It's well known that the Ornstein-Uhlenbeck semigroup defined by
$$
P\_tf(x)=\int\_{\mathbb{R}}f\left(xe^{-t}+\sqrt{1-e^{-2t}}z\right)\frac{e^{-z^2/2}}{\sqrt{2\pi}}\,dz
$$
is **not** strongly continuous on the space $C(\mathbb{R})$ of continuous functions on $\mathbb{R}$ with the supremum norm. I was wondering if the Ornstein-Uhlenbeck semigroup might be strongly continuous if we consider strong continuity over a Banach space of functions bounded with respect to a weighted supremum instead, i.e.
$$
\lVert f\rVert\_k=\sup\_{x\in\mathbb{R}}\frac{|f(x)|}{1+|x|^k},
$$
for some positive integer $k$.



Strong continuity of the Ornstein-Uhlenbeck operator
Nope. To see this, let $X\_t(x)$ denote the OU process at time $t>0$ with initial condition $x$; set $f(x) = g(x) (1+|x|^k)$ where $g \in C\_b(\mathbb{R})$; and consider:
\begin{align\*}
\frac{P\_t f(x) - f(x)}{1+|x|^k} &= \frac{E\{ g(X\_t(x)) (1+|X(t)|^k) - g(x) (1+|x|^k) \}}{1+|x|^k} \\
&= \frac{E\{ g(X\_t(x)) (1+|X\_t(x)|^k) - g(x) (1+|x|^k) \}}{1+|x|^k} \\
&= E\{ g(X\_t(x)) - g(x) \} + \frac{E\{ g(X\_t(x)) (1+|X\_t(x)|^k - 1-|x|^k) \}}{ 1+|x|^k} 
\end{align\*}
The second term is nicely uniformly continuous. However, the first term is not always so nice: when $x$ is very large then $X\_t(x) \approx e^{-t} x$ (the effect of the noise is negligible) and the first term behaves like $g(e^{-t} x) - g(x)$, which does not in general converge to zero uniformly in $x$ as $t \to 0$. Take, e.g., $g(x) = \cos(x^2)$.


