CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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

Statement

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: A normed space $X$ over $R$ or $C$ and a vector $x \in X$ .

[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

1.1

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 $∥ x ∥$ .

L1given

2.1

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 $∣ g (x) ∣ = ∥ x ∥$ .

L2givenchoose

3.1

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

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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)