Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
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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 residual and derivative error give a local root certificate

Statement

For a linear map A between Euclidean spaces, write ∥A∥op:=sup⁡∥u∥2=1∥Au∥2. Let m≥1, let V⊆Rm be open, and let G:V→Rm be C1. Let y∈V, R>0, and suppose the closed ball B‾2(y,R) lies in V. Assume that for some L≥0, ∥DG(u)−DG(v)∥op≤L∥u−v∥2(u,v∈B‾2(y,R)). Let g^∈Rm and A^∈Rm×m come with certified bounds ∥g^−G(y)∥2≤ϵg,∥A^−DG(y)∥op≤ϵJ, where ϵg,ϵJ≥0. Suppose A^ is invertible and ∥A^−1∥op≤β^, with β^ϵJ<1. Define β:=β^1−β^ϵJ,ρ:=∥g^∥2+ϵg,q:=βLR. If q<1 and βρ≤(1−q)R, then G has exactly one zero y∗ in B‾2(y,R) and ∥y∗−y∥2≤βρ1−q.

Facts & Assumptions

Given: G,V,y,R,L,g^,A^,ϵg,ϵJ and β^ as in the Statement.

[F1]
[F2]

R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R: Euclidean Rm is complete. A closed Euclidean ball is complete because every Cauchy sequence in it converges in Rm and continuity of the norm keeps the limit in the ball.

[F3]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): the derivative of a composite is the composite of its derivatives.

[F7]

Continuously differentiable maps, local inverses, and local diffeomorphisms: C1 means total differentiability at every point and continuity of the first derivative matrix entries.

[F8]

The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder: the total derivative is the linear first-order approximation to a map at each point.

Proof

Given: The data and hypotheses in the Statement.

1.1givenalgebra

For matrices X,Y, the definition of the operator norm gives ∥XYu∥2≤∥X∥op∥Y∥op∥u∥2 for every u; taking the supremum over unit u proves submultiplicativity. Put A:=DG(y), E:=A−A^, P:=A^−1E, and a:=β^ϵJ<1. Then A=A^(I+P) and ∥Pk∥op≤ak by induction.

2.1F2F5F6step 1.1givenalgebra

The finite sums SN:=∑j=0N(−P)j satisfy (I+P)SN=SN(I+P)=I+(−1)NPN+1. By [F6], the tails are bounded by aN+1/(1−a)→0. Each matrix entry of SN−SM is bounded in absolute value by ∥SN−SM∥op, by [F5] applied to standard unit vectors. Thus the entries are Cauchy; completeness [F2] gives a matrix limit S. Taking entrywise limits in the finite matrix products gives (I+P)S=S(I+P)=I. Also the finite-sum bound in [F6] gives ∥S∥op≤(1−a)−1 by taking limits on each vector. Hence A−1=SA^−1 and ∥A−1∥op≤β.

3.1F3F4F7F8givenstep 2.1algebra

Define T(x):=x−A−1G(x) on the ball. Its derivative is DT(x)=A−1(A−DG(x)). For x in the open ball, the Lipschitz hypothesis gives ∥DT(x)∥op≤βL∥x−y∥2≤q. The open ball is convex, so [F4] gives ∥T(u)−T(v)∥2≤q∥u−v∥2 there; continuity extends this inequality to the closed ball.

4.1F1F2step 2.1step 3.1givenalgebra

Since ∥T(y)−y∥2=∥A−1G(y)∥2≤βρ, every x∈B‾2(y,R) satisfies ∥T(x)−y∥2≤∥T(x)−T(y)∥2+∥T(y)−y∥2≤qR+βρ≤R. Thus T maps the nonempty complete ball into itself. By [F1] it has a unique fixed point y∗ there, and invertibility of A makes T(y∗)=y∗ equivalent to G(y∗)=0. Any zero of G is a fixed point, so this zero is unique.

5.1F1step 3.1step 4.1givenalgebra∎

The contraction inequality and fixed-point identity give ∥y∗−y∥2≤∥T(y∗)−T(y)∥2+∥T(y)−y∥2≤q∥y∗−y∥2+βρ, which rearranges to the stated error bound because 1−q>0.

Remarks

For an augmented pseudo-arclength residual, use this certificate only when the ball lies in the local curve-chart neighborhood and the corrector hyperplane is transverse to the curve. Failure of either strict contraction condition gives no certificate; a finite Newton budget alone certifies neither existence nor nonexistence of a root.

Depends on

Used by

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Sources