I am self-studying Pazy's book "Semigroups of Linear Operators and Applications to PDE's." In Theorem 2.6, $T(t)$ and $S(t)$ are $C_{0}$-semigroups, $A$ generates $T(t)$, and $B$ generates $S(t)$. Moreover, $(A,D(A))=(B,D(B))$. We are supposed to show that $T(t)=S(t)$ for all $t\geq 0$.
Pazy's strategy is basically to let $x\in D(A)=D(B)$, define the function $s\mapsto T(t-s)S(s)x$, show that this function is constant, and then make use of the fact that $D(A)=D(B)$ is dense in overall space, $X$. In taking the derivative of the defined function, we have:
$\frac{d}{ds}\Big{[}T(t-s)S(s)x\Big{]}= -AT(t-s)S(s)x+T(t-s)BS(s)x$
Intuitively, this theorem should clearly be true. My question is simply how does Pazy arrive at this derivative. It looks like he's using the product rule, and I know I'm just missing something really simple, but could someone point me in the right direction here?
Yes, he is using the product rule and you can use it in the same way it is proven in classical calculus: \begin{align*} \frac{d}{ds} [T(t-s)S(s)x] &= \lim_{h \rightarrow 0} \frac{1}{h}[ T(t-(s+h))S(s+h)x - T(t-s)S(s) ] \\ &= \lim_{h \rightarrow 0} \frac{1}{h}[ T(t-(s+h))S(s+h)x \\&- T(t-s)S(s+h)x + T(t-s)S(s+h)x - T(t-s)S(s) ] \\ &= \lim_{h \rightarrow 0} \frac{1}{h}[ T(t-(s+h)) \\&- T(t-s)]S(s+h)x + T(t-s)\lim_{h \rightarrow 0}\frac{1}{h}[S(s+h) - S(s) ] x . \end{align*} The first term in square brackets converges to $-AT(t-s)$, the second one to $BS(s)$. That this is true follows from the semigroup property: $$ T(t-(s+h)) - T(t-s) = [I - T(h)]T(t-s-h) = - [T(h) - I]T(t - s - h). $$ Upon multiplying by $S(s+h)x$, dividing by $h$ and sending $h \rightarrow 0$, the first term converges to A, and the second, due to strong continuity, to $T(t-s)S(s)$.
Hope that helps!