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 $x$ .

Depends on

Used by

The dual space separates points of a normed space Corollary
The norm of a vector is the supremum of |f(x)| over the dual unit ball Corollary
Banach mean value estimate on a convex set Lemma
Elementary kernel and range annihilator identities Lemma
The transpose is bounded with the same norm Lemma
Atkinson Theorem
The canonical bidual map is an isometry Theorem

Dependency tree · two levels

16 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, Corollary 26.5 (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, Section 4.2 (standard reference, not scraped)