# Proof of the integral representation of the resolvent

Let

• $$E$$ 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?