Hölder-continuity of a function

Question

Consider $f:[-1,1]\rightarrow \mathbb{R} $ given by $f(x) =\sqrt{|x|} $. Show that $$ \sup_{x, y\in [-1,1] x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{1/2}} $$ is finite.

I started by writing the definition but got stuck at first step only.

Answer 1

Let $x,y \ge 0$, $x \neq y$, then $|\sqrt{x} - \sqrt{y}| \le |\sqrt{x} + \sqrt{y}|$ and hence $$|\sqrt{x} - \sqrt{y}|^2 \le |\sqrt{x} - \sqrt{y}| |\sqrt{x} + \sqrt{y}| = |x - y|^2.$$ This shows that $$\sup_{x\neq y \in[0,1]}\frac{|\sqrt{x} - \sqrt{y}|}{\sqrt{|x - y|}} \le 1.$$ If $x,y \le 0$ you can use the same argument, applied to $-x, -y$. If $x < 0 < y$ then $|x| > 0$ and hence $||x| - y| \le |x - y|$. In particular, since the square root is increasing on $[0,2]$, $$|\sqrt{|x|} - \sqrt{y}| \le \sqrt{||x| - y|} \le \sqrt{|x - y|}.$$