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Convergence to a fixed point [duplicate]

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 $-x$.

4

$begingroup$

This question already has an answer here:

• 
  Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  
  3 answers
  
  

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 17 hours ago

Grace

asked 17 hours ago

Grace Grace

255

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Grace is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos Santos 2 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another related question.
  $endgroup$
  – rtybase
  17 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
  15 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

This question already has an answer here:

• 
  Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  
  3 answers
  
  

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 17 hours ago

Grace

asked 17 hours ago

Grace Grace

255

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.

$endgroup$

marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos Santos 2 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another related question.
  $endgroup$
  – rtybase
  17 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
  15 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Contraction mapping in the context of $f(x_n)=x_{n+1}$.
  
  3 answers
  
  

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 17 hours ago

Grace

asked 17 hours ago

Grace Grace

255

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 17 hours ago

Grace

asked 17 hours ago

Grace Grace

255

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.

share|cite|improve this question

edited 17 hours ago

Grace

asked 17 hours ago

Grace Grace

255

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.

asked 17 hours ago

Grace Grace

255

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.

asked 17 hours ago

Grace Grace

255

3 Answers
3

active

oldest

votes

3

$begingroup$

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 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! This is very helpful :-)
  $endgroup$
  – Grace
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 17 hours ago

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

$endgroup$

add a comment | 

2

$begingroup$

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 17 hours ago

Sean LeeSean Lee

493211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

3

$begingroup$

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 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! This is very helpful :-)
  $endgroup$
  – Grace
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! This is very helpful :-)
  $endgroup$
  – Grace
  7 hours ago
  

  
  
  
  

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 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

share|cite|improve this answer

answered 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

answered 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

answered 16 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

3

$begingroup$

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 17 hours ago

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

$endgroup$

add a comment | 

3

$begingroup$

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 17 hours ago

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

$endgroup$

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 17 hours ago

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

$endgroup$

share|cite|improve this answer

edited 17 hours ago

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

answered 17 hours ago

Robert ShoreRobert Shore

2,079116

2

$begingroup$

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 17 hours ago

Sean LeeSean Lee

493211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

2

$begingroup$

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 17 hours ago

Sean LeeSean Lee

493211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore Yes, you are right, thanks!
  $endgroup$
  – Sean Lee
  17 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 17 hours ago

Sean LeeSean Lee

493211

$endgroup$

share|cite|improve this answer

answered 17 hours ago

Sean LeeSean Lee

493211

answered 17 hours ago

Sean LeeSean Lee

493211

answered 17 hours ago

Sean LeeSean Lee

493211