Where is the mistake in my counter-example to the statement? (sequences of functions, uniform convergence)

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Statement:

Let $f,f_n:[a,b]tomathbbR$ for $ninmathbbN$ and for all $xin[a,b]$ let $lim_ntoinftyf_n(x) = f(x)$.

If $f$ and all $f_n$ are continuous, then $f_n$ converges uniformly against $f$.

Counter example is supposed to be the function
$$f_n( x) ≔ begincases
g(n·x) &, 0<x<frac 1 n\
0 &, textelse
endcases$$

Where $g(x)≔ 8·x^4 - 16·x^3 + 8·x^2$ .

Here is a plot of it for $n$ increasing:

With this function, all requirements of the statement should be fullfilled:

We have $f(x) = lim_ntoinfty f_n(x) = 0$, so $f$ is continuous.

Further, all $f_n$ are continuous, as $lim_xto 0 f_n(x) = 0 = lim_xto frac 1 n f_n(x) $.

However, $f_n$ doesn't converge uniformly against $f$, as for $epsilon = frac 1 2$, no matter how high we choose $ninmathbbN$, we have $f_n(frac 1 2n) = frac 1 2$.

Now, I've found two text books claiming the statement to be true, so somewhere I've got to have made an error in my reasoning. I can't find it though: Where is it?

---

Both text books are German (and rather old), the statement is however as faithful translated as possible.

First is: Landers Rogge, Nichtstandardanalysis, page 132, exercise 1 (from 1994).

Second is (cited from wikipedia): F. Hausdorff: Grundzüge der Mengenlehre. 1914, Chelsea Publishing Co., New York 1949, Kap. IX, § 4

real-analysis sequence-of-function

share|cite|improve this question

edited Jul 20 at 21:52

asked Jul 20 at 21:38

Sudix

7911316

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are $a$ and $b$ for the counterexample?
  – Adrian Keister
  Jul 20 at 21:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are absolutely correct (there is no mistake to be found). Which textbooks are claiming such a thing to be true?
  – Severin Schraven
  Jul 20 at 21:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is required that the sequence $(f_n)_n$ is monotone: Dini's theorem
  – mechanodroid
  Jul 20 at 21:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't see that statement in Hausdorff's book. It's easy to misread II. in chapter IX §4 in this way, which says "Sind alle $f_n(x)$ in $a$ stetig, so ist $f(x)$ in $a$ stetig dann und nur dann, wenn $a$ ein Punkt uniformer Konvergenz ist." However, the term "uniforme Konvergenz" used by Hausdorff is different from (strictly weaker than) uniform convergence (= gleichmäßige Konvergenz). It was an unfortunate choice of term. Hausdorff was aware of its problematic status (to some extent at least).
  – Daniel Fischer♦
  Jul 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For a new edition with corrections, two things are necessary. First, the authors must have been made aware of the mistake, which may not have happened. Second, there must be enough demand for the book, that the publisher expects profit from a new edition. Which is easier in mainstream topics.
  – Daniel Fischer♦
  Jul 20 at 23:23
  

  
  
  
  

 | 
show 9 more comments

up vote
4
down vote

favorite

Statement:

Let $f,f_n:[a,b]tomathbbR$ for $ninmathbbN$ and for all $xin[a,b]$ let $lim_ntoinftyf_n(x) = f(x)$.

If $f$ and all $f_n$ are continuous, then $f_n$ converges uniformly against $f$.

Counter example is supposed to be the function
$$f_n( x) ≔ begincases
g(n·x) &, 0<x<frac 1 n\
0 &, textelse
endcases$$

Where $g(x)≔ 8·x^4 - 16·x^3 + 8·x^2$ .

Here is a plot of it for $n$ increasing:

With this function, all requirements of the statement should be fullfilled:

We have $f(x) = lim_ntoinfty f_n(x) = 0$, so $f$ is continuous.

Further, all $f_n$ are continuous, as $lim_xto 0 f_n(x) = 0 = lim_xto frac 1 n f_n(x) $.

However, $f_n$ doesn't converge uniformly against $f$, as for $epsilon = frac 1 2$, no matter how high we choose $ninmathbbN$, we have $f_n(frac 1 2n) = frac 1 2$.

Now, I've found two text books claiming the statement to be true, so somewhere I've got to have made an error in my reasoning. I can't find it though: Where is it?

---

Both text books are German (and rather old), the statement is however as faithful translated as possible.

First is: Landers Rogge, Nichtstandardanalysis, page 132, exercise 1 (from 1994).

Second is (cited from wikipedia): F. Hausdorff: Grundzüge der Mengenlehre. 1914, Chelsea Publishing Co., New York 1949, Kap. IX, § 4

real-analysis sequence-of-function

share|cite|improve this question

edited Jul 20 at 21:52

asked Jul 20 at 21:38

Sudix

7911316

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are $a$ and $b$ for the counterexample?
  – Adrian Keister
  Jul 20 at 21:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are absolutely correct (there is no mistake to be found). Which textbooks are claiming such a thing to be true?
  – Severin Schraven
  Jul 20 at 21:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is required that the sequence $(f_n)_n$ is monotone: Dini's theorem
  – mechanodroid
  Jul 20 at 21:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't see that statement in Hausdorff's book. It's easy to misread II. in chapter IX §4 in this way, which says "Sind alle $f_n(x)$ in $a$ stetig, so ist $f(x)$ in $a$ stetig dann und nur dann, wenn $a$ ein Punkt uniformer Konvergenz ist." However, the term "uniforme Konvergenz" used by Hausdorff is different from (strictly weaker than) uniform convergence (= gleichmäßige Konvergenz). It was an unfortunate choice of term. Hausdorff was aware of its problematic status (to some extent at least).
  – Daniel Fischer♦
  Jul 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For a new edition with corrections, two things are necessary. First, the authors must have been made aware of the mistake, which may not have happened. Second, there must be enough demand for the book, that the publisher expects profit from a new edition. Which is easier in mainstream topics.
  – Daniel Fischer♦
  Jul 20 at 23:23
  

  
  
  
  

 | 
show 9 more comments

up vote
4
down vote

favorite

up vote
4
down vote

favorite

Statement:

Let $f,f_n:[a,b]tomathbbR$ for $ninmathbbN$ and for all $xin[a,b]$ let $lim_ntoinftyf_n(x) = f(x)$.

If $f$ and all $f_n$ are continuous, then $f_n$ converges uniformly against $f$.

Counter example is supposed to be the function
$$f_n( x) ≔ begincases
g(n·x) &, 0<x<frac 1 n\
0 &, textelse
endcases$$

Where $g(x)≔ 8·x^4 - 16·x^3 + 8·x^2$ .

Here is a plot of it for $n$ increasing:

With this function, all requirements of the statement should be fullfilled:

We have $f(x) = lim_ntoinfty f_n(x) = 0$, so $f$ is continuous.

Further, all $f_n$ are continuous, as $lim_xto 0 f_n(x) = 0 = lim_xto frac 1 n f_n(x) $.

However, $f_n$ doesn't converge uniformly against $f$, as for $epsilon = frac 1 2$, no matter how high we choose $ninmathbbN$, we have $f_n(frac 1 2n) = frac 1 2$.

Now, I've found two text books claiming the statement to be true, so somewhere I've got to have made an error in my reasoning. I can't find it though: Where is it?

---

Both text books are German (and rather old), the statement is however as faithful translated as possible.

First is: Landers Rogge, Nichtstandardanalysis, page 132, exercise 1 (from 1994).

Second is (cited from wikipedia): F. Hausdorff: Grundzüge der Mengenlehre. 1914, Chelsea Publishing Co., New York 1949, Kap. IX, § 4

real-analysis sequence-of-function

share|cite|improve this question

edited Jul 20 at 21:52

asked Jul 20 at 21:38

Sudix

7911316

Statement:

Let $f,f_n:[a,b]tomathbbR$ for $ninmathbbN$ and for all $xin[a,b]$ let $lim_ntoinftyf_n(x) = f(x)$.

If $f$ and all $f_n$ are continuous, then $f_n$ converges uniformly against $f$.

Counter example is supposed to be the function
$$f_n( x) ≔ begincases
g(n·x) &, 0<x<frac 1 n\
0 &, textelse
endcases$$

Where $g(x)≔ 8·x^4 - 16·x^3 + 8·x^2$ .

Here is a plot of it for $n$ increasing:

With this function, all requirements of the statement should be fullfilled:

We have $f(x) = lim_ntoinfty f_n(x) = 0$, so $f$ is continuous.

Further, all $f_n$ are continuous, as $lim_xto 0 f_n(x) = 0 = lim_xto frac 1 n f_n(x) $.

However, $f_n$ doesn't converge uniformly against $f$, as for $epsilon = frac 1 2$, no matter how high we choose $ninmathbbN$, we have $f_n(frac 1 2n) = frac 1 2$.

Now, I've found two text books claiming the statement to be true, so somewhere I've got to have made an error in my reasoning. I can't find it though: Where is it?

---

Both text books are German (and rather old), the statement is however as faithful translated as possible.

First is: Landers Rogge, Nichtstandardanalysis, page 132, exercise 1 (from 1994).

Second is (cited from wikipedia): F. Hausdorff: Grundzüge der Mengenlehre. 1914, Chelsea Publishing Co., New York 1949, Kap. IX, § 4

real-analysis sequence-of-function

share|cite|improve this question

edited Jul 20 at 21:52

asked Jul 20 at 21:38

Sudix

7911316

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 21:52

edited Jul 20 at 21:52

edited Jul 20 at 21:52

asked Jul 20 at 21:38

Sudix

7911316

asked Jul 20 at 21:38

Sudix

7911316

asked Jul 20 at 21:38

Sudix

7911316

7911316

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are $a$ and $b$ for the counterexample?
  – Adrian Keister
  Jul 20 at 21:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are absolutely correct (there is no mistake to be found). Which textbooks are claiming such a thing to be true?
  – Severin Schraven
  Jul 20 at 21:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is required that the sequence $(f_n)_n$ is monotone: Dini's theorem
  – mechanodroid
  Jul 20 at 21:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't see that statement in Hausdorff's book. It's easy to misread II. in chapter IX §4 in this way, which says "Sind alle $f_n(x)$ in $a$ stetig, so ist $f(x)$ in $a$ stetig dann und nur dann, wenn $a$ ein Punkt uniformer Konvergenz ist." However, the term "uniforme Konvergenz" used by Hausdorff is different from (strictly weaker than) uniform convergence (= gleichmäßige Konvergenz). It was an unfortunate choice of term. Hausdorff was aware of its problematic status (to some extent at least).
  – Daniel Fischer♦
  Jul 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For a new edition with corrections, two things are necessary. First, the authors must have been made aware of the mistake, which may not have happened. Second, there must be enough demand for the book, that the publisher expects profit from a new edition. Which is easier in mainstream topics.
  – Daniel Fischer♦
  Jul 20 at 23:23
  

  
  
  
  

 | 
show 9 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What are $a$ and $b$ for the counterexample?
  – Adrian Keister
  Jul 20 at 21:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You are absolutely correct (there is no mistake to be found). Which textbooks are claiming such a thing to be true?
  – Severin Schraven
  Jul 20 at 21:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Perhaps it is required that the sequence $(f_n)_n$ is monotone: Dini's theorem
  – mechanodroid
  Jul 20 at 21:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't see that statement in Hausdorff's book. It's easy to misread II. in chapter IX §4 in this way, which says "Sind alle $f_n(x)$ in $a$ stetig, so ist $f(x)$ in $a$ stetig dann und nur dann, wenn $a$ ein Punkt uniformer Konvergenz ist." However, the term "uniforme Konvergenz" used by Hausdorff is different from (strictly weaker than) uniform convergence (= gleichmäßige Konvergenz). It was an unfortunate choice of term. Hausdorff was aware of its problematic status (to some extent at least).
  – Daniel Fischer♦
  Jul 20 at 22:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For a new edition with corrections, two things are necessary. First, the authors must have been made aware of the mistake, which may not have happened. Second, there must be enough demand for the book, that the publisher expects profit from a new edition. Which is easier in mainstream topics.
  – Daniel Fischer♦
  Jul 20 at 23:23
  

  
  
  
  

What are $a$ and $b$ for the counterexample?
– Adrian Keister
Jul 20 at 21:44

What are $a$ and $b$ for the counterexample?
– Adrian Keister
Jul 20 at 21:44

1

1

You are absolutely correct (there is no mistake to be found). Which textbooks are claiming such a thing to be true?
– Severin Schraven
Jul 20 at 21:46

You are absolutely correct (there is no mistake to be found). Which textbooks are claiming such a thing to be true?
– Severin Schraven
Jul 20 at 21:46

1

1

Perhaps it is required that the sequence $(f_n)_n$ is monotone: Dini's theorem
– mechanodroid
Jul 20 at 21:59

Perhaps it is required that the sequence $(f_n)_n$ is monotone: Dini's theorem
– mechanodroid
Jul 20 at 21:59

1

1

I don't see that statement in Hausdorff's book. It's easy to misread II. in chapter IX §4 in this way, which says "Sind alle $f_n(x)$ in $a$ stetig, so ist $f(x)$ in $a$ stetig dann und nur dann, wenn $a$ ein Punkt uniformer Konvergenz ist." However, the term "uniforme Konvergenz" used by Hausdorff is different from (strictly weaker than) uniform convergence (= gleichmäßige Konvergenz). It was an unfortunate choice of term. Hausdorff was aware of its problematic status (to some extent at least).
– Daniel Fischer♦
Jul 20 at 22:56

I don't see that statement in Hausdorff's book. It's easy to misread II. in chapter IX §4 in this way, which says "Sind alle $f_n(x)$ in $a$ stetig, so ist $f(x)$ in $a$ stetig dann und nur dann, wenn $a$ ein Punkt uniformer Konvergenz ist." However, the term "uniforme Konvergenz" used by Hausdorff is different from (strictly weaker than) uniform convergence (= gleichmäßige Konvergenz). It was an unfortunate choice of term. Hausdorff was aware of its problematic status (to some extent at least).
– Daniel Fischer♦
Jul 20 at 22:56

1

1

For a new edition with corrections, two things are necessary. First, the authors must have been made aware of the mistake, which may not have happened. Second, there must be enough demand for the book, that the publisher expects profit from a new edition. Which is easier in mainstream topics.
– Daniel Fischer♦
Jul 20 at 23:23

For a new edition with corrections, two things are necessary. First, the authors must have been made aware of the mistake, which may not have happened. Second, there must be enough demand for the book, that the publisher expects profit from a new edition. Which is easier in mainstream topics.
– Daniel Fischer♦
Jul 20 at 23:23

 | 
show 9 more comments

1 Answer
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2
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Theorem 7.9 in Rudin's Principles of Mathematical Analysis states the following:

Suppose $$lim_ntoinftyf_n(x)=f(x) qquad (xin E).$$
Put $$M_n=sup_xin E|f_n(x)-f(x)|.$$
Then $f_nto f$ uniformly on $E$ if and only if $M_nto 0$ as $nto infty.$

You can see that for your counterexample, $M_nnotto 0$ as $ntoinfty$. Also note the "if and only if".

share|cite|improve this answer

edited Jul 21 at 1:50

answered Jul 20 at 22:04

Adrian Keister

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Theorem 7.9 in Rudin's Principles of Mathematical Analysis states the following:

Suppose $$lim_ntoinftyf_n(x)=f(x) qquad (xin E).$$
Put $$M_n=sup_xin E|f_n(x)-f(x)|.$$
Then $f_nto f$ uniformly on $E$ if and only if $M_nto 0$ as $nto infty.$

You can see that for your counterexample, $M_nnotto 0$ as $ntoinfty$. Also note the "if and only if".

share|cite|improve this answer

edited Jul 21 at 1:50

answered Jul 20 at 22:04

Adrian Keister

3,61721533

add a comment | 

up vote
2
down vote

Theorem 7.9 in Rudin's Principles of Mathematical Analysis states the following:

Suppose $$lim_ntoinftyf_n(x)=f(x) qquad (xin E).$$
Put $$M_n=sup_xin E|f_n(x)-f(x)|.$$
Then $f_nto f$ uniformly on $E$ if and only if $M_nto 0$ as $nto infty.$

You can see that for your counterexample, $M_nnotto 0$ as $ntoinfty$. Also note the "if and only if".

share|cite|improve this answer

edited Jul 21 at 1:50

answered Jul 20 at 22:04

Adrian Keister

3,61721533

add a comment | 

up vote
2
down vote

up vote
2
down vote

Theorem 7.9 in Rudin's Principles of Mathematical Analysis states the following:

Suppose $$lim_ntoinftyf_n(x)=f(x) qquad (xin E).$$
Put $$M_n=sup_xin E|f_n(x)-f(x)|.$$
Then $f_nto f$ uniformly on $E$ if and only if $M_nto 0$ as $nto infty.$

You can see that for your counterexample, $M_nnotto 0$ as $ntoinfty$. Also note the "if and only if".

share|cite|improve this answer

edited Jul 21 at 1:50

answered Jul 20 at 22:04

Adrian Keister

3,61721533

Theorem 7.9 in Rudin's Principles of Mathematical Analysis states the following:

Suppose $$lim_ntoinftyf_n(x)=f(x) qquad (xin E).$$
Put $$M_n=sup_xin E|f_n(x)-f(x)|.$$
Then $f_nto f$ uniformly on $E$ if and only if $M_nto 0$ as $nto infty.$

You can see that for your counterexample, $M_nnotto 0$ as $ntoinfty$. Also note the "if and only if".

share|cite|improve this answer

edited Jul 21 at 1:50

answered Jul 20 at 22:04

Adrian Keister

3,61721533

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 1:50

edited Jul 21 at 1:50

edited Jul 21 at 1:50

answered Jul 20 at 22:04

Adrian Keister

3,61721533

answered Jul 20 at 22:04

Adrian Keister

3,61721533

answered Jul 20 at 22:04

Adrian Keister

3,61721533

3,61721533

add a comment | 

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