proving that a function is $\alpha$-Holder Prove that the function $ f(x)=\sqrt{x}$ , is $\alpha$-Holder, with $0&lt;\alpha\le \frac{1}{2} $ , on the set $[0,\infty)$

i.e there exist a constant $K$, such that $|\sqrt{x}-\sqrt{y}| \leqslant K|x-y|^{\alpha} $ for every $x,y \in [0,\infty)$.
It's not true for $\alpha &lt; \frac{1}{2}$. Fix $y = 1$ and see that if $x \geq 1$, $$\frac{\sqrt{x} - 1}{(x-1)^\alpha} \sim x^{\frac{1}{2} - \alpha} \rightarrow \infty.$$
Assuming $x,y\in[0,+\infty)$, with $x\not=y$, then $|\sqrt{x}-\sqrt{y}|=\Big|\frac{|x-y|}{\sqrt{x}+\sqrt{y}}\Big|\leq\frac{\sqrt{|x|+|y|}}{\sqrt{x}+\sqrt{y}}|x-y|^{1/2}\leq \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}|x-y|^{1/2}=|x-y|^{1/2}.$
Thus the mapping $x\mapsto \sqrt{x} $ is H\"{o}lder continuous of order $1/2$ on interval $[0,+\infty)$.