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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 $f: \mathbb C \to \mathbb C$

  • so, those strips are lines ? – mercio Mar 29 '17 at 15:51
  • @mercio : yes .. for now .. I am only asking for lines ... –  Mar 29 '17 at 16:31
  • Every what function? Complex valued? Complex vector valued? There are lots of kinds of functions. – Stella Biderman Mar 29 '17 at 16:32
  • @StellaBiderman : I hope it is clear now ... –  Mar 29 '17 at 16:34
  • Hmm, you really mean lines? Just today, I learned of a very strange function. Sorry no references! This function is entire, non-constant, yet for every real line in the complex plane, $f(z)$ goes to zero as $z\to\infty$ along that line. The existence of such wild functions makes me doubt that you can prove a result along the lines (no pun intended) you suggest. – Harald Hanche-Olsen Mar 29 '17 at 20:00
  • Can you give some examples of how this conjectured decomposition works? I'm curious about $f(z)=e^{(1+i)z}$. – Chris Culter Mar 29 '17 at 21:30
  • @HaraldHanche-Olsen : but how does your example go against the spirit of such a decomposition ? –  Mar 30 '17 at 06:29
  • The point of the example is that the behaviour of functions along real lines provides very little information about the function's general behaviour. This is, of course, a bit vague. But I have a much easier time imagining a theorem where the hypotheses deal with boundedness along strips of positive width. I have no idea how to proceed even in that case, though. – Harald Hanche-Olsen Mar 30 '17 at 12:45
  • It can be done with Arakelian's approximation theorem in a similar way as in my answer here http://math.stackexchange.com/a/1195779/1321. In fact you can decompose into entire functions bounded on the regions $\lvert\Re(z)\rvert\ge\lvert\Im(z)\rvert+1$ and $\lvert\Im(z)\rvert\ge\lvert\Re(z)\rvert+1$ respectively. – George Lowther Apr 05 '17 at 10:59

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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.

George Lowther
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  • What about $f(z) = z^n$ ? – reuns Apr 06 '17 at 04:40
  • Nothing special, the result applies to $z^n$, but it doesn't give any simple expression for the decomposition. – George Lowther Apr 06 '17 at 08:36
  • It's not difficult to explicitly construct decompositions for polynomial or any functions which do not grow too quickly. – George Lowther Apr 06 '17 at 09:53
  • Investigating the solution $G_n(z)$ for $f(z)= z^n$ might provide an explicit solution for any entire function, if $G_n$ doesn't increase too fast as $n \to \infty$. And I got stuck finding what $G_1$ can be – reuns Apr 06 '17 at 20:45
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    If $N(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^xe^{-u^2/2}du$ is the standard normal distribution function then $N(-x^2)$ tends to 0 along horizontal strips and 1 along vertical strips. You can take $G_n(x)=N(-x^2)x^n$. – George Lowther Apr 07 '17 at 10:32
  • Of course, great. For $f(z)$ not growing too fast the solution is simply $N(-z^2)f(z) $. And we can generalize this : replace $\frac{1}{2\pi}e^{-u^2/2}$ by $C e^{-h(u^2)}$ where $h(u)$ is an entire function positive for $> 0$ and growing fast enough as $u \to \infty$ – reuns Apr 07 '17 at 12:09