$C_0$ Semigroups with same infinitesimal generator

Question

I was dealing with this:

Let $S(t)$ and $T(t)$ be $C_0$-semigroups with infinitesimal generators $A:D(A) \subset X \rightarrow X$ and $B:D(B) \subset X \rightarrow X$ respectively. Show that \begin{equation} A = B \hspace{5mm} \Rightarrow \hspace{4mm}S(t) = T(t) \hspace{5mm}\forall t \ge 0 \end{equation}

I've already shown that the infinitesimal generator coincides with derivate (right and left). So I think that to prove this it would be sufficient to show something like

$\frac{d}{dt}S(t)u = \frac{d}{dt}T(t)u \hspace{7mm} \forall u \in X, \forall t \ge 0$

I would just need some help to formalize this..

Answer 1

Hints:

1. Recall (or show) that $S(t)f \in D(A)$ and $$\frac{d}{dt} S(t) f = AS(t)f$$ for any $t \geq 0$ and $f \in D(A)$. (An analogous statement holds for $T(t)$ and $B$.)
2. For fixed $f \in D(A)=D(B)$ and $t>0$ show that $$[0,t] \ni u \mapsto S(u)T(t-u)f$$ is differentiable and that the derivative equals zero.
3. Apply the fundamental theorem of calculus to deduce that $S(t)f=T(t)f$ for all $f \in D(A)$.
4. Use the fact that $D(A)$ is dense in $C_0$ to conclude that $S(t)f=T(t)f$ for all $f \in C_0$.

Answer 2

IF $T(t)$ is a bounded $C_0$ semigroup on a Banach space $X$, with generator $A$, then $$ (\lambda I-A)^{-1}x=\int_0^{\infty}e^{-\lambda t}T(t)x dt,\;\;\;\Re\lambda >0,\; x\in X. $$ You can use this to relate the resolvents of $A$ and $B$ to each other, which leads to the desired conclusion.