Provedalphabaguettebaguette-zoptimizer

With clipped gradients the first moment stays in [−c, c] forever

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

\htmlData{mx=6}{\forall t,\ \htmlData{mx=3}{\left\lvert \htmlData{mx=2}{\htmlData{mx=0}{m}(\htmlData{mx=1}{t})} \right\rvert} \mathrel{\htmlData{mx=5}{\leq}} \htmlData{mx=4}{c}}

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, m range over ℕ → ℝ: the statement must hold for every choice. ((g m : 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)
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: m(t + 1) and β · m(t) + (1 − β) · g(t) are the same. (m (t + 1) = β * m t + (1 - β) * g t)
f(x)Function application: g applied to t. Lean writes f x; papers write f(x). (g t)
+Addition: t plus 1. (t + 1)
·Multiplication: β times m(t). (β * m t)
-Subtraction: 1 minus β. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (1 - β)

theorem bz-first-moment-box (g m : ℕ → ℝ) (β c : ℝ) (hβ0 : 0 ≤ β) (hβ1 : β ≤ 1) (hg : ∀ t, |g t| ≤ c) (hm0 : |m 0| ≤ c) (hm : ∀ t, m (t + 1) = β * m t + (1 - β) * g t) :
    ∀ t, |m t| ≤ c

m(t+1) = β·m(t) + (1−β)·g(t) with |g| ≤ c and |m(0)| ≤ c gives |m(t)| ≤ c for all t. The first-moment bank never needs more than the clip range. 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: 21 lines of Lean

With clipped gradients the first moment stays in [−c, c] forever

How it was proved

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