1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

How does the lemma follow by the Hahn-Banach Theorem?

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.