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I am studying the compactness of some convolution operators. Let the convolution $$ \Gamma: X\longrightarrow X; x\mapsto\int_0^t T(t-s)B(s)x\mathrm{d}s. $$ Here $T(\cdot)$ is a $C_0$-semigroup on some Banach space $X$, $B(s)\in\mathcal{L}(X)$ is a compact operator for all $s\in [0,t]$ and $t\ge 0$.

$\star$ If $B(s)=B$ are independent of $s$, it is not difficult to prove the compactness of $\Gamma$ ( I did it using Riemann summation)

$\star$ I have a positive answer if $s\mapsto B(s)$ is immediately continuous ( continuous in the norm of $\mathcal{L}(X)$).

${\color{blue}\star}$ Now if $s\mapsto B(s)$ is strongly continuous what can we say about the compactness in this case?

${\color{red}\star}$ Another problem which I am facing right now is if we consider $$ \Gamma: X\longrightarrow X; x\mapsto\int_0^t T_{-1}(t-s)B(s)x\mathrm{d}s. $$ This time $B(s)$ arrives in the Favard space $\mathbb{F}_1$ associated with the $C_0-$semigroup (so that the convolution makes sense). $T_{-1}$ is the extrapolated semigroup. Do we still recover compactness?

Many thanks.

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    $\begingroup$ Integrals over strongly continuous functions with values in the compact operators are compact. That's a result by Jürgen Voigt (from the 80s or 90s) if I remember correctly. $\endgroup$ Sep 29, 2022 at 21:08
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    $\begingroup$ Here you go. $\endgroup$ Sep 29, 2022 at 21:14
  • $\begingroup$ Many thanks! That's what I was looking for $\endgroup$ Sep 29, 2022 at 21:48
  • $\begingroup$ Please note that if you have an additional question, the recommended procedure on MathOverflow is to ask a new question rather than to edit an existing one. $\endgroup$ Oct 14, 2022 at 22:01
  • $\begingroup$ I copied my comments that answer the original question into an answer box, as comments are not archived (at least not publically available) in the post history. $\endgroup$ Oct 14, 2022 at 22:06

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Integrals over strongly continuous functions with values in the compact operators are compact. That's a result by Jürgen Voigt.

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