Square root x is not Lipschitz on [0,1]

f(x)=\sqrt x is not Lipschitz on [0,1]:

Suppose there exists C\geq 0 such that for all x,y\in [0,1], x\neq y, \displaystyle \frac{|f(x)-f(y)|}{|x-y|}\leq C.

By Mean Value Theorem, this means that |f'(\xi)|\leq C for some \xi between x and y.

However, f'(x)=\frac{1}{2}x^{-1/2} is unbounded on [0,1], a contradiction.

Note however, that \sqrt x is absolutely continuous on [0,1]. So Lipschitz is a stronger condition than absolutely continuous.

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