Question

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.

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.