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

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The norm of a vector is the supremum of |f(x)| over the dual unit ball

Statement

Assume the Axiom of Choice. Let $X$ be a normed space over $R$ or $C$ . Then for every $x \in X$ ,

$∥ x ∥ = \sup {∣ f (x) ∣ : f \in X^{*}, ∥ f ∥ \leq 1} .$

Facts & Assumptions

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

[A1]

AC is used through the norming functional [L2] in step 2.1.

[L1]

The dual norm on $X^{*}$ is the operator norm, so $∣ f (x) ∣ \leq ∥ f ∥ ∥ x ∥$ for every $f \in X^{*}$ (The dual space X^* of a normed space and its dual norm).

[L2]

Every nonzero vector admits a norming functional (Every nonzero vector has a norming functional).

Proof

technique · direct

1.1L1given

If $f \in X^{*}$ and $∥ f ∥ \leq 1$ , then [L1] gives $∣ f (x) ∣ \leq ∥ f ∥ ∥ x ∥ \leq ∥ x ∥ .$ So the displayed supremum is at most $∥ x ∥$ .

2.1L2givenchoose

If $x = 0$ , step 1.1 already shows that the supremum is $0$ , so the formula holds. Assume now that $x \neq 0$ . By [L2], choose $g \in X^{*}$ with $∥ g ∥ = 1$ and $g (x) = ∥ x ∥$ . Then the displayed supremum is at least $∣ g (x) ∣ = ∥ x ∥$ .

3.1step 1.1step 2.1∎

Step 1.1 gives an upper bound of $∥ x ∥$ , and step 2.1 gives equality. So $∥ x ∥ = \sup {∣ f (x) ∣ : f \in X^{*}, ∥ f ∥ \leq 1} .$

Depends on

Used by

Spectral radius formula Theorem

Dependency tree · two levels

10 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, Remark 25.2(c) (standard reference, not scraped)
Daniel Daners, Introduction to Functional Analysis, Corollary 26.5 (standard reference, not scraped)