A modified Paley–Wiener theorem with weaker condition

Question

Let's consider the following argument: let $f$ be a function in $L^2(\mathbb R)$ such that $\hat f$ extends to an entire function on $\mathbb C.$ Assume that for each $t>0$ and $x \in \mathbb R$ $$ \int_0^t \hat f(k+is) e^{ix(k+is)} ds \to 0 $$ as $k \to \infty.$ Then $$ f(x) = \int_{-\infty}^\infty \hat f(k) e^{ikx} dk= \int_{-\infty}^\infty \hat f(k+is) e^{i(k+is)x} dk $$

for all $s.$

Now also assume that $|\hat f(z)|\leq g(\Re z) e^{\lambda |\Im z|}$ for some $g\in L^1(\mathbb R),\lambda.$ Then, we have the estimate $$ \left|\int_{-\infty}^\infty \hat f(k+is) e^{i(k+is)x} dk\right | \leq e^{(\lambda-|x|)|s|} \int_{-\infty}^\infty g(k) dk $$ This $\to 0$ as $s\to \infty$ if $|x|\geq \lambda.$ Hence $f$ is compactly supported in $[-\lambda, \lambda].$

Paley–Wiener theorem requires the stronger estimate $$ |\hat f(z)|\leq C(1+|z|)^{-N}e^{\lambda |\Im z|} $$ so that the above argument works. However, this seems way stronger than what we need to make it work. Can we weaken the condition?

For instance, a condition like $\hat f|_\mathbb R \in L^p(\mathbb R).$

Does anyone know any existing results of this kind?

Answer 1

You are right: assumptions of the (converse part) of the Wiener-Paley theorem can be substantially relaxed.

The usual formulation of this converse part says that if $F$ is an entire function of exponential type (say $\lambda>0$), and the restriction of $F$ on the real line is in $L^2$, then $F$ is a Fourier transform of an $L^2$ function with bounded support (bounded by $\lambda$).

In your problem, you know a priori that $f\in L^2$, and thus $F=\hat{f}\in L^2$, so the only condition needed is that $F$ is of exponential type. This means $$\log|F(z)|\leq\lambda|z|+o(z),\quad z\to\infty$$ Or, which is equivalent, $$\frac{1}{2\pi}\int_0^{2\pi}\log^+|F(re^{i\theta})|d\theta\leq (\lambda+o(1))r, \quad r\to\infty.$$ So, for example, if you know $|F(z)|\leq g(|x|)e^{\lambda|y|}$, where $g>0$, then it is sufficient to check $$\int_0^\pi\log^+ g(r\cos\theta)d\theta=o(r),\quad r\to\infty,$$ or even that this holds for some sequence $r_k\to\infty,$ to ensure that the support of $f$ is bounded by $\lambda$.

This can be further relaxed by using the Phragmen-Lindelof Principle. If you know that $f\in L^2$ on the real line, and also know that $\log|F(iy)|\leq \lambda|y|$, then it is enough to assume additionally that $$\log|F(z)|=O(|z|^{2-\epsilon}), \quad |z|=r_k\to\infty,$$ for some sequence $r_k$ to conclude that $F$ is in fact of exponential type $\lambda$.

For all these things I refer to the books of B. Y. Levin:

1. Lectures on entire functions, and

2. Distribution of zeros of entire functions.

Both are available on the web.

Remark. The stronger condition in Wiener-Paley that you cited is needed to assure that the inverse transform is in $L^2$ (rather than some distribution or hyperfunction). But in your problem this is known from the beginning. If we only care of support of $f$ (and allow $f$ to be a hyperfunction) then the neseccary and sufficient condition for support to be on $[-\lambda,\lambda]$ is that $$\log|F(re^{i\theta})|\leq \lambda r|\sin\theta|+o(r).$$