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.