TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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 complex case by Real and complex scalar conventions for normed spaces. 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 (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]

The complex case is read with the same norm axioms and with scalar absolute value replaced by the complex modulus (Real and complex scalar conventions for normed spaces).

Proof

technique · direct

1.1

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.

1.2

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

givenalgebra

1.3

In the complex case $K = C$ , write $Ψ : C^{n} \to R^{2 n}$ for the coordinatewise real-imaginary-part map from [L4]. Define $N_{R} (u, v) : = ∥ T (u_{0} + i v_{0}, \dots, u_{n - 1} + i v_{n - 1}) ∥ .$ By [L4] and [L5] this is a real norm on $R^{2 n}$ . If $n = 0$ the inverse is again bounded trivially. If $n \geq 1$ , [L3] applied to $R^{2 n}$ gives $c > 0$ with $c ∥ (u, v) ∥_{2} \leq N_{R} (u, v)$ for every $(u, v) \in R^{2 n}$ . Also $∣ u_{j} + i v_{j} ∣ \leq ∣ u_{j} ∣ + ∣ v_{j} ∣ \leq \sqrt{2} (u_{j}^{2} + v_{j}^{2})^{1 / 2}$ , so $∥ (u_{0} + i v_{0}, \dots, u_{n - 1} + i v_{n - 1}) ∥_{1} \leq \sqrt{2 n} ∥ (u, v) ∥_{2} .$ Hence $∥ a ∥_{1} \leq \frac{\sqrt{2 n}}{c} ∥ T (a) ∥ (a \in C^{n}),$ which is the boundedness of $T^{- 1}$ .

L3L4L5algebra

2.1

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.

step 1.2givenalgebra

2.2

In the real case $K = R$ , 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.

step 1.1step 1.2L3choose

3.1

Steps 1.1, 1.2, 1.3, 2.1, and 2.2 verify the three clauses of [L2]. Therefore $T$ is a topological isomorphism of normed spaces.

L2step 1.1step 1.2step 1.3step 2.1step 2.2∎

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.

Depends on

Used by

Dependency tree · two levels

42 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

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