How to show square root of absolute of x, √|x|\sqrt{|x|}, is not Lipschitz continuous?

f(x)=|x| is a famous example of a function which is not Lipschitz continuous but is uniformly continuous. This link shows detailed explanation of it.

Here provides the figure of this function:

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However, I am still confused about how to show |x| is not Lipschitz continuous?

  1. Consider [a,a], which is compact. f(y)f(x)yx should be bounded by L. This method is also used to prove f(x) is a uniformly continuous.
  2. On [a,) and (,a], f(x) has a bounded derivative.

So, based on 1 and 2, f(x)=|x| is Lipschitz continuous.

I have no idea how to prove it is not a Lipschitz continuous; clearly speaking I do not know how to distinguish the proof of Lipschitz continuity from uniformly continuity.

Answer

The derivative of |x| is sgn(x)2|x|. Let x0=14L2. The derivative at x0 is L, so, the derivative is unbounded.

Attribution
Source : Link , Question Author : sleeve chen , Answer Author : lisyarus

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