TITLE: Is $\sqrt{x}$ Lipschitz continuous on $(0,\infty)$?
QUESTION [1 upvotes]: Is $\sqrt{x}$ Lipschitz continuous on $(0,\infty)$?
I wonder that because I know it is on $[a,\infty)$ for all $a>0$.

REPLY [1 votes]: No. If it were, then there would be a constant $K > 0$ such that $|\sqrt{x} - \sqrt{y}| \le K|x - y|$ for all $x,y\in (0,\infty)$. Then for $0 < y < x < \infty$, $$\frac{\sqrt{x} - \sqrt{y}}{x - y} \le K$$
or 
$$\frac{1}{\sqrt{x} + \sqrt{y}} \le K\qquad (0 < y < x < \infty).$$
It follows that 
$$\frac{1}{2\sqrt{x}} \le K$$
for all $x > 0$. Taking $x = 1/(16K^2)$ results in the inequality $2K \le K$, a contradiction.
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