# Math Help - Lipschity function proof

1. ## Lipschity function proof

Show that $f(x)=\sqrt{a^2+x^2}$ is a Lipschity function.

My proof:

Let u,v be in R.

then $f(u)=\sqrt{a^2+u^2}$ and $f(v)=\sqrt{a^2+v^2}$

Now $|\sqrt{a^2+u^2}-\sqrt{a^2+v^2}|$
$\leq |\sqrt{a^2+u^2}-\sqrt{a^2+v^2}||\sqrt{a^2+u^2}+\sqrt{a^2+v^2}|$
$= |a^2+u^2-a^2-v^2| = |u^2-v^2| = |u-v| |u+v|$

So pick $C \geq |u+v|$, then we have $|f(u)-f(v)| \leq C|u-v|$

Is that right?

Thanks.

2. Say that $x,y\not = 0$ (if they are there is nothing to proof so it is safe to say there are not).

$|f(x)-f(y)| = |\sqrt{a^2+x^2} - \sqrt{a^2+y^2}| \cdot \frac{\sqrt{a^2+x^2}+\sqrt{a^2+y^2}}{\sqrt{a^2+x^2 }+\sqrt{a^2+y^2}} = \frac{|x^2-y^2|}{\sqrt{a^2+x^2}+\sqrt{a^2+y^2}}$ $\leq \frac{|x-y||x+y|}{|x|+|y|} \leq \frac{|x-y||x+y|}{|x+y|} = |x-y|$.