What is wrong with this proof? (lipschitz functions)

  • #1

Main Question or Discussion Point

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.
QED

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.
 
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Answers and Replies

  • #2
HallsofIvy
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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.
QED

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?
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.

If not, how do I prove this theorem?)

Thank you.
 

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