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

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A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $X$ be a normed space, let $D \subseteq X$ be a dense normed subspace, let $Y$ be a Banach space, and let $T : D \to Y$ be a bounded linear operator. Then there is a unique bounded linear operator $\tilde{T} : X \to Y$ such that

$\tilde{T} ∣_{D} = T,$

and $∥ \tilde{T} ∥ = ∥ T ∥$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, a normed space $X$ , a dense normed subspace $D \subseteq X$ , a Banach space $Y$ , and a bounded linear operator $T : D \to Y$ .

[L0]

Countable Choice is assumed (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

A bounded linear operator has a constant $C \geq 0$ with $∥ T u ∥ \leq C ∥ u ∥$ for all $u$ , and it is continuous (A bounded linear operator between normed spaces, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent).

[L2]

A Banach space is complete for its norm metric (Banach space).

[L3]

A normed subspace carries the restricted norm, and density means every ball in $X$ meets $D$ (Normed subspace, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[L4]

Limits in a metric space are unique, and addition and scalar multiplication are continuous in normed spaces (A sequence in a metric space has at most one limit, Vector addition and scalar multiplication are continuous in a normed space).

Proof

technique · direct

1.1

Fix $x \in X$ . By [L0] and density in [L3], for each $n \geq 1$ choose $d_{n} (x) \in D$ with $∥ d_{n} (x) - x ∥ < 1 / n$ . This is the selected ${A C}_{ω}$ step: one approximating sequence for each fixed point $x$ .

L0L3choose

2.1

Let $C$ be a bound for $T$ from [L1]. Then $∥ T d_{n} (x) - T d_{m} (x) ∥ \leq C ∥ d_{n} (x) - d_{m} (x) ∥ \leq C (∥ d_{n} (x) - x ∥ + ∥ d_{m} (x) - x ∥)$ , so $(T d_{n} (x))$ is Cauchy in $Y$ . By [L2] it converges. Define $\tilde{T} (x) : = \lim_{n} T d_{n} (x)$ .

step 1.1L1L2choose

3.1

The value in step 2.1 is independent of the chosen approximating sequence. If $e_{n} (x) \in D$ also satisfies $e_{n} (x) \to x$ , then $∥ T d_{n} (x) - T e_{n} (x) ∥ \leq C ∥ d_{n} (x) - e_{n} (x) ∥ \leq C (∥ d_{n} (x) - x ∥ + ∥ e_{n} (x) - x ∥) \to 0$ , so the two image sequences have the same limit by [L4].

step 2.1L1L4algebra

3.2

Uniqueness: if $S : X \to Y$ is another bounded linear extension of $T$ , then $S$ is continuous by [L1]. For every $x \in X$ , the sequence $d_{n} (x)$ of step 1.1 lies in $D$ , so $S d_{n} (x) = T d_{n} (x) \to \tilde{T} (x)$ by step 2.1 and also $S d_{n} (x) \to S x$ by continuity of $S$ . By [L4], $S x = \tilde{T} (x)$ . Thus $S = \tilde{T}$ .

step 1.1step 2.1L1L4

4.1

If $x \in D$ , choose the constant approximating sequence $d_{n} (x) = x$ . Then step 2.1 gives $\tilde{T} (x) = T x$ , so $\tilde{T}$ extends $T$ .

step 2.1step 3.1

4.2

To prove linearity, let $x, y \in X$ and choose the approximating sequences of step 1.1 for them. Then $d_{n} (x) + d_{n} (y) \to x + y$ and $λ d_{n} (x) \to λ x$ by [L4]. Using step 3.1 to replace the chosen sequence at $x + y$ by $d_{n} (x) + d_{n} (y)$ , and similarly at $λ x$ , we get $\tilde{T} (x + y) = \lim_{n} T (d_{n} (x) + d_{n} (y)) = \lim_{n} (T d_{n} (x) + T d_{n} (y))$ and $\tilde{T} (λ x) = \lim_{n} λ T d_{n} (x)$ . Continuity of addition and scalar multiplication from [L4] lets the limit pass through, so $\tilde{T} (x + y) = \tilde{T} (x) + \tilde{T} (y)$ and $\tilde{T} (λ x) = λ \tilde{T} (x)$ .

step 1.1step 3.1L4construct

5.1

The same bound $C$ works for $\tilde{T}$ . Indeed, with the sequence of step 1.1, $∥ T d_{n} (x) ∥ \leq C ∥ d_{n} (x) ∥ \leq C (∥ x ∥ + ∥ d_{n} (x) - x ∥) < C (∥ x ∥ + 1 / n)$ . Given $ε > 0$ , choose $n$ large enough that $∥ \tilde{T} (x) - T d_{n} (x) ∥ < ε$ and $1 / n < ε / C$ when $C > 0$ ; then $∥ \tilde{T} (x) ∥ \leq C ∥ x ∥ + 2 ε$ . Hence $∥ \tilde{T} (x) ∥ \leq C ∥ x ∥$ for all $x$ , so $\tilde{T}$ is bounded and $∥ \tilde{T} ∥ \leq ∥ T ∥$ . Since $\tilde{T}$ agrees with $T$ on $D$ , also $∥ T ∥ \leq ∥ \tilde{T} ∥$ . Therefore $∥ \tilde{T} ∥ = ∥ T ∥$ .

step 2.1step 4.1L1algebra

6.1

Steps 4.1, 4.2, 5.1, and 4.1 prove that $\tilde{T}$ is the unique bounded linear extension of $T$ and that it has the same norm.

step 4.1step 4.2step 5.1step 3.2∎

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Sources

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