# 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 $\mathrm{Re}(z)^2 > 1$ ?

• you mean for $|x| > 1 , z = x + iy$ ? – mike Dec 23 '13 at 13:17
• @arbautjc Please make that an answer, since it clearly is. – WimC Dec 23 '13 at 13:52
• @arbautjc It's to get it off the unanswered list. Copy paste of the comment would be an honest answer. – WimC Dec 23 '13 at 17:32
• @WimC Ok, done :-) – Jean-Claude Arbaut Dec 23 '13 at 18:15
• I remark that there is an elementary answer to this as well. I mean there is an elementary function $f(z)$. I will get back to that later. – mick Feb 10 '14 at 22:29

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, $f\circ \theta$.

$$f(z)=\int_0^\infty \frac{e^{zt}}{t^t}\mathrm{d}t$$

• Could you transcibe the definition here? The page doesn't display on google books. – Avi Steiner Dec 23 '13 at 18:45
• Can you prove the boundedness of $f(z)$ ? Merry X-mas !! – mick Dec 26 '13 at 12:16
• Thank you arbautjc , Joseph Bak and Donald J. Newman !! – mick Dec 26 '13 at 21:01

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.

• so for any path to $\infty$ there is an entire function unbounded only in the neighborhood of that path ? – reuns Feb 13 '16 at 0:57
• Yes, that is correct. – user147263 Feb 13 '16 at 1:14