Let
- $(E,\mathcal E)$ be a measurable space
- $\mathcal E_b:=\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\}$
- $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $E$
- $f\in\mathcal E_b$ with $$(\kappa_th-f)(y)\xrightarrow{h\to0+}f(x)\;\;\;\text{for all }y\in E\tag1$$
Are we able to show that $$g(t):=(\kappa_tf)(x)\;\;\;\text{for }t\ge0$$ is continuous?
Let $t\ge0$. Noting that $(\kappa_t)_{t\ge0}$ is a contraction semigroup on $\mathcal E_b$, we see that $$\left|(\kappa_th-f)(y)\right|\le2\|f\|_\infty\;\;\;\text{for all }y\in E\tag2$$ and hence $$(\kappa_{t+h}f-\kappa_tf)(x)=(\kappa_t(\kappa_hf-f))(x)=\int\kappa_t(x,{\rm d}y)(\kappa_hf-f)(y)\xrightarrow{h\to0+}0\tag3$$ by Lebesgue's dominated convergence theorem. Thus, $g$ is right-continuous at $t$.
Now we may write $$(\kappa_{t-h}f-\kappa_tf)(x)=(\kappa_{t-h}(f-f\kappa_hf))(x)=\int\kappa_{t-h}(x,{\rm d}y)(f-\kappa_hf)(y)\tag4$$ for all $h\in[0,t]$, but this time the dependence of $h$ on the right-hand side prevents us from a simple application of Lebesgue's dominated convergence theorem.
Can we fix this?
Maybe $\kappa_{t-h}$ tends to $\kappa_t$ in a suitable sense ...