Dual Spaces Adjoint Operators and Annihilators

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The continuous dual, its completeness, and evaluation

Remark

Let $K = R$ or $C$ . For a normed $X$ , its continuous dual is $X^{*} = B (X, K)$ with the operator norm, as in The dual space X^* of a normed space and its dual norm. Since the scalar field is complete, If (Y) is Banach then (\mathcal B(X,Y)) is Banach makes $X^{*}$ Banach even when $X$ is incomplete. The pairing $⟨ f, x ⟩ = f (x)$ is bilinear; it is not an inner product on $X$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Annihilator notation and the preannihilator

Definition

Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is The dual space X^* of a normed space and its dual norm. The first notation agrees with Continuous annihilator of a linear subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Annihilators and preannihilators are norm closed

Statement

Let $K = R$ or $C$ . For any normed $X$ and arbitrary $M \subseteq X$ , $N \subseteq X^{*}$ , both $M^{⊥} \subseteq X^{*}$ and $^{⊥} N \subseteq X$ are norm-closed linear subspaces. Moreover ${\overline{N}}^{∥ \cdot ∥} \subseteq (^{⊥} N)^{⊥}$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F2]

From The dual space X^* of a normed space and its dual norm, with its stated hypotheses: Let $X$ be a normed space over the scalar field $K$ , where $K = R$ in the literal definition and $K = C$ by the convention of rem-real-and-complex-normed-space-convention. The dual space of $X$ is $X^{*} : = B (X, K),$ the space of bounded linear functionals on $X$ (def-space-of-bounded-linear-operators). Each $f \in X^{*}$ is in particular a linear functional in the algebraic sense, so $X^{*}$ is a subspace of the algebraic dual from def-algebraic-dual-and-linear-functional. The dual norm on $X^{*}$ is the operator norm: $∥ f ∥_{X^{*}} : = ∥ f ∥ = \sup {∣ f (x) ∣ : ∥ x ∥ \leq 1} .$

Proof

1.1

For fixed $x \in X$ , evaluation $E_{x} (f) = f (x)$ is linear and satisfies $∣ E_{x} (f) ∣ \leq ∥ x ∥ ∥ f ∥$ , hence has a norm-closed linear kernel. Each $f \in X^{*}$ likewise has a closed linear kernel in $X$ .

F2

2.1

By definition $M^{⊥} = ⋂_{x \in M} \ker E_{x}$ and $^{⊥} N = ⋂_{f \in N} \ker f$ . Intersections of closed sets are closed and these equations preserve zero, addition, and scalar multiplication. The empty intersection is the whole ambient space, including when $X = {0}$ .

F1step 1.1

3.1

Every $g \in N$ vanishes on $^{⊥} N$ . Thus $N \subseteq (^{⊥} N)^{⊥}$ ; the latter is norm closed by step 2.1, so it contains the norm closure of $N$ .

F1step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The dual of a quotient is its annihilator

Statement

Let $K = R$ or $C$ . Let $X$ be normed and $M \leq X$ closed. With quotient norm $∥ x + M ∥ = \inf_{m \in M} ∥ x + m ∥$ and $q (x) = x + M$ , the map $Q : (X / M)^{*} ⟶ M^{⊥}, Q h = h \circ q$ is a linear isometric bijection.

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The dual space X^* of a normed space and its dual norm, with its stated hypotheses: Let $X$ be a normed space over the scalar field $K$ , where $K = R$ in the literal definition and $K = C$ by the convention of rem-real-and-complex-normed-space-convention. The dual space of $X$ is $X^{*} : = B (X, K),$ the space of bounded linear functionals on $X$ (def-space-of-bounded-linear-operators). Each $f \in X^{*}$ is in particular a linear functional in the algebraic sense, so $X^{*}$ is a subspace of the algebraic dual from def-algebraic-dual-and-linear-functional. The dual norm on $X^{*}$ is the operator norm: $∥ f ∥_{X^{*}} : = ∥ f ∥ = \sup {∣ f (x) ∣ : ∥ x ∥ \leq 1} .$

[F2]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F3]

From A bounded operator that vanishes on a subspace factors uniquely through the normed quotient, with its stated hypotheses: 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 $∥ \overline{T} ∥ = ∥ T ∥$ .

Proof

1.1

For $h \in (X / M)^{*}$ , $∣ h (q x) ∣ \leq ∥ h ∥ ∥ q x ∥ \leq ∥ h ∥ ∥ x ∥$ , and $h (q m) = 0$ for $m \in M$ . Thus $Q h \in M^{⊥}$ and $Q$ is linear.

F1F2

1.2

If $f \in M^{⊥}$ , define $h (x + M) = f (x)$ . Equality of cosets means $x - x^{'} \in M$ , so this value is independent of the representative. The quotient universal property gives a bounded linear $h$ with $h \circ q = f$ and $∥ h ∥ = ∥ f ∥$ .

F2F3

2.1

Surjectivity of $q$ makes its pullback injective. Step 1.2 applied to $f = Q h$ returns the original $h$ by uniqueness, giving $∥ Q h ∥ = ∥ h ∥$ . If $M = X$ , both spaces are zero; if $M = {0}$ , the same formulas apply.

step 1.1step 1.2∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The dual of a closed subspace is a dual quotient

Statement

Let $K = R$ or $C$ . Let $X$ be normed and $M \leq X$ closed. Restriction $R : X^{*} \to M^{*}$ induces a linear isometric bijection $\tilde{R} : X^{*} / M^{⊥} ⟶ M^{*}, f + M^{⊥} ⟼ f ∣_{M} .$ Also $∥ R ∥ \leq 1$ ; its norm is $1$ when $M \neq {0}$ and $0$ when $M = {0}$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F2]

From Annihilators and preannihilators are norm closed, with its stated hypotheses: Let $K = R$ or $C$ . For any normed $X$ and arbitrary $M \subseteq X$ , $N \subseteq X^{*}$ , both $M^{⊥} \subseteq X^{*}$ and $^{⊥} N \subseteq X$ are norm-closed linear subspaces. Moreover ${\overline{N}}^{∥ \cdot ∥} \subseteq (^{⊥} N)^{⊥}$ .

[F3]

From A bounded operator that vanishes on a subspace factors uniquely through the normed quotient, with its stated hypotheses: 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 $∥ \overline{T} ∥ = ∥ T ∥$ .

[F4]

From A bounded linear functional on an arbitrary subspace extends with the same norm, without assuming the subspace is closed, with its stated hypotheses: Let $X$ be a normed space over $R$ or $C$ , let $M \subseteq X$ be a linear subspace, and let $f_{0} : M \to R$ or $f_{0} : M \to C$ be a bounded linear functional over the ambient scalar field. Then there exists a bounded linear extension $F$ of $f_{0}$ to all of $X$ such that $∥ F ∥ = ∥ f_{0} ∥$ . No closedness hypothesis on $M$ is needed.

Proof

1.1

The inequality $∥ f ∣_{M} ∥ \leq ∥ f ∥$ makes restriction bounded. Its kernel is $M^{⊥}$ , which is closed. Thus equal cosets have equal restrictions, and the quotient universal property gives a bounded linear induced map; the kernel calculation makes it injective.

F1F2F3

2.1

For $h \in M^{*}$ , norm-preserving extension gives $f \in X^{*}$ with $f ∣_{M} = h$ and $∥ f ∥ = ∥ h ∥$ , proving surjectivity. Every other representative $f + a$ , $a \in M^{⊥}$ , restricts to $h$ , so $∥ f + a ∥ \geq ∥ h ∥$ . Taking the infimum and using this extension proves $∥ f + M^{⊥} ∥ = ∥ h ∥$ .

F4step 1.1

3.1

If $M \neq {0}$ , fix $m \in M ∖ {0}$ . The functional $λ m \mapsto λ ∥ m ∥$ on its line has norm one. Extend first to $M$ and then to $X$ by norm-preserving extension; its restriction and extension both have norm one, so $∥ R ∥ \geq 1$ . If $M = {0}$ , $R = 0$ . For $M = X$ the induced map is the identity, including the zero ambient space.

F4step 1.1step 2.1∎

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Distance to an annihilator is the restriction norm

Statement

Let $K = R$ or $C$ . If $M$ is a closed linear subspace of a normed $X$ and $f \in X^{*}$ , then $dist (f, M^{⊥}) = ∥ f ∣_{M} ∥ .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The dual of a closed subspace is a dual quotient, with its stated hypotheses: Let $K = R$ or $C$ . Let $X$ be normed and $M \leq X$ closed. Restriction $R : X^{*} \to M^{*}$ induces a linear isometric bijection $\tilde{R} : X^{*} / M^{⊥} ⟶ M^{*}, f + M^{⊥} ⟼ f ∣_{M} .$ Also $∥ R ∥ \leq 1$ ; its norm is $1$ when $M \neq {0}$ and $0$ when $M = {0}$ .

Proof

1.1

The quotient norm is $∥ f + M^{⊥} ∥ = \inf_{a \in M^{⊥}} ∥ f + a ∥ = dist (f, M^{⊥})$ , since $M^{⊥}$ is a linear subspace.

givenalgebra

2.1

The restriction isometry identifies this quotient norm with $∥ f ∣_{M} ∥$ . For $f = 0$ both sides vanish; $M = 0$ gives zero and $M = X$ gives $∥ f ∥$ .

F1step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The transpose of a bounded operator

Definition

Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are The dual space X^* of a normed space and its dual norm. Composition is bounded by Composition satisfies |ST|\le|S|,|T|, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The transpose is bounded with the same norm

Statement

Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $T^{*} : Y^{*} \to X^{*}$ is bounded linear and $∥ T^{*} ∥ = ∥ T ∥$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The transpose of a bounded operator, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

[F2]

From Every nonzero vector has a norming functional, with its stated hypotheses: Let $X$ be a normed space over $R$ or $C$ , and let $x \in X$ be nonzero. Then there exists $f \in X^{*}$ such that $∥ f ∥ = 1 and f (x) = ∥ x ∥ .$

Proof

1.1

Composition is linear in $g$ , and $∣ (T^{*} g) (x) ∣ = ∣ g (T x) ∣ \leq ∥ g ∥ ∥ T ∥ ∥ x ∥$ . Taking the two unit-ball suprema gives $∥ T^{*} ∥ \leq ∥ T ∥$ .

F1

2.1

If $T x \neq 0$ , choose a unit functional $g \in Y^{*}$ with $g (T x) = ∥ T x ∥$ . Then $∥ T x ∥ = ∣ (T^{*} g) (x) ∣ \leq ∥ T^{*} ∥ ∥ x ∥$ . If $T x = 0$ this inequality holds directly. Taking the supremum for $∥ x ∥ \leq 1$ gives the reverse norm bound, including $T = 0$ and zero spaces.

F2step 1.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Transposition reverses composition

Statement

Let $K = R$ or $C$ . For bounded linear $T : X \to Y$ , $S : Y \to Z$ between normed spaces, $(S T)^{*} = T^{*} S^{*}, I_{X}^{*} = I_{X^{*}} .$ For bounded linear $T, U : X \to Y$ and $a, b \in K$ , $(a T + b U)^{*} = a T^{*} + b U^{*}$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The transpose of a bounded operator, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

Proof

1.1

For every $g \in Z^{*}$ and $x \in X$ , $((S T)^{*} g) (x) = g (S (T x)) = (T^{*} (S^{*} g)) (x)$ . Equality on every $x$ and then every $g$ proves the first operator identity with domain $Z^{*}$ and codomain $X^{*}$ .

F1

2.1

For $f \in X^{*}$ , $I_{X}^{*} f = f \circ I_{X} = f$ . For $g \in Y^{*}$ , $((a T + b U)^{*} g) (x) = g (a T x + b U x) = a (T^{*} g) (x) + b (U^{*} g) (x)$ . These identities also hold for zero maps, zero scalars, and zero spaces.

F1algebra∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Elementary kernel and range annihilator identities

Statement

Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $(ran T)^{⊥} = \ker T^{*},^{⊥} (ran T^{*}) = \ker T, {\overline{ran T}}^{∥ \cdot ∥} =^{⊥} (\ker T^{*}) .$ The closure in the last identity is in $Y$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The transpose of a bounded operator, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

[F2]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F3]

From The annihilator detects the closure of a subspace, with its stated hypotheses: For every linear subspace $M \subseteq X$ , $\overline{M} = ⋂_{f \in M^{⊥}} \ker f .$

[F4]

From Every nonzero vector has a norming functional, with its stated hypotheses: Let $X$ be a normed space over $R$ or $C$ , and let $x \in X$ be nonzero. Then there exists $f \in X^{*}$ such that $∥ f ∥ = 1 and f (x) = ∥ x ∥ .$

Proof

1.1

A functional $g \in Y^{*}$ vanishes on $ran T$ exactly when $g (T x) = 0$ for every $x$ , exactly when $T^{*} g = 0$ .

F1F2

1.2

A vector $x$ belongs to $^{⊥} (ran T^{*})$ exactly when $g (T x) = 0$ for every $g \in Y^{*}$ . This holds if $T x = 0$ ; if $T x \neq 0$ , a norming functional has $g (T x) = ∥ T x ∥ > 0$ , so it fails.

F1F2F4

2.1

Apply the primal annihilator-closure identity to the linear subspace $ran T \leq Y$ and substitute step 1.1. If $T = 0$ , the three identities read $Y^{*} = Y^{*}$ , $X = X$ , and ${0} =^{⊥} Y^{*}$ ; the last equality follows from the same norming separation.

F3F4step 1.1step 1.2∎

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Dense range is equivalent to injectivity of the transpose

Statement

Let $K = R$ or $C$ . For a bounded linear map $T : X \to Y$ between normed spaces, ${\overline{ran T}}^{∥ \cdot ∥} = Y ⟺ \ker T^{*} = {0} .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $(ran T)^{⊥} = \ker T^{*},^{⊥} (ran T^{*}) = \ker T, {\overline{ran T}}^{∥ \cdot ∥} =^{⊥} (\ker T^{*}) .$ The closure in the last identity is in $Y$ .

Proof

1.1

If the range is dense and $g \in \ker T^{*}$ , then $g$ vanishes on the range by the elementary identity. Continuity makes it vanish on $Y$ , so $g = 0$ .

F1

2.1

If $\ker T^{*} = {0}$ , the third elementary identity gives $\overline{ran T} =^{⊥} {0} = Y$ . This also handles $Y = {0}$ and the zero operator whenever the criterion holds.

F1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The canonical evaluation map into the bidual

Definition

Let $K = R$ or $C$ . For a normed $X$ , define $J_{X} : X \to X^{* *}, (J_{X} x) (f) = f (x) (f \in X^{*}) .$ With the dual norm from The dual space X^* of a normed space and its dual norm, evaluation is linear in $f$ and $∣ (J_{X} x) (f) ∣ \leq ∥ x ∥ ∥ f ∥$ , so $J_{X} x$ is a bounded functional on $X^{*}$ . The map is canonical and uses no chosen basis or conjugation.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The canonical bidual map is an isometry

Statement

Let $K = R$ or $C$ . For every normed $X$ , the map $J_{X} : X \to X^{* *}$ is linear and $∥ J_{X} x ∥ = ∥ x ∥$ for every $x \in X$ . In particular it is injective.

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The canonical evaluation map into the bidual, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ , define $J_{X} : X \to X^{* *}, (J_{X} x) (f) = f (x) (f \in X^{*}) .$ With the dual norm from def-dual-space-of-a-normed-space, evaluation is linear in $f$ and $∣ (J_{X} x) (f) ∣ \leq ∥ x ∥ ∥ f ∥$ , so $J_{X} x$ is a bounded functional on $X^{*}$ . The map is canonical and uses no chosen basis or conjugation.

[F2]

From Every nonzero vector has a norming functional, with its stated hypotheses: Let $X$ be a normed space over $R$ or $C$ , and let $x \in X$ be nonzero. Then there exists $f \in X^{*}$ such that $∥ f ∥ = 1 and f (x) = ∥ x ∥ .$

Proof

1.1

For $a, b \in K$ , evaluation gives $J_{X} (a x + b y) (f) = a f (x) + b f (y)$ for every $f$ . Also $∣ f (x) ∣ \leq ∥ f ∥ ∥ x ∥$ gives $∥ J_{X} x ∥ \leq ∥ x ∥$ .

F1

2.1

If $x \neq 0$ , a norming $f$ satisfies $∥ f ∥ = 1$ and $f (x) = ∥ x ∥$ , so $∥ J_{X} x ∥ \geq ∥ x ∥$ . For $x = 0$ both norms are zero. Equality of norms now forces $J_{X} x = 0$ only for $x = 0$ , including when $X$ is the zero space.

F2step 1.1∎

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Distance to a closed subspace via unit annihilators

Statement

Let $K = R$ or $C$ . For a closed linear subspace $M$ of a normed $X$ and $x \in X$ , $dist (x, M) = \sup {∣ f (x) ∣ : f \in M^{⊥}, ∥ f ∥ \leq 1} .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The dual of a quotient is its annihilator, with its stated hypotheses: Let $K = R$ or $C$ . Let $X$ be normed and $M \leq X$ closed. With quotient norm $∥ x + M ∥ = \inf_{m \in M} ∥ x + m ∥$ and $q (x) = x + M$ , the map $Q : (X / M)^{*} ⟶ M^{⊥}, Q h = h \circ q$ is a linear isometric bijection.

[F2]

From The canonical bidual map is an isometry, with its stated hypotheses: Let $K = R$ or $C$ . For every normed $X$ , the map $J_{X} : X \to X^{* *}$ is linear and $∥ J_{X} x ∥ = ∥ x ∥$ for every $x \in X$ . In particular it is injective.

Proof

1.1

In the normed quotient $Z = X / M$ , the bidual isometry gives $∥ x + M ∥ = \sup_{h \in Z^{*}, ∥ h ∥ \leq 1} ∣ h (x + M) ∣$ . The quotient norm on the left is $dist (x, M)$ .

F2

2.1

The quotient-dual isometry sends its unit ball onto the unit ball of $M^{⊥}$ , with $f (x) = h (x + M)$ . Substitution proves the formula. The unit balls always contain zero; if $x \in M$ or $M = X$ both sides are zero, and $M = 0$ recovers the dual norm formula.

F1step 1.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The canonical map is natural

Statement

Let $K = R$ or $C$ . If $T : X \to Y$ is bounded linear between normed spaces, then $T^{* *} J_{X} = J_{Y} T : X ⟶ Y^{* *} .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The canonical evaluation map into the bidual, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ , define $J_{X} : X \to X^{* *}, (J_{X} x) (f) = f (x) (f \in X^{*}) .$ With the dual norm from def-dual-space-of-a-normed-space, evaluation is linear in $f$ and $∣ (J_{X} x) (f) ∣ \leq ∥ x ∥ ∥ f ∥$ , so $J_{X} x$ is a bounded functional on $X^{*}$ . The map is canonical and uses no chosen basis or conjugation.

[F2]

From The transpose of a bounded operator, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

[F3]

From The transpose is bounded with the same norm, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $T^{*} : Y^{*} \to X^{*}$ is bounded linear and $∥ T^{*} ∥ = ∥ T ∥$ .

Proof

1.1

The transpose is bounded, so its transpose $T^{* *} : X^{* *} \to Y^{* *}$ is defined. For $x \in X$ and $g \in Y^{*}$ , $(T^{* *} J_{X} x) (g) = (J_{X} x) (T^{*} g) = (T^{*} g) (x)$ .

F1F2F3

2.1

The last expression is $g (T x) = (J_{Y} T x) (g)$ . Equality at every $g$ proves equality in $Y^{* *}$ and then equality of operators for every $x$ . With $x = 0$ or $T = 0$ every expression vanishes; the same calculation covers zero spaces.

F1F2step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Reflexivity is surjectivity of the canonical map

Definition

Let $K = R$ or $C$ . A Banach space $X$ is reflexive if its canonical map $J_{X} : X \to X^{* *}$ from The canonical evaluation map into the bidual is surjective. By The canonical bidual map is an isometry this map is already an isometric embedding. Surjectivity means that every bounded linear functional on $X^{*}$ is evaluation at a vector of $X$ . Merely specifying some isomorphism between $X$ and $X^{* *}$ is not this definition.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Membership in the transpose range by an operator estimate

Statement

Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded linear between normed spaces and $f \in X^{*}$ . Then $f \in ran T^{*} ⟺ \exists C \geq 0 \forall x \in X : ∣ f (x) ∣ \leq C ∥ T x ∥ .$ For any such $C$ , a representing $g \in Y^{*}$ can be chosen with $∥ g ∥ \leq C$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The transpose of a bounded operator, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

[F2]

From A bounded linear functional on an arbitrary subspace extends with the same norm, without assuming the subspace is closed, with its stated hypotheses: Let $X$ be a normed space over $R$ or $C$ , let $M \subseteq X$ be a linear subspace, and let $f_{0} : M \to R$ or $f_{0} : M \to C$ be a bounded linear functional over the ambient scalar field. Then there exists a bounded linear extension $F$ of $f_{0}$ to all of $X$ such that $∥ F ∥ = ∥ f_{0} ∥$ . No closedness hypothesis on $M$ is needed.

Proof

1.1

If $f = T^{*} g$ , then $∣ f (x) ∣ = ∣ g (T x) ∣ \leq ∥ g ∥ ∥ T x ∥$ for all $x$ . Take $C = ∥ g ∥ \geq 0$ .

F1

1.2

Conversely suppose the domination bound holds. It forces $f$ to vanish on $\ker T$ . Therefore $h (T x) = f (x)$ defines a linear functional on $ran T$ : if $T x = T x^{'}$ , the difference is in the kernel and $f (x) = f (x^{'})$ . The bound says $∣ h (y) ∣ \leq C ∥ y ∥$ there.

given

2.1

Extend $h$ to $g \in Y^{*}$ with $∥ g ∥ = ∥ h ∥ \leq C$ , without assuming the range is closed. Then $T^{*} g = f$ . If $C = 0$ , $f = 0$ and $g = 0$ works; for $T = 0$ the bound holds precisely for $f = 0$ .

F1F2step 1.2∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Closed range is equivalent to a quotient estimate

Statement

Let $K = R$ or $C$ . Assume DC. For a bounded linear map $T : X \to Y$ between Banach spaces, $ran T is norm closed ⟺ \exists C > 0 \forall x \in X : dist (x, \ker T) \leq C ∥ T x ∥ .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F2]

From A quotient of a Banach space by a closed subspace is Banach, with its stated hypotheses: Assume the Axiom of Countable Choice (def-countable-choice). 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.

[F3]

From A bounded operator that vanishes on a subspace factors uniquely through the normed quotient, with its stated hypotheses: 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 $∥ \overline{T} ∥ = ∥ T ∥$ .

[F4]

From A closed subspace of a Banach space is Banach, with its stated hypotheses: Let $V$ be a Banach space and let $W \subseteq V$ be a closed linear subspace, equipped with the restricted norm. Then $W$ is a Banach space.

[F5]

From Bounded inverse theorem, with its stated hypotheses: Assume DC. A bounded bijective linear map $T : X \to Y$ between Banach spaces has a bounded linear inverse $T^{- 1} : Y \to X$ .

[F6]

From Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range, with its stated hypotheses: Assume the Axiom of Dependent Choice (def-dependent-choice). 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.

Proof

1.1

The kernel $N = \ker T$ is closed: $x_{j} \to x$ with $T x_{j} = 0$ implies $T x = 0$ by boundedness. DC supplies the countable choice required for quotient completeness, so $Z = X / N$ is Banach. The quotient universal property gives a bounded injective $A : Z \to Y$ , $A (x + N) = T x$ , with range $ran T$ .

F2F3

2.1

If that range is closed, it is Banach. View $A$ as a bounded bijection onto this range and use bounded inverse to obtain $∥ x + N ∥ \leq C ∥ T x ∥$ with a positive $C$ (enlarge a zero bound if necessary). This is the required distance estimate.

F4F5step 1.1

3.1

Conversely the estimate is $∥ z ∥ \leq C ∥ A z ∥$ for every $z \in Z$ , so $A$ is bounded below. Since $Z$ and $Y$ are Banach, the bounded-below criterion makes its range closed. If $T = 0$ , then $Z = 0$ and the estimate is $0 \leq 0$ for any positive $C$ ; all steps cover this case.

F6step 1.1given∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A lower bound for the transpose forces a dense image of a ball

Statement

Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded linear between normed spaces, and let $C > 0$ satisfy $∥ g ∥ \leq C ∥ T^{*} g ∥$ for every $g \in Y^{*}$ . With open balls, $B_{Y} (0, 1 / C) \subseteq {\overline{T (B_{X} (0, 1))}}^{∥ \cdot ∥} .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The transpose of a bounded operator, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{*} : Y^{*} ⟶ X^{*}, (T^{*} g) (x) = g (T x) .$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $K$ . No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

[F2]

From The transpose is bounded with the same norm, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $T^{*} : Y^{*} \to X^{*}$ is bounded linear and $∥ T^{*} ∥ = ∥ T ∥$ .

[F3]

From Strong separation of a closed and a compact convex set, with its stated hypotheses: Let $C, K \subseteq X$ be disjoint nonempty convex sets, where $C$ is closed and $K$ is compact. Then they are strongly separated by a nonzero functional in $X^{*}$ .

Proof

1.1

Put $D = \overline{T (B_{X} (0, 1))}$ . It contains zero and is closed, convex and balanced: the image ball has these last two algebraic properties, and continuity of linear combinations preserves them on taking closure. For $g \in Y^{*}$ , continuity and phase rotation in the unit ball give $\sup_{y \in D} Re g (y) = \sup_{∥ x ∥ < 1} ∣ g (T x) ∣ = ∥ T^{*} g ∥$ . The open-ball supremum equals the closed-ball supremum by scaling vectors by real numbers tending to one.

F1F2

2.1

If $y \notin D$ , strong separation of the nonempty closed convex set $D$ and compact singleton ${y}$ gives a nonzero $g \in Y^{*}$ , oriented so that $Re g (y) > \sup_{D} Re g$ . Thus $∥ g ∥ ∥ y ∥ \geq Re g (y) > ∥ T^{*} g ∥ \geq ∥ g ∥ / C$ , hence $∥ y ∥ > 1 / C$ .

F3step 1.1given

3.1

Consequently every point of norm less than $1 / C$ lies in $D$ . Zero was already in $D$ . If $T = 0$ , the hypothesis forces $Y^{*} = 0$ ; separation in step 2.1 rules out any point outside $D = {0}$ , so the argument remains valid even in that degenerate case.

step 1.1step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Surjectivity is equivalent to a lower bound for the transpose

Statement

Let $K = R$ or $C$ . Assume DC. For a bounded linear $T : X \to Y$ between Banach spaces, $T is onto ⟺ \exists C > 0 \forall g \in Y^{*} : ∥ g ∥ \leq C ∥ T^{*} g ∥ .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From A lower bound for the transpose forces a dense image of a ball, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded linear between normed spaces, and let $C > 0$ satisfy $∥ g ∥ \leq C ∥ T^{*} g ∥$ for every $g \in Y^{*}$ . With open balls, $B_{Y} (0, 1 / C) \subseteq {\overline{T (B_{X} (0, 1))}}^{∥ \cdot ∥} .$

[F2]

From Successive approximation turns a closure-ball inclusion into an actual preimage, with its stated hypotheses: Assume DC. Let $T : X \to Y$ be bounded linear, with $X$ Banach. If $B_{Y} (0, r) \subseteq \overline{T (B_{X} (0, 1))}$ for some $r > 0$ , then $B_{Y} (0, r / 2) \subseteq T (B_{X} (0, 1)) .$

[F3]

From Quantitative lifting form of the open mapping theorem, with its stated hypotheses: Assume DC. For a surjective bounded linear $T : X \to Y$ between Banach spaces, some $c > 0$ satisfies $c B_{Y} (0, 1) \subseteq T (B_{X} (0, 1))$ .

Proof

1.1

If $T$ is onto, quantitative open mapping supplies $c > 0$ with $c B_{Y} (0, 1) \subseteq T (B_{X} (0, 1))$ . Taking the supremum of $∣ g ∣$ over these balls gives $c ∥ g ∥ \leq ∥ T^{*} g ∥$ . Hence $C = 1 / c$ works, also for $g = 0$ .

F3

1.2

Conversely the dual estimate yields $B_{Y} (0, 1 / C) \subseteq \overline{T (B_{X} (0, 1))}$ . Under DC, successive approximation in the Banach domain gives $B_{Y} (0, 1 / (2 C)) \subseteq T (B_{X} (0, 1))$ .

F1F2

2.1

For any nonzero $y \in Y$ , choose the explicit scale $a = 4 C ∥ y ∥ > 0$ . Then $y / a$ lies in that image ball, so multiplying a preimage by $a$ yields a preimage of $y$ . Zero has preimage zero. Thus $T$ is onto, including the case $Y = 0$ .

step 1.2∎

Remark

The estimate is tested on a given dual input $g$ . The reverse direction then establishes existence of primal solutions; an a priori bound alone must not be described as an already constructed solution.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Banach closed-range theorem

Statement

Let $K = R$ or $C$ . Assume DC and let $T : X \to Y$ be bounded linear between Banach spaces. The following are equivalent: $ran T$ is norm closed; $ran T^{*}$ is norm closed; and there is $C > 0$ such that $dist (x, \ker T) \leq C ∥ T x ∥$ for all $x \in X$ . In that case $ran T =^{⊥} (\ker T^{*}), ran T^{*} = (\ker T)^{⊥} .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Closed range is equivalent to a quotient estimate, with its stated hypotheses: Let $K = R$ or $C$ . Assume DC. For a bounded linear map $T : X \to Y$ between Banach spaces, $ran T is norm closed ⟺ \exists C > 0 \forall x \in X : dist (x, \ker T) \leq C ∥ T x ∥ .$

[F2]

From Membership in the transpose range by an operator estimate, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded linear between normed spaces and $f \in X^{*}$ . Then $f \in ran T^{*} ⟺ \exists C \geq 0 \forall x \in X : ∣ f (x) ∣ \leq C ∥ T x ∥ .$ For any such $C$ , a representing $g \in Y^{*}$ can be chosen with $∥ g ∥ \leq C$ .

[F3]

From Surjectivity is equivalent to a lower bound for the transpose, with its stated hypotheses: Let $K = R$ or $C$ . Assume DC. For a bounded linear $T : X \to Y$ between Banach spaces, $T is onto ⟺ \exists C > 0 \forall g \in Y^{*} : ∥ g ∥ \leq C ∥ T^{*} g ∥ .$

[F4]

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $(ran T)^{⊥} = \ker T^{*},^{⊥} (ran T^{*}) = \ker T, {\overline{ran T}}^{∥ \cdot ∥} =^{⊥} (\ker T^{*}) .$ The closure in the last identity is in $Y$ .

[F5]

From The dual of a closed subspace is a dual quotient, with its stated hypotheses: Let $K = R$ or $C$ . Let $X$ be normed and $M \leq X$ closed. Restriction $R : X^{*} \to M^{*}$ induces a linear isometric bijection $\tilde{R} : X^{*} / M^{⊥} ⟶ M^{*}, f + M^{⊥} ⟼ f ∣_{M} .$ Also $∥ R ∥ \leq 1$ ; its norm is $1$ when $M \neq {0}$ and $0$ when $M = {0}$ .

[F6]

From Distance to an annihilator is the restriction norm, with its stated hypotheses: Let $K = R$ or $C$ . If $M$ is a closed linear subspace of a normed $X$ and $f \in X^{*}$ , then $dist (f, M^{⊥}) = ∥ f ∣_{M} ∥ .$

[F7]

From Annihilators and preannihilators are norm closed, with its stated hypotheses: Let $K = R$ or $C$ . For any normed $X$ and arbitrary $M \subseteq X$ , $N \subseteq X^{*}$ , both $M^{⊥} \subseteq X^{*}$ and $^{⊥} N \subseteq X$ are norm-closed linear subspaces. Moreover ${\overline{N}}^{∥ \cdot ∥} \subseteq (^{⊥} N)^{⊥}$ .

[F8]

From The transpose is bounded with the same norm, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $T^{*} : Y^{*} \to X^{*}$ is bounded linear and $∥ T^{*} ∥ = ∥ T ∥$ .

[F9]

From If (Y) is Banach then (\mathcal B(X,Y)) is Banach, with its stated hypotheses: 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.

[F10]

From A closed subspace of a Banach space is Banach, with its stated hypotheses: Let $V$ be a Banach space and let $W \subseteq V$ be a closed linear subspace, equipped with the restricted norm. Then $W$ is a Banach space.

Proof

1.1

The quotient-estimate lemma equates closedness of $ran T$ with the stated estimate. Suppose these hold, and write $N = \ker T$ . If $f \in N^{⊥}$ , then for every $n \in N$ , $∣ f (x) ∣ = ∣ f (x + n) ∣ \leq ∥ f ∥ ∥ x + n ∥$ ; infimizing gives $∣ f (x) ∣ \leq C ∥ f ∥ ∥ T x ∥$ .

F1

1.2

For the reverse implication, assume $ran T^{*}$ is norm closed. Both duals are Banach because the scalar field is Banach, and $T^{*}$ is bounded. The quotient-estimate lemma applied to $T^{*} : Y^{*} \to X^{*}$ therefore supplies $C > 0$ with $dist (g, \ker T^{*}) \leq C ∥ T^{*} g ∥$ .

F1F8F9

1.3

Let $Y_{0} = \overline{ran T} \leq Y$ ; it is Banach as a closed subspace. The bounded map $S : X \to Y_{0}$ , $S x = T x$ , has dense range. The elementary identity and continuity give $Y_{0}^{⊥} = \ker T^{*}$ .

F4F10

2.1

Domination yields $f \in ran T^{*}$ . Conversely every $T^{*} g$ vanishes on $N$ by evaluation. Thus $ran T^{*} = N^{⊥}$ , which is norm closed.

F2F4F7step 1.1

2.2

For any $h \in Y_{0}^{*}$ , the restriction quotient isometry supplies an extension $g \in Y^{*}$ with $g ∣_{Y_{0}} = h$ . The distance formula gives $∥ h ∥ = dist (g, Y_{0}^{⊥})$ . Direct evaluation gives $S^{*} h = T^{*} g$ , so step 1.2 implies $∥ h ∥ \leq C ∥ S^{*} h ∥$ .

F5F6step 1.2step 1.3

3.1

The separately proved surjectivity criterion makes $S$ onto $Y_{0}$ . Hence $ran T = Y_{0}$ is norm closed, proving the reverse implication. The elementary primal closure identity now yields $ran T =^{⊥} \ker T^{*}$ ; step 2.1 supplies the dual identity.

F3F4step 2.1step 2.2

4.1

If $T = 0$ , its two ranges are zero and the primal estimate is $dist (x, X) = 0$ . Both identities reduce to the same zero spaces by the elementary identities. The quotient and restriction steps allow $Y_{0} = 0$ , so no nonzero-range assumption has entered either implication.

F4step 1.1step 3.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Bounded below is equivalent to surjectivity of the transpose

Statement

Let $K = R$ or $C$ . Assume DC. If $T : X \to Y$ is bounded linear between Banach spaces, then $(\exists c > 0 \forall x \in X : ∥ T x ∥ \geq c ∥ x ∥) ⟺ T^{*} : Y^{*} \to X^{*} is onto .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Banach closed-range theorem, with its stated hypotheses: Let $K = R$ or $C$ . Assume DC and let $T : X \to Y$ be bounded linear between Banach spaces. The following are equivalent: $ran T$ is norm closed; $ran T^{*}$ is norm closed; and there is $C > 0$ such that $dist (x, \ker T) \leq C ∥ T x ∥$ for all $x \in X$ . In that case $ran T =^{⊥} (\ker T^{*}), ran T^{*} = (\ker T)^{⊥} .$

[F2]

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $(ran T)^{⊥} = \ker T^{*},^{⊥} (ran T^{*}) = \ker T, {\overline{ran T}}^{∥ \cdot ∥} =^{⊥} (\ker T^{*}) .$ The closure in the last identity is in $Y$ .

[F3]

From Membership in the transpose range by an operator estimate, with its stated hypotheses: Let $K = R$ or $C$ . Let $T : X \to Y$ be bounded linear between normed spaces and $f \in X^{*}$ . Then $f \in ran T^{*} ⟺ \exists C \geq 0 \forall x \in X : ∣ f (x) ∣ \leq C ∥ T x ∥ .$ For any such $C$ , a representing $g \in Y^{*}$ can be chosen with $∥ g ∥ \leq C$ .

[F4]

From Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range, with its stated hypotheses: Assume the Axiom of Dependent Choice (def-dependent-choice). 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.

Proof

1.1

If $T$ is bounded below with constant $c > 0$ , every $f \in X^{*}$ satisfies $∣ f (x) ∣ \leq ∥ f ∥ ∥ x ∥ \leq (∥ f ∥ / c) ∥ T x ∥$ . Domination gives $f \in ran T^{*}$ , proving surjectivity.

F3

1.2

If $T^{*}$ is onto, its range $X^{*}$ is closed. Closed range duality makes $ran T$ closed. Also $\ker T =^{⊥} X^{*} = {0}$ : the elementary identity applied to the identity operator on $X$ gives the last equality. Thus $T$ is injective with closed range, and the Banach bounded-below criterion applies.

F1F2F4

2.1

When $X = 0$ , the lower bound holds for any positive $c$ and $T^{*}$ maps onto $X^{*} = 0$ . The preceding arguments cover this case without choosing a unit vector or dividing by its norm.

step 1.1step 1.2∎

CorollaryStatement: AI-adaptedProof: AI-adaptedaudited 2026-09-07Open item page →

Invertibility and the inverse of the transpose

Statement

Let $K = R$ or $C$ . Assume DC. A bounded linear $T : X \to Y$ between Banach spaces is bijective if and only if $T^{*}$ is bijective. In that case $(T^{*})^{- 1} = (T^{- 1})^{*} .$ Furthermore, $T$ is a surjective linear isometry if and only if $T^{*}$ is a surjective linear isometry.

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Surjectivity is equivalent to a lower bound for the transpose, with its stated hypotheses: Let $K = R$ or $C$ . Assume DC. For a bounded linear $T : X \to Y$ between Banach spaces, $T is onto ⟺ \exists C > 0 \forall g \in Y^{*} : ∥ g ∥ \leq C ∥ T^{*} g ∥ .$

[F2]

From Bounded below is equivalent to surjectivity of the transpose, with its stated hypotheses: Let $K = R$ or $C$ . Assume DC. If $T : X \to Y$ is bounded linear between Banach spaces, then $(\exists c > 0 \forall x \in X : ∥ T x ∥ \geq c ∥ x ∥) ⟺ T^{*} : Y^{*} \to X^{*} is onto .$

[F3]

From Transposition reverses composition, with its stated hypotheses: Let $K = R$ or $C$ . For bounded linear $T : X \to Y$ , $S : Y \to Z$ between normed spaces, $(S T)^{*} = T^{*} S^{*}, I_{X}^{*} = I_{X^{*}} .$ For bounded $T, U : X \to Y$ and $a, b \in K$ , $(a T + b U)^{*} = a T^{*} + b U^{*}$ .

[F4]

From The transpose is bounded with the same norm, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $T^{*} : Y^{*} \to X^{*}$ is bounded linear and $∥ T^{*} ∥ = ∥ T ∥$ .

[F5]

From Bounded inverse theorem, with its stated hypotheses: Assume DC. A bounded bijective linear map $T : X \to Y$ between Banach spaces has a bounded linear inverse $T^{- 1} : Y \to X$ .

[F6]

From If (Y) is Banach then (\mathcal B(X,Y)) is Banach, with its stated hypotheses: 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.

Proof

1.1

If $T$ is bijective, its inverse is bounded, so $T$ is bounded below. The two dual criteria give surjectivity and bounded-belowness of $T^{*}$ , hence its bijectivity.

F1F2F5

1.2

If $T^{*}$ is bijective, the dual spaces are Banach by the completeness of bounded-operator spaces with scalar target, so bounded inverse applies. Hence $T^{*}$ is bounded below, so $T$ is onto; surjectivity of $T^{*}$ also makes $T$ bounded below, hence injective.

F1F2F5F6

2.1

For bijective $T$ , put $U = T^{- 1}$ , which is bounded. Transpose $T U = I_{Y}$ and $U T = I_{X}$ to get $U^{*} T^{*} = I_{Y^{*}}$ and $T^{*} U^{*} = I_{X^{*}}$ . Thus $(T^{*})^{- 1} = U^{*}$ .

F3F5step 1.1step 1.2

3.1

A bounded bijection $A$ is an isometry exactly when $∥ A ∥ \leq 1$ and $∥ A^{- 1} ∥ \leq 1$ : the two bounds give $∥ A x ∥ \leq ∥ x ∥ \leq ∥ A x ∥$ , and the converse follows by taking suprema. Transpose norm equality and step 2.1 transfer these two bounds between $T$ and $T^{*}$ . Using inequalities covers the unique bijection between zero spaces, whose operator norms are zero.

F4step 2.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Finite truncations approximate null and summable sequences

Statement

Let $K = R$ or $C$ . Use coordinates indexed by $N = {0, 1, \dots}$ . Define $ℓ^{1} (K) = {a : \sum_{n = 0}^{\infty} ∣ a_{n} ∣ < \infty}$ , with coordinatewise operations and norm $∥ a ∥_{1} = \sum_{n} ∣ a_{n} ∣$ . Let $P_{N}$ retain coordinates $0, \dots, N$ and set all others to zero. Then $∥ x - P_{N} x ∥_{\infty} ⟶ 0 (x \in c_{0} (K)), ∥ a - P_{N} a ∥_{1} ⟶ 0 (a \in ℓ^{1} (K)) .$

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The sequence spaces c_0 and ell-infinity, with its stated hypotheses: Let $K = R$ or $C$ , with absolute value in the real case and modulus $∣ a + i b ∣ = \sqrt{a^{2} + b^{2}}$ in the complex case. A scalar sequence here is a function $x : N \to K$ , including index zero. Let $ℓ^{\infty} = {x = (x_{n})_{n \in N} : \sup_{n \in N} ∣ x_{n} ∣ < \infty}, c_{0} = {x \in ℓ^{\infty} : x_{n} \to 0},$ both equipped with $∥ x ∥_{\infty} = \sup_{n \in N} ∣ x_{n} ∣$ . Thus $c_{0}$ is a specified linear subspace of the bounded-sequence space $ℓ^{\infty}$ . Here $x_{n} \to 0$ means that for every real $ε > 0$ there is $N \in N$ such that $∣ x_{n} ∣ < ε$ for all $n \geq N$ . Addition and scalar multiplication are coordinatewise. The scalar triangle inequality makes bounded sequences and null sequences linear spaces and gives the triangle inequality for the displayed supremum norm. Absolute homogeneity follows coordinatewise, and a zero supremum forces every coordinate to vanish.

[F2]

From $ℓ^{p}$ is the $L^{p}$ space of counting measure, with its stated hypotheses: On $(N, P (N), #)$ with counting measure, every function $f : N \to R$ is measurable. Writing $a_{k} : = f (k)$ , one has $\int ∣ f ∣^{p} d # = \sum_{k = 0}^{\infty} ∣ a_{k} ∣^{p} (0 < p < \infty),$ by the counting-measure integral dictionary, so $L^{p} (#)$ is exactly the usual sequence class $ℓ^{p}$ . Also $∥ f ∥_{\infty} = \sup_{k \in N} ∣ a_{k} ∣,$ because a subset of $N$ has counting measure zero only when it is empty. Hence the quotient by almost-everywhere equality does nothing: for counting measure on $N$ , equality almost everywhere means equality everywhere.

Proof

1.1

The real absolute-sum model agrees with the counting-measure dictionary at $p = 1$ . For either scalar field, $\sum ∣ a_{n} + b_{n} ∣ \leq \sum ∣ a_{n} ∣ + \sum ∣ b_{n} ∣$ , absolute homogeneity holds termwise, and a zero sum forces every coordinate to vanish; thus the stated model is a normed linear space.

F2given

2.1

For $x \in c_{0}$ , $∥ x - P_{N} x ∥_{\infty} = \sup_{n > N} ∣ x_{n} ∣$ , which tends to zero by the definition of convergence to zero. For $a \in ℓ^{1}$ , the error is $\sum_{n > N} ∣ a_{n} ∣$ , the tail of a convergent nonnegative series, so it also tends to zero. Both formulas hold at $N = 0$ and for the zero sequence.

F1step 1.1

3.1

Each truncation has finite support, so these limits establish finite-support density in both norms. For a sequence already supported in ${0, \dots, N}$ the corresponding error is exactly zero.

step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The continuous dual of c0 is ell-one

Statement

Let $K = R$ or $C$ . With coordinates starting at zero, the map $ℓ^{1} (K) ⟶ c_{0} (K)^{*}, a ⟼ f_{a}, f_{a} (x) = \sum_{n = 0}^{\infty} a_{n} x_{n}$ is a linear isometric bijection. The pairing is bilinear, including over $C$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The dual space X^* of a normed space and its dual norm, with its stated hypotheses: Let $X$ be a normed space over the scalar field $K$ , where $K = R$ in the literal definition and $K = C$ by the convention of rem-real-and-complex-normed-space-convention. The dual space of $X$ is $X^{*} : = B (X, K),$ the space of bounded linear functionals on $X$ (def-space-of-bounded-linear-operators). Each $f \in X^{*}$ is in particular a linear functional in the algebraic sense, so $X^{*}$ is a subspace of the algebraic dual from def-algebraic-dual-and-linear-functional. The dual norm on $X^{*}$ is the operator norm: $∥ f ∥_{X^{*}} : = ∥ f ∥ = \sup {∣ f (x) ∣ : ∥ x ∥ \leq 1} .$

[F2]

From The sequence spaces c_0 and ell-infinity, with its stated hypotheses: Let $K = R$ or $C$ , with absolute value in the real case and modulus $∣ a + i b ∣ = \sqrt{a^{2} + b^{2}}$ in the complex case. A scalar sequence here is a function $x : N \to K$ , including index zero. Let $ℓ^{\infty} = {x = (x_{n})_{n \in N} : \sup_{n \in N} ∣ x_{n} ∣ < \infty}, c_{0} = {x \in ℓ^{\infty} : x_{n} \to 0},$ both equipped with $∥ x ∥_{\infty} = \sup_{n \in N} ∣ x_{n} ∣$ . Thus $c_{0}$ is a specified linear subspace of the bounded-sequence space $ℓ^{\infty}$ . Here $x_{n} \to 0$ means that for every real $ε > 0$ there is $N \in N$ such that $∣ x_{n} ∣ < ε$ for all $n \geq N$ . Addition and scalar multiplication are coordinatewise. The scalar triangle inequality makes bounded sequences and null sequences linear spaces and gives the triangle inequality for the displayed supremum norm. Absolute homogeneity follows coordinatewise, and a zero supremum forces every coordinate to vanish.

[F3]

From Finite truncations approximate null and summable sequences, with its stated hypotheses: Let $K = R$ or $C$ . Use coordinates indexed by $N = {0, 1, \dots}$ . Define $ℓ^{1} (K) = {a : \sum_{n = 0}^{\infty} ∣ a_{n} ∣ < \infty}$ , with coordinatewise operations and norm $∥ a ∥_{1} = \sum_{n} ∣ a_{n} ∣$ . Let $P_{N}$ retain coordinates $0, \dots, N$ and set all others to zero. Then $∥ x - P_{N} x ∥_{\infty} ⟶ 0 (x \in c_{0} (K)), ∥ a - P_{N} a ∥_{1} ⟶ 0 (a \in ℓ^{1} (K)) .$

Proof

1.1

For $a \in ℓ^{1}$ and $x \in c_{0}$ , $\sum ∣ a_{n} x_{n} ∣ \leq ∥ a ∥_{1} ∥ x ∥_{\infty}$ . Hence the scalar series is absolutely convergent, defines a linear functional, and gives $∥ f_{a} ∥ \leq ∥ a ∥_{1}$ ; the map $a \mapsto f_{a}$ is linear.

F1F2F3

2.1

For each finite prefix set $u_{n} = \overline{a_{n}} / ∣ a_{n} ∣$ when $n \leq N$ and $a_{n} \neq 0$ , and $u_{n} = 0$ otherwise (real conjugation does nothing). Then $u \in c_{0}$ , $∥ u ∥_{\infty} \leq 1$ , and $f_{a} (u) = \sum_{n \leq N} ∣ a_{n} ∣$ . Taking $N \to \infty$ proves $∥ f_{a} ∥ \geq ∥ a ∥_{1}$ . This includes $a = 0$ without a unit-vector assumption.

step 1.1

3.1

Given $f \in c_{0}^{*}$ , define $a_{n} = f (e_{n})$ , where $e_{n}$ is the coordinate unit sequence. The same finite phase test gives $\sum_{n \leq N} ∣ a_{n} ∣ = f (u) \leq ∥ f ∥$ for every $N$ . Thus $a \in ℓ^{1}$ . No simultaneous arbitrary sign selections occur: every phase is specified by a formula.

F1F2step 2.1

4.1

By truncation density and continuity, $f (x) = \lim_{N} f (P_{N} x) = \lim_{N} \sum_{n \leq N} a_{n} x_{n} = f_{a} (x)$ . Thus the map is onto. Evaluation at each $e_{n}$ determines $a_{n}$ uniquely, proving injectivity as well as uniqueness of the representation.

F3step 1.1step 3.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The complex continuous dual of ell-one is ell-infinity

Statement

For complex sequence spaces, with indices starting at zero, $ℓ^{\infty} (C) ⟶ ℓ^{1} (C)^{*}, b ⟼ h_{b}, h_{b} (a) = \sum_{n = 0}^{\infty} b_{n} a_{n}$ is a complex-linear isometric bijection. There is no conjugation in this pairing.

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The dual space X^* of a normed space and its dual norm, with its stated hypotheses: Let $X$ be a normed space over the scalar field $K$ , where $K = R$ in the literal definition and $K = C$ by the convention of rem-real-and-complex-normed-space-convention. The dual space of $X$ is $X^{*} : = B (X, K),$ the space of bounded linear functionals on $X$ (def-space-of-bounded-linear-operators). Each $f \in X^{*}$ is in particular a linear functional in the algebraic sense, so $X^{*}$ is a subspace of the algebraic dual from def-algebraic-dual-and-linear-functional. The dual norm on $X^{*}$ is the operator norm: $∥ f ∥_{X^{*}} : = ∥ f ∥ = \sup {∣ f (x) ∣ : ∥ x ∥ \leq 1} .$

[F2]

From The sequence spaces c_0 and ell-infinity, with its stated hypotheses: Let $K = R$ or $C$ , with absolute value in the real case and modulus $∣ a + i b ∣ = \sqrt{a^{2} + b^{2}}$ in the complex case. A scalar sequence here is a function $x : N \to K$ , including index zero. Let $ℓ^{\infty} = {x = (x_{n})_{n \in N} : \sup_{n \in N} ∣ x_{n} ∣ < \infty}, c_{0} = {x \in ℓ^{\infty} : x_{n} \to 0},$ both equipped with $∥ x ∥_{\infty} = \sup_{n \in N} ∣ x_{n} ∣$ . Thus $c_{0}$ is a specified linear subspace of the bounded-sequence space $ℓ^{\infty}$ . Here $x_{n} \to 0$ means that for every real $ε > 0$ there is $N \in N$ such that $∣ x_{n} ∣ < ε$ for all $n \geq N$ . Addition and scalar multiplication are coordinatewise. The scalar triangle inequality makes bounded sequences and null sequences linear spaces and gives the triangle inequality for the displayed supremum norm. Absolute homogeneity follows coordinatewise, and a zero supremum forces every coordinate to vanish.

[F3]

From Finite truncations approximate null and summable sequences, with its stated hypotheses: Let $K = R$ or $C$ . Use coordinates indexed by $N = {0, 1, \dots}$ . Define $ℓ^{1} (K) = {a : \sum_{n = 0}^{\infty} ∣ a_{n} ∣ < \infty}$ , with coordinatewise operations and norm $∥ a ∥_{1} = \sum_{n} ∣ a_{n} ∣$ . Let $P_{N}$ retain coordinates $0, \dots, N$ and set all others to zero. Then $∥ x - P_{N} x ∥_{\infty} ⟶ 0 (x \in c_{0} (K)), ∥ a - P_{N} a ∥_{1} ⟶ 0 (a \in ℓ^{1} (K)) .$

Proof

1.1

For bounded $b$ and $a \in ℓ^{1} (C)$ , $\sum ∣ b_{n} a_{n} ∣ \leq ∥ b ∥_{\infty} ∥ a ∥_{1}$ . The series therefore defines a complex-linear functional with $∥ h_{b} ∥ \leq ∥ b ∥_{\infty}$ , and depends complex-linearly on $b$ .

F1F2F3

2.1

Since $∥ e_{n} ∥_{1} = 1$ , $∥ h_{b} ∥ \geq ∣ h_{b} (e_{n}) ∣ = ∣ b_{n} ∣$ for every $n$ . Taking the supremum gives equality of norms; no maximizing coordinate is required. For $b = 0$ this reads $0 = 0$ .

step 1.1

3.1

Given $h \in ℓ^{1} (C)^{*}$ , let $b_{n} = h (e_{n})$ . Then $∣ b_{n} ∣ \leq ∥ h ∥$ , so $b \in ℓ^{\infty}$ . Truncation density gives $h (a) = \lim_{N} h (P_{N} a) = \sum_{n} b_{n} a_{n}$ . Hence the map is onto, and evaluation at $e_{n}$ makes the coefficient sequence unique.

F1F2F3step 1.1∎

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The published Lp duality theorem in abstract notation

Remark

For a sigma-finite measure space and real $L^{p}$ , with $1 \leq p < \infty$ and $1 / p + 1 / q = 1$ (so $q = \infty$ for $p = 1$ ), On a sigma-finite measure space, every bounded linear functional on $L^{p}$ is integration against a unique $L^{q}$ function identifies every bounded functional uniquely as $[f] \mapsto \int f g d μ$ , with norm $∥ g ∥_{q}$ . In the language of The dual space X^* of a normed space and its dual norm, this is a linear isometric identification $(L^{p})^{*} ≅ L^{q}$ . In particular Counting measure specializes the representation theorem to $ℓ^{p}$ and $ℓ^{q}$ at $p = 1$ gives $(ℓ^{1} (R))^{*} ≅ ℓ^{\infty} (R)$ . This remark asserts neither a $p = \infty$ representation nor a complex or arbitrary-measure extension of the cited theorem.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The weak-star topology from finite evaluations

Definition

Let $K = R$ or $C$ . For a normed $X$ with continuous dual $X^{*}$ from The dual space X^* of a normed space and its dual norm, the weak-star topology $σ (X^{*}, X)$ is the initial topology of all evaluations $f \mapsto f (x)$ into $K$ with its usual topology. At $f_{0}$ , a neighbourhood basis consists of $U (f_{0}; x_{1}, \dots, x_{n}; ε) = {f \in X^{*} : ∣ (f - f_{0}) (x_{j}) ∣ < ε (1 \leq j \leq n)},$ where $n$ is finite and $ε > 0$ . For $n = 0$ the set is all of $X^{*}$ . Finite intersections of inverse images of scalar open sets form the initial-topology basis; at the given point, finitely many disks can be refined using their smallest positive radius. Weak-star closure means closure in this topology, not merely sequential closure.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Finite evaluations separate a functional from a dual subspace

Statement

Let $K = R$ or $C$ . Let $X$ be normed, $N \leq X^{*}$ a linear subspace, and $x_{1}, \dots, x_{n} \in X$ with $n \geq 1$ . Define $E (f) = (f (x_{1}), \dots, f (x_{n}))$ . If $f_{0} \in X^{*}$ satisfies $E (f_{0}) \notin E (N)$ , there is $x \in span {x_{1}, \dots, x_{n}}$ such that $g (x) = 0$ for all $g \in N$ and $f_{0} (x) = 1$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F2]

From A finite-dimensional normed subspace is closed, with its stated hypotheses: Let $V$ be a normed space and let $W \subseteq V$ be a normed subspace. If $W$ admits an ordered basis of finite length, then $W$ is closed in $V$ .

[F3]

From Geometric Hahn--Banach theorem for subspaces, with its stated hypotheses: For a linear subspace $M \subseteq X$ and $x \notin \overline{M}$ , there is $f \in M^{⊥}$ with $f (x) = 1$ .

Proof

1.1

The image $E (N)$ is a linear subspace of the finite-dimensional normed space $K^{n}$ , hence admits a finite basis and is closed. Geometric Hahn–Banach applied to $E (f_{0}) \notin E (N)$ supplies a linear functional $h$ on $K^{n}$ with $h ∣_{E (N)} = 0$ and $h (E (f_{0})) = 1$ .

F2F3

2.1

For the standard coordinate vectors $e_{j}$ , set $a_{j} = h (e_{j})$ and $x = \sum_{j = 1}^{n} a_{j} x_{j}$ . Expanding in that basis gives $h (E (f)) = \sum_{j} a_{j} f (x_{j}) = f (x)$ for every $f \in X^{*}$ . Thus $g (x) = 0$ for $g \in N$ and $f_{0} (x) = 1$ ; in particular $x \in^{⊥} N$ .

F1step 1.1

3.1

Linear dependence among the $x_{j}$ , or $E (N) = 0$ , does not affect either step. For $n = 1$ the same sum has one term. The hypothesis excludes $f_{0} = 0$ and excludes all $x_{j}$ being zero; the conclusion never demands $f_{0} (0) = 1$ .

step 1.1step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Double annihilators give norm and weak-star closures

Statement

Let $K = R$ or $C$ . For a normed $X$ and a linear subspace $N \leq X^{*}$ , $(^{⊥} N)^{⊥} = {\overline{N}}^{σ (X^{*}, X)} .$ Consequently $N$ is weak-star closed if and only if $N = (^{⊥} N)^{⊥}$ , and weak-star dense in $X^{*}$ if and only if $^{⊥} N = {0}$ . For a linear subspace $M \leq X$ , the primal formula is $^{⊥} (M^{⊥}) = {\overline{M}}^{∥ \cdot ∥}$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The weak-star topology from finite evaluations, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ with continuous dual $X^{*}$ from def-dual-space-of-a-normed-space, the weak-star topology $σ (X^{*}, X)$ is the initial topology of all evaluations $f \mapsto f (x)$ into $K$ with its usual topology. At $f_{0}$ , a neighbourhood basis consists of $U (f_{0}; x_{1}, \dots, x_{n}; ε) = {f \in X^{*} : ∣ (f - f_{0}) (x_{j}) ∣ < ε (1 \leq j \leq n)},$ where $n$ is finite and $ε > 0$ . For $n = 0$ the set is all of $X^{*}$ . Finite intersections of inverse images of scalar open sets form the initial-topology basis; at the given point, finitely many disks can be refined using their smallest positive radius. Weak-star closure means closure in this topology, not merely sequential closure.

[F2]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F3]

From Finite evaluations separate a functional from a dual subspace, with its stated hypotheses: Let $K = R$ or $C$ . Let $X$ be normed, $N \leq X^{*}$ a linear subspace, and $x_{1}, \dots, x_{n} \in X$ with $n \geq 1$ . Define $E (f) = (f (x_{1}), \dots, f (x_{n}))$ . If $f_{0} \in X^{*}$ satisfies $E (f_{0}) \notin E (N)$ , there is $x \in span {x_{1}, \dots, x_{n}}$ such that $g (x) = 0$ for all $g \in N$ and $f_{0} (x) = 1$ .

[F4]

From The annihilator detects the closure of a subspace, with its stated hypotheses: For every linear subspace $M \subseteq X$ , $\overline{M} = ⋂_{f \in M^{⊥}} \ker f .$

Proof

1.1

If $x \in^{⊥} N$ , evaluation at $x$ vanishes on $N$ . Its kernel is weak-star closed by the definition of that topology. Intersecting these kernels shows ${\overline{N}}^{σ (X^{*}, X)} \subseteq (^{⊥} N)^{⊥}$ .

F1F2

1.2

Let $f_{0}$ lie outside the weak-star closure. Choose a basic neighbourhood $U (f_{0}; x_{1}, \dots, x_{n}; ε)$ disjoint from $N$ . Since $0 \in N$ , this neighbourhood cannot have $n = 0$ . Its finite-coordinate map satisfies $E (f_{0}) \notin E (N)$ , since equality with $E (g)$ would put that $g \in N$ in $U$ .

F1

2.1

Finite-evaluation separation supplies $x \in^{⊥} N$ with $f_{0} (x) = 1$ . Hence $f_{0} \notin (^{⊥} N)^{⊥}$ . Together with step 1.1, this proves equality.

F2F3step 1.1step 1.2

3.1

A set is closed exactly when it equals its closure, so the first equivalence follows in both directions. For density, if $^{⊥} N = 0$ , the equality gives closure $X^{*}$ . Conversely if the closure is $X^{*}$ , any $x \in^{⊥} N$ is annihilated by all of $X^{*}$ . The published primal formula with $M = 0$ says $^{⊥} X^{*} = 0$ , so $x = 0$ .

F2F4step 2.1

4.1

The primal formula in the statement is exactly the published annihilator-closure identity, with the preannihilator notation unpacked. It also checks the extremes $N = 0$ and $N = X^{*}$ : their closures and double annihilators are respectively $0$ and $X^{*}$ . For $X = 0$ these coincide.

F2F4step 2.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Kernel-range identities and the weak-star closure of the transpose range

Statement

Let $K = R$ or $C$ . If $T : X \to Y$ is bounded linear between normed spaces, then $(ran T)^{⊥} = \ker T^{*}, (\ker T)^{⊥} = {\overline{ran T^{*}}}^{σ (X^{*}, X)} .$ The second closure is weak-star closure, with no norm-closure substitution.

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From Elementary kernel and range annihilator identities, with its stated hypotheses: Let $K = R$ or $C$ . For a bounded linear $T : X \to Y$ between normed spaces, $(ran T)^{⊥} = \ker T^{*},^{⊥} (ran T^{*}) = \ker T, {\overline{ran T}}^{∥ \cdot ∥} =^{⊥} (\ker T^{*}) .$ The closure in the last identity is in $Y$ .

[F2]

From The weak-star topology from finite evaluations, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ with continuous dual $X^{*}$ from def-dual-space-of-a-normed-space, the weak-star topology $σ (X^{*}, X)$ is the initial topology of all evaluations $f \mapsto f (x)$ into $K$ with its usual topology. At $f_{0}$ , a neighbourhood basis consists of $U (f_{0}; x_{1}, \dots, x_{n}; ε) = {f \in X^{*} : ∣ (f - f_{0}) (x_{j}) ∣ < ε (1 \leq j \leq n)},$ where $n$ is finite and $ε > 0$ . For $n = 0$ the set is all of $X^{*}$ . Finite intersections of inverse images of scalar open sets form the initial-topology basis; at the given point, finitely many disks can be refined using their smallest positive radius. Weak-star closure means closure in this topology, not merely sequential closure.

[F3]

From Double annihilators give norm and weak-star closures, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and a linear subspace $N \leq X^{*}$ , $(^{⊥} N)^{⊥} = {\overline{N}}^{σ (X^{*}, X)} .$ Consequently $N$ is weak-star closed if and only if $N = (^{⊥} N)^{⊥}$ , and weak-star dense in $X^{*}$ if and only if $^{⊥} N = {0}$ . For a linear subspace $M \leq X$ , the primal formula is $^{⊥} (M^{⊥}) = {\overline{M}}^{∥ \cdot ∥}$ .

Proof

1.1

The elementary identities give $(ran T)^{⊥} = \ker T^{*}$ and $^{⊥} (ran T^{*}) = \ker T$ . The range of the linear operator $T^{*}$ is a linear subspace of $X^{*}$ .

F1

2.1

Apply the bipolar closure theorem to $N = ran T^{*}$ in the topology $σ (X^{*}, X)$ . It gives ${\overline{ran T^{*}}}^{σ (X^{*}, X)} = (^{⊥} ran T^{*})^{⊥} = (\ker T)^{⊥}$ . For $T = 0$ this is ${\overline{{0}}}^{σ (X^{*}, X)} = X^{⊥} = 0$ , so zero ranges are included.

F2F3step 1.1∎

Dual Spaces Adjoint Operators and Annihilators

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The continuous dual, its completeness, and evaluation

Remark

Annihilator notation and the preannihilator

Definition

Annihilators and preannihilators are norm closed

Statement

Facts & Assumptions

Proof

The dual of a quotient is its annihilator

Statement

Facts & Assumptions

Proof

The dual of a closed subspace is a dual quotient

Statement

Facts & Assumptions

Proof

Distance to an annihilator is the restriction norm

Statement

Facts & Assumptions

Proof

The transpose of a bounded operator

Definition

The transpose is bounded with the same norm

Statement

Facts & Assumptions

Proof

Transposition reverses composition

Statement

Facts & Assumptions

Proof

Elementary kernel and range annihilator identities

Statement

Facts & Assumptions

Proof

Dense range is equivalent to injectivity of the transpose

Statement

Facts & Assumptions

Proof

The canonical evaluation map into the bidual

Definition

The canonical bidual map is an isometry

Statement

Facts & Assumptions

Proof

Distance to a closed subspace via unit annihilators

Statement

Facts & Assumptions

Proof

The canonical map is natural

Statement

Facts & Assumptions

Proof

Reflexivity is surjectivity of the canonical map

Definition

Membership in the transpose range by an operator estimate

Statement

Facts & Assumptions

Proof

Closed range is equivalent to a quotient estimate

Statement

Facts & Assumptions

Proof

A lower bound for the transpose forces a dense image of a ball

Statement

Facts & Assumptions

Proof

Surjectivity is equivalent to a lower bound for the transpose

Statement

Facts & Assumptions

Proof

Remark

Banach closed-range theorem

Statement

Facts & Assumptions

Proof