Gradient of $1/|x|$ is Lipschitz continuous if $|x|>1$

Question

How to show when $|x| > 1$, $\nabla\left(\frac{1}{|x|}\right)$ is Lipschitz continuous on $\Bbb{R}^3$?!

I have $$\left|\nabla\left(\frac{1}{|x|}\right)-\nabla\left(\frac{1}{|y|}\right)\right| = 3 \left|\frac{x}{|x|^3}-\frac{y}{|y|^3}\right|$$

I wanted to use $|x|>1$, or $\dfrac{1}{|x|}<1$, to get rid of the denominators I have, but the fractions are stuck in the absolute value! I want to get a constant times $|x-y|$ but I can't! I can't use triangle inequality because then that minus will turn into a plus!

$\nabla\left(\frac{1}{|x|}\right) = -\frac{x}{|x|^3}$, so I'm not sure why there's a $3$ in front. — Joey Zou, Aug 23 '16 at 06:25
Joey Zou, you can change it if you would like. Apparently users can edit questions here! — user362922, Aug 23 '16 at 06:34
Indeed, and since this is your own question, I'm pretty sure you can edit it as well. — Joey Zou, Aug 23 '16 at 06:36

score 2 · Accepted Answer · answered Aug 23 '16 at 06:27

2

Hint: Write $\frac{x}{|x|^3}-\frac{y}{|y|^3}= \frac{x-y}{|x|^3} + y \frac{|y|^3-|x|^3}{|x|^3|y|^3}$ and factor the last numerator. Finally use $\left||x|-|y|\right| \leq |x-y|$.

answered Aug 23 '16 at 06:27

H. H. Rugh

35,236

You added and subtracted y/|x|^3? – user362922 Aug 23 '16 at 06:44
yes, indeed what I did. – H. H. Rugh Aug 23 '16 at 07:09

Gradient of $1/|x|$ is Lipschitz continuous if $|x|>1$

1 Answers1