1
$\begingroup$

One of the variations of the Paley-Wiener theorem yields: If $f\in L^2(\mathbb R_+)$, then the Fourier transform $F$, defined by $$F(\zeta)=\int_0^\infty f(x)e^{ix\zeta}dx$$ is a holomorphic function in the upper half-plane $\{\zeta:0<\arg(\zeta)<\pi\}$.

What I'd like to state is that there exists $\epsilon>0$ such that the function $F$ is analytic in the sector $S_\epsilon=\{\zeta:-\epsilon<\arg(\zeta)<\pi+\epsilon\}$.

Is there any restrictions, I can make on $f$ in order to say that $F$ is holomorphic in $S_\epsilon$ for some $\epsilon>0$?

$\endgroup$
1
$\begingroup$

If $f$ is supported on $[-N,N]$, then the other Paley-Wiener theorem says $F(\zeta) = \int_{-N}^N f(z)\; e^{ix\zeta}\; dx$ is an entire function. So that's certainly one restriction you can put on.

An obstruction to the kind of conclusion you want: if $f(x) = \exp(-bx)$ with $\text{Re}(b) > 0$, then $\int_0^\infty f(x)e^{ix\zeta}\; dx $ diverges for $\text{Im}(\zeta) = - \text{Re}(b)$, with a pole at $\zeta = -ib$, so even if you allow analytic continuation rather than requiring convergence of the integral, these functions can't satisfy your property if $\text{Im}(b)/\text{Re}(b)$ is large enough.

$\endgroup$
  • $\begingroup$ Thank you for the comment. I like your example but if the analytic continuation of $\int_0^\infty f(x)e^{ix\zeta}dx$ has only one pole at $\zeta=-ib$ then I can always find $\epsilon>0$ small enough so that it is holomorphic in $S_\epsilon$. What I am worried about is the case when $F(\zeta)$ has infinitely many poles. For instance, $F(\zeta)=\tan(\zeta+i)$ $\endgroup$ – Aleksandr Aug 14 '14 at 0:58

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.