proofs
Provedalphabaguettebaguette-zcapstone

Every step's Adam direction is bounded: |m(t)/(√v(t) + ε)| ≤ C

For all g,m,v:N→R\htmlData{mx=3}{g, m, v : \htmlData{mx=0}{\mathbb{N}} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}}, β1,β2∈R\htmlData{mx=1}{\beta_{1}, \beta_{2} \in \htmlData{mx=0}{\mathbb{R}}} and ε∈R\htmlData{mx=1}{\varepsilon \in \htmlData{mx=0}{\mathbb{R}}}, if 0<β20 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{\beta_{2}}, β2<1\htmlData{mx=0}{\beta_{2}} \mathrel{\htmlData{mx=1}{<}} 1, β12<β2\htmlData{mx=1}{\htmlData{mx=0}{\beta_{1}}^{2}} \mathrel{\htmlData{mx=3}{<}} \htmlData{mx=2}{\beta_{2}}, m(0)=0\htmlData{mx=1}{\htmlData{mx=0}{m}(0)} \mathrel{\htmlData{mx=2}{=}} 0, v(0)=0\htmlData{mx=1}{\htmlData{mx=0}{v}(0)} \mathrel{\htmlData{mx=2}{=}} 0, ∀t, m(t+1)=β1⋅m(t)+(1−β1)⋅g(t)\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_{1}} \cdot \htmlData{mx=7}{\htmlData{mx=5}{m}(\htmlData{mx=6}{t})}} + \htmlData{mx=14}{(\htmlData{mx=10}{1 - \htmlData{mx=9}{\beta_{1}}}) \cdot \htmlData{mx=13}{\htmlData{mx=11}{g}(\htmlData{mx=12}{t})}}}}, ∀t, v(t+1)=β2⋅v(t)+(1−β2)⋅g(t)2\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_{2}} \cdot \htmlData{mx=7}{\htmlData{mx=5}{v}(\htmlData{mx=6}{t})}} + \htmlData{mx=15}{(\htmlData{mx=10}{1 - \htmlData{mx=9}{\beta_{2}}}) \cdot \htmlData{mx=14}{\htmlData{mx=13}{\htmlData{mx=11}{g}(\htmlData{mx=12}{t})}^{2}}}}} and 0<ε0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{\varepsilon}, then

∀t,∣m(t)v(t)+ε∣≤(1−β1)2(1−β2)⋅(1−β12β2)\begin{aligned} & \htmlData{mx=0}{\forall} t, \\ & \quad \htmlData{mx=12}{\left\lvert \htmlData{mx=11}{\frac{\htmlData{mx=4}{\htmlData{mx=2}{m}(\htmlData{mx=3}{t})}}{\htmlData{mx=10}{\htmlData{mx=8}{\sqrt{\htmlData{mx=7}{\htmlData{mx=5}{v}(\htmlData{mx=6}{t})}}} + \htmlData{mx=9}{\varepsilon}}}} \right\rvert} \\ & \quad \quad \htmlData{mx=1}{\leq} \htmlData{mx=25}{\sqrt{\htmlData{mx=24}{\frac{\htmlData{mx=15}{(\htmlData{mx=14}{1 - \htmlData{mx=13}{\beta_{1}}})^{2}}}{\htmlData{mx=23}{(\htmlData{mx=17}{1 - \htmlData{mx=16}{\beta_{2}}}) \cdot \left(\htmlData{mx=22}{1 - \htmlData{mx=21}{\frac{\htmlData{mx=19}{\htmlData{mx=18}{\beta_{1}}^{2}}}{\htmlData{mx=20}{\beta_{2}}}}}\right)}}}}} \end{aligned}
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, v range over ℕ → ℝ: the statement must hold for every choice. ((g m v : ℕ → ℝ))
x ∈ TVariables
β₁, β₂ range over real numbers: the statement must hold for every choice. ((β₁ β₂ : ℝ))
<Less than
0 is strictly smaller than β₂. (0 < β₂)
aᵇPower
β₁ raised to the power 2. (β₁ ^ 2)
f(x)Function application
m applied to 0. Lean writes f x; papers write f(x). (m 0)
=Equality
m(0) and 0 are the same. (m 0 = 0)
∀For all
The statement after the comma holds for every t. (∀ t, m (t + 1) = β₁ * m t + (1 - β₁) * …)
RReal numbers
The real numbers: every point on the number line. (ℝ)
|x|Absolute value
The size of m(t)/(√(v(t)) + ε) without its sign. (|m t / (Real.sqrt (v t) + ε)|)
≤At most
|m(t)/(√(v(t)) + ε)| is less than or equal to √((1 − β₁)^2/((1 − β₂) · (1 − β₁^2/β₂))). (|m t / (Real.sqrt (v t) + ε)| ≤ Real.sq…)
a/bDivision
m(t) divided by √(v(t)) + ε. In Lean, dividing by 0 gives 0. (m t / (Real.sqrt (v t) + ε))
+Addition
t plus 1. (t + 1)
√Square root
The non-negative number whose square is v(t). (Real.sqrt (v t))
·Multiplication
β₁ times m(t). (β₁ * m t)
-Subtraction
1 minus β₁. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (1 - β₁)

Composes the moment-ratio capstone with the update bound: from the raw recurrences alone, every coordinate's Adam direction is at most C = √((1−β₁)²/((1−β₂)(1−β₁²/β₂))) at every step. For Baguette C < 7.3. 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
105 lines of Lean
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