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If $f$ is Lipschitz of order 1 at $x$, is it differentiable at $x$?

A function $f$ is Lipschitz of order $\beta$ at $x$ if there is a constant D such that

$$|f(x)-f(y)|\le D \,|x-y|^\beta$$

for all $y$ in the interval containing $x$.

If yes, can anyone motivate a proof for me?

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what does it mean Lipschitz of order 1? What is the order of a Lipschitz function? – Emanuele Paolini Jan 25 at 0:12
That is typically called Holder continuity, but we get the idea. – icurays1 Jan 25 at 0:16

2 Answers

up vote 3 down vote accepted

To answer your question, no (as $f(x)=\vert x\vert$ demonstrates). However, you should take a look at Rademacher's theorem, which says that Lipschitz functions are almost everywhere differentiable.

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I was think of a $f(x) = |x|$, but couldn't quite flesh the entire thing out. Don't know why this condition is giving me such an issue. It has been very hard to gather intuition. Thanks for the theorem. I will take a look at it. Would there be a way you can walk me through that counter example? – user43901 Jan 25 at 0:15
Which part? The function is Lipschitz via reverse triangle inquality, and clearly not differentiable at zero. – icurays1 Jan 25 at 0:18
Darn it! The reverse triangle inequality, my arch nemesis! LOL. Thanks for pointing that out. My friend wrote that, but he told me it's a formula, but could not name it and I had sat down to prove it wrong =) – user43901 Jan 25 at 0:29
one quick question, is the converse of the statement true? – user43901 Jan 25 at 3:25
Only locally, i.e. $f(x)$ continuously differentiable implies $f$ is Lipschitz on every compact subset of its domain. $f(x)=x^2$ is smooth but not globally Lipschitz. – icurays1 Jan 25 at 3:59
up vote 1 down vote

Consider the function $f(x) = |x|$.

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