Hölder condition: $g: [0, \infty)$, $g(x)=\sqrt{x}$ with $L=1$ and $\alpha=1/2$

Question

Hölder condition: $g: [0, \infty)$, $g(x)=\sqrt{x}$ with $L=1$ and $\alpha=1/2$

i tried proving it like this:

$|\sqrt{x}-\sqrt{y}|\leq L \sqrt{|x-y|}$ with $L=1$

$\Leftrightarrow |\sqrt{x}-\sqrt{y}|^2 \leq |x-y|$

$\Leftrightarrow x-y \leq |x-y|$

hence proved. Is that correct?

Answer 1

$|\sqrt{x} - \sqrt{y}|^{2} = |\sqrt{x} - \sqrt{y}||\sqrt{x} - \sqrt{y}| \leqslant |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}| = |x - y|$

Answer 2

Going from second line to the third isn’t correct. Observe:

$$|\sqrt{x}-\sqrt{y}|^2 = |x+y-2 \sqrt{x}\sqrt{y}|$$

Which isn’t equal to what you have written.