I will formulate this question in the language of Jacobi operators and spectral measures although it could be entirely rewritten in terms of orthogonal polynomials and measures of orthogonality.

Objects of interest: Let $J$ be a semi-infinite Jacobi matrix of the form $$J=\begin{pmatrix} b_{1} & a_{1}\\ a_{1} & b_{2} & a_{2}\\ & a_{2} & b_{3} & a_{3}\\ & & \ddots & \ddots & \ddots \end{pmatrix},$$ where $b_{n}\in\mathbb{R}$ and $a_{n}>0$. Let $\mathcal{M}$ denotes the set of Jacobi matrices $J$ such that $b_{n}\to 0$ and $a_{n}\to 1$ as $n\to\infty$ (the so-called Szegö class in OG polynomials terminology).

Let $J_{0}$ be the Jacobi matrix with $b_{n}=0$ and $a_{n}=1$ for all $n\in\mathbb{N}$. It is well-known that the spectrum of $J_{0}$ is $$\sigma(J_{0})=[-2,2].$$ A Jacobi matrix $J\in\mathcal{M}$ iff $J-J_{0}$ is compact.

For any self-adjoint operator $J$ (assume $J$ is bounded), there exists unique measure $\mu$ (the spectral measure) determined by the equality $$\int_{\mathbb{R}}\frac{d\mu(x)}{x-z}=\langle e_{1},(J-z)e_{1}\rangle, \quad \forall z\in\mathbb{C}\setminus\mathbb{R},$$ where $e_{1}$ is the first vector of the standard basis of $\ell^{2}(\mathbb{N})$. It holds that $\mbox{supp}\mu=\sigma(J)$.

Question: Assume $\mu$ to be a probability measure with $\mbox{supp}\mu=[-2,2]$. Then there exists a unique Jacobi operator $J$ whose spectral measure coincides with $\mu$. I am interested in what additional conditions (to $\mbox{supp}\mu=[-2,2]$) is to be imposed on the measure $\mu$ to guarantee that $J\in\mathcal{M}$. The condition (if any) should be weaker than the Szegö condition which gives us more, see below.

What is known: There is an answer to this kind of question known for a long time but it implies even more than just $J\in\mathcal{M}$. Namely, if $\mbox{supp}\mu=[-2,2]$ and the density $\rho$ of the the Lebesgue absolutely continuous component of $\mu$ satisfies the so-caled Szegö condition: $$\int_{-2}^{2}\frac{\log\rho(x)}{\sqrt{4-x^{2}}}dx>-\infty,$$ then $$\sum_{n=1}^{\infty}b_{n}^{2}<\infty \quad \mbox{ and } \quad \sum_{n=1}^{\infty}(a_{n}-1)^{2}<\infty.$$ In particular, $J\in\mathcal{M}$. Proof and references for these claims (and much more) can be found here.


1 Answer 1


One such condition (which is weaker, though not dramatically so perhaps) is that $\rho(x)>0$ a.e. on $(-2,2)$. This is usually called the Denisov-Rakhmanov theorem; see here. In fact, it will give the conclusion also if $J$ has spectrum outside $[-2,2]$ as long as this part of the spectrum is discrete.

This condition is clearly not necessary for $J-J_0$ to be compact; a compact perturbation can easily have purely singular spectrum. I very much doubt that a characterization of compact perturbations (as in the case of Hilbert-Schmidt) is possible.

However, if you keep the $a_n$'s equal to $1$, then the obvious necessary condition from Weyl's theorem (namely, $\sigma_{ess}=[-2,2]$) becomes sufficient also. This lovely result is due to Damanik, Hundertmark, Killip, and Simon.

  • $\begingroup$ Thank you, Christian Remling, this info is very helpful! $\endgroup$
    – Twi
    Dec 29, 2016 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.