CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-07

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A pointwise limit of bounded operators is bounded with the liminf norm bound

Statement

Assume DC. Let $X$ be Banach, $Y$ normed, and $T_{n} : X \to Y$ bounded linear (A bounded linear operator between normed spaces). If $T_{n} x \to T x$ in $Y$ for every $x \in X$ , then $T$ is bounded linear and

$∥ T ∥ \leq {lim inf}_{n \to \infty} ∥ T_{n} ∥,$

with the operator norm of The operator norm as the least bound and as the unit-sphere or unit-ball supremum.

Facts & Assumptions

Given: DC and $X, Y, (T_{n}), T$ with pointwise convergence as in the statement.

Proof

technique · direct

1.1given

Pointwise convergence makes $(T_{n} x)_{n}$ bounded for each $x$ . Thus Uniform boundedness principle gives $M : = \sup_{n} ∥ T_{n} ∥ < \infty$ .

2.1step 1.1

Passing $T_{n} (x + y) = T_{n} x + T_{n} y$ and $T_{n} (λ x) = λ T_{n} x$ to limits shows that $T$ is linear.

2.2step 1.1

For each $x$ , continuity of the norm gives $∥ T x ∥ = \lim_{n} ∥ T_{n} x ∥ \leq M ∥ x ∥$ , so $T$ is bounded.

3.1step 2.2∎

If ${lim inf}_{n} ∥ T_{n} ∥ = \infty$ , the asserted bound is immediate. Otherwise select a subsequence whose norms tend to the finite liminf. The preceding inequality along that subsequence gives $∥ T x ∥ \leq ({lim inf}_{n} ∥ T_{n} ∥) ∥ x ∥$ for every $x$ , hence the asserted norm bound.

Depends on

Used by

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Sources

Buhler--Salamon, Functional Analysis, Theorem 2.5 (standard reference, not scraped)