TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Sokal's gliding-hump proof of uniform boundedness

Statement

Assume ${A C}_{ω}$ and DC. If $X$ is Banach, $Y$ is normed, and a family $F$ of bounded linear maps $X \to Y$ is pointwise bounded, then $\sup_{T \in F} ∥ T ∥ < \infty$ .

Facts & Assumptions

Given: The stated choice principles, $X, Y, F$ , and pointwise boundedness.

Proof

technique · constructive

1.1

Suppose the norms are unbounded. Countable choice selects $T_{n} \in F$ with $∥ T_{n} ∥ \geq 4^{n}$ for $n \geq 1$ . Set $x_{0} = 0$ .

givenconstruct

2.1

Recursively, apply A nonzero bounded linear operator is large on one of two nearby points with centre $x_{n - 1}$ and radius $3^{- n}$ to choose $x_{n}$ with $∥ x_{n} - x_{n - 1} ∥ < 3^{- n}$ and $∥ T_{n} x_{n} ∥ > (2 / 3) 3^{- n} ∥ T_{n} ∥$ . DC licenses these dependent choices.

step 1.1construct

3.1

The sequence $(x_{n})$ is Cauchy, since its tails are bounded by a tail of $\sum_{n \geq 1} 3^{- n}$ ; completeness gives $x_{n} \to x \in X$ . Moreover $∥ x - x_{n} ∥ \leq \sum_{k > n} 3^{- k} = 3^{- n} / 2$ .

step 2.1

4.1

Hence $∥ T_{n} x ∥ \geq ∥ T_{n} x_{n} ∥ - ∥ T_{n} ∥ ∥ x - x_{n} ∥ > \frac{1}{6} 3^{- n} ∥ T_{n} ∥ \geq \frac{1}{6} (4 / 3)^{n},$ which contradicts pointwise boundedness at $x$ .

step 1.1step 2.1step 3.1algebradischarge-construct∎

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Dependency tree · two levels

15 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

Sokal, A Really Simple Elementary Proof of the Uniform Boundedness Theorem, pp. 1--3 (standard reference, not scraped)