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Given $g:[1,\infty) \rightarrow (0,\infty)$ with $g(t) = o(t)$, does there exist $f:\mathbb{R} \rightarrow [0,\infty)$ with support contained in $[-1,1]$ such that $$ \widehat{f}(y) = \int_{\mathbb{R}} f(x)\exp(-2\pi i x y) dx = O(\exp(-g(|y|)))? $$

Motivation:

If $f$ is compactly supported and $\widehat{f}(y) = O(\exp(-c|y|))$, then $f$ is everywhere $0$. (See the question Compactly supported function whose Fourier transform decays exponentially?.)

The function $$ f = \left\{ \begin{array}{ll} \exp(-1/(1-x^2)) & \text{if } x\in (-1,1) \\ 0 & \text{otherwise} \end{array}\right. $$ has $\widehat{f}(y) = O(\exp(-\sqrt{|y|}))$. (See http://math.mit.edu/~stevenj/bump-saddle.pdf. By the way, does anyone have a textbook or published reference as an alternative to this note?)

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  • $\begingroup$ GummyWorm, Does $g$ decay or grow sub-exponentially? Are we describing its behavior about the origin, or as $y$ grows to infinity? $\endgroup$ – Jonathan Y. Sep 4 '13 at 21:57
  • $\begingroup$ Jonathan Y., it decays sub-exponentially as $y$ goes to infinity. I've corrected the question. $\endgroup$ – GummyWorm Sep 5 '13 at 1:44

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