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   <subfield code="a">Iterative functional equations related to a competition equation</subfield>
   <subfield code="h">[Elektronische Daten]</subfield>
   <subfield code="c">[Peter Kahlig, Janusz Matkowski]</subfield>
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   <subfield code="a">The diagonalization of a two-variable functional equation (related to competition) leads to the iterative equation $$f\left( \frac{2x}{1-x^{2}}\right) =\frac{2f( x)}{1+f( x) ^{2}},\quad x\in {\mathbb{R}},\, x^{2}\neq 1.$$ 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 $${f:{\mathbb{R}}\rightarrow {\mathbb{R}}}$$ f : R → R , such that f(0) =0, satisfies this equation for all $${x\in (-1,1),}$$ x ∈ ( - 1 , 1 ) , and is twice differentiable at the point 0, then $${f=\tanh \circ (p\,\tan ^{-1}) }$$ f = tanh ∘ ( p tan - 1 ) for some real p. In this paper we prove the following stronger result. A function $${f:{\mathbb{R}} \rightarrow {\mathbb{R}},\;f(0) =0}$$ 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 $${f=\tanh \circ (p\,\tan ^{-1})}$$ 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 $${(n=5, 6, 7, \ldots)}$$ ( n = 5 , 6 , 7 , ... ) .</subfield>
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