEs.
