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Lipschitz functions

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

    1. Let 0 < a < b <= 1. Prove that the set of all Lipschitz functions of order
    b is contained in the set of all Lipschitz functions of order a.

    2. Is the set of all Lipschitz functions of order b a closed subspace of those
    of order a?

    2. Relevant equations

    I know that a function f: [a,b] -> R is Lipschitz of order a if there exists a constant K
    such that |f(x) - f(y)| <= K |x-y|^a and for all x,y in [a,b].

    3. The attempt at a solution

    Assume f is a Lipschitz function of order b then there exists some constant K such that
    |f(x)-f(y)|<= K |x-y|^b. Then I need to prove that we can find some constant say C
    such that |f(x) - f(y)| <= C |x-y|^a , where 0 < a < b=1.

    Then I don't know how to proceed. Can you please help?
     
  2. jcsd
  3. May 30, 2009 #2
    Fix a and b. Have you considered the equality K|x - y|b = K|x - y|a|x - y|b - a ?
     
  4. May 30, 2009 #3
    OK, thanks for your reply.
    So |f(x)-f(y)|<= K |x-y|^b implies |f(x)-f(y)|<= K |x-y|^a |x-y|^(b-a).

    Therefore: |f(x) - f(y)| /|x-y|^a <= |x-y|^(b-a)

    But x and y are both in [a,b] so |x-y| <= |x|+|y| = b + b = 2b.

    Therefore |f(x)-f(y)|/|x-y|^a <= (2b)^(b-a).

    So our constant C is then (2b)^(b-a). Is this OK?

    How to show the closedness part? I know I have to take a sequence and show its closed
    under the limit but really I have no clue how to proceed.
     
  5. May 30, 2009 #4
    What happened to K?

    The first equality does not make sense, x and y are variables. Add some more rigor to your statements.
     
  6. May 30, 2009 #5
    OK, thanks again.
    My try:

    Since f is Lipschitz of order b then there exists a constant K >0 such that
    for all x, y in [a,b] we have: |f(x)-f(y)|<= K |x-y|^b.

    Observe K|x-y|^b = K|x-y|^a |x-y|^(b-a).

    Therefore |f(x)-f(y)| <= K |x-y|^a |x-y|^(b-a) and thus:

    |f(x)-f(y)|/|x-y|^a <= K |x-y|^(b-a).

    Since x, y are points in [a,b] then |x-y|^(b-a) <= (2b)^(b-a).

    Therefore |f(x)-f(y)|/|x-y|^a <= K (2b)^(b-a) and hence:

    |f(x)-f(y)|<= K (2b)^(b-a) |x-y|^a so f is Lipschitz of order a with constant
    C = K (2b)^(b-a).

    OK?
     
  7. May 30, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You appear to be using a and b with two different meanings here. I think more important is that since a< b< 1, 0< b-a< 1.

     
  8. May 30, 2009 #7
    Halls: Sorry, I do not follow your hint/suggestion, what do you mean?
     
  9. May 30, 2009 #8
    Two things that Hall mentioned:
    1) You are denoting the endpoints of the closed interval with the same constants you are using to denote the exponents of the Lipschitz inequality. The two are not related; if necessary use different letters, like [c, d].
    2) The fact that 0 < b-a < 1 is important.
     
  10. May 30, 2009 #9
    Gotcha guys.

    |f(x)-f(y)| / |x-y|^a<= K |x-y|^(b-a).

    Since 0 < b-a< 1 then |x-y|^(b-a) < |x-y|.

    Actually I meant: then |f(x)-f(y)|/|x-y|^a <= K |x-y|^a |x-y|. But x,y are both points in [c,d] so |x-y| <= d.

    Thus we get |f(x)-f(y)| <= K |x-y|^a * d so the constant is C = k*d, correct?
     
    Last edited: May 30, 2009
  11. May 31, 2009 #10
    Well since a and b are endpoints, and x,y are any points in [a,b], the inequality |x-y| =< |b -a| holds, so I don't think you need any extra variables beyond what's given in the problem statement.
     
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