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The generator an OU process is given by

$$A = \operatorname{tr}(QD^2)+\langle Bx,D\rangle.$$

This one possesses an invariant measure given by $$d\mu(x) = b(x) \ dx \text{ with } b(x) = \frac{1}{(4\pi)^{N/2} \vert Q_{\infty} \vert^{1/2}} e^{-\langle Q_{\infty}^{-1}x,x \rangle /4},$$ where $Q_{\infty}= \int_0^{\infty} e^{sB} Q e^{sB^*} \ ds.$

I read that $A$ is self-adjoint on $L^2(\mathbb R^n, d\mu)$ if and only if $QA^*=AQ,$ but I have never seen an argument for this.

In particular, what I am mainly interested in, is there anywhere an explicit representation of the adjoint operator of $A$?

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The argument is a bit tricky. Assume for simplicity that $Q=I$, then $BQ_\infty+Q_\infty B^*=-I$. Next decompose $A$ as a self-adjoint part plus a remainder, namely introduce the form $a(u,v)=\int_{\mathbb R^n} \nabla u \nabla v \, d\mu$ which corresponds to $A_1=\Delta-\frac 12 Q_\infty ^{-1} x \cdot \nabla$ and write $A=A_1+C$ with $C=(Bx+\frac 12 Q_\infty^{-1} x) \cdot \nabla$.

A computation shows that $C$ is skew-adjoint in $L^2_\mu$, so that $A$ is self-adjoint iff $C=0$ or $B=-\frac 12 Q_\infty^{-1}$, in particular $B=B^*$. On the other hand, if $B=B^*$, we can diagonalize $B$ and check that $B=-\frac12 Q_\infty^{-1}$ so that $A=A_1$ is self-adjoint. This answers to the first question. Concerning the second, $A^*=A_1-C$, but I dooubt that one can compute it explicitely without knowing $Q_\infty$.

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  • $\begingroup$ thanks a lot for this proper explanation, but just to be sure. What does this assumption $Q=I$ you make in the beginning mean? Is the adjoint otherwise still given by $A_1-C$ where $\Delta$ is replaced by $tr(QD^2)$ in $A_1$? $\endgroup$
    – Kung Yao
    Commented Jan 30, 2022 at 2:32
  • $\begingroup$ When $Q$ is non degenerate, through a linear change of variables one reduces to the case above. However one can use the same argument in the answer, modifying the form to $a(u,v)=\int Q\nabla u\nabla v$ which gives $A_1=tr(QD^2)-\frac 12 QQ_\infty^{-1}x \cdot \nabla $ so that $C=(Bx+\frac 12 Q Q_\infty^{-1} x)\cdot \nabla$. The $A$ is self adjoint iff $B=-\frac 12 QQ_\infty^{-1}$. Note that if $Q$ is degenerate this cannot happen. In fact the determinant of $B$ would be zero and $Q_\infty$ would not exist. $\endgroup$
    – Giorgio Metafune
    Commented Jan 30, 2022 at 9:22
  • $\begingroup$ thank you. It seems that the degenerate case is quite interesting, too. I read that if $\ker(Q)$ does not contain any non-trivial subspace invariant under $B^T$ , see Corollary 12 here arxiv.org/pdf/1510.05936.pdf, that there exists still an invariant measure. This situation allow even for $Q$ to be degenerate. Do you know if one can understand the situation of adjoints in that case, too? Or is this condition precisely such that $Q_{\infty}$ is still invertible and the invariant measure is as in my question?-The author of that paper does not seem to specify the invariant measure. $\endgroup$
    – Kung Yao
    Commented Jan 31, 2022 at 3:14
  • $\begingroup$ My previous comment applies also to the degenerate case, but in this situation $A$ cannot be self-adjoint, this I meant in the final part. There is a nice paper by A. Lunardi and M. Geissert on the domain of OU operators in $L^2$ of the invariant measure in the degenerate case. I think there you find more basic information. $\endgroup$
    – Giorgio Metafune
    Commented Jan 31, 2022 at 8:50

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