1
$\begingroup$

Let $A=\{ z \in \mathbb C : |\mathrm{Im}(z)| < a \}$ for some $a >0$. Further let $F:A \to \mathbb C$ be a holomorphic function on $A$ with the following properties:

  1. $F$ is bounded on $A$
  2. For every $y \in (-a,a)$ the map $x \mapsto |F(x + iy)|$ has a certain decay to zero as $|x| \to \infty$, for instance an exponential decay.

I'm wondering if for such functions certain uniqueness results holds true, in the sense that if $F$ vanishes on a certain discrete set on the real axis (say the integers $\mathbb Z$) then $F$ vanishes identically. Is anyone of you familiar with results in this direction?

$\endgroup$
1
  • 3
    $\begingroup$ Maybe the discrete set $\mathbb Z$ is not enough, as there is the counterexample $F(x)=\exp(-x^2)\sin(x)$. $\endgroup$ Jan 3, 2022 at 11:40

1 Answer 1

3
$\begingroup$

Decrease on horizontal lines and density of zeros are two independent things.

  1. A bounded function cannot have too many zeros. This is a consequence of Jensen's inequality which implies the Blaschke condition. The Blaschke condition is usually stated for the unit disk or for the upper half-plane. Use a conformal map of your strip onto the upper half-plane $z\mapsto\exp(\pi z/(2a))$. In your example, let $z_k=x_k+iy_k$ be the sequence of zeros; if $f\neq 0$ then the Blaschke condition becomes $$\exp\left(-\frac{\pi}{2a}|x_k|\right)\sin\left(\frac{\pi}{2a}|y_k|\right)<\infty.$$ If this series is divergent then $f=0$.

  2. A bounded function cannot decrease too fast. This is a form of Phragmen-Lindelof theorem. In your situation, if $$\exp\left(\frac{\pi}{2a}|x|\right)\log|f(x+iy)|\to-\infty,$$ as $x\to+\infty$ or $x\to-\infty$, then $f=0$.

A good reference is is B. Ya. Levin, Lectures on entire functions, AMS, 1996. The mentioned form of the Phragmen-Lindelof theorem is Exercise 1 on p. 40, and Blaschke's condition in formula (6) on p. 219.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.