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Which functions $f$ differentiable in all $mathbb R$ satisfy $|f(x)| geq |f'(x)|$ for all $x$ with $f(0)=0$?

0

$begingroup$

Which functions $f$ differentiable in all $mathbb R$ satisfy $|f(x)| geq |f'(x)|$ for all $x$ with $f(0)=0$?

I'm pretty stuck right now. Any hint would be very helpfull.
Thank you.

calculus

share|cite|improve this question

edited Mar 19 at 12:34

Rodrigo Sango

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

$endgroup$

add a comment | 

0

$begingroup$

Which functions $f$ differentiable in all $mathbb R$ satisfy $|f(x)| geq |f'(x)|$ for all $x$ with $f(0)=0$?

I'm pretty stuck right now. Any hint would be very helpfull.
Thank you.

calculus

share|cite|improve this question

edited Mar 19 at 12:34

Rodrigo Sango

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

$endgroup$

add a comment | 

$begingroup$

Which functions $f$ differentiable in all $mathbb R$ satisfy $|f(x)| geq |f'(x)|$ for all $x$ with $f(0)=0$?

I'm pretty stuck right now. Any hint would be very helpfull.
Thank you.

calculus

share|cite|improve this question

edited Mar 19 at 12:34

Rodrigo Sango

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

$endgroup$

share|cite|improve this question

edited Mar 19 at 12:34

Rodrigo Sango

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

share|cite|improve this question

edited Mar 19 at 12:34

Rodrigo Sango

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

asked Mar 19 at 11:57

Rodrigo SangoRodrigo Sango

1326

1 Answer
1

active

oldest

votes

2

$begingroup$

Let $x_0 in mathbbR$ such that $f(x_0) neq 0$. Because $f$ is continuous, there exists an maximal interval $]a,b[$ containing $x_0$ (with $a$, $b$ eventually $-infty$ or $+infty$), such that $f(x) neq 0$ for all $x in ]a,b[$. Moreover, because $f(0) = 0$, you can always choose that $f(a)=0$ or $f(b)=0$.

Let's suppose that $f(a)=0$, and that $f$ is positive on $]a,b[$ (the other cases are similar). In particular, $f(x_0) > 0$.

You have then that for all $x in [a,b]$,
$$-f(x) leq f'(x) leq f(x)$$

Consider the function $g : [a,b]rightarrow mathbbR$ defined by
$$g(x)=f(x)e^-x$$

$g$ is differentiable and its derivative is given by $g'(x)=(f'(x)-f(x))e^-x$, which is always negative by asumption. So $g$ decreases, so for all $x in [a,b]$, you have $g(x) leq g(a)=0$. In particular $g(x_0) leq 0$, so $f(x_0) leq 0$.

But by asumption, you have $f(x_0) > 0$. Contradiction.

The other cases are similar, you get finally that there cannot exist such a $x_0$, so $f$ is the constant function equal to $0$.

share|cite|improve this answer

edited Mar 19 at 12:42

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, TheSilverDoe.
  $endgroup$
  – Rodrigo Sango
  Mar 23 at 14:47
  

  
  
  
  

add a comment | 

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2

$begingroup$

Let $x_0 in mathbbR$ such that $f(x_0) neq 0$. Because $f$ is continuous, there exists an maximal interval $]a,b[$ containing $x_0$ (with $a$, $b$ eventually $-infty$ or $+infty$), such that $f(x) neq 0$ for all $x in ]a,b[$. Moreover, because $f(0) = 0$, you can always choose that $f(a)=0$ or $f(b)=0$.

Let's suppose that $f(a)=0$, and that $f$ is positive on $]a,b[$ (the other cases are similar). In particular, $f(x_0) > 0$.

You have then that for all $x in [a,b]$,
$$-f(x) leq f'(x) leq f(x)$$

Consider the function $g : [a,b]rightarrow mathbbR$ defined by
$$g(x)=f(x)e^-x$$

$g$ is differentiable and its derivative is given by $g'(x)=(f'(x)-f(x))e^-x$, which is always negative by asumption. So $g$ decreases, so for all $x in [a,b]$, you have $g(x) leq g(a)=0$. In particular $g(x_0) leq 0$, so $f(x_0) leq 0$.

But by asumption, you have $f(x_0) > 0$. Contradiction.

The other cases are similar, you get finally that there cannot exist such a $x_0$, so $f$ is the constant function equal to $0$.

share|cite|improve this answer

edited Mar 19 at 12:42

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, TheSilverDoe.
  $endgroup$
  – Rodrigo Sango
  Mar 23 at 14:47
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $x_0 in mathbbR$ such that $f(x_0) neq 0$. Because $f$ is continuous, there exists an maximal interval $]a,b[$ containing $x_0$ (with $a$, $b$ eventually $-infty$ or $+infty$), such that $f(x) neq 0$ for all $x in ]a,b[$. Moreover, because $f(0) = 0$, you can always choose that $f(a)=0$ or $f(b)=0$.

Let's suppose that $f(a)=0$, and that $f$ is positive on $]a,b[$ (the other cases are similar). In particular, $f(x_0) > 0$.

You have then that for all $x in [a,b]$,
$$-f(x) leq f'(x) leq f(x)$$

Consider the function $g : [a,b]rightarrow mathbbR$ defined by
$$g(x)=f(x)e^-x$$

$g$ is differentiable and its derivative is given by $g'(x)=(f'(x)-f(x))e^-x$, which is always negative by asumption. So $g$ decreases, so for all $x in [a,b]$, you have $g(x) leq g(a)=0$. In particular $g(x_0) leq 0$, so $f(x_0) leq 0$.

But by asumption, you have $f(x_0) > 0$. Contradiction.

The other cases are similar, you get finally that there cannot exist such a $x_0$, so $f$ is the constant function equal to $0$.

share|cite|improve this answer

edited Mar 19 at 12:42

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, TheSilverDoe.
  $endgroup$
  – Rodrigo Sango
  Mar 23 at 14:47
  

  
  
  
  

add a comment | 

$begingroup$

Let $x_0 in mathbbR$ such that $f(x_0) neq 0$. Because $f$ is continuous, there exists an maximal interval $]a,b[$ containing $x_0$ (with $a$, $b$ eventually $-infty$ or $+infty$), such that $f(x) neq 0$ for all $x in ]a,b[$. Moreover, because $f(0) = 0$, you can always choose that $f(a)=0$ or $f(b)=0$.

Let's suppose that $f(a)=0$, and that $f$ is positive on $]a,b[$ (the other cases are similar). In particular, $f(x_0) > 0$.

You have then that for all $x in [a,b]$,
$$-f(x) leq f'(x) leq f(x)$$

Consider the function $g : [a,b]rightarrow mathbbR$ defined by
$$g(x)=f(x)e^-x$$

$g$ is differentiable and its derivative is given by $g'(x)=(f'(x)-f(x))e^-x$, which is always negative by asumption. So $g$ decreases, so for all $x in [a,b]$, you have $g(x) leq g(a)=0$. In particular $g(x_0) leq 0$, so $f(x_0) leq 0$.

But by asumption, you have $f(x_0) > 0$. Contradiction.

The other cases are similar, you get finally that there cannot exist such a $x_0$, so $f$ is the constant function equal to $0$.

share|cite|improve this answer

edited Mar 19 at 12:42

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215

$endgroup$

share|cite|improve this answer

edited Mar 19 at 12:42

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215

answered Mar 19 at 12:16

TheSilverDoeTheSilverDoe

4,894215