TheoremStatement: Literature-sourcedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A bounded complex linear functional on a subspace of a complex normed space extends with the same norm

Statement

Assume the Axiom of Choice. Let $X$ be a complex normed space, let $M \subseteq X$ be a linear subspace, and let $f_{0} : M \to C$ be a bounded complex linear functional. Then there exists a bounded complex linear functional $F : X \to C$ such that $F ∣_{M} = f_{0}$ and $∥ F ∥ = ∥ f_{0} ∥$ .

Facts & Assumptions

Given: The Axiom of Choice, A complex normed space $X$ , a linear subspace $M \subseteq X$ , and a bounded complex linear functional $f_{0} : M \to C$ .

[A1]

AC is used through the real extension supplier [L1] in step 2.1.

[L1]

A bounded real linear functional on a real normed subspace extends with the same norm (A bounded real linear functional on a subspace of a real normed space extends with the same norm).

[L2]

A complex linear functional is recovered from its real part by $f (x) = u (x) - i u (i x)$ , and conversely every such formula defines a complex linear functional (A complex linear functional is recovered from its real part by f(x)=u(x)-iu(ix)).

[L3]

The complex norm axioms are those of Complex norm and complex normed space. The same norm is a real norm on realification (Realification preserves a complex norm and its metric); a complex-linear subspace carries its restricted complex norm (A complex normed linear subspace) and realifies to a real normed subspace with the same norm and restricted metric (A complex normed subspace and its finite basis realify).

Proof

technique · direct

1.1L2givenalgebra

Let $u : = Re f_{0}$ . By [L2], $u$ is real linear on the underlying real subspace $M$ . Also $∣ u (m) ∣ \leq ∣ f_{0} (m) ∣ \leq ∥ f_{0} ∥ ∥ m ∥ (m \in M),$ so $u$ is bounded with $∥ u ∥ \leq ∥ f_{0} ∥$ .

2.1L1L2L3step 1.1construct

Apply [L1] to the underlying real normed spaces. This yields a bounded real linear functional $U : X \to R$ extending $u$ and satisfying $∥ U ∥ = ∥ u ∥$ . Define $F (x) : = U (x) - i U (i x) (x \in X) .$ By [L2], $F$ is complex linear and $Re F = U$ .

3.1L2step 2.1

For $m \in M$ , the equality $U ∣_{M} = u$ and [L2] give $F (m) = u (m) - i u (i m) = f_{0} (m) .$ So $F$ extends $f_{0}$ .

3.2L2L3step 1.1step 2.1choosealgebra

Fix $x \in X$ . Choose $θ \in R$ so that $e^{i θ} F (x) = ∣ F (x) ∣$ is a nonnegative real number. Since $Re F = U$ and $F$ is complex linear, $∣ F (x) ∣ = Re (e^{i θ} F (x)) = Re F (e^{i θ} x) = U (e^{i θ} x) .$ Therefore $∣ F (x) ∣ \leq ∥ U ∥ ∥ e^{i θ} x ∥ = ∥ U ∥ ∥ x ∥ \leq ∥ f_{0} ∥ ∥ x ∥,$ where the last inequality uses step 1.1 and $∥ U ∥ = ∥ u ∥$ . Hence $∥ F ∥ \leq ∥ f_{0} ∥$ .

4.1step 3.1step 3.2algebra∎

Since $F ∣_{M} = f_{0}$ , every $m \in M$ with $∥ m ∥ \leq 1$ satisfies $∣ f_{0} (m) ∣ = ∣ F (m) ∣ \leq ∥ F ∥$ . Taking the supremum over the unit ball of $M$ gives $∥ f_{0} ∥ \leq ∥ F ∥$ . Combined with step 3.2, this yields $∥ F ∥ = ∥ f_{0} ∥$ .

A bounded complex linear functional on a subspace of a complex normed space extends with the same norm

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources