Is every entire function is a sum of an entire function bounded on every horizontal strip and an entire function bounded on every vertical strip?

Is it true that every entire function is a sum of an entire function bounded on every horizontal strip (horizontal strip is a set of the form $H_y:=\{x+iy : x \in \mathbb R \}$ ) and an entire function bounded on every vertical strip (vertical strip is a set of the form $V_x:=\{x+iy:y\in \mathbb R \}$) ? I see no way of rigorously deciding it anyway.

NOTE : By entire function , I mean any holomorphic function $f: \mathbb C \to \mathbb C$

Solution 1:

The existence of a decomposition can be proved using Arakelian's approximation theorem. Consider the two regions in the complex plane \begin{align} S_1 &=\left\{z\in\mathbb C\colon \lvert\Re(z)\rvert\ge\lvert\Im(z)\rvert+1\right\}.\\ S_2 &=\left\{z\in\mathbb C\colon \lvert\Im(z)\rvert\ge\lvert\Re(z)\rvert+1\right\}. \end{align} These are disjoint closed sets with $S_1$ containing each horizontal strip up to a bounded set and $S_2$ containing each vertical strip up to a bounded set.

Their union $S=S_1\cup S_2$ is an Arakelian set. That is, it is a closed subset of $\mathbb{C}$ whose complement does not contain any bounded connected components and such that, for each closed disk $D$, the union of the bounded components of $\mathbb{C}\setminus(S\cup D)$ is bounded (it is empty, so trivially bounded).

For an entire function $f$, define $G\colon S\to\mathbb C$ by $G(z)=f(z)$ on $S_2$ and $G(z)=0$ on $S_1$. Then, Arakelian's approximation theorem states that, for any $\epsilon > 0$, there is an entire function $g$ with $\lvert g-G\rvert\le\epsilon$ on $S$. Setting $h=f-g$ decomposes $f=g+h$, with $g$ bounded on $S_1$ and hence on horizontal strips, and $h$ bounded on $S_2$ and hence on vertical strips.