Show that $f(x):=\sqrt{\lvert x\rvert}$ belongs to $C^{0,\frac{1}{2}}(\mathbb{R})$

Question

Show that $f(x):=\sqrt{\lvert x\rvert}$ belongs to $C^{0,\frac{1}{2}}(\mathbb{R})$.

Hello, when I got it right, I have to show four things:

(1) $f\in C(\mathbb{R})$

(2) $f\in C(\overline{B_R(0)})$ for all $R>0$

(3) $$ \sup\limits_{x,y\in\overline{B_R(0)}), x\neq y}\left\{\frac{\lvert f(x)-f(y)\rvert}{\lvert x-y\rvert^{\frac{1}{2}}}\right\}<\infty $$

(4) $f$ uniform continious in $B_R(0)$ for all $R>0$

Am I right?

Answer 1

$$\frac{\left||x|^{1/2}-|y|^{1/2}\right|}{|x-y|^{1/2}}=\frac{||x|-|y||}{|x-y|^{1/2}\left(|x|^{1/2}+|y|^{1/2}\right)}\leq\frac{|x-y|^{1/2}}{|x|^{1/2}+|y|^{1/2}}=\sqrt{\frac{|x-y|}{|x|+|y|+2|x|^{1/2}|y|^{1/2}}}\leq\sqrt{\frac{|x|+|y|}{|x|+|y|}}=1.$$ This gives that $\sqrt{|x|}$ is uniformly Holder-continuous with exponent $\frac{1}{2}$. (1),(2) and (4) follow.