Linear Mappings are Lipschitz Continuous .... D&K Example 1.8.14 .... ....

[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 1.8.14 ... ...

The start of Duistermaat and Kolk's Example 1.8.14 reads as follows:https://www.physicsforums.com/attachments/7753In the above example we read the following:

"... ... any linear mapping \(\displaystyle A \ : \ \mathbb{R}^n \rightarrow \mathbb{R}^p\), whether bijective or not, is Lipschitz continuous and therefore uniformly continuous ... ... "I am somewhat unsure regarding proving this statement ... but I think the proof goes like the following:


For \(\displaystyle A \ : \ \mathbb{R}^n \rightarrow \mathbb{R}^p\) to be Lipschitz continuous we require

\(\displaystyle \mid \mid Ax - Ax' \mid \mid \le k \mid \mid x - x' \mid \mid \text{ where } x, x' \in \text{dom} (A)\) ... ... ... (1)now we have:

\(\displaystyle \mid \mid Ax \mid \mid \le k \mid \mid x \mid \mid\) ... ... ... (2)Now ... maybe put \(\displaystyle x = x' - x''\)

... then (2) becomes \(\displaystyle \mid \mid A(x' - x'') \mid \mid = \mid \mid Ax' - Ax'' \mid \mid \le k \mid \mid x' - x'' \mid \mid\)

which is the required result ...Is that correct?

Peter

 

[S.G. Janssens]

Yes, and you can be a bit more self-confident (Happy)
(Remark which may be superfluous: Your $k$ is independent of $x$, $x'$, $x''$.)

 

[Math Amateur]

Yes, and you can be a bit more self-confident (Happy)
(Remark which may be superfluous: Your $k$ is independent of $x$, $x'$, $x''$.)

Indeed, Krylov... you are probably right :)

Thanks for your reply ...

Peter