Question about derivative in Pazy's Semigroup book

Question

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?

Answer 1

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!