If $T(t)$ is a semigroup and $x$ satisfies $\left\|T(h)x-x\right\|\to0$ as $h\to0$, are we able to conclude that $t\mapsto T(t)x$ is continous?

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Let $E$ be a $\mathbb R$-Banach space and $(T(t))_{t\ge0}$ be a semigroup on $E$.

Definition:

  1. If $x\in E$, then $(T(t))_{t\ge0}$ is called strongly continuous at $x$ iff $$[0,\infty)\to E\;,\;\;\;t\mapsto T(t)x\tag1$$ is continuous
  2. $(T(t))_{t\ge0}$ is called locally bounded$^1$ iff $$\exists t>0:\sup_{s\in[0,\:t)}\left\|T(s)\right\|_{\mathfrak L(E)}<\infty\tag2$$

Question: Let $x\in E$ with $$\left\|T(h)x-x\right\|_E\xrightarrow{h\to0+}0\tag4.$$ Are able to show that $(T(t))_{t\ge0}$ is strongly continuous at $x$?

It's easy to see that $(1)$ is right-continuous. Moreover, I'm able to show that $(1)$ is left-coninuous, as long as I assume that $(T(t))_{t\ge0}$ is locally bounded.

Are we able to drop the local boundedness assumption and still conclude the left-continuity?

My idea is as follows: By $(4)$, $$\exists t>0:\forall h\in[0,t):\left\|T(h)x\right\|_E<1+\left\|x\right\|_E\tag5$$ and maybe we are able to show local boundedness by the uniform boundedness principle in some way.


$^1$ By the semigroup property, $(1)$ implies that $$\sup_{s\in I}\left\|T(s)\right\|_{\mathfrak L(E)}<\infty$$ for each bounded interval $I\subseteq[0,\infty)$.