Show that if $(κ_t)_{t≥0}$ is the transition semigroup of a strong solution to an SDE, $t↦(κ_tf)(x)$ is continuous for all $x$ and suitable $f:ℝ→ℝ$

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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $b,\sigma:\mathbb R\to\mathbb R$ be Lipschitz continuous and $$Lf:=bf'+\frac12\sigma^2f''\;\;\;\text{for }f\in C^2(\mathbb R)$$
  • $W$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$

We know that for any $X_0:\Omega\to\mathbb R$ there is an unique (up to indistinguishability) process $X$ on $(\Omega,\mathcal A,\operatorname P)$ with $$X_t=X_0+\int_0^tb(X_s)\:{\rm d}s+\int_0^t\sigma(X_s)\:{\rm d}W_s\;\;\;\text{for all }t\ge0\text{ almost surely}\tag1.$$ Moreover, there is a continuous process $(X^x_t)_{(t,\:x)\in[0,\:\infty)\times\mathbb R}$ on $(\Omega,\mathcal A,\operatorname P)$ with $$X_t^x=x+\int_0^tb(X_s)\:{\rm d}s+\int_0^t\sigma(X_s)\:{\rm d}W_s\;\;\;\text{for all }t\ge0\text{ almost surely for all }x\in\mathbb R\tag2.$$

Now, let $$\kappa_t(x,B):=\operatorname P\left[X_t^x\in B\right]\;\;\;\text{for all }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\text{ and }t\ge0.$$ Note that $$[0,\infty)\times\mathbb R\ni(t,x)\mapsto\kappa_t(x,B)\tag3$$ is Borel measurable for all $B\in\mathcal B(\mathbb R)$. In particular, $\kappa_t$ is a Markov kernel on $(\mathbb R,\mathcal B(\mathbb R))$ for all $t\ge0$. Let $\mathcal F^X$ denot the filtration generated by $X$. It can be shown that $$\operatorname E\left[f(X_{s+t})\mid\mathcal F^X_s\right]=\left.\operatorname E\left[f(X^x_t)\right]\right|_{x\:=\:X_s}\;\;\;\text{almost surely}\tag4$$ for all bounded Borel measurable $f:\mathbb R\to\mathbb R$ and $s,t\ge0$. So, $X$ is a time-homogeneous Markov process with transition semigroup $(\kappa_t)_{t\ge0}$.

Let $$\kappa_tf:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)$$ for bounded Borel measurable $f:\mathbb R\to\mathbb R$. Suppose we want that $$[0,\infty)\ni t\mapsto(\kappa_tf)(x)\tag5$$ is continuous for all $x\in\mathbb R$ and $f$ belonging to a suitable class $\mathcal C$ of bounded Borel measurable $f:\mathbb R\to\mathbb R$. By time-homogeneity, it should suffice to consider the continuity at $t=0$.

Usually, this is shown for $\mathcal C=C_0(\mathbb R)$ in the following way: By the Itō formula, $$(\kappa_tf)(x)=\operatorname E\left[f(X^x_t)\right]=f(x)+\operatorname E\left[\int_0^t(Lf)(X_s)\:{\rm d}s\right]\;\;\;\text{for all }(t,x)\in[0,\infty)\times\mathbb R\tag6$$ and hence $$\left\|\kappa_tf-f\right\|_\infty\le\left(\sup_K|bf'|+\frac12\sup_K\sigma^2|f''|\right)t\xrightarrow{t\to0+}0\tag7$$ with $K:=\operatorname{supp}f$ for all $f\in C_c^2(\mathbb R)$. Now, $C_c^2(\mathbb R)$ is dense in $C_0(\mathbb R)$ and hence we obtain $\left\|\kappa_tf-f\right\|_\infty\xrightarrow{t\to0+}0$ for all $f\in C_0(\mathbb R)$. We have shown that $(5)$ is even uniformly continuous at $t=0$ for all $f\in C_0(\mathbb R)$.

On the other hand, there is the general claim (cf. my other question) that if $X$ is a general right-continuous time-homogeneous Markov process with transition semigroup $(\kappa_t)_{t\ge0}$, $(1)$ is continuous for all $x\in\mathbb R$ and bounded continuous $f:\mathbb R\to\mathbb R$.

So, the question is: What's the point of proving the continuity for our special $X$ only for $\mathcal C=C_0(\mathbb R)$? Is it to obtain the uniform continuity which cannot be shown for all bounded continuous $f:\mathbb R\to\mathbb R$? And how can we prove that $(1)$ is continuous for all bounded continuous $f:\mathbb R\to\mathbb R$?