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ON A SYMMETRIC FUNCTIONAL EQUATION

  • Received : 2012.06.12
  • Accepted : 2012.07.01
  • Published : 2012.09.25

Abstract

We find a general solution $f:G{\rightarrow}G$ of the symmetric functional equation $$x+f(y+f(x))=y+f(x+f(y)),\;f(0)=0$$ where G is a 2-divisible abelian group. We also prove that there exists no measurable solution $f:\mathbb{R}{\rightarrow}\mathbb{R}$ of the equation. We also find the continuous solutions $f:\mathbb{C}{\rightarrow}\mathbb{C}$ of the equation.

Acknowledgement

Supported by : National Research Foundation of Korea(NRF)

References

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