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## 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.

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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