# Integral operator (compactness)

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.

• 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. Sep 29, 2022 at 21:08
• Here you go. Sep 29, 2022 at 21:14
• Many thanks! That's what I was looking for Sep 29, 2022 at 21:48
• 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. Oct 14, 2022 at 22:01
• 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. Oct 14, 2022 at 22:06