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Sufficient condition for a Lipschitz function

0

$begingroup$

Let $f:[a,b] rightarrow mathbbR$ be continuous and $x_1<x_2$ in $[a,b] Rightarrow f(x_2)-f(x_1) geq k(x_2-x_1)$ for some fixed positive constant k.
Is the function $f$ Lipschitz on $[a,b]$?

Note: The above condition implies $f$ is strictly increasing on $[a,b]$

real-analysis

share|cite|improve this question

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily, consider $f(x)=sqrtx$ on $[0,1]$
  $endgroup$
  – user408856
  Mar 31 at 19:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $f:[a,b] rightarrow mathbbR$ be continuous and $x_1<x_2$ in $[a,b] Rightarrow f(x_2)-f(x_1) geq k(x_2-x_1)$ for some fixed positive constant k.
Is the function $f$ Lipschitz on $[a,b]$?

Note: The above condition implies $f$ is strictly increasing on $[a,b]$

real-analysis

share|cite|improve this question

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily, consider $f(x)=sqrtx$ on $[0,1]$
  $endgroup$
  – user408856
  Mar 31 at 19:04
  

  
  
  
  

add a comment | 

$begingroup$

Let $f:[a,b] rightarrow mathbbR$ be continuous and $x_1<x_2$ in $[a,b] Rightarrow f(x_2)-f(x_1) geq k(x_2-x_1)$ for some fixed positive constant k.
Is the function $f$ Lipschitz on $[a,b]$?

Note: The above condition implies $f$ is strictly increasing on $[a,b]$

real-analysis

share|cite|improve this question

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

$endgroup$

share|cite|improve this question

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

share|cite|improve this question

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

asked Mar 31 at 18:56

Murtaza WaniMurtaza Wani

313

1 Answer
1

active

oldest

votes

0

$begingroup$

I am wondering if the inequality is right. I mean everything works fine if $f(x_2)-f(x_1)leq k(x_2-x_1)$ for some positive costant $k$.

In fact the definition of Lipschitz continuity is that for any $x,yin [a,b]$ exists a positive constant $L$ such that
beginequation
|f(x)-f(y)|leq L|x_2-x_1|.
endequation

Recall that $Lip((a,b)) subset C^0([a,b])$, hence if the function is Lipschitz is also continous. In particular if $f$ is stricly increasing there's no need of the absolute value and you get the desired conclusion.

share|cite|improve this answer

edited Apr 1 at 6:59

answered Mar 31 at 19:09

GiovanniGiovanni

459

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inequality is right. I was wondering if the class of functions as described by my hypothesis was contained in the class of Lipschitz functions.
  $endgroup$
  – Murtaza Wani
  Apr 1 at 6:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my comment, then you know it is not
  $endgroup$
  – user408856
  Apr 1 at 8:07
  

  
  
  
  

add a comment | 

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0

$begingroup$

I am wondering if the inequality is right. I mean everything works fine if $f(x_2)-f(x_1)leq k(x_2-x_1)$ for some positive costant $k$.

In fact the definition of Lipschitz continuity is that for any $x,yin [a,b]$ exists a positive constant $L$ such that
beginequation
|f(x)-f(y)|leq L|x_2-x_1|.
endequation

Recall that $Lip((a,b)) subset C^0([a,b])$, hence if the function is Lipschitz is also continous. In particular if $f$ is stricly increasing there's no need of the absolute value and you get the desired conclusion.

share|cite|improve this answer

edited Apr 1 at 6:59

answered Mar 31 at 19:09

GiovanniGiovanni

459

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inequality is right. I was wondering if the class of functions as described by my hypothesis was contained in the class of Lipschitz functions.
  $endgroup$
  – Murtaza Wani
  Apr 1 at 6:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my comment, then you know it is not
  $endgroup$
  – user408856
  Apr 1 at 8:07
  

  
  
  
  

add a comment | 

0

$begingroup$

I am wondering if the inequality is right. I mean everything works fine if $f(x_2)-f(x_1)leq k(x_2-x_1)$ for some positive costant $k$.

In fact the definition of Lipschitz continuity is that for any $x,yin [a,b]$ exists a positive constant $L$ such that
beginequation
|f(x)-f(y)|leq L|x_2-x_1|.
endequation

Recall that $Lip((a,b)) subset C^0([a,b])$, hence if the function is Lipschitz is also continous. In particular if $f$ is stricly increasing there's no need of the absolute value and you get the desired conclusion.

share|cite|improve this answer

edited Apr 1 at 6:59

answered Mar 31 at 19:09

GiovanniGiovanni

459

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inequality is right. I was wondering if the class of functions as described by my hypothesis was contained in the class of Lipschitz functions.
  $endgroup$
  – Murtaza Wani
  Apr 1 at 6:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my comment, then you know it is not
  $endgroup$
  – user408856
  Apr 1 at 8:07
  

  
  
  
  

add a comment | 

$begingroup$

I am wondering if the inequality is right. I mean everything works fine if $f(x_2)-f(x_1)leq k(x_2-x_1)$ for some positive costant $k$.

In fact the definition of Lipschitz continuity is that for any $x,yin [a,b]$ exists a positive constant $L$ such that
beginequation
|f(x)-f(y)|leq L|x_2-x_1|.
endequation

Recall that $Lip((a,b)) subset C^0([a,b])$, hence if the function is Lipschitz is also continous. In particular if $f$ is stricly increasing there's no need of the absolute value and you get the desired conclusion.

share|cite|improve this answer

edited Apr 1 at 6:59

answered Mar 31 at 19:09

GiovanniGiovanni

459

$endgroup$

share|cite|improve this answer

edited Apr 1 at 6:59

answered Mar 31 at 19:09

GiovanniGiovanni

459

answered Mar 31 at 19:09

GiovanniGiovanni

459

answered Mar 31 at 19:09

GiovanniGiovanni

459