Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?

Question

Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $||u|^n - |v|^n| \leq |u - v|^n$?

I want to use this property as an intermediate step for something else but I'm having trouble proving it. Intuition tells me it should be true. If $n=1$, it is simply the triangle inequality. What about for $n > 1$?

Generally speaking, the values of a non-constant function can't be that close together. Related: http://math.stackexchange.com/questions/164804/show-f-is-constant-if-fx-fy-leq-x-y2 — , Apr 28 '16 at 20:18

nikola · Answer 1 · 2016-04-28T20:08:04.033

2

It is not true. Take $u = \frac{1}{3}, v= \frac{1}{2}$ and n = 2. Now your inequality tells $\frac{5}{36} = |\frac{1}{9}-\frac{1}{4}| \leq |\frac{1}{3}-\frac{1}{2}|^2$ or $\frac{5}{36} \leq \frac{1}{36}$ which is not.

edited Apr 28 '16 at 20:08

answered Apr 28 '16 at 19:59

nikola

1,045

Thank you for pointing out this easy counter-example. – mathjacks Apr 28 '16 at 20:10
You are welcome. Ask whatever, whenever. There is only one rule when you try to learn something: There are no stupid questions. – nikola Apr 28 '16 at 20:19

score 1 · Accepted Answer · answered Apr 28 '16 at 20:05

1

Take $u = ( \frac{1}{2}, \frac{1}{2})$ and $v = ( \frac{1}{2}, \frac{1}{3})$, then $|u|^2 = \frac{1}{2}$, $|v|^2 = \frac{13}{36}$ and $|u - v|^2 = \frac{1}{36}$

answered Apr 28 '16 at 20:05

Reinaldo R.

338

Thanks for this. I will have to do my proof another way. – mathjacks Apr 28 '16 at 20:11

Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?

2 Answers2