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].

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources