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4

1

$begingroup$

Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.

Take a real-valued function $f$ which is continuous and differentiable on $[a,b]$ and $f(x) ge 0$ for all $x$ in $[a,b]$, that has the following properties:

1. $f(a)=0$




2. $f′(x) le f(x)$ for all $x$ in $[a,b]$

Show that $f(x)=0$ for all $x$ in $[a,b]$.

What if the second property instead is that $f(x) le int_a^x f(t) ,dt$ for all $x$ in $[a,b]$? Is the conclusion still true?

real-analysis integration analysis derivatives

share|cite|improve this question

edited Apr 30 at 9:57

leftaroundabout

3,6741630

asked Apr 30 at 6:11

jacksonfjacksonf

17011

$endgroup$

closed as off-topic by RRL, José Carlos Santos, Cesareo, Carsten S, Martin R Apr 30 at 9:55

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, Cesareo, Carsten S, Martin R

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
  $endgroup$
  – Trebor
  Apr 30 at 6:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Trebor updated!
  $endgroup$
  – jacksonf
  Apr 30 at 6:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I am seeing lots of rectangles in place of characters.
  $endgroup$
  – badjohn
  Apr 30 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @badjohn sorry, could you specify where? I can correct it now!
  $endgroup$
  – jacksonf
  Apr 30 at 6:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My problems were when using a Samsung tablet running Android. Now on a Windows laptop, I can see the intention. The problem seemed to be the fancy characters that you used in many places. I just edited your question and replaced these characters with plain ones and surrounded then with $$$. It now looks better on the tablet. More could be done if you wished e.g. use symbols such as $in$ and $forall$.
  $endgroup$
  – badjohn
  Apr 30 at 8:19
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.

Take a real-valued function $f$ which is continuous and differentiable on $[a,b]$ and $f(x) ge 0$ for all $x$ in $[a,b]$, that has the following properties:

1. $f(a)=0$




2. $f′(x) le f(x)$ for all $x$ in $[a,b]$

Show that $f(x)=0$ for all $x$ in $[a,b]$.

What if the second property instead is that $f(x) le int_a^x f(t) ,dt$ for all $x$ in $[a,b]$? Is the conclusion still true?

real-analysis integration analysis derivatives

share|cite|improve this question

edited Apr 30 at 9:57

leftaroundabout

3,6741630

asked Apr 30 at 6:11

jacksonfjacksonf

17011

$endgroup$

closed as off-topic by RRL, José Carlos Santos, Cesareo, Carsten S, Martin R Apr 30 at 9:55

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, Cesareo, Carsten S, Martin R

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
  $endgroup$
  – Trebor
  Apr 30 at 6:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Trebor updated!
  $endgroup$
  – jacksonf
  Apr 30 at 6:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I am seeing lots of rectangles in place of characters.
  $endgroup$
  – badjohn
  Apr 30 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @badjohn sorry, could you specify where? I can correct it now!
  $endgroup$
  – jacksonf
  Apr 30 at 6:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My problems were when using a Samsung tablet running Android. Now on a Windows laptop, I can see the intention. The problem seemed to be the fancy characters that you used in many places. I just edited your question and replaced these characters with plain ones and surrounded then with $$$. It now looks better on the tablet. More could be done if you wished e.g. use symbols such as $in$ and $forall$.
  $endgroup$
  – badjohn
  Apr 30 at 8:19
  
  

  
  
  
  

add a comment | 

$begingroup$

Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.

Take a real-valued function $f$ which is continuous and differentiable on $[a,b]$ and $f(x) ge 0$ for all $x$ in $[a,b]$, that has the following properties:

1. $f(a)=0$




2. $f′(x) le f(x)$ for all $x$ in $[a,b]$

Show that $f(x)=0$ for all $x$ in $[a,b]$.

What if the second property instead is that $f(x) le int_a^x f(t) ,dt$ for all $x$ in $[a,b]$? Is the conclusion still true?

real-analysis integration analysis derivatives

share|cite|improve this question

edited Apr 30 at 9:57

leftaroundabout

3,6741630

asked Apr 30 at 6:11

jacksonfjacksonf

17011

$endgroup$

share|cite|improve this question

edited Apr 30 at 9:57

leftaroundabout

3,6741630

asked Apr 30 at 6:11

jacksonfjacksonf

17011

share|cite|improve this question

edited Apr 30 at 9:57

leftaroundabout

3,6741630

asked Apr 30 at 6:11

jacksonfjacksonf

17011

edited Apr 30 at 9:57

leftaroundabout

3,6741630

edited Apr 30 at 9:57

leftaroundabout

3,6741630

asked Apr 30 at 6:11

jacksonfjacksonf

17011

asked Apr 30 at 6:11

jacksonfjacksonf

17011

2 Answers
2

active

oldest

votes

5

$begingroup$

Let us use a classical trick.

Write your inequality under the form $(f'(x)-f(x))leq 0$.

Multiply by the positive quantity $e^-x$ ; you obtain :

$(e^-xf(x))'leq 0 tag1$

Thus function

$$g(x):=e^-xf(x)tag2$$

is decreasing.

As $f(a)=0$ implies $g(a)=0$, $g$ is a decreasing function always negative on $[a,b]$. As $e^-x>0$, (2) gives $forall x in [a,b], f(x) leq 0$,
But with the hypothesis (you have added) $f(x)geq 0$, we must have $f=0$ identically on $[a,b]$.

For your second question that I understand now : yes you can have the same conclusion. Let us define

$$h(x):=int_a^x f(t)dt.$$

We have $h'(x)=f(x)$ by the fundamental theorem of calculus.

Thus, we can transform your second inequality into this one :

$$h'(x) leq h(x), textwith, moreover h(a)=0$$

And thus we are back to the first issue with $h$ replacing $f$.

Remarks :

1) See a kind of generalization, i.e., Grönwall inequality https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality where exponential "comes in" in a natural way.

2) A similar issue : Does $f(0)=0$ and $left|f^prime(x)right|leqleft|f(x)right|$ imply $f(x)=0$?

share|cite|improve this answer

edited Apr 30 at 7:47

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kavi Rama Murthy You are right, under the new conditions given by the asker.
  $endgroup$
  – Jean Marie
  Apr 30 at 6:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this also work for the second question?
  $endgroup$
  – jacksonf
  Apr 30 at 6:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jacksonf Take look at the similar issue I just gave as a second remark.
  $endgroup$
  – Jean Marie
  Apr 30 at 7:48
  

  
  
  
  

add a comment | 

8

$begingroup$

The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.

The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^-xf(x))'leq 0$. So $e^-xf(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.

The result is also true under the second hypothesis. Simply apply the first part to the function $int_a^xf(t), dt$.

share|cite|improve this answer

edited Apr 30 at 6:34

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think I understand your proof... how do you decide to use (𝑒−𝑥𝑓(𝑥))′?
  $endgroup$
  – jacksonf
  Apr 30 at 6:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is a standard trick. Whenever you see $f'(x)-f(x)$ you think of the fact that $(e^-xf(x))'= e^-x(f'(x)-f(x))$. You should remember this from now on.
  $endgroup$
  – Kavi Rama Murthy
  Apr 30 at 6:43
  

  
  
  
  

add a comment | 

5

$begingroup$

Let us use a classical trick.

Write your inequality under the form $(f'(x)-f(x))leq 0$.

Multiply by the positive quantity $e^-x$ ; you obtain :

$(e^-xf(x))'leq 0 tag1$

Thus function

$$g(x):=e^-xf(x)tag2$$

is decreasing.

As $f(a)=0$ implies $g(a)=0$, $g$ is a decreasing function always negative on $[a,b]$. As $e^-x>0$, (2) gives $forall x in [a,b], f(x) leq 0$,
But with the hypothesis (you have added) $f(x)geq 0$, we must have $f=0$ identically on $[a,b]$.

For your second question that I understand now : yes you can have the same conclusion. Let us define

$$h(x):=int_a^x f(t)dt.$$

We have $h'(x)=f(x)$ by the fundamental theorem of calculus.

Thus, we can transform your second inequality into this one :

$$h'(x) leq h(x), textwith, moreover h(a)=0$$

And thus we are back to the first issue with $h$ replacing $f$.

Remarks :

1) See a kind of generalization, i.e., Grönwall inequality https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality where exponential "comes in" in a natural way.

2) A similar issue : Does $f(0)=0$ and $left|f^prime(x)right|leqleft|f(x)right|$ imply $f(x)=0$?

share|cite|improve this answer

edited Apr 30 at 7:47

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kavi Rama Murthy You are right, under the new conditions given by the asker.
  $endgroup$
  – Jean Marie
  Apr 30 at 6:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this also work for the second question?
  $endgroup$
  – jacksonf
  Apr 30 at 6:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jacksonf Take look at the similar issue I just gave as a second remark.
  $endgroup$
  – Jean Marie
  Apr 30 at 7:48
  

  
  
  
  

add a comment | 

5

$begingroup$

Let us use a classical trick.

Write your inequality under the form $(f'(x)-f(x))leq 0$.

Multiply by the positive quantity $e^-x$ ; you obtain :

$(e^-xf(x))'leq 0 tag1$

Thus function

$$g(x):=e^-xf(x)tag2$$

is decreasing.

As $f(a)=0$ implies $g(a)=0$, $g$ is a decreasing function always negative on $[a,b]$. As $e^-x>0$, (2) gives $forall x in [a,b], f(x) leq 0$,
But with the hypothesis (you have added) $f(x)geq 0$, we must have $f=0$ identically on $[a,b]$.

For your second question that I understand now : yes you can have the same conclusion. Let us define

$$h(x):=int_a^x f(t)dt.$$

We have $h'(x)=f(x)$ by the fundamental theorem of calculus.

Thus, we can transform your second inequality into this one :

$$h'(x) leq h(x), textwith, moreover h(a)=0$$

And thus we are back to the first issue with $h$ replacing $f$.

Remarks :

1) See a kind of generalization, i.e., Grönwall inequality https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality where exponential "comes in" in a natural way.

2) A similar issue : Does $f(0)=0$ and $left|f^prime(x)right|leqleft|f(x)right|$ imply $f(x)=0$?

share|cite|improve this answer

edited Apr 30 at 7:47

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kavi Rama Murthy You are right, under the new conditions given by the asker.
  $endgroup$
  – Jean Marie
  Apr 30 at 6:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does this also work for the second question?
  $endgroup$
  – jacksonf
  Apr 30 at 6:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jacksonf Take look at the similar issue I just gave as a second remark.
  $endgroup$
  – Jean Marie
  Apr 30 at 7:48
  

  
  
  
  

add a comment | 

$begingroup$

Let us use a classical trick.

Write your inequality under the form $(f'(x)-f(x))leq 0$.

Multiply by the positive quantity $e^-x$ ; you obtain :

$(e^-xf(x))'leq 0 tag1$

Thus function

$$g(x):=e^-xf(x)tag2$$

is decreasing.

As $f(a)=0$ implies $g(a)=0$, $g$ is a decreasing function always negative on $[a,b]$. As $e^-x>0$, (2) gives $forall x in [a,b], f(x) leq 0$,
But with the hypothesis (you have added) $f(x)geq 0$, we must have $f=0$ identically on $[a,b]$.

For your second question that I understand now : yes you can have the same conclusion. Let us define

$$h(x):=int_a^x f(t)dt.$$

We have $h'(x)=f(x)$ by the fundamental theorem of calculus.

Thus, we can transform your second inequality into this one :

$$h'(x) leq h(x), textwith, moreover h(a)=0$$

And thus we are back to the first issue with $h$ replacing $f$.

Remarks :

1) See a kind of generalization, i.e., Grönwall inequality https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality where exponential "comes in" in a natural way.

2) A similar issue : Does $f(0)=0$ and $left|f^prime(x)right|leqleft|f(x)right|$ imply $f(x)=0$?

share|cite|improve this answer

edited Apr 30 at 7:47

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

$endgroup$

share|cite|improve this answer

edited Apr 30 at 7:47

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

answered Apr 30 at 6:32

Jean MarieJean Marie

32.3k42356

8

$begingroup$

The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.

The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^-xf(x))'leq 0$. So $e^-xf(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.

The result is also true under the second hypothesis. Simply apply the first part to the function $int_a^xf(t), dt$.

share|cite|improve this answer

edited Apr 30 at 6:34

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think I understand your proof... how do you decide to use (𝑒−𝑥𝑓(𝑥))′?
  $endgroup$
  – jacksonf
  Apr 30 at 6:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is a standard trick. Whenever you see $f'(x)-f(x)$ you think of the fact that $(e^-xf(x))'= e^-x(f'(x)-f(x))$. You should remember this from now on.
  $endgroup$
  – Kavi Rama Murthy
  Apr 30 at 6:43
  

  
  
  
  

add a comment | 

8

$begingroup$

The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.

The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^-xf(x))'leq 0$. So $e^-xf(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.

The result is also true under the second hypothesis. Simply apply the first part to the function $int_a^xf(t), dt$.

share|cite|improve this answer

edited Apr 30 at 6:34

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think I understand your proof... how do you decide to use (𝑒−𝑥𝑓(𝑥))′?
  $endgroup$
  – jacksonf
  Apr 30 at 6:41
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  This is a standard trick. Whenever you see $f'(x)-f(x)$ you think of the fact that $(e^-xf(x))'= e^-x(f'(x)-f(x))$. You should remember this from now on.
  $endgroup$
  – Kavi Rama Murthy
  Apr 30 at 6:43
  

  
  
  
  

add a comment | 

$begingroup$

The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.

The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^-xf(x))'leq 0$. So $e^-xf(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.

The result is also true under the second hypothesis. Simply apply the first part to the function $int_a^xf(t), dt$.

share|cite|improve this answer

edited Apr 30 at 6:34

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673

$endgroup$

share|cite|improve this answer

edited Apr 30 at 6:34

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673

answered Apr 30 at 6:27

Kavi Rama MurthyKavi Rama Murthy

81.5k53673