Proving an identity of the resolvent set $\{T(t)\}_{t\ge 0}$ is a $C_{0}$-semigroup with infinitesimal operator $A$.
I'm trying to prove that the set $\{ z|\text{ Re }z>\omega_{0}\}$ belongs to $\rho(A)$, and for $z$ in this set, the identity $R(z)=(zI-A)^{-1}$ holds; where $\omega_{0}=\inf{\{\omega|\exists M_{\omega}\ge 1\text{ such that }\|T(t)\|\le M_{\omega}e^{\omega t}\}}$
I am given that $R(z)x=\int_{0}^{\infty}e^{-zs}T(s)x ds$ and this integral exists $\forall z\in\mathbb{C}$.
I define $$\tilde{R}(z)x=\int_{0}^{t}e^{-zs}T(s)xds$$
Then, for $h>0$, $x\in X$, consider
$$\frac{1}{h}(T(h)-I)\tilde{R}(z)x=\frac{1}{h}(T(h)-I)\int_{0}^{t}e^{-zs}T(s)xds$$
After some more formula manipulation, I want to let $h\downarrow 0$ and $t\to\infty$; and also use the fact that $A$ is closed.
I know that
$$\frac{1}{h}\int_{t}^{t+h}T(s)xds\to T(t)x\text{ as $h\downarrow 0$} $$
But I don't know how to deal with the $T(h)-I$.
EDIT: After consulting some literature it seems that this could be the next step, $$\frac{1}{h}(T(h)-I)\int_{0}^{t}e^{-zs}T(s)x ds=\frac{1}{h}\left\{\int_{0}^{t}e^{-zs}[T(s+h)x-T(s)x]ds\right\}$$
Is the above correct, and if so, could someone offer an explanation of how the equality is derived?
 A: $$\begin{aligned}
\frac{1}{h}(T(h)-I)\tilde{R}(z)x
  &=\frac{1}{h}(T(h)-I)\int_{0}^{t}e^{-zs}T(s)xds \\
  &=\frac{1}{h}\int_{0}^{t}e^{-zs}T(s+h)xds-\frac{1}{h}\int_{0}^{t}e^{-zs}T(s)xds \\
  &=\frac{1}{h}\int_{h}^{t+h}e^{-z(s-h)}T(s)xds-\frac{1}{h}\int_{0}^{t}e^{-zs}T(s)xds\\
  &=e^{zh}\frac{1}{h}\int_{t}^{t+h}e^{-zs}T(s)xds-\frac{1}{h}\int_{0}^{h}e^{-zs}T(s)xds \\
  &+\frac{e^{zh}-1}{h}\int_{h}^{t}e^{-zs}T(s)xds
\end{aligned}$$
Added: The steps for the last equality:
\begin{align}
  \frac{1}{h}\int_{h}^{t+h}e^{-z(s-h)}T(s)xds & = \frac{e^{zh}}{h}\int_{h}^{t}z^{-zs}T(s)xds+\frac{e^{zh}}{h}\int_{t}^{t+h}e^{-zs}T(s)xds \\
 \frac{1}{h}\int_{0}^{t}e^{-zs}T(s)xds & = \frac{1}{h}\int_{h}^{t}e^{-zs}T(s)xds+\frac{1}{h}\int_{0}^{h}e^{-zs}T(s)xds
\end{align}
Subtract the second equation from the first, and you get the final equality.
Now, when you take the limit as $h\downarrow 0$, you get
$$
     A\tilde{R}(z)x = e^{-zt}T(t)x-x+z\int_{0}^{t}e^{-zs}T(s)xds\\
      (A-zI)\tilde{R}(z)x = e^{-zt}T(t)x-x
$$
If $\Re z$ is large enough, then you can take the limit of the above and use the fact that $A$ is closed in order to conclude that
$$
         (A-zI)\int_{0}^{\infty}e^{-zs}T(s)xds = -x.
$$
You could have just started with the integral over $[0,\infty)$ from the beginning, though, and that would have simplified the expressions a bit.
