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A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space

Statement

Let $X$ be a normed space over $K \in {R, C}$ , read in the real case by A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and the complex case by Complex norm and complex normed space. Let $e : n \to X$ be an ordered basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and write $e_{j} : = e (j)$ . Give $K^{n}$ its coordinate $ℓ^{1}$ norm

$∥ (a_{0}, \dots, a_{n - 1}) ∥_{1} : = \sum_{j < n} ∣ a_{j} ∣ .$

Define

$T : K^{n} \to X, T (a_{0}, \dots, a_{n - 1}) = \sum_{j < n} a_{j} e_{j} .$

Then $T$ is a topological isomorphism of normed spaces in the sense of A topological isomorphism of normed spaces.

Facts & Assumptions

Given: A normed space $X$ over $K \in {R, C}$ and an ordered basis $e : n \to X$ .

[L1]

An ordered basis is a finite list whose image is a basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis identifies its span with exactly the vectors of the form $\sum_{j < n} a_{j} e_{j}$ , with those coordinates unique.

[L2]

A topological isomorphism of normed spaces is a bounded linear bijection whose inverse is bounded (A topological isomorphism of normed spaces).

[L3]

For $m \geq 1$ , every norm on $R^{m}$ is equivalent to the Euclidean norm, and any two norms on $R^{m}$ are equivalent (For $n \geq 1$ all norms on $R^{n}$ are equivalent).

[L4]

The map $Φ : C \to R^{2}$ , $Φ (a + b i) = (a, b)$ , is a bijection with the stated coordinate arithmetic ( $C$ is the real coordinate plane, with coordinate arithmetic).

[L5]

Complex norms have full complex absolute homogeneity (Complex norm and complex normed space). Their realifications carry the same norm and metric (Realification of a complex vector space by restriction of scalars, Complex modulus restricts to real absolute value, Realification preserves a complex norm and its metric). The coordinate $ℓ^{1}$ function is a complex norm, including $n = 0$ (The complex coordinate sum norm, including the zero space).

[F1]

Real absolute value is definite, multiplicative and subadditive (Basic properties of the absolute value, The triangle inequality).

[F2]

Finite real sums distribute over addition and scaling and preserve inequalities (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

Proof

technique · direct

1.1L1

By [L1], every $x \in X$ has exactly one coordinate list $a = (a_{0}, \dots, a_{n - 1}) \in K^{n}$ with $x = \sum_{j < n} a_{j} e_{j}$ . Therefore the displayed map $T$ is well defined, surjective, and injective.

1.2givenalgebra

$T$ is linear, because finite sums and scalar multiplication distribute over the coordinate formula: $T (λ a + b) = \sum_{j < n} (λ a_{j} + b_{j}) e_{j} = λ \sum_{j < n} a_{j} e_{j} + \sum_{j < n} b_{j} e_{j} = λ T (a) + T (b)$ .

2.1L1L3L4L5step 1.1step 1.2algebra

In the complex case, [L5] makes both $P (a) : = ∥ a ∥_{1}$ and $N (a) : = ∥ T (a) ∥$ real norms on $(C^{n})_{R}$ : for $N$ , separation uses injectivity in step 1.1, and homogeneity and the triangle inequality use step 1.2 and the norm on $X$ . The coordinate map $Ψ (a) = (Re a_{0}, Im a_{0}, \dots, Re a_{n - 1}, Im a_{n - 1})$ is a real-linear bijection by [L4], so $P \circ Ψ^{- 1}$ and $N \circ Ψ^{- 1}$ are real norms on $R^{2 n}$ . If $n \geq 1$ , the pairwise equivalence clause of [L3] gives a positive real $c$ such that $c P (a) \leq N (a)$ for every $a$ . Thus $∥ T^{- 1} x ∥_{1} \leq c^{- 1} ∥ x ∥$ . If $n = 0$ , both spaces consist only of zero by [L1], and the inverse is bounded with bound zero.

2.2step 1.2givenalgebra

Put $C : = \sum_{j < n} ∥ e_{j} ∥$ , a finite real. Then $∥ T (a) ∥ = ∥ \sum_{j < n} a_{j} e_{j} ∥ \leq \sum_{j < n} ∣ a_{j} ∣ ∥ e_{j} ∥ \leq C \sum_{j < n} ∣ a_{j} ∣ = C ∥ a ∥_{1},$ so $T$ is bounded.

2.3step 1.1step 1.2L3F1F2choose

In the real case $K = R$ , the coordinate $P (a) = \sum_{j < n} ∣ a_{j} ∣$ is a real norm: absolute-value definiteness, multiplicativity, and the triangle inequality give the three axioms by finite summation, including the empty list. The pullback $N (a) : = ∥ T (a) ∥$ is a norm on $R^{n}$ : definiteness uses step 1.1, and the triangle and homogeneity axioms come from the norm axioms on $X$ and the linearity of $T$ . If $n = 0$ , then $R^{0} = {0}$ and the inverse of $T$ is the zero map, hence bounded. If $n \geq 1$ , [L3] gives $c > 0$ with $c ∥ a ∥_{1} \leq N (a) = ∥ T (a) ∥$ for every $a \in R^{n}$ , so $∥ T^{- 1} x ∥_{1} \leq c^{- 1} ∥ x ∥$ for every $x \in X$ . Thus $T^{- 1}$ is bounded in the real case.

3.1L2step 1.1step 1.2step 2.1step 2.2step 2.3∎

Steps 1.1, 1.2, 2.1, 2.2, and 2.3 verify the three clauses of [L2]. Therefore $T$ is a topological isomorphism of normed spaces.

Remarks

The proof uses the coordinate $ℓ^{1}$ norm because it makes the boundedness of $T$ immediate. Any other standard coordinate norm would do, and on a fixed finite-dimensional coordinate space all of them are equivalent.
The finite-dimensional language in the title is implemented here by the actual datum the page uses: a chosen ordered basis of finite length.

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Sources

Daniel Daners, Introduction to Functional Analysis (standard reference, not scraped)
Tomasz Kochanek, Functional analysis, Lecture 1 (standard reference, not scraped)