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✓ 4 results · all verified · 1 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 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Complex Norms and Realification

1 · Prerequisites

2 · Summary

This page builds on complexification-realification-and-real-structures and rn-as-a-normed-space. It defines complex norms with absolute homogeneity for every complex scalar, identifies complex modulus on the real embedding, and proves that realification preserves the norm and its induced metric. It also verifies the finite coordinate sum norm, defines restricted complex norms on subspaces, and fixes the real-valued comparison convention for equivalent complex norms.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Complex norm and complex normed space

Definition

Let V be a vector space over C (Vector space over a field, with the complex field of The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i). A complex norm on V is a function N:V→R such that, for all u,v∈V and every λ∈C:

A complex normed space is a pair (V,N) with V a complex vector space and N a complex norm on V. Since the codomain is R, every norm value is a finite real number; extended-real values are not allowed. The homogeneity clause is for every complex scalar, not merely real scalars.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

Complex modulus restricts to real absolute value

Statement

For every real r, write λr for its canonical image in C. Then

∣λr∣C=∣r∣R.

Facts & Assumptions

Given: A real number r and the complex modulus and real absolute value in the cited definitions. Write λr for its constant-class image in C.

[F1]

The canonical real embedding sends r to its constant class in C (The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i).

[F2]

The imaginary unit i is the class of the polynomial x in C=R[x]/(x2+1) (The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i).

[F3]

Every complex number has unique real coordinates a+bi (Real and imaginary parts, complex conjugation, and modulus).

[F4]

The complex modulus is ∣a+bi∣C=a2+b2 (Real and imaginary parts, complex conjugation, and modulus).

[F5]

Real absolute value is nonnegative and equals r when r≥0 and −r when r<0 (Absolute value in an ordered field, Basic properties of the absolute value).

[F6]

If r≠0, then r2>0 (Squares of nonzero elements are positive); if r=0, then r2=0 by 0⋅a=0 in a field (Multiplication by zero: 0⋅a=0).

Proof

1.1F1F2F3algebra

The canonical image of r is the class of the constant polynomial r by [F1]. Since i is the class of x by [F2], this is also the class of r+0x, namely r+0i. Uniqueness of complex coordinates in [F3] identifies that image with r+0i.

1.2F6F7cases

If r=0 then r2=0, and if r≠0 then [F6] gives r2>0. Thus r2≥0 in either case, so [F7] guarantees that its nonnegative square root exists.

1.3F5algebra

By [F5], ∣r∣R≥0. If r≥0, then ∣r∣=r; if r<0, then ∣r∣=−r. In either case ∣r∣R2=r2, including r=0.

2.1F4step 1.1step 1.2algebra

By [F4] and step 1.1, the modulus of the canonical image of r is r2+02=r2, whose radicand is nonnegative by step 1.2.

3.1F7step 2.1step 1.3∎

Step 2.1 and step 1.3 identify the complex modulus of the canonical image of r and ∣r∣R as nonnegative square roots of the same r2. By uniqueness in [F7], they are equal.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

Realification preserves a complex norm and its metric

Statement

Let (V,N) be a complex normed space and define dN(u,v):=N(u−v) for u,v∈V. On the realification VR, the same function N is a real norm. The underlying set, addition, norm values, and induced metric dN are unchanged.

Facts & Assumptions

Given: A complex normed space (V,N) and its realification VR. For each r∈R, write λr for its canonical complex image.

[F1]

N:V→R separates zero, is homogeneous for every complex scalar by the complex modulus, and satisfies the triangle inequality (Complex norm and complex normed space).

[F2]

Realification has the same underlying set and addition, and its real scalar action is the complex scalar action restricted along the canonical real embedding (Realification of a complex vector space by restriction of scalars).

[F3]

For every real r, the complex modulus of its canonical image λr is ∣r∣R (Complex modulus restricts to real absolute value).

[F4]

A real norm separates zero, is absolutely homogeneous for real scalars, and satisfies the triangle inequality (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[F5]

The induced real norm metric is the function dN(u,v)=N(u−v), which is a metric (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

1.1F1F2

Since realification preserves the set, zero vector, and addition, the separation and triangle conditions in [F1] hold unchanged on VR. The codomain remains R.

2.1F1F2F3F4step 1.1

For r∈R and v∈V, [F2] gives r⋅VRv=λrv. Hence [F1] and [F3] give N(r⋅VRv)=N(λrv)=∣λr∣CN(v)=∣r∣RN(v). Together with step 1.1, this is exactly the real norm criterion [F4].

3.1F2F5step 2.1∎

The set and addition are unchanged by [F2], so additive inverses and differences are unchanged as well. The realified norm's induced metric is therefore N(u−v) by [F5], the same function dN defined in the statement, with the same values on the same pairs.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The complex coordinate sum norm, including the zero space

Statement

For every n∈N, on the complex function space Cn with pointwise operations, define

N1(z):=∑j<n∣zj∣C.

Then N1 is a complex norm, including at n=0, when C0 is the singleton zero space.

Facts & Assumptions

Given: n∈N and z,w∈Cn, with finite sums over the coordinates j<n.

[F2]

Addition and scalar multiplication on Cn are pointwise, and its zero vector is the constant zero function (The vector space FX of all functions X→F with pointwise operations, and Fn as the case X=n={0,1,…,n−1}).

[F5]

Finite sums over a finite coordinate list use the recursively defined sum after zero extension (Finite sums and finite products, by recursion).

[F6]

The empty finite sum is zero (Finite sums and finite products, by recursion).

[F7]

Finite sums preserve addition and real scaling (Laws of finite sums and finite products).

[F8]

Finite sums preserve pointwise inequalities, and a sum of nonnegative terms is zero only when every term is zero (Laws of finite sums and finite products).

[F9]

A complex norm separates zero, is homogeneous for every complex scalar by the complex modulus, and satisfies the triangle inequality (Complex norm and complex normed space).

Proof

1.1F1F2F3F4F5F6F8F9

If n=0, [F3], [F5] and [F6] say that the only vector is zero and its empty sum is 0. If n>0 and z=0, pointwise equality in [F2] makes each coordinate zero; [F4] makes each summand zero, and the finite sum in [F5] is then zero. Conversely, if N1(z)=0, the nonnegative summands from [F4] satisfy the zero-sum condition in [F8], so every ∣zj∣C=0 and hence every zj=0 by [F4]. Pointwise equality [F1] identifies z with the zero function [F2]. Thus N1(z)=0 iff z=0 for every n, as required in [F9].

1.2F2F4F5F7F9

For every λ∈C, the finite-list sum in [F5], pointwise scalar multiplication from [F2], and multiplicativity from [F4] give N1(λz)=∑j<n∣λzj∣C=∑j<n∣λ∣C∣zj∣C=∣λ∣CN1(z), where the last equality is finite-sum scaling by the real scalar ∣λ∣C from [F7]. This is the complex absolute homogeneity clause in [F9].

2.1F2F4F5F7F8F9step 1.1step 1.2∎

Pointwise addition from [F2], the finite-list sum from [F5], and complex modulus subadditivity from [F4] give ∣zj+wj∣C≤∣zj∣C+∣wj∣C for every j<n. Finite-sum monotonicity and additivity [F7, F8] therefore give N1(z+w)=∑j<n∣zj+wj∣C≤∑j<n(∣zj∣C+∣wj∣C)=N1(z)+N1(w). Together with steps 1.1 and 1.2, this is the triangle clause in [F9], so all three complex norm criteria hold.

DefinitionDefinition: AI-adaptedProof: AI-generatedprecheck passOpen item page →

A complex normed linear subspace

Definition

Let (V,N) be a complex normed space (Complex norm and complex normed space) and let W⊆V be a complex-linear subspace (Linear subspace of a vector space over C). Define the restricted norm NW:W→R by NW(w):=N(w). The pair (W,NW) is the complex normed subspace determined by W.

Facts & Assumptions

Given: A complex normed space (V,N) and a complex-linear subspace W⊆V.

[F1]

N separates zero, satisfies N(λv)=∣λ∣CN(v) for every λ∈C, and satisfies the triangle inequality (Complex norm and complex normed space).

[F2]

A linear subspace contains zero, is closed under addition, and is closed under multiplication by every scalar in its field; here that field is C (Linear subspace of a vector space).

Proof

1.1F1

For w∈W, NW(w)=0 iff N(w)=0 iff w=0 by [F1], so the restriction separates zero.

2.1F1F2step 1.1∎

If w∈W and λ∈C, then λw∈W by [F2] and NW(λw)=N(λw)=∣λ∣CN(w)=∣λ∣CNW(w) by [F1]. If u,w∈W, then u+w∈W and NW(u+w)=N(u+w)≤N(u)+N(w)=NW(u)+NW(w), again by [F1]. Thus the restricted function has the full complex homogeneity and triangle properties as well.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Equivalent complex norms

Definition

Let M and N be complex norms (Complex norm and complex normed space) on the same complex vector space V. They are equivalent complex norms if there are positive real constants c,C∈R such that

cM(v)≤N(v)≤CM(v)for every v∈V.

Both inequalities and the positivity of c,C are in the real complete ordered field (Complete ordered field (least-upper-bound property), with completeness established in The Cauchy-sequence reals have the least-upper-bound property); the vector scalar field remains C. No order comparison of complex numbers is intended.

5 · Examples, counterexamples and false statements

None yet.

Sources