A semigroup $S(t)$ on a Banach space $E$ is a family of bounded linear operators $\{S(t)\}_{t\ge 0}$ with the property that $S(t)S(s)=S(t+s)$ for any $s,t\ge 0$ and that $S(0)=I$. A semigroup is furthermore called strongly continuous if the map $(x,t)\mapsto S(t)x$ is continuous.
Now I want to show the following
Being strongly continuous, i.e. $(x,t)\mapsto S(t)x$, is equivalent to $t\mapsto S(t)x$ being continuous at $t=0$ for every $x\in E$.
The operator norm of $S(t)$ being bounded by $M\,e^{at}$ for some constants $M$ and $a$.
Show then that the first condition can be relaxed to $t\mapsto S(t)x$ being continuous for all $x$ in some dense subset of $E$.
For my 1), if $S(t)$ to be strongly continuous, it follows that $t\mapsto S(t)x$ is continuous for all $x$, one can tell this by looking at both left and right sided continuity in the following manner:
let $x\in X$, $t>0$. Then $T(t+h)x-T(t)x=T(t)(T(h)x-x)\to 0$ as $h\to 0^{+}$, then \begin{equation*} \|T(t+h)x-T(t)x\|\le \left(\sup_{0\le s\le t}\|T(s)\|\right)(T(h)x-x)\to 0 \end{equation*} as $h\to 0^+$. This proves the right-sided continuity of $T(\cdot)x$ of $T(\cdot)x$.
In order to prove the left-sided continuity, we let $-t\le h<0$ and write $T(t+h)x-T(t)x=T(t+h)(x-T(-h)x).$ Then one obtains \begin{equation*} \|T(t+h)x-T(t)x\|\le \left(\sup_{0\le s\le t}\|T(s)\|\right)(x-T(-h)x)\to 0 \end{equation*} as $h\to 0^-$.
However, I am lost on how the join continuity of $(x,t)\mapsto S(t)x$ could define strong continuity. By this I mean, if $S(t)$ is strongly continuous, then $(x,t)\mapsto S(t)x$ is strongly continuous, but I cannot write down like the above.So $(x,t)\mapsto S(t)x$ is continuous for???
Please help providing a complete answer for 1).
Thanks