I am currently learning about the inverse problem for orthogonal polynomials for orthogonality measures supported on the real line.
My question is not about finding the orthogonality measure from the moments but finding the moments from the recurrence coefficients.
Background. This is the problem of finding the orthogonality measure
$\mu$ for the sequence of orthogonal polynomials $(P_n(x))_n$ such
that $$ \int_{supp(\mu)}P_n(x)P_m(x)d\mu(x) = h_n \delta_{n,m}, $$
given the recurrence coefficients $(a_n)_{n\geq 0}$ and
$(b_n)_{n\geq 1}$ that the polynomials satisfy
$$ x P_n(x) = P_{n+1}(x) + a_n P_n(x) + b_n P_{n-1}(x), \quad P_{-1}(x):=0. $$
where we have chosen monic polynomials so that our recursion leads to a unique sequence of
polynomials where $\deg(P_n)=n$.Now if we have the moments of the measure $\mu$,
$$ m_k := \int_{supp(\mu)} x^k d\mu(x), $$ with $k\geq 0$,
we can determine the
monic polynomials and equivalently their recurrence coefficients
uniquely. However given the moments we might not be able to determine
the measure. We might be dealing with an indeterminate measure
problem.For simplicity we should restrict ourselves to the case of absolutely
continuous measures. This means there exists a continuous function $w(x)$
such that $d\mu(x)=w(x)dx$.
Question
Given the recurrence coefficients (or the polynomials or the square norms $h_n$) can we uniquely determine the moments? And if not, what restrictions
can we use to make them unique?
What I have tried so far
I have two points. The first of which seems to indicate that the moments
are not determined. The second seems to restrict the apparent freedom shown
in point 1.
1
Given all the square norms $h_n$, we seem to have some freedom in how to take the moments since
\begin{equation}
\begin{aligned}
h_0 &= m_0 \\
h_1 &= m_2 + v_{1,1}m_1 + v_{1,0}m_0 \\
h_2 &= m_4 + v_{2,3}m_3 + v_{2,2}m_2 + v_{2,1}m_1 + v_{2,0}m_0,\\
& \text{etc.}
\end{aligned}
\end{equation}
where the $v_{n,k}$ are just some constants determined by the monic polynomial coefficients.
This leads me to believe that for a given $h_n$, both $(m_{2n}, m_{2n-1})$ as
well as a distinct $(\widetilde{m}_{2n},\widetilde{m}_{2n-1})$ could lead to the same
$h_n$ as long as
$$
m_{2n} + v_{n,n-1} m_{2n-1} = \widetilde{m}_{2n} + v_{n,n-1} \widetilde{m}_{2n-1}.
$$
So it seems that given all $h_n$, we need to fix one extra parameter
for each $n$ to dispense of the ambiguity of the moments $(m_{2n},m_{2n-1})$.
2
However there seems to be less freedom than is immediately apparent. We cannot, for example, just take $m_{2k}=0$ for $k>0$. This would imply that
the orthogonality measure is anti-symmetric. This can't be true in general, right?
Many thanks in advance.
Best Answer
It is conventional to normalise the polynomials ($h_n = 1$) so that the recurrence coefficients (not the same $a$'s and $b$'s as yours) are the elements of a Jacobi Matrix $$A= \begin{bmatrix} a_0 & b_0 & 0 & 0 & \cdots \\ b_0 & a_1 & b_1 & 0 & \cdots\\ 0 & b_1 & a_2 & b_2 & \cdots \\ 0 & 0 & b_2 & a_3 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}. $$ The moments are then $(\delta_0, A^n\delta_0)$ where $\{\delta_i\}$ is the natural basis. See Barry Simon's The Classical Moment Problem as a Self-Adjoint Finite Difference Operator , around eq. (1.16).