Equivalence between Hölder norms

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Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?

calculus real-analysis functional-analysis holder-spaces holder-inequality

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edited Jul 28 at 17:37

copper.hat

122k557156

asked Jul 28 at 17:07

JCF

18411

add a comment | 

up vote
0
down vote

favorite

Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?

calculus real-analysis functional-analysis holder-spaces holder-inequality

share|cite|improve this question

edited Jul 28 at 17:37

copper.hat

122k557156

asked Jul 28 at 17:07

JCF

18411

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?

calculus real-analysis functional-analysis holder-spaces holder-inequality

share|cite|improve this question

edited Jul 28 at 17:37

copper.hat

122k557156

asked Jul 28 at 17:07

JCF

18411

Let us consider the following two norms:
$$
leftlVert frightrVert_alpha = leftlVert frightrVert_infty + displaystylesup_substackx,y in U \ x neq y fracleft^alpha,
$$
$$
leftlVert frightrVert_alpha • = |f(0)|+ displaystylesup_substackx,y in U \ x neq y fracleft^alpha.
$$
I need to show that those two norms are equivalent on $mathcalC^alpha(U,mathbbR)$, where $U=[0,T]$ that is the space of $alpha$-Hölder functions from $U$ to $mathbbR$, then I have to find two constants, for instance let us call them $a,b in mathbbR$, such that:
$$
a leftlVert frightrVert_alpha • leq leftlVert frightrVert_alpha leq b leftlVert frightrVert_alpha • .
$$
I have issues in handling the $infty$-norm and readapting the constant to the Hölder seminorm. Any suggestions?

calculus real-analysis functional-analysis holder-spaces holder-inequality

share|cite|improve this question

edited Jul 28 at 17:37

copper.hat

122k557156

asked Jul 28 at 17:07

JCF

18411

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 17:37

copper.hat

122k557156

edited Jul 28 at 17:37

copper.hat

122k557156

edited Jul 28 at 17:37

copper.hat

122k557156

122k557156

asked Jul 28 at 17:07

JCF

18411

asked Jul 28 at 17:07

JCF

18411

asked Jul 28 at 17:07

JCF

18411

18411

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.

The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.

share|cite|improve this answer

answered Jul 28 at 17:30

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, maybe I should do math when I am awake :)
  – JCF
  Jul 28 at 18:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JCF: Begin awake doesn't help me :-).
  – copper.hat
  Jul 28 at 18:20
  

  
  
  
  

add a comment | 

up vote
2
down vote

It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.

For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.

share|cite|improve this answer

edited Jul 28 at 18:42

answered Jul 28 at 17:30

Rhys Steele

5,5651828

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.

The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.

share|cite|improve this answer

answered Jul 28 at 17:30

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, maybe I should do math when I am awake :)
  – JCF
  Jul 28 at 18:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JCF: Begin awake doesn't help me :-).
  – copper.hat
  Jul 28 at 18:20
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.

The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.

share|cite|improve this answer

answered Jul 28 at 17:30

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, maybe I should do math when I am awake :)
  – JCF
  Jul 28 at 18:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JCF: Begin awake doesn't help me :-).
  – copper.hat
  Jul 28 at 18:20
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.

The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.

share|cite|improve this answer

answered Jul 28 at 17:30

copper.hat

122k557156

begineqnarray
|f(x)| &le& |f(0)| + |f(x)-f(0)| \
&le& |f(0)| + |f|_alpha • |x-0|^alpha \
&le& |f|_alpha • + T^alpha |f|_alpha • \
&=& (1+T^alpha)|f|_alpha •
endeqnarray
And so $|f|_infty le (1+T^alpha) |f|_alpha • $ and then
$|f|_alpha le (2+T^alpha) |f|_alpha • $.

The other direction is immediate since $|f(0)| le |f|_infty$ and so $|f|_alpha • le |f|_alpha$.

share|cite|improve this answer

answered Jul 28 at 17:30

copper.hat

122k557156

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 17:30

copper.hat

122k557156

answered Jul 28 at 17:30

copper.hat

122k557156

answered Jul 28 at 17:30

copper.hat

122k557156

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, maybe I should do math when I am awake :)
  – JCF
  Jul 28 at 18:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JCF: Begin awake doesn't help me :-).
  – copper.hat
  Jul 28 at 18:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, maybe I should do math when I am awake :)
  – JCF
  Jul 28 at 18:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JCF: Begin awake doesn't help me :-).
  – copper.hat
  Jul 28 at 18:20
  

  
  
  
  

Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03

Thank you, maybe I should do math when I am awake :)
– JCF
Jul 28 at 18:03

@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20

@JCF: Begin awake doesn't help me :-).
– copper.hat
Jul 28 at 18:20

add a comment | 

up vote
2
down vote

It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.

For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.

share|cite|improve this answer

edited Jul 28 at 18:42

answered Jul 28 at 17:30

Rhys Steele

5,5651828

add a comment | 

up vote
2
down vote

It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.

For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.

share|cite|improve this answer

edited Jul 28 at 18:42

answered Jul 28 at 17:30

Rhys Steele

5,5651828

add a comment | 

up vote
2
down vote

up vote
2
down vote

It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.

For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.

share|cite|improve this answer

edited Jul 28 at 18:42

answered Jul 28 at 17:30

Rhys Steele

5,5651828

It is clear that $|f|_alpha • leq |f|_alpha$. Essentially the only difficult part in showing the other inequality is controlling $|f|_infty$ in terms of $|f|_alpha • $ so I will do this.

For $x in [0,T]$ we have
beginalign*
|f(x)| &leq |f(0)| + |f(x) - f(0)|
\ & leq |f(0)| + |f|_alpha • |x|^alpha \&
leq |f|_alpha • + |f|_alpha • T^alpha \&
= (1+ T^alpha) |f|_alpha •
endalign*
Taking the $sup$ over $U$ gives an inequality of the desired form.

share|cite|improve this answer

edited Jul 28 at 18:42

answered Jul 28 at 17:30

Rhys Steele

5,5651828

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 18:42

edited Jul 28 at 18:42

edited Jul 28 at 18:42

answered Jul 28 at 17:30

Rhys Steele

5,5651828

answered Jul 28 at 17:30

Rhys Steele

5,5651828

answered Jul 28 at 17:30

Rhys Steele

5,5651828

5,5651828

add a comment | 

add a comment | 

 

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