# Searching for $f(z) \ne -1$, $\lim\limits_{t→+\infty} f(z-t) = 0$ [closed]

I am looking for a real-entire function* $$f(z)$$ such that for any (finite) complex $$z$$ we have both properties at once:

Property 1)

$$\lim_{t → +\infty} |f(z+t)| = \infty,\quad \lim_{t → +\infty} f(z-t) = 0,$$ where $$t$$ is real and

Property 2)

$$-1$$ is not in the range of $$f$$: $$f(z) \ne -1$$.

(* real-entire means a function is entire and maps the reals to a subset of the reals. In other words an entire function where the Maclaurin series has all real coefficients )

Any help would be appreciated.

*** Edit ***

Entire function $f(z)$ bounded for $\mathrm{Re}(z)^2 > 1$?

• posted it at mathoverflow too ; mathoverflow.net/questions/394336/searching-for-fz-ne-1 – mick Jun 2 at 11:27
• According to Weierstrass, $f(z)+1=e^{g(z)}$ where $g$ is entire. $f(\Bbb{R}) \subseteq \Bbb{R}$ implies $e^{g(\Bbb{R})} \subseteq \Bbb{R}$, so $g(\Bbb{R}) \subseteq \cup_{k \in \Bbb{Z}} (\Bbb{R}+k \pi i)$. Due to the connectness of $\Bbb{R}$, there exists $k_0$ such that $g(\Bbb{R}) \subseteq \Bbb{R}+k_0 \pi i$. Therefore $g-k_0 \pi i$ is a real-analytic function. – Zerox Jun 2 at 12:42
• @Zerox yes that is certainly true. However note that $f(z)$ is independent of your choice of $k_0$ since those are just the log branches from taking a log. – mick Jun 2 at 14:49
• It would be better if you could edit in the range interpretation of the third condition. – Lutz Lehmann Jun 2 at 15:05
• $f(z) = e^{e^z} - 1$ almost works except that the very first property fails if $\mathrm{Im} (z) = \frac{\pi}{2} + \pi k$. – Random Jun 2 at 19:37

I guess this is easy! Define $$g(z)\equiv f(z) + 1$$ This has no roots, therefore we may write $$g(z) = e^{h(z)}$$ Now, $$h(z)$$ must be an entire-real function that goes to zeros when $$|z|\rightarrow\infty$$ and $$\arg(z)\rightarrow\{0, \pi\}$$. There are many examples for this, as an example take $$h(z) = e^{-z^2}$$ which is equivalent to $$\boxed{f(z) = e^{e^{-z^2}}-1}$$