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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?

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    $\begingroup$ Any operator of the form $I - B$ with $\|B\| < 1$ is invertible. So $G$ will be invertible if $\int_0^t \|T(t-s)A(s)\|\, ds \leq \alpha < 1$ for all $t$. I don't think the fact that $T$ is a semigroup has any particular bearing on the question. $\endgroup$
    – Nik Weaver
    Commented Apr 5, 2019 at 14:52
  • $\begingroup$ What if $\|B\|>1$, then it's never invertible? $\endgroup$
    – Saj_Eda
    Commented Apr 5, 2019 at 14:59
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    $\begingroup$ Of course not. You can find an easy counterexample to that statement. $\endgroup$
    – Nik Weaver
    Commented Apr 5, 2019 at 15:12

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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.

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  • $\begingroup$ Very nice argument! $\endgroup$
    – Nik Weaver
    Commented Apr 5, 2019 at 16:33
  • $\begingroup$ @Jochen can we explicit the inverse in case of semigroups? and can we get a like invertible operator on $X$. Thanks in advance. $\endgroup$
    – S. Maths
    Commented Jul 23, 2019 at 0:38
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    $\begingroup$ @S.Maths: We can compute the inverse of $G = \operatorname{id} - B$ by means of the von Neumann series, but I think in most cases it will not be possible to explicitly compute the powers of $B$ that occur in this series, even if $T$ is a semigroup. Concerning your second question, I'm not sure what you mean by "a like invertible operator on $X$". $\endgroup$
    – Jochen Glueck
    Commented Jul 24, 2019 at 19:54
  • $\begingroup$ @Jochen, Thanks. I calculated $B^2$ in terms of $B$ but I don't know if this help. For the last question, I mean $G$ is invertible in $\mathcal{L}(C(0,\tau;X))$, can we deduce an invertible operator in $\mathcal{L}(X)$. $\endgroup$
    – S. Maths
    Commented Jul 24, 2019 at 20:55
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    $\begingroup$ @S.Maths: Well, to find an explicit formula for all powers of $B$ - even for the case where $T$ is a semigroup and $A$ is constant - would imply to have an explicit formula (whatever "explicit" means) for all terms that occur in the Dyson-Phillips series for perturbed semigroups. I think this is much more than what we can expect. Concerning your second question again, I still have difficulties to follow; what does it mean to "deduce" an operator? How should this invertible operator on $\mathcal{L}(X)$ that you seek be related to $G$? $\endgroup$
    – Jochen Glueck
    Commented Jul 24, 2019 at 21:25

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