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Eq. (24) — identity linking zk+1z^{k+1}zk+1, zˉk+1\bar z^{k+1}zˉk+1 and zˉk\bar z^kzˉk

Proved
GoldenRatioVI.Explicit.zbar_identity

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

golden-ratio-algorithmp2o-batch-p200ap2o-gran-per-chapterp2o-plan-paperp2o-v1

Consider a run of Algorithm 1 (EGRAAL) with parameter ϕ∈(1,φ]\phi\in(1,\varphi]ϕ∈(1,φ] and iterates (zk)(z^k)(zk), (zˉk)(\bar z^k)(zˉk). For every point z∈Ez\in\mathcal Ez∈E and every k≥0k\ge 0k≥0,

∥zk+1−z∥2=ϕϕ−1∥zˉk+1−z∥2−1ϕ−1∥zˉk−z∥2+1ϕ∥zk+1−zˉk∥2.(24)\|z^{k+1}-z\|^2 = \frac{\phi}{\phi-1}\|\bar z^{k+1}-z\|^2 - \frac1{\phi-1}\|\bar z^k-z\|^2 + \frac1\phi\|z^{k+1}-\bar z^k\|^2. \tag{24}∥zk+1−z∥2=ϕ−1ϕ​∥zˉk+1−z∥2−ϕ−11​∥zˉk−z∥2+ϕ1​∥zk+1−zˉk∥2.(24)

The identity follows from the averaging step (16) and lets the analysis measure progress through the averaged iterates zˉk\bar z^kzˉk.

Preamble
import Mathlib
import Definitions.Def_GoldenRatioVI_Explicit_egraalRun
Formal statement
namespace GoldenRatioVI.Explicit

/-- Eq. (24) of Malitsky (p. 6): along a run of Algorithm 1, for every point `z` of `E` and
every `k ≥ 0`,
`‖z^{k+1} − z‖² = (ϕ/(ϕ−1))‖z̄^{k+1} − z‖² − (1/(ϕ−1))‖z̄^k − z‖² + (1/ϕ)‖z^{k+1} − z̄^k‖²`. -/
theorem zbar_identity {E : Type*} [NormedAddCommGroup E] [InnerProductSpace ℝ E]
    [FiniteDimensional ℝ E] (g : E → EReal) (F : E → E) (ϕ lamBar : ℝ)
    (z zbar : ℕ → E) (lam theta : ℕ → ℝ)
    (hrun : IsEGRAALRun g F ϕ lamBar z zbar lam theta) (u : E) (k : ℕ) :
    ‖z (k + 1) - u‖ ^ 2 =
      ϕ / (ϕ - 1) * ‖zbar (k + 1) - u‖ ^ 2 - 1 / (ϕ - 1) * ‖zbar k - u‖ ^ 2
        + 1 / ϕ * ‖z (k + 1) - zbar k‖ ^ 2 := by sorry

end GoldenRatioVI.Explicit
Source
Malitsky, Golden Ratio Algorithms for Variational Inequalities, preprint (Optimization Online 6598, 2018), p. 6, Eq. (24)
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).

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