proving that a function is $\alpha$-Holder Prove that the function $ f(x)=\sqrt{x}$ , is $\alpha$-Holder, with $0<\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 $\alpha < \frac{1}{2}$. Fix $y = 1$ and see that if $x \geq 1$, $$\frac{\sqrt{x} - 1}{(x-1)^\alpha} \sim x^{\frac{1}{2} - \alpha} \rightarrow \infty.$$
