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Let

  • $E$ be a $\mathbb R$-Banach space
  • $S:[0,\infty)\to E$ be a $C^0$-semigroup
  • $T>0$
  • $f\in C^0([0,T],E)$

Since $S$ is a $C^0$-semigroup, $$[0,\infty)\to E\;,\;\;\;t\mapsto S(t)x\tag1$$ is continuous for all $x\in E$. However, unless $S$ is even uniformly continous, this shouldn't imply the continuity of $$[0,t]\to E\;,\;\;\;s\mapsto S(t-s)f(s)\tag2$$ for all $t\in(0,T]$.

However, how is then the fundamental theorem of calculus used in equation (12.28) of the book An Introduction to Partial Differential Equations by Michael Renardy and Robert C. Rogers?

Excerpt of the mentioned book (the authors use the symbol $T$ for both the maximal time and the semigroup $S$):

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I think everything is okay. Because $S$ is uniformly bounded in operator norm on any finite interval $[0,T]$, then \begin{align} & \|S(t-s)f(s)-S(t-s')f(s')\| \\ &= \|\{S(t-s)-S(t-s')\}f(s)+S(t-s')\{f(s)-f(s')\}\| \\ &\le \|\{S(t-s)-S(t-s')\}f(s)\|+M\|f(s)-f(s')\|. \end{align} Hence, $$ \lim_{s'\rightarrow s}S(t-s')f(s')=S(t-s)f(s). $$

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    Do you see the reason why the authors are defining a classical solution to be in $C^0([0,T],\mathcal D(A))$ (instead of $C^0([0,T],\mathcal D(A))$)? – 2017-01-20
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    @0xbadf00d : In the proof of the theorem, he's deriving the necessity of the integral solution. There he needs $C^1([0,T],X)\cap C^0([0,T],\mathcal{D}(A))$ to carry out the operations needed to derive the necessity of the stated form of solution. At least that's what it looks like at first glance. – 2017-01-20