I try to solve the following problem:
Let $X$ locally compact and $a \in C(X)$ such that $\text{Re } a \leq w$. Further let $T(t)f := e^{ta}f$ for all $f \in C_0(X)$ and $t \geq 0$. Determine the generator of $(T(t))_{t \geq 0}$.
I figured out that the generator must be the multiplication operator of $a$ defined by $M_a f = a f$ for all $f \in C_0(X)$. So I need to show that
$$ \lim_{t \to 0} \left\Vert \frac{T(t)f - f}{t} - M_a f \right\Vert_\infty = 0 \qquad \text{for all } f \in C_0(X).$$ So let $\epsilon > 0$, $(t_n)_{n \in \mathbb N}$ in $\mathbb R_+$ with $t_n \to 0$ and $f \in C_0(X)$. So I need a $N \in \mathbb N$ with
$$ \left\Vert \frac{T(t_n)f - f}{t} - M_a f \right\Vert_\infty = \sup_{x \in X} \left\vert \frac{1}{t_n}(e^{t_n a(x)} - 1) - a(x) \right\vert \vert f \vert < \epsilon \qquad \text{for all } n \geq N$$
So its enough to show that $\left\vert \frac{1}{t_n}(e^{t_n a(x)} - 1) - a(x) \right\vert < \epsilon/\Vert f \Vert_\infty$ for all $n \geq N$ and $x \in X$. By L'Hopital I know that $\frac{1}{t}(e^{t a(x)} - 1) \to a(x)$ for $t \to 0$. Thus for a $x \in X$ I get by continuity:
$$\forall \epsilon > 0\ \exists \delta > 0\ \forall \vert t \vert < \delta: \left\vert\frac{1}{t}(e^{t a(x)} - 1) - a(x) \right\vert < \epsilon. $$
At this point I don't know how to proceed because I think that $\delta$ depends on the $x$, so that I don't get the estimate for all $x \in X$. Another idea of mine is to show $$ \lim_{t \to 0} \left\Vert \frac{T(t)f - f}{t} - M_a f \right\Vert_\infty = 0 \qquad \text{for all } f \in C_c(X)$$ and use the fact that $C_c(X)$ is dense in $C_0(X)$. But I figured out that I would have the same issues with that idea... I would appreciate some hints on the topic :)
It's easier if you write things in terms of an integral: $$ \frac{1}{t}(e^{ta(x)}-1)-a(x)=\left(\frac{1}{t}\int_{0}^{t}e^{sa(x)}ds-1\right)a(x) = \frac{1}{t}\int_{0}^{t}(e^{sa(x)}-1)ds\cdot a(x). $$