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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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    @Sivaram, @Arturo: Thanks. Yes, I know there was an edit. My $f$irst comment was nearly simultaneous with the closure, and my second comment was right a$f$terward, and before TCL made the inte$g$ral more explicit. What I hadn't realized at the time was that the integrals had actually been interpreted by many as indefinite integrals, and it's definitely a good thing that TCL cleared it up.2011-01-30

1 Answers 1

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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).