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Updated-Jacobian Newton has local quadratic convergence at a simple root

Statement

Let n≥1, let U⊆Rn be open, and let F:U→Rn be C2, meaning F is C1 and its first derivative is itself continuously differentiable. Write J(x)=DF(x) as a matrix in the standard basis. Let x∗∈U satisfy F(x∗)=0 and suppose J(x∗) is invertible. Suppose there are R>0 and L≥0 such that the closed Euclidean ball B‾2(x∗,R):={x∈Rn:∥x−x∗∥2≤R} lies in U and ∥J(x)−J(y)∥op≤L∥x−y∥2(x,y∈B‾2(x∗,R)), where ∥A∥op:=sup⁡∥v∥2=1∥Av∥2. Then there are r∈(0,R] and K>0 such that every J(x) on B‾2(x∗,r) is invertible with ∥J(x)−1∥op≤K. Put C:=KL/2 and choose ρ∈(0,r] with Cρ≤1/2 (if L=0, any ρ≤r works).

For every x0∈B‾2(x∗,ρ), each update attempted by Finite-budget Newton for a square vector residual is defined, remains in this ball, and satisfies ∥xk+1−x∗∥2≤C∥xk−x∗∥22. If the ordinary exact-Jacobian Newton recurrence is continued indefinitely, these estimates imply xk→x∗ with a quadratic error bound. A finite budget may stop before the residual tolerance is met; root status records only that tolerance, not an exact-zero certificate. This result gives no arbitrary-start or global-convergence guarantee.

Facts & Assumptions

Given: n,U,F,x∗,R,L and the Jacobian conditions in the Statement; a residual tolerance ε≥0, update budget N∈N0, and an initial value x0∈B‾2(x∗,ρ).

[F1]

Newton maps are uniform contractions near a point with invertible derivative: if F is C1 on an open subset of Rn and DF(x∗) is invertible, there is a closed ball about x∗ on which every DF(x) is invertible and its inverse has a uniform operator-norm bound. Its additional contraction conclusion concerns the fixed-inverse map Ty(x)=x+B(y−F(x)) with B=DF(x∗)−1.

[F2]

Vector Newton remainder under a Lipschitz Jacobian: for x in a convex region containing x∗, ∥F(x)−F(x∗)−J(x)(x−x∗)∥2≤(L/2)∥x−x∗∥22 when J is L-Lipschitz there.

[F3]

Newton correction for a square vector residual: if J(xk) is invertible, the unique correction solves J(xk)sk=−F(xk); otherwise the correction is unavailable.

[F4]

Finite-budget Newton for a square vector residual: the exact-Jacobian procedure tests the residual and update budget before a correction, and checks trial membership in U before updating.

[F5]

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

[F6]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: the Jacobian entries in the standard basis are the first partial derivatives; A total derivative computes every directional derivative, and its matrix is the Jacobian identifies this matrix with the matrix of DF.

Proof

1.1givenF1F5choose

Apply [F1] at x∗. It supplies a radius r0>0 and a finite inverse bound K>0 on B‾2(x∗,r0). Set r:=min⁡{R,r0}, so the ball remains inside U, the Jacobian remains L-Lipschitz, and ∥J(x)−1∥op≤K throughout it. Only this inverse-bound conclusion of [F1] is used; the contraction there is for a fixed inverse matrix, not for the updated-Jacobian recurrence.

2.1step 1.1givenchoosealgebra

Let C:=KL/2. If C>0, choose 0<ρ≤min⁡{r,(2C)−1}; if C=0, choose any 0<ρ≤r. Then Cρ≤1/2.

3.1givenF2F3F5F6step 1.1step 2.1algebra

Suppose an update is attempted at xk∈B‾2(x∗,ρ) and put ek:=xk−x∗. The inverse bound makes the correction in [F3] unique. Since F(x∗)=0, [F2] and J(xk)ek−F(xk)=J(xk)ek−[F(xk)−F(x∗)] give ek+1=J(xk)−1(J(xk)ek−F(xk)) and ∥ek+1∥2≤C∥ek∥22. The full trial is in U because ∥ek+1∥2≤Cρ∥ek∥2≤∥ek∥2/2≤ρ/2.

4.1givenF4F7step 3.1induction∎

Starting from x0 in the stated ball, induction using step 3.1 shows every attempted update is valid, remains in the ball, and obeys the quadratic bound. It also gives ∥xk−x∗∥2≤2−k∥x0−x∗∥2 for every generated iterate; [F7] makes this tend to zero if the recurrence is continued indefinitely. The finite-budget algorithm may stop earlier under its own residual or budget rule.

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