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If $g circ f=g$, prove that $g$ is constant

1

$begingroup$

Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.

This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.

real-analysis functions

share|cite|improve this question

edited Mar 31 at 18:50

TonyK

44.1k358137

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

$endgroup$

add a comment | 

1

$begingroup$

Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.

This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.

real-analysis functions

share|cite|improve this question

edited Mar 31 at 18:50

TonyK

44.1k358137

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

$endgroup$

add a comment | 

$begingroup$

Let $f:mathbbR to [0,1]$ be a monotonic function so that $|f(x) - f(y) | <|x-y|$, $forall x, y in mathbbR $, $xneq y$. If $g:mathbbR to mathbbR $ is a continuous function and $g circ f=g$, prove that $g$ is constant.

This problem also previously asked to prove that $f$ has a unique fixed point and I could show this by proving that $f$ is continuous and considering the function $h(x) =f(x) - x$. Yet, I don't know how to use this to prove that $g$ is constant.

real-analysis functions

share|cite|improve this question

edited Mar 31 at 18:50

TonyK

44.1k358137

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

$endgroup$

share|cite|improve this question

edited Mar 31 at 18:50

TonyK

44.1k358137

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

share|cite|improve this question

edited Mar 31 at 18:50

TonyK

44.1k358137

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

edited Mar 31 at 18:50

TonyK

44.1k358137

edited Mar 31 at 18:50

TonyK

44.1k358137

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

asked Mar 31 at 18:45

MathEnthusiastMathEnthusiast

47913

1 Answer
1

active

oldest

votes

4

$begingroup$

Denote by $c$ the fixed point of $f$.

Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$

The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$

$$g(a)=g(x_n) to g(c)$$

This shows that $g(a)=g(c)$ for all $a in mathbb R$.

share|cite|improve this answer

answered Mar 31 at 18:49

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
  $endgroup$
  – MathEnthusiast
  Mar 31 at 18:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
  $endgroup$
  – Xander Henderson
  Mar 31 at 18:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
  $endgroup$
  – N. S.
  Mar 31 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
  $endgroup$
  – MathEnthusiast
  Mar 31 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
  $endgroup$
  – MathEnthusiast
  Mar 31 at 19:12
  
  

  
  
  
  

 | 
show 4 more comments

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4

$begingroup$

Denote by $c$ the fixed point of $f$.

Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$

The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$

$$g(a)=g(x_n) to g(c)$$

This shows that $g(a)=g(c)$ for all $a in mathbb R$.

share|cite|improve this answer

answered Mar 31 at 18:49

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
  $endgroup$
  – MathEnthusiast
  Mar 31 at 18:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
  $endgroup$
  – Xander Henderson
  Mar 31 at 18:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
  $endgroup$
  – N. S.
  Mar 31 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
  $endgroup$
  – MathEnthusiast
  Mar 31 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
  $endgroup$
  – MathEnthusiast
  Mar 31 at 19:12
  
  

  
  
  
  

 | 
show 4 more comments

4

$begingroup$

Denote by $c$ the fixed point of $f$.

Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$

The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$

$$g(a)=g(x_n) to g(c)$$

This shows that $g(a)=g(c)$ for all $a in mathbb R$.

share|cite|improve this answer

answered Mar 31 at 18:49

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, nice proof! Could you explain to me why $x_n$ converges to $c$?
  $endgroup$
  – MathEnthusiast
  Mar 31 at 18:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MathEnthusiast Try Googling the "Banach fixed point theorem" (or the "contraction mapping principle").
  $endgroup$
  – Xander Henderson
  Mar 31 at 18:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MathEnthusiast Didn't you prove it in part 1? If not, First show that there exists some $D<1$ such that $|f(x)-f(y)| <D |x-y|$ for all $x,y in mathbb R$... Then $$|x_n+1-c|=|f(x_n)-f(c)| leq D |x_n -c|$$ and hence, by induction $$|x_n-c| < D^n-1 |x_1-c|$$ Since $D<1$, it follows immediately that $x_n to c$.
  $endgroup$
  – N. S.
  Mar 31 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Xander Henderson You are right, $f$ is a contraction from the initial constraint. Yet, I don't know how the fact that $f$ is monotonous helps.
  $endgroup$
  – MathEnthusiast
  Mar 31 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @N. S. Now I understand, I could have used Banach's fixed point theorem directly. However, this doesn't use the fact that $f$ is monotonous. Is that condition redundant?
  $endgroup$
  – MathEnthusiast
  Mar 31 at 19:12
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Denote by $c$ the fixed point of $f$.

Let $a in mathbb R$ be arbitrary. Then
$$g(a)=g(f(a))=g(f^2(a))=....=g(f^n(a))=...$$

The sequence $x_n =f^n(a)$ converges to the fixed point $c$ of $f$. Therefore by continuity of $g$

$$g(a)=g(x_n) to g(c)$$

This shows that $g(a)=g(c)$ for all $a in mathbb R$.

share|cite|improve this answer

answered Mar 31 at 18:49

N. S.N. S.

105k7115210

$endgroup$

share|cite|improve this answer

answered Mar 31 at 18:49

N. S.N. S.

105k7115210

answered Mar 31 at 18:49

N. S.N. S.

105k7115210

answered Mar 31 at 18:49

N. S.N. S.

105k7115210