Is the transition semigroup of the solution of an SDE with Lipschitz coefficients strongly continuous on $C_b$?

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $b,\sigma:\mathbb R\to\mathbb R$ be Lipschitz continuous (and hence at most of linear growth) and $$Lf:=bf'+\frac12\sigma^2f''\;\;\;\text{for }f\in C^2(\mathbb R)$$
• $W$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
• $(X^x_t)_{(t,\:x)\in[0,\:\infty)\times\mathbb R}$ be a continuous process on $(\Omega,\mathcal A,\operatorname P)$ with $$X_t^x=x+\int_0^tb(X^x_s)\:{\rm d}s+\int_0^t\sigma(X^x_s)\:{\rm d}W_s\;\;\;\text{for all }t\ge0\text{ almost surely for all }x\in\mathbb R\tag1$$ and $$(\kappa_tf)(x):=\operatorname E\left[f(X^x_t)\right]\;\;\;\text{for }x\in\mathbb R$$ for any bounded Borel measurable $f:\mathbb R\to\mathbb R$ and $t\ge0$

If $f\in C_b(\mathbb R)$, are we able to conclude $\left\|\kappa_tf-f\right\|_\infty\xrightarrow{t\to0+}0$?

Assume first that $f\in C^2_b$. Fix $(t,x)$. Since $f'$ is bounded, $$(\kappa_tf)(x)=f(x)+\text E\left[\int_0^t(Lf)(X^x_s)\:{\rm d}s\right].\tag2$$ The crucial question seems to be whether we're allowed to apply Fubini's theorem to the second term on the rhs. By Jensen's inequality, $$\text E\left[\sup_{s\le t}|X^x_s|\right]^2\le\text E\left[\sup_{s\le t}|X^x_s|^2\right]<\infty\tag3$$ (where the finiteness of the rhs follows from the Lipschitz assumption) and by the linear growth assumption for some $$\text E\left[\int_0^t|(Lf)(X^x_s)|\:{\rm d}s\right]\le ct\left(1+\text E\left[\sup_{s\le t}|X^x_s|\right]\right)\left\|f'\right\|_\infty+\frac c2t\left(1+\text E\left[\sup_{s\le t}|X^x_s|^2\right]\right)\left\|f''\right\|_\infty<\infty\tag4$$ for some $c\ge0$. So, we should be able to apply Fubini's theorem and obtain $$(\kappa_tf)(x)=f(x)+\int_0^t(\kappa_s(Lf))(x)\:{\rm d}s.\tag5$$ So, we should have $$|(\kappa_tf)(x)-f(x)|\le t\left\|Lf\right\|_\infty.\tag6$$ The only problem is that $Lf$ might be unbounded. So, I guess we need to assume $f\in C_c^2(\mathbb R)$ (since I don't see that a larger class ensures that $Lf$ is bounded).

This allows us to conclude the claim for such a $f$ and, by density, for $f\in C_0(\mathbb R)$. Is it possible to extend the result to $f\in C_b(\mathbb R)$?

No, the convergence does, in general, not hold true for bounded continuous functions. Strong continuity on $C_b(\mathbb{R}^d)$ is a pretty strong assumption on a Markov semigroup.

Example: For fixed $b \neq 0$ consider the semigroup $$\kappa_t f(x) := f(x+bt)$$ associated with the deterministic process $X_t := bt$. Clearly, $(X_t)_{t \geq 0}$ satisfies the SDE $$dX_t = b \, dt.$$ Since $$\kappa_t f(x)-f(x) = f(x+bt)-f(x)$$ it follows that $$\|\kappa_t f-f\|_{\infty} \xrightarrow[]{t \to 0} 0 \iff \text{$f$ is uniformly continuous}.$$ So, if we pick some bounded continuous function $f$ which fails to be uniformly continuous, then $\|\kappa_t f-f\|_{\infty}$ fails to converge to $0$ as $t \to 0$.

Remark: It is possible to show that the semigroup $(\kappa_t)_{t \geq 0}$ associated with the one-dimensional Brownian motion fails to be strongly continuous on $C_b(\mathbb{R})$, i.e. there exists $f \in C_b(\mathbb{R})$ such that $\|\kappa_t f-f\|_{\infty}$ does not converge to $0$ as $t \to 0$; see e.g. Example 1.7(d) in Lévy Matters III by Böttcher & Schilling & Wang for details.