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 ℂ𝑛 and ℝ𝑛 is simpler.

Complex vector space

Any two norms ‖ ⋅‖𝑎 and ‖ ⋅‖𝑏 on a finite-dimensional complex vector space 𝑉 are equivalent. vec

Proof

Let (𝑒𝑖)𝑛𝑖=1 be a vector basis of 𝑉, so that any 𝑥 ∈𝑉 may be uniquely written as 𝑥 =∑𝑛𝑖=1𝑥𝑖𝑒𝑖. Let ‖ ⋅‖1 denote the 1-norm ‖𝑥‖1 =∑𝑛𝑖=1𝑥𝑖. Now we will show that any norm ‖ ⋅‖𝑎 is continuous under ‖ ⋅‖1, i.e. for all 𝜖 >0 there exists 𝛿 >0 such that

‖𝑥−𝑦‖1<𝛿⟹|‖𝑥‖𝑎−‖𝑦‖𝑎|<𝜖

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

‖𝑥−𝑦‖1<𝛿⟹‖𝑥−𝑦‖𝑎<𝜖

Letting 𝑥 =∑𝑛𝑖=1𝑥𝑖𝑒𝑖 and 𝑦 =∑𝑛𝑖=1𝑦𝑖𝑒𝑖, it follows from the triangle inequality of ‖ ⋅‖𝑎 that

‖𝑥−𝑦‖𝑎≤𝑛∑𝑖=1|𝑥𝑖−𝑦𝑖|‖𝑒𝑖‖𝑎≤‖𝑥−𝑦‖1𝑛max𝑖=1‖𝑒𝑖‖𝑎

so

‖𝑥−𝑦‖1<𝜖max𝑛𝑖=1‖𝑒𝑖‖𝑎⟹‖𝑥−𝑦‖𝑎<𝜖

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

𝑎<‖𝑣‖𝑎<𝑏

for all 𝑣 ∈𝑉 with ‖𝑣‖1 =1. Therefore all norms on 𝑉 are equivalent to ‖ ⋅‖1, and by transitivity equivalent.

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Footnotes

1. See these lecture notes. ↩