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Is there a smooth funtion $f:\mathbb{R}\to \mathbb{C}$ such that

(a) $f(x)$ decreases faster than $e^{-e^x}$ when $x\to \infty$,

(b) $\widehat{f}(t)$ decreases faster than $e^{-|t|}$ when $t\to \pm\infty$?

Note there is no restriction on $f(x)$ for $x\to -\infty$ except that it decay fast enough for the Fourier transform $\widehat{f}$ to be well defined. The function $f_\epsilon(x) = e^{\epsilon x - e^x}$, $\epsilon>0$, decays almost a fast as $e^{-e^x}$ for $x\to \infty$ and $\widehat{f_\epsilon}(t)$ decays roughly as $e^{-|t|}$ for $t\to \pm \infty$ -- so I am really asking whether one can defeat $f_\epsilon$ on both the physical and Fourier aspect.

PS. Yes, this does come from a Mellin transform, as the double exponential probably gives away.

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I can, by the way, do much better than (a) while doing not uch worse than (b): take $f(x) = e^{-x^2}$. –  H A Helfgott Nov 8 '12 at 16:58
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How is $e^{-x^2}$ better than (a)? Also, $\hat f(t)$ decaying exponentially means that $f(x)$ extends to an analytic function on a horizontal strip and is $L^2$ on horizontal lines. Then the question comes down to finding a function analytic on a strip such that it decays very quickly on the positive real axis, and Phragmen Lindelof gives you bounds on how quickly it can decay (and how wide the strip may be) –  Ralph Furmaniak Nov 8 '12 at 19:55
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Sorry, forgot to change variables. Meant $f(x)=e^{e^{−2x}}$. Agreed, it's not much better than (a), and the Fourier transform is slightly worse than (b). – H A Helfgott 0 secs ago –  H A Helfgott Nov 9 '12 at 2:21
    
I know Phragmen-Lindelof only as a tool to get subconvexity, so I am not familiar with the argument you allude to - could you give us a reference (or a one-line explanation)? –  H A Helfgott Nov 9 '12 at 11:15
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There are some nice consequences of Phragmen-Lindelof (and its proof) along these lines in Titchmarsh's Theory of Functions. For example there's the following result of Carlson. Suppose $f(z)$ is holomorphic in some sector of interior angle $\theta$, is exponentially bounded $|f(z)|\ltlt e^{k|z|}$ and exponentially decays on the boundary. $\exp$ gives an example for $\theta<\pi$ and the result is that if $\theta=\pi$ then $f(z)\equiv0$. In fact, having such a non-trivial function would give you a stronger Phragmen-Lindelof principle that could prove that the exponential function does not e –  Ralph Furmaniak Nov 12 '12 at 1:36

1 Answer 1

How about $$\frac{e^{-e^x}}{1+\epsilon x^2}$$ If you compute the Fourier transform you can shift the contour to height $\pm\pi/2$ to get an $e^{-|t|}$ times something decaying to 1, by Riemann-Lebesgue lemma

Edit.

Or you can look at a shift of your original example: $f(x+\log A)$ to get something on the lines of $$e^{-A e^x} e^{-B x^2} e^{C x}$$

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This (roughly speaking) matches (a) and (b), perhaps a little more precisely than the example I gave. Do you think this is optimal? –  H A Helfgott Nov 9 '12 at 11:12
    
You can also replace the $1+x^2$ with any function growing in the strip, the best example being along the lines of $e^{A e^x}$. In fact, if you translate your original example you get something decaying much more quickly, without changing the modulus of the Fourier transform: $$e^{-A e^x}e^{B x}$$ –  Ralph Furmaniak Nov 9 '12 at 18:48
    
Thanks for this, but it's like the example I gave ($e^{e^{Cx}}$) - these variations due to rescaling improve one of the two aspects, but not the other (though one should be happy if they don't degrade it). I was wondering whether one can improve both... –  H A Helfgott Nov 10 '12 at 16:01
    
$e^{-e^{C x}}$ ends up hurting one of the bounds, but I think with $e^{-A e^x}$ you help the first super-exponentially without hurting the second at all, and when you add in the $e^{-x^2}$ you end up beating the second bound by something like $e^{-(\log t)^2}$ –  Ralph Furmaniak Nov 10 '12 at 20:03
    
I should really add "shifts not allowed", since they give no gain in the original problem, but perhaps that calls for a change in the setup to reflect that. –  H A Helfgott Nov 12 '12 at 9:52

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