I know that if $\{T_t, t>0 \}$ is a conservative Markov semigroup on E, and $f \in D(A)$ has an absolute maximum in x then $Af(x) \le 0$. Where $D(A)$ is the infinitesimal generator of $T_t$.
I have to prove in $\mathbb{R}$, that the operator $Af=f^{(k)}$, the k-derivative of f, cannot be a generator of a Markov semigroup if $k \ge 3$.
I start with $Af(x)= \lim_{t \to 0}\frac{T_tf(x)-f(x)}{t}=f^{(3)}=\lim_{t \to 0}\frac{f^{''}(x+t)-f^{''}(x)}{t}$.
$T_tf(x)=\int{f(y)p(t,x,dy)}$ where p() is the transition function. I would like to show that the previous is greater than zero but I don't know how to proced.