Let $E$ be a locally compact Hausdorff space, $A$ be a bounded linear operator on the space $B(E)$ of bounded functions $E\to\mathbb R$ and $$T(t)f:=e^{tA}f\;\;\;\text{for }f\in B(E)\text{ and }t\ge0,$$ where $e^{tA}:=\sum_{n=0}^\infty\frac{t^n}{n!}A^n$.
If $f\in C_b(E)$ and $t\ge0$, can we show that $T(t)f\in C_b(E)$? What if $f\in C_0(E)$, can we conclude $T(t)f\in C_0(E)$?
I wasn't able to prove the claim, but I can't find a counter-example.
What do you mean by $B(E)$, is this space of all bounded functions or borel measurable bounded functions?
Nevertheless, I think the claim is false: consider $E=[0,2]$, and $A\colon B([0,2])\ni f\mapsto \chi_{[0,1]} f\in B([0,2])$, where $\chi_{[0,1]}$ is the characteristic function of $[0,1]$. Then $e^{tA}f=f+\sum_{n=1}^{\infty} \frac{t^n}{n!} \chi_{[0,1]}^{n} f=f+(e^{t}-1) \chi_{[0,1]}f$ which is not continuous for $t>0$ and $f(x)=x$.
The problem is that you don't assume that $A$ preserves $C_b(E)$.