TITLE: Entire function bounded on every horizontal and vertical line , then is it bounded on every horizontal and vertical strip?
QUESTION [5 upvotes]: Let $f:\mathbb C \to \mathbb C$ be an entire function such that $f$ is bounded on every horizontal and every vertical line , then is it true that $f$ is bounded on any set of the form $V_{[a,b]}:=\{x+iy : y\in \mathbb R , a \le x \le b\}$ and any set of the form $H_{[a,b]}:=\{x+iy : x\in \mathbb R , a \le y \le b\}$ ?

REPLY [3 votes]: This is not true. Let $G\subset \mathbb{C}$ be the set $\{x+iy:|x|<\pi/2, y> -1, |y-\tan x|<1\}$. This is a connected open set that does not contain any line, or even a half-line. Let $E=\mathbb{C}\setminus G$. The function $f(z)=1/z$ is holomorphic on $E$. By Arakelyan's approximation theorem there exists an entire function $F$ such that $|F-f|<1/3$ on $E$. Consequently, 

$F$ is bounded on every line (not just on the vertical and horizontal ones)
$F$ is nonconstant, since a constant $c$ cannot satisfy $|c-f|<1/3$ on $E$.
Being nonconstant, $F$ is not bounded on $\mathbb{C}$. Yet it is bounded on $E$; thus, it is not bounded on $G$.   
Since the set $\{x+iy:y\in\mathbb{R},|x|\le \pi/2\}$ contains $G$, the function $F$ is not bounded on this set.

Remarks

To check the assumptions of Arakelyan's theorem, as stated on Wikipedia, take $\Omega=\mathbb{C}$, so that $\Omega^*\setminus E = G\cup\{\infty\}$, which is a connected set. It's important that $G$ stretches out to infinity.
Stronger tangential approximation is possible, where $|F(z)-f(z)|<1/|z|$ on $E$. The references in Wikipedia article should have this; in any case, Lectures on Complex Approximation by Gaier presents this and many other approximation theorems. In this case, $F$ tends to zero along every line in the complex plane.
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