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Let $f\in S_\infty\subset L_1(\mathbb{R},\mu)$ with $\mu$ as the Lebesgue linear measure be a Lebesgue-summable function such that $$\forall (p,q)\in\mathbb{N}^2_{\ge 0}\quad\exists C_{pq}>0: \Bigg|x^p\frac{d^q}{d x^q}f(x)\Bigg|< C_{pq}$$ I was wondering whether, if $\forall p\in\mathbb{N_{\ge 0}}\quad\int_{\mathbb{R}}x^pf(x)d\mu=0$, then $f$ is constantly, or almost everywhere, null. I cannot find a counterexample and therefore I think that the implication might well hold, but I cannot prove it either.

Since $x^p f(x)$ belongs to $S_\infty\subset L_1$, thanks to the fact that $f$ belongs to it, and is continuous I think that the Lebesgue integral and the Riemann improper integral $\mathscr{R}\int_{-\infty}^\infty t^pf(t)dt$ are the same. Please correct if I am wrong.

Nevertheless I cannot use calculus facts to prove the desired implication... I have also tried using the fact that the Fourier transform induces a bijection $S_\infty\to S_\infty$, but with no result. What do you think about it? Has anybody got a counterexample or proof? Thank you very much!

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Hint: Up to a constant, you have $\int x^p f(x) \, dx = \widehat{x^p \cdot f}(0)$.

Now use that $\mathcal{F}(x^p \cdot f) = \partial^p \widehat{f}$ (again up to a constant, depending on your definition of the Fourier transform) to construct some $f$ for which $\int x^p f\, dx$ vanishes for all $p$, but $f \not \equiv 0$.

(Essentially, construct $\widehat{f} \in S_\infty$ and define $f$ as $\mathcal{F}^{-1} \widehat{f}$.)

EDIT: Further comment: The condition $\int x^p \cdot f =0$ for a range of $p$ is (at least in the theory of Wavelets) known as a vanishing moment condition and the above considerations imply that vanishing moments are more or less equivalent to the vanishing of a range of derivatives of the Fourier transform at the origin.

See also this: What is a "vanishing moment"?

EDIT 2: If $f$ has compact support, then one can show that the Fourier transform is an analytic function. Hence, if $\partial^p \widehat{f}(0) = 0$ for all $p$, then $\widehat{f} \equiv 0$ and hence $f \equiv 0$.

Hence, functions with infinitely many vanishing moments can not be compactly supported.

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