Existence of a strictly increasing transformation between two functions [closed]

2

$begingroup$

Assume $f$ and $g$ are two differentiable functions defined on a compact interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . I want to proof or disproof the following statement

$ forall x in X: operatorname{sign}(f'(x))=operatorname{sign}(g'(x));; implies exists ;; m: mathbb{R} to mathbb{R}$, strictly increasing s.t. $f=m circ g$

My attempts raised the elementary question which conditions on arbitrary $f,g$ are in general sufficient for the existence of an $m$ such that $f=m circ g$.

Any suggestions?

derivatives transformation monotone-functions

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edited Jan 29 at 14:31

Arthur

121k7121207

asked Jan 29 at 14:24

StMaStMa

134

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closed as off-topic by Frpzzd, max_zorn, Alexander Gruber♦ Jan 30 at 1:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Frpzzd, max_zorn, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We usually put $forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols.
  $endgroup$
  – Arthur
  Jan 29 at 14:31
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Assume $f$ and $g$ are two differentiable functions defined on a compact interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . I want to proof or disproof the following statement

$ forall x in X: operatorname{sign}(f'(x))=operatorname{sign}(g'(x));; implies exists ;; m: mathbb{R} to mathbb{R}$, strictly increasing s.t. $f=m circ g$

My attempts raised the elementary question which conditions on arbitrary $f,g$ are in general sufficient for the existence of an $m$ such that $f=m circ g$.

Any suggestions?

derivatives transformation monotone-functions

share|cite|improve this question

edited Jan 29 at 14:31

Arthur

121k7121207

asked Jan 29 at 14:24

StMaStMa

134

$endgroup$

closed as off-topic by Frpzzd, max_zorn, Alexander Gruber♦ Jan 30 at 1:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Frpzzd, max_zorn, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We usually put $forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols.
  $endgroup$
  – Arthur
  Jan 29 at 14:31
  
  
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Assume $f$ and $g$ are two differentiable functions defined on a compact interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . I want to proof or disproof the following statement

$ forall x in X: operatorname{sign}(f'(x))=operatorname{sign}(g'(x));; implies exists ;; m: mathbb{R} to mathbb{R}$, strictly increasing s.t. $f=m circ g$

My attempts raised the elementary question which conditions on arbitrary $f,g$ are in general sufficient for the existence of an $m$ such that $f=m circ g$.

Any suggestions?

derivatives transformation monotone-functions

share|cite|improve this question

edited Jan 29 at 14:31

Arthur

121k7121207

asked Jan 29 at 14:24

StMaStMa

134

$endgroup$

Assume $f$ and $g$ are two differentiable functions defined on a compact interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . I want to proof or disproof the following statement

$ forall x in X: operatorname{sign}(f'(x))=operatorname{sign}(g'(x));; implies exists ;; m: mathbb{R} to mathbb{R}$, strictly increasing s.t. $f=m circ g$

My attempts raised the elementary question which conditions on arbitrary $f,g$ are in general sufficient for the existence of an $m$ such that $f=m circ g$.

Any suggestions?

derivatives transformation monotone-functions

derivatives transformation monotone-functions

share|cite|improve this question

edited Jan 29 at 14:31

Arthur

121k7121207

asked Jan 29 at 14:24

StMaStMa

134

share|cite|improve this question

edited Jan 29 at 14:31

Arthur

121k7121207

asked Jan 29 at 14:24

StMaStMa

134

share|cite|improve this question

share|cite|improve this question

edited Jan 29 at 14:31

Arthur

121k7121207

edited Jan 29 at 14:31

Arthur

121k7121207

edited Jan 29 at 14:31

Arthur

121k7121207

121k7121207

asked Jan 29 at 14:24

StMaStMa

134

asked Jan 29 at 14:24

StMaStMa

134

asked Jan 29 at 14:24

StMaStMa

134

134

closed as off-topic by Frpzzd, max_zorn, Alexander Gruber♦ Jan 30 at 1:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Frpzzd, max_zorn, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Frpzzd, max_zorn, Alexander Gruber♦ Jan 30 at 1:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Frpzzd, max_zorn, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We usually put $forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols.
  $endgroup$
  – Arthur
  Jan 29 at 14:31
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We usually put $forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols.
  $endgroup$
  – Arthur
  Jan 29 at 14:31
  
  
  

  
  
  
  

$begingroup$
We usually put $forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols.
$endgroup$
– Arthur
Jan 29 at 14:31

$begingroup$
We usually put $forall x$ in front of whatever statement it applies to. Not when using words, but when using symbols.
$endgroup$
– Arthur
Jan 29 at 14:31

add a comment | 

3 Answers
3

active

oldest

votes

1

$begingroup$

Counter-example:
Let $f(x)=frac{1}{3}x^{3}-pi x^{2}+frac{3}{4}pi^{2} x$ and let $g(x)=sin(x)$ on the interval [0,2pi].
Note that $f'(x)=x^{2}-2pi x+frac{3}{4}pi^{2}=(x-frac{1}{2}pi)(x-frac{3}{2}pi)$ so $f'(x)<0$ if $0leq x<frac{1}{2}pi$ and $frac{3}{2}pi<xleq 2pi$ and $f'(x)>0$ if $frac{1}{2}pi<x<frac{3}{2}pi$ which corresponds with $g'(x)$.

Now suppose a transformation function exists, then $0=f(0)=m(g(0))=m(0)$ and $frac{1}{12}pi^{3}=f(pi)=m(g(pi))=m(0)$. This is a contradiction.

I believe it should work if $f$ and $g$ are both strictly increasing.

share|cite|improve this answer

edited Jan 29 at 14:58

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, thank you. I've adjusted my example to ensure the signs are the same.
  $endgroup$
  – Floris Claassens
  Jan 29 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! I got the intuition why the statement cannot hold
  $endgroup$
  – StMa
  Jan 29 at 16:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $I=[-{pi over 3} , {pi over 3}]$.

Let $f = cos$, and let $g(x) = begin{cases} -x^2,& x le 0 \
-2x^2, & text{otherwise}end{cases}$.

The sign of derivative condition is satisfied, but $f$ is even and $g$ is not.

share|cite|improve this answer

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
  $endgroup$
  – Bertrand
  Jan 29 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
  $endgroup$
  – copper.hat
  Jan 29 at 15:55
  

  
  
  
  

add a comment | 

0

$begingroup$

To summarize this discussion, now that the initial claim has been disproved, it may be interesting to compare it with a case which works. May be this one:

Assume that $f$ and $g$ are one to one, defined on an interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . Then

$ forall x in X, exists ;; m: mathbb{R} to mathbb{R}$ s.t. $f=m circ g$

Proof. If $g$ is one to one, then $g^{-1}$ exists and $m equiv f circ g^{-1}$ satisfies $m circ g = f$.

share|cite|improve this answer

answered Jan 29 at 20:43

BertrandBertrand

45815

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add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Counter-example:
Let $f(x)=frac{1}{3}x^{3}-pi x^{2}+frac{3}{4}pi^{2} x$ and let $g(x)=sin(x)$ on the interval [0,2pi].
Note that $f'(x)=x^{2}-2pi x+frac{3}{4}pi^{2}=(x-frac{1}{2}pi)(x-frac{3}{2}pi)$ so $f'(x)<0$ if $0leq x<frac{1}{2}pi$ and $frac{3}{2}pi<xleq 2pi$ and $f'(x)>0$ if $frac{1}{2}pi<x<frac{3}{2}pi$ which corresponds with $g'(x)$.

Now suppose a transformation function exists, then $0=f(0)=m(g(0))=m(0)$ and $frac{1}{12}pi^{3}=f(pi)=m(g(pi))=m(0)$. This is a contradiction.

I believe it should work if $f$ and $g$ are both strictly increasing.

share|cite|improve this answer

edited Jan 29 at 14:58

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, thank you. I've adjusted my example to ensure the signs are the same.
  $endgroup$
  – Floris Claassens
  Jan 29 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! I got the intuition why the statement cannot hold
  $endgroup$
  – StMa
  Jan 29 at 16:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Counter-example:
Let $f(x)=frac{1}{3}x^{3}-pi x^{2}+frac{3}{4}pi^{2} x$ and let $g(x)=sin(x)$ on the interval [0,2pi].
Note that $f'(x)=x^{2}-2pi x+frac{3}{4}pi^{2}=(x-frac{1}{2}pi)(x-frac{3}{2}pi)$ so $f'(x)<0$ if $0leq x<frac{1}{2}pi$ and $frac{3}{2}pi<xleq 2pi$ and $f'(x)>0$ if $frac{1}{2}pi<x<frac{3}{2}pi$ which corresponds with $g'(x)$.

Now suppose a transformation function exists, then $0=f(0)=m(g(0))=m(0)$ and $frac{1}{12}pi^{3}=f(pi)=m(g(pi))=m(0)$. This is a contradiction.

I believe it should work if $f$ and $g$ are both strictly increasing.

share|cite|improve this answer

edited Jan 29 at 14:58

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, thank you. I've adjusted my example to ensure the signs are the same.
  $endgroup$
  – Floris Claassens
  Jan 29 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! I got the intuition why the statement cannot hold
  $endgroup$
  – StMa
  Jan 29 at 16:16
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Counter-example:
Let $f(x)=frac{1}{3}x^{3}-pi x^{2}+frac{3}{4}pi^{2} x$ and let $g(x)=sin(x)$ on the interval [0,2pi].
Note that $f'(x)=x^{2}-2pi x+frac{3}{4}pi^{2}=(x-frac{1}{2}pi)(x-frac{3}{2}pi)$ so $f'(x)<0$ if $0leq x<frac{1}{2}pi$ and $frac{3}{2}pi<xleq 2pi$ and $f'(x)>0$ if $frac{1}{2}pi<x<frac{3}{2}pi$ which corresponds with $g'(x)$.

Now suppose a transformation function exists, then $0=f(0)=m(g(0))=m(0)$ and $frac{1}{12}pi^{3}=f(pi)=m(g(pi))=m(0)$. This is a contradiction.

I believe it should work if $f$ and $g$ are both strictly increasing.

share|cite|improve this answer

edited Jan 29 at 14:58

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

$endgroup$

Counter-example:
Let $f(x)=frac{1}{3}x^{3}-pi x^{2}+frac{3}{4}pi^{2} x$ and let $g(x)=sin(x)$ on the interval [0,2pi].
Note that $f'(x)=x^{2}-2pi x+frac{3}{4}pi^{2}=(x-frac{1}{2}pi)(x-frac{3}{2}pi)$ so $f'(x)<0$ if $0leq x<frac{1}{2}pi$ and $frac{3}{2}pi<xleq 2pi$ and $f'(x)>0$ if $frac{1}{2}pi<x<frac{3}{2}pi$ which corresponds with $g'(x)$.

Now suppose a transformation function exists, then $0=f(0)=m(g(0))=m(0)$ and $frac{1}{12}pi^{3}=f(pi)=m(g(pi))=m(0)$. This is a contradiction.

I believe it should work if $f$ and $g$ are both strictly increasing.

share|cite|improve this answer

edited Jan 29 at 14:58

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

share|cite|improve this answer

share|cite|improve this answer

edited Jan 29 at 14:58

edited Jan 29 at 14:58

edited Jan 29 at 14:58

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

answered Jan 29 at 14:40

Floris ClaassensFloris Claassens

1,21527

1,21527

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, thank you. I've adjusted my example to ensure the signs are the same.
  $endgroup$
  – Floris Claassens
  Jan 29 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! I got the intuition why the statement cannot hold
  $endgroup$
  – StMa
  Jan 29 at 16:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, thank you. I've adjusted my example to ensure the signs are the same.
  $endgroup$
  – Floris Claassens
  Jan 29 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot! I got the intuition why the statement cannot hold
  $endgroup$
  – StMa
  Jan 29 at 16:16
  

  
  
  
  

$begingroup$
You're right, thank you. I've adjusted my example to ensure the signs are the same.
$endgroup$
– Floris Claassens
Jan 29 at 14:59

$begingroup$
You're right, thank you. I've adjusted my example to ensure the signs are the same.
$endgroup$
– Floris Claassens
Jan 29 at 14:59

$begingroup$
Thanks a lot! I got the intuition why the statement cannot hold
$endgroup$
– StMa
Jan 29 at 16:16

$begingroup$
Thanks a lot! I got the intuition why the statement cannot hold
$endgroup$
– StMa
Jan 29 at 16:16

add a comment | 

1

$begingroup$

Let $I=[-{pi over 3} , {pi over 3}]$.

Let $f = cos$, and let $g(x) = begin{cases} -x^2,& x le 0 \
-2x^2, & text{otherwise}end{cases}$.

The sign of derivative condition is satisfied, but $f$ is even and $g$ is not.

share|cite|improve this answer

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
  $endgroup$
  – Bertrand
  Jan 29 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
  $endgroup$
  – copper.hat
  Jan 29 at 15:55
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $I=[-{pi over 3} , {pi over 3}]$.

Let $f = cos$, and let $g(x) = begin{cases} -x^2,& x le 0 \
-2x^2, & text{otherwise}end{cases}$.

The sign of derivative condition is satisfied, but $f$ is even and $g$ is not.

share|cite|improve this answer

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
  $endgroup$
  – Bertrand
  Jan 29 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
  $endgroup$
  – copper.hat
  Jan 29 at 15:55
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Let $I=[-{pi over 3} , {pi over 3}]$.

Let $f = cos$, and let $g(x) = begin{cases} -x^2,& x le 0 \
-2x^2, & text{otherwise}end{cases}$.

The sign of derivative condition is satisfied, but $f$ is even and $g$ is not.

share|cite|improve this answer

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

$endgroup$

Let $I=[-{pi over 3} , {pi over 3}]$.

Let $f = cos$, and let $g(x) = begin{cases} -x^2,& x le 0 \
-2x^2, & text{otherwise}end{cases}$.

The sign of derivative condition is satisfied, but $f$ is even and $g$ is not.

share|cite|improve this answer

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

share|cite|improve this answer

share|cite|improve this answer

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

answered Jan 29 at 15:02

copper.hatcopper.hat

128k560161

128k560161

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
  $endgroup$
  – Bertrand
  Jan 29 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
  $endgroup$
  – copper.hat
  Jan 29 at 15:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
  $endgroup$
  – Bertrand
  Jan 29 at 15:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
  $endgroup$
  – copper.hat
  Jan 29 at 15:55
  

  
  
  
  

$begingroup$
Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
$endgroup$
– Bertrand
Jan 29 at 15:30

$begingroup$
Why should $m$ not exist in this case? It can be defined differently for $x leq 0$ and $x>0$.
$endgroup$
– Bertrand
Jan 29 at 15:30

1

1

$begingroup$
@Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
$endgroup$
– copper.hat
Jan 29 at 15:55

$begingroup$
@Bertrand: If such a function existed we would have $g(-1) = -1, g(1) = -2$. $f(-1) = m(-1) = f(1) = m(-2)$. Hence $m$ is not injective.
$endgroup$
– copper.hat
Jan 29 at 15:55

add a comment | 

0

$begingroup$

To summarize this discussion, now that the initial claim has been disproved, it may be interesting to compare it with a case which works. May be this one:

Assume that $f$ and $g$ are one to one, defined on an interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . Then

$ forall x in X, exists ;; m: mathbb{R} to mathbb{R}$ s.t. $f=m circ g$

Proof. If $g$ is one to one, then $g^{-1}$ exists and $m equiv f circ g^{-1}$ satisfies $m circ g = f$.

share|cite|improve this answer

answered Jan 29 at 20:43

BertrandBertrand

45815

$endgroup$

add a comment | 

0

$begingroup$

To summarize this discussion, now that the initial claim has been disproved, it may be interesting to compare it with a case which works. May be this one:

Assume that $f$ and $g$ are one to one, defined on an interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . Then

$ forall x in X, exists ;; m: mathbb{R} to mathbb{R}$ s.t. $f=m circ g$

Proof. If $g$ is one to one, then $g^{-1}$ exists and $m equiv f circ g^{-1}$ satisfies $m circ g = f$.

share|cite|improve this answer

answered Jan 29 at 20:43

BertrandBertrand

45815

$endgroup$

add a comment | 

0

0

0

$begingroup$

To summarize this discussion, now that the initial claim has been disproved, it may be interesting to compare it with a case which works. May be this one:

Assume that $f$ and $g$ are one to one, defined on an interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . Then

$ forall x in X, exists ;; m: mathbb{R} to mathbb{R}$ s.t. $f=m circ g$

Proof. If $g$ is one to one, then $g^{-1}$ exists and $m equiv f circ g^{-1}$ satisfies $m circ g = f$.

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answered Jan 29 at 20:43

BertrandBertrand

45815

$endgroup$

To summarize this discussion, now that the initial claim has been disproved, it may be interesting to compare it with a case which works. May be this one:

Assume that $f$ and $g$ are one to one, defined on an interval $X subseteq mathbb{R}$ mapping into $mathbb{R}$ . Then

$ forall x in X, exists ;; m: mathbb{R} to mathbb{R}$ s.t. $f=m circ g$

Proof. If $g$ is one to one, then $g^{-1}$ exists and $m equiv f circ g^{-1}$ satisfies $m circ g = f$.

share|cite|improve this answer

answered Jan 29 at 20:43

BertrandBertrand

45815

share|cite|improve this answer

share|cite|improve this answer

answered Jan 29 at 20:43

BertrandBertrand

45815

answered Jan 29 at 20:43

BertrandBertrand

45815

answered Jan 29 at 20:43

BertrandBertrand

45815

45815

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