Show there is $Cin [0,1)$ s.t. $|f(x)-f(y)|leq C|x-y|$ when $|x-y|geq a$.

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Let $f:[0,1]to mathbb R$ s.t. there is $ain (0,1)$ s.t.
$$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(y)|<|x-y|.$$

Show there is $Cin [0,1)$ s.t. $$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(x)|leq C|x-y|.$$

As Hint I have that : Set $E=x-y$. Prove that $$F:Eto mathbb R, (x,y)longmapsto left|fracf(x)-f(y)x-yright|,$$
is continuous. I really don't understand why $F$ is continuous... so

Q1) Why is $F$ continuous ?

Attempt

I tried to do an other way. Suppose that for all $n$, there is $x_n,y_nin [0,1]$ s.t. $$|x_n-y_n|geq aquad textandquad |f(x_n)-f(x_n)|geq left(1-frac1nright)|x_n-y_n|.$$

I suppose WLOG that $(x_n)$ and $(y_n)$ converge to $x,yin [0,1]$. Then $|x-y|geq a$ and since
$$|x_n-y_n|>|f(x_n)-f(y_n)|geq left(1-frac1nright)|x_n-y_n|,$$
we get $$lim_nto infty |f(x_n)-f(y_n)|=|x-y|geq a,$$
but I don't see in what it's a contradiction.

real-analysis

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asked Jul 21 at 19:54

user386627

714214

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up vote
3
down vote

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Let $f:[0,1]to mathbb R$ s.t. there is $ain (0,1)$ s.t.
$$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(y)|<|x-y|.$$

Show there is $Cin [0,1)$ s.t. $$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(x)|leq C|x-y|.$$

As Hint I have that : Set $E=x-y$. Prove that $$F:Eto mathbb R, (x,y)longmapsto left|fracf(x)-f(y)x-yright|,$$
is continuous. I really don't understand why $F$ is continuous... so

Q1) Why is $F$ continuous ?

Attempt

I tried to do an other way. Suppose that for all $n$, there is $x_n,y_nin [0,1]$ s.t. $$|x_n-y_n|geq aquad textandquad |f(x_n)-f(x_n)|geq left(1-frac1nright)|x_n-y_n|.$$

I suppose WLOG that $(x_n)$ and $(y_n)$ converge to $x,yin [0,1]$. Then $|x-y|geq a$ and since
$$|x_n-y_n|>|f(x_n)-f(y_n)|geq left(1-frac1nright)|x_n-y_n|,$$
we get $$lim_nto infty |f(x_n)-f(y_n)|=|x-y|geq a,$$
but I don't see in what it's a contradiction.

real-analysis

share|cite|improve this question

asked Jul 21 at 19:54

user386627

714214

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Let $f:[0,1]to mathbb R$ s.t. there is $ain (0,1)$ s.t.
$$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(y)|<|x-y|.$$

Show there is $Cin [0,1)$ s.t. $$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(x)|leq C|x-y|.$$

As Hint I have that : Set $E=x-y$. Prove that $$F:Eto mathbb R, (x,y)longmapsto left|fracf(x)-f(y)x-yright|,$$
is continuous. I really don't understand why $F$ is continuous... so

Q1) Why is $F$ continuous ?

Attempt

I tried to do an other way. Suppose that for all $n$, there is $x_n,y_nin [0,1]$ s.t. $$|x_n-y_n|geq aquad textandquad |f(x_n)-f(x_n)|geq left(1-frac1nright)|x_n-y_n|.$$

I suppose WLOG that $(x_n)$ and $(y_n)$ converge to $x,yin [0,1]$. Then $|x-y|geq a$ and since
$$|x_n-y_n|>|f(x_n)-f(y_n)|geq left(1-frac1nright)|x_n-y_n|,$$
we get $$lim_nto infty |f(x_n)-f(y_n)|=|x-y|geq a,$$
but I don't see in what it's a contradiction.

real-analysis

share|cite|improve this question

asked Jul 21 at 19:54

user386627

714214

Let $f:[0,1]to mathbb R$ s.t. there is $ain (0,1)$ s.t.
$$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(y)|<|x-y|.$$

Show there is $Cin [0,1)$ s.t. $$forall x,yin [0,1], |x-y|geq aimplies |f(x)-f(x)|leq C|x-y|.$$

As Hint I have that : Set $E=x-y$. Prove that $$F:Eto mathbb R, (x,y)longmapsto left|fracf(x)-f(y)x-yright|,$$
is continuous. I really don't understand why $F$ is continuous... so

Q1) Why is $F$ continuous ?

Attempt

I tried to do an other way. Suppose that for all $n$, there is $x_n,y_nin [0,1]$ s.t. $$|x_n-y_n|geq aquad textandquad |f(x_n)-f(x_n)|geq left(1-frac1nright)|x_n-y_n|.$$

I suppose WLOG that $(x_n)$ and $(y_n)$ converge to $x,yin [0,1]$. Then $|x-y|geq a$ and since
$$|x_n-y_n|>|f(x_n)-f(y_n)|geq left(1-frac1nright)|x_n-y_n|,$$
we get $$lim_nto infty |f(x_n)-f(y_n)|=|x-y|geq a,$$
but I don't see in what it's a contradiction.

real-analysis

share|cite|improve this question

asked Jul 21 at 19:54

user386627

714214

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 19:54

user386627

714214

asked Jul 21 at 19:54

user386627

714214

asked Jul 21 at 19:54

user386627

714214

714214

add a comment | 

add a comment | 

1 Answer
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up vote
1
down vote

I think we need more assumptions on $f$ (such as continuity), because $F$ is not necessarily continuous.

Otherwise, if $F: Erightarrow mathbbR$
$$F(x,y) = left| fracf(x) - f(y)x-yright|$$
is continuous, note that its domain $E$ is a closed subset of the unit square thus compact, so $F$ attains a maximum. Lets call it $C:= max_E F(x,y)$, then for some $(x^*, y^*)$ the maximum is attained, and together with our assumption, we have

$$C = F(x^*, y^*) = left| fracf(x^*) - f(y^*)x^*-y^*right| < 1.$$ This $C$ will satisfy the constant we are trying to find in the question.

share|cite|improve this answer

edited Jul 21 at 21:22

answered Jul 21 at 21:14

Xiao

4,38811333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. They probably forgot to say that $f$ is continuous.
  – user386627
  Jul 21 at 21:40
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

I think we need more assumptions on $f$ (such as continuity), because $F$ is not necessarily continuous.

Otherwise, if $F: Erightarrow mathbbR$
$$F(x,y) = left| fracf(x) - f(y)x-yright|$$
is continuous, note that its domain $E$ is a closed subset of the unit square thus compact, so $F$ attains a maximum. Lets call it $C:= max_E F(x,y)$, then for some $(x^*, y^*)$ the maximum is attained, and together with our assumption, we have

$$C = F(x^*, y^*) = left| fracf(x^*) - f(y^*)x^*-y^*right| < 1.$$ This $C$ will satisfy the constant we are trying to find in the question.

share|cite|improve this answer

edited Jul 21 at 21:22

answered Jul 21 at 21:14

Xiao

4,38811333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. They probably forgot to say that $f$ is continuous.
  – user386627
  Jul 21 at 21:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

I think we need more assumptions on $f$ (such as continuity), because $F$ is not necessarily continuous.

Otherwise, if $F: Erightarrow mathbbR$
$$F(x,y) = left| fracf(x) - f(y)x-yright|$$
is continuous, note that its domain $E$ is a closed subset of the unit square thus compact, so $F$ attains a maximum. Lets call it $C:= max_E F(x,y)$, then for some $(x^*, y^*)$ the maximum is attained, and together with our assumption, we have

$$C = F(x^*, y^*) = left| fracf(x^*) - f(y^*)x^*-y^*right| < 1.$$ This $C$ will satisfy the constant we are trying to find in the question.

share|cite|improve this answer

edited Jul 21 at 21:22

answered Jul 21 at 21:14

Xiao

4,38811333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. They probably forgot to say that $f$ is continuous.
  – user386627
  Jul 21 at 21:40
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

I think we need more assumptions on $f$ (such as continuity), because $F$ is not necessarily continuous.

Otherwise, if $F: Erightarrow mathbbR$
$$F(x,y) = left| fracf(x) - f(y)x-yright|$$
is continuous, note that its domain $E$ is a closed subset of the unit square thus compact, so $F$ attains a maximum. Lets call it $C:= max_E F(x,y)$, then for some $(x^*, y^*)$ the maximum is attained, and together with our assumption, we have

$$C = F(x^*, y^*) = left| fracf(x^*) - f(y^*)x^*-y^*right| < 1.$$ This $C$ will satisfy the constant we are trying to find in the question.

share|cite|improve this answer

edited Jul 21 at 21:22

answered Jul 21 at 21:14

Xiao

4,38811333

I think we need more assumptions on $f$ (such as continuity), because $F$ is not necessarily continuous.

Otherwise, if $F: Erightarrow mathbbR$
$$F(x,y) = left| fracf(x) - f(y)x-yright|$$
is continuous, note that its domain $E$ is a closed subset of the unit square thus compact, so $F$ attains a maximum. Lets call it $C:= max_E F(x,y)$, then for some $(x^*, y^*)$ the maximum is attained, and together with our assumption, we have

$$C = F(x^*, y^*) = left| fracf(x^*) - f(y^*)x^*-y^*right| < 1.$$ This $C$ will satisfy the constant we are trying to find in the question.

share|cite|improve this answer

edited Jul 21 at 21:22

answered Jul 21 at 21:14

Xiao

4,38811333

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 21:22

edited Jul 21 at 21:22

edited Jul 21 at 21:22

answered Jul 21 at 21:14

Xiao

4,38811333

answered Jul 21 at 21:14

Xiao

4,38811333

answered Jul 21 at 21:14

Xiao

4,38811333

4,38811333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. They probably forgot to say that $f$ is continuous.
  – user386627
  Jul 21 at 21:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. They probably forgot to say that $f$ is continuous.
  – user386627
  Jul 21 at 21:40
  

  
  
  
  

Thank you. They probably forgot to say that $f$ is continuous.
– user386627
Jul 21 at 21:40

Thank you. They probably forgot to say that $f$ is continuous.
– user386627
Jul 21 at 21:40

add a comment | 

 

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