Definition: Let $X$ be a Banach space and $I$ the identity operator on $X$. A family $\{T(t)\}_{t\geq 0}$ of bounded linear operators from $X$ into $X$ is a semigroup of bounded linear operator on $X$ if
(i) $T(0)=I$;
(ii) $T(t + s)= T(t)T(s)$ for every $t,s\geq 0$.A semigroup $\{T(t)\}_{t\geq0}$ of bounded linear operators is uniformly continuous if $$\lim_{t\downarrow 0}\|T(t)-I\|=0.\;\;\;\;(*)$$
The Pazy's book says "from the definition it's clear that if $\{T(t)\}_{t\geq 0}$ is an uniformly continuous semigroup of bounded linear operators then $\lim_{s\to t}\|T(s)-T(t)\|=0$" (page 1).
For me, this equality it's not so clear. Could someone help me to prove it? I'm trying to show that $$\lim_{s\to t^+}\|T(s)-T(t)\|=\lim_{s\to t^-}\|T(s)-T(t)\|=0.$$
If I'm not wrong, from the definition we can conclude that, for all $h>0$,
$$0\leq \|T(t+h)-T(t)\| \overset{(ii)} =\|T(t)T(h)-T(t)\| \leq \|T(t)\|\|T(h)-I\|$$
Now, notice that
$$\lim_{h\to 0^+}\|T(t)\|\|T(h)-I\|=\|T(t)\|\lim_{h\downarrow 0}\|T(h)-I\| \overset{(*)}=\|T(t)\|\;0=0$$
Thus (by Squeeze theorem),
$$\lim_{s\to t^+}\|T(s)-T(t)\| =\lim_{h\to 0^+}\|T(t+h)-T(t)\| = 0$$
My question is: how to prove that $\lim_{s\to t^-}\|T(s)-T(t)\|=0$?
Thanks.
Since $\Vert T(t)-I\Vert\to 0\,$ as $t\to 0^+$, one can find $\delta >0$ such that $\Vert T(t)\Vert\leq 2$ for all $t\in [0,\delta]$. By the semigroup property, it follows that $(T(t))$ is uniformly bounded on any compact interval $[0,A]$. Indeed, if $n\in\mathbb N$ is such that $n\delta>A$, then $\Vert T(t)\Vert\leq 2^n$ on $[0,n\delta]$ and hence on $[0,A]$.
Now, fix $t_0>0$, and choose $C$ such that $\Vert T(t)\Vert\leq C$ on $[0,t_0]$. For any $t\in [0, t_0]$ you can write $T(t)-T(t_0)=T(t)\left (I-T(t_0-t)\right)\, ,$ so that $$\Vert T(t)-T(t_0)\Vert\leq \Vert T(t)\Vert\, \Vert I-T(t_0-t)\Vert\, \leq C\, \Vert I-T(t_0-t)\Vert\, .$$ Since $T(t_0-t)\to I$ as $t\to t_0^-$, this shows continuity on the left at $t_0$.