**Question.** Is there an entire function $F$ satisfying first two or all three of the following assertions:


* $F(z)\neq 0$ for all $z\in \mathbb{C}$;
* $1/F - 1\in H^2(\mathbb{C}\_+)$ -- the classical Hardy space in the upper half-plane;
* $F$ is bounded in every horizontal half-plane $\{z\colon \text{Im}(z) > \delta\}$?


**Thoughts**. Let $G= 1/F$. Then we have $G(z) = 1 + \int\_0^{\infty} h(x)e^{izx}\, dx$ for some $h\in L^2[0,\infty)$ and all $z\in \mathbb{C}\_+$. For nice functions $h$ (e.g., for super-exponentially decreasing) this integral representation can be extended to the whole complex plane and probably the example can be constructed in terms of $h$. However, I don't know if it is possible to find $h$ such that $G$ is non-zero for every $z$.



Existence of nonzero entire function with restrictions of growth
There is a zero-free entire function bounded in every left half-plane, and such that $f-1$ is in $H^2$ in every left half-plane.


Let $\gamma$ be the boundary of the region $$D=\left\{ x+iy: |y|<2\pi/3, x>0\right\}
.$$ Consider the function
$$g(z)=\int\_\gamma \frac{\exp e^\zeta}{\zeta-z}d\zeta,\quad z\in {\mathbf{C}}\backslash D.$$ The integral evidently converges and $g(z)=O(1/z)$ in ${\mathbf{C}}\backslash D.$ Now, $g$ has an analytic continuation to an entire function: deforming the contour to $\partial D\_t$,
where $D\_t=\{ x+iy:|y|<2\pi/3, x>t\},\; t>0$ does not change $g$ in $D$,
and shows that $g$ has an analytic continuation to ${\mathbf{C}}\backslash D\_t$, and this is for every $t>0$, so $g$ is entire. Now $f(z)=e^{g(z)}$ is the desired function. If you want upper half-planes take $f(iz)$.


Remark. You can improve the estimate $g(z)=O(1/z)$.
Evidently, $g$ has infinitely many zeros $z\_1,z\_2,\ldots$. Then $g\_k(z)=g(z)/((z-z\_1)\ldots(z-z\_k))$
satisfies $g\_k(z)=O(z^{-k-1})$ as $z\to\infty$ outside $D\_t$.


Remark 2. This construction is standard in the theory of entire functions, see, for example, [Entire function bounded at every line](https://mathoverflow.net/questions/190837/entire-function-bounded-at-every-line/190874#190874)
Sometimes this $g$ is called the Mittag-Leffler function.


