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Let $f\in C([a,b],\mathbb{R})$ such that $\displaystyle\int_{a}^{b} t^nf(t)dt=0$ for all integer n.

We know that $f\equiv 0$. It's call Hausdorff theorem.

  • This theorem is wrong on $\mathbb{R^+}$, a counter example is : $$f(x)=\exp(-x^{\frac{1}{4}})\sin(x^\frac{1}{4})$$

In fact this exercice was posted in MSE and actually I don't understand how someone can construct a such example ? Can we find it by ourselves ? Is there exist some reference of this theorem (History perhaps..) ?

Reference

  1. The exercise in Mathematics Stack Exchange

Thank you in advance for your time,

Julien.

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    $\begingroup$ See Fourier Analysis by T. W. Körner. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 6, 2014 at 23:43
  • $\begingroup$ @AndresCaicedo Thank you but 80$ for a book is a bit expensive .. $\endgroup$
    – user46896
    Commented Mar 7, 2014 at 0:04
  • $\begingroup$ @NeilHoffman What is your $k$ ? I did not really understand everything. Can you elaborate a little more please ? $\endgroup$
    – user46896
    Commented Mar 7, 2014 at 0:08
  • $\begingroup$ Perhaps I missing an implied assumption, but I think this works. Consider a function $g_0(t)=\{0 \mbox{ on } [0,1], 1/k \mbox{ on } [2k-1,2k], -1/k \mbox{ on } (2k,2k+1)\}$. Let $g_1(t)$ be a continuous function that approximates $g_0(t)$ such that $g_1(t)$ also has the property that $\int_{x=2k} ^{2k+2} g_1(t)dt=0$. Now, let $f(t)=g_1(t)/t^n$ if $t\ne 0$ and $f(0)=0$. $\endgroup$
    – Neil Hoffman
    Commented Mar 7, 2014 at 0:09
  • $\begingroup$ I just cleaned the comment up a little, so hopefully its more readable. $k$ is the floor of (t+1)/2. $\endgroup$
    – Neil Hoffman
    Commented Mar 7, 2014 at 0:12

1 Answer 1

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The standard reference is N. I. Akhiezer, The Classical moment problem and some related questions of analysis, MR0184042.

There is also a paper: MR1627806 Simon, Barry The classical moment problem as a self-adjoint finite difference operator. Adv. Math. 137 (1998), no. 1, 82–203.

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    $\begingroup$ Available here. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 7, 2014 at 2:11
  • $\begingroup$ I didn't know this reference. Thank you. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 7, 2014 at 2:11
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    $\begingroup$ Andres, thanks for the reference:-) I did not know about the free English translation. Almost all Russian (Soviet) books are available free. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 7, 2014 at 2:15