Iterative functional equations related to a competition equation

• Published in: Aequationes mathematicae Vol. 89; no. 3; pp. 613 - 624
• Main Authors: Kahlig, Peter, Matkowski, Janusz
• Format: Journal Article
• Language: English
• Published: Basel Springer Basel 01.06.2015; Springer Nature B.V
• Subjects: Analysis; Combinatorics; Mathematical functions; Mathematical problems; Mathematics; Mathematics and Statistics; Primary 39B12; 39B22; Functional equation; solution depending on an arbitrary function; competition equation; iterative functional equation; differentiable solution
• ISSN: 0001-9054; 1420-8903
• DOI: 10.1007/s00010-013-0248-0

The diagonalization of a two-variable functional equation (related to competition) leads to the iterative equation f 2 x 1 - x 2 = 2 f ( x ) 1 + f ( x ) 2 , x ∈ R , x 2 ≠ 1 . It was shown in Kahlig (Appl Math 39:293–303, 2012 ) that if a function f : R → R , such that f (0) = 0, satisfies this equation for all x ∈ ( - 1 , 1 ) , and is twice differentiable at the point 0, then f = tanh ∘ ( p tan - 1 ) for some real p . In this paper we prove the following stronger result. A function f : R → 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 ∘ ( 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 , … ) .