Pointwise and Uniformly Convergent Series [closed]

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Does there exist a pointwise convergent series on $mathbbR$, but not uniformly convergent at any interval?

real-analysis sequences-and-series

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edited Jul 22 at 22:52

mechanodroid

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asked Jul 22 at 20:54

ArMaRm

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closed as off-topic by Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel Jul 23 at 7:54

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." – Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel

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

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  What does it mean for a series to be uniformly convergent 'at a point'?
  – Fimpellizieri
  Jul 22 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry , it just a mistake , I have modified it
  – ArMaRm
  Jul 22 at 21:14
  

  
  
  
  

add a comment | 

up vote
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1

Does there exist a pointwise convergent series on $mathbbR$, but not uniformly convergent at any interval?

real-analysis sequences-and-series

share|cite|improve this question

edited Jul 22 at 22:52

mechanodroid

22.2k52041

asked Jul 22 at 20:54

ArMaRm

959

closed as off-topic by Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel Jul 23 at 7:54

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." – Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel

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

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What does it mean for a series to be uniformly convergent 'at a point'?
  – Fimpellizieri
  Jul 22 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry , it just a mistake , I have modified it
  – ArMaRm
  Jul 22 at 21:14
  

  
  
  
  

add a comment | 

up vote
0
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favorite

1

up vote
0
down vote

favorite

1

1

Does there exist a pointwise convergent series on $mathbbR$, but not uniformly convergent at any interval?

real-analysis sequences-and-series

share|cite|improve this question

edited Jul 22 at 22:52

mechanodroid

22.2k52041

asked Jul 22 at 20:54

ArMaRm

959

Does there exist a pointwise convergent series on $mathbbR$, but not uniformly convergent at any interval?

real-analysis sequences-and-series

share|cite|improve this question

edited Jul 22 at 22:52

mechanodroid

22.2k52041

asked Jul 22 at 20:54

ArMaRm

959

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 22:52

mechanodroid

22.2k52041

edited Jul 22 at 22:52

mechanodroid

22.2k52041

edited Jul 22 at 22:52

mechanodroid

22.2k52041

22.2k52041

asked Jul 22 at 20:54

ArMaRm

959

asked Jul 22 at 20:54

ArMaRm

959

asked Jul 22 at 20:54

ArMaRm

959

959

closed as off-topic by Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel Jul 23 at 7:54

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." – Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel

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

closed as off-topic by Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel Jul 23 at 7:54

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." – Leucippus, Shailesh, Xander Henderson, Taroccoesbrocco, Parcly Taxel

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

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What does it mean for a series to be uniformly convergent 'at a point'?
  – Fimpellizieri
  Jul 22 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry , it just a mistake , I have modified it
  – ArMaRm
  Jul 22 at 21:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  What does it mean for a series to be uniformly convergent 'at a point'?
  – Fimpellizieri
  Jul 22 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry , it just a mistake , I have modified it
  – ArMaRm
  Jul 22 at 21:14
  

  
  
  
  

3

3

What does it mean for a series to be uniformly convergent 'at a point'?
– Fimpellizieri
Jul 22 at 20:55

What does it mean for a series to be uniformly convergent 'at a point'?
– Fimpellizieri
Jul 22 at 20:55

Sorry , it just a mistake , I have modified it
– ArMaRm
Jul 22 at 21:14

Sorry , it just a mistake , I have modified it
– ArMaRm
Jul 22 at 21:14

add a comment | 

3 Answers
3

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up vote
4
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Let $mathbbQ = r_1, r_2, ldots$ be an enumeration of the rational numbers and take

$$f_n(x) = begincases1, &x = r_n \ 0, & textotherwise endcases$$

We have pointwise convergence of the series

$$sum_n=1^infty f_n(x) = f(x) =begincases1, &x in mathbbQ \ 0, & textotherwise endcases$$

The rational numbers in any interval $[a,b]$ can be enumerated by a subsequence $(r_n_k)$. The series does not converge uniformly on the interval because for any $k in mathbbN$ we have $n_k geqslant k$ such that

$$left|sum_n=1^n_k f_n(r_n_k+1) - f(r_n_k+1)right| = 1, $$

and it is not possible to find for any $0 < epsilon < 1$ an integer $k$ such that for all $m >k$ and all $x in [a,b],$

$$left|sum_n=1^m f_n(x) - f(x)right| < epsilon$$

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answered Jul 22 at 21:56

RRL

43.6k42260

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up vote
2
down vote

If you mean "Does there exist a sequence of functions $f_n:Bbb RtoBbb R$ which converges pointwise, but doesn't converge uniformly on any interval?", then yes, there are such sequences.

For instance, let $f$ be the Conway base 13 function. It is a function which attains all real numbers as value on any (non-trivial) interval. Then set $f_n(x)=frac1n f(x)$. Each $f_n$ still has the property that on any interval it attains any real number as value, so the sequence cannot possibly converge uniformly on any interval. But the sequence clearly converges pointwise to the zero function.

share|cite|improve this answer

answered Jul 22 at 21:29

Arthur

98.5k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
  – ArMaRm
  Jul 22 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
  – Arthur
  Jul 22 at 22:04
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

This answer constructs a sequence $(f_n)_n$ of continuous functions $f_n : mathbbR to mathbbR$ which converges pointwise to Thomae's function $f: mathbbR to mathbbR$ given by

$$f(x)= begincases
frac1q &: text if $x = fracpq$ with $p inmathbbZ, q in mathbbN$ coprime\
0 &: text if $x$ is irrational
endcases$$

The set of discontinuities of $f$ is precisely $mathbbQ$.

If $f_n|_[a,b] to f|_[a,b]$ uniformly for some interval $[a,b]$ then it would follow that $f|_[a,b]$ is continuous, which is impossible because the set of discontinuities $mathbbQ$ is dense in $mathbbR$.

Therefore the convergence $f_n to f$ is not uniform on any interval.

share|cite|improve this answer

edited Jul 23 at 9:30

answered Jul 22 at 22:48

mechanodroid

22.2k52041

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

Let $mathbbQ = r_1, r_2, ldots$ be an enumeration of the rational numbers and take

$$f_n(x) = begincases1, &x = r_n \ 0, & textotherwise endcases$$

We have pointwise convergence of the series

$$sum_n=1^infty f_n(x) = f(x) =begincases1, &x in mathbbQ \ 0, & textotherwise endcases$$

The rational numbers in any interval $[a,b]$ can be enumerated by a subsequence $(r_n_k)$. The series does not converge uniformly on the interval because for any $k in mathbbN$ we have $n_k geqslant k$ such that

$$left|sum_n=1^n_k f_n(r_n_k+1) - f(r_n_k+1)right| = 1, $$

and it is not possible to find for any $0 < epsilon < 1$ an integer $k$ such that for all $m >k$ and all $x in [a,b],$

$$left|sum_n=1^m f_n(x) - f(x)right| < epsilon$$

share|cite|improve this answer

answered Jul 22 at 21:56

RRL

43.6k42260

add a comment | 

up vote
4
down vote

Let $mathbbQ = r_1, r_2, ldots$ be an enumeration of the rational numbers and take

$$f_n(x) = begincases1, &x = r_n \ 0, & textotherwise endcases$$

We have pointwise convergence of the series

$$sum_n=1^infty f_n(x) = f(x) =begincases1, &x in mathbbQ \ 0, & textotherwise endcases$$

The rational numbers in any interval $[a,b]$ can be enumerated by a subsequence $(r_n_k)$. The series does not converge uniformly on the interval because for any $k in mathbbN$ we have $n_k geqslant k$ such that

$$left|sum_n=1^n_k f_n(r_n_k+1) - f(r_n_k+1)right| = 1, $$

and it is not possible to find for any $0 < epsilon < 1$ an integer $k$ such that for all $m >k$ and all $x in [a,b],$

$$left|sum_n=1^m f_n(x) - f(x)right| < epsilon$$

share|cite|improve this answer

answered Jul 22 at 21:56

RRL

43.6k42260

add a comment | 

up vote
4
down vote

up vote
4
down vote

Let $mathbbQ = r_1, r_2, ldots$ be an enumeration of the rational numbers and take

$$f_n(x) = begincases1, &x = r_n \ 0, & textotherwise endcases$$

We have pointwise convergence of the series

$$sum_n=1^infty f_n(x) = f(x) =begincases1, &x in mathbbQ \ 0, & textotherwise endcases$$

The rational numbers in any interval $[a,b]$ can be enumerated by a subsequence $(r_n_k)$. The series does not converge uniformly on the interval because for any $k in mathbbN$ we have $n_k geqslant k$ such that

$$left|sum_n=1^n_k f_n(r_n_k+1) - f(r_n_k+1)right| = 1, $$

and it is not possible to find for any $0 < epsilon < 1$ an integer $k$ such that for all $m >k$ and all $x in [a,b],$

$$left|sum_n=1^m f_n(x) - f(x)right| < epsilon$$

share|cite|improve this answer

answered Jul 22 at 21:56

RRL

43.6k42260

Let $mathbbQ = r_1, r_2, ldots$ be an enumeration of the rational numbers and take

$$f_n(x) = begincases1, &x = r_n \ 0, & textotherwise endcases$$

We have pointwise convergence of the series

$$sum_n=1^infty f_n(x) = f(x) =begincases1, &x in mathbbQ \ 0, & textotherwise endcases$$

The rational numbers in any interval $[a,b]$ can be enumerated by a subsequence $(r_n_k)$. The series does not converge uniformly on the interval because for any $k in mathbbN$ we have $n_k geqslant k$ such that

$$left|sum_n=1^n_k f_n(r_n_k+1) - f(r_n_k+1)right| = 1, $$

and it is not possible to find for any $0 < epsilon < 1$ an integer $k$ such that for all $m >k$ and all $x in [a,b],$

$$left|sum_n=1^m f_n(x) - f(x)right| < epsilon$$

share|cite|improve this answer

answered Jul 22 at 21:56

RRL

43.6k42260

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 21:56

RRL

43.6k42260

answered Jul 22 at 21:56

RRL

43.6k42260

answered Jul 22 at 21:56

RRL

43.6k42260

43.6k42260

add a comment | 

add a comment | 

up vote
2
down vote

If you mean "Does there exist a sequence of functions $f_n:Bbb RtoBbb R$ which converges pointwise, but doesn't converge uniformly on any interval?", then yes, there are such sequences.

For instance, let $f$ be the Conway base 13 function. It is a function which attains all real numbers as value on any (non-trivial) interval. Then set $f_n(x)=frac1n f(x)$. Each $f_n$ still has the property that on any interval it attains any real number as value, so the sequence cannot possibly converge uniformly on any interval. But the sequence clearly converges pointwise to the zero function.

share|cite|improve this answer

answered Jul 22 at 21:29

Arthur

98.5k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
  – ArMaRm
  Jul 22 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
  – Arthur
  Jul 22 at 22:04
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

If you mean "Does there exist a sequence of functions $f_n:Bbb RtoBbb R$ which converges pointwise, but doesn't converge uniformly on any interval?", then yes, there are such sequences.

For instance, let $f$ be the Conway base 13 function. It is a function which attains all real numbers as value on any (non-trivial) interval. Then set $f_n(x)=frac1n f(x)$. Each $f_n$ still has the property that on any interval it attains any real number as value, so the sequence cannot possibly converge uniformly on any interval. But the sequence clearly converges pointwise to the zero function.

share|cite|improve this answer

answered Jul 22 at 21:29

Arthur

98.5k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
  – ArMaRm
  Jul 22 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
  – Arthur
  Jul 22 at 22:04
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

If you mean "Does there exist a sequence of functions $f_n:Bbb RtoBbb R$ which converges pointwise, but doesn't converge uniformly on any interval?", then yes, there are such sequences.

For instance, let $f$ be the Conway base 13 function. It is a function which attains all real numbers as value on any (non-trivial) interval. Then set $f_n(x)=frac1n f(x)$. Each $f_n$ still has the property that on any interval it attains any real number as value, so the sequence cannot possibly converge uniformly on any interval. But the sequence clearly converges pointwise to the zero function.

share|cite|improve this answer

answered Jul 22 at 21:29

Arthur

98.5k793174

If you mean "Does there exist a sequence of functions $f_n:Bbb RtoBbb R$ which converges pointwise, but doesn't converge uniformly on any interval?", then yes, there are such sequences.

For instance, let $f$ be the Conway base 13 function. It is a function which attains all real numbers as value on any (non-trivial) interval. Then set $f_n(x)=frac1n f(x)$. Each $f_n$ still has the property that on any interval it attains any real number as value, so the sequence cannot possibly converge uniformly on any interval. But the sequence clearly converges pointwise to the zero function.

share|cite|improve this answer

answered Jul 22 at 21:29

Arthur

98.5k793174

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 21:29

Arthur

98.5k793174

answered Jul 22 at 21:29

Arthur

98.5k793174

answered Jul 22 at 21:29

Arthur

98.5k793174

98.5k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
  – ArMaRm
  Jul 22 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
  – Arthur
  Jul 22 at 22:04
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
  – ArMaRm
  Jul 22 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
  – Arthur
  Jul 22 at 22:04
  
  

  
  
  
  

Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
– ArMaRm
Jul 22 at 21:37

Let $f_n$ be a sequence of functions , and $S_n$ = $f_1$ + $f_2$ + .... be the partial sum of the series $Z$ . If the limit of $S_n$ converges P.W then $Z$ converges P.W , similarly for U.C .
– ArMaRm
Jul 22 at 21:37

From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
– Arthur
Jul 22 at 22:04

From any sequence, you can easily make a series with that sequence as partial sums. So the existence of one implies the existence of the other.
– Arthur
Jul 22 at 22:04

add a comment | 

up vote
2
down vote

This answer constructs a sequence $(f_n)_n$ of continuous functions $f_n : mathbbR to mathbbR$ which converges pointwise to Thomae's function $f: mathbbR to mathbbR$ given by

$$f(x)= begincases
frac1q &: text if $x = fracpq$ with $p inmathbbZ, q in mathbbN$ coprime\
0 &: text if $x$ is irrational
endcases$$

The set of discontinuities of $f$ is precisely $mathbbQ$.

If $f_n|_[a,b] to f|_[a,b]$ uniformly for some interval $[a,b]$ then it would follow that $f|_[a,b]$ is continuous, which is impossible because the set of discontinuities $mathbbQ$ is dense in $mathbbR$.

Therefore the convergence $f_n to f$ is not uniform on any interval.

share|cite|improve this answer

edited Jul 23 at 9:30

answered Jul 22 at 22:48

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

This answer constructs a sequence $(f_n)_n$ of continuous functions $f_n : mathbbR to mathbbR$ which converges pointwise to Thomae's function $f: mathbbR to mathbbR$ given by

$$f(x)= begincases
frac1q &: text if $x = fracpq$ with $p inmathbbZ, q in mathbbN$ coprime\
0 &: text if $x$ is irrational
endcases$$

The set of discontinuities of $f$ is precisely $mathbbQ$.

If $f_n|_[a,b] to f|_[a,b]$ uniformly for some interval $[a,b]$ then it would follow that $f|_[a,b]$ is continuous, which is impossible because the set of discontinuities $mathbbQ$ is dense in $mathbbR$.

Therefore the convergence $f_n to f$ is not uniform on any interval.

share|cite|improve this answer

edited Jul 23 at 9:30

answered Jul 22 at 22:48

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

up vote
2
down vote

This answer constructs a sequence $(f_n)_n$ of continuous functions $f_n : mathbbR to mathbbR$ which converges pointwise to Thomae's function $f: mathbbR to mathbbR$ given by

$$f(x)= begincases
frac1q &: text if $x = fracpq$ with $p inmathbbZ, q in mathbbN$ coprime\
0 &: text if $x$ is irrational
endcases$$

The set of discontinuities of $f$ is precisely $mathbbQ$.

If $f_n|_[a,b] to f|_[a,b]$ uniformly for some interval $[a,b]$ then it would follow that $f|_[a,b]$ is continuous, which is impossible because the set of discontinuities $mathbbQ$ is dense in $mathbbR$.

Therefore the convergence $f_n to f$ is not uniform on any interval.

share|cite|improve this answer

edited Jul 23 at 9:30

answered Jul 22 at 22:48

mechanodroid

22.2k52041

This answer constructs a sequence $(f_n)_n$ of continuous functions $f_n : mathbbR to mathbbR$ which converges pointwise to Thomae's function $f: mathbbR to mathbbR$ given by

$$f(x)= begincases
frac1q &: text if $x = fracpq$ with $p inmathbbZ, q in mathbbN$ coprime\
0 &: text if $x$ is irrational
endcases$$

The set of discontinuities of $f$ is precisely $mathbbQ$.

If $f_n|_[a,b] to f|_[a,b]$ uniformly for some interval $[a,b]$ then it would follow that $f|_[a,b]$ is continuous, which is impossible because the set of discontinuities $mathbbQ$ is dense in $mathbbR$.

Therefore the convergence $f_n to f$ is not uniform on any interval.

share|cite|improve this answer

edited Jul 23 at 9:30

answered Jul 22 at 22:48

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 9:30

edited Jul 23 at 9:30

edited Jul 23 at 9:30

answered Jul 22 at 22:48

mechanodroid

22.2k52041

answered Jul 22 at 22:48

mechanodroid

22.2k52041

answered Jul 22 at 22:48

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment |