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NOTE ON ROBINSON’S FUNCTIONAL EQUATION

NOTE ON ROBINSON’S FUNCTIONAL EQUATION Krzysztof PiejkoNOTE ON ROBINSON'S FUNCTIONAL EQUATIONAbstract. The purpose of this note is to give a new proof of the fact, that theonly entire solutions of the Robinson's functional equation are given by f(z) = Az orf(z) = A sin az, where A, a are complex constants and a is real or purely imaginary.Robinson in [3] solved the following functional equation(1)\f(s + it)\ = \f(s) +f(it)\,where / is entire function of a complex variable, and s, t are real variables.Hille [2] found all entire solutions of the functional equation(2)|/(s + zi)|2 = |/(s)|2 + |/(ii)|2.Haruki [1] showed that equations (1) and (2) are equivalent.Robinson's result from the paper [3] may be presented in the followingform:THEOREM. The only entire solutions of (I) aref(z)= Azandf(z)= A sin az,where A is an arbitrary complex constant and a is an arbitrary real or purelyimaginary constant.This paper gives a new proof of the above theorem - different to theproof we can find in [3].First we will prove the followingLEMMA 1. The only entire solutions of the functional equation(3)f{x + y)f(x-y)= /(χ) 2 - f(y)2,x,y e Care1991 Mathematics Subject Classification: 39B32, 30D05.Key words and phrases: entire functions, Hille's functional equation, Robinson's functional equation.714Κ. P i e http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Demonstratio Mathematica de Gruyter

NOTE ON ROBINSON’S FUNCTIONAL EQUATION

Piejko, Krzysztof
Demonstratio Mathematica , Volume 32 (4): 4 – Jan 1, 1999
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Publisher
de Gruyter
Copyright
© by Krzysztof Piejko
ISSN
0420-1213
eISSN
2391-4661
DOI
10.1515/dema-1999-0405
Publisher site
See Article on Publisher Site

Abstract

Krzysztof PiejkoNOTE ON ROBINSON'S FUNCTIONAL EQUATIONAbstract. The purpose of this note is to give a new proof of the fact, that theonly entire solutions of the Robinson's functional equation are given by f(z) = Az orf(z) = A sin az, where A, a are complex constants and a is real or purely imaginary.Robinson in [3] solved the following functional equation(1)\f(s + it)\ = \f(s) +f(it)\,where / is entire function of a complex variable, and s, t are real variables.Hille [2] found all entire solutions of the functional equation(2)|/(s + zi)|2 = |/(s)|2 + |/(ii)|2.Haruki [1] showed that equations (1) and (2) are equivalent.Robinson's result from the paper [3] may be presented in the followingform:THEOREM. The only entire solutions of (I) aref(z)= Azandf(z)= A sin az,where A is an arbitrary complex constant and a is an arbitrary real or purelyimaginary constant.This paper gives a new proof of the above theorem - different to theproof we can find in [3].First we will prove the followingLEMMA 1. The only entire solutions of the functional equation(3)f{x + y)f(x-y)= /(χ) 2 - f(y)2,x,y e Care1991 Mathematics Subject Classification: 39B32, 30D05.Key words and phrases: entire functions, Hille's functional equation, Robinson's functional equation.714Κ. P i e

Journal

Demonstratio Mathematica – de Gruyter

Published: Jan 1, 1999

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