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Proving that a function is additive in a functional equation
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I have the equation
$$h(x,k(y,z))=f(y,g(y,x)+t(y,z)), tag*labelequation$$
where
- $h,k,f,g,t$ are continuous real valued functions.
- $h,f$ are strictly monotone in their second argument.
- all functions are "sensitive" to each of the arguments - in some natural sense (if $x,y,z,$ are real numbers, just suppose that all functions are strictly monotone in each of their arguments).
- $x,y,z$ are from some "well behaved" topological space (connected, ...) - you can assume that they are from some real connected interval.
I want to show that $g$ is additive, in the sense that: $g(y,x)=g_y(y)+g_x(x)$ (and that $h$ and $f$ are strictly monotone transformations of an additive function).
The reason I think this is true is that in the right-hand side of eqrefequation, $x$ is only "tied to" $y$ not $z$, while on the left hand side it is "tied to" a function of both $y$ and $z$.
More specifically, by eqrefequation we have:
$$k(y,z)=h^-1(x,f(y,g(y,x)+t(y,z))) tag**labelequation_a$$
(where $h^-1$ is the the inverse of $h$ on the second argument, that is: $h(a,h^-1(a,v))=v$).
So, the right-hand side of eqrefequation_a must be independent of $x$. Is there any option but that $g$ is additive? I do not know how to go about attacking this. What theorems/tools are available?
Thanks,
functional-equations
asked Jul 20 at 10:05
mike
703
add a comment |Â
up vote
2
down vote
favorite
I have the equation
$$h(x,k(y,z))=f(y,g(y,x)+t(y,z)), tag*labelequation$$
where
- $h,k,f,g,t$ are continuous real valued functions.
- $h,f$ are strictly monotone in their second argument.
- all functions are "sensitive" to each of the arguments - in some natural sense (if $x,y,z,$ are real numbers, just suppose that all functions are strictly monotone in each of their arguments).
- $x,y,z$ are from some "well behaved" topological space (connected, ...) - you can assume that they are from some real connected interval.
I want to show that $g$ is additive, in the sense that: $g(y,x)=g_y(y)+g_x(x)$ (and that $h$ and $f$ are strictly monotone transformations of an additive function).
The reason I think this is true is that in the right-hand side of eqrefequation, $x$ is only "tied to" $y$ not $z$, while on the left hand side it is "tied to" a function of both $y$ and $z$.
More specifically, by eqrefequation we have:
$$k(y,z)=h^-1(x,f(y,g(y,x)+t(y,z))) tag**labelequation_a$$
(where $h^-1$ is the the inverse of $h$ on the second argument, that is: $h(a,h^-1(a,v))=v$).
So, the right-hand side of eqrefequation_a must be independent of $x$. Is there any option but that $g$ is additive? I do not know how to go about attacking this. What theorems/tools are available?
Thanks,
functional-equations
asked Jul 20 at 10:05
mike
703
I've changed the formatting of your question a bit, so that typingeqrefequationshould give a link to the equation like this eqrefequation. It should also work in answers as well.
– Arnaud D.
Jul 20 at 10:16
I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation.
– mike
Jul 22 at 12:20
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the equation
$$h(x,k(y,z))=f(y,g(y,x)+t(y,z)), tag*labelequation$$
where
- $h,k,f,g,t$ are continuous real valued functions.
- $h,f$ are strictly monotone in their second argument.
- all functions are "sensitive" to each of the arguments - in some natural sense (if $x,y,z,$ are real numbers, just suppose that all functions are strictly monotone in each of their arguments).
- $x,y,z$ are from some "well behaved" topological space (connected, ...) - you can assume that they are from some real connected interval.
I want to show that $g$ is additive, in the sense that: $g(y,x)=g_y(y)+g_x(x)$ (and that $h$ and $f$ are strictly monotone transformations of an additive function).
The reason I think this is true is that in the right-hand side of eqrefequation, $x$ is only "tied to" $y$ not $z$, while on the left hand side it is "tied to" a function of both $y$ and $z$.
More specifically, by eqrefequation we have:
$$k(y,z)=h^-1(x,f(y,g(y,x)+t(y,z))) tag**labelequation_a$$
(where $h^-1$ is the the inverse of $h$ on the second argument, that is: $h(a,h^-1(a,v))=v$).
So, the right-hand side of eqrefequation_a must be independent of $x$. Is there any option but that $g$ is additive? I do not know how to go about attacking this. What theorems/tools are available?
Thanks,
functional-equations
asked Jul 20 at 10:05
mike
703
I have the equation
$$h(x,k(y,z))=f(y,g(y,x)+t(y,z)), tag*labelequation$$
where
- $h,k,f,g,t$ are continuous real valued functions.
- $h,f$ are strictly monotone in their second argument.
- all functions are "sensitive" to each of the arguments - in some natural sense (if $x,y,z,$ are real numbers, just suppose that all functions are strictly monotone in each of their arguments).
- $x,y,z$ are from some "well behaved" topological space (connected, ...) - you can assume that they are from some real connected interval.
I want to show that $g$ is additive, in the sense that: $g(y,x)=g_y(y)+g_x(x)$ (and that $h$ and $f$ are strictly monotone transformations of an additive function).
The reason I think this is true is that in the right-hand side of eqrefequation, $x$ is only "tied to" $y$ not $z$, while on the left hand side it is "tied to" a function of both $y$ and $z$.
More specifically, by eqrefequation we have:
$$k(y,z)=h^-1(x,f(y,g(y,x)+t(y,z))) tag**labelequation_a$$
(where $h^-1$ is the the inverse of $h$ on the second argument, that is: $h(a,h^-1(a,v))=v$).
So, the right-hand side of eqrefequation_a must be independent of $x$. Is there any option but that $g$ is additive? I do not know how to go about attacking this. What theorems/tools are available?
Thanks,
functional-equations
asked Jul 20 at 10:05
mike
703
edited Jul 22 at 12:18
asked Jul 20 at 10:05
mike
703
asked Jul 20 at 10:05
mike
703
asked Jul 20 at 10:05
mike
703
703
I've changed the formatting of your question a bit, so that typingeqrefequationshould give a link to the equation like this eqrefequation. It should also work in answers as well.
– Arnaud D.
Jul 20 at 10:16
I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation.
– mike
Jul 22 at 12:20
add a comment |Â
I've changed the formatting of your question a bit, so that typingeqrefequationshould give a link to the equation like this eqrefequation. It should also work in answers as well.
– Arnaud D.
Jul 20 at 10:16
I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation.
– mike
Jul 22 at 12:20
I've changed the formatting of your question a bit, so that typing
eqrefequation should give a link to the equation like this eqrefequation. It should also work in answers as well.– Arnaud D.
Jul 20 at 10:16
I've changed the formatting of your question a bit, so that typing
eqrefequation should give a link to the equation like this eqrefequation. It should also work in answers as well.– Arnaud D.
Jul 20 at 10:16
I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation.
– mike
Jul 22 at 12:20
I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation.
– mike
Jul 22 at 12:20
add a comment |Â
1 Answer
1
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votes
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0
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The statement is false. Counter example:
- $h(a,b)=k(a,b)=a+b$
- $g(a,b)=t(a,b)=ab$
- $f(a,b)=b/a+a$
Then, both sides of $(*)$ are $x+y+z$.
(this is not a counter example to what I was really trying to prove, so I need to go back and rephrase the proposition).
answered Jul 22 at 12:38
mike
703
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The statement is false. Counter example:
- $h(a,b)=k(a,b)=a+b$
- $g(a,b)=t(a,b)=ab$
- $f(a,b)=b/a+a$
Then, both sides of $(*)$ are $x+y+z$.
(this is not a counter example to what I was really trying to prove, so I need to go back and rephrase the proposition).
answered Jul 22 at 12:38
mike
703
add a comment |Â
up vote
0
down vote
The statement is false. Counter example:
- $h(a,b)=k(a,b)=a+b$
- $g(a,b)=t(a,b)=ab$
- $f(a,b)=b/a+a$
Then, both sides of $(*)$ are $x+y+z$.
(this is not a counter example to what I was really trying to prove, so I need to go back and rephrase the proposition).
answered Jul 22 at 12:38
mike
703
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The statement is false. Counter example:
- $h(a,b)=k(a,b)=a+b$
- $g(a,b)=t(a,b)=ab$
- $f(a,b)=b/a+a$
Then, both sides of $(*)$ are $x+y+z$.
(this is not a counter example to what I was really trying to prove, so I need to go back and rephrase the proposition).
answered Jul 22 at 12:38
mike
703
The statement is false. Counter example:
- $h(a,b)=k(a,b)=a+b$
- $g(a,b)=t(a,b)=ab$
- $f(a,b)=b/a+a$
Then, both sides of $(*)$ are $x+y+z$.
(this is not a counter example to what I was really trying to prove, so I need to go back and rephrase the proposition).
answered Jul 22 at 12:38
mike
703
answered Jul 22 at 12:38
mike
703
answered Jul 22 at 12:38
mike
703
answered Jul 22 at 12:38
mike
703
703
add a comment |Â
add a comment |Â
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I've changed the formatting of your question a bit, so that typing
eqrefequationshould give a link to the equation like this eqrefequation. It should also work in answers as well.– Arnaud D.
Jul 20 at 10:16
I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation.
– mike
Jul 22 at 12:20