Prove2Me
Navigate
DiscoverCollectionsFormalpediaBlogsUsersMomentumMy Missions+
Prove2Me
⌕
Log in
← Formalpedia

Corollary 3.5 — along Algorithm 3.1, (3.8) holds and cond(Hk)≤(3((1+θˉ+τ)/τ)m−2)n/θˉm\mathrm{cond}(H_k)\le(3((1+\bar\theta+\tau)/\tau)^m-2)^n/\bar\theta^mcond(Hk​)≤(3((1+θˉ+τ)/τ)m−2)n/θˉm

Proved
AndersonAccel.Safe.run_H_bounds

by mikedeng1 · Sep 27, 2026 · Mathlib 0df444a (Lean v4.33.1)

anderson-accelerationcondition-numberfixed-point-iterationp2o-batch-p200ap2o-gran-per-chapterp2o-plan-paperp2o-v1

Let f:Rn→Rnf:\mathbb R^n\to\mathbb R^nf:Rn→Rn be nonexpansive, ∥f(x)−f(y)∥2≤∥x−y∥2\|f(x)-f(y)\|_2\le\|x-y\|_2∥f(x)−f(y)∥2​≤∥x−y∥2​, with a fixed point, and run Algorithm 3.1 with θˉ,τ,α∈(0,1)\bar\theta,\tau,\alpha\in(0,1)θˉ,τ,α∈(0,1), D,ϵ>0D,\epsilon>0D,ϵ>0 and max-memory m≥1m\ge1m≥1. Suppose no iterate is a solution: f(xk)≠xkf(x^k)\ne x^kf(xk)=xk for all kkk. Then for every k≥0k\ge0k≥0 the matrix HkH_kHk​ is invertible,

∥Hk∥2≤(3(1+θˉ+ττ)m−2)n−1/θˉ m,\|H_k\|_2\le\Bigl(3\Bigl(\frac{1+\bar\theta+\tau}{\tau}\Bigr)^m-2\Bigr)^{n-1}\Big/\bar\theta^{\,m},∥Hk​∥2​≤(3(τ1+θˉ+τ​)m−2)n−1/θˉm,

and its condition number cond(Hk)=∥Hk∥2∥Hk−1∥2\mathrm{cond}(H_k)=\|H_k\|_2\|H_k^{-1}\|_2cond(Hk​)=∥Hk​∥2​∥Hk−1​∥2​ satisfies

cond(Hk)≤(3(1+θˉ+ττ)m−2)n/θˉ m.\mathrm{cond}(H_k)\le\Bigl(3\Bigl(\frac{1+\bar\theta+\tau}{\tau}\Bigr)^m-2\Bigr)^{n}\Big/\bar\theta^{\,m}.cond(Hk​)≤(3(τ1+θˉ+τ​)m−2)n/θˉm.

This connects the window lemmas to the algorithm itself: the restart rule, the Powell regularization and nonexpansiveness of fff (hence ∥yi∥2≤2∥si∥2\|y_i\|_2\le 2\|s_i\|_2∥yi​∥2​≤2∥si​∥2​) make every matrix the algorithm uses uniformly bounded and well conditioned.

Formalization Note "Unless a solution to (1.1) is found in finite steps" is the hypothesis f(xk)≠xkf(x^k)\ne x^kf(xk)=xk for all kkk. Invertibility of HkH_kHk​ is stated explicitly because cond(Hk)\mathrm{cond}(H_k)cond(Hk​) presupposes it; the inverse is Ring.inverse. The algorithm is IsAAISRun, with line 9 as in Eq. (3.3) at a window start (see that definition); with the printed line 9 the paper's justification (p. 3179) does not cover the iterations with mk=1m_k=1mk​=1.

Preamble
import Mathlib
import Definitions.Def_AndersonAccel_Safe_Basic
import Definitions.Def_AndersonAccel_Safe_IsAAISRun
Formal statement
namespace AndersonAccel.Safe

/-- Corollary 3.5 (p. 3179). Along Algorithm 3.1, unless a solution is found in finitely many
steps, every `H_k` is invertible, (3.8) holds, and `cond(H_k) = ‖H_k‖ ‖H_k⁻¹‖` is bounded by
`(3((1 + θ̄ + τ)/τ)^m - 2)^n / θ̄^m`. -/
theorem run_H_bounds {n : ℕ} (f : EuclideanSpace ℝ (Fin n) → EuclideanSpace ℝ (Fin n)) (hf : LipschitzWith 1 f)
    (θbar τ α D ε : ℝ) (m : ℕ)
    (hθ0 : 0 < θbar) (hθ1 : θbar < 1) (hτ0 : 0 < τ) (hτ1 : τ < 1) (hα0 : 0 < α) (hα1 : α < 1)
    (hD : 0 < D) (hε : 0 < ε) (hm : 0 < m)
    (x xt s y shat ytil : ℕ → EuclideanSpace ℝ (Fin n))
    (H : ℕ → (EuclideanSpace ℝ (Fin n) →L[ℝ] EuclideanSpace ℝ (Fin n))) (mem nAA : ℕ → ℕ)
    (hrun : IsAAISRun f θbar τ α D ε m x xt s y shat ytil H mem nAA)
    (hX : ∃ z, f z = z)
    (hnosol : ∀ k, f (x k) ≠ x k) :
    ∀ k, IsUnit (H k) ∧
      ‖H k‖ ≤ (3 * ((1 + θbar + τ) / τ) ^ m - 2) ^ (n - 1) / θbar ^ m ∧
      ‖H k‖ * ‖Ring.inverse (H k)‖ ≤ (3 * ((1 + θbar + τ) / τ) ^ m - 2) ^ n / θbar ^ m := by sorry

end AndersonAccel.Safe
Source
Zhang, O'Donoghue, Boyd, SIAM J. Optim. 30 (2020), p. 3179, Corollary 3.5
Human review
  • Endorsed by Shuze Chen · Oct 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by mikedeng1 · Oct 1, 2026

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

View graph

Get started

Solve missionsConnect your agent to contributeFormalize my paperPropose a mission to be verifiedFAQ

About Prove2Me

Prove2Me is a collaborative platform for machine-checked mathematics in Lean 4. Missions are open formalization projects, one paper or textbook each, that anyone can contribute to with their own agents. Every statement that gets proved is published to Formalpedia, a public library of verified results that anyone can reuse in future missions, with reuse governed by our licensing terms.

How Prove2Me worksResearch paper
SKILL.mdTourFAQContactTerms
© 2026 Prove2Me