Showing that $\sqrt{x}$ is Lipschitz. - Proof Verification

Question

Show that $\sqrt{x}$ is Lipschitz.

Defn: A function $f: S \subset \mathbb{R^n} \rightarrow \mathbb{R^m}$ is Lipschitz if $\exists$ a constant $C$ such that $ \|f(x) - f(y)\| \leq C\|x - y\|$ for all $x,y \in S$.

Attempt:

$$\|\sqrt{x} - \sqrt{y}\| = \frac{\|x-y\|}{\|\sqrt{x} + \sqrt{y}\|}$$

By the completeness of the Reals, there exists a number $C' < \sqrt{x} + \sqrt{y}$ which would imply:

$$\frac{\|x-y\|}{\|\sqrt{x} + \sqrt{y}\|} < \frac{1}{C'} \|x-y\|$$

So if we let $C = \frac{1}{C'}$ then we have our needed constant.

Answer 1

That function is not a Lipschitz function. If it was, there would be a non-negative constant $C$ such that$$\bigl(\forall x\in[0,\infty)\bigr):\bigl\lvert\sqrt x-\sqrt0\bigr\rvert\leqslant C\lvert x-0\rvert,$$which is equivalent to$$\bigl(\forall x\in[0,\infty)\bigr):\sqrt x\leqslant Cx.$$Therefore, $\frac{\sqrt x}x$ would be a bounded function near $0$. But$$\lim_{x\to0}\frac{\sqrt x}x=\lim_{x\to0}\frac1{\sqrt x}=\infty.$$

Concerning your attempted proof, note that you wrote that “By the completeness of the Reals, there exists a number $C′<\sqrt x+\sqrt y$”. Why? The numbers of the form $\sqrt x+\sqrt y$ can be as small as we want, and therefore there is no number $C'>0$ for which that condition holds.

Answer 2

The statement "By the completeness of the Reals, there exists a number $C′< \sqrt {x} + \sqrt {y}$ is not correct. The completeness of reals does not imply the claimed result.

Note that $x$ and $y$ could be very small positive numbers so the sum of their square roots could be smaller than $C'$.