Suppose that $f$ is a many-times integrable function on $[-1, 1]$. We can consider integral moments of $f$, given by $$ I_n(f) := \int_{-1}^1 \big( f(x) \big)^n dx.$$ My question is: to what extent do the moments $\{I_n(f)\}_{n \in \mathbb{N}}$ determine $f(x)$?

Clearly this is hard to answer if some conditions are not imposed on $f$. For instance, changing $f$ at any single value doesn't affect any of the moments. So a better question is to ask to what extend do the moments determine a nice function $f$ for suitable choices of niceness.

In particular, does the sequence of moments $\{ I_n(f)\}_{n \in \mathbb{N}}$ determine $f$ completely if $f$ is continuous and positive? Or perhaps if $f$ is smooth and positive? Or perhaps if $f$ is analytic and positive?

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    $\begingroup$ It might be better to use a different word. The standard usage would define the $n$th moment as $\int_{-1}^1 x^n f(x)dx$. $\endgroup$ – Ben Crowell Nov 22 '16 at 23:30
  • $\begingroup$ @BenCrowell Yes, I had thought of that. But I'm not quite sure what word to use. My background is in analytic number theory, where one considers moments of the zeta function (or $L$-functions) defined analagously as in my question. But I am familiar with the standard moments as in probability, which are as you define. $\endgroup$ – davidlowryduda Nov 22 '16 at 23:32
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    $\begingroup$ If you don't require that $f$ be positive, then clearly if $f$ is odd, you can't distinguish $f$ from $-f$ by their moments. If $f$ is periodic with period 2, then you can't distinguish $f$ from $f$ precomposed with a phase shift. You also need restrictions that prevent you from surgically moving parts of the domain of $f$ around. $\endgroup$ – Ben Crowell Nov 22 '16 at 23:38
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    $\begingroup$ It is not enough to assume $f\in L^1$ for all your "moments" to exist. And the answer to your question is clearly "no" even when $f$ is positive and smooth. $\endgroup$ – Alexandre Eremenko Nov 23 '16 at 2:32
  • $\begingroup$ @AlexandreEremenko Yes, mentioning only $L^1$ was a careless error on my part. Thank you for mentioning that. As to the heart of the question, I would point out the excellent answer by Robert Israel. The point is to identify characteristics of $f$ that allow one to determine $f$ to some extent. The bounds of this question are fuzzy, but I think this is a fundamental enough question that people have thought about it before. $\endgroup$ – davidlowryduda Nov 23 '16 at 2:49

Since you can't distinguish between $f$ and its decreasing or increasing rearrangement, let's suppose $f$ is known to be increasing. In fact, let's suppose $f$ is differentiable with derivative bounded away from $0$. Then the moments determine the sum of the convergent series

$$ \sum_{n=0}^\infty I_n(f) (iz)^n/n! = \int_{-1}^1 \exp(iz f(x))\; dx$$ The change of variables $x = f^{-1}(t)$ makes this integral into $$\int_{f(-1)}^{f(1)} \exp(izt) (f^{-1})'(t)\; dt$$ By uniqueness of Fourier transforms, we can determine $(f^{-1})'$ and the interval $[f(-1),f(1)]$, and thus $f^{-1}$ and $f$.

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  • $\begingroup$ This is an excellent insight, and is exactly the sort of answer I'm looking for. Thank you very much. $\endgroup$ – davidlowryduda Nov 23 '16 at 2:37

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