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.
- L. Fonda, G. C. Ghirardi and A. Rimini. Decay theory of unstable quantum systems. Rep. Prog. Phys. 41, pp. 587-631 (1978). Page 592.
- R. Paley and N. Wiener. Fourier Transforms in the Complex Domain (Providence, Rhode Island: American Mathematical Society, 1934). Theorem XII, p16.
Best Answer
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.