TITLE: Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$?
QUESTION [8 upvotes]: Let $z$ be a complex number and $\mathrm{Re}$ denote the real part.
Does there exist a nonconstant entire function $f(z)$ such that $f(z)$ is bounded for  $\mathrm{Re}(z)^2 > 1$ ?

REPLY [3 votes]: A more general statement follows from an approximation theorem due to Alice Roth, which can be found in the book Lectures on Complex Approximation by Dieter Gaier.  

Theorem. Suppose $F\subset \mathbb{C}$ is a closed set such that its complement is connected and locally connected at $\infty$. Let $f$ be a function that is holomorphic on an open subset containing $F$. Then for every $\epsilon>0$ there exists an entire function $g$ such that 
  $$|f(z)-g(z)| < \min(\epsilon, |z|^{-1}),\quad z\in F$$

For example, letting  $F=\{z: |\operatorname{Re} z|\ge 1\}$ and $f(z)=1/z$ yields a nonconstant entire function that is bounded on $F$ (and also tends to zero away from a horizontal strip).  
But one can go further and let, for example, $F=\{x+iy: |y|\ge \exp(-x^2)\}$; that is, the infinite strip can be narrowing at infinity, with any rate one wishes.