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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Adaptive gradient algorithms and regret limits

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

RMSProp with an exponential second-moment estimate

Definition

Fix a dimension d≥1, an initial vector x1∈Rd, a finite or infinite sequence of finite gradient vectors gt∈Rd, positive step sizes ηt>0, a decay parameter ρ∈[0,1), and a denominator constant ϵ>0. Use one-based coordinate notation zi:=z(i−1) for every vector z∈Rd and 1≤i≤d. Set r0=0. At each integer round t≥1, update the second-moment estimate coordinatewise by

rt,i=ρrt−1,i+(1−ρ)gt,i2(1≤i≤d),

and then set

xt+1,i=xt,i−ηtgt,irt,i+ϵ(1≤i≤d).

For a finite gradient sequence of length T≥0, perform rounds 1≤t≤T and return the trajectory (x1,…,xT+1); when T=0, return (x1). For an infinite gradient sequence, perform every round t≥1, thereby defining and returning the iterate sequence (xt)t≥1. The root, addition, and division in the update are coordinatewise. The procedure makes no convergence or regret claim.

Definition references

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: vectors in Rd have the stated real coordinates, and vector updates are interpreted coordinatewise.

Squares of nonzero elements are positive: every nonzero squared gradient coordinate is positive.

Multiplication by zero: 0⋅a=0: a zero gradient coordinate has square zero.

Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}: each nonnegative real has a unique nonnegative square root.

Ordered field

Sign rules for products and monotonicity of multiplication: the decay coefficients and their complements are nonnegative, so the moment recurrence preserves nonnegativity.

Bottou, Curtis, and Nocedal, RMSprop update

Reddi, Kale, and Kumar, adaptive-method setup

Remarks

This is the no-momentum RMSProp variant with a positive constant added outside the square root. It has no bias correction and defines only an update procedure.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Uncorrected Adam for online convex losses

Definition

Fix a finite horizon T≥1, an initial scalar x1∈[−1,1], a step parameter α>0, moment parameters β1∈[0,1) and β2∈(0,1), a denominator parameter ϵ≥0, and a sequence of scalar gradients g1,…,gT. Set m0=v0=0. At round t, update

mt=β1mt−1+(1−β1)gt,vt=β2vt−1+(1−β2)gt2.

The input is admissible only when vt+ϵ>0 at every update. Set αt=α/t, compute yt+1=xt−αtmt/vt+ϵ, and set xt+1=max⁡(−1,min⁡(1,yt+1)). Thus the iterate is projected onto [−1,1] after the update. Return the trajectory (x1,…,xT+1), whose final iterate is xT+1. The denominator parameter is added inside the square root. No moment bias correction is applied.

Definition references

The natural numbers N (von Neumann): the finite round indices are natural numbers.

Intervals of R: the nine order-convex forms, nondegeneracy, and length: [−1,1] is the closed scalar interval used for initialization and clipping.

Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}: every nonnegative real radicand has a unique nonnegative square root.

Maximum and minimum of a set

Every nonempty finite set of reals has a maximum and a minimum: the displayed clipping operation is the maximum and minimum of finite real sets.

Bias-corrected Adam update: the published bias-corrected Adam procedure instead divides by v^t+ϵ; it is a distinct update convention.

Reddi, Kale, and Kumar, Algorithm 1 and the inside-root epsilon variant

Kingma and Ba, Adam Algorithm 1

Remarks

This is the uncorrected scalar online variant used for the cited regret counterexamples. Its ϵ=0 mode is defined only on gradient sequences with positive vt at every round; the counterexample sequences in the theorem meet that condition. The fixed-ϵ mode uses the source's addition inside the square root, not the published bias-corrected denominator convention.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Uncorrected AMSGrad with a coordinatewise historical maximum

Definition

Fix a dimension d≥1, a finite horizon T≥1, a nonempty compact convex set F⊆Rd, an initial point x1∈F, positive step sizes αt>0, a fixed parameter β2∈(0,1), and parameters β1,t∈[0,1) for 1≤t≤T. Use one-based coordinate notation zi:=z(i−1) for every vector z∈Rd and 1≤i≤d. At round t, receive a finite subgradient vector gt∈Rd. Set m0=v0=v^0=0 and update coordinatewise by

mt,i=β1,tmt−1,i+(1−β1,t)gt,i,vt,i=β2vt−1,i+(1−β2)gt,i2,

v^t,i=max⁡{v^t−1,i,vt,i}.

If v^t,i>0, set yt+1,i=xt,i−αtmt,i/v^t,i; if v^t,i=0, set yt+1,i=xt,i. Choose any minimizer

xt+1∈argmin⁡x∈F∑i=1dv^t,i(xi−yt+1,i)2.

Return the trajectory through xT+1. The procedure uses the uncorrected first and second moments and the coordinatewise running maximum of the raw second moment. It contains no denominator stabilizer.

Definition references

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: vectors use the stated real coordinates and coordinatewise operations.

A convex subset of Rm contains every line segment between two of its points: the feasible set contains every segment between two of its points.

Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}: each nonnegative second-moment coordinate has a unique nonnegative square root.

Maximum and minimum of a set

Every nonempty finite set of reals has a maximum and a minimum: the maximum of the two real second-moment values exists.

For a nonempty subset of Rn with n≥1, compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: a continuous real-valued function on the nonempty compact feasible set attains its minimum.

Bias-corrected AMSGrad update: the published procedure instead maxes bias-corrected second moments; it is not this raw-moment update.

Reddi, Kale, and Kumar, AMSGrad Algorithm 2 and Theorem 4

Remarks

This is the source-matched online AMSGrad procedure for the regret theorem on this page. It is distinct from the published bias-corrected procedure. The moment well-definedness result justifies the zero-coordinate convention and existence of each finite minimization step.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Polynomial factors times a geometric rate decay to zero

Statement

For every real q∈[0,1), every p∈{1,2}, and every real δ>0, there is an integer N≥1 such that npqn<δ for every integer n≥N.

Statement references

Facts & Assumptions

Given: q∈[0,1), p∈{1,2}, and δ>0.

[F1]

For ∣r∣<1 the sequence rk is null, and for ∣r∣>1 the sequence ∣r∣k diverges to +∞: if 0≤s<1, then for every positive tolerance there is an index after which sk is below that tolerance.

[F2]

Every complete ordered field is Archimedean: for every real z there is an integer N≥1 with z<N.

[F3]

The natural numbers N (von Neumann) and Integer powers am: the powers and indices use natural-number recursion, with qn+1=qnq.

Proof

Given: q,p,δ as in the Statement.

1.1givenF4algebracases

If q=0, then npqn=0<δ for every n≥1, so take N=1. If q>0, set s=(q+1)/2, so q<s<1, and set A=2p−1∈{1,3}.

2.1givenF2F3F4step 1.1choosealgebra

For every integer n≥1, (1+1/n)p≤1+A/n: for p=1 this is equality; for p=2, expand to 1+2/n+1/n2 and use 1/n2≤1/n. By [F2] choose N0≥1 so large that q(1+A/N0)≤s. Then for every n≥N0, the ratio ((n+1)pqn+1)/(npqn)=q(1+1/n)p≤s.

3.1F1F3F4step 1.1step 2.1inductionchoosealgebra∎

Induction on k≥0 using step 2.1 gives (N0+k)pqN0+k≤N0pqN0sk. By [F1] choose K with N0pqN0sK<δ. The ratio bound in step 2.1 is at most s<1, so the sequence npqn is nonincreasing for n≥N0; hence every n≥N0+K satisfies npqn≤N0pqN0sK<δ. This proves the Statement in the q>0 case, and step 1.1 proves the q=0 case.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Clipping a negative prefix followed by a positive suffix

Statement

Let n∈N with n≥1, let x0∈[−1,1], and let k∈{0,…,n}. Let real increments δ1,…,δn satisfy δj≤0 for j≤k and δj≥0 for j>k. Define c(y)=max⁡{−1,min⁡{1,y}} and xj=c(xj−1+δj). Then

xn≥min⁡{1,x0+∑j=1nδj}.

Statement references

Maximum and minimum of a set: maximum and minimum of finite real sets define the clipping map c.

Intervals of R: the nine order-convex forms, nondegeneracy, and length: [−1,1] is the closed interval with endpoints −1 and 1.

Reddi, Kale, and Kumar, Appendix B

Facts & Assumptions

Given: n,x0,k,δ1,…,δn as in the Statement.

[F1]

Maximum and minimum of a set: min⁡{a,b} is the lesser element and max⁡{a,b} is the greater element of a two-element real set.

[F2]

Ordered field and Sign rules for products and monotonicity of multiplication: the real order is total, and adding the same real to both sides preserves order.

[F3]

The natural numbers N (von Neumann) and The principle of mathematical induction: the finite sequence indices are natural numbers, and induction proves assertions across all of them.

[F4]

Finite sums and finite products, by recursion and Laws of finite sums and finite products: the displayed partial sums are finite sums and obey the termwise addition laws.

Proof

Given: The data and sign conditions in the Statement.

1.1givenF1F2F3F4induction

Put sj=x0+∑i=1jδi and s0=x0. For 0≤j≤k, induction gives xj≥sj and xj≤1: the base is x0=s0∈[−1,1]; at a next nonpositive increment, xj−1+δj≤1, so clipping replaces it only by its maximum with −1, which is no smaller than xj−1+δj≥sj.

2.1givenF1F2F3F4step 1.1inductioncases∎

For k≤j≤n, induction gives xj≥min⁡{1,sj}. The base follows from step 1.1 and xk≤1. If sj−1≤1, then xj−1≥sj−1 and δj≥0, so y:=xj−1+δj≥sj. If sj<−1, then c(y)≥−1>sj. If −1≤sj≤1, both 1 and y are at least sj, so min⁡{1,y}≥sj and then c(y)=max⁡{−1,min⁡{1,y}}≥sj. If sj>1, then y>1 and c(y)=1. In all cases c(y)≥min⁡{1,sj}. If instead sj−1>1, the induction bound and xj−1≤1 imply xj−1=1; a nonnegative increment keeps the clipped value at 1. Taking j=n proves the Statement.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

RMSProp, Adam, and AMSGrad moment updates are well-defined

Statement

For every finite admissible run of the RMSProp, uncorrected Adam, or uncorrected AMSGrad procedures defined on this page, all moment coordinates are finite and all second-moment coordinates are nonnegative. Every denominator specified by the procedure is positive. For AMSGrad, a zero historical-maximum coordinate has zero current and previous first moments, and the raw second moment is positive whenever its historical maximum is positive; the zero-ratio convention is therefore consistent. Each weighted minimization over the compact feasible set attains a minimum.

Statement references

RMSProp with an exponential second-moment estimate: defines the exponential second-moment recurrence and positive stabilizer.

Uncorrected Adam for online convex losses: defines the uncorrected scalar moment recurrences, admissible denominators, and clipping update.

Uncorrected AMSGrad with a coordinatewise historical maximum: defines the raw-moment recurrence, historical maximum, zero-coordinate convention, and compact-set minimization.

Reddi, Kale, and Kumar, adaptive-moment procedures

Bottou, Curtis, and Nocedal, RMSprop recurrence

Facts & Assumptions

Given: Finite gradient inputs for the three procedures, their parameter ranges and initial zero moments, and the admissibility conditions in their Definitions.

[F1]

RMSProp with an exponential second-moment estimate: r0=0 and rt,i=ρrt−1,i+(1−ρ)gt,i2, with 0≤ρ<1 and ϵ>0.

[F2]

Uncorrected Adam for online convex losses: m0=v0=0, vt=β2vt−1+(1−β2)gt2, and each used radicand vt+ϵ is positive.

[F3]

Uncorrected AMSGrad with a coordinatewise historical maximum: m0=v0=v^0=0, β2∈(0,1), vt,i=β2vt−1,i+(1−β2)gt,i2, and v^t,i=max⁡{v^t−1,i,vt,i}.

[F4]

The principle of mathematical induction: a property true at the initial index and preserved at each finite successor holds at every finite round.

[F5]

Squares of nonzero elements are positive, Multiplication by zero: 0⋅a=0, and Sign rules for products and monotonicity of multiplication: coordinate squares are nonnegative, and nonnegative coefficients preserve nonnegativity under sums and products.

[F7]
[F8]

For a nonempty subset of Rn with n≥1, compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: a compact subset of Rd is bounded, and a continuous real-valued function on a nonempty compact subset attains its minimum.

Proof

Given: The data and hypotheses in the Statement.

1.1F1F2F3F4F5F9giveninductionalgebra

For each procedure, induction over its finite horizon shows that every moment is a finite real and each second moment is nonnegative. The initial moments are zero. At a successor round, each recurrence is a finite sum of products of finite reals by [F1]–[F3] and [F9]; the coefficients ρ,1−ρ,β2,1−β2 are nonnegative, and each squared gradient is nonnegative by [F5], so the next second moment remains nonnegative.

2.1F1F2F6step 1.1givenalgebra

For RMSProp, [F6] supplies rt,i≥0, so rt,i+ϵ>0 by [F1]. For Adam, [F6] supplies vt+ϵ because vt≥0 and ϵ≥0; its positivity is the explicit admissibility condition in [F2]. Thus each stated quotient has a positive denominator.

2.2F3F5F7step 1.1giveninductionalgebra

In AMSGrad, both entries in each maximum are finite and nonnegative by [F3], [F5], and step 1.1. The maximum exists by [F7] and is nonnegative. If v^t,i=0, then v^t−1,i=vt,i=0. Since β2>0, 1−β2>0, and both summands defining vt,i are nonnegative, vt−1,i=0 and gt,i=0. Repeating this argument back to time zero gives every earlier gs,i=0; the first-moment recurrence then gives mt,i=mt−1,i=0. Conversely, if v^t,i>0, some vs,i>0 for s≤t, and then vu+1,i≥β2vu,i>0 for each u≥s, so vt,i>0. The zero-ratio convention in [F3] is therefore consistent with the recurrence.

3.1F3F8F9step 2.2givenalgebra∎

Fix a round and its finite vector y and nonnegative weights qi=v^t,i. Since F is compact, [F8] bounds ∥x∥2 by some B for x∈F. For x,z∈F, coordinatewise expansion and [F9] give ∣∑iqi(xi−yi)2−∑iqi(zi−yi)2∣≤∥x−z∥2∑iqi(2B+2∥y∥2). Thus the objective is continuous on F. Since F is nonempty and compact, [F8] gives a minimizer. The objective can be constant along zero-weight coordinates, so allowing any minimizer is necessary and sufficient. This proves the remaining well-definedness claim.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

One-step regret inequality for AMSGrad

Statement

At an AMSGrad round, let F⊆Rd be nonempty, compact, and convex, let xt,xt+1∈F, let ℓt:F→R be convex, and let gt∈∂ℓt(xt). Take αt>0 and β1,t∈[0,1). Write qt,i=v^t,i, Qt=diag⁡(qt,1,…,qt,d), and ∥z∥Qt2:=∑iqt,izi2,Et(z):=∑i:qt,i>0zi2qt,i, with zero contribution from coordinates where qt,i=0. For the AMSGrad update, every such zero coordinate has mt,i=mt−1,i=0. Then for every comparator u∈F, ℓt(xt)−ℓt(u)≤∥xt−u∥Qt2−∥xt+1−u∥Qt22αt(1−β1,t)+αt2(1−β1,t)Et(mt)+β1,tαt2(1−β1,t)Et(mt−1)+β1,t2αt(1−β1,t)∥xt−u∥Qt2.

Statement references

Uncorrected AMSGrad with a coordinatewise historical maximum: defines the raw-moment recurrence, weighted minimization, and zero-coordinate convention.

RMSProp, Adam, and AMSGrad moment updates are well-defined: zero historical maximum implies zero current and previous first moments.

Subgradients and the subdifferential of a convex function: a subgradient gives the supporting inequality for the loss.

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: Cauchy–Schwarz and the Euclidean norm identities used in the weighted Young bound.

Reddi, Kale, and Kumar, Appendix D

Facts & Assumptions

Given: The AMSGrad state and update, a convex loss, a subgradient at xt, a comparator u∈F, and αt>0, β1,t∈[0,1).

[F1]

Uncorrected AMSGrad with a coordinatewise historical maximum: mt,i=β1,tmt−1,i+(1−β1,t)gt,i, and xt+1 minimizes the weighted squared distance from the specified yt+1.

[F2]

RMSProp, Adam, and AMSGrad moment updates are well-defined: Qt has finite nonnegative diagonal entries; if qt,i=0, then mt,i=mt−1,i=0.

[F3]

Subgradients and the subdifferential of a convex function: ℓt(xt)−ℓt(u)≤⟨gt,xt−u⟩.

[F4]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: the inner product and squared weighted norms are finite coordinate sums.

[F5]

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: ∣⟨a,b⟩∣≤∥a∥2∥b∥2; in particular the finite weighted-coordinate form follows by applying it to ai=mt−1,i/qt,i and bi=qt,i(xt,i−ui) on positive-weight coordinates.

Proof

Given: The state, update, convex loss, subgradient, comparator, and positive step size in the Statement.

1.1F1F2F4F6givenalgebra

Put yi=xt,i−αtmt,i/qt,i when qt,i>0 and yi=xt,i otherwise. Let p=xt+1, D=∥u−p∥Qt2≥0, and c=∑iqt,i(pi−yi)(ui−pi). Convexity of F makes p+s(u−p)∈F for 0<s≤1. Minimality of p gives 0≤2sc+s2D. If c<0, set a=−c>0 and s=a/(D+a+1); then 0<s<1 and sD<a, so 2sc+s2D=s(−2a+sD)<−sa<0, a contradiction. Thus c≥0, and expansion of ∥u−y∥Qt2 yields ∥p−u∥Qt2≤∥y−u∥Qt2.

2.1F1F2F4F6step 1.1givenalgebra

Expanding the right side from step 1.1 and using mt,i=0 whenever qt,i=0 gives ⟨mt,xt−u⟩≤∥xt−u∥Qt2−∥xt+1−u∥Qt22αt+αt2Et(mt). This follows coordinatewise from yi=xt,i−αtmt,i/qt,i on positive weights; zero-weight coordinates contribute zero.

3.1F1F2F3F5F6step 2.1givenalgebra∎

Rearranging the first-moment recurrence in [F1] gives ⟨gt,xt−u⟩=⟨mt,xt−u⟩−β1,t⟨mt−1,xt−u⟩1−β1,t. By [F2] zero-weight coordinates vanish; by [F5], weighted Cauchy–Schwarz on positive weights gives ∣⟨mt−1,xt−u⟩∣≤Et(mt−1)1/2∥xt−u∥Qt. For nonnegative a,b, 0≤(αta−b/αt)2 expands to ab≤(αta2+b2/αt)/2, which bounds the negative cross term. Substituting step 2.1 and the subgradient inequality [F3] gives the stated four terms.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

AMSGrad momentum energy is bounded by a coordinatewise weighted gradient sum

Statement

Run uncorrected AMSGrad for a finite horizon T≥1 with αt=α/t, α>0, β2∈(0,1), 0≤β1,t≤β1<β2, and γ=β1/β2. For each coordinate, interpret mt,i2/v^t,i as zero when v^t,i=0. Then ∑t=1Tαt∑i=1dmt,i2v^t,i≤α(1−β1)(1−γ)1−β2∑i=1d∑t=1T∣gt,i∣t.

Statement references

Uncorrected AMSGrad with a coordinatewise historical maximum: gives the moment recurrences, historical maximum, and time-varying first-moment weights.

RMSProp, Adam, and AMSGrad moment updates are well-defined: all moment coordinates are finite and all second-moment coordinates are nonnegative; a zero second-moment coordinate has zero first moment.

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: finite-coordinate Cauchy–Schwarz.

Reddi, Kale, and Kumar, Appendix D

Facts & Assumptions

Given: The finite gradient vectors and AMSGrad recurrences in the Statement, with the stated parameter ranges.

[F1]

Uncorrected AMSGrad with a coordinatewise historical maximum: mt,i=β1,tmt−1,i+(1−β1,t)gt,i and vt,i=β2vt−1,i+(1−β2)gt,i2.

[F2]

RMSProp, Adam, and AMSGrad moment updates are well-defined: 0≤vt,i≤v^t,i; because β2>0, vt,i=0 exactly when v^t,i=0, and then all corresponding moment and gradient entries are zero.

[F4]

Integer powers am, The natural numbers N (von Neumann), Finite sums and finite products, by recursion, and Laws of finite sums and finite products: β0=1, βn+1=βnβ, finite sums distribute over addition, and finite sums may be reindexed.

Proof

Given: The hypotheses in the Statement.

1.1F1F6giveninductionalgebra

Expanding the first-moment recurrence [F1] recursively gives mt,i=∑j=1tat,jgj,i, where at,t=1−β1,t and at,j=β1,tat−1,j for j<t. Induction gives 0≤at,j≤β1t−j: the terminal coefficient is at most 1, and each earlier coefficient acquires one factor at most β1.

2.1F3F4F5F6step 1.1givenalgebra

By [F3] applied to at,j and at,jgj,i, mt,i2≤(∑j=1tat,j)(∑j=1tat,jgj,i2)≤11−β1∑j=1tβ1t−jgj,i2, because ∑k=0t−1β1k=(1−β1t)/(1−β1)≤1/(1−β1). The identity follows by multiplying the finite sum by 1−β1 and telescoping ∑k=0t−1(β1k−β1k+1)=1−β1t using [F4]; the denominator is positive by [F6].

3.1F1F2F5F6step 2.1giveninductionalgebra

Iterating the second-moment recurrence [F1] gives vt,i=(1−β2)∑j=1tβ2t−jgj,i2. By [F2] and [F5], v^t,i≥vt,i. If this value is zero, [F2] makes the energy term zero. Otherwise, for each nonzero gj,i, vt,i≥1−β2 β2t−j∣gj,i∣. The root recursion gives β2n=(β2)n for every integer n≥0; induction proves this because both sides are nonnegative and have the same square. Also β1n/(β2)n=(β1/β2)n by the integer-power recursion, so β1n/β2n=γn. Using step 2.1 termwise in the finite sum therefore gives mt,i2v^t,i≤1(1−β1)1−β2∑j=1tγt−j∣gj,i∣. Zero gradient terms contribute zero.

4.1F4F6step 2.1step 3.1givenalgebra∎

Multiply step 3.1 by α/t, sum over t and coordinates, and reorder the finite sums. Since 1/t≤1/j for t≥j and (1−γ)∑k=0T−jγk=1−γT−j+1≤1 by the same finite telescoping identity from step 2.1, ∑t=1Tαt∑imt,i2v^t,i≤α(1−β1)1−β2∑i∑j=1T∣gj,i∣∑t=jTγt−jt≤α(1−β1)(1−γ)1−β2∑i∑j=1T∣gj,i∣j. This is the stated inequality.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Adam can have nonzero average regret on a one-dimensional convex sequence

Statement

For the scalar Adam update on [−1,1] with step sizes αt=α/t, fixed parameters β1∈[0,1) and β2∈(0,1), no bias correction, and clipping after every update, define the best-comparator regret by RT=∑t=1Tℓt(xt)−min⁡u∈[−1,1]∑t=1Tℓt(u). For every α>0 and every β1<β2, there is an even integer C≥2 such that, with ϵ=0, x1=1, and the periodic linear losses ℓkC+1(x)=Cx,ℓkC+j(x)=−x(2≤j≤C, k≥0), the algorithm receives their slopes gkC+1=C and gkC+j=−1, and its regret satisfies lim inf⁡T→∞RT/T≥2/C>0.

For every fixed ϵ>0, there is also a counterexample for the update with denominator vt+ϵ: set α=1/10, β1=0, β2=1/45, x1=1, and repeat the losses ℓ3k+1(x)=3ϵ x,ℓ3k+2(x)=ℓ3k+3(x)=−ϵ x(k≥0). The algorithm receives the corresponding slopes g3k+1=3ϵ and g3k+2=g3k+3=−ϵ. Then lim inf⁡T→∞RT/T≥2ϵ/3>0. In both cases the losses are convex on [−1,1], and the comparator is selected separately for each finite horizon.

Statement references

Uncorrected Adam for online convex losses: defines the uncorrected moment recurrences, inside-root denominator parameter, step schedule, and clipping update used in both constructions.

Polynomial factors times a geometric rate decay to zero: ensures that a sufficiently large even period satisfies the finitely many geometric-moment inequalities in the zero-epsilon construction.

Clipping a negative prefix followed by a positive suffix: bounds the clipped endpoint of a period whose update increments have a nonpositive prefix and a nonnegative suffix.

Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾: defines the limit inferior in the two average-regret conclusions.

Intervals of R: the nine order-convex forms, nondegeneracy, and length: [−1,1] is the closed interval on which the losses and clipping operate.

Reddi, Kale, and Kumar, Appendix B and Appendix F

Adam diverges on a periodic bounded-gradient loss sequence: the published bias-corrected periodic example is a distinct update convention and is not used as a proof of either local Adam branch.

Facts & Assumptions

Given: The Adam recurrence and clipping rule, the parameters in the Statement, and the periodic gradients defining the losses.

[F1]

Uncorrected Adam for online convex losses: mt=β1mt−1+(1−β1)gt, vt=β2vt−1+(1−β2)gt2, and xt+1=max⁡(−1,min⁡(1,xt−αtmt/vt+ϵ)) on admissible inputs.

[F2]

RMSProp, Adam, and AMSGrad moment updates are well-defined: the stated recurrences have finite nonnegative second moments and every denominator used in the constructions is positive.

[F3]

Polynomial factors times a geometric rate decay to zero: for q∈[0,1) and p∈{1,2}, npqn→0; hence all finitely many eventual inequalities required below hold for every sufficiently large even C.

[F4]

Clipping a negative prefix followed by a positive suffix: if clipped scalar updates have nonpositive increments through an index j and nonnegative increments afterward, their final iterate is at least the minimum of 1 and the initial iterate plus the sum of the increments.

[F5]

The principle of mathematical induction, The natural numbers N (von Neumann), and Integer powers am: finite recurrences may be iterated, and powers use q0=1, qn+1=qnq.

[F7]

Convex and strictly convex functions on Euclidean convex sets and A convex subset of Rm contains every line segment between two of its points: a set is convex when it contains each segment between its points, and a function is convex when it lies below each chord.

[F9]

Every complete ordered field is Archimedean: the positive integers are unbounded in R.

[F13]

Intervals of R: the nine order-convex forms, nondegeneracy, and length: the domain [−1,1] is the closed interval with endpoints −1 and 1.

[F10]

Finite sums and finite products, by recursion and Laws of finite sums and finite products: finite sums may be reindexed and distributed; multiplying ∑r=0Nqr by 1−q yields the telescoping sum ∑r=0N(qr−qr+1).

Proof

Given: The assumptions in the Statement.

1.1F1F2F3F5F6F9givenchoosealgebra

First take ϵ=0. Every input gradient is either C or −1, so v1=(1−β2)C2>0 and vt+1≥β2vt>0; thus every no-epsilon denominator is positive. Put γ=β1/β2<1 and H=(1−β1)(1+γ/(1−γ))1−β2. Choose an even C≥2 so large that C≥β11−β1,β1C/2≤12,(1−β1)Cβ1C/2≤14,C2β2C/2≤1, (1−β1)Cβ1C−1≤1−β1C−1,C16>H,C82>H. Such an even integer exists by [F3], using n=C/2 for the half-period powers and then increasing C to satisfy the remaining linear and square-root bounds.

1.2F1F3F5F6giveninductionalgebra

Let s=kC be a period start and let ms be its first moment. Induction over periods gives −1≤ms≤0: it holds at m0=0; if it holds at s, then ms+1=β1ms+(1−β1)C lies in [0,(1−β1)C], while after the following C−1 gradients equal to −1, ms+C=β1C−1ms+1−(1−β1C−1). The last choice inequality gives ms+C≤0, and ms+1≥0 gives ms+C≥−(1−β1C−1)≥−1. Thus every high-gradient update has ms+1≥0. At the later positions j=2,…,C, ms+j=β1j−1(ms+1+1)−1, which is nonincreasing in j. Hence the update increments δs+j=−αs+jms+j/vs+j have a nonpositive prefix and a nonnegative suffix.

1.3F1F5F6F10givenalgebra

At the high-gradient round, vs+1≥(1−β2)C2 and ms+1≤(1−β1)C, so its negative increment has magnitude at most α(1−β1)1−β2s+1. For a low-gradient position j≥2, the high-gradient contribution gives vs+j≥(1−β2)β2j−1C2. If its increment is negative, then ms+j>0 and the displayed moment formula gives ms+j≤(1−β1)Cβ1j−1. Therefore the sum of all negative increments at low-gradient positions is at most α(1−β1)1−β2s+1∑j=2Cγj−1≤α(1−β1)γ(1−γ)1−β2s+1. Indeed, multiplying the finite geometric sum by 1−γ>0 gives ∑r=1C−1(γr−γr+1)=γ−γC≤γ, by [F5], [F6], and [F10]. The total magnitude of negative increments in a period is thus at most αH/s+1.

1.4F1F3F5F6givenalgebra

For j≥C/2+1, the moment formula and the choice of C give ms+j≤(1−β1)Cβ1C/2−(1−β1C/2)≤−14. Also vs+1≤C2, since the second-moment recurrence averages values in [0,C2], and subsequent unit-gradient updates give vs+j=β2j−1vs+1+1−β2j−1≤1+C2β2C/2≤2. There are C/2 such updates. Their positive increments sum to at least αC/(82s+C). For k≥1, s+C≤2(s+1), so this is at least αC/(16s+1); for k=0 it is at least αC/(82). The final two conditions on C therefore show that the sum Dk=∑j=1Cδs+j of the period's increments is positive for every k, and for k≥1 it satisfies Dk≥α(C/16−H)kC+1.

1.5F1F5F6giveninductionalgebra

Now fix ϵ>0 and use the second construction, with b=1/45, c=44/45, and wt=vt/ϵ. Since gt2/ϵ is either 9 or 1, the recurrence wt=bwt−1+c(gt2/ϵ) gives 0≤wt≤9 by induction. At a high-gradient time 3k+1, w3k+1≥9c=44/5, so the magnitude of the negative update is at most 3α3k+149/5=35 α73k+1. At the following two low-gradient times, the recurrence and w3k+1≤9 give w3k+2≤53/45<5/4 and w3k+3≤2033/2025<5/4. Their denominator roots are therefore less than 3/2, so their positive increments sum to at least 2α3(13k+2+13k+3).

2.1F4F6F9step 1.2step 1.4giveninductionalgebra

Apply [F4] to each period using the sign pattern from step 1.2. It gives x(k+1)C+1≥min⁡{1,xkC+1+Dk}. The positive lower bounds in step 1.4 have divergent sum: for every N≥1, ∑k=1N1kC+1≥NNC+1≥NC+1, which is unbounded by [F9]. Iterating the endpoint inequality, the boundary iterate therefore reaches 1 after finitely many periods and equals 1 at every later period boundary.

3.1F6F7F8F9F13step 2.1givenalgebra

For x,y∈[−1,1] and θ∈[0,1], the nonnegative coefficients 1−θ and θ give −1=(1−θ)(−1)+θ(−1)≤(1−θ)x+θy≤(1−θ)1+θ1=1, so the interval is convex by [F7]. For every real slope a, the loss ℓ(x)=ax satisfies ℓ((1−θ)x+θy)=(1−θ)ℓ(x)+θℓ(y); hence it is convex by [F7]. Every finite-horizon gradient sum is positive: a complete period sums to 1, and a nonempty proper prefix of a period sums to one of C,C−1,…,2. Thus u=−1 minimizes the cumulative linear loss on [−1,1]. In each complete period after the boundary iterate reaches 1, the first round has regret 2C and each of the other C−1 rounds has regret at least −2, since every iterate lies in [−1,1]; each such period therefore contributes at least 2. Each incomplete period contributes at least −2C, and the finitely many initial periods contribute a fixed finite amount. Hence there is a constant K independent of T such that RT≥2⌊T/C⌋−K≥2T/C−(2+K). For every N, all T≥N consequently satisfy RT/T≥2/C−(2+K)/N. By [F8] and the unboundedness in [F9], taking the supremum of these tail infima gives lim inf⁡T→∞RT/T≥2/C.

3.2F4F6F9step 2.1step 1.5givenalgebra

For k≥1, 3k+2≤54(3k+1) and 3k+3≤32(3k+1). Also 2/5>8/9, 2/3>4/5, and 35/7<27/28, so the low-gradient sum in step 1.5 exceeds the high-gradient magnitude by at least α3k+1(152135−2728)>0. For the first cycle, its high update has magnitude 35α/7<α=1/10, so x2>0.9 and the two positive low updates ensure x4>0.9. The displayed cycle lower bounds have divergent sum, since ∑k=1N(3k+1)−1/2≥N/3N+1≥N/2. The three update increments in every cycle have a nonpositive prefix and a nonnegative suffix. By [F4], their positive cumulative drift makes the boundary iterates reach 1 and remain 1 at cycle boundaries, as in step 2.1.

4.1F7F8F9step 3.1step 3.2givenalgebra∎

The convexity of the interval and of all linear losses follows from step 3.1, which covers every real slope. Every finite-horizon gradient sum is positive because every complete cycle sums to ϵ and the proper nonempty prefixes sum to 3ϵ or 2ϵ. Hence u=−1 is a best comparator. A full cycle after the boundary iterate reaches 1 has regret at least 6ϵ−4ϵ=2ϵ; every incomplete cycle has regret at least −4ϵ, and the initial finite segment contributes a fixed amount. Thus there is a constant Kϵ independent of T with RT≥2ϵ⌊T/3⌋−Kϵ≥2ϵT/3−(2ϵ+Kϵ). Taking tail infima and then their supremum as in step 3.1 gives lim inf⁡T→∞RT/T≥2ϵ/3.

Remarks

The positive-epsilon branch fixes β1=0 and β2=1/45; it does not claim failure for every fixed parameter pair. The zero-epsilon branch applies to every fixed pair satisfying β1<β2. Both are online convex-loss statements for the stated uncorrected update and do not assert a claim about bias-corrected Adam.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

AMSGrad has an online convex regret bound under decaying first-moment weights

Statement

Let d,T∈N with d,T≥1. Let F⊆Rd be nonempty, compact, and convex, choose x1∈F, and suppose a finite D∞≥0 satisfies ∣xi−ui∣≤D∞ for every x,u∈F and coordinate i. Let ℓ1,…,ℓT:F→R be continuous convex losses, choose gt∈∂ℓt(xt), and assume ∣gt,i∣≤G∞ for all t,i, where G∞≥0. Run uncorrected AMSGrad with αt=α/t for α>0, fixed β2∈(0,1), β1=β1,1∈[0,1), 0≤β1,t≤β1, and γ=β1/β2<1. For a comparator u∈F, define RT(u)=∑t=1T(ℓt(xt)−ℓt(u)); the best-comparator regret is RT=sup⁡u∈FRT(u). Then RT(u)≤D∞2T2α(1−β1)∑i=1dv^T,i+D∞2(1−β1)2∑t=1T∑i=1dβ1,tv^t,iαt+α(1−β1)2(1−γ)1−β2∑i=1d∑t=1T∣gt,i∣t. If, in addition, β1,t=β1λt−1 for a fixed λ∈[0,1) or β1,t=β1/t, then with the other displayed parameters fixed, the best-comparator regret satisfies RT=O(T).

Statement references

Uncorrected AMSGrad with a coordinatewise historical maximum: defines the uncorrected raw-moment update and coordinatewise historical maximum used here.

Convex and strictly convex functions on Euclidean convex sets: convexity supplies the subgradient regret inequality.

Subgradients and the subdifferential of a convex function: gt∈∂ℓt(xt) means ℓt(u)≥ℓt(xt)+⟨gt,u−xt⟩.

One-step regret inequality for AMSGrad: gives the per-round projection and momentum inequality.

AMSGrad momentum energy is bounded by a coordinatewise weighted gradient sum: bounds the accumulated preconditioned momentum term.

For a nonempty subset of Rn with n≥1, compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: continuous losses on the nonempty compact set have bounded sums, so the best-comparator regret is finite.

Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined: finite sums of continuous real-valued losses are continuous.

Reddi, Kale, and Kumar, Theorem 4 and Appendix D

Hazan, Introduction to Online Convex Optimization, §5.6.1

Facts & Assumptions

Given: The finite horizon, compact convex domain, losses, subgradients, coordinate bounds, AMSGrad update, and parameters in the Statement.

[F1]

Uncorrected AMSGrad with a coordinatewise historical maximum: vt,i=β2vt−1,i+(1−β2)gt,i2, v^t,i=max⁡{v^t−1,i,vt,i}, qt,i:=v^t,i is nondecreasing in t, and the weighted projection uses ∑iqt,i(xi−yt+1,i)2.

[F2]

RMSProp, Adam, and AMSGrad moment updates are well-defined: all moment coordinates are finite and all second-moment coordinates are nonnegative; zero historical-maximum coordinates have zero first moments.

[F3]

One-step regret inequality for AMSGrad: for each comparator u, the one-step loss difference is bounded by a potential difference, two momentum-energy terms, and a first-moment correction term.

[F4]

AMSGrad momentum energy is bounded by a coordinatewise weighted gradient sum: ∑tαtEt(mt)≤α(1−β1)(1−γ)1−β2∑i,t∣gt,i∣/t.

[F5]

Subgradients and the subdifferential of a convex function: ℓt(xt)−ℓt(u)≤⟨gt,xt−u⟩.

[F6]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn and Laws of finite sums and finite products: weighted squared distances and sums over coordinates expand termwise and telescope.

[F7]

Squaring is monotone on the nonnegatives and Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}: v^t,i≥v^t−1,i≥0 implies qt,i≥qt−1,i.

[F8]

Ordered field, Sign rules for products and monotonicity of multiplication, and Inverses of positives are positive, and reciprocation reverses order: positive denominators preserve inequalities, and 1−β1>0, 1−γ>0, and 1−β2>0.

[F10]

The natural numbers N (von Neumann), Integer powers am, Finite sums and finite products, by recursion, and Laws of finite sums and finite products: integer powers obey rn+1=rnr, and finite sums may be telescoped and reindexed.

[F11]

The principle of mathematical induction: induction propagates a base bound through every finite moment update.

Proof

Given: All data and hypotheses in the Statement. Fix an arbitrary comparator u∈F.

1.1F3F5F6givenalgebra

By [F5] and the one-step inequality [F3], sum over t=1,…,T. With qt,i=v^t,i, Φt=∑iqt,i(xt,i−ui)2, and Ψt=∑iqt,i(xt+1,i−ui)2, the resulting bound is RT(u)≤∑t=1T[Φt−Ψt2αt(1−β1,t)+αtEt(mt)2(1−β1,t)+β1,tαtEt(mt−1)2(1−β1,t)+β1,tΦt2αt(1−β1,t)].

1.2F1F2F7F8F11giveninductionalgebra

The moment recurrence and ∣gt,i∣≤G∞ give 0≤vt,i≤G∞2 by induction; since v^t,i is the maximum of the raw second moments through round t, qt,i≤G∞. Also ∑t=1T∣gt,i∣t≤G∞∑t=1T1t≤2G∞T, because 1/t≤2(t−t−1) for t≥1. Thus the first and third terms in the bound are O(T).

2.1F1F2F8step 1.1givenalgebra

Since qt,i is nondecreasing and αt is nonincreasing, Et(mt−1)≤Et−1(mt−1) and αt≤αt−1 for t≥2; the t=1 previous-moment term is zero. Also β1,t/(1−β1,t)≤β1/(1−β1). Thus the two energy sums in step 1.1 are at most (12(1−β1)+β12(1−β1))∑t=1TαtEt(mt)≤11−β1∑t=1TαtEt(mt).

2.2F1F6F7F8step 1.1givenalgebra

In the potential sum of step 1.1, drop the final nonpositive term −ΨT/(2αT(1−β1,T)), since ΨT≥0. Fix coordinate i and put rt,i=qt,i/αt, with r0,i=0. The unpaired first positive potential is at most D∞2r1,i/(2(1−β1)). For each t≥2, pair the positive potential at round t with the negative potential from round t−1. Its coefficient difference is qt,iαt(1−β1,t)−qt−1,iαt−1(1−β1,t−1)=rt,i−rt−1,i1−β1,t−1+rt,i(11−β1,t−11−β1,t−1). The first difference rt,i−rt−1,i is nonnegative by [F1], [F7], and αt=α/t. The reciprocal difference is nonpositive when β1,t≤β1,t−1; otherwise it equals (β1,t−β1,t−1)/((1−β1,t)(1−β1,t−1))≤β1,t/(1−β1)2. Thus the sum of all first coefficient differences, including the first-round term, is at most D∞2rT,i/(2(1−β1)), because r1,i+∑t=2T(rt,i−rt−1,i)=rT,i. Summing coordinates gives D∞22αT(1−β1)∑iqT,i. The coefficient correction and the explicit last term of step 1.1 together contribute at most D∞22(1−β1)2∑t,iβ1,tqt,iαt+D∞22(1−β1)∑t,iβ1,tqt,iαt≤D∞2(1−β1)2∑t,iβ1,tqt,iαt, because 2−β1≤2.

3.1F4step 2.1step 2.2givenalgebra

Apply [F4] to the energy bound in step 2.1 and combine with step 2.2. Since 1/αT=T/α, this gives the displayed regret inequality.

4.1F4F6F8F9F10F12step 1.2givenalgebra∎

The cumulative loss u↦∑t=1Tℓt(u) is continuous by [F12], so [F9] gives a minimizer on F. The bound proved for arbitrary u is uniform in u, hence it also holds for the best comparator. For any r∈[0,1), [F10] gives the finite geometric identity (1−r)∑k=0Nrk=∑k=0N(rk−rk+1)=1−rN+1≤1. For β1,t=β1λt−1, use t≤t and reindex the finite sum to obtain ∑t=1Ttλt−1=∑j=1T∑t=jTλt−1≤11−λ∑j=1Tλj−1≤1(1−λ)2 so ∑tβ1,tt is bounded independently of T. For β1,t=β1/t, it is at most β1∑t=1T1/t≤2β1T by step 1.2. Therefore the second term is bounded by a constant for the geometric schedule and by a constant times T for the reciprocal schedule; both give RT=O(T).

Remarks

The theorem concerns online convex loss regret under the stated schedules. It does not imply convergence of a nonconvex neural-network training objective. A constant positive β1,t=β1 is not one of the two schedules in the O(T) conclusion.

4 · Examples, counterexamples and false statements

None yet.

Sources