ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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Quantitative separation of a norm ball from an exterior point

Example

Assume HB. Let $X$ be a real or complex normed space, $a, z \in X$ , and $r \geq 0$ with $d = ∥ z - a ∥ > r$ . There is $f \in X^{*}$ with $∥ f ∥ = 1$ and $f (z - a) = d$ . For every $x$ in the closed ball $∥ x - a ∥ \leq r$ , $Re f (x) \leq Re f (a) + r < Re f (z) .$ The gap between the displayed upper bound and exterior-point value is $d - r > 0$ . Its midpoint $b = Re f (a) + (r + d) / 2$ gives uniform margin $ε = (d - r) / 2$ . For the open ball with $r > 0$ , the left bound is strict.

Facts & Assumptions

[F1]

Under HB every nonzero vector $v$ has a norm-one functional with value $∥ v ∥$ (Relative dual norming, point separation, and recovery of the norm).

[F2]

Writing $u = Re f$ , a separator with uniform margin $ε > 0$ satisfies $u (x) \leq b - ε < b + ε \leq u (y)$ (Convex sets and continuous real-hyperplane separation in a normed space).

Verification

Given: HB, $a, z \in X$ , $r \geq 0$ , and $d = ∥ z - a ∥ > r$ .

1.1

Since $d = ∥ z - a ∥ > r \geq 0$ , $z - a \neq 0$ . Apply dual norming to this vector to get $∥ f ∥ = 1$ and $f (z - a) = d$ , a positive real. Put $u = Re f$ . Linearity gives $u (z) = u (a) + d$ .

givenF1algebra

2.1

If $∥ x - a ∥ \leq r$ , then $u (x) - u (a) = Re f (x - a) \leq ∣ f (x - a) ∣ \leq ∥ f ∥ ∥ x - a ∥ \leq r$ . Thus $u (x) \leq u (a) + r < u (a) + d = u (z)$ . If $∥ x - a ∥ < r$ with $r > 0$ , the same chain gives $u (x) < u (a) + r$ .

step 1.1F2algebra

3.1

Set $b = u (a) + (r + d) / 2$ and $ε = (d - r) / 2 > 0$ . Direct subtraction and addition give $b - ε = u (a) + r$ and $b + ε = u (a) + d = u (z)$ . Hence step 2.1 gives $u (x) \leq b - ε < b + ε = u (z)$ , the prescribed uniform margin. For $r = 0$ , the closed ball is ${a}$ by norm definiteness and these formulas give margin $d / 2 > 0$ .

step 1.1step 2.1F2algebra

4.1

For a numerical instance, take $X = R$ , $a = 0$ , $r = 1$ , $z = 3$ , and $f (t) = t$ . Then $∥ f ∥ = \sup_{∣ t ∣ \leq 1} ∣ t ∣ = 1$ , $d = 3$ , $b = 2$ , and $ε = 1$ . Every $x \in [- 1, 1]$ satisfies $f (x) = x \leq 1 = b - ε < 3 = b + ε = f (3)$ ; at $x = 1$ the left bound is attained. This explicitly realizes gap two and margin one.

step 3.1algebra∎

Source notes

Brezis Corollary 1.3 and Theorem 1.7, pp.3,7 (quantitative specialization); Teschl Theorem 4.20 proof and Corollary 5.4, pp.116,140.

Depends on

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Sources

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, §§1.1–1.2 and §1.3 evaluation paragraph (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, Theorems 4.13–4.20 and §5.1 (2018 university-hosted copy) (standard reference, not scraped)