CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Hahn-Banach norm-preserving extensions need not be unique

Statement refuted

For a bounded linear functional on a subspace of a normed space, a norm-preserving Hahn-Banach extension to the whole space is unique.

Facts & Assumptions

Given: The diagonal subspace $M = span {(1, 1)} \subseteq (R^{2}, ∥ \cdot ∥_{\infty})$ and the functional $f_{0} (t, t) = t$ on $M$ .

[L1]

For every $c \in [- 1, 1]$ , the functional $F_{c} (a, b) = \frac{1 + c}{2} a + \frac{1 - c}{2} b$ is a norm-preserving extension of $f_{0}$ (A codimension-one subspace can admit many norm-preserving Hahn-Banach extensions).

Counterexample

technique · direct

1.1

By [L1], the functionals $F_{1} (a, b) = a and F_{- 1} (a, b) = b$ are both norm-preserving extensions of $f_{0}$ .

L1given

2.1

They agree on $M$ , since $F_{1} (t, t) = t = F_{- 1} (t, t)$ for every $(t, t) \in M$ , but they differ on $(1, - 1)$ : $F_{1} (1, - 1) = 1 \neq - 1 = F_{- 1} (1, - 1) .$

step 1.1givenalgebra

3.1

Thus the same functional on the same subspace has two distinct norm-preserving Hahn-Banach extensions. Therefore the uniqueness claim is false.

step 1.1step 2.1∎

Depends on

A codimension-one subspace can admit many norm-preserving Hahn-Banach extensions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Gerald Teschl, Topics in Real and Functional Analysis, Corollary 4.15 (standard reference, not scraped)