Feller continuity of the stochastic kernel

Given a metric space $X$ with a Borel sigma-algebra, the stochastic kernel $K(x,B)$ is such that $x\mapsto K(x,B)$ is a measurable function and a $B\mapsto K(x,B)$ is a probability measure on $X$ for each $x$

Let $f:X\to \mathbb R$. We say that $f\in \mathcal C(B)$ if $f$ is continuous and bounded on $B$.

Weak Feller continuity of $K$ means that if $f\in\mathcal C(X)$ then $F\in\mathcal C(X)$ where
$$
F(x):=\int\limits_X f(y)K(x,dy).
$$

I wonder if it implies that if $g\in \mathcal C(B)$ then
$$
G(x):=\int\limits_Bg(y)K(x,dy)
$$
also belongs to $\mathcal C(B)$?

Solutions Collecting From Web of "Feller continuity of the stochastic kernel"

No. Take $X = \mathbb{R}$ and $K(x,B) = 1_B(x+1)$. (This corresponds to a process that jumps 1 unit to the right at each step.) Note that $\int f(y) K(x,dy) = f(x+1)$, so $K$ is certainly weak Feller.

However, take $f = 1_{(0, \infty)}$ to be a step function and $B = \mathbb{R} \backslash \{0\}$. Then $f \in C(B)$ by your notation. But I’ll let you check that $\int_B f(y) K(x, dy) = 1_{(-1, \infty)}(x)$ which is not in $C(B)$.

To make it clearer, you can re-write $ G(x) = \int\limits_X g(y)\mathbf{1}_B(y) K(x,{\rm d} y)$. Now in general $g{\mathbf 1}_B$ is not continuous anymore so as Nate pointed out you should not expect $G$ to be such.

However, if you take a continuous $g $
with $\overline{{\rm support}(g)} \subsetneq B$ then (I let you 🙂 show this) $g{\mathbf 1}_B$ is still continuous and so is $G$ then.