• $E$ be a $\mathbb R$-Banach space
  • $(T(t))_{t\ge0}$ be a $C^0$-semigroup on $E$
  • $(\mathcal D(A),A)$ denote the infinitesimal generator of $(T(t))_{t\ge0}$
  • $A_\lambda:=\lambda\operatorname{id}_{\mathcal D(A)}-A$ for $\lambda\in\mathbb R$
  • $\rho(A)$ denote the resolvent set of $(\mathcal D(A),A)$
  • $R_\lambda(A):=A_\lambda^{-1}$ for $\lambda\in\rho(A)$

Now, let $$\tilde\rho:=\left\{\lambda\in\mathbb R:\int_0^\infty e^{-\lambda t}T(s)x\:{\rm d}s\text{ exists for all }x\in E\right\}$$ and $$\tilde R_\lambda x:=\int_0^\infty e^{-\lambda s}T(s)x\:{\rm d}s\;\;\;\text{for }x\in E$$ for $\lambda\in\tilde\rho$.

I want to show that $$\tilde\rho\subseteq\rho(A)\tag1$$ and $$R_\lambda(A)=\tilde R_\lambda\;\;\;\text{for all }\lambda\in\tilde\rho\tag2.$$

So, let $\lambda\in\tilde\rho$. I was able to prove the claim assuming $\lambda=0$. How are we able to conclude the general case by a rescaling argument?

Clearly, we can define $$S(t):=e^{\mu t}T(\alpha t)\;\;\;\text{for }t\ge0$$ for $\mu\in\mathbb R$ and $\alpha>0$ to obtain a semigroup similar to $(T(t))_{t\ge0}$.

I guess we simply need to choose $(\mu,\alpha)=(\lambda,1)$, but why does this yield the claim for the general case?


Assume that $y=\int_{0}^{\infty}e^{-\lambda s} T(s)xds$ converges as an improper integral for a given $x$. Then $$ \frac{1}{h}(T(h)-I)y=\frac{1}{h}(T(h)-I)\int_{0}^{\infty}e^{-\lambda s}T(s)xds \\ = \frac{1}{h}\int_{0}^{\infty}e^{-\lambda s}T(s+h)x-e^{-\lambda s}T(s)xds \\ = \frac{e^{\lambda h}}{h}\int_{h}^{\infty} e^{-\lambda s}T(s)xds-\frac{1}{h}\int_{0}^{\infty}e^{-\lambda s}T(s)x ds \\ = \frac{e^{\lambda h}-1}{h}\int_{h}^{\infty}e^{-\lambda s}T(s)xds+\frac{1}{h}\int_{0}^{h}e^{-\lambda s}T(s)xds. $$

The limit as $h\downarrow 0$ exists, and that limit is

$$ \lambda\int_{0}^{\infty}e^{-\lambda s}T(s)xds+x. $$

Consequently, $y\in \mathcal{D}(A)$ and

$$ Ay = \lambda y+x, \\ (A-\lambda I)\int_{0}^{\infty}e^{-\lambda s}T(s)xds = x. $$ Can you take it from there?


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