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. 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.

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Footnotes

1. See these lecture notes. ↩