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The n-th moment of a real valued function $f$ is defined as: $m_n(f)=\int_{-\infty}^{+\infty}x^nf(x)dx$. I heard that a function $f$ is uniquely determined by its moments. I would be quite surprised if this worked for any $f$. Is it true? And in that case, does anyone know how to re-build $f$ from its moments?

Thanks

Chevallier
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  • For $m_n(f)$ to be defined for all $n$, certainly $f$ must decay faster than the reciprocal of any polynomial, which is a very strong condition. Certainly restricting $f$ to functions in the Schwartz space is sufficient: https://en.wikipedia.org/wiki/Schwartz_space . – Travis Willse Jul 20 '15 at 23:22
  • I'm not sure about this if you integrate on all of $(-\infty,\infty)$, but if you are integrating over some finite interval, $[a,b]$, and $f$ is continuous, then yes, $f$ is determined by its moments.

    This is a classic problem in undergraduate analysis: to use the Weierstrass approximation theorem to show that if $f$ is continuous on $[a,b]$ and $\int_a^b x^n f(x)\ dx=0$ for all natural numbers $n$, then $f=0$. (Can you see that this implies that the moments of a function are unique, by linearity of integration?)

     –  Jul 20 '15 at 23:32
  • Ok. Is there a way to reconstruct $f$ then? – Chevallier Jul 20 '15 at 23:38
  • This answer seems to give all the necessary information https://math.stackexchange.com/a/1029178/269624 – Yuriy S Dec 28 '17 at 09:00

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One example of nonuniqueness was found (I think) by Stieltjes in 1894: The function $$ f(x) = \cases{x^{-\ln(x)} \sin(2\pi \ln(x))& $x > 0$\cr 0 & otherwise} $$
has all its moments $0$.

EDIT: The proof of this is a nice exercise in integration and symmetry. Start with the change of variables $x = \exp(t)$...

Robert Israel
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