Many Paley-Weiner theorems are variations on the theme "the faster a function $f(x)$ falls off as $x \rightarrow \infty$, the smoother its Fourier transform $\tilde{f}(k)$ is as $k \rightarrow 0$." In particular, we know that if $f(x)$ decays exponentially at large $x$ then $\tilde{f}(k)$ is analytic, and if $f(x) = 1/x^n$ (with $n \in \mathbb{Z}$) then $\tilde{f}(k) \propto k^{n-1} \mathrm{sgn}( k)$. But I'm wondering about the (inverse) FT of $\tilde{f}(k) = 1 - e^{-1/k^2}$. This function is smooth and falls off as $1/k^2$ at large $k$, so its inverse FT $f(x)$ should exist and not be too pathological. It should fall off faster than any power-law at large $x$, or else some finite derivative of $\tilde{f}(k)$ would be discontinuous, but $\tilde{f}(k)$ is smooth. But it cannot fall off as fast as an exponential, or else $\tilde{f}(k)$ would be analytic, which it isn't. So its large-$x$ falloff must lie somewhere in between power-law and exponential. But what is it? I have no idea how to evaluate the Fourier transform because of the essential singularity at $k = 0$.

Edit: Follow-up question: Is it true that any smooth function that falls off faster than any power-law but slower than any exponential has a non-analytic smooth FT?

  • 6
    $\begingroup$ A charming question, although I might disagree with the appraisal of Paley-Wiener theorems... but just for quibbly technical reasons. If no one else has much to say on this, I will respond tomorrow. A fun question, for sure! :) $\endgroup$ – paul garrett Aug 1 '16 at 23:03
  • 1
    $\begingroup$ Essentially, the answer is yes. If your function is holomorphic on a strip $|\textrm{Im}\, z |<a$, then the FT will have exponential decay at rate $e^{-(a-\epsilon)|x|}$ (and conversely). $\endgroup$ – user138530 Aug 1 '16 at 23:47
  • $\begingroup$ @ChristianRemling If it's holomorphic in that strip and satisfies some sort of boundedness on horizontal lines. I imagine that's what you meant - regardless, now we need a counterexample to what you actually said. Heh... $\endgroup$ – David C. Ullrich Aug 1 '16 at 23:58
  • $\begingroup$ @ChristianRemling Heh. Say $f(z)=\exp(ie^z)$. Then $f$ is bounded on the real axis but blows up exponentially on the line $y=-\epsilon$. (Then, say, $(f(z)-(zf'(0)+f(0)))/z^2$ is $L^1$ on the real axis but still very bad on the line $y=-\epsilon$...) $\endgroup$ – David C. Ullrich Aug 2 '16 at 0:20
  • 1
    $\begingroup$ @paulgarrett It seems that no one else has much to say on this ... ;) $\endgroup$ – tparker Aug 4 '16 at 1:31

I found a paper that uses the saddle-point approximation to derive the asymptotic form for large $k$ of the Fourier transform of another non-analytic smooth function: http://arxiv.org/abs/1508.04376. It falls off like $k^{-3/4} e^{-\sqrt{k}}$ - indeed faster than any power-law but slower than any exponential, as I expected.

If we use their method to compute $f(x) = \int_{-\infty}^\infty dk\, e^{i k x - 1/k^2} = 2\, \text{Re} \left[ \int_0^\infty dk\, e^{i k x - 1/k^2} \right]$, we get that the saddle point is $k_0 = \left( \frac{2 i}{x} \right)^{1/3}$. Taylor expanding the exponent about the saddle point, $$ i k x - \frac{1}{k^2} = i k_0 x - \frac{1}{k_0^2} + \frac{1}{2}\left( -\frac{6}{k_0^4} \right) (k - k_0)^2 + o \left( (k-k_0)^3 \right).$$ For large $x$, the integral oscillates wildly and mostly interferes destructively unless $k \ll 1$, and in this regime we also see that $k_0 \ll 1$, so the higher terms in the expansion are negligible for large $x$, and $$ f(x) \sim 2\, \text{Re} \left[ \exp \left( i k_0 x - \frac{1}{k_0^2} \right) \int_0^\infty dk\, \exp \left(-\frac{3}{k_0^4}(k - k_0)^2 \right) \right].$$ We now need to deform the contour so that it runs through the saddle point $k_0 = (2/x)^{1/3} i^{1/3}$, so we make the change of variable $k = u\, i^{1/3},\ dk = du\, i^{1/3}$, which rotates the contour by $-\pi/6$ in the complex plane: $$ f(x) \sim 2\, \text{Re} \left[ \exp \left( i k_0 x - \frac{1}{k_0^2} \right) \int_0^{i^{-1/3} \infty} du\, i^{1/3} \exp \left[ -\frac{3}{k_0^4} i^{2/3} \left( u - \left( \frac{2}{x} \right)^{1/3} \right)^2 \right] \right].$$ The only pole is at the origin, so we can deform the contour back onto the positive real line without affecting the integral. For large $x$, the integral from $-\infty$ to $0$ is negligible so we can extend the range of integration over the whole real line. (One might have expected that since the Gaussian is peaked near zero, the integral over the negative ray would contribute equally to the integral over the positive ray so we'd need a factor of 1/2 to compensate; see the paper for why this is not the case.) Since $\text{Re}\left( 3 i^{2/3}/k_0^4 \right) = \text{Re}\left( (2/x)^{1/3} e^{-i \pi/3} \right) = (4x)^{-1/3} > 0$, we can use the Gaussian integral identity $\int_{-\infty}^\infty dk\, e^{-a k^2} = \sqrt{\pi/a}$ to get $$f(x) \sim 2\, \text{Re} \left[ \exp \left( i k_0 x - \frac{1}{k_0^2} \right) i^{1/3} \sqrt{\frac{\pi k_0^4}{3 i^{2/3}}} \right] = 2 \sqrt{\frac{\pi}{3}} \text{Re} \left[ k_0^2 \exp \left( i k_0 x - \frac{1}{k_0^2} \right) \right].$$ Plugging in $k_0 = \left( \frac{2 i}{x} \right)^{1/3}$, we finally get after some algebra $$f(x) \sim 2^{5/2} \sqrt{\frac{\pi}{3}} x^{-2/3} \exp \left( -\frac{3}{2^{5/3}} x^{2/3} \right) \cos \left( \frac{\pi}{3} + \frac{3 \sqrt{3}}{2^{5/3}} x^{2/3} \right).$$ We do indeed find that the Fourier transform falls off faster than any power-law but slower than any exponential. But I'm not sure how to prove this result for an arbitrary nonanalytic smooth function.

  • $\begingroup$ Nice one...this deserves a lot more upvotes (+1) $\endgroup$ – tired Aug 22 '16 at 16:05
  • $\begingroup$ @tired Sounds like a job for a huge bounty ... ;) (just kidding). $\endgroup$ – tparker Aug 23 '16 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.