CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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A linear map from a finite-dimensional normed space is bounded

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field, and assume $X$ admits an ordered basis of finite length. Then every linear map $S : X \to Y$ is a bounded linear operator in the sense of A bounded linear operator between normed spaces.

Facts & Assumptions

Given: Normed spaces $X$ and $Y$ , a linear map $S : X \to Y$ , and an ordered basis $e : n \to X$ .

[L1]

The basis map $T : K^{n} \to X$ is a topological isomorphism (A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space).

[L2]

A bounded linear operator is a linear map satisfying one global norm bound (A bounded linear operator between normed spaces).

[L3]

Linearity means $S (a u + b v) = a S (u) + b S (v)$ (Linear map between vector spaces over the same field).

Proof

technique · direct

1.1L1choose

Let $T : K^{n} \to X$ be the basis map from [L1]. Since $T^{- 1}$ is bounded, there is $B > 0$ such that $∥ T^{- 1} x ∥_{1} \leq B ∥ x ∥ (x \in X) .$

1.2L3givenalgebra

Put $M : = \sum_{j < n} ∥ S (e_{j}) ∥$ , a finite real. If $x = \sum_{j < n} a_{j} e_{j} = T (a_{0}, \dots, a_{n - 1})$ , then by [L3] $S x = \sum_{j < n} a_{j} S (e_{j}),$ so $∥ S x ∥ \leq \sum_{j < n} ∣ a_{j} ∣ ∥ S (e_{j}) ∥ \leq M \sum_{j < n} ∣ a_{j} ∣ = M ∥ T^{- 1} x ∥_{1} .$

2.1L2L3step 1.1step 1.2∎

Combining steps 1.1 and 1.2 gives $∥ S x ∥ \leq M B ∥ x ∥ (x \in X) .$ Therefore $S$ is bounded, and with [L3] this makes $S$ a bounded linear operator by [L2].

Depends on

Used by

Exact one-qubit gates and rotations Definition
A compact group with no faithful continuous finite-dimensional representation Example
Bounded operators form a Banach algebra, noncommutative in dimension at least two Example
Finite-dimensional duals and matrix transposes Example
A split surjective derivative parametrises its level set Theorem
Continuous dual of a weak topology Theorem
Fredholm index is locally constant Theorem
Pauli formula for an axis rotation Theorem

Dependency tree · two levels

29 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)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)