Why is $(f_n)_{ninmathbb{N}}$ with $f_n :[0,1]rightarrowmathbb{R}, f_n(x)=x^n$ not uniformly convergent?...

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  Prove $x^n$ is not uniformly convergent
  
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If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?

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edited Jan 8 at 21:48

Bernard

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asked Jan 8 at 21:46

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Jan 8 at 23:19

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.

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  Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
  $endgroup$
  – copper.hat
  Jan 8 at 21:58
  

  
  
  
  

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0

$begingroup$

This question already has an answer here:

• 
  Prove $x^n$ is not uniformly convergent
  
  2 answers
  
  

If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?

real-analysis

share|cite|improve this question

edited Jan 8 at 21:48

Bernard

120k740113

asked Jan 8 at 21:46

RM777RM777

33112

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Jan 8 at 23:19

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
  $endgroup$
  – copper.hat
  Jan 8 at 21:58
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Prove $x^n$ is not uniformly convergent
  
  2 answers
  
  

If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?

real-analysis

share|cite|improve this question

edited Jan 8 at 21:48

Bernard

120k740113

asked Jan 8 at 21:46

RM777RM777

33112

$endgroup$

This question already has an answer here:

• 
  Prove $x^n$ is not uniformly convergent
  
  2 answers
  
  

If I choose $f:[0,1]rightarrow mathbb{R}$ with $f(1)=1$ and otherwise $f(x)=0$ then $||f_n-f||_{[0,1]}rightarrow0$ therefore it is uniformly convergent and because it is uniformly convergent $f$ would have to be continuous but it is not, where is the mistake?

This question already has an answer here:

• 
  Prove $x^n$ is not uniformly convergent
  
  2 answers
  
  

real-analysis

real-analysis

share|cite|improve this question

edited Jan 8 at 21:48

Bernard

120k740113

asked Jan 8 at 21:46

RM777RM777

33112

share|cite|improve this question

edited Jan 8 at 21:48

Bernard

120k740113

asked Jan 8 at 21:46

RM777RM777

33112

share|cite|improve this question

share|cite|improve this question

edited Jan 8 at 21:48

Bernard

120k740113

edited Jan 8 at 21:48

Bernard

120k740113

edited Jan 8 at 21:48

Bernard

120k740113

120k740113

asked Jan 8 at 21:46

RM777RM777

33112

asked Jan 8 at 21:46

RM777RM777

33112

asked Jan 8 at 21:46

RM777RM777

33112

33112

marked as duplicate by RRL real-analysis
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Jan 8 at 23:19

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.

marked as duplicate by RRL real-analysis
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Jan 8 at 23:19

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
  $endgroup$
  – copper.hat
  Jan 8 at 21:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
  $endgroup$
  – copper.hat
  Jan 8 at 21:58
  

  
  
  
  

$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58

$begingroup$
Since $lim_n f_n$ is not continuous, it cannot be the uniform limit of any sequence of continuous functions.
$endgroup$
– copper.hat
Jan 8 at 21:58

add a comment | 

1 Answer
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It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.

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answered Jan 8 at 21:52

MarkMark

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.

share|cite|improve this answer

answered Jan 8 at 21:52

MarkMark

6,848416

$endgroup$

add a comment | 

2

$begingroup$

It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.

share|cite|improve this answer

answered Jan 8 at 21:52

MarkMark

6,848416

$endgroup$

add a comment | 

2

2

2

$begingroup$

It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.

share|cite|improve this answer

answered Jan 8 at 21:52

MarkMark

6,848416

$endgroup$

It is not uniformly convergent. For each $ninmathbb{N}$ you can choose $x=frac{1}{sqrt[n]{2}}$ and get $|f_n(x)-f(x)|geqfrac{1}{2}$. So it doesn't matter how far you go in the sequence, the distance will never be less than $frac{1}{2}$ for all $x$.

share|cite|improve this answer

answered Jan 8 at 21:52

MarkMark

6,848416

share|cite|improve this answer

share|cite|improve this answer

answered Jan 8 at 21:52

MarkMark

6,848416

answered Jan 8 at 21:52

MarkMark

6,848416

answered Jan 8 at 21:52

MarkMark

6,848416

6,848416

add a comment | 

add a comment | 

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