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Find an increasing continuous function such that : $mid p(x) - p(y) mid leq w(mid x - y mid)$

Increasing, bounded and continuous is uniformly continuousStrictly increasing continuous functionContinuous function on discrete values.Show that the function $M(x,y)=xy$ is continuous on any disk; $mid (x,y) mid leq r$Prove that a bijective, strictly increasing function is continuous.Find a sequence of continuous functionsShow that a monotonically increasing derivative must be continuousFind an interval over which a continuous function is strictly increasingConfusion about definition of uniformly continuous functionProve that if a function is increasing on [a,b] and satisfied the intermediate value property, then that function is continuous on [a,b].

0

$begingroup$

Let $p : [0,1] to mathbbR$ be a continuous function. Prove the existence of an increasing and continuous function $w : [0,1] to mathbbR$ such that : $lim_x to 0 w(x) = 0$ and such that :

$$ forall t, s in [0,1], mid p(t)-p(s) mid leq w(mid t-smid)$$

I don't know how to find a function $w$ which fullfils thse conditions. I know that I can find a $delta$ such that if $mid t-s mid leq delta$ then $mid p(t)-p(s) mid leq epsilon$. Yet the fact that $w$ is continuous an dincreasing makes it hard.

Thank you !

real-analysis sequences-and-series continuity

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asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

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bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you trying to show the existence? Or, do you have to show the existence? In the first case, what is the reason to believe that such a function exist?
  $endgroup$
  – mfl
  Mar 11 at 20:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are a lot of questions (and answers) on that subject: try searching Modulus of continuity.
  $endgroup$
  – user539887
  Mar 11 at 20:09
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $p : [0,1] to mathbbR$ be a continuous function. Prove the existence of an increasing and continuous function $w : [0,1] to mathbbR$ such that : $lim_x to 0 w(x) = 0$ and such that :

$$ forall t, s in [0,1], mid p(t)-p(s) mid leq w(mid t-smid)$$

I don't know how to find a function $w$ which fullfils thse conditions. I know that I can find a $delta$ such that if $mid t-s mid leq delta$ then $mid p(t)-p(s) mid leq epsilon$. Yet the fact that $w$ is continuous an dincreasing makes it hard.

Thank you !

real-analysis sequences-and-series continuity

share|cite|improve this question

asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

New contributor

bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you trying to show the existence? Or, do you have to show the existence? In the first case, what is the reason to believe that such a function exist?
  $endgroup$
  – mfl
  Mar 11 at 20:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are a lot of questions (and answers) on that subject: try searching Modulus of continuity.
  $endgroup$
  – user539887
  Mar 11 at 20:09
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $p : [0,1] to mathbbR$ be a continuous function. Prove the existence of an increasing and continuous function $w : [0,1] to mathbbR$ such that : $lim_x to 0 w(x) = 0$ and such that :

$$ forall t, s in [0,1], mid p(t)-p(s) mid leq w(mid t-smid)$$

I don't know how to find a function $w$ which fullfils thse conditions. I know that I can find a $delta$ such that if $mid t-s mid leq delta$ then $mid p(t)-p(s) mid leq epsilon$. Yet the fact that $w$ is continuous an dincreasing makes it hard.

Thank you !

real-analysis sequences-and-series continuity

share|cite|improve this question

asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

New contributor

bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

New contributor

bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

New contributor

bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

New contributor

bonjoufhajl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 19:30

bonjoufhajlbonjoufhajl

324

1 Answer
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1

$begingroup$

Since $[0,1]$ is compact and $p$ is a continuous function then for all $epsilon > 0$ there is $delta $ such that :

$$forall x, y, mid x-y mid leq delta, mid p(x)-p(y) mid leq epsilon$$

Thus you can just take :

$$w(x) = sup mid p(a)-p(b) mid mid a, b in [0,1], mid a-b mid leq x $$

share|cite|improve this answer

edited Mar 12 at 7:26

answered Mar 11 at 21:32

ThinkingThinking

1,22716

$endgroup$

add a comment | 

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1

$begingroup$

Since $[0,1]$ is compact and $p$ is a continuous function then for all $epsilon > 0$ there is $delta $ such that :

$$forall x, y, mid x-y mid leq delta, mid p(x)-p(y) mid leq epsilon$$

Thus you can just take :

$$w(x) = sup mid p(a)-p(b) mid mid a, b in [0,1], mid a-b mid leq x $$

share|cite|improve this answer

edited Mar 12 at 7:26

answered Mar 11 at 21:32

ThinkingThinking

1,22716

$endgroup$

add a comment | 

1

$begingroup$

Since $[0,1]$ is compact and $p$ is a continuous function then for all $epsilon > 0$ there is $delta $ such that :

$$forall x, y, mid x-y mid leq delta, mid p(x)-p(y) mid leq epsilon$$

Thus you can just take :

$$w(x) = sup mid p(a)-p(b) mid mid a, b in [0,1], mid a-b mid leq x $$

share|cite|improve this answer

edited Mar 12 at 7:26

answered Mar 11 at 21:32

ThinkingThinking

1,22716

$endgroup$

add a comment | 

$begingroup$

Since $[0,1]$ is compact and $p$ is a continuous function then for all $epsilon > 0$ there is $delta $ such that :

$$forall x, y, mid x-y mid leq delta, mid p(x)-p(y) mid leq epsilon$$

Thus you can just take :

$$w(x) = sup mid p(a)-p(b) mid mid a, b in [0,1], mid a-b mid leq x $$

share|cite|improve this answer

edited Mar 12 at 7:26

answered Mar 11 at 21:32

ThinkingThinking

1,22716

$endgroup$

share|cite|improve this answer

edited Mar 12 at 7:26

answered Mar 11 at 21:32

ThinkingThinking

1,22716

answered Mar 11 at 21:32

ThinkingThinking

1,22716

answered Mar 11 at 21:32

ThinkingThinking

1,22716