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All norms on a finite-dimensional complex normed space are equivalent

Statement

Let X be a complex vector space carrying two complex norms ∥⋅∥a and ∥⋅∥b, and suppose X admits an ordered basis of finite length. Then ∥⋅∥a and ∥⋅∥b are equivalent in the sense of Equivalent complex norms.

Facts & Assumptions

Given: A complex vector space X with two complex norms ∥⋅∥a and ∥⋅∥b, and an ordered basis e:n→X.

[L1]

For either norm on X, the basis map from Cn with the coordinate ℓ1 norm is a topological isomorphism (A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space).

[L2]

Equivalent norms are exactly those satisfying c∥x∥a≤∥x∥b≤C∥x∥a for some positive real constants c,C (Equivalent complex norms).

Proof

technique · direct
1.1L1choose

Let T:Cn→X be the common algebraic basis map T(a0,…,an−1)=∑j<najej. Applied to the norm ∥⋅∥a, [L1] gives constants A,B>0 such that ∥T(a)∥a≤A∥a∥1and∥a∥1≤B∥T(a)∥a for every a∈Cn.

1.2L1choose

Applied to the norm ∥⋅∥b, [L1] gives constants A′,B′>0 such that ∥T(a)∥b≤A′∥a∥1and∥a∥1≤B′∥T(a)∥b for every a∈Cn.

2.1step 1.1step 1.2L1algebra

Let x∈X and write x=T(a), which is possible and unique by [L1]. Then ∥x∥b=∥T(a)∥b≤A′∥a∥1≤A′B∥T(a)∥a=A′B∥x∥a. Interchanging a and b gives ∥x∥a≤AB′∥x∥b.

3.1L2step 2.1∎

Taking c=(AB′)−1>0 and C=A′B>0, step 2.1 is exactly the two-sided estimate of [L2], so the two norms are equivalent.

Remarks

Depends on

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Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources