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

Aiming for a contradiction, suppose $\sqrt {\size x}$ 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 $x = \dfrac 1 {n^2}$ with $n \in \N$ and $y = 0$.

Then:

$\dfrac 1 n \le \dfrac L {n^2}$

In other words:

$\forall n \in \N : n \le L$

However, $L$ was assumed to be finite.

Hence the contradiction.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces