# Show a function is not Lipschitz Continuous [duplicate]

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?

## marked as duplicate by Clayton, user91500, Tom-Tom, Davide Giraudo real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 7 '15 at 9:25

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 $x,y\to 0$.
$|f(x) - f(y)|=\frac{| x - y|}{|\sqrt x + \sqrt y|}>L\vert x-y\vert$