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Hahn-Banach dominated extension theorem for real vector spaces

Statement

Assume the Axiom of Choice. Let $X$ be a real vector space, let $M \subseteq X$ be a linear subspace, let $p : X \to R$ be sublinear, and let $f : M \to R$ be linear with $f (m) \leq p (m)$ for every $m \in M$ . Then there exists a linear functional $F : X \to R$ such that $F ∣_{M} = f$ and $F (x) \leq p (x)$ for every $x \in X$ .

Facts & Assumptions

Given: The Axiom of Choice, a real vector space $X$ , a linear subspace $M \subseteq X$ , a sublinear functional $p : X \to R$ , and a linear functional $f : M \to R$ with $f \leq p$ on $M$ .

[L1]

The one-step extension problem over $M + R x_{0}$ has a nonempty interval of admissible values for $F (x_{0})$ (The admissible values in a one-step Hahn-Banach extension form a nonempty interval).

[L2]

The union of a chain of dominated extensions is again a well-defined dominated extension (The union of a chain of dominated extensions is a well-defined dominated linear functional).

[L3]

Assuming the Axiom of Choice, a nonempty poset in which every chain has an upper bound has a maximal element (Zorn's lemma).

Proof

technique · direct

1.1

Let $E$ be the set of all pairs $(N, g)$ such that $M \subseteq N \subseteq X$ , the set $N$ is a linear subspace of $X$ , the map $g : N \to R$ is linear, $g ∣_{M} = f$ , and $g \leq p$ on $N$ . Order $E$ by extension: $(N_{1}, g_{1}) ⪯ (N_{2}, g_{2}) ⟺ N_{1} \subseteq N_{2} and g_{2} ∣_{N_{1}} = g_{1} .$ The pair $(M, f)$ lies in $E$ , so this poset is nonempty.

givenconstruct

2.1

Let $C \subseteq E$ be a chain. If $C = \emptyset$ , then $(M, f)$ is an upper bound for it. If $C \neq \emptyset$ , [L2] applies to the union of its domains and yields a well-defined linear functional dominated by $p$ ; because every chain element extends $f$ , that union functional still extends $f$ . Hence every chain in $E$ has an upper bound in $E$ .

L2step 1.1given

3.1

By [L3], choose a maximal element $(N, F)$ of $E$ . If $N \neq X$ , choose $x_{0} \in X ∖ N$ . Applying [L1] to the dominated functional $F$ on the subspace $N$ produces a dominated linear extension $\tilde{F}$ on $N + R x_{0}$ . Then $(N, F) ≺ (N + R x_{0}, \tilde{F})$ , contradicting maximality. Therefore $N = X$ .

L1L3step 2.1choose

4.1

Since the maximal domain is all of $X$ , the corresponding functional $F$ is the required dominated extension of $f$ .

step 3.1∎

Depends on

Used by

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Sources

Daniel Daners, Introduction to Functional Analysis, Theorem 26.1 (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, Theorem 4.13 (standard reference, not scraped)