# 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 $x,y \in [0,\infty)$.

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–  Pragabhava Oct 3 '12 at 3:05
I don't know how to use that, there are proving that the function is continuous –  Andy Oct 3 '12 at 3:14
If you look carefully, you can see that it has been proven that $|\sqrt{x}-\sqrt{y}|^2 \le |x - y|$ for $x,\,y \in [0,\infty)$ –  Pragabhava Oct 3 '12 at 3:16
Ok , and how can I do the other cases D:? when $\alpha < \frac{1}{2}$ –  Andy Oct 3 '12 at 3:24

## 2 Answers

It's not true for $\alpha < \frac{1}{2}$. Fix $y = 1$ and see that if $x \geq 1$, $$\frac{\sqrt{x} - 1}{(x-1)^\alpha} \sim x^{\frac{1}{2} - \alpha} \rightarrow \infty.$$

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It seems like you are right O= , so Wikipedia is wrong in this example en.wikipedia.org/wiki/H%C3%B6lder_condition#Examples , example number 2 and also number 3. –  Andy Oct 3 '12 at 3:43
Well, the Wikipedia entry restricts the function to [0,1]. The Hölder condition can sort of fail in two ways: it can fail "locally" as the two points get close together, or it can fail "globally" as they get far apart. Here, everything works fine locally but fails globally. –  Zach L. Oct 3 '12 at 3:49

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 $[0,+\infty)$.

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