Must T be a contraction? [on hold]

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Let $X$ be a complete metric space . Suppose $T:Xto X$ is a function and $T^n$ is a contraction for some positive integer n.
Here $T^n$ is the composition of $T$ with itself $n$ times. Must $T$ have a fixed point? Must $T$ be a contraction ?

metric-spaces

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edited 2 days ago

Cornman

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asked 2 days ago

Gül

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put on hold as off-topic by Mostafa Ayaz, Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos 5 hours ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Clayton, José Carlos Santos

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

add a comment | 

up vote
1
down vote

favorite

1

Let $X$ be a complete metric space . Suppose $T:Xto X$ is a function and $T^n$ is a contraction for some positive integer n.
Here $T^n$ is the composition of $T$ with itself $n$ times. Must $T$ have a fixed point? Must $T$ be a contraction ?

metric-spaces

share|cite|improve this question

edited 2 days ago

Cornman

2,24521027

asked 2 days ago

Gül

113

put on hold as off-topic by Mostafa Ayaz, Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos 5 hours ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Clayton, José Carlos Santos

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

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Let $X$ be a complete metric space . Suppose $T:Xto X$ is a function and $T^n$ is a contraction for some positive integer n.
Here $T^n$ is the composition of $T$ with itself $n$ times. Must $T$ have a fixed point? Must $T$ be a contraction ?

metric-spaces

share|cite|improve this question

edited 2 days ago

Cornman

2,24521027

asked 2 days ago

Gül

113

Let $X$ be a complete metric space . Suppose $T:Xto X$ is a function and $T^n$ is a contraction for some positive integer n.
Here $T^n$ is the composition of $T$ with itself $n$ times. Must $T$ have a fixed point? Must $T$ be a contraction ?

metric-spaces

share|cite|improve this question

edited 2 days ago

Cornman

2,24521027

asked 2 days ago

Gül

113

share|cite|improve this question

share|cite|improve this question

edited 2 days ago

Cornman

2,24521027

edited 2 days ago

Cornman

2,24521027

edited 2 days ago

Cornman

2,24521027

2,24521027

asked 2 days ago

Gül

113

asked 2 days ago

Gül

113

asked 2 days ago

Gül

113

113

put on hold as off-topic by Mostafa Ayaz, Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos 5 hours ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Clayton, José Carlos Santos

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

put on hold as off-topic by Mostafa Ayaz, Jendrik Stelzner, Shailesh, Clayton, José Carlos Santos 5 hours ago

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, Clayton, José Carlos Santos

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

add a comment | 

add a comment | 

1 Answer
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up vote
2
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$T$ doesn't have to have fixed point. Let's consider space $X=1 times I cup 2 times I$, where $I$ is a unit compact interval, and $T colon X rightarrow X$, which acts on $X$ as follows: it takes each interval, shrinks it to $[0,frac12]$ and moves to the other interval, such that $T(1 times I)=2 times[0,frac12]$ and $T(2 times I)=1 times[0,frac12]$, it does not have fixed point, but $T^2$ does have one, for example $x=(1,0)$.

share|cite|improve this answer

answered 2 days ago

jon.sand

365

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
  – jon.sand
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
  – jon.sand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
  – jon.sand
  2 days ago
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

$T$ doesn't have to have fixed point. Let's consider space $X=1 times I cup 2 times I$, where $I$ is a unit compact interval, and $T colon X rightarrow X$, which acts on $X$ as follows: it takes each interval, shrinks it to $[0,frac12]$ and moves to the other interval, such that $T(1 times I)=2 times[0,frac12]$ and $T(2 times I)=1 times[0,frac12]$, it does not have fixed point, but $T^2$ does have one, for example $x=(1,0)$.

share|cite|improve this answer

answered 2 days ago

jon.sand

365

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
  – jon.sand
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
  – jon.sand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
  – jon.sand
  2 days ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

$T$ doesn't have to have fixed point. Let's consider space $X=1 times I cup 2 times I$, where $I$ is a unit compact interval, and $T colon X rightarrow X$, which acts on $X$ as follows: it takes each interval, shrinks it to $[0,frac12]$ and moves to the other interval, such that $T(1 times I)=2 times[0,frac12]$ and $T(2 times I)=1 times[0,frac12]$, it does not have fixed point, but $T^2$ does have one, for example $x=(1,0)$.

share|cite|improve this answer

answered 2 days ago

jon.sand

365

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
  – jon.sand
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
  – jon.sand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
  – jon.sand
  2 days ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

$T$ doesn't have to have fixed point. Let's consider space $X=1 times I cup 2 times I$, where $I$ is a unit compact interval, and $T colon X rightarrow X$, which acts on $X$ as follows: it takes each interval, shrinks it to $[0,frac12]$ and moves to the other interval, such that $T(1 times I)=2 times[0,frac12]$ and $T(2 times I)=1 times[0,frac12]$, it does not have fixed point, but $T^2$ does have one, for example $x=(1,0)$.

share|cite|improve this answer

answered 2 days ago

jon.sand

365

$T$ doesn't have to have fixed point. Let's consider space $X=1 times I cup 2 times I$, where $I$ is a unit compact interval, and $T colon X rightarrow X$, which acts on $X$ as follows: it takes each interval, shrinks it to $[0,frac12]$ and moves to the other interval, such that $T(1 times I)=2 times[0,frac12]$ and $T(2 times I)=1 times[0,frac12]$, it does not have fixed point, but $T^2$ does have one, for example $x=(1,0)$.

share|cite|improve this answer

answered 2 days ago

jon.sand

365

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

jon.sand

365

answered 2 days ago

jon.sand

365

answered 2 days ago

jon.sand

365

365

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
  – jon.sand
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
  – jon.sand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
  – jon.sand
  2 days ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
  – Cornman
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
  – jon.sand
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
  – jon.sand
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
  – jon.sand
  2 days ago
  

  
  
  
  

Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
– Cornman
2 days ago

Maybe it is just me, but I do not really get, what your function does. How do you calculate $T^2$. It would be cool, if you try to give more insight. :)
– Cornman
2 days ago

1

1

We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
– jon.sand
2 days ago

We can say, that $T((1,x))=(2,fracx2)$ and $T((2,x))=(1,fracx2)$, for $x in I$, then $T^2((i,x))=(i,fracx4)$ for $i=1,2$.
– jon.sand
2 days ago

Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
– jon.sand
2 days ago

Well, there is mistake somewhere, look here: math.stackexchange.com/questions/1447255/…
– jon.sand
2 days ago

1

1

Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
– jon.sand
2 days ago

Yes, sorry, $T^2$ is a contracion on each interval, but it is not a contracion on whole $X$ - consider points $(1,0)$ and $(2,0)$. Problem is solved in link above, look there.
– jon.sand
2 days ago

add a comment |