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

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Uniform boundedness principle

Statement

Assume DC. Let $X$ be a Banach space, $Y$ a normed space, and let $F$ be a family of bounded linear operators $X \to Y$ (A bounded linear operator between normed spaces). If $\sup_{T \in F} ∥ T x ∥ < \infty$ for every $x \in X$ , then $\sup_{T \in F} ∥ T ∥ < \infty$ , where the norm is The operator norm as the least bound and as the unit-sphere or unit-ball supremum.

Facts & Assumptions

Given: DC, $X, Y, F$ as in the statement, and pointwise boundedness.

Proof

technique · direct

1.1given

Put $E_{n} = {x : \sup_{T \in F} ∥ T x ∥ \leq n}$ . Each $E_{n}$ is closed (an intersection of inverse images of closed balls), and pointwise boundedness gives $X = ⋃_{n \geq 1} E_{n}$ .

2.1step 1.1

By Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior, some $E_{N}$ contains a ball $B (x_{0}, r)$ .

3.1step 2.1

For $∥ h ∥ < r$ , both $x_{0}$ and $x_{0} + h$ lie in that ball, so $∥ T h ∥ \leq ∥ T (x_{0} + h) ∥ + ∥ T x_{0} ∥ \leq 2 N$ for every $T$ .

4.1step 3.1algebra∎

Rescaling a nonzero $x$ to $h = r x / (2 ∥ x ∥)$ yields $∥ T x ∥ \leq 4 N ∥ x ∥ / r$ ; the same is clear for $x = 0$ . Thus $∥ T ∥ \leq 4 N / r$ for every $T$ .

Depends on

Dependency tree · 2 levels

Sources

Buhler--Salamon, Functional Analysis, Theorem 2.1