Showing that the generator of the semigroup $(e^{tA})_{t\in\mathbb{R}}$ is $A$.

Question

Hello. I am studying semigroup theory. In "one-parameter semigroups of positive operators" it is stated that $A$ is the generator. Why?

Given the definition, it should be $$Af=\lim_{h\to 0} \frac{e^{hA}f-f}{h}$$ but $\displaystyle \lim_{h\to 0} \frac{e^{hA}f-f}{h}=0$ for all $f$

Why is $A$ the generator?

Answer 1

Thanks for IAN

$$\lim_{h\to 0} \frac{ (e^{hA}-I)f}{h}=\lim_{h\to 0} \frac{(hA+(h^2A^2)/2!+\cdots)f}{h}=Af$$