TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-24 (gpt-6-sol)

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Every nonzero vector has a norming functional

Statement

Assume the Axiom of Choice. Let $X$ be a normed space over $R$ or $C$ , and let $x \in X$ be nonzero. Then there exists $f \in X^{*}$ such that

$∥ f ∥ = 1 and f (x) = ∥ x ∥ .$

Facts & Assumptions

Given: The Axiom of Choice, A normed space $X$ over $R$ or $C$ and a vector $x \in X$ with $x \neq 0$ .

[A1]

AC is used through real or complex extension [L2] or [L3] in steps 2.1 and 2.2.

[L1]

The dual space $X^{*}$ is the space of bounded linear functionals on $X$ (The dual space X^* of a normed space and its dual norm).

[L2]

In the real case, a bounded linear functional extends with the same norm (A bounded real linear functional on a subspace of a real normed space extends with the same norm).

[L3]

In the complex case, a bounded linear functional extends with the same norm (A bounded complex linear functional on a subspace of a complex normed space extends with the same norm).

Proof

technique · direct

1.1givenconstructalgebra

Let $M : = span {x}$ . Define $f_{0} : M \to R$ or $f_{0} : M \to C$ by $f_{0} (λ x) : = λ ∥ x ∥ .$ Because $x \neq 0$ , each vector of $M$ has a unique representation $λ x$ . Moreover, $∣ f_{0} (λ x) ∣ = ∣ λ ∣ ∥ x ∥ = ∥ λ x ∥,$ so $f_{0}$ is bounded and $∥ f_{0} ∥ = 1$ .

2.1L2step 1.1

If the scalar field is $R$ , [L2] extends $f_{0}$ to a bounded real linear functional $f$ on $X$ with $∥ f ∥ = ∥ f_{0} ∥ = 1$ . Since $x = 1 \cdot x$ , $f (x) = f_{0} (x) = ∥ x ∥$ .

2.2L3step 1.1

If the scalar field is $C$ , [L3] extends $f_{0}$ to a bounded complex linear functional $f$ on $X$ with $∥ f ∥ = ∥ f_{0} ∥ = 1$ . Again $f (x) = f_{0} (x) = ∥ x ∥$ .

3.1L1step 2.1step 2.2∎

In either scalar case, the extension produced in step 2.1 or step 2.2 is a bounded linear functional on $X$ , hence an element of $X^{*}$ by [L1], with norm $1$ and value $∥ x ∥$ at $x$ .

Every nonzero vector has a norming functional

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources