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Fix a smooth $\mathcal{C}^{\infty}$ compactly supported function $f$ with the support of $f$ being the unit interval $(-1,1)$ and with $\hat{f} \geq 0$. Is it true that as $k$ goes to infinity $$ \int_{\mathbb R} x^{2k} \cdot \hat{f}(x) dx \gg (2k)! \cdot A^{-k} $$ for some large fixed $A > 0$? I think the answer is affirmative, but I haven't been able to fully justify it to myself. I'm posting this in the hope that there is a slick and short argument.

guest0101
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  • Bernstein's inequality for bandlimited functions (in your case $\hat{f}$ is bandlimited) might be relevant to you (gives you $L^p$ bounds on the derivatives of $\hat{f}$... okay maybe not relevant at all, actually), and it is an upper bound... Also, $f=0$ is a trivial counterexample, so something needs tweaking (perhaps the bump satisfies $|f|_1 = 1$ or some other normalization) – Evan Sep 06 '13 at 00:16
  • Support equal to (-1,1) means that the function is non-zero on this open interval, so $f = 0$ is not an admissible counterexample... – guest0101 Sep 06 '13 at 00:23
  • Since $\widehat{f}$ is entire, I was thinking of deforming the contour but haven't gotten very far so far... – guest0101 Sep 06 '13 at 00:29
  • Ah okay. My thought was to use repeated integration by parts to give you something of that form? You can probably use the fact that $\hat{f}$ is a Schwartz class function (compactly supported F.T. shows smoothness, smoothness of F.T. shows exponential decay). I do not know the role of $k^{-A}$ though. – Evan Sep 06 '13 at 00:30
  • Hm... I'm actually starting to doubt this lower bound... Let me think some more about it. – guest0101 Sep 06 '13 at 00:39

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