The Euclidean Norm is Lipschitz Continuous .... D&K Example 1.3.5 .... ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of Example 1.3.5 ... ...

The start of Duistermaat and Kolk's Example 1.3.5 reads as follows:https://www.physicsforums.com/attachments/7752In the above example we read the following:

" ... ... The norm function $$x \mapsto \mid \mid x \mid \mid$$ is Lipschitz continuous on $$\mathbb{R}^n$$ with Lipschitz constant 1"To rigorously prove this statement we need to show that:

$$\mid \mid \ \mid \mid x \mid \mid \ - \ \mid \mid x' \mid \mid \ \mid \mid \ \le \ \mid \mid x - x' \mid \mid$$ ...Can someone help me to formally and rigorously show this ... ?Help will be much appreciated ...

Peter

 

[castor28]

Hi Peter,

The norm is a real number; this means that what you want to prove is about absolute values:
$$\left|\Vert x\Vert - \Vert x'\Vert\right|\le \Vert x-x'\Vert$$

If $\Vert x\Vert\ge\Vert x'\Vert$, this gives $\Vert x\Vert\le\Vert x-x'\Vert+\Vert x'\Vert$, which is true because of the triangle inequality; if $\Vert x\Vert<\Vert x'\Vert$, you can interchange $x$ and $x'$.

 

[Math Amateur]

Hi Peter,

The norm is a real number; this means that what you want to prove is about absolute values:
$$\left|\Vert x\Vert - \Vert x'\Vert\right|\le \Vert x-x'\Vert$$

If $\Vert x\Vert\ge\Vert x'\Vert$, this gives $\Vert x\Vert\le\Vert x-x'\Vert+\Vert x'\Vert$, which is true because of the triangle inequality; if $\Vert x\Vert<\Vert x'\Vert$, you can interchange $x$ and $x'$.

Thanks castor28 ... for a most helpful post ...

Peter