0
$\begingroup$

I'm just going through some examples in my Analysis notes and we're looking at Holder continuity of $\sqrt{x}$ on $[0,1]$. One of the steps is

$$\left|\sqrt{x}-\sqrt{y} \right|=\frac{\left|x-y \right|}{\sqrt{x}+\sqrt{y}}$$

I've been fiddling around with the left hand side for half an hour and cannot figure out how that works. I know that you can define the absolute value as $|x| = \sqrt{x^2}$, but I can't get it to work.

$\endgroup$
0
$\begingroup$

Hint:

Squares' difference: $\;a^2-b^2=(a-b)(a+b)\;$ , and now put $\;a=\sqrt x\;,\;\;b=\sqrt y\;$ .

$\endgroup$
0
$\begingroup$

$\left|\sqrt{x}-\sqrt{y} \right|= \left|\sqrt{x}-\sqrt{y} \right|\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}} = \frac{\left|x-y\right|}{\sqrt{x}+\sqrt{y}}$, as required.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.