2
$\begingroup$

We have trace map as $tr : M_n(F) \to F.$

Then using trace map is linear map I get $||tr(A)-tr(B)||=||tr(A-B)|| \le ||tr||||A-B||.$

Since I know that trace is a continuous map, I have $||tr||$ as Lipschitz constant. (Because $||tr||=\sup_{||A||=1}||tr(A)||$ and as the set $||A||=1$ is compact, the image $tr(A)$ is also compact hence obtaining maximum and minimum.)

By Lipschitz condition, I get trace map is uniformly continuous.

Am I right?

  • 2
    More simply, trace is a linear mapping on a finite dimensional normed vector space, so it is continuous. Any continuous linear map is Lipschitz by default, $\therefore$ uniformly continuous. – 2017-01-29
  • 0
    @OpenBall did you forget to put the word 'linear' after the word 'trace'? – 2017-01-29
  • 0
    Yes; I added it. – 2017-01-29
  • 1
    @OpenBall So general result is "Any continuous linear map on a finite dimensional normed vector space is uniformly continuous." Thanks. +1. – 2017-01-29

0 Answers 0