If $(T_t)$ is a semi group, why $\lim_{s\to 0}T_t\frac{T_su-u}{s}=T_t\lim_{s\to 0}\frac{T_su-u}{s}$?

Question

Let $(T_t)_t$ is a continuous semi group (i.e. $\lim_{t\to 0}T_tu=u$ for all $u$). $T_t$ is a contraction for all $t$. We suppose that $\lim_{t\to 0}\frac{T_tu-u}{t}$ exist. Why $$\lim_{s\to 0}T_t\frac{T_su-u}{s}=T_t\lim_{s\to 0}\frac{T_su-u}{s}\ \ ?$$ i.e. why we can permute $T_t$ and the limit ? I know that $t\mapsto T_t$ is continuous on $[0,\infty )$ but there is a priori no reason that $T_t$ is continuous (i.e. if $u_n\to u$, there is no reason that $\lim_{n\to \infty }T_tu_n=T_tu$)

Answer 1

I assume you are studying one-parameter semigroups of operators. As you say, $T_t$ is a contraction, hence $\|T_tx\|\leq\|x\|$ (or $<$, or $\leq c\|\cdot\|$ with $c\in(0,1)$, depending on your definition of contraction). This is enough to obtain $T_t$'s continuity: if $(x_n)$ converges to $x$, then $\|T_tx_n-T_tx\|=\|T_t(x_n-x)\|\leq\|x_n-x\|\to0$. Now as $s\to0^+$, the element $\displaystyle{\frac{T_su-u}{s}}$ of your (Banach, I presume) space $X$ converges to something, by the infinitesimal Generator. Therefore the limit passes inside the argument of $T_t$, by its proven continuity.