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This question already has an answer here:

First I constructed the negation to Lipschitz continuity:

$ \forall L > 0, \exists x, y \in [a,b]\ |f(x) - f(y)| > L|x - y| $

For $f(x) = \sqrt x$, notice $$|f(x) - f(y)| = |\sqrt x -\sqrt y| \cdot\frac{|\sqrt x + \sqrt y |}{|\sqrt x + \sqrt y |} = \frac{| x - y|}{|\sqrt x + \sqrt y|} \ge L |x - y| .$$

The problem is $I = [0,1]$ with $ \frac{1}{|\sqrt x + \sqrt y|}$ assuming values between $ \left(\frac{1}{2},\infty\right)$ and $ \frac{1}{|\sqrt x + \sqrt y|} \ge L$. So for sufficiently large $L$, the desired inequality for a function not being Lipschitz continuous cannot hold. Can someone explain the issue?

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marked as duplicate by Clayton, user91500, Tom-Tom, Davide Giraudo real-analysis Sep 7 '15 at 9:25

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You only need to interpret what you did:

Let $L>0$ be given. Then, choose $x,y\in (0,\infty )$ so that $\frac{1}{\sqrt{x}+\sqrt{y}}>L$. This is possible since $\sqrt x,\sqrt y\to 0$ as $x,y\to 0$.

Then, according to what you wrote

$|f(x) - f(y)|=\frac{| x - y|}{|\sqrt x + \sqrt y|}>L\vert x-y\vert $

and you are done.

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