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ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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A codimension-one subspace can admit many norm-preserving Hahn-Banach extensions

Example

Let X=(R2,), let M:=span{(1,1)}, and define f0:MR by f0(t,t)=t. Then f0=1, and for every c[1,1] the formula

Fc(a,b):=1+c2a+1c2b

defines a norm-preserving extension of f0 to all of X. So a codimension-one subspace can have infinitely many Hahn-Banach extensions of the same norm.

Facts & Assumptions

Given: The normed space X=(R2,), the diagonal subspace M=span{(1,1)}, and the functional f0(t,t)=t.

[L1]

In a one-step Hahn-Banach extension, the admissible values form a nonempty interval (The admissible values in a one-step Hahn-Banach extension form a nonempty interval).

[L2]

Verification

technique · direct
1.1

For (t,t)M one has f0(t,t)=t=(t,t), so f0=1. Also every (a,b)R2 decomposes as (a,b)=a+b2(1,1)+ab2(1,1), so X=MR(1,1).

givenalgebra
2.1

If F is a linear extension of f0 and c:=F(1,1), then step 1.1 forces F(a,b)=a+b2F(1,1)+ab2F(1,1)=1+c2a+1c2b. Conversely, the displayed formula defines a linear functional extending f0.

step 1.1givenconstructalgebra
3.1

Let φ(a,b):=αa+βb on (R2,). If (a,b)1, then φ(a,b)αa+βbα+β. Choosing a=sgnα and b=sgnβ (interpreting sgn0=0) gives (a,b)1 and φ(a,b)=α+β, so φ=α+β. Applying this to the formula from step 2.1 yields Fc=1+c+1c2. This equals 1 exactly when 1c1.

step 2.1algebra
4.1

Therefore every c[1,1] gives a norm-preserving extension of f0. This computes explicitly the admissible interval predicted abstractly by [L1], and in particular is consistent with the existence statement of [L2].

L1L2step 3.1

Depends on

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