# Accuracy of the truncated Hausdorff moment problem

For a sequence of real numbers $s = (s_i)_{i \in n}$ let $M_s$ be the collection of functions $f:[0,1] \to [0,1]$ such that

$$(\forall i \leq n) \int_0^1 x^i f(x) dx = s_i$$

In other words, $M_s$ is the collection of density functions whose first $n$ moments are $(s_i)_{i \leq n}$.

I am interested in how much information knowing the first $n$ moments tells us about the function. Specifically, if we have the first $n$ moments of a function $f$ and we wish to approximate $f$ by $g$ with the same first $n$ moments, how badly could we be wrong?

To make this precise, for $p \in \{1, 2, \infty\}$ let $\alpha_p$ be the function of $s$

$$\alpha_p(s) = \sup_{f, g \in M_s} ||f - g||_p$$

So $\alpha(s)$ is an upper bound for how far off we could be if we approximate a function with first $n$ moments $s$ by another function with the first $n$ moments $s$.

Are there known lower bounds for the functions $n \to \sup_{|s| = n} \alpha_p(s)$ or $n \to \inf_{|s| = n} \alpha_p(s)$?

I feel this must have been studied so references are also welcome.

• "truncation error in moment problem" - about a million references in google. Nothing on the topic? – Sergei May 24 '15 at 9:21
• @Sergei: Most of what I found regarding the error seemed to be giving upper bounds on the error (i.e. a given method won't be worse than X), when I am looking for lower bounds on the error (i.e. you can't hope to do better than X). That said I would bet that among the many references there probably is one which answers my question. However being very far from an expert in this area I wasn't able to find an appropriate result after several hours of looking. Although I am hoping it would be easy/immediate to an expert. – Nate Ackerman May 24 '15 at 9:36
• Not a specialist too. Not useful?:m.iopscience.iop.org/0266-5611/3/3/016/pdf/… – Sergei May 24 '15 at 9:50

For $p=\infty$, there is a best possible lower bound of $$\inf_{|s| = n} \alpha_p(s)=\infty$$ since you can change the maximum of a (continuous, say) function without changing its integrals much.

That is, $(\forall\varepsilon, n)(\exists\delta)$ such that you can make $f-g\ge 1/\varepsilon$ on an interval of size $\le\delta$ and have the first $n$ moments of $f$ and $g$ differ by at most $\varepsilon$.

• Thanks. However I am specifically interested in the case when the range is contained in a fixed bounded set (e.g. $[0,1]$) – Nate Ackerman May 24 '15 at 9:03
• Ah yes. But then it's just 1 instead of $\infty$, for the same reason. – Bjørn Kjos-Hanssen May 24 '15 at 9:09

As already indicated by Bjorn, this is hopeless. Whether or not a solution $\mu$ to a moment problem has an absolutely continuous part is decided exclusively at infinity; it does not depend on finitely many moments.

So you can always have two measures, one absolutely continuous and the other one purely singular, that have the same moments on an arbitrarily long initial piece. It is perhaps best to think of this in terms of the recurrence $$a_n p_{n+1}(z) + a_{n-1} p_{n-1}(z) + b_n p_n(z) = zp_n(z)$$ satisfied by the polynomials orthogonal with respect to your measure. Knowing the first $N$ moments is the same as knowing the first $N$ coefficients $a_n,b_n$. (This also shows that it's easy to make the supports subsets of $[0,1]$, by controlling the $a$'s and $b$'s.)

So $\alpha_p=\infty$ for any $p>1$ and $\alpha_1=2s_0$. (Even if you insist that both measures are absolutely continuous, the same conclusions hold by approximation.)

Some background information (digression): (1) The collection of measures with (the first) $N$ prescribed moments can be described. Search for Nevanlinna parametrization.

(2) A related question that has a good answer is: Fix an interval $I\subset\mathbb R$ (or $I\subset [0,1]$ in your setting). What can we say about $\mu(I)$ if I'm given the first $N$ moments of $\mu$? There is classical work of Chebyshev and Markov on this. I wrote a paper on the continuous analog, which you are probably not interested in, but it also points to the relevant literature. See item #16 here.