TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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Bounded linear maps extend uniquely across the completion

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $X$ be a normed space, let $(\hat{X}, i)$ be its completion, and let $Y$ be a Banach space. Suppose $T : X \to Y$ is linear and bounded in the concrete sense that some constant $C \geq 0$ satisfies $∥ T x ∥ \leq C ∥ x ∥ (x \in X) .$ Then there is a unique bounded linear map $\hat{T} : \hat{X} \to Y$ with $\hat{T} \circ i = T,$ and the same constant $C$ satisfies $∥ \hat{T} u ∥ \leq C ∥ u ∥ (u \in \hat{X}) .$

Facts & Assumptions

Given: The Axiom of Countable Choice; a normed space $X$ , its completion $(\hat{X}, i)$ , a Banach space $Y$ , a linear map $T : X \to Y$ , and a constant $C \geq 0$ with $∥ T x ∥ \leq C ∥ x ∥$ for all $x \in X$ .

[A1]

Countable Choice selects one point from each member of a countable family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

The completion $(\hat{X}, i)$ is a Banach space and $i [X]$ is dense in $\hat{X}$ (The metric completion of a normed space carries a unique compatible Banach-space structure, Completion of a normed space).

[L2]

A bound $∥ T x ∥ \leq C ∥ x ∥$ implies $∥ T x - T y ∥ \leq C ∥ x - y ∥$ , so $T$ is Lipschitz and hence uniformly continuous (Linear map between vector spaces over the same field, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent).

[L3]

Assuming Countable Choice, a uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a continuous map on the whole space (A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space).

[L4]

Addition and scalar multiplication are continuous in every normed space, hence in the Banach spaces $\hat{X}$ and $Y$ (Vector addition and scalar multiplication are continuous in a normed space).

Proof

technique · direct

1.1L2

For $x, y \in X$ , linearity gives $T x - T y = T (x - y)$ , so the bound in [L2] yields $∥ T x - T y ∥ \leq C ∥ x - y ∥$ . Thus $T$ is Lipschitz and uniformly continuous.

2.1step 1.1L1

Define $S : i [X] \to Y$ by $S (i (x)) : = T (x)$ . This is well defined because $i$ is injective as a linear isometry, and $S$ is uniformly continuous with the same constant $C$ .

3.1step 2.1A1L1L3

Since $i [X]$ is dense in the Banach space $\hat{X}$ and $Y$ is complete, [A1] and [L3] give a unique continuous map $\hat{T} : \hat{X} \to Y$ with $\hat{T} ⁣ ↾_{i [X]} = S$ , equivalently $\hat{T} \circ i = T$ .

4.1step 3.1A1L1L4construct

Let $u, v \in \hat{X}$ . Density and [A1] supply sequences $i (x_{n}) \to u$ and $i (y_{n}) \to v$ from the dense copy. By [L4], $i (x_{n} + y_{n}) = i (x_{n}) + i (y_{n}) \to u + v$ in $\hat{X}$ , so continuity of $\hat{T}$ gives $\hat{T} (u + v) = \lim_{n} \hat{T} (i (x_{n} + y_{n})) = \lim_{n} T (x_{n} + y_{n}) = \lim_{n} (T x_{n} + T y_{n}) .$ Using linearity of $T$ and continuity of addition in $Y$ from [L4], the right-hand side is $\lim_{n} T x_{n} + \lim_{n} T y_{n} = \hat{T} (u) + \hat{T} (v) .$ Hence $\hat{T} (u + v) = \hat{T} (u) + \hat{T} (v)$ .

4.2step 1.1step 3.1A1L1

For $u \in \hat{X}$ , density and [A1] supply a sequence $i (x_{n}) \to u$ from the dense copy. Continuity gives $\hat{T} (u) = \lim_{n} T (x_{n})$ , and step 1.1 yields $∥ T (x_{n}) ∥ \leq C ∥ x_{n} ∥ = C ∥ i (x_{n}) ∥$ for every $n$ ; passing to the limit gives $∥ \hat{T} (u) ∥ \leq C ∥ u ∥$ . So $\hat{T}$ is bounded with the same constant.

5.1step 3.1A1L1L4construct

Let $λ$ be a scalar and let $u \in \hat{X}$ , and use [A1] with density to choose a sequence $i (x_{n}) \to u$ from the dense copy. By [L4], $i (λ x_{n}) = λ i (x_{n}) \to λ u$ in $\hat{X}$ , so continuity of $\hat{T}$ gives $\hat{T} (λ u) = \lim_{n} \hat{T} (i (λ x_{n})) = \lim_{n} T (λ x_{n}) = \lim_{n} λ T (x_{n}) = λ \lim_{n} T (x_{n}) = λ \hat{T} (u) .$ Thus $\hat{T}$ is homogeneous, and with step 4.1 it is linear.

6.1step 3.1step 5.1∎

The continuity extension in step 3.1 was unique, so no second bounded linear map can also extend $T$ . Therefore $\hat{T}$ is the unique bounded linear extension of $T$ across the completion.

Depends on

Used by

Dependency tree · two levels

53 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

Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)