# Strong continuity of the Ornstein-Uhlenbeck operator

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 $$k$$.

• 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. – 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 - 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., $g(x) = \cos(x^2)$.
• 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 pre-factor 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. – Nawaf Bou-Rabee Oct 11 '16 at 3:22