Provedalphabaguettebaguette-zoptimizer

With clipped gradients the second moment stays in [0, c²] forever

For all $\htmlData{mx=3}{g, v : \htmlData{mx=0}{\mathbb{N}} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}}$ and $\htmlData{mx=1}{\beta, c \in \htmlData{mx=0}{\mathbb{R}}}$ , if $0 \mathrel{\htmlData{mx=1}{\leq}} \htmlData{mx=0}{\beta}$ , $\htmlData{mx=0}{\beta} \mathrel{\htmlData{mx=1}{\leq}} 1$ , $\htmlData{mx=6}{\forall t,\ \htmlData{mx=3}{\left\lvert \htmlData{mx=2}{\htmlData{mx=0}{g}(\htmlData{mx=1}{t})} \right\rvert} \mathrel{\htmlData{mx=5}{\leq}} \htmlData{mx=4}{c}}$ , $0 \mathrel{\htmlData{mx=2}{\leq}} \htmlData{mx=1}{\htmlData{mx=0}{v}(0)} \mathbin{\htmlData{mx=8}{\land}} \htmlData{mx=4}{\htmlData{mx=3}{v}(0)} \mathrel{\htmlData{mx=7}{\leq}} \htmlData{mx=6}{\htmlData{mx=5}{c}^{2}}$ and $\htmlData{mx=18}{\forall t,\ \htmlData{mx=3}{\htmlData{mx=0}{v}(\htmlData{mx=2}{\htmlData{mx=1}{t} + 1})} \mathrel{\htmlData{mx=17}{=}} \htmlData{mx=16}{\htmlData{mx=8}{\htmlData{mx=4}{\beta} \cdot \htmlData{mx=7}{\htmlData{mx=5}{v}(\htmlData{mx=6}{t})}} + \htmlData{mx=15}{(\htmlData{mx=10}{1 - \htmlData{mx=9}{\beta}}) \cdot \htmlData{mx=14}{\htmlData{mx=13}{\htmlData{mx=11}{g}(\htmlData{mx=12}{t})}^{2}}}}}$ , then

\htmlData{mx=11}{\forall t,\ 0 \mathrel{\htmlData{mx=3}{\leq}} \htmlData{mx=2}{\htmlData{mx=0}{v}(\htmlData{mx=1}{t})} \mathbin{\htmlData{mx=10}{\land}} \htmlData{mx=6}{\htmlData{mx=4}{v}(\htmlData{mx=5}{t})} \mathrel{\htmlData{mx=9}{\leq}} \htmlData{mx=8}{\htmlData{mx=7}{c}^{2}}}

What the notation means · or point at any symbol

NNatural numbers: The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
⟹Implies: If ℕ, then ℝ. (Between types, A → B is the type of functions from A to B.) (ℕ → ℝ)
x : TVariables: g, v range over ℕ → ℝ: the statement must hold for every choice. ((g v : ℕ → ℝ))
x ∈ TVariables: β, c range over real numbers: the statement must hold for every choice. ((β c : ℝ))
≤At most: 0 is less than or equal to β. (0 ≤ β)
∀For all: The statement after the comma holds for every t. (∀ t, |g t| ≤ c)
∧And: Both hold: 0 ≤ v(0), and v(0) ≤ c^2. (0 ≤ v 0 ∧ v 0 ≤ c ^ 2)
RReal numbers: The real numbers: every point on the number line. (ℝ)
|x|Absolute value: The size of g(t) without its sign. (|g t|)
=Equality: v(t + 1) and β · v(t) + (1 − β) · g(t)^2 are the same. (v (t + 1) = β * v t + (1 - β) * g t ^ 2)
f(x)Function application: g applied to t. Lean writes f x; papers write f(x). (g t)
+Addition: t plus 1. (t + 1)
aᵇPower: c raised to the power 2. (c ^ 2)
·Multiplication: β times v(t). (β * v t)
-Subtraction: 1 minus β. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (1 - β)

theorem bz-second-moment-box (g v : ℕ → ℝ) (β c : ℝ) (hβ0 : 0 ≤ β) (hβ1 : β ≤ 1) (hg : ∀ t, |g t| ≤ c) (hv0 : 0 ≤ v 0 ∧ v 0 ≤ c ^ 2) (hv : ∀ t, v (t + 1) = β * v t + (1 - β) * g t ^ 2) :
    ∀ t, 0 ≤ v t ∧ v t ≤ c ^ 2

v(t+1) = β·v(t) + (1−β)·g(t)² with |g| ≤ c and 0 ≤ v(0) ≤ c² gives 0 ≤ v(t) ≤ c² for all t. Baguette's recorded failure grew v to 6·10³² before clipping existed; with clipping that is impossible. Part of the plan 'Baguette-Z: closed-loop certified integer-state training'.

How it was proved

Proved: 3 October 2026
By: Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by: Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms: Classical.choice, Quot.sound, propext
Proof: 25 lines of Lean

With clipped gradients the second moment stays in [0, c²] forever

How it was proved

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