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✓ 16 results · all verified · 8 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Bounded Linear Operators and Quotient Spaces

1 · Prerequisites

2 · Summary

This page packages the first functional-analysis uses of norm completeness: bounded linear and bilinear maps, the operator norm and the Banach space $B (X, Y)$ , and quotient norms with their factorization property. Under Countable Choice it proves dense extension into Banach targets and quotient completeness; under Dependent Choice it proves the one-sided inverse criteria. The quotient proofs keep the $ε$ -minimizer route explicit, so no best-approximation theorem is silently assumed.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-01Open item page →

A bounded linear operator between normed spaces

Definition

Let $X$ and $Y$ be normed spaces over the same scalar field $K$ , read in the real case from A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and in the complex case from Real and complex scalar conventions for normed spaces. A linear map $T : X \to Y$ (Linear map between vector spaces over the same field) is a bounded linear operator when there is a real constant $C \geq 0$ such that

$∥ T x ∥ \leq C ∥ x ∥ for all x \in X .$

Any such $C$ is called a bound for $T$ .

Remarks

The zero operator is bounded with bound $0$ .
A bound is not unique: if $C$ works and $C^{'} \geq C$ , then $C^{'}$ works as well.
When $X = {0}$ , every linear map $X \to Y$ is bounded with bound $0$ .

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field, and let $T : X \to Y$ be linear. Then the following are equivalent:

$T$ is bounded.
$T$ is continuous at $0$ .
$T$ is continuous on $X$ .
$T$ is Lipschitz.

Facts & Assumptions

Given: Normed spaces $X$ and $Y$ , a linear map $T : X \to Y$ , a real $ε > 0$ , and a vector $x_{0} \in X$ .

[L1]

A bounded linear operator has a constant $C \geq 0$ with $∥ T x ∥ \leq C ∥ x ∥$ for every $x \in X$ (A bounded linear operator between normed spaces).

[L2]

A map is Lipschitz when one constant controls all distances, and every Lipschitz map between metric spaces is continuous (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]

Continuity at a point in a metric space is the $ε$ - $δ$ condition of Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form.

Proof

technique · direct

1.1L1L2

Assume $T$ is bounded, with constant $C$ from [L1]. Then $∥ T x - T y ∥ = ∥ T (x - y) ∥ \leq C ∥ x - y ∥$ for all $x, y \in X$ , so $T$ is Lipschitz.

1.2L3

If $T$ is continuous on $X$ , then in particular it is continuous at $0$ , so $(3) \Rightarrow (2)$ .

1.3L3choose

Assume $T$ is continuous at $0$ . Applying [L3] with $ε = 1$ gives $δ > 0$ such that $∥ u ∥ < δ$ implies $∥ T u ∥ < 1$ .

2.1step 1.1L2

Step 1.1 proves $(1) \Rightarrow (4)$ , and [L2] gives $(4) \Rightarrow (3)$ .

2.2step 1.3L1algebra

Let $x \in X$ with $x \neq 0$ and put $u : = \frac{δ}{2 ∥ x ∥} x$ . Then $∥ u ∥ = δ / 2 < δ$ , so $∥ T u ∥ < 1$ by step 1.3. By linearity, $T u = \frac{δ}{2 ∥ x ∥} T x$ , hence $∥ T x ∥ < \frac{2}{δ} ∥ x ∥$ . The same inequality is trivial at $x = 0$ , so $T$ is bounded.

3.1step 2.1step 1.2step 2.2∎

Thus $(2) \Rightarrow (1)$ . Combining steps 2.1, 1.2, and 2.2 gives all four equivalences.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-09-09 (codex)Open item page →

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

Definition

Let $X$ and $Y$ be normed spaces over the same scalar field, and let $T : X \to Y$ be bounded in the sense of A bounded linear operator between normed spaces. The operator norm of $T$ is

$∥ T ∥ : = \sup {∥ T x ∥ : x \in X, ∥ x ∥ \leq 1} .$

The set on the right contains $0$ and every bound $C$ for $T$ bounds it above. Thus its finite real supremum exists by the proved least-upper-bound property of the library's real field (The Cauchy-sequence reals have the least-upper-bound property). In particular every number strictly below the supremum is exceeded by a member of this set; otherwise it would be a smaller upper bound. No choice principle is used for this assertion.

The same number is the least bound of $T$ :

$∥ T ∥ = \inf {C \geq 0 : ∥ T x ∥ \leq C ∥ x ∥ for all x \in X} .$

If $X \neq {0}$ , positive homogeneity also gives

$∥ T ∥ = \sup {∥ T x ∥ : ∥ x ∥ = 1} .$

When $X = {0}$ , the unit sphere is empty and the unit-ball supremum is $0$ , so $∥ T ∥ = 0$ .

Remarks

The inequality $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ holds for every $x \in X$ by the unit-ball definition and rescaling.
The unit-ball formula is the one used uniformly below, because it also covers the zero-space case without a separate convention.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The operator norm is a norm on the space of bounded linear operators

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field. On the vector space $B (X, Y)$ of bounded linear operators, the operator norm of The operator norm as the least bound and as the unit-sphere or unit-ball supremum is a norm.

Facts & Assumptions

Given: Bounded linear operators $S, T \in B (X, Y)$ , a scalar $λ$ , and a vector $x \in X$ .

[L1]

The operator norm is the supremum of $∥ T x ∥$ over the unit ball of $X$ , and it satisfies $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x \in X$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L2]

$B (X, Y)$ is the space of bounded linear operators, with pointwise addition and scalar multiplication (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

Proof

technique · direct

1.1L1algebra

By [L1], $∥ T ∥ \geq 0$ for every $T \in B (X, Y)$ . If $∥ T ∥ = 0$ , then $∥ T x ∥ \leq 0$ for every $x$ with $∥ x ∥ \leq 1$ , so $T x = 0$ there. If $x \neq 0$ , apply this to $u : = x / ∥ x ∥$ to get $T u = 0$ , hence $T x = ∥ x ∥ T u = 0$ . Thus $∥ T ∥ = 0$ implies $T = 0$ .

1.2L1

Conversely, if $T = 0$ then $∥ T x ∥ = 0$ for every $x$ in the unit ball, so the supremum in [L1] is $0$ . This proves definiteness.

1.3L1L2algebra

For every $x$ with $∥ x ∥ \leq 1$ , [L2] and [L1] give $∥ (λ T) x ∥ = ∣ λ ∣ ∥ T x ∥ \leq ∣ λ ∣ ∥ T ∥$ . Taking the supremum over the unit ball yields $∥ λ T ∥ \leq ∣ λ ∣ ∥ T ∥$ . The reverse inequality is immediate when $λ = 0$ , and for $λ \neq 0$ the same estimate applied to $T = λ^{- 1} (λ T)$ gives $∥ T ∥ \leq ∣ λ ∣^{- 1} ∥ λ T ∥$ . Hence $∥ λ T ∥ = ∣ λ ∣ ∥ T ∥$ .

1.4L1L2algebra

For every $x$ with $∥ x ∥ \leq 1$ , [L2] and [L1] give $∥ (S + T) x ∥ \leq ∥ S x ∥ + ∥ T x ∥ \leq ∥ S ∥ + ∥ T ∥$ . Taking the supremum over the unit ball gives $∥ S + T ∥ \leq ∥ S ∥ + ∥ T ∥$ .

2.1step 1.1step 1.2step 1.3step 1.4∎

Steps 1.1, 1.2, 1.3, and 1.4 are the norm axioms, so the operator norm is a norm on $B (X, Y)$ .

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

Composition satisfies |ST|\le|S|,|T|

Statement

Let $X$ , $Y$ , and $Z$ be normed spaces over the same scalar field. If $T \in B (X, Y)$ and $S \in B (Y, Z)$ , then

$∥ S T ∥ \leq ∥ S ∥ ∥ T ∥ .$

Facts & Assumptions

Given: Bounded linear operators $T : X \to Y$ and $S : Y \to Z$ .

[L1]

The operator norm is the unit-ball supremum and satisfies $∥ T u ∥ \leq ∥ T ∥ ∥ u ∥$ for every vector $u$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L2]

Bounded linear operators compose to a linear map, and $B (X, Y)$ denotes the bounded ones (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

Proof

technique · direct

1.1L1L2

Let $x \in X$ satisfy $∥ x ∥ \leq 1$ . Then [L1] gives $∥ T x ∥ \leq ∥ T ∥$ , and applying [L1] again to $S$ yields $∥ S T x ∥ \leq ∥ S ∥ ∥ T x ∥ \leq ∥ S ∥ ∥ T ∥$ .

2.1step 1.1L1∎

Step 1.1 holds for every $x$ in the unit ball of $X$ , so taking the supremum over that ball gives $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators

Definition

Let $X$ and $Y$ be normed spaces over the same scalar field. Write

$B (X, Y) : = {T : X \to Y : T is a bounded linear operator} .$

This is a subspace of the vector space $L (X, Y)$ of The space $L (V, W)$ of linear maps with pointwise addition and scalar multiplication, with the same pointwise operations:

$(S + T) (x) : = S x + T x, (λ T) (x) : = λ T x .$

If $X = Y$ , write

$B (X) : = B (X, X) .$

Remarks

Boundedness is preserved by the two pointwise operations, because the triangle inequality and absolute homogeneity combine the corresponding bounds.
The operator norm of The operator norm as the least bound and as the unit-sphere or unit-ball supremum is defined on $B (X, Y)$ , not on all of $L (X, Y)$ .

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

If (Y) is Banach then (\mathcal B(X,Y)) is Banach

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field. If $Y$ is Banach, then $B (X, Y)$ is Banach for the operator norm.

Facts & Assumptions

Given: A Banach space $Y$ and an operator-norm Cauchy sequence $(T_{n})$ in $B (X, Y)$ .

[L1]

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

[L2]

For a bounded operator, the operator norm is the unit-ball supremum and satisfies $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L3]

$B (X, Y)$ is the vector space of bounded linear operators (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

Proof

technique · direct

1.1L1L2choose

Fix $x \in X$ . Since $(T_{n})$ is Cauchy in operator norm, [L2] gives $∥ T_{n} x - T_{m} x ∥ \leq ∥ T_{n} - T_{m} ∥ ∥ x ∥$ , so $(T_{n} x)$ is a Cauchy sequence in $Y$ . Because $Y$ is Banach, there is $T x \in Y$ with $T_{n} x \to T x$ .

2.1step 1.1L3algebra

Step 1.1 defines a map $T : X \to Y$ . If $x, y \in X$ , then $T_{n} (x + y) = T_{n} x + T_{n} y$ for every $n$ , so passing to the limit gives $T (x + y) = T x + T y$ . The same argument with $λ x$ gives $T (λ x) = λ T x$ . Thus $T$ is linear.

2.2step 1.1L2L3

Choose $N$ such that $∥ T_{n} - T_{m} ∥ \leq 1$ for all $m, n \geq N$ . Fix $n \geq N$ and $x \in X$ . For every $m \geq N$ , $∥ T_{m} x ∥ \leq ∥ T_{N} x ∥ + ∥ (T_{m} - T_{N}) x ∥ \leq (∥ T_{N} ∥ + 1) ∥ x ∥$ by [L2]. Letting $m \to \infty$ in step 1.1 gives $∥ T x ∥ \leq (∥ T_{N} ∥ + 1) ∥ x ∥$ , so $T$ is bounded and hence $T \in B (X, Y)$ .

2.3step 1.1L2

Let $ε > 0$ . Since $(T_{n})$ is operator-norm Cauchy, choose $N$ so that $∥ T_{n} - T_{m} ∥ < ε$ for all $m, n \geq N$ . Fix $n \geq N$ and $x$ with $∥ x ∥ \leq 1$ . Step 1.1 gives $T_{m} x \to T x$ , so $∥ (T_{n} - T) x ∥ = \lim_{m} ∥ (T_{n} - T_{m}) x ∥ \leq ε$ by [L2]. Taking the supremum over the unit ball yields $∥ T_{n} - T ∥ \leq ε$ .

3.1step 2.2step 2.3L1∎

Step 2.3 shows $T_{n} \to T$ in operator norm, with $T \in B (X, Y)$ by step 2.2. Therefore every operator-norm Cauchy sequence converges in $B (X, Y)$ , so $B (X, Y)$ is Banach by [L1].

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

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

1.1L0L3choose

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 $x$ .

2.1step 1.1L1L2choose

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 $\tilde{T} (x) : = \lim_{n} T d_{n} (x)$ .

3.1step 2.1L1L4algebra

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].

3.2step 1.1step 2.1L1L4

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 $S = \tilde{T}$ .

4.1step 2.1step 3.1

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 $T$ .

4.2step 1.1step 3.1L4construct

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 $\tilde{T} (λ x) = λ \tilde{T} (x)$ .

5.1step 2.1step 4.1L1algebra

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 $∥ \tilde{T} ∥ = ∥ T ∥$ .

6.1step 4.1step 4.2step 5.1step 3.2∎

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.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

A topological isomorphism of normed spaces

Definition

Let $X$ and $Y$ be normed spaces over the same scalar field. A map $T : X \to Y$ is a topological isomorphism of normed spaces when

$T$ is a bounded linear operator (A bounded linear operator between normed spaces);
$T$ is bijective (Injection, surjection, bijection);
the inverse map $T^{- 1} : Y \to X$ is also bounded.

Remarks

The definition does not hide item-level theorems: boundedness of the inverse is part of the data, not an automatic consequence here.
A topological isomorphism is an algebraic isomorphism and a homeomorphism for the norm topologies.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-01Open item page →

A bounded bilinear map between normed spaces

Definition

Let $X$ , $Y$ , and $Z$ be normed spaces over the same scalar field $K$ . A map $B : X \times Y \to Z$ is bilinear when $x \mapsto B (x, y)$ is linear for each fixed $y \in Y$ and $y \mapsto B (x, y)$ is linear for each fixed $x \in X$ .

It is a bounded bilinear map when there is a real constant $C \geq 0$ such that

$∥ B (x, y) ∥ \leq C ∥ x ∥ ∥ y ∥ for all x \in X, y \in Y .$

Remarks

If either variable is $0$ , the displayed estimate forces $B (x, 0) = 0$ and $B (0, y) = 0$ .
The constant $C$ is not unique, exactly as for bounded linear operators.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

For a bilinear map, boundedness is equivalent to joint continuity

Statement

Let $X$ , $Y$ , and $Z$ be normed spaces over the same scalar field, and let $B : X \times Y \to Z$ be bilinear. Then the following are equivalent:

$B$ is bounded.
$B$ is continuous at $(0, 0)$ .
$B$ is jointly continuous on $X \times Y$ for the product norm $∥ (x, y) ∥_{\max} : = \max {∥ x ∥, ∥ y ∥}$ .

Facts & Assumptions

Given: A bilinear map $B : X \times Y \to Z$ , points $x_{0} \in X$ , $y_{0} \in Y$ , and perturbations $h \in X$ , $k \in Y$ .

[L1]

A bounded bilinear map has a constant $C \geq 0$ with $∥ B (x, y) ∥ \leq C ∥ x ∥ ∥ y ∥$ for all $x, y$ (A bounded bilinear map between normed spaces).

[L2]

The finite-product maximum norm is a norm on $X \times Y$ (The standard product norms on a finite product of normed spaces).

[L3]

Continuity on metric spaces is the $ε$ - $δ$ condition of Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, and addition and scalar multiplication in normed spaces are continuous (Vector addition and scalar multiplication are continuous in a normed space).

Proof

technique · direct

1.1

Assume $B$ is bounded, with constant $C$ from [L1]. Bilinearity gives $B (x_{0} + h, y_{0} + k) - B (x_{0}, y_{0}) = B (h, k) + B (h, y_{0}) + B (x_{0}, k)$ . If $\max {∥ h ∥, ∥ k ∥} < δ \leq 1$ , then $∥ B (h, k) ∥ \leq C δ^{2}$ , $∥ B (h, y_{0}) ∥ \leq C δ ∥ y_{0} ∥$ , and $∥ B (x_{0}, k) ∥ \leq C ∥ x_{0} ∥ δ$ . Hence

$∥ B (x_{0} + h, y_{0} + k) - B (x_{0}, y_{0}) ∥ \leq C δ (δ + ∥ x_{0} ∥ + ∥ y_{0} ∥) .$

[L1, L2, algebra]

1.2L3

The implication $(3) \Rightarrow (2)$ is immediate by specializing the point of continuity to $(0, 0)$ .

1.3L3choose

Assume $B$ is continuous at $(0, 0)$ . Applying [L3] with $ε = 1$ gives $δ > 0$ such that $\max {∥ u ∥, ∥ v ∥} < δ$ implies $∥ B (u, v) ∥ < 1$ .

2.1step 1.1L3choose

Given $ε > 0$ , choose $δ \leq 1$ so that the bound in step 1.1 is below $ε$ . Then [L3] shows that $B$ is continuous at $(x_{0}, y_{0})$ . Since $(x_{0}, y_{0})$ was arbitrary, $(1) \Rightarrow (3)$ .

2.2step 1.3L1algebra

If $x = 0$ or $y = 0$ , bilinearity gives $B (x, y) = 0$ . Otherwise put $u : = \frac{δ}{2 ∥ x ∥} x$ and $v : = \frac{δ}{2 ∥ y ∥} y$ . Then $\max {∥ u ∥, ∥ v ∥} = δ / 2 < δ$ , so $∥ B (u, v) ∥ < 1$ by step 1.3. By bilinearity, $B (u, v) = \frac{δ^{2}}{4 ∥ x ∥ ∥ y ∥} B (x, y)$ , hence $∥ B (x, y) ∥ < \frac{4}{δ^{2}} ∥ x ∥ ∥ y ∥$ . Therefore $B$ is bounded.

3.1step 2.1step 1.2step 2.2∎

Steps 2.1, 1.2, and 2.2 prove $(1) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$ , so the three conditions are equivalent.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)

Definition

Let $X$ be a normed space and let $M \leq X$ be a linear subspace. The underlying quotient vector space is the one already defined in The quotient vector space $V / W$ and its canonical projection, with cosets written

$x + M : = {x + m : m \in M} .$

On this page the canonical projection is written

$q : X \to X / M, q (x) : = x + M .$

By Coset equality, well-defined quotient operations, and the canonical projection with kernel $W$ , $q$ is a surjective linear map and $\ker q = M$ .

Remarks

The quotient is algebraic at this stage; its norm is introduced next.
Different representatives of the same coset differ by an element of $M$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-01Open item page →

The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))

Definition

Let $X$ be a normed space and let $M \leq X$ be a linear subspace. For a coset $x + M \in X / M$ , define

$∥ x + M ∥_{X / M} : = \inf_{m \in M} ∥ x + m ∥ .$

Equivalently, this is the distance from $x$ to $M$ inside the ambient normed space:

$∥ x + M ∥_{X / M} = dist (x, M) .$

The formula is representative-independent by The quotient seminorm is independent of the chosen coset representative ↗, so it is a well-defined seminorm on the quotient vector space.

Remarks

The word seminorm is deliberate: definiteness is the next theorem, and it requires closedness of $M$ .
No nearest point is assumed to exist. The definition uses an infimum only.

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The quotient seminorm is independent of the chosen coset representative

Statement

Let $X$ be a normed space and let $M \leq X$ . If $x + M = x^{'} + M$ in $X / M$ , then

$\inf_{m \in M} ∥ x + m ∥ = \inf_{m \in M} ∥ x^{'} + m ∥ .$

Therefore the quotient seminorm of The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)) is well defined.

Facts & Assumptions

Given: A normed space $X$ , a linear subspace $M \leq X$ , and representatives $x, x^{'} \in X$ with $x + M = x^{'} + M$ .

[L1]

The quotient seminorm is defined by $∥ x + M ∥_{X / M} : = \inf_{m \in M} ∥ x + m ∥$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L2]

Two cosets are equal exactly when their representatives differ by an element of $M$ (Coset equality, well-defined quotient operations, and the canonical projection with kernel $W$ ).

Proof

technique · direct

1.1L2

By [L2], there is $m_{0} \in M$ with $x^{'} = x + m_{0}$ . For every $m \in M$ , $x^{'} + m = x + (m_{0} + m)$ , and $m_{0} + m$ still lies in $M$ . Thus ${∥ x^{'} + m ∥ : m \in M} \subseteq {∥ x + n ∥ : n \in M}$ .

2.1step 1.1L1L2∎

The same argument with the roles of $x$ and $x^{'}$ reversed gives the reverse inclusion, so the two sets of admissible norms are equal. Their infima are therefore equal, which is exactly the claim of [L1].

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

The quotient seminorm satisfies the triangle inequality

Statement

Let $X$ be a normed space and let $M \leq X$ . Then for all $x, y \in X$ ,

$∥ (x + y) + M ∥_{X / M} \leq ∥ x + M ∥_{X / M} + ∥ y + M ∥_{X / M} .$

Facts & Assumptions

Given: A normed space $X$ , a linear subspace $M \leq X$ , vectors $x, y \in X$ , and a real $ε > 0$ .

[L1]

The quotient seminorm is $∥ x + M ∥_{X / M} = \inf_{m \in M} ∥ x + m ∥$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L2]

The quotient seminorm is representative-independent, so $(x + y) + (m_{1} + m_{2}) + M = (x + y) + M$ may be read with any $m_{1}, m_{2} \in M$ (The quotient seminorm is independent of the chosen coset representative).

Proof

technique · direct

1.1L1choose

By [L1], choose $m_{1}, m_{2} \in M$ such that $∥ x + m_{1} ∥ < ∥ x + M ∥_{X / M} + ε / 2$ and $∥ y + m_{2} ∥ < ∥ y + M ∥_{X / M} + ε / 2$ .

2.1step 1.1L1L2algebra

Since $m_{1} + m_{2} \in M$ , [L2] lets us evaluate the quotient seminorm of $(x + y) + M$ at the representative $x + y + m_{1} + m_{2}$ . Hence $∥ (x + y) + M ∥_{X / M} \leq ∥ x + y + m_{1} + m_{2} ∥ \leq ∥ x + m_{1} ∥ + ∥ y + m_{2} ∥ < ∥ x + M ∥_{X / M} + ∥ y + M ∥_{X / M} + ε$ by step 1.1.

3.1step 2.1given∎

Since $ε > 0$ was arbitrary, the displayed strict inequality of step 2.1 implies the stated triangle inequality.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The quotient seminorm is a norm exactly when the subspace is closed

Statement

Let $X$ be a normed space and let $M \leq X$ . The quotient seminorm on $X / M$ is a norm if and only if $M$ is closed in $X$ .

Facts & Assumptions

Given: A normed space $X$ , a linear subspace $M \leq X$ , and a vector $x \in X$ .

[L1]

The quotient seminorm is $∥ x + M ∥_{X / M} = dist (x, M)$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L2]

A linear subspace contains $0$ , so $M$ is nonempty (Linear subspace of a vector space).

[L3]

For a nonempty subset $A$ of a metric space, the closure of $A$ is exactly ${u : dist (u, A) = 0}$ , and a set is closed exactly when it equals its closure (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

Proof

technique · direct

1.1L1L2L3

By [L2] and [L3], $dist (x, M) = 0$ exactly when $x \in \overline{M}$ . Therefore [L1] gives $∥ x + M ∥_{X / M} = 0$ exactly when $x \in \overline{M}$ .

2.1step 1.1L3

If $M$ is closed and $∥ x + M ∥_{X / M} = 0$ , then step 1.1 gives $x \in \overline{M} = M$ . Hence $x + M = M$ , the zero coset. So the quotient seminorm is definite and therefore a norm.

2.2step 1.1L3assume-contradischarge-contradiction

Conversely, assume the quotient seminorm is a norm. If $M$ were not closed, then [L3] would give some $x \in \overline{M} ∖ M$ . Step 1.1 would then give $∥ x + M ∥_{X / M} = 0$ , while $x + M \neq M$ because $x \notin M$ , contradicting definiteness. Therefore $M$ is closed.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the equivalence.

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The quotient map sends every open ball onto a set containing the corresponding quotient ball

Statement

Let $X$ be a normed space, let $M \leq X$ be closed, and let $q : X \to X / M$ be the quotient map. Then for every $x_{0} \in X$ and every $r > 0$ ,

$q (B_{X} (x_{0}, r)) \supseteq B_{X / M} (q (x_{0}), r) .$

In particular, $q$ is an open map.

Facts & Assumptions

Given: A normed space $X$ , a closed linear subspace $M \leq X$ , a vector $x_{0} \in X$ , a real $r > 0$ , and a coset $ξ \in X / M$ .

[L1]

The quotient map is $q (x) = x + M$ , and addition of cosets is inherited from the vector-space quotient (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)).

[L2]

The quotient norm is $∥ x + M ∥_{X / M} = \inf_{m \in M} ∥ x + m ∥$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L3]

Because $M$ is closed, the quotient seminorm is an honest norm on $X / M$ (The quotient seminorm is a norm exactly when the subspace is closed).

Proof

technique · direct

1.1L1L2L3choose

First take $x_{0} = 0$ . Let $ξ = x + M$ satisfy $∥ ξ ∥_{X / M} < r$ . By [L2], choose $m \in M$ with $∥ x + m ∥ < r$ . Then $x + m \in B_{X} (0, r)$ and $q (x + m) = x + m + M = x + M = ξ$ by [L1]. Hence $B_{X / M} (0, r) \subseteq q (B_{X} (0, r))$ .

2.1step 1.1L1

For general $x_{0}$ , a coset $ξ$ lies in $B_{X / M} (q (x_{0}), r)$ exactly when $ξ - q (x_{0}) \in B_{X / M} (0, r)$ . By step 1.1 there is $u \in B_{X} (0, r)$ with $q (u) = ξ - q (x_{0})$ . Then $q (x_{0} + u) = q (x_{0}) + q (u) = ξ$ , and $x_{0} + u \in B_{X} (x_{0}, r)$ . Therefore $q (B_{X} (x_{0}, r)) \supseteq B_{X / M} (q (x_{0}), r)$ .

3.1step 2.1L3∎

Every open ball in $X$ has image containing an open ball in $X / M$ , so $q$ is open.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

A quotient of a Banach space by a closed subspace is Banach

Statement

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

Let $X$ be a Banach space and let $M \leq X$ be a closed linear subspace. Then $X / M$ is Banach for the quotient norm.

Facts & Assumptions

Given: The Axiom of Countable Choice, a Banach space $X$ , a closed linear subspace $M \leq X$ , and a Cauchy sequence $(ξ_{n})$ in $X / M$ .

[L0]

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

[L1]

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

[L2]

The quotient norm is $∥ x + M ∥_{X / M} = \inf_{m \in M} ∥ x + m ∥$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L3]

Because $M$ is closed, the quotient seminorm is a norm on $X / M$ (The quotient seminorm is a norm exactly when the subspace is closed).

[L4]

In a Banach space, every absolutely convergent series converges (Series criterion for Banach spaces).

Proof

technique · direct

1.1givenchoose

Since $(ξ_{n})$ is Cauchy in $X / M$ , choose a strictly increasing sequence $(n_{k})$ such that $∥ ξ_{n_{k + 1}} - ξ_{n_{k}} ∥_{X / M} < 2^{- k}$ for every $k \geq 0$ .

2.1L0step 1.1L2choose

For each $k$ , choose $u_{k} \in X$ representing $ξ_{n_{k + 1}} - ξ_{n_{k}}$ and satisfying $∥ u_{k} ∥ < 2^{- k} + 2^{- 2 k}$ . This is possible by [L0], [L2], and step 1.1.

3.1step 2.1L1L4

The series $\sum_{k} u_{k}$ is absolutely convergent because $\sum_{k} (2^{- k} + 2^{- 2 k})$ converges. Since $X$ is Banach, [L4] gives a vector $u \in X$ with $\sum_{k} u_{k} = u$ .

4.1step 3.1L2

Let $s_{j} : = \sum_{k = 0}^{j - 1} u_{k}$ . Because each $u_{k}$ represents $ξ_{n_{k + 1}} - ξ_{n_{k}}$ , the coset $q (s_{j})$ equals $ξ_{n_{j}} - ξ_{n_{0}}$ . Therefore $ξ_{n_{j}} = ξ_{n_{0}} + q (s_{j})$ . Since $s_{j} \to u$ in $X$ , the tails satisfy $∥ u - s_{j} ∥ \to 0$ , and [L2] gives $∥ q (u) - q (s_{j}) ∥_{X / M} \leq ∥ u - s_{j} ∥$ . Hence $ξ_{n_{j}} \to η : = ξ_{n_{0}} + q (u)$ in $X / M$ .

5.1

The whole sequence $(ξ_{n})$ converges to $η$ . Given $ε > 0$ , choose $J$ so that $∥ ξ_{n} - ξ_{m} ∥_{X / M} < ε / 2$ for all $m, n \geq J$ , and also choose $k$ with $n_{k} \geq J$ and $∥ ξ_{n_{k}} - η ∥_{X / M} < ε / 2$ from step 4.1. Then for every $n \geq J$ ,

$∥ ξ_{n} - η ∥_{X / M} \leq ∥ ξ_{n} - ξ_{n_{k}} ∥_{X / M} + ∥ ξ_{n_{k}} - η ∥_{X / M} < ε .$

So $(ξ_{n})$ converges in $X / M$ . [step 4.1, given, choose]

6.1step 5.1L1L3∎

Every Cauchy sequence in $X / M$ converges, so $X / M$ is Banach by [L1].

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

A bounded operator that vanishes on a subspace factors uniquely through the normed quotient

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field, let $M \leq X$ be a closed linear subspace, let $q : X \to X / M$ be the quotient map, and let $T : X \to Y$ be a bounded linear operator with $M \subseteq \ker T$ . Then there is a unique bounded linear operator $\overline{T} : X / M \to Y$ such that

$\overline{T} \circ q = T,$

and moreover $∥ \overline{T} ∥ = ∥ T ∥$ .

Facts & Assumptions

Given: A closed linear subspace $M \leq X$ , the quotient map $q : X \to X / M$ , and a bounded linear operator $T : X \to Y$ with $M \subseteq \ker T$ .

[L1]

The algebraic quotient universal property gives a unique linear map $\overline{T} : X / M \to Y$ with $\overline{T} (q x) = T x$ (Universal property of the quotient vector space).

[L2]

The quotient norm is $∥ x + M ∥_{X / M} = \inf_{m \in M} ∥ x + m ∥$ (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L3]

A bounded operator has a concrete bound, and $q (x) = x + M$ (A bounded linear operator between normed spaces, The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)).

[L4]

The quotient map sends the open unit ball of $X$ onto a set containing the open unit ball of $X / M$ (The quotient map sends every open ball onto a set containing the corresponding quotient ball).

Proof

technique · direct

1.1L1

By [L1], there is a unique linear map $\overline{T} : X / M \to Y$ with $\overline{T} (q x) = T x$ for every $x \in X$ . This proves representative independence before any norm estimate.

2.1step 1.1L2L3

Let $ξ = x + M$ and let $m \in M$ . Since $T m = 0$ , $\overline{T} (ξ) = T x = T (x + m)$ . If $C$ is a bound for $T$ from [L3], then $∥ \overline{T} (ξ) ∥ \leq C ∥ x + m ∥$ for every $m \in M$ . Taking the infimum over $m$ and using [L2] gives $∥ \overline{T} (ξ) ∥ \leq C ∥ ξ ∥_{X / M}$ . Hence $\overline{T}$ is bounded and $∥ \overline{T} ∥ \leq ∥ T ∥$ .

3.1step 2.1L2L3

For every $x \in X$ , $∥ q (x) ∥_{X / M} \leq ∥ x ∥$ because the infimum in [L2] can be evaluated at $m = 0$ . Therefore $∥ T x ∥ = ∥ \overline{T} (q x) ∥ \leq ∥ \overline{T} ∥ ∥ x ∥$ by step 2.1. Taking the supremum over $∥ x ∥ \leq 1$ gives $∥ T ∥ \leq ∥ \overline{T} ∥$ .

4.1step 2.1step 3.1L4

Step 2.1 gave $∥ \overline{T} ∥ \leq ∥ T ∥$ , and step 3.1 gave the reverse inequality, so $∥ \overline{T} ∥ = ∥ T ∥$ . The unit-ball content of [L4] is the same geometric reason that no larger quotient bound is needed.

5.1step 1.1step 4.1∎

The linear map in step 1.1 is therefore the unique bounded factor of $T$ through the normed quotient, and it has the same operator norm.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

A complemented closed subspace of a normed space

Definition

Let $X$ be a normed space and let $M \leq X$ be a closed linear subspace. We say that $M$ is complemented when there is a closed linear subspace $N \leq X$ such that every $x \in X$ admits a unique decomposition

$x = m + n (m \in M, n \in N),$

and the coordinate maps

$P_{M} (x) : = m, P_{N} (x) : = n$

are bounded linear operators on $X$ .

Remarks

The decomposition is written $X = M \oplus N$ .
The theorem below shows that this is equivalent to $M$ being the range of a bounded projection.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01Open item page →

A closed subspace is complemented exactly when it is the range of a bounded projection

Statement

Let $X$ be a normed space and let $M \leq X$ . Then $M$ is complemented if and only if there is a bounded linear operator $P : X \to X$ such that

$P^{2} = P, ran (P) = M .$

Facts & Assumptions

Given: A normed space $X$ , a linear subspace $M \leq X$ , and a bounded linear operator $P : X \to X$ .

[L1]

If $M$ is complemented, then $X = M \oplus N$ for some closed subspace $N$ and the coordinate maps $P_{M}$ and $P_{N}$ are bounded linear operators (A complemented closed subspace of a normed space).

[L2]

A bounded linear operator is linear and satisfies norm estimates (A bounded linear operator between normed spaces).

Proof

technique · direct

1.1L1

Assume $M$ is complemented, and write $x = m + n$ with $m \in M$ , $n \in N$ as in [L1]. Let $P : = P_{M}$ . Then $P (x) = m$ , so $P^{2} (x) = P (m) = m = P (x)$ and $ran (P) = M$ . Thus a complemented subspace is the range of a bounded projection.

1.2

Conversely, assume $P^{2} = P$ and $ran (P) = M$ . For every $x \in X$ ,

$x = P x + (x - P x) .$

Here $P x \in M$ , and $P (x - P x) = P x - P^{2} x = 0$ , so $x - P x \in \ker P$ . [L2, algebra]

1.3L2choose

The kernel $\ker P$ is a closed linear subspace of $X$ . It is linear because $P$ is linear by [L2]. If $x_{n} \in \ker P$ and $x_{n} \to x$ , let $C$ be a bound for $P$ from [L2]. Then $∥ P x ∥ = ∥ P (x - x_{n}) ∥ \leq C ∥ x - x_{n} ∥ \to 0$ , so $P x = 0$ and $x \in \ker P$ .

2.1step 1.2L2algebra

The sum in step 1.2 is direct: if $z \in M \cap \ker P$ , then $z = P w$ for some $w$ and also $P z = 0$ , so $z = P z = P^{2} w = 0$ . Therefore $X = M \oplus \ker P$ .

3.1step 2.1step 1.3L1L2algebra

The coordinate projections for the direct sum $X = M \oplus \ker P$ are $P$ and $I - P$ . The first is bounded by hypothesis, and the second is bounded because $∥ (I - P) x ∥ \leq ∥ x ∥ + ∥ P x ∥$ for every $x$ . Together with steps 2.1 and 1.3, this is exactly the complemented-subspace condition of [L1].

4.1step 1.1step 3.1∎

Steps 1.1 and 3.1 prove the equivalence.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

A bounded operator that is bounded below

Definition

Let $T : X \to Y$ be a bounded linear operator. It is bounded below when there is a real constant $c > 0$ such that

$∥ T x ∥ \geq c ∥ x ∥ for all x \in X .$

Remarks

The inequality forces injectivity, because $∥ T x ∥ = 0$ then implies $∥ x ∥ = 0$ .
A bounded-below operator need not be surjective.

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Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Let $X$ and $Y$ be Banach spaces over the same scalar field, and let $T : X \to Y$ be a bounded linear operator. Then $T$ is bounded below if and only if it is injective and has closed range.

Facts & Assumptions

Given: Banach spaces $X$ and $Y$ , and a bounded linear operator $T : X \to Y$ .

[L0]

Dependent Choice is assumed (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[L1]

Being bounded below means that some $c > 0$ satisfies $∥ T x ∥ \geq c ∥ x ∥$ for every $x \in X$ (A bounded operator that is bounded below).

[L2]

A Banach space is complete, and a closed subspace of a Banach space is Banach (Banach space, A closed subspace of a Banach space is Banach).

[L3]

In a nonempty complete metric space, a countable union of closed sets with empty interior cannot be the whole space (Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior).

[L4]

A map is injective when equal outputs force equal inputs (Injection, surjection, bijection).

Proof

technique · direct

1.1L1L4

Assume $T$ is bounded below, with constant $c > 0$ from [L1]. If $T x = 0$ , then $0 = ∥ T x ∥ \geq c ∥ x ∥$ , so $∥ x ∥ = 0$ and $x = 0$ . Thus $T$ is injective.

1.2L1L2choose

Let $(T x_{n})$ be a sequence in $ran (T)$ converging to $y \in Y$ . Then $∥ x_{n} - x_{m} ∥ \leq c^{- 1} ∥ T x_{n} - T x_{m} ∥$ by [L1], so $(x_{n})$ is Cauchy in $X$ and hence converges to some $x \in X$ by [L2]. If $C$ is any bound for $T$ , then $∥ T x_{n} - T x ∥ \leq C ∥ x_{n} - x ∥ \to 0$ , so $T x_{n} \to T x$ . Since limits are unique in normed spaces, $y = T x$ . Therefore $ran (T)$ is closed.

1.3L2L4

Conversely, assume $T$ is injective and $ran (T)$ is closed. Then $Z : = ran (T)$ is Banach by [L2], and $T : X \to Z$ is a bounded linear bijection.

1.4L0L2L3choose

Let $B : = {x \in X : ∥ x ∥ < 1}$ . Since $Z = ⋃_{n \geq 1} \overline{T (n B)}$ , [L0] and [L3] yield an integer $N \geq 1$ such that $\overline{T (N B)}$ has nonempty interior in $Z$ . So there exist $z_{0} \in Z$ and $r > 0$ with $B_{Z} (z_{0}, r) \subseteq \overline{T (N B)}$ . Because $0 \in \overline{T (N B)}$ as well, subtraction gives $B_{Z} (0, r) \subseteq \overline{T (2 N B)}$ .

2.1step 1.4L2chooseconstruct

We claim that every $y \in Z$ with $∥ y ∥ < r$ has a preimage $x$ with $T x = y$ and $∥ x ∥ \leq 4 N$ . Start with $e_{0} : = y$ . If $∥ e_{k} ∥ < r 2^{- k}$ , then $∥ 2^{k} e_{k} ∥ < r$ , so step 1.4 gives $v_{k} \in 2 N B$ with $∥ 2^{k} e_{k} - T v_{k} ∥ < r / 2$ . Put $u_{k} : = 2^{- k} v_{k}$ and $e_{k + 1} : = e_{k} - T u_{k}$ . Then $∥ u_{k} ∥ \leq 2 N 2^{- k}$ and $∥ e_{k + 1} ∥ < r 2^{- k - 1}$ . Inductively this constructs $(u_{k})$ with $∥ e_{k} ∥ < r 2^{- k}$ for every $k$ . The series $\sum_{k} u_{k}$ is absolutely convergent because $\sum_{k} 2 N 2^{- k} < \infty$ , so [L2] gives $x : = \sum_{k} u_{k} \in X$ with $∥ x ∥ \leq 4 N$ . Also $y - T (\sum_{j < m} u_{j}) = e_{m} \to 0$ , hence $T x = y$ .

3.1step 2.1algebra

Now let $z \in Z$ with $z \neq 0$ . Put $λ : = r / (2 ∥ z ∥)$ , so $∥ λ z ∥ = r / 2 < r$ . Step 2.1 gives $u \in X$ with $T u = λ z$ and $∥ u ∥ \leq 4 N$ . Then $x : = λ^{- 1} u$ satisfies $T x = z$ and $∥ x ∥ \leq (8 N / r) ∥ z ∥$ . The same inequality is trivial at $z = 0$ , so the inverse $T^{- 1} : Z \to X$ is bounded by $8 N / r$ .

4.1step 3.1L1

Applying step 3.1 to $z = T x$ gives $∥ x ∥ \leq (8 N / r) ∥ T x ∥$ for every $x \in X$ , that is, $∥ T x ∥ \geq (r / 8 N) ∥ x ∥$ . Therefore $T$ is bounded below.

5.1step 1.1step 1.2step 1.3step 4.1∎

Step 1.2 proves that bounded below implies injective with closed range, and steps 1.3 through 4.1 prove the converse.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

Bounded left inverses and bounded right inverses

Definition

Let $T : X \to Y$ be a bounded linear operator.

A bounded linear operator $S : Y \to X$ is a bounded left inverse for $T$ when $S T = I_{X} .$
A bounded linear operator $S : Y \to X$ is a bounded right inverse for $T$ when $T S = I_{Y} .$

Here $I_{X}$ and $I_{Y}$ denote the identity operators on $X$ and $Y$ .

Remarks

A left inverse forces injectivity, and a right inverse forces surjectivity.
Neither condition implies the other without extra hypotheses.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

Under Dependent Choice, a surjective bounded operator between Banach spaces has a bounded right inverse exactly when its kernel is complemented

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Let $X$ and $Y$ be Banach spaces over the same scalar field, and let $T : X \to Y$ be a surjective bounded linear operator. Then $T$ has a bounded right inverse if and only if $\ker T$ is complemented in $X$ .

Facts & Assumptions

Given: Banach spaces $X$ and $Y$ , a surjective bounded linear operator $T : X \to Y$ , and a bounded linear operator $S : Y \to X$ .

[L0]

Dependent Choice is assumed (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[L1]

A bounded right inverse means $T S = I_{Y}$ (Bounded left inverses and bounded right inverses).

[L2]

Complemented subspaces are exactly the ranges of bounded projections (A closed subspace is complemented exactly when it is the range of a bounded projection).

[L3]

For bounded operators between Banach spaces, injective with closed range is equivalent to bounded below (Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range).

[L4]

A closed subspace of a Banach space is Banach (A closed subspace of a Banach space is Banach).

Proof

technique · direct

1.1L1givenalgebra

Assume $S$ is a bounded right inverse of $T$ , so $T S = I_{Y}$ by [L1]. Define $P : = I_{X} - S T$ . Then $P^{2} = I_{X} - 2 S T + S T S T = I_{X} - S T = P$ , because $T S T = T$ . Also, if $∥ S y ∥ \leq C_{S} ∥ y ∥$ and $∥ T x ∥ \leq C_{T} ∥ x ∥$ , then $∥ S T x ∥ \leq C_{S} C_{T} ∥ x ∥$ and therefore $∥ P x ∥ \leq (1 + C_{S} C_{T}) ∥ x ∥$ for every $x \in X$ ; so $P$ is a bounded projection. Finally, $T P = T - T S T = 0$ , so $ran (P) \subseteq \ker T$ .

1.2L4given

Conversely, assume $\ker T$ is complemented. Then there is a closed subspace $L \leq X$ with $X = \ker T \oplus L$ . The restriction $T ∣_{L} : L \to Y$ is injective, because $L \cap \ker T = {0}$ , and it is surjective because every $x \in X$ decomposes as $k + ℓ$ with $T x = T ℓ$ . Since $L$ is closed in the Banach space $X$ , [L4] makes $L$ Banach.

2.1step 1.1L2

If $x \in \ker T$ , then $P x = x - S T x = x$ . Hence $\ker T \subseteq ran (P)$ , and step 1.1 gives $ran (P) = \ker T$ . Therefore [L2] makes $\ker T$ complemented.

2.2step 1.2L0L3

The map $T ∣_{L}$ is a bounded bijection from the Banach space $L$ onto the Banach space $Y$ , so [L0] and [L3] make it bounded below. Hence its inverse $R : Y \to L$ is bounded, because $∥ R (y) ∥ \leq c^{- 1} ∥ y ∥$ when $∥ T ℓ ∥ \geq c ∥ ℓ ∥$ . The inclusion $L ↪ X$ now gives a bounded linear map $S : Y \to X$ with $T S = I_{Y}$ . Thus $S$ is a bounded right inverse.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the equivalence.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

Under Dependent Choice, an injective bounded operator between Banach spaces has a bounded left inverse exactly when its range is closed and complemented

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Let $X$ and $Y$ be Banach spaces over the same scalar field, and let $T : X \to Y$ be an injective bounded linear operator. Then $T$ has a bounded left inverse if and only if $ran (T)$ is closed and complemented in $Y$ .

Facts & Assumptions

Given: Banach spaces $X$ and $Y$ , an injective bounded linear operator $T : X \to Y$ , and a bounded linear operator $S : Y \to X$ .

[L0]

Dependent Choice is assumed (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[L1]

A bounded left inverse means $S T = I_{X}$ (Bounded left inverses and bounded right inverses).

[L2]

Complemented subspaces are exactly the ranges of bounded projections (A closed subspace is complemented exactly when it is the range of a bounded projection).

[L3]

Proof

technique · direct

1.1L1given

Assume $S$ is a bounded left inverse of $T$ , so $S T = I_{X}$ by [L1]. If $T x_{n} \to y$ in $Y$ , then $x_{n} = S (T x_{n}) \to S (y)$ because $S$ is bounded. Since $T$ is bounded as well, $T x_{n} \to T (S y)$ . Hence $y = T (S y)$ , so $ran (T)$ is closed.

1.2L0L3

Conversely, assume $ran (T)$ is closed and complemented in $Y$ . Since $T$ is injective and has closed range, [L0] and [L3] make it bounded below. Thus the inverse $R : ran (T) \to X$ defined by $R (T x) = x$ is bounded.

2.1step 1.1L1givenL2algebra

With $P : = T S$ , one has $P^{2} = T (S T) S = T S = P$ . If $∥ T x ∥ \leq C_{T} ∥ x ∥$ and $∥ S y ∥ \leq C_{S} ∥ y ∥$ , then $∥ P y ∥ = ∥ T S y ∥ \leq C_{T} C_{S} ∥ y ∥$ , so $P$ is bounded. For every $y \in Y$ , $P y = T S y$ lies in $ran (T)$ . If $y = T x$ is already in the range, then $P y = T S T x = T x = y$ . So $ran (P) = ran (T)$ , and [L2] shows that the range is complemented.

2.2step 1.2L2algebra

Let $P : Y \to Y$ be a bounded projection onto $ran (T)$ , given by [L2]. Then $S : = R \circ P : Y \to X$ is bounded and $S T x = R (P (T x)) = R (T x) = x$ for every $x \in X$ . Hence $S$ is a bounded left inverse of $T$ .

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the equivalence.

5 · Examples, counterexamples and false statements

None yet.