Aequationes mathematicae

, Volume 89, Issue 3, pp 613–624

# Iterative functional equations related to a competition equation

Article

## Abstract

The diagonalization of a two-variable functional equation (related to competition) leads to the iterative equation
$$f\left( \frac{2x}{1-x^{2}}\right) =\frac{2f( x)}{1+f( x) ^{2}},\quad x\in {\mathbb{R}},\, x^{2}\neq 1.$$
It was shown in Kahlig (Appl Math 39:293–303, 2012) that if a function $${f:{\mathbb{R}}\rightarrow {\mathbb{R}}}$$, such that f(0) = 0, satisfies this equation for all $${x\in (-1,1),}$$ and is twice differentiable at the point 0, then $${f=\tanh \circ (p\,\tan ^{-1}) }$$ for some real p. In this paper we prove the following stronger result. A function $${f:{\mathbb{R}} \rightarrow {\mathbb{R}},\;f(0) =0}$$, differentiable at the point 0, satisfies this functional equation if, and only if, there is a real p such that $${f=\tanh \circ (p\,\tan ^{-1})}$$. We also show that the assumption of the differentiability of f at 0 cannot be replaced by the continuity of f. The corresponding result for the iterative equation coming from a three- respectively four-variable competition equation is also proved. Our conjecture is that analogous results hold true for the diagonalization of any n-variable competition equation $${(n=5, 6, 7, \ldots)}$$.

## Mathematics Subject Classification (1991)

Primary 39B12 39B22

## Keywords

Functional equation competition equation iterative functional equation differentiable solution solution depending on an arbitrary function

## References

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