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I have a question about semigroups of one-dimensional diffusions.

Let $X$ be the Ornstein Uhlenbeck process on $\mathbb{R}$. The generator is expresses as $$\frac{d^2}{dx^2}-x\frac{d}{dx}.$$ It is known that the $L^2$ semigroup $\{T_t\}$ of $X$ is a compact operator on $L^{2}(\mathbb{R},m)$, where $m$ is the speed measure of $X$. $\{T_t\}$ is extended to strongly continuous contraction semigroup on $L^{p}(\mathbb{R},m)$, $1\le p<\infty$. Moreover, $\{T_t\}$ becomes a compact operator on $L^{p}(\mathbb{R},m)$ for any $1<p<\infty$. $\{T_t\}$ is not a compact operator on $L^{1}(\mathbb{R},m)$.

My question

  • Is there are nontrivial diffusion on $\mathbb{R}$ whose semigroup is compact on $L^{p}(\mathbb{R})$ for any $1 \le p \le \infty$? Consider the following differential operator: \begin{equation*} \frac{d^2}{dx^2}-x^{3}\frac{d}{dx}. \end{equation*} and the diffusion $Y$ associated with the above operator. The semigroups associated with $Y$ is compact on $L^{p}(\mathbb{R},m)$ for any $1 \le p \le \infty$? Here, $m$ is the speed measure of $Y$.

Dirichlet form of $Y$

The Dirichlet form of $Y$ is expresses as follows: \begin{align*} \mathcal{E}(f,g)&=\int_{\mathbb{R}}f'g'd\mu,\quad f,\ g \in \mathcal{F}\\ \mathcal{F}&=\{f \in L^{2}(\mathbb{R},\mu): f' \in L^{2}(\mathbb{R}.\mu)\}, \end{align*} where $d\mu=\exp(-x^4/2)\,dx$ and $dx$ is the Lebesgue measure on $\mathbb{R}$, and $f'$ is the distributional derivative of $f$.

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  • $\begingroup$ What is $m$ for this $Y$ ? And are you sure you mean $\frac{d^2}{dx^2}-x^2\frac{d}{dx}$ rather than $\frac{d^2}{dx^2}-|x|x\frac{d}{dx}$ ? $\endgroup$ – Jean Duchon Apr 30 '18 at 14:17
  • $\begingroup$ Thank you for your comment. $m$ is the speed measure for $Y$. Also, I changed the definition of generator. $\endgroup$ – sharpe Apr 30 '18 at 18:14
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    $\begingroup$ Concerning your original question "Is there are nontrivial diffusion on ℝ whose semigroup is compact on $L^p(\mathbb R)$ for any $p$?": if you switch to quadratic forms and allow for weighed spaces, then the semigroup on $L^2$ is compact if and only if the form domain is compactly embedded in $L^2$. This will be most likely be related to the weights. $\endgroup$ – Delio Mugnolo Apr 30 '18 at 23:26
  • $\begingroup$ Please tell me related papers. $\endgroup$ – sharpe May 1 '18 at 2:27
  • $\begingroup$ Do you want $\frac{d^2}{dx^2}+x^3\frac{d}{dx}$? $\endgroup$ – Michael Renardy May 3 '18 at 14:41

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