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Convergence to a fixed point

Contraction mapping in the context of $f(x_n)=x_{n+1}$.Confused about fixed point method conditionShrinking Map and Fixed Point via Iteration MethodWhich negation of the definition of a null sequence is correct?Relation between two different definitions of quadratic convergenceFixed point, bounded derivativeProving Cauchy when given a sequenceBanach fixed point questionHelp me understand this proof of Implicit Function Theorem on Banach spacesConvergence of functions (in sense of distributions)Banach fixed-point theoremIf the odd function $f:mathbb Rtomathbb R$ letting $x>0$ is continuous at $x$, prove the function is continuous at $-x$.

3

$begingroup$

Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.

Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

analysis convergence numerical-methods fixed-point-theorems fixedpoints

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edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another related question.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.

Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

analysis convergence numerical-methods fixed-point-theorems fixedpoints

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another related question.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  $endgroup$
  – rtybase
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. Prove that for any point $x_0$ in $[a,b]$ the sequence defined by $$ x_n=f(x_{n-1}), n>0$$
converges to one unique fixed point.

Attempt:
Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $x$ in$ [a,b]$ s.t. $f(x)=x$ but how do I relate the sequence to this $x$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

analysis convergence numerical-methods fixed-point-theorems fixedpoints

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

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Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 3 hours ago

Grace

asked 4 hours ago

Grace Grace

164

New contributor

Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago

Grace Grace

164

New contributor

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asked 4 hours ago

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3 Answers
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2

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

add a comment | 

1

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
  $endgroup$
  – Robert Shore
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

add a comment | 

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2

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

add a comment | 

2

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

add a comment | 

$begingroup$

The Mean Value Theorem tells you that the sequence ${x_n }$ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

answered 4 hours ago

Robert ShoreRobert Shore

2,044116

1

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
  $endgroup$
  – Robert Shore
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$.
  $endgroup$
  – Robert Shore
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $F$ such that the distance between any two points $x$ and $y$ (when scaled by a constant $q$) is always larger than the distance between the corresponding images ($F(x)$ and $F(y)$), then the sequence
$$x_n = F(x_{n-1}) $$
will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that
$$|f'(x)| leq t $$
This implies that

$$Big|frac{f(x) - f(y)}{x - y}Big| leq t$$

for all possible points $x,y in [a,b]$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y| $$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $[a,b]$ for the sequence

$$ x_n = f(x_{n-1})$$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Sean LeeSean Lee

483211

answered 4 hours ago

Sean LeeSean Lee

483211

answered 4 hours ago

Sean LeeSean Lee

483211

1

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

add a comment | 

1

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

add a comment | 

$begingroup$

To prove this from scratch note that by MVT $$|x_n-x_m| leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+cdots+|x_{n}-x_{n-1}|leq |x_{m+1}-x_m| (1+t+t^{2}+cdots+t^{n+m-1})$$ for $n >m$. Also $|x_{m+1}-x_m| leq t^{m-1} |x_2-x_1|$. Using the convergence of the geometric series $sum t^{n}$ conclude that ${x_n}$ is Cauchy. If $x =lim x_n$ then the definition of $x_n$'s and continuity of $f$ tells you that $f(x)=x$. Uniqueness follows easily by MVT.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766

answered 3 hours ago

Kavi Rama MurthyKavi Rama Murthy

65k42766