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Let $f(x)$ be a real-valued function on $\mathbb{R}$ such that $x^nf(x), n=0,1,2,\ldots$ are Lebesgue integrable.

Suppose $$\int_{-\infty}^\infty x^n f(x) dx=0$$ for all $n=0,1,2,\ldots. $

Does it follow that $f(x)=0$ almost everywhere?

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This is a duplicate question. –  PEV Jan 29 '11 at 23:49
    
Here: math.stackexchange.com/questions/16831/… –  PEV Jan 29 '11 at 23:51
    
@TCL: Note also that the indefinite integral cannot equal $0$... –  Arturo Magidin Jan 29 '11 at 23:59
    
This is more apt I think: math.stackexchange.com/questions/17026/… –  Aryabhata Jan 29 '11 at 23:59
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Neither of the questions linked in the comments above is a duplicate, but Hans Lundmark's answer on the first question shows that the answer to this one is no: math.stackexchange.com/questions/16831/… –  Jonas Meyer Jan 30 '11 at 0:08

1 Answer 1

No. This can be seen via Fourier transform.

Let $f$ be such that for the Fourier transform $\hat f$ it holds that for all $n$ \begin{equation*} \hat f^{(n)}(0) = 0. \end{equation*} Note that this does not imply that $f$ itself is zero since there are infinitely flat functions. By using the rules for Fourier transforms and derivatives you see that $\int x^n f(x) dx = 0$ for all $n$.

One application of such functions with infinitely many vanishing moment is the construction of infinitely regualel orthonomal wavelet bases (see, e.g., the Meyer wavelet).

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