Let $(X_n)_{n>0}$ be a Markov chain with general state space $\mathcal{X}=[0,1]$. The chain is specified by a transition operator $P$ which defines transition probabilities
$$\mathrm{Pr}(X_{n+1}\in A | X_n=x)=\int_A P(x, dy),$$
for any $x$ in $\mathcal{X}$ and any measurable set $A\subset\mathcal{X}$.
The operator $P$ can be regarded as an operator on functions in the following way. For $f:\mathcal{X}\to\mathbb{R}$ the function $Pf:\mathcal{X}\to\mathbb{R}$ is defined as
$$Pf(x)=\int_{\mathcal{X}}f(y)P(x, dy).$$
My question is as follows: supposing $f$ is differentiable and in $L^2(\mathcal{X})$, does it follow that $Pf$ is also differentiable?
So this operator is linear, so when computing the limit can I simply do $$\lim_{h\to0}\frac{(Pf)(a+h)-(Pf)(a)}{h} = \lim_{h\to0}\frac{P(f(a+h)-f(a))}{h} = \lim_{h\to0}P\left( \frac{f(a+h)-f(a))}{h}\right). $$
And then pass the limit?