On the functional equation x + f(y + f(x))=y + f(x + f(y))

For an abelian group (G, + ,0) we consider the functional equation $$f : G \to G, x + f(y + f(x)) = y + f(x + f(y)) \quad (\forall x, y \in G), \quad\quad\qquad (1)$$ most times together with the condition $$f(0) = 0.\qquad\qquad\qquad\qquad\qquad (0)$$ Our main question is whether a solution of ${(1) \wedge (0)}$ must be additive, i.e., an endomorphism of G. We shall answer this question in the negative (Example 3.14) Rätz (Aequationes Math 81:300, 2011).     

On the functional equation $$f(x+y)=f(x)+f(y)$$f(x+y)=f(x)+f(y)

Monatshefte für Mathematik - An extension of the linearity of the continuous solutions of the functional equation $$f(x+y)=f(x)+f(y)$$ to measurable functions is presented.     

On the system of functional equationsf(x + y) = f(x) + f(y) andf(xy) = p (x)f(y) + q(y)f(x)

Aequationes mathematicae -     

On the functional equationf(x) g(y) = p(x + y) q(x/y)

Aequationes mathematicae -     

Über die Funktionalgleichung f(f(x)*f(y))=f(x*y)

In this paper, the general solution of three functional equations occuring in interpolation theory are given in the general situation when variables belong to a semi-group with unit. We then specialize to some particular cases by specifying either the semi-group, or adding some regular properties like continuity or measurability.     

Sur l'équationf(x)f(y)f(−x−y)=g(x)g(y)g(−x−y)

Sans résumé

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