# Proof of an analogue of the Fundamental Theorem of Calculus for infinitesimal generators.

This is a proof of the below proposition from Davies' One-Parameter Semigroups.

If $f\in Dom(Z)$, where $Z$ is the infinitesimal generator, then $$T_tf-f=\int_0^t T_x Zf dx.$$

Proof. If $f\in Dom(Z)$ and $\phi$ lies in the Banach dual space $B^*$ of $B$, we define the complex-valued function $F(t)$ by $$F(t)=\left\langle T_tf-f-\int_0^t T_x Zfdx, \phi\right\rangle.$$

Its right hand derivative $D^+F(t)$ is given by $$D^+F(t)=\langle ZT_tf-T_tZf, \phi\rangle=0.$$

Since $F(0)=0$ and $F$ is continuous, we see that $F(t)=0$ for all $t\in [0,\infty)$. Since $\phi\in B^*$ is arbitrary, the lemma follows by an application of the Hahn-Banach Theorem.

My questions are: First, why do we have $F(t)=0$ for all $t$ by the continuity of $F$ and $F(0)=0$? Second, how does the lemma follow by the Hahn-Banach Theorem? I would greatly appreciate any help to understand these points.

Why do we have $F(t)=0$ for all $t$ by the continuity of $F$ and $F(0)=0$?
Because in the previous step you proved that $D^+F=0$. This result with the continuity of $F$ implies that $F$ is constant (see here the proof).
As a consequence of the Hahn-Banach Theorem, the dual of a Banach space $B$ separates points of $B$. Thus, if the equality were not true, there would be $\phi\in B^*$ such that $$\left\langle T_tf-f-\int_0^t T_x Zfdx, \phi\right\rangle\neq 0.$$ But, as proved, this is not the case.