Why does the semigroup commute with integration?

Question

I have a question about Theorem 7.4.2 in Evan's PDEs book. If $S(t)$ is a contraction semigroup on a Banach space $X$. He uses $$S(r)\int_0^t S(s)u\,ds = \int_0^t S(r+s)u\,ds$$ and I don't understand how $S(r)$ and integration commute. Could anyone explain why this is true?

Answer 1

Because $S$ is continuous, the following vector limit exists as a Riemann integral: $$ \int_{0}^{t}S(s)u ds = \lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{\mathscr{P}}S(s_j^{\star})u\Delta s_j. $$ Because $S(r)$ is a bounded linear operator, then $S(r)$ is continous, which gives $$ \begin{align} S(r)\int_{0}^{t}S(s)uds & = S(r)\left(\lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{\mathscr{P}}S(s_j^{\star})u\Delta s_j\right) \\ & =\lim_{\|\mathscr{P}\|\rightarrow 0}S(r)\sum_{\mathscr{P}}S(s_j^{\star})u\Delta s_j \\ & =\lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{\mathscr{P}}S(r)S(s_j^{\star})u\Delta s_j \\ & =\lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{\mathscr{P}}S(r+s_j^{\star})u\Delta s_j \\ & = \int_{0}^{t}S(r+s)uds \end{align} $$

Answer 2

This is a particular case of the following result due to Hille.

Theorem. Let $A:D(A)\subseteq X\to Y$ be a linear closed operator and let $x : \Omega\to D(A)$. Then $$\int_{\Omega} A x(\omega)\;d\mu(\omega)=A\int_{\Omega} x(\omega)\;d\mu(\omega)$$ whenever both sides of the above equality are well-defined.

Proof: see Vrabie's book (Theorem 1.2.2, page 8) or Diestel's book (Theorem 6, page 47) or Hille's book (Theorem 3.7.12, page 83).