Do there exist three non-constant entire functions $f,g,h:\mathbb{C}\to\mathbb{C}$ such that for any $z\in\mathbb{C}$, at least two of $f(z)$, $g(z)$ and $h(z)$ belong to the closed unit disk?
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2$\begingroup$ Yes, but I doubt that there are three such where two of them are not constant. Gerhard "Is That All There Is?" Paseman, 2017.03.14. $\endgroup$– Gerhard PasemanCommented Mar 15, 2017 at 3:02
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5$\begingroup$ The answer is positive. You can e.g. use Arakelyan's theorem to construct entire functions that are large only on a very thin half-strip (say). Just use three pairwise disjoint half strips. $\endgroup$– Lasse RempeCommented Mar 15, 2017 at 3:46
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$\begingroup$ @Gerhard Paseman I edited the question so that it excludes constant functions. $\endgroup$– Erfan SalavatiCommented Mar 15, 2017 at 4:25
2 Answers
To elaborate on my comment, choose three pairwise disjoint Jofrdan $U_1$, $U_2$, $U_3$ whose boundary passes through infinity, and let $\gamma_i\subset U_i$ be a curve to infinity for each $i$. Fix $\newcommand{\eps}{\varepsilon}\eps>0$ (e.g. $\eps=1$).
By Arakelyan's approximation theorem (see, e.g. this exposition by Rosay and Rudin), there are non-constant entire functions $f_i$ such that $|f_i(z)|\leq \eps$ for $z\not\in U_i$. (Simply approximate the function $g$ defined by $g(z)=0$ on $\mathbb{C}\setminus U_i$, and $g(z)=|z|$ on $\gamma_i$ up to an error of at most $\eps$.)
Since the $U_i$ were chosen pairwise disjoint, they satisfy your requirement. Clearly you can use the same argument to even construct countably many entire functions $f_1, f_2, \dots$ such that, for all $z$, only one of the $f_i$ takes a value of modulus greater than $1$.
Remark 1. There are other ways of constructing functions that are small outside some large sets, without invoking Arakelyan's theorem. These can also be used to give an answer to your question. But I think this argument shows that nothing along the lines of what you seem to be looking for is likely to be true.
Remark 2. Arakelyan's theorem tells you nothing about the order of growth of the functions in question. By (the argument used to prove) the Denjoy-Carleman-Ahlfors theorem, if you have $n$ functions with the desired properties, at least one thereof must have order at least $n/2$. In particular, you cannot make the infinite collection mentioned above using finite-order functions. Using e.g. Cauchy integrals, it is possible to show that, for each $n$, such a collection of functions does indeed exist. (Alternatively, you can explicitly use Mittag-Leffler functions.)
The simplest example is constructed using Mittag-Leffler functions which are bounded outside a the sector $|\arg z|<\alpha$. They are defined by the integral $$f(z)=\int_\gamma\frac{e^{\zeta^{\alpha}}d\zeta}{\zeta-z}$$ with an appropriately chosen contour $\gamma$. As $f$ is bounded (and even tends to $0$ outside the sector $|\arg z|<\alpha$), the pair $f(z),f(-z)$ has the required property when $\alpha<\pi/2$.
Taking the angle less than $2\pi/3$ you construct three functions, or any number you want, so that at every point all but one are less than $1$ in absolute value.
A modification of this is $$\int\frac{e^{e^\zeta}d\zeta}{\zeta-z}$$ which is bounded outside of the strip $\{ z=x+iy:|y|<\pi\}$.