It's well known that the OrnsteinUhlenbeck semigroup defined by $$ P_tf(x)=\int_{\mathbb{R}}f\left(xe^{t}+\sqrt{1e^{2t}}z\right)\frac{e^{z^2/2}}{\sqrt{2\pi}}\,dz $$ 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 OrnsteinUhlenbeck 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 $k$.

3$\begingroup$ For what it's worth, the O.U. semigroup is strongly continuous on $C_0(\Bbb R)$, the space of continuous functions on $\Bbb R$ that tends to $0$ at infinity, with the uniform norm. $\endgroup$ – John Dawkins Oct 11 '16 at 16:50
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  1x^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., $g(x) = \cos(x^2)$.

$\begingroup$ This answer gives a family of continuous functions whose weighted norm does not vanish at infinity because these functions grow at the rate of $1+x^k$ with a prefactor that can be any function in $C_b(\mathbb{R})$. This reduces the original problem to whether or not the OU semigroup is strongly continuous on the space $C_b(\mathbb{R})$ with the supremum norm, and the example provided shows that it is not. $\endgroup$ – Nawaf BouRabee Oct 11 '16 at 3:22

$\begingroup$ Thanks for you answer! Indeed, I agree with the statement you wrote. I'm just surprised I have seen (even classical) works where strong continuity of the OU semigroup in such norms is asserted, and stated as an obvious result. $\endgroup$ – RadonNikodym Oct 11 '16 at 14:35

$\begingroup$ FYI: Daprato, Giuseppe, and Alessandra Lunardi. "On the OrnsteinUhlenbeck operator in spaces of continuous functions." Journal of Functional Analysis 131.1 (1995): 94114. From the abstract: We study the realization of the OrnsteinUhlenbeck operator A in the space of the uniformly continuous and bounded functions in Rn. We prove that it generates a semigroup which is neither strongly continuous nor analytic, but enjoys nice smoothing properties. $\endgroup$ – Nawaf BouRabee Oct 11 '16 at 22:58