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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.

Bounded linear maps extend uniquely across the completion

Statement

Facts & Assumptions

Proof

Depends on

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