What does “finite but unbounded” mean?

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In the following example, Gelbaum and Olmsted (Counterexamples in Analysis, 1964, p. 37) speak of a derivative being "finite but unbounded".

What does this mean and how is it possible that a quantity is both finite and unbounded?

real-analysis

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asked 3 hours ago

dtcm840

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up vote
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In the following example, Gelbaum and Olmsted (Counterexamples in Analysis, 1964, p. 37) speak of a derivative being "finite but unbounded".

What does this mean and how is it possible that a quantity is both finite and unbounded?

real-analysis

share|cite|improve this question

asked 3 hours ago

dtcm840

1418

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

In the following example, Gelbaum and Olmsted (Counterexamples in Analysis, 1964, p. 37) speak of a derivative being "finite but unbounded".

What does this mean and how is it possible that a quantity is both finite and unbounded?

real-analysis

share|cite|improve this question

asked 3 hours ago

dtcm840

1418

In the following example, Gelbaum and Olmsted (Counterexamples in Analysis, 1964, p. 37) speak of a derivative being "finite but unbounded".

What does this mean and how is it possible that a quantity is both finite and unbounded?

real-analysis

real-analysis

share|cite|improve this question

asked 3 hours ago

dtcm840

1418

share|cite|improve this question

asked 3 hours ago

dtcm840

1418

share|cite|improve this question

share|cite|improve this question

asked 3 hours ago

dtcm840

1418

asked 3 hours ago

dtcm840

1418

asked 3 hours ago

dtcm840

1418

1418

add a comment | 

add a comment | 

3 Answers
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The term "finite" in this case is stressing the fact that it is "defined", "well defined" or "exists as a real number".

To better understand why this is important, consider the following naive attempt to give an example of a function with unbounded derivative on a closed interval:

$$f(x) = sqrtx$$

The derivative, which is $$dfrac12sqrtx$$ is clearly unbounded in $(0,1]$ but it is not even defined in zero.

The goal of the book is to disallow such examples. It technically could say "defined and unbounded", or even simply "unbounded", but to avoid any ambiguity, it is usual to say "finite and unbounded" here. Some might argue that the derivative is "defined as $infty$" in zero (as opposed to "utterly undefined" when for example the function is not even continuous at the point), and to avoid any ambiguity the usual wording is "finite" (i.e. it is defined as a real number).

This is actually why that example in your book is interesting. By ruling out the naive examples such as $sqrtx$, one might wonder if there could be any function with unbounded derivative in a closed interval.

I remember perfectly that when my teacher asked this in class, I thought that if such example would exist, surely it would involve a function that gets steeper and steeper, and that it would "end up being infinite in a closed interval" and I answered that it was impossible.

But guess what, it's possible, as the example in your book shows.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Pedro A

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The sequence $1,2,...$ is finite in the sense each term is finite, but the sequence is not bounded in the sense there is no fixed real number $m$ such that all the terms are bouded by this number. Similarly, $f(x)=frac 1 x$ is finite in $(0,1)$ but it is not bounded. In the above example take $x =frac 1 sqrt 2npi$ to see that $f'$ is not bounded.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So when (if ever) can a function be said to be infinite?
  – dtcm840
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
  – Kavi Rama Murthy
  3 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

The theorem indicates that the derivative of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$

It is obvious that a finite set is bounded, therefore the author's intention is to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].

share|cite|improve this answer

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

The term "finite" in this case is stressing the fact that it is "defined", "well defined" or "exists as a real number".

To better understand why this is important, consider the following naive attempt to give an example of a function with unbounded derivative on a closed interval:

$$f(x) = sqrtx$$

The derivative, which is $$dfrac12sqrtx$$ is clearly unbounded in $(0,1]$ but it is not even defined in zero.

The goal of the book is to disallow such examples. It technically could say "defined and unbounded", or even simply "unbounded", but to avoid any ambiguity, it is usual to say "finite and unbounded" here. Some might argue that the derivative is "defined as $infty$" in zero (as opposed to "utterly undefined" when for example the function is not even continuous at the point), and to avoid any ambiguity the usual wording is "finite" (i.e. it is defined as a real number).

This is actually why that example in your book is interesting. By ruling out the naive examples such as $sqrtx$, one might wonder if there could be any function with unbounded derivative in a closed interval.

I remember perfectly that when my teacher asked this in class, I thought that if such example would exist, surely it would involve a function that gets steeper and steeper, and that it would "end up being infinite in a closed interval" and I answered that it was impossible.

But guess what, it's possible, as the example in your book shows.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Pedro A

1,7991824

add a comment | 

up vote
2
down vote

The term "finite" in this case is stressing the fact that it is "defined", "well defined" or "exists as a real number".

To better understand why this is important, consider the following naive attempt to give an example of a function with unbounded derivative on a closed interval:

$$f(x) = sqrtx$$

The derivative, which is $$dfrac12sqrtx$$ is clearly unbounded in $(0,1]$ but it is not even defined in zero.

The goal of the book is to disallow such examples. It technically could say "defined and unbounded", or even simply "unbounded", but to avoid any ambiguity, it is usual to say "finite and unbounded" here. Some might argue that the derivative is "defined as $infty$" in zero (as opposed to "utterly undefined" when for example the function is not even continuous at the point), and to avoid any ambiguity the usual wording is "finite" (i.e. it is defined as a real number).

This is actually why that example in your book is interesting. By ruling out the naive examples such as $sqrtx$, one might wonder if there could be any function with unbounded derivative in a closed interval.

I remember perfectly that when my teacher asked this in class, I thought that if such example would exist, surely it would involve a function that gets steeper and steeper, and that it would "end up being infinite in a closed interval" and I answered that it was impossible.

But guess what, it's possible, as the example in your book shows.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Pedro A

1,7991824

add a comment | 

up vote
2
down vote

up vote
2
down vote

The term "finite" in this case is stressing the fact that it is "defined", "well defined" or "exists as a real number".

To better understand why this is important, consider the following naive attempt to give an example of a function with unbounded derivative on a closed interval:

$$f(x) = sqrtx$$

The derivative, which is $$dfrac12sqrtx$$ is clearly unbounded in $(0,1]$ but it is not even defined in zero.

The goal of the book is to disallow such examples. It technically could say "defined and unbounded", or even simply "unbounded", but to avoid any ambiguity, it is usual to say "finite and unbounded" here. Some might argue that the derivative is "defined as $infty$" in zero (as opposed to "utterly undefined" when for example the function is not even continuous at the point), and to avoid any ambiguity the usual wording is "finite" (i.e. it is defined as a real number).

This is actually why that example in your book is interesting. By ruling out the naive examples such as $sqrtx$, one might wonder if there could be any function with unbounded derivative in a closed interval.

I remember perfectly that when my teacher asked this in class, I thought that if such example would exist, surely it would involve a function that gets steeper and steeper, and that it would "end up being infinite in a closed interval" and I answered that it was impossible.

But guess what, it's possible, as the example in your book shows.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Pedro A

1,7991824

The term "finite" in this case is stressing the fact that it is "defined", "well defined" or "exists as a real number".

To better understand why this is important, consider the following naive attempt to give an example of a function with unbounded derivative on a closed interval:

$$f(x) = sqrtx$$

The derivative, which is $$dfrac12sqrtx$$ is clearly unbounded in $(0,1]$ but it is not even defined in zero.

The goal of the book is to disallow such examples. It technically could say "defined and unbounded", or even simply "unbounded", but to avoid any ambiguity, it is usual to say "finite and unbounded" here. Some might argue that the derivative is "defined as $infty$" in zero (as opposed to "utterly undefined" when for example the function is not even continuous at the point), and to avoid any ambiguity the usual wording is "finite" (i.e. it is defined as a real number).

This is actually why that example in your book is interesting. By ruling out the naive examples such as $sqrtx$, one might wonder if there could be any function with unbounded derivative in a closed interval.

I remember perfectly that when my teacher asked this in class, I thought that if such example would exist, surely it would involve a function that gets steeper and steeper, and that it would "end up being infinite in a closed interval" and I answered that it was impossible.

But guess what, it's possible, as the example in your book shows.

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Pedro A

1,7991824

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 2 hours ago

Pedro A

1,7991824

answered 2 hours ago

Pedro A

1,7991824

answered 2 hours ago

Pedro A

1,7991824

1,7991824

add a comment | 

add a comment | 

up vote
0
down vote

The sequence $1,2,...$ is finite in the sense each term is finite, but the sequence is not bounded in the sense there is no fixed real number $m$ such that all the terms are bouded by this number. Similarly, $f(x)=frac 1 x$ is finite in $(0,1)$ but it is not bounded. In the above example take $x =frac 1 sqrt 2npi$ to see that $f'$ is not bounded.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So when (if ever) can a function be said to be infinite?
  – dtcm840
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
  – Kavi Rama Murthy
  3 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

The sequence $1,2,...$ is finite in the sense each term is finite, but the sequence is not bounded in the sense there is no fixed real number $m$ such that all the terms are bouded by this number. Similarly, $f(x)=frac 1 x$ is finite in $(0,1)$ but it is not bounded. In the above example take $x =frac 1 sqrt 2npi$ to see that $f'$ is not bounded.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So when (if ever) can a function be said to be infinite?
  – dtcm840
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
  – Kavi Rama Murthy
  3 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

The sequence $1,2,...$ is finite in the sense each term is finite, but the sequence is not bounded in the sense there is no fixed real number $m$ such that all the terms are bouded by this number. Similarly, $f(x)=frac 1 x$ is finite in $(0,1)$ but it is not bounded. In the above example take $x =frac 1 sqrt 2npi$ to see that $f'$ is not bounded.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

The sequence $1,2,...$ is finite in the sense each term is finite, but the sequence is not bounded in the sense there is no fixed real number $m$ such that all the terms are bouded by this number. Similarly, $f(x)=frac 1 x$ is finite in $(0,1)$ but it is not bounded. In the above example take $x =frac 1 sqrt 2npi$ to see that $f'$ is not bounded.

share|cite|improve this answer

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

share|cite|improve this answer

share|cite|improve this answer

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

answered 3 hours ago

Kavi Rama Murthy

31.7k31543

31.7k31543

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So when (if ever) can a function be said to be infinite?
  – dtcm840
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
  – Kavi Rama Murthy
  3 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  So when (if ever) can a function be said to be infinite?
  – dtcm840
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
  – Kavi Rama Murthy
  3 hours ago
  

  
  
  
  

1

1

So when (if ever) can a function be said to be infinite?
– dtcm840
3 hours ago

So when (if ever) can a function be said to be infinite?
– dtcm840
3 hours ago

Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
– Kavi Rama Murthy
3 hours ago

Any real valued function is finite. If you allow the values $infty$/$-infty$ at some points then the function is not finite valued.
– Kavi Rama Murthy
3 hours ago

add a comment | 

up vote
0
down vote

The theorem indicates that the derivative of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$

It is obvious that a finite set is bounded, therefore the author's intention is to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].

share|cite|improve this answer

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

add a comment | 

up vote
0
down vote

The theorem indicates that the derivative of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$

It is obvious that a finite set is bounded, therefore the author's intention is to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].

share|cite|improve this answer

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

add a comment | 

up vote
0
down vote

up vote
0
down vote

The theorem indicates that the derivative of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$

It is obvious that a finite set is bounded, therefore the author's intention is to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].

share|cite|improve this answer

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

The theorem indicates that the derivative of f(x) is finite at $x=0$ and it is unbounded on $[-1,1]$

It is obvious that a finite set is bounded, therefore the author's intention is to show a function which has a finite derivative at $x=0$ but unbounded derivative on the interval [0,1].

share|cite|improve this answer

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

answered 1 hour ago

Mohammad Riazi-Kermani

35.3k41855

35.3k41855

add a comment | 

add a comment | 

 

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