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Newton quadratic-phase contraction

Proved
ConvexOptimization.newton_quadratic_phase_contraction

by Shuze Chen · Aug 13, 2026 · Mathlib c5ea003 (Lean v4.30.0)

convexoptimizationnewtonmethodoptimizationalgorithms

The quadratically convergent phase: the scaled gradient norm squares at every step — inequality (9.33) of Boyd & Vandenberghe.

Let f:Rn→Rf : \mathbb{R}^n \to \mathbb{R}f:Rn→R be twice differentiable and assume, for constants 0<m≤M0 < m \le M0<m≤M and L>0L > 0L>0,

mI⪯∇2f(x)⪯MI,∥∇2f(x)−∇2f(y)∥≤L∥x−y∥2,m I \preceq \nabla^2 f(x) \preceq M I, \qquad \lVert \nabla^2 f(x) - \nabla^2 f(y)\rVert \le L\lVert x - y\rVert_2 ,mI⪯∇2f(x)⪯MI,∥∇2f(x)−∇2f(y)∥≤L∥x−y∥2​,

fix backtracking parameters α∈(0,1/2)\alpha \in (0,1/2)α∈(0,1/2), β∈(0,1)\beta \in (0,1)β∈(0,1), and set η=min⁡{1,3(1−2α)} m2/L\eta = \min\{1, 3(1-2\alpha)\}\,m^{2}/Lη=min{1,3(1−2α)}m2/L. Let xxx satisfy ∥∇f(x)∥2<η\lVert \nabla f(x)\rVert_2 < \eta∥∇f(x)∥2​<η and let Δ\DeltaΔ solve the Newton system ∇2f(x) Δ=−∇f(x)\nabla^2 f(x)\,\Delta = -\nabla f(x)∇2f(x)Δ=−∇f(x). Then the unit step passes the Armijo test, so backtracking accepts t=1t = 1t=1, and the next iterate x+=x+Δx^{+} = x + \Deltax+=x+Δ satisfies

f(x+Δ)≤f(x)+α⟨∇f(x),Δ⟩andL2m2∥∇f(x+)∥2  ≤  (L2m2∥∇f(x)∥2)2.f(x + \Delta) \le f(x) + \alpha \langle \nabla f(x), \Delta\rangle \qquad\text{and}\qquad \frac{L}{2m^{2}}\lVert \nabla f(x^{+})\rVert_2 \;\le\; \Bigl(\frac{L}{2m^{2}}\lVert \nabla f(x)\rVert_2\Bigr)^{2}.f(x+Δ)≤f(x)+α⟨∇f(x),Δ⟩and2m2L​∥∇f(x+)∥2​≤(2m2L​∥∇f(x)∥2​)2.

Once the gradient drops below η\etaη it never rises above it again, the method takes full Newton steps from then on, and the scaled quantity L2m2∥∇f∥2\frac{L}{2m^2}\lVert\nabla f\rVert_22m2L​∥∇f∥2​ squares at each iteration — the number of correct digits doubles per step. This is the source of the log⁡2log⁡2(ε0/ε)\log_2\log_2(\varepsilon_0/\varepsilon)log2​log2​(ε0​/ε) term in the mission's goal theorem, and the precise reason Newton's method is qualitatively different from any first-order method.

Formalization Note The conclusion is a conjunction: acceptance of the unit step is stated as the Armijo inequality at t=1t = 1t=1 rather than by invoking the backtracking predicate, since acceptance is exactly what has to be proved. Hessian bounds are in quadratic-form guise and the Lipschitz condition uses the operator norm. Source: B&V §9.5.3, pp. 488–491, eq. (9.33).

Preamble
import Mathlib

open scoped RealInnerProductSpace ENNReal
open MeasureTheory

Formal statement
theorem ConvexOptimization.newton_quadratic_phase_contraction {n : ℕ} (m M L α β η : ℝ)
    (hm : 0 < m) (hmM : m ≤ M) (hL : 0 < L)
    (hα0 : 0 < α) (hα : α < 1 / 2) (hβ0 : 0 < β) (hβ1 : β < 1)
    (hη : η = min 1 (3 * (1 - 2 * α)) * m ^ 2 / L)
    (f : EuclideanSpace ℝ (Fin n) → ℝ)
    (g : EuclideanSpace ℝ (Fin n) → EuclideanSpace ℝ (Fin n))
    (hg : ∀ x, HasGradientAt f (g x) x)
    (H : EuclideanSpace ℝ (Fin n) →
      EuclideanSpace ℝ (Fin n) →L[ℝ] EuclideanSpace ℝ (Fin n))
    (hH : ∀ x, HasFDerivAt g (H x) x)
    (hHm : ∀ x v, m * ‖v‖ ^ 2 ≤ ⟪H x v, v⟫)
    (hHM : ∀ x v, ⟪H x v, v⟫ ≤ M * ‖v‖ ^ 2)
    (hHL : ∀ x y, ‖H x - H y‖ ≤ L * ‖x - y‖)
    (x Δ : EuclideanSpace ℝ (Fin n))
    (hΔ : H x Δ = -g x) (hgx : ‖g x‖ < η) :
    (f (x + Δ) ≤ f x + α * ⟪g x, Δ⟫) ∧
    L / (2 * m ^ 2) * ‖g (x + Δ)‖ ≤ (L / (2 * m ^ 2) * ‖g x‖) ^ 2 := by
  sorry
Source
Boyd & Vandenberghe 2004, Convex Optimization, Cambridge University Press (seventh printing with corrections, 2009), https://web.stanford.edu/~boyd/cvxbook/, pp. 488-491, §9.5.3 eq. (9.33) (quadratically convergent phase: the scaled gradient norm squares and the unit step is accepted)
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What the Lean code literally says, in plain math · claude-fable-5

For every n∈Nn \in \mathbb{N}n∈N (including 000; at n=0n = 0n=0 all gradients are 000 and the gradient hypothesis ∥g(x)∥<η\lVert g(x)\rVert < \eta∥g(x)∥<η holds automatically since η>0\eta > 0η>0) and reals m,M,L,α,β,ηm, M, L, \alpha, \beta, \etam,M,L,α,β,η with 0<m0 < m0<m, m≤Mm \le Mm≤M, 0<L0 < L0<L, 0<α<120 < \alpha < \tfrac120<α<21​, 0<β<10 < \beta < 10<β<1, and η=min⁡(1, 3(1−2α))⋅m2/L\eta = \min\big(1,\ 3(1-2\alpha)\big) \cdot m^2 / Lη=min(1, 3(1−2α))⋅m2/L (a defining hypothesis; note β\betaβ is assumed but plays no role in either conclusion), given: fff with gradient field ggg (for every xxx, g(x)g(x)g(x) is the gradient of fff at xxx); a Hessian field HHH (for every xxx, ggg is Fréchet-differentiable at xxx with derivative HxH_xHx​); the bounds m∥v∥2≤⟨Hx(v),v⟩≤M∥v∥2m\lVert v\rVert^2 \le \langle H_x(v), v\rangle \le M \lVert v \rVert^2m∥v∥2≤⟨Hx​(v),v⟩≤M∥v∥2 for all x,vx, vx,v (two separate hypotheses); the operator-norm Lipschitz condition ∥Hx−Hy∥≤L∥x−y∥\lVert H_x - H_y \rVert \le L \lVert x - y \rVert∥Hx​−Hy​∥≤L∥x−y∥ for all x,yx, yx,y; and a point xxx and vector Δ\DeltaΔ with Hx(Δ)=−g(x)H_x(\Delta) = -g(x)Hx​(Δ)=−g(x) (some solution, not assumed unique) and ∥g(x)∥<η\lVert g(x) \rVert < \eta∥g(x)∥<η (strict — the quadratic phase); then both of: (1) the unit step satisfies the Armijo inequality: f(x+Δ)≤f(x)+α ⟨g(x),Δ⟩f(x + \Delta) \le f(x) + \alpha\, \langle g(x), \Delta \ranglef(x+Δ)≤f(x)+α⟨g(x),Δ⟩ (so a backtracking search would accept t=1t = 1t=1; the theorem does not itself mention ttt); and (2) the scaled gradient norm contracts quadratically:

L2m2 ∥g(x+Δ)∥  ≤  (L2m2 ∥g(x)∥)2,\frac{L}{2m^2}\, \lVert g(x + \Delta) \rVert \;\le\; \Big(\frac{L}{2m^2}\, \lVert g(x) \rVert\Big)^{2},2m2L​∥g(x+Δ)∥≤(2m2L​∥g(x)∥)2,

with the exact scaling constant L/(2m2)L / (2m^2)L/(2m2) on both sides.

Human review
  • Endorsed by Community (Bot) · Aug 13, 2026

  • Endorsed by Shuze Chen · Aug 13, 2026

    Confirmed by the mission captain (proposal self-audit).

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