Consider a function whose Fourier transform is supported on a half-ray: $$ A(t)=\int_0^\infty \omega(E) e^{-iEt}d E, $$ where I can suppose $\omega(E)\geq 0$ and any suitable regularity conditions on $\omega(E)$ in the limit $E\rightarrow\infty$. I am interested in results constraining the rate of decay of $A(t)$ in the limit $t\rightarrow\infty$. Specifically, I would like to rule out the asymptote $\mathbf{|A(t)|\sim e^{-\Gamma t}}$.

The reference I have [1] uses the original 1934 Paley-Wiener theorem [2], which states that under these (or similar) assumptions for $A$ the integral $$ \int_{-\infty}^\infty \frac{\left|\ln|A(t)|\right|}{1+t^2}dt<\infty $$ must converge. This is strong enough to rule out the asymptote.

However, I have looked up the proof in Paley and Wiener and I find it far too technical and non-self-contained for me to follow with any ease; it also has an air of old mathematics that has probably been replaced with cleaner arguments by now. I do get some of the intuition behind the appearance of the $1/(1+t^2)$ factor. (Namely, a unitary transform of the upper half-plane $z$ space into the unit circle in $\zeta=i\frac{z+1}{z-1}$, where the measure transforms as $\frac{d\zeta}{\zeta}\approx\frac{dz}{1+z^2}$.) I still don't find, however, any intuition into how the can't-be-too-fast decay of $A$ correlates with the support of $\omega$, or at least no intuition that can be turned into a rigorous argument.

I am looking for references or arguments that prove in a clearer fashion that exponential decay of $A$ is impossible with such a Fourier domain, and particularly for ones that have clear intuition behind them that can be turned into a solid argument, even if the rigorous details are fiddly.

  1. L. Fonda, G. C. Ghirardi and A. Rimini. Decay theory of unstable quantum systems. Rep. Prog. Phys. 41, pp. 587-631 (1978). Page 592.
  2. R. Paley and N. Wiener. Fourier Transforms in the Complex Domain (Providence, Rhode Island: American Mathematical Society, 1934). Theorem XII, p16.
  • $\begingroup$ See p.158-160 of Dym and McKean (there is, however, antecedent material in that book that is necessary). $\endgroup$ Feb 22, 2013 at 15:01
  • $\begingroup$ Have you looked into Katznelson's book? $\endgroup$
    – Asaf
    Feb 22, 2013 at 15:11
  • 1
    $\begingroup$ Koosis, The logarithmic integral part I has a good treatment of this theory (part II treats the somewhat more complicated Beurling-Malliavin theorem). $\endgroup$ Feb 22, 2013 at 15:24
  • $\begingroup$ Hi Alexandre, Your contribution was very instructive but I have a couple of questions about what you wrote : - Can you comment on the difference between the integral $\int_{-\infty}^{+\infty}\mathrm{d}t\frac{|\log|f(t)||}{1+t^2}$ written by Emilio and the one you wrote which is $\int_0^{+\infty}\mathrm{d}t\frac{\log|f(t)|}{1+t^2}$ ? - Can you explain why $t$ is the argument of $f$ in your last integral ? It seems to me that the conformal mapping would map $\mathrm{e}^{\mathrm{i}\theta}$ onto $\left(t-\mathrm{i}\right)/\left(t+\mathrm{i}\right)$. Thanks in advance. V.D. $\endgroup$
    – user35276
    Jun 24, 2013 at 17:39
  • $\begingroup$ Related: Fastest decay of Fourier transform for generalized functions of compact support. $\endgroup$ Aug 19, 2013 at 19:50

1 Answer 1


The idea behind this theorem is actually simple.

For the Fourier transform to make sense (with usual understanding of the integral) the function $\omega$ must be summable, that is in $L^1(0,+\infty)$. This immediately implies that the Fourier integral converges not only on the real line but also in the lower half-plane, so $f$ is an analytic function in the lower half-plane which is bounded. Indeed, $$|f(t)|\leq\int_0^\infty|\omega(t)|e^{\Im t}dt\leq\int_0^\infty|\omega(t)|dt,$$ because $\Im t\leq 0$.

Now, there is a general principle that a bounded analytic function cannot be too small. This is easier to see in the unit disc. $\log|f(z)|$ is subharmonic which means that $$\log|f(0)|\leq\frac{1}{2\pi}\int_0^{2\pi}\log|f(e^{i\theta})|d\theta.$$ The integrand is bounded from above, so the integral of the NEGATIVE part of $\log|f(e^{i\theta})|$ must converge. (If $f(0)=0$ this is still true: just divide $f$ by an appropriate power of $z$; this will not change the right hand side). This is called Jensen's inequality.

Now, this fact that the negative part of $\log|f |$ is integrable, when transfered from the disc to the half-plane via conformal mapping, means exactly that the logarithmic integral $$\int_0^\infty\frac{\log|f(t)|}{1+t^2}dt$$ cannot diverge to $-\infty$, that is the Wiener-Paley theorem.

Of course, one can argue directly in the half-plane without the reference to the unit disc. Boundedness of $\log|f|$ from above implies that the Poisson integral of its boundary value on the real line must converge, which is exactly the same condition.

Now Beurling-Malliavin theorem says essentially that the condition of convergence of the log integral is best possible (subject to some technical regularity assumptions). Even if you consider $\omega$ of arbitrarily small compact support.

A good modern reference for all these things is the books of P. Koosis.

  • 1
    $\begingroup$ Nice answer! But which book? He has two books $\endgroup$
    – John
    Jun 25, 2019 at 3:59
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    $\begingroup$ @John: He has more than 2. But one of them is called Lecons sur les theoremes de Beurling-Malliavin (in French), and another The Logarithmic integral. Both contain detailed expositions of the topic. $\endgroup$ Jun 26, 2019 at 4:07

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