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Pexider's (/ Cauchy's) functional equation over a bounded domain

1

$begingroup$

I am looking at Pexider's equation $f(x+y)=g(x)+h(y)$, where $f,g,h$ are continuous functions but are defined over bounded domains. Specifically, $f,g,h$ each is defined on a real interval (of length $>0$), but not necessarily the entire real line. (The domains are such that the functional equation holds throughout the respective domains.) It seems that uniqueness of the Pexider's solution still holds. Is this true? any references?

If necessary one may assume:

• all three domains contain $0$ in their interior.


• 
  $f$ is strictly monotone.

Thanks.

functional-equations

share|cite|improve this question

asked Nov 14 '18 at 20:22

user154729user154729

124

$endgroup$

add a comment | 

1

$begingroup$

I am looking at Pexider's equation $f(x+y)=g(x)+h(y)$, where $f,g,h$ are continuous functions but are defined over bounded domains. Specifically, $f,g,h$ each is defined on a real interval (of length $>0$), but not necessarily the entire real line. (The domains are such that the functional equation holds throughout the respective domains.) It seems that uniqueness of the Pexider's solution still holds. Is this true? any references?

If necessary one may assume:

• all three domains contain $0$ in their interior.


• 
  $f$ is strictly monotone.

Thanks.

functional-equations

share|cite|improve this question

asked Nov 14 '18 at 20:22

user154729user154729

124

$endgroup$

add a comment | 

$begingroup$

I am looking at Pexider's equation $f(x+y)=g(x)+h(y)$, where $f,g,h$ are continuous functions but are defined over bounded domains. Specifically, $f,g,h$ each is defined on a real interval (of length $>0$), but not necessarily the entire real line. (The domains are such that the functional equation holds throughout the respective domains.) It seems that uniqueness of the Pexider's solution still holds. Is this true? any references?

If necessary one may assume:

• all three domains contain $0$ in their interior.


• 
  $f$ is strictly monotone.

Thanks.

functional-equations

share|cite|improve this question

asked Nov 14 '18 at 20:22

user154729user154729

124

$endgroup$

share|cite|improve this question

asked Nov 14 '18 at 20:22

user154729user154729

124

share|cite|improve this question

asked Nov 14 '18 at 20:22

user154729user154729

124

asked Nov 14 '18 at 20:22

user154729user154729

124

asked Nov 14 '18 at 20:22

user154729user154729

124

1 Answer
1

active

oldest

votes

0

$begingroup$

First, you can do substitutions $f(x) = g(x) + h(0)$ to reduce the question to a Cauchy functional equation.

The resulting functional equation $bar h(x + y) = bar h(x)+bar h(y)$, $(x,y)in (-a,b)times(-c,d)$ can be extended to the entire real line. Notice that $x+y$ will lie outside the original domain for sufficiently small/large $x$ and $y$. Afterwards, you can simply use the standard Cauchy solution.

share|cite|improve this answer

answered Apr 1 at 14:39

HRSEHRSE

233110

$endgroup$

add a comment | 

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0

$begingroup$

First, you can do substitutions $f(x) = g(x) + h(0)$ to reduce the question to a Cauchy functional equation.

The resulting functional equation $bar h(x + y) = bar h(x)+bar h(y)$, $(x,y)in (-a,b)times(-c,d)$ can be extended to the entire real line. Notice that $x+y$ will lie outside the original domain for sufficiently small/large $x$ and $y$. Afterwards, you can simply use the standard Cauchy solution.

share|cite|improve this answer

answered Apr 1 at 14:39

HRSEHRSE

233110

$endgroup$

add a comment | 

0

$begingroup$

First, you can do substitutions $f(x) = g(x) + h(0)$ to reduce the question to a Cauchy functional equation.

The resulting functional equation $bar h(x + y) = bar h(x)+bar h(y)$, $(x,y)in (-a,b)times(-c,d)$ can be extended to the entire real line. Notice that $x+y$ will lie outside the original domain for sufficiently small/large $x$ and $y$. Afterwards, you can simply use the standard Cauchy solution.

share|cite|improve this answer

answered Apr 1 at 14:39

HRSEHRSE

233110

$endgroup$

add a comment | 

$begingroup$

First, you can do substitutions $f(x) = g(x) + h(0)$ to reduce the question to a Cauchy functional equation.

The resulting functional equation $bar h(x + y) = bar h(x)+bar h(y)$, $(x,y)in (-a,b)times(-c,d)$ can be extended to the entire real line. Notice that $x+y$ will lie outside the original domain for sufficiently small/large $x$ and $y$. Afterwards, you can simply use the standard Cauchy solution.

share|cite|improve this answer

answered Apr 1 at 14:39

HRSEHRSE

233110

$endgroup$

share|cite|improve this answer

answered Apr 1 at 14:39

HRSEHRSE

233110

answered Apr 1 at 14:39

HRSEHRSE

233110

answered Apr 1 at 14:39

HRSEHRSE

233110