Square Root is not Lipschitz Continuous

Theorem
Let $\sqrt {\size x} : \R \to \R_{\ge 0}$ be a real function.

$\sqrt {\size x}$ is not Lipschitz continuous.

Proof
$\sqrt {\size x}$ is Lipschitz continuous.

Then:


 * $\exists L \in \R_{> 0} : \forall x,y \in \R : \size {\sqrt {\size x} - \sqrt {\size y} } \le L \size {x - y}$

Suppose $\ds x = \frac 1 {n^2}$ with $n \in \N$ and $y = 0$.

Then:


 * $\ds \frac 1 n \le \frac L {n^2}$

In other words:


 * $\forall n \in \N : n \le L$

However, $L$ was assumed to be finite.

Hence the contradiction.