**Question.** Does there exist an entire function $h$ satisfying three following assertions:


* $h$ belongs to the $H^2$ Hardy space in every horizontal upper half-plane;
* $zh - 1$ belongs to $H^2(\mathbb{C}\_+)$, where $\mathbb{C}\_+ = \{\text{Im}(z) > 0\}$;
* $h$ has infinitely many zeroes in some horizontal strip $\{\text{Im}(z) \in [y\_1, y\_2] \}$?


Does the answer change if we additionally assume that $h$ is of finite order?


**On the first assertion.** Assume that the function $H$ belongs to the Hardy space in every upper horizontal half-plane. and let $G = \mathcal{F}H\in L^2[0, +\infty)$. In terms of $G$ the first assertion is equivalent to the fact that $G e^{\delta x}\in L^2[0, \infty)$ for every $\delta > 0$ and $H(z) = \int\_0^{\infty} G(x)e^{izx}dx$ for all $z\in \mathbb{C}$.


In this situation $G$ can be expressed as $G = G\_1 + G\_2$ where $G\_1\in C^1[0,\gamma]$ for some $\gamma > 0$ and $\|G\_2\|\_{L^1[0,\infty)} < \varepsilon$. Via this decomposition one can show that $H(z)$ tends to $0$ as $|z|\to \infty$ uniformly in every horizontal strip (consequently $H(z) = w$ has only finitely many roots for all $\omega\in \mathbb{C}\setminus\{0\}$).


**Construction attempts.** To fulfil the first two assertions one can take function $H$ from the previous paragraph and put $h = (1 + H)/(z - z\_0)$ where $H(z\_0)= -1$. However in this situation $h(z) = 0 \Longleftrightarrow H(z) = -1$, which has only a finite number of roots in any horizontal half-plane.


Furthermore, by the Paley-Wiener theorem, the third assertion cannot hold if $\mathcal{F}h$ has a compact support.



Existence of the special entire Hardy space function with infinitely many zeros in the strip
Let
$$f(z)=\frac{e^{iz}-1}{iz}.$$
This function is in the Hardy class for any upper half-plane,
and has these properties: $f(0)=1,$ $f(2\pi n)=0$,
$$|f(z)|\leq C\frac{e^y+1}{|z|+1},$$
(this evidently holds for large and small $|z|$, therefore there is a constant $C$ so that this holds everywhere).


Since the $L^2$ norm has the property $\| f(kx)\|^2=\| f(x)\|^2/k,$ and the norm is invariant under real translations, the following function $g$ belongs to the Hardy spaces for upper half-planes:
$$g(z)=\sum\_{n=0}^\infty f(n^2(z-a\_n)),$$
where $a\_n\in\{2\pi k: k\in {\mathbf{Z}}\}$.
Moreover, it is entire, if the sequence $a\_n$ grows fast enough. (It follows from the estimate for $f$ that the series is uniformly convergent on any compact subset of the plane).
Now $g(a\_n)=1$, so $h(z)=(g(z-i)-1)/(z-i)$, where has all properties that you required.


Some extra labor is needed to show that $h$ can be made of finite order.


