Equivalence of norms

All norms on a finite dimensional space are equivalent

The theorem as stated above holds in general1, but is currently beyond me. The case for and is simpler.

Complex vector space

Any two norms and on a finite-dimensional complex vector space are equivalent. #m/thm/anal/vec

Proof

Let be a vector basis of , so that any may be uniquely written as . Let denote the 1-norm . Now we will show that any norm is continuous under , i.e. for all there exists such that

By the Reverse triangle inequality, it is sufficient to show that

Letting and , it follows from the triangle inequality of that

so

Hence is 1-continuous. Since the unit sphere is compact, by the Extreme Value Theorem has an absolute minimum and maximum on this domain. Hence

for all with . Therefore all norms on are equivalent to , and by transitivity equivalent.

#state/develop | #lang/en | #SemBr

Footnotes

  1. See these lecture notes. ↩