I am sorry for any inconveniences arising from my inexperience in writing questions on this site.
First I will give some context for my then succeeding question. I am working with the following definition of Feller transition kernels:
A probability transition kernel $p(t,x,A)$ is said to be a $Feller$ $transition$ $kernel$ if the following conditions hold:
(1) $p(t,x,\mathbb{R}^d)=1$ $\forall$ $t,x$ (some say this is superfluous, are they correct?)
(2) $\int_{\mathbb{R}^d}p(t,\cdot,dy)f(y) \in {C}_0(\mathbb{R}^d)\text{ for } f \in {C}_0(\mathbb{R}^d)$
(3) $\text{lim}_{t\downarrow0}\int_{\mathbb{R}^d}p(t,x,dy)f(y)=f(x),\text{ for every }f \in {C}_0(\mathbb{R}^d), x \in \mathbb{R}^d$.
Given a Feller transition kernel $p(t,x,A)$, one may define
$(P_tf)(x):=\int_{\mathbb{R}^d}f(y)p(t,x,dy)$.
This gives rise to a semigroup $\{P_t|t\geq0\}$ acting on $C_0(\mathbb{R}^d)$. Now letting $\alpha>0$ one may define
$f_\alpha(x):=\int_0^\infty\alpha e^{-\alpha t}(P_tf)(x)dt$,
then $f_\alpha$ $\in$ $C_0(\mathbb{R}^d)$ by the definition of Feller transition kernels as $f$ is assumed to be in $C_0(\mathbb{R}^d)$. Furthermore one may define $f_{\alpha,h}(x):=\int_h^\infty\alpha e^{-\alpha t}(P_tf)(x)dt$. Then clearly $f_{\alpha,h}(x)\uparrow f_{\alpha}(x)$ for $h\downarrow0$ by monotone convergence.
My question now lies in showing that
$\lim_{t\rightarrow0}(P_tf_{\alpha,h}(x))=\lim_{t\rightarrow0}P_t\int_h^\infty\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}(x)$.
The second equality to be precise. My own thoughts go as follows: A defining property of Feller transition kernels is that $\text{lim}_{t\downarrow0}\int_{\mathbb{R}^d}f(y)p(t,x,dy)=f(x),\text{ for every }f \in {C}_0^{\infty}(\mathbb{R}^d), x \in \mathbb{R}^d$. This implies, if one is able to show that $\int_h^\infty\alpha e^{-\alpha s}(P_sf)(\cdot)ds\in C_0(\mathbb{R}^d)$ by the definition of Feller transition kernels (I am not that sure of this actually, maybe someone else can convince me), that $\lim_{t\rightarrow0}P_t\int_h^{\infty}\alpha e^{-\alpha s}(P_sf)(x)ds=\int_h^{\infty}\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}$, which, at least in my eyes, would already answer the question. I am however led to believe that the proof is not as simple as seen above.
Yours truly!
Best Answer
I am now more convinced that the proof is quite immediate. I will supply the necessary details in order to conclude what I have already alluded to in my finalizing remarks above.
Let $\alpha \geq 0$ and $f_\alpha(x) :=\int_0^{\infty}\alpha e^{-\alpha t}P_tf(x)dt$. Since $P_tf\in{C}_0(\mathbb{R}^d)$ by the definition above (2), as $f \in {C}_0(\mathbb{R}^d)$ and since every function continuous and tending to $0$ at $\infty$ is bounded, one has $\alpha e^{-\alpha t}||P_tf(\cdot)||_\infty<C\cdot\alpha e^{-\alpha t}$ for some constant $C>0$. Thus one may apply dominated convergence to obtain
$\lim_{x\rightarrow0}f_{\alpha}(x)=\int_0^{\infty}\alpha e^{-\alpha t}\lim_{x\rightarrow0}P_tf(x)dt=0$
showing $f_\alpha \in {C}_0(\mathbb{R}^d)$. Now let $f\geq0$ and define $f_{\alpha,h}(x)=\int_h^{\infty}\alpha e^{-\alpha t}P_tf(x)dt$. Setting $f(t,x):=\alpha e^{-\alpha t}(P_tf)(x)$, one has that $\lim_{x\rightarrow0}f(t,x)=0$ $\forall$ $t\in(h,\infty)$ by the definition above point (2) and $f(t,x)$ is measurable in $t$ $\forall$ $x\in\mathbb{R}^d$. Thus one has
$\lim_{x\rightarrow\infty}f_{\alpha,h}(x)=\lim_{x\rightarrow\infty}\int_h^{\infty}\alpha e^{-\alpha t}(P_tf)(x)dt=\int_h^{\infty}\lim_{x\rightarrow\infty}\big(\alpha e^{-\alpha t}(P_tf)(x)\big)dt$
by Lebesgues theorem as the integrand is bounded in $x$, showing that $f_{\alpha,h}\in{C}_0(\mathbb{R}^d)$. Now by the definition point (3)
$\lim_{t\downarrow0}(P_tf_{\alpha,h})(x)=\lim_{t\downarrow0}P_t\int_{h}^\infty\alpha e^{-\alpha s}(P_sf)(x)ds=f_{\alpha,h}(x)$,
finally showing the claim.