Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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.

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.

Depends on

Used by

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources