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Consider a sequence of real numbers $s=(s_0,s_1,\ldots)$. When is there a Borel measure $\mu$ supported on $[-1,1]$ so that $$ s_k = \int_{[-1,1]} x^k\,\mathrm{d}\mu,\quad \forall k\in\mathbb N\;? $$

The well known Hausdorff moment problem asks the same question on the interval $[0,1]$. In that case, $s$ is a moment sequence of a measure supported on $[0,1]$ if and only if $s$ is completely monotone. It follows that, for our problem, the even terms of $s$ must be completely monotone. What is known about the odd terms?

Question: Are the necessary and sufficient conditions for $s$ to be a moment sequence on $[-1,1]$ known? What are the simplest known necessary conditions?

The following question maybe easier to solve.

Assume now that $s$ is a sequence of nonnegative reals. When is there measure supported on $[-1,1]$ with the moment sequence $s$?

There is a well known way of obtaining necessary conditions: Let $P(x) = \sum a_i x^i$ be a polynomial that is nonnegative on $[-1,1]$. Form the Hankel matrix $H_P$ $$ H_P(j,k):= \sum_i a_i s_{i+j+k},\quad j,k\in \mathbb N. $$ Then we observe

Claim: Suppose $P(x)\ge 0$ on $[-1,1]$. Then $H_P$ is positive semi-definite.

Pf. Let $c_0,c_1,\ldots$ be a complex sequence with finitely many nonzero elements. We have \begin{align*} cH_Pc^* &= \sum_{j,k}c_j\bar{c_k}\int \sum_i a_i x^{i+j+k}\,\mathrm{d}\mu\\ &=\int_{[-1,1]} P(x)\left|\sum_j c_j x^j\right|^2\,\mathrm{d}\mu \ge 0. \end{align*}

A result due to Riesz implies that if $H_P$ is PSD for all polynomials $P$ that are nonnegative on $[-1,1]$ then $s$ is a moment sequence. However if the interval is $[0,1]$ this PSD criterion is equivalent to a much much simpler condition: that the sequence is completely monotone. Is there a simpler characterization for $[-1,1]$?

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    $\begingroup$ I think it's easy to show that your condition for all $P$ with $P\ge 0$ on $[-1,1]$ implies existence of a measure as desired. You wouldn't expect a very easy characterization really since it's not straightforward to tell from the moments what the support of the measure is. $\endgroup$ Apr 6, 2016 at 23:16
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    $\begingroup$ Another comment is that a (much simpler) positivity condition characterizes solvability of the moment problem, but with no restriction on the support. You're probably well aware of this. $\endgroup$ Apr 6, 2016 at 23:18
  • $\begingroup$ I vaguely remember seeing that if $H_P$ is PSD for all $P\ge 0$ on $[-1,1]$ then $s$ is a moment sequence (Potentially named after Riesz). The reason I am hoping a smaller family may suffice is that for the interval $[0,1]$, considering polynomials $P(x) = x^p(1-x)^q$ for all $p,q \in \mathbb N$ is sufficient as per Hausdorff problem. $\endgroup$
    – MERTON
    Apr 6, 2016 at 23:27
  • $\begingroup$ Sure. In this setup, we may say that the polynomial $P(x)=1$ is positive on $(-\infty, \infty)$ and $H_1(j,k) = x^{j+k}$ coincides with Hamburger's matrix. $\endgroup$
    – MERTON
    Apr 6, 2016 at 23:32

2 Answers 2

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If $\mu$ is a measure supported on $[-1,1]$, the change of variables $y = (1+x)/2$ gives you a measure $\rho$ supported on $[0,1]$, and the corresponding moments are related by

$$\int_0^1 y^k\; d\rho(y) = \int_{-1}^1 ((1+x)/2)^k \; d\mu(x) = 2^{-k} \sum_{j=0}^k {k \choose j} \int_{-1}^1 x^j \; d\mu(x)$$

Moreover, we can also go in the opposite direction to give $\int_{-1}^1 x^k\; d\mu(x)$ in terms of $\int_0^1 y^j\; d\rho(y)$ for $j = 0 \ldots k$.

So the necessary and sufficient condition is that $r_k = 2^{-k} \sum_{j=0}^k {k \choose j} s_j$ are completely monotone.

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The following paper answers a more general version of this question, namely the Hausdorff moment problem for compact sets in $\mathbb{R}^n$.

This is a fairly broad topic, and you may enjoy learning more by looking at these notes or this really nice book.

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    $\begingroup$ Of course, for answering the OP, doing a change of variables to the 1D-case as noted by Robert Israel works! $\endgroup$
    – Suvrit
    Apr 13, 2016 at 1:47

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