I'm stuck with the following claim:

Let $(T(t))_{t \geq 0}$ be a $C_0$ semigroup with generator $A$. The generator $A$ is not necessarily bounded. Let $s>0$ such that the range of $T(s)$ is dense in $D(A^\infty)$. Define an operator $S_n$ for $x \in D(A^\infty)$ by \begin{equation} S_n x = \int_0^t (t-s)^n A^{n+1} T(s) x ds \end{equation} Then $S_n$ is bounded: $\Vert S_n \Vert \leq (n+1)^{-1} \Vert t^{n+1} A^{n+1} T(s)\Vert$.

I guess the whole claim reduces to showing that $\Vert A^{n+1} T(u) \Vert \leq \Vert A^{n+1} T(s) \Vert$ for all $u \in [0,t]$, which then yields the bound. Is it necessary to assume that the semigroup is analytic to get there? What is the significance of the range condition?

Thanks in advance for your help!


  • 1
    $\begingroup$ Do you have $s$ and $t$ mixed up in the statement? $\endgroup$ – Keith McClary Jun 17 '17 at 4:15
  • $\begingroup$ No, this is how it is stated in my reference. $\endgroup$ – ChrisT Jun 17 '17 at 7:50

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