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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 $∥ \cdot ∥_{a}$ and $∥ \cdot ∥_{b}$ , and suppose $X$ admits an ordered basis of finite length. Then $∥ \cdot ∥_{a}$ and $∥ \cdot ∥_{b}$ are equivalent in the sense of Equivalent complex norms.

Facts & Assumptions

Given: A complex vector space $X$ with two complex norms $∥ \cdot ∥_{a}$ and $∥ \cdot ∥_{b}$ , and an ordered basis $e : n \to X$ .

[L1]

For either norm on $X$ , the basis map from $C^{n}$ with the coordinate $ℓ^{1}$ 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 $c ∥ x ∥_{a} \leq ∥ x ∥_{b} \leq C ∥ x ∥_{a}$ for some positive real constants $c, C$ (Equivalent complex norms).

Proof

technique · direct

1.1L1choose

Let $T : C^{n} \to X$ be the common algebraic basis map $T (a_{0}, \dots, a_{n - 1}) = \sum_{j < n} a_{j} e_{j}$ . Applied to the norm $∥ \cdot ∥_{a}$ , [L1] gives constants $A, B > 0$ such that $∥ T (a) ∥_{a} \leq A ∥ a ∥_{1} and ∥ a ∥_{1} \leq B ∥ T (a) ∥_{a}$ for every $a \in C^{n}$ .

1.2L1choose

Applied to the norm $∥ \cdot ∥_{b}$ , [L1] gives constants $A^{'}, B^{'} > 0$ such that $∥ T (a) ∥_{b} \leq A^{'} ∥ a ∥_{1} and ∥ a ∥_{1} \leq B^{'} ∥ T (a) ∥_{b}$ for every $a \in C^{n}$ .

2.1step 1.1step 1.2L1algebra

Let $x \in X$ and write $x = T (a)$ , which is possible and unique by [L1]. Then $∥ x ∥_{b} = ∥ T (a) ∥_{b} \leq A^{'} ∥ a ∥_{1} \leq A^{'} B ∥ T (a) ∥_{a} = A^{'} B ∥ x ∥_{a} .$ Interchanging $a$ and $b$ gives $∥ x ∥_{a} \leq A B^{'} ∥ x ∥_{b} .$

3.1L2step 2.1∎

Taking $c = (A B^{'})^{- 1} > 0$ and $C = A^{'} B > 0$ , step 2.1 is exactly the two-sided estimate of [L2], so the two norms are equivalent.

Remarks

The proof does not need a separate norm-comparison theorem on $C^{n}$ : one coordinate isomorphism for each norm already supplies the comparison.
The published theorem For $n \geq 1$ all norms on $R^{n}$ are equivalent remains the real base case on which A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space rests.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources