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

被引：3

作者：

Raetz, Juerg ^{[1
]}

机构：

[1] Univ Bern, Math Inst, CH-3012 Bern, Switzerland

来源：

AEQUATIONES MATHEMATICAE | 2013年 / 86卷 / 1-2期

关键词：

Abelian groups; composite functional equations;

D O I：

10.1007/s00010-013-0188-8

中图分类号：

O29 [应用数学];

学科分类号：

070104 ;

摘要：

For an abelian group (G, + ,0) we consider the functional equation f : G -> G, x + f(y + f(x)) = y + f(x + f(y)) (for all x, y is an element of G), (1) most times together with the condition f(0) = 0. (0) Our main question is whether a solution of (1) boolean AND (0) must be additive, i.e., an endomorphism of G. We shall answer this question in the negative (Example 3.14) Ratz (Aequationes Math 81:300, 2011).

引用

页码：187 / 200

页数：14

共 6 条

[1] On the functional equation x plus f(y plus f(x)) = y plus f(x plus f(y)) [J].