Let $M$ be a compact metric space and $$\mathcal C^0(M) :=\{f:M\to \mathbb R;\ f\ \text{is continuous}\}.$$
Consider the semigroup of contractive bounded linear positive operators $$\{P^t : (\mathcal C^0(M),\|\cdot \|_{\infty}) \to \left(\mathcal C^0(M),\|\cdot\|_{\infty}\right)\}_{t\geq 0}, $$ i.e. $$\|P^t\|\leq 1,\ P^0 = \mathrm{Id}\ \ \text{and} \ P^{s+t} = P^t \circ P^s,\ \forall \ s,t\geq 0,$$ and $$ f\in \mathcal C^0(M),\ f\geq 0 \Rightarrow \ P^t f \geq 0,\ \forall\ t\geq 0.$$
Assume that there exist $K,\alpha >0$, such that $$ \|P^t\|\leq Ke^{-\alpha t}.$$ and
\begin{align} G: (\mathcal C^0(M,\|\cdot\|_{\infty}) &\to \left(\mathcal C^0(M),\|\cdot\|_{\infty}\right)\\ f&\mapsto \left(x\mapsto \int_0^\infty \left(P^t f\right)(x)\ \mathrm{d}t\right) \end{align} is a compact operator.
Question: Do $G$ being a compact and $P^t$ presenting exponential decay imply that there exist a $t_0>0$ such that $P^{t_0}$ is a compact operator?
In my head, it is harder to $G$ be compact than to $P^t$ be compact, but I was not able to prove it. Can anyone help me?