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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 MX be a linear subspace, let p:XR be sublinear, and let f:MR be linear with f(m)p(m) for every mM. Fix x0XM and put N:=M+Rx0.

For cR define

Fc(m+tx0):=f(m)+tc(mM, tR).

This is well defined, and with

α:=supyM(f(y)p(yx0)),β:=infzM(p(z+x0)f(z)),

one has αβ. Moreover, Fc(n)p(n) for every nN if and only if αcβ. 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 MX, a sublinear functional p:XR, a linear functional f:MR with fp on M, and a point x0XM.

[L1]

A sublinear functional satisfies p(u+v)p(u)+p(v) and p(tu)=tp(u) for every real t0 (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
1.1

If m+tx0=m+tx0 with m,mM and t,tR, then mm=(tt)x0M. If tt, closure under scalar multiplication from [L3] gives x0M, 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.

L3givenconstruct
1.2

For y,zM, one has Fc(yx0)=f(y)candFc(z+x0)=f(z)+c. Therefore Fc(yx0)p(yx0)    f(y)p(yx0)c, and Fc(z+x0)p(z+x0)    cp(z+x0)f(z).

L2givenalgebra
1.3

Let y,zM. Since y+zM by [L3], linearity and domination on M give f(y)+f(z)=f(y+z)p(y+z). Also y+z=(yx0)+(z+x0), so subadditivity from [L1] yields p(y+z)p(yx0)+p(z+x0). Combining these inequalities gives f(y)p(yx0)p(z+x0)f(z). Hence every lower endpoint is at most every upper endpoint.

L1L2L3givenalgebra
2.1

Suppose first that the two inequalities from step 1.2 hold for every y,zM. Let m+tx0N. 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)(mtx0), so the lower-bound half of step 1.2 applied to y:=m/t gives Fc(m+tx0)=(t)Fc(mtx0)(t)p(mtx0)=p(m+tx0). If t=0, then m+tx0=mM, so the hypothesis fp gives Fc(m)=f(m)p(m). Thus Fcp on N. Conversely, if Fcp on N, then applying that inequality to yx0 and z+x0 yields the two inequalities in step 1.2.

step 1.2L1L2givenalgebra
3.1

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 αβ 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 αcβ. Therefore the admissible values form the nonempty interval [α,β].

step 2.1step 1.3

Depends on

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