Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$? 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  $\mathrm{Re}(z)^2 > 1$ ?
 A: Look at the function f defined in section 12.2 here, in Complex Analysis by Joseph Bak and Donald J. Newman.
It defines an entire function $f$ bounded outside the strip $|\mathrm{Im}(z)|\leq \pi$, and by a suitable linear transformation, namely $\theta(z)=i\pi z$, it's possible to get the function you want, $f\circ \theta$.
$$f(z)=\int_0^\infty \frac{e^{zt}}{t^t}\mathrm{d}t$$
A: 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.
