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Proof of Theorem 3.18, p. 293 — v_s − x_s = √κ(x_s − y_s)

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ConvexOptAlg.NesterovStrong.thm_3_18_coupling

by mikedeng1 · Oct 5, 2026 · Mathlib 0df444a (Lean v4.33.1)

accelerated-gradientconvex-optimizationestimate-sequencep2o-batch-pfp2ap2o-gran-per-chapterp2o-plan-bookp2o-v1

Let α,β>0\alpha,\beta>0α,β>0, κ=β/α\kappa=\beta/\alphaκ=β/α, let g:Rn→Rng:\mathbb R^n\to\mathbb R^ng:Rn→Rn be any map (in the role of ∇f\nabla f∇f), let (xt),(yt)(x_t),(y_t)(xt​),(yt​) be a run of Nesterov's accelerated gradient descent with this map, and let vsv_svs​ be defined from the points xsx_sxs​ by (3.21) with v1=x1v_1=x_1v1​=x1​. Then for every s≥1s\ge1s≥1,

vs−xs=κ (xs−ys).v_s-x_s=\sqrt\kappa\,(x_s-y_s).vs​−xs​=κ​(xs​−ys​).

This ties the centre vsv_svs​ of the model Φs\Phi_sΦs​ to the two sequences of the method; it is the identity that turns (3.22) into (3.20), and it explains the momentum coefficient (κ−1)/(κ+1)(\sqrt\kappa-1)/(\sqrt\kappa+1)(κ​−1)/(κ​+1).

Formalization Note The identity is algebraic: it uses only the recursions of the method and of vsv_svs​, so no convexity or smoothness of a function is assumed.

Preamble
import Mathlib
import Definitions.Def_OnlineConvexOpt_ConvexBasics_StronglyConvexOn
import Definitions.Def_ConvexOptAlg_NesterovStrong_Defs

open scoped InnerProductSpace
Formal statement
namespace ConvexOptAlg.NesterovStrong

/-- Bubeck, proof of Theorem 3.18, p. 293 ("Finally we show by induction that …"): for
`α, β > 0`, any gradient map `g` and a run `(x, y)` of Nesterov's accelerated gradient descent,
the centres `v_s` of (3.21) satisfy `v_s − x_s = √κ (x_s − y_s)` for every `s ≥ 1`. -/
theorem thm_3_18_coupling {n : ℕ}
    (g : EuclideanSpace ℝ (Fin n) → EuclideanSpace ℝ (Fin n)) (α β : ℝ)
    (hα : 0 < α) (hβ : 0 < β)
    (x y : ℕ → EuclideanSpace ℝ (Fin n)) (hrun : IsNesterovSCRun g α β x y)
    (s : ℕ) (hs : 1 ≤ s) :
    v g α β x s - x s = Real.sqrt (kappa α β) • (x s - y s) := by sorry

end ConvexOptAlg.NesterovStrong
Source
Bubeck, arXiv:1405.4980v2, proof of Theorem 3.18, p. 293, "Finally we show by induction that v_s − x_s = √κ(x_s − y_s)"

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