Results 1 to 7 of 7

Math Help - Square root x satisfies holders condition of alpha >= 1/2 ... How?

  1. #1
    Junior Member
    Joined
    Oct 2009
    From
    London
    Posts
    42

    Holders Condition Reference?

    Hello
    Can anyone tell me a good reference for holders condition?
    Last edited by aukie; October 29th 2009 at 11:28 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Why did you delete the original question? If you want to ask another question simply make another thread.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    From
    London
    Posts
    42
    Because nobody replied so i didn't think anyone was interested. More so I didn't want anyone to waste their energy (i have answered questions on this forum only to find the author of the post is no longer interested). For any one interested here is a proof to my original question (the title of the post).

    \sqrt{x} is lipschitz of order 1/2 on [0, +inf)

    proof:

    |\sqrt{x} - \sqrt{y}| <= \sqrt{x}+ \sqrt{y} (by the triangle inequality)

    => |\sqrt{x} - \sqrt{y}|^2 <= |x - y| (some simple algebra)
    => |\sqrt{x} - \sqrt{y}| <= |x - y|^{1/2} (i.e. Lipschitz condition).

    If anyone knows how to prove the more general case i.e. prove x^r, 0<r<1, satisfies the lipschitz conditon of some order. i would be greatfull if you could share your proof.

    ps: lipschitz condition of order \alpha is equivalent to saying holder's condition of order \alpha
    Last edited by aukie; October 31st 2009 at 04:46 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by aukie View Post
    If anyone knows how to prove the more general case i.e. prove x^r, 0<r<1, satisfies the lipschitz conditon of some order. i would be greatfull if you could share your proof.
    Hi,
    I just devised the following simple proof:

    Let 0<r<1. Let 0<a\leq b.

    For any 0\leq x\leq 1, we have x\leq x^r, hence x^r+(1-x)^r\geq x+(1-x)=1.

    Taking x=\frac{a}{b} and multiplying both sides by b^r, we get a^r+(b-a)^r\geq b^r, hence b^r-a^r\leq (b-a)^r. QED.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    From
    London
    Posts
    42
    Hi Laurent, thanks for your input.

    If I have interpreted your proof correctly. It does not prove the problem statement completely, i.e.

    |x^r - y^r| \leq K|x-y|^r, \forall x,y \in [0, \infty) and some constant K>0.

    but your proof only proves this condition  \forall x,y \in \{ q \in \mathbb{Q}: 0 \leq q \leq 1 \}, i.e rational numbers in the unit interval.

    Here is what I have come up with so far. We want to prove the equivalent statement:

    \frac{|x^r - y^r|}{|x-y|^r} \leq K \label{eq1} , for some constant K>0. (1)

    If y = 0, it is obvious the lipschitz constant is K = 1. Thus assume y>0. Now if I consider the limit of the l.h.s as x \rightarrow y, L'Hopitals rule tells me

    \lim_{x \to y} \frac{x^r - y^r}{(x-y)^r} = 0 ... (here i am assuming x > y, hopefully without loss of generality)

    Now this tells me \forall \epsilon >0, \exists \delta > 0 , s.t. if |x-y|<\delta then \frac{x^r - y^r}{(x-y)^r} \leq \epsilon. i.e. equation (1) holds in some neighbourhood of y.

    But I am not quite sure on how or if I can extend this result to the the entire half line, [0, \infty) ... ???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by aukie View Post
    but your proof only proves this condition  \forall x,y \in \{ q \in \mathbb{Q}: 0 \leq q \leq 1 \}, i.e rational numbers in the unit interval.
    Where did you see this ?? I took any 0<a\leq b and proved b^r-a^r\leq (b-a)^r... I don't know where you see rationals and unit interval. I needed a first inequality where x\in[0,1], and then this is \frac{a}{b} (which obviously is in [0,1]) that played the role of x.

    Note also that b^r-a^r\leq (b-a)^r implies |b^r-a^r|\leq |b-a|^r since both sides are positive. The last inequality is symmetric in a,b so that it holds even if a>b.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2009
    From
    London
    Posts
    42
    hello laurent, spare me my blushes, for some reason i was assuming a, b were integers. very elegantly done ... thank you. i cant believe how over complicated i was making the proof in my attempt.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square root inside square root equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 10th 2011, 05:17 PM
  2. alpha is a square of an element in Fq star
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: March 17th 2011, 11:40 AM
  3. Replies: 12
    Last Post: November 22nd 2008, 01:41 PM
  4. Replies: 1
    Last Post: October 10th 2008, 03:51 AM
  5. Replies: 2
    Last Post: April 29th 2006, 02:13 AM

Search Tags


/mathhelpforum @mathhelpforum