Csdrhrt

Let

• 
  $(Omega,mathcal A,operatorname P)$ be a probability space


• 
  $(X_t)_{tge0}$ be a real-valued process on $(Omega,mathcal A,operatorname P)$
  


• 
  $E$ be a closed subspace of $left{f:mathbb Rtomathbb Rmid ftext{ is bounded and Borel measurable}right}$ equipped with the supremum norm

Note that there is a Markov kernel $kappa_{s,:t}$ (a regular version of the conditional probability distribution of $X_t$ given $X_s$ under $operatorname P$) on $(mathbb R,mathcal B(mathbb R))$ with $$kappa_{s,:t}(x,B)=operatorname Pleft[X_tin Bmid X_s=xright];;;text{for all }(x,B)inmathbb Rtimesmathcal B(mathbb R)tag1$$ for all $s,tge0$. Moreover, $$(T_{s,:t}f)(x):=intkappa_{s,:t}(x,{rm d}y)f(y)=operatorname Eleft[f(X_t)mid X_s=xright];;;text{for }xinmathbb Rtext{ and }fin E$$ is a contractive linear operator on $E$ for all $s,tge0$.

Assuming that $X$ is continuous and $Esubseteq C(mathbb R)$, we easily obtain $$(T_{s,:t}f)(x)xrightarrow{tto s}f(x);;;text{for all }xinmathbb Rtext{ and }fin Etag2$$ by the dominated convergence theorem.

Which further condition on $X$ do we need if we want that $T_{s,:t}C_0(mathbb R)subseteq C_0(mathbb R)$.

I think a necessary condition would be $$operatorname Pleft[|X_t|<rmid X_s=xright]xrightarrow{|x|toinfty}0;;;text{for all }r>0tag3,$$ since then $$operatorname Eleft[f(X_t)mid X_s=xright]xrightarrow{|x|toinfty}inftytag4$$ as long as $fin E$ with $$f(x)xrightarrow{|x|toinfty}infty.$$ So, let $s,tge0$, $r>0$ and $xinmathbb R$ with $|x|>r$. By Markov's inequality, $$operatorname Pleft[|X_t|<rmid X_s=xright]lefrac{operatorname Eleft[|X_t-x|^2mid X_s=xright]}{(|x|-r)^2}.tag5$$

I'm primarily interested in the case where $X$ is the strong solution of an SDE. So, let $(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$, $W$ be a $mathcal F$-Brownian motion on $(Omega,mathcal A,operatorname P)$ and $b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable. We now assume that $$X_t=X_s+int_s^tb(r,X_r):{rm d}r+int_s^tsigma(s,X_s):{rm d}W_s;;;text{almost surely}tag6.$$ I'll assume that $$operatorname Eleft[int_s^tb^2(r,X_r)+sigma^2(r,X_r):{rm d}rright]<inftytag7.$$

By Jensen's inequality and the Itō isometry, we obtain $$operatorname Eleft[|X_t-x|^2mid X_s=xright]le2operatorname Eleft[int_s^t(t-s)b^2(r,X_r)+sigma^2(r,X_r):{rm d}rright]tag8.$$

If we additionally assume that $b$ and $sigma$ are bounded, we are able to conclude that there is a $Cge0$ with $$operatorname Pleft[|X_t|<rmid X_s=xright]lefrac C{(|x|-r)^2}xrightarrow{|x|toinfty}0tag9.$$ Please tell me when I made a mistake. In any case, are we able to drop the assumption $(7)$ and/or find an assumption milder than boundedness on $b,sigma$? I could imagine that we could assume square-integrability of $X$ and linear growth of $b,sigma$ instead, for example.