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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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The admissible values in a one-step Hahn-Banach extension form a nonempty interval

Statement

Let X be a real vector space, let M⊆X be a linear subspace, let p:X→R be sublinear, and let f:M→R be linear with f(m)≤p(m) for every m∈M. Fix x0∈X∖M and put N:=M+Rx0.

For c∈R define

Fc(m+tx0):=f(m)+tc(m∈M, t∈R).

This is well defined, and with

α:=sup⁡y∈M(f(y)−p(y−x0)),β:=inf⁡z∈M(p(z+x0)−f(z)),

one has α≤β. Moreover, Fc(n)≤p(n) for every n∈N if and only if α≤c≤β. In particular the admissible values of Fc(x0)=c form the nonempty interval [α,β].

Facts & Assumptions

Given: A real vector space X, a linear subspace M⊆X, a sublinear functional p:X→R, a linear functional f:M→R with f≤p on M, and a point x0∈X∖M.

[L1]

A sublinear functional satisfies p(u+v)≤p(u)+p(v) and p(tu)=tp(u) for every real t≥0 (A sublinear functional on a real vector space).

[L2]

A linear functional is additive and homogeneous over the scalar field (Linear functionals and the algebraic dual V∗=L(V,F)).

[L3]

A linear subspace is closed under addition and scalar multiplication (Linear subspace of a vector space).

Proof

technique · direct
1.1L3givenconstruct

If m+tx0=m′+t′x0 with m,m′∈M and t,t′∈R, then m−m′=(t′−t)x0∈M. If t≠t′, closure under scalar multiplication from [L3] gives x0∈M, contradicting the hypothesis. So t=t′ and then m=m′. Therefore every element of N has a unique representation m+tx0, and Fc is well defined.

1.2L2givenalgebra

For y,z∈M, one has Fc(y−x0)=f(y)−candFc(z+x0)=f(z)+c. Therefore Fc(y−x0)≤p(y−x0)  ⟺  f(y)−p(y−x0)≤c, and Fc(z+x0)≤p(z+x0)  ⟺  c≤p(z+x0)−f(z).

1.3L1L2L3givenalgebra

Let y,z∈M. Since y+z∈M by [L3], linearity and domination on M give f(y)+f(z)=f(y+z)≤p(y+z). Also y+z=(y−x0)+(z+x0), so subadditivity from [L1] yields p(y+z)≤p(y−x0)+p(z+x0). Combining these inequalities gives f(y)−p(y−x0)≤p(z+x0)−f(z). Hence every lower endpoint is at most every upper endpoint.

2.1step 1.2L1L2givenalgebra

Suppose first that the two inequalities from step 1.2 hold for every y,z∈M. Let m+tx0∈N. If t>0, then m+tx0=t(mt+x0), so by linearity and positive homogeneity, Fc(m+tx0)=tFc(mt+x0)≤tp(mt+x0)=p(m+tx0). If t<0, then m+tx0=(−t)(−mt−x0), so the lower-bound half of step 1.2 applied to y:=−m/t gives Fc(m+tx0)=(−t)Fc(−mt−x0)≤(−t)p(−mt−x0)=p(m+tx0). If t=0, then m+tx0=m∈M, so the hypothesis f≤p gives Fc(m)=f(m)≤p(m). Thus Fc≤p on N. Conversely, if Fc≤p on N, then applying that inequality to y−x0 and z+x0 yields the two inequalities in step 1.2.

3.1step 2.1step 1.3∎

Step 1.3 shows that the set of lower endpoints is bounded above by every upper endpoint, and the set of upper endpoints is bounded below by every lower endpoint. Completeness of R therefore gives real numbers α≤β with the displayed formulas in the statement. By step 2.1, a real number c is admissible exactly when it lies between every lower endpoint and every upper endpoint, that is, exactly when α≤c≤β. Therefore the admissible values form the nonempty interval [α,β].

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