CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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The c0 unit ball has no extreme points

Statement

Over either $R$ or $C$ , the closed unit ball of $c_{0}$ has no extreme points.

Facts & Assumptions

Given: The real or complex space $c_{0}$ and an arbitrary $x$ in its closed unit ball.

[F1]

A point is extreme only if every strict two-term convex representation is trivial (Extreme point and face).

[F2]

The elements of $c_{0}$ are bounded scalar sequences tending to zero, with the supremum norm (The sequence spaces c_0 and ell-infinity).

Proof

technique · direct finite-coordinate perturbation

1.1

Since $x_{n} \to 0$ , there is an index $N$ with $∣ x_{N} ∣ < 1$ ; take the least such index if a canonical witness is desired. Put $ε = (1 - ∣ x_{N} ∣) / 2 > 0$ and let $e_{N}$ be the sequence equal to one at $N$ and zero elsewhere.

F2given

2.1

The sequences $y = x + ε e_{N}$ and $z = x - ε e_{N}$ still tend to zero and satisfy $∣ y_{N} ∣, ∣ z_{N} ∣ \leq ∣ x_{N} ∣ + ε < 1$ , while every other coordinate is unchanged. Hence $y, z \in B_{c_{0}}$ by [F2]; they are distinct and $x = (y + z) / 2$ .

F2step 1.1

3.1

By [F1], the nontrivial midpoint representation in step 2.1 shows that the arbitrary $x \in B_{c_{0}}$ is not extreme. Therefore the ball has no extreme points over either scalar field.

F1step 2.1∎

Depends on

Used by

c0 is not isometrically a dual space Corollary

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)