Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is wrong with this proof? (lipschitz functions)

  1. Sep 5, 2011 #1
    Suppose f:[0,1]->R is absolutely continuous and |f'(x)|<M almost everywhere in [0,1]. Prove that f is Lipschitz.

    I wrote up the following proof and got significant deductions:

    By the Mean Value Theorem, for all x,y in [0,1], there exists c between x and y such that |f(x)-f(y)/(x-y)|=|f'(c)|<M.

    That is, for all x,y in [0,1], |f(x)-f(y)/(x-y)|<M.
    Then |f(x)-f(y)|<M|x-y|, thus f is Lipschitz.

    What is wrong with my proof? (Since f is AC, I know that f is differentiable almost everywhere in [0,1]. Is that not enough to invoke the mean value theorem? If not, how do I prove this theorem?)

    Thank you.
    Last edited by a moderator: Sep 6, 2011
  2. jcsd
  3. Sep 6, 2011 #2


    User Avatar
    Science Advisor

    No, it is not. For example, the function [itex]f(x)= x^2[/itex] is differentiable everywhere in [0, 1] and the mean value theorem says there must be some c in [0, 1] such that f'(c)= (f(1)- f(0))/(1- 0)= 1. Obviously, that point is x= 1/2. But [itex]g(x)= x^2[/itex] for all x except 1/2 and g(1/2)= 0 is differentiable "almost everywhere" (everywhere except x= 1/2 and a singleton set has measure 0) but there is no c such that g'(c)= (g(1)- g(0))/(1- 0)= 1.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook