Alphabeta Math
Pipeline-generated
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.

✓ 8 results · all verified · 6 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full by a delegated reviewing agent on the owner's instruction; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Open frontiers in calibrated and robust learned inversion — Examples and Counterexamples

1 · Prerequisites

None. This page is self-contained.

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Exact Gaussian credible-ball coverage with a score-error budget of zero

Example

Let n≥1, σ,τ>0, and 0<γ<1. Given a scalar truth θ∈R, observe independent Yi∼N(θ,σ2), i=1,…,n, and use the proper prior θ∼N(0,τ2). Write Yˉ=n−1∑iYi, vn=(τ−2+nσ−2)−1, an=vnnσ−2, and zγ=Φ−1(1−γ/2), where Φ is the standard normal cumulative distribution function. The exact posterior is N(anYˉ,vn) and its central (1−γ) credible interval is Cn(Yˉ)=[anYˉ−zγvn,anYˉ+zγvn]. Its frequentist coverage at truth θ is Pθ{θ∈Cn(Yˉ)}=Φ ⁣(zγvn+(1−an)θanσ/n)−Φ ⁣(−zγvn+(1−an)θanσ/n).

Facts & Assumptions

Given: The normal observation model, the proper N(0,τ2) prior, the posterior credible interval, n≥1, σ,τ>0, and 0<γ<1 from the Example.

[A1]

The likelihood is proportional to exp⁡(−12σ2∑i=1n(Yi−θ)2) and the prior density is proportional to exp⁡(−θ2/(2τ2)).

[A2]

Φ(t)=P(Z≤t) for Z∼N(0,1), and zγ is its (1−γ/2) quantile.

Verification

1.1givenA1A2algebra

Multiplying the likelihood by the prior and completing the square in θ gives posterior precision τ−2+nσ−2 and posterior mean (τ−2+nσ−2)−1nσ−2Yˉ=anYˉ. Thus the posterior variance is vn and the stated interval has posterior mass 1−γ.

2.1givenA1A2step 1.1algebra∎

Under the fixed truth θ, anYˉ−θ is normal with mean −(1−an)θ and standard deviation anσ/n. The event θ∈Cn(Yˉ) is exactly −zγvn≤anYˉ−θ≤zγvn; standardizing the two endpoints gives the displayed difference of normal cumulative probabilities.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A proper Gaussian prior gives nonuniform finite-sample coverage on an unbounded truth class

Example

Fix n≥1, σ,τ>0, and 0<γ<1. Observe independent Yi∼N(θ,σ2) under truth θ∈R and use prior N(0,τ2). Let vn=(τ−2+nσ−2)−1, an=vnnσ−2, and zγ=Φ−1(1−γ/2), where Φ is the standard normal cumulative distribution function. The exact posterior credible interval is [anYˉ−zγvn,anYˉ+zγvn]. Since 0<an<1, its repeated-sampling coverage at fixed n tends to zero as ∣θ∣→∞, despite exact posterior computation and exact sampling.

Facts & Assumptions

Given: The scalar normal observation model and proper prior in the Example, with n≥1, σ,τ>0, and 0<γ<1.

[A1]

Completing the Gaussian likelihood and prior gives posterior mean anYˉ, variance vn=(τ−2+nσ−2)−1, and central credible radius rn=zγvn.

[A2]

The standard normal cumulative distribution has vanishing tails: Φ(t)→0 as t→−∞ and Φ(t)→1 as t→+∞.

Verification

1.1givenA1algebra

At every fixed n, an=nτ2/(σ2+nτ2) lies strictly between zero and one, and rn is finite and independent of the observation. Under truth θ, anYˉ−θ has mean −(1−an)θ and fixed positive standard deviation anσ/n.

1.2A1algebra

Standardizing the event in the Example gives cn(θ)=Φ ⁣(rn+(1−an)θanσ/n)−Φ ⁣(−rn+(1−an)θanσ/n).

2.1A1A2step 1.2algebra∎

As θ→+∞, both arguments in step 1.2 tend to +∞, so both cumulative probabilities tend to one by [A2] and their difference tends to zero. As θ→−∞, both arguments tend to −∞ and their difference again tends to zero. Thus the infimum coverage over θ∈R is zero at each fixed n.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A scalar inverse with a conditioning-dependent shift modulus

Example

Fix R,a0,σ>0 and 0≤ρA<a0/2. The working inverse is g(y)=y/a0. Under a test operator a=a0+δ with ∣δ∣≤ρA, the observation is Y=aθ+ε for θ∈[−R,R] and independent ε∼N(0,σ2). For any two truth laws μ0,μ on [−R,R], define W1(μ,μ0) as the infimum of E∣Θ−Θ0∣ over their couplings. The risk is Rμ,δ(g)=Eθ∼μ,ε∣g((a0+δ)θ+ε)−θ∣2=δ2a02Eμθ2+σ2a02, and it satisfies Rμ,δ(g)≤(ρA2R2+σ2)/a02 and ∣Rμ,δ(g)−Rμ0,δ(g)∣≤2δ2Ra02W1(μ,μ0).

Facts & Assumptions

Given: R,a0,σ>0, 0≤ρA<a0/2, ∣δ∣≤ρA, θ∈[−R,R], ε∼N(0,σ2) independent of θ, and the working inverse g(y)=y/a0.

[A1]

The test law μ and source law μ0 are Borel probability measures on [−R,R]; W1 is the infimum of expected absolute distance over their couplings.

Verification

1.1givenA1algebra

Substitution gives g(Y)−θ=((a0+δ)θ+ε)/a0−θ=δθ/a0+ε/a0. Independence and Eε=0 make the cross term vanish, so the expected squared error is δ2Eμθ2/a02+σ2/a02. Since ∣θ∣≤R and ∣δ∣≤ρA, this is at most (ρA2R2+σ2)/a02.

2.1A1step 1.1algebra∎

For any coupling (Θ,Θ0) of μ,μ0, ∣Θ2−Θ02∣=∣Θ−Θ0∣∣Θ+Θ0∣≤2R∣Θ−Θ0∣. Taking expectations, then the infimum over couplings, gives ∣Eμθ2−Eμ0θ2∣≤2RW1(μ,μ0). Multiplication by δ2/a02 and step 1.1 yields the claimed risk-transfer bound.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Small absolute operator error does not give a uniform inverse bound without conditioning

Example

Fix a truth bound R>0. For every absolute operator-shift radius ρA>0, take the working scalar operator A0=ρA/2, the true operator A=0, the noiseless datum Y=Aθ at truth θ=R, and the working inverse g(y)=y/A0. Then ∣A−A0∣=ρA/2≤ρA but ∣g(Y)−θ∣2=R2. Thus no recovery modulus tending to zero with the absolute operator error can hold uniformly when the baseline inverse conditioning is allowed to vanish.

Facts & Assumptions

Given: R>0, any ρA>0, working operator A0=ρA/2, true operator A=0, truth θ=R, noiseless data Y=Aθ, and working inverse g(y)=y/A0.

[A1]

In the scalar model the operator norm is absolute value, and the recovery loss is squared absolute error.

Verification

1.1givenA1algebra

The absolute operator discrepancy is ∣A−A0∣=∣0−ρA/2∣=ρA/2≤ρA.

2.1givenA1step 1.1algebra∎

By step 1.1 this is an allowed perturbation. The true noiseless datum is Y=Aθ=0, so g(Y)=0/A0=0. The squared reconstruction error is ∣0−R∣2=R2, independent of ρA. Since R>0 is fixed while ρA can be arbitrarily small, a uniform modulus depending only on absolute operator shift cannot tend to zero.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Exact iid samples estimate two-mode posterior weights

Example

Let b>0, 0<w<1, and let the two-mode posterior be πw=wδ−b+(1−w)δb on R. From M≥1 independent exact posterior draws X1,…,XM, define w^=M−1∑i1{Xi=−b} and π^M=w^δ−b+(1−w^)δb. The one-Wasserstein error and the probability that at least one posterior mode is absent from the sample are W1(π^M,πw)=2b∣w^−w∣,P{w^∈{0,1}}=(1−w)M+wM, and EW1(π^M,πw)≤2bw(1−w)/M.

Facts & Assumptions

Given: b>0, 0<w<1, M≥1, posterior πw=wδ−b+(1−w)δb, and independent draws Xi∼πw.

[A1]

W1 is the infimum of E∣X−X′∣ over couplings of the two probability measures.

Verification

1.1givenA1construct

Put w^=M−1∑i1{Xi=−b}. In any coupling of π^M and πw, mass min⁡{w^,w} at −b and min⁡{1−w^,1−w} at b can be matched at zero cost; the remaining mass is ∣w^−w∣ and must move between the atoms, a distance 2b. This coupling is feasible and no coupling can move less unmatched mass, so it is optimal and gives W1=2b∣w^−w∣.

2.1givenstep 1.1algebra∎

The count Mw^ is binomial (M,w), so Ew^=w and Var⁡(w^)=w(1−w)/M. Cauchy–Schwarz gives E∣w^−w∣≤w(1−w)/M; step 1.1 yields the expected Wasserstein bound. The event of missing −b has probability (1−w)M, and the event of missing b has probability wM; for M≥1 these events are disjoint, giving the displayed miss probability.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

No fixed sample budget uniformly protects arbitrarily rare modes

Example

For the two-mode posterior πw=wδ−b+(1−w)δb with b>0 and 0<w<1/2, M≥1 independent exact draws miss the rare mode −b with probability (1−w)M. For any fixed M, this probability tends to one as w↓0. If instead w∈[wmin⁡,1−wmin⁡] with 0<wmin⁡≤1/2, then the probability of missing either mode is at most 2(1−wmin⁡)M, so M≥log⁡(2/α)/[−log⁡(1−wmin⁡)] suffices to make it at most α∈(0,1).

Facts & Assumptions

Given: b>0, a mixture weight 0<w<1/2, M≥1 independent draws from πw=wδ−b+(1−w)δb, and, for the final bound, w∈[wmin⁡,1−wmin⁡] with 0<wmin⁡≤1/2.

[A1]

Each draw independently equals the rare atom −b with probability w and the other atom with probability 1−w.

Verification

1.1givenA1algebra

The event that no draw equals −b is the intersection of M independent events, each of probability 1−w, hence has probability (1−w)M. Its limit as w↓0 is one for every fixed finite M.

2.1givenA1step 1.1algebra∎

The event that at least one of the two atoms is absent is contained in the union of the event “no −b draw” and the event “no b draw.” Their probabilities are (1−w)M and wM, each at most (1−wmin⁡)M under the displayed weight interval. The union bound gives at most 2(1−wmin⁡)M. If M≥log⁡(2/α)/[−log⁡(1−wmin⁡)], then (1−wmin⁡)M≤α/2, proving the sufficient budget.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

A diagonal inverse has an explicit Fourier truncation error

Example

Write ℓ2=ℓ2(N>0;R), where N>0={1,2,…}, so all sequence coordinates below have positive indices. Let p,s,R>0, let m≥1 be an integer, let θ∈ℓ2 satisfy ∑j≥1j2sθj2≤R2, and let (Apθ)j=j−pθj. Given data y=Apθ+η with ∥η∥ℓ2≤δ, the mesh-m truncated inverse is gm(y)=(jpyj)j≤m with zero output coordinates for j>m. It satisfies the continuum error bound ∥gm(y)−θ∥ℓ2≤mpδ+R(m+1)−s. This separates measurement-noise amplification from the mesh truncation tail; it makes no statistical or neural-operator claim.

Facts & Assumptions

Given: p,s,R>0, an integer m≥1, ∑j≥1j2sθj2≤R2, data y=Apθ+η, and ∥η∥ℓ2≤δ.

[A1]

gm(y) retains the first m coordinates jpyj and sets all later coordinates to zero.

Verification

1.1givenA1algebra

For j≤m, jpyj=θj+jpηj. The norm of the first-m noise vector is at most mp∥η∥ℓ2≤mpδ.

1.2givenA1algebra

The omitted truth tail satisfies ∑j>mθj2≤(m+1)−2s∑j>mj2sθj2≤R2(m+1)−2s. Its norm is at most R(m+1)−s.

2.1A1step 1.1step 1.2algebra∎

The error vector is the sum of the retained-coordinate noise error and the omitted truth tail. The triangle inequality with steps 1.1 and 1.2 gives ∥gm(y)−θ∥ℓ2≤mpδ+R(m+1)−s.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite samples cannot identify every Lipschitz function uniformly

Example

Let N≥1 and let x1,…,xN∈[0,1] be arbitrary training inputs. For every deterministic learning rule that returns a prediction function g:[0,1]→R from the observed pairs (xi,f(xi)), there are two 1-Lipschitz functions f0,f1:[0,1]→R agreeing with each other and taking the common value 0 at every training input but satisfying ∥f1−f0∥∞≥12(N+1). Consequently, at the midpoint of a largest sample-free interval, the returned prediction function g has absolute prediction error at least 1/(4(N+1)) for at least one of these two indistinguishable target functions.

Facts & Assumptions

Given: N≥1, arbitrary sample sites x1,…,xN∈[0,1], exact labels f(xi), and a deterministic learning rule depending only on these observed pairs.

[A1]

A function is 1-Lipschitz when ∣f(x)−f(y)∣≤∣x−y∣ for all x,y∈[0,1].

Verification

1.1givenA1construct

Sort the distinct sample sites together with endpoints 0 and 1. The at most N+1 consecutive gaps have total length one, so one open gap (a,b) has length ℓ=b−a≥1/(N+1). Let x∗=(a+b)/2, r=ℓ/2, f0(x)=0, and f1(x)=max⁡{0,r−∣x−x∗∣}. The tent has slopes of magnitude at most one, vanishes at every sample site, and has height r≥1/(2(N+1)) at x∗. Thus f0,f1 are both 1-Lipschitz and yield identical training pairs.

2.1givenstep 1.1algebra∎

Because the training pairs are identical, the deterministic rule returns the same value g(x∗) for both target functions. Since ∣g(x∗)−f0(x∗)∣+∣g(x∗)−f1(x∗)∣≥∣f1(x∗)−f0(x∗)∣=r, one of the two errors is at least r/2≥1/(4(N+1)). Their uniform distance is r≥1/(2(N+1)), proving both claims.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Integrated score error can miss every Euler evaluation

Example

In the finite-dimensional Gaussian credible-ball question reproduced below, choose d=n=1, R=1, σ=1, T=1, β(t)=1, and γ=1/4, with arbitrary positive p,s,a. Take the numerical operator equal to the true operator, so δA=0. For every integer K≥1 there is a jointly Borel finite score field with integrated score error exactly zero, whose Euler terminal law is N(1,K−1) for every observation. With M=K independent terminal draws, the resulting empirical ball has frequentist coverage tending to zero at truth θ=0, although the stated uniform boundary-margin condition holds with exponent κ=1. The exact credible-ball baseline satisfies δ1,1/40<3/4. Consequently no bound in that question can have Φ→0 uniformly over its original score class as (δA,εs,K−1,M−1/2)→(0,0,0,0). The original formulation is retained verbatim under Original question.

Original question

Fix integers d,n≥1, parameters p,s,a,R,σ>0, a credibility error γ∈(0,1), and a continuous schedule β:[0,T]→(0,∞) with T>0. Let Θs(R)={θ∈Rd:∑j=1dj2sθj2≤R2},Ap=diag⁡(1−p,2−p,…,d−p). Observe Yi=Apθ+εi, 1≤i≤n, with independent εi∼N(0,σ2Id). Use the proper Gaussian prior Π0=N(0,diag⁡(j−1−2a)j=1d), and let Πn(⋅∣Y) be its exact posterior under the stated forward operator and noise law. Let mn(Y) be the posterior mean and let rn,γ(Y) be the unique radius for which the Euclidean ball Cn,γ(Y)={u∈Rd:∥u−mn(Y)∥2≤rn,γ(Y)} has posterior probability 1−γ. Define the exact frequentist coverage cn,γ(θ)=Pn,θ{θ∈Cn,γ(Y)} and its baseline error δn,γ0=sup⁡θ∈Θs(R)∣cn,γ(θ)−(1−γ)∣.

For a deterministic numerical matrix A^n∈Rd×d with δA≥0 and ∥A^n−Ap∥op≤δA, let Π^n(⋅∣Y) be the Gaussian posterior formed with A^n and the same prior and noise variance. Write p^n,tY for the density obtained by applying the forward variance-preserving diffusion dXt=−12β(t)Xt dt+β(t) dWt to X0∼Π^n(⋅∣Y), and set s^n,tY=∇log⁡p^n,tY. A jointly Borel learned score field sn,tY is required to satisfy sup⁡θ∈Θs(R)EY∼Pn,θ[1T∫0T∫Rd∥sn,tY(x)−s^n,tY(x)∥22p^n,tY(x) dx dt]≤εs2. For h=T/K, define the reverse-time Euler chain by Z0∼p^n,TY and, for k=0,…,K−1, Zk+1=Zk+h[12β(T−kh)Zk+β(T−kh)sn,T−khY(Zk)]+hβ(T−kh) ζk, where the ζk are independent N(0,Id) vectors independent of Z0 and the observations. Draw M independent copies of this chain conditionally on Y and denote their terminal values by X(1),…,X(M). Let Xˉ=M−1∑iX(i), let r^ be the empirical 1−γ quantile of {∥X(i)−Xˉ∥2}i=1M (the smallest order statistic whose empirical cumulative mass is at least 1−γ), and put C^n,K,M(Y)={u:∥u−Xˉ∥2≤r^}. Suppose the exact credible-ball boundary satisfies the uniform margin condition sup⁡θ∈Θs(R)Pn,θ(∣∥θ−mn(Y)∥2−rn,γ(Y)∣≤u)≤Cacuκ(0<u≤u0) for fixed Cac,u0>0 and κ>0. Determine whether a minimax-sharp nonasymptotic function Φ exists uniformly over all such score fields and, if it does, find it as a function of d,n,p,s,a,R,σ,T,γ,Cac,u0,κ,δA,εs,K−1, and M−1/2 such that sup⁡θ∈Θs(R)∣Pn,θ,sampler{θ∈C^n,K,M(Y)}−(1−γ)∣≤δn,γ0+Φ, with matching lower bounds that identify when Φ→0. The probability includes observation noise and sampler randomness. The target concerns the additional coverage error introduced by the approximate forward operator, learned score, finite reverse discretization, and finite posterior sample; it does not assume that the exact Bayesian credible ball is already calibrated on every truth class.

Facts & Assumptions

Given: The scalar parameter choices in the Example and the original score-error, Euler and empirical-ball definitions reproduced below it.

[A1]

Write ϕ(x)=(2π)−1/2e−x2/2 and Φ(x)=∫−∞xϕ(z) dz. A standard Gaussian has density ϕ, mean zero, variance one and tail probability tending to zero at infinity.

[A2]

Let q=3/4 and let r>0 solve P{∣G∣≤2r}=q for a standard Gaussian G. Continuity and strict positivity of ϕ give existence and uniqueness of r.

Verification

Given: The scalar model and [A1]–[A2].

1.1A1A2algebra

Completing the square in e−u2/2e−(Y−u)2/2 gives posterior N(Y/2,1/2), so its exact credible interval is [Y/2−r,Y/2+r]. Under truth θ∈[−1,1], Y=θ+G and the exact coverage is c(θ)=P{∣θ/2−G/2∣≤r}. This is a continuous strictly positive function of θ, and is at most one. By compactness of [−1,1] it has a positive minimum c∗>0. Therefore δ1,1/40≤max⁡{q−c∗,1−q}<q.

2.1givenstep 1.1constructalgebra

Put h=1/K and let GK={1−kh:0≤k<K}. Away from this finite set of times, use the exact noised-posterior score. At each t∈GK set stY(x)=(1−x)/h−x/2. This finite-valued jointly Borel field has zero integrated score error: its discrepancy from the exact score is supported on the finite, Lebesgue-null set GK in time. The noised posterior has mean e−t/2Y/2 and variance 1−e−t/2: the integrating-factor solution of the linear forward equation is e−t/2X0+e−t/2∫0teu/2 dWu, whose independent Gaussian integral has variance 1−e−t (use dyadic Brownian-increment sums and pass their Gaussian characteristic functions to the mean-square limit). Its exact score is therefore affine in x, so the inner Gaussian error integral is finite at every time. The zero-error conclusion holds for every observation and hence uniformly over all truths.

2.2A1step 1.1algebra

The variable θ/2−G/2 has a density bounded by 2/2π uniformly in θ. The event ∣∣θ/2−G/2∣−r∣≤u is contained in two intervals of total length at most 4u. Thus its probability is at most 8u/2π. The required margin holds, for example with κ=1, u0=1 and Cac=8/2π.

3.1A1step 2.1constructalgebra

At every Euler step the deterministic update becomes x+h(x/2+stY(x))=1. Thus each new state is 1+h ζk, independently of its input and of Y, and the terminal law is N(1,h). For M=K, write the independent terminal draws as 1+K−1/2Gi. If VK=max⁡i≤K∣Gi∣, then the empirical center differs from one by at most K−1/2VK and every empirical distance to that center is at most 2K−1/2VK. Hence the empirical radius has that same upper bound.

4.1A1step 1.1step 2.2step 3.1algebra∎

The Gaussian density gives P{∣G∣>v}≤2e−v2/2 for v≥1: integrate xϕ(x)/v on [v,∞) and use 1/(v2π)≤1. The union bound then gives P{VK>K1/4}≤2Ke−K/2→0. On the complementary event the center is at least 1−K−1/4 and the radius at most 2K−1/4, so for all sufficiently large K the empirical interval excludes truth zero. Its coverage therefore tends to zero, and its deviation from q tends to q>δ1,1/40. Since both error norms are zero and K−1,M−1/2→0, a uniformly vanishing Φ would contradict this strict gap.

Scope of the resolution

This resolves the requested uniformly vanishing-error guarantee negatively under the original score hypothesis. Time-integrated error does not control pointwise values at a deterministic solver grid. A constant worst-case coverage bound remains possible; no impossibility of all numerical functions is asserted. No Bayesian calibration claim is imported from a conformal theorem.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Worst-law ellipsoid recovery has no training-size rate

Example

For the robust minimax value RN(ρX,ρA) in the original question reproduced below, fix p,s>0, R,σ>0, N≥1, ρX≥0 and 0≤ρA<1/2. Put Qp,s(R,σ)=R2min⁡{1,(σ/R)4s/(2s+2p+1)}. There is cp,s>0, depending only on p,s, such that cp,s(Qp,s(R,σ)+R2ρA2)≤RN(ρX,ρA)≤min⁡{R2,2Qp,s(R,σ)+R2ρA2}. These bounds hold for every training size and truth-law shift radius. For fixed positive p,s,R, along any sequence of the remaining parameters the risk tends to zero if and only if both σ→0 and ρA→0. In particular, increasing N with fixed positive noise does not force the worst-law risk to zero. The original question, including both distinct shift classes, is retained verbatim under Original question.

Original question

Fix an integer N≥1, parameters p,s,R,σ>0, a shift radius ρX≥0, and an operator radius 0≤ρA<1/2. Let H=ℓ2(N), with norm ∥θ∥22=∑j≥1θj2, and define the compact truth class Θs(R)={θ∈H:∑j≥1j2sθj2≤R2},(Apθ)j=j−pθj. Let Y=RN have its product Borel sigma-algebra. A sequence observation with forward operator A is Yj=(Aθ)j+σξj, where the ξj are independent N(0,1) variables; this defines a probability law on Y even though the white-noise sequence need not lie in ℓ2. Write P(K) for the Borel probability laws on a Borel set K. For a Borel probability law μ0 on Θs(R), the training data are DN=((θi,Yi))i=1N with independent θi∼μ0 and Yi=Apθi+σξi. At test time, θ∼μ for another Borel probability law μ on Θs(R) and the unknown forward operator is Ap,E=Ap(I+E), where E:ℓ2→ℓ2 is bounded linear and ∥E∥op≤ρA. Write L(ℓ2) for the bounded linear operators from ℓ2 to itself. The learner receives DN and the test observation Y, but not E.

For probability laws μ,μ0 on Θs(R), define their Wasserstein distance by W1(μ,μ0)=inf⁡Γ∈Π(μ,μ0)∫Θs(R)×Θs(R)∥θ−θ0∥2 Γ(dθ,dθ0), where Π(μ,μ0) is the set of couplings with those marginals. A learner is a Borel map GN:(Θs(R)×Y)N×Y→ℓ2. For fixed training data its test risk is Rμ,E(GN∣DN)=Eθ∼μ, ξ∥GN(DN,Ap(I+E)θ+σξ)−θ∥22, where ξ is an independent standard Gaussian sequence. Define the robust minimax value RN(ρX,ρA)=inf⁡GNsup⁡μ0,μ∈P(Θs(R)), W1(μ,μ0)≤ρXE∈L(ℓ2), ∥E∥op≤ρAEDN∼(μ0,Ap)Rμ,E(GN∣DN). Determine matching upper and lower bounds for RN(ρX,ρA) as a function of N,ρX,ρA,p,s,R,σ, and identify necessary and sufficient joint regimes in which it tends to zero. In particular, the bound must distinguish source-to-test truth-law shift from perturbation of the conditional measurement law induced by E; no access to E is available at deployment.

Facts & Assumptions

Given: All data spaces, training laws, test laws, learners and shift classes in the original question. Let ej denote the jth coordinate vector. Let Φ be the standard normal distribution function and put b=Φ(−1)>0.

[A1]

A scalar observation V=δU+σG with U uniform on {−1,1} and independent standard Gaussian G has minimum sign-testing error Φ(−δ/σ) for δ>0: the two Gaussian densities cross at zero, and choosing the larger density pointwise minimizes the error integral. Their ratio is exp⁡(2δV/σ2).

[A2]

For vectors x,a,b in a real Hilbert space, ∥x−a∥2+∥x−b∥2=2∥x−(a+b)/2∥2+∥a−b∥2/2. This identity follows by expansion of the squares.

Verification

Given: The original robust minimax model and [A1]–[A2].

1.1givenconstructalgebra

The zero estimator gives risk at most R2 for every allowed law and operator. If σ<R, put t=(R/σ)2/(2s+2p+1)>1 and k=⌊t⌋≥1. Use the Borel finite-coordinate estimator θ^j=jpYj for j≤k and zero otherwise; it ignores the training data. Under any test truth its retained error is PkEθ+(σjpξj)j≤k and its omitted error is −(I−Pk)θ. These components are orthogonal, and the centered noise is independent of the truth. Hence its expected squared error is at most R2ρA2+R2(k+1)−2s+σ2∑j=1kj2p. Because k+1>t and ∑j≤kj2p≤k2p+1≤t2p+1, the last two terms are at most 2R2t−2s=2Qp,s. If σ≥R, Qp,s=R2 and the zero estimator already gives the asserted upper bound. Taking the better of the zero and cutoff rules proves the displayed upper bound uniformly over both shift classes.

1.2A1givenconstructalgebra

For the noise lower bound first suppose σ≥R. Set E=0 and μ=μ0=(δRe1+δ−Re1)/2. These laws have Wasserstein distance zero and are admissible at every ρX. The training sample is independent of the fresh test sign, and all other test coordinates are independent pure noise. Given those ancillary variables, testing the fresh sign reduces to [A1] with δ=R. Thus any prediction of its first coordinate has squared error at least R2 whenever its induced sign test is wrong, and its integrated risk is at least R2Φ(−R/σ)≥bR2=bQp,s. At a zero predicted coordinate, choose either sign; the squared error is still R2.

1.3A2givenconstructalgebra

Put r=ρA. If r>0, define a=R/(1+r), b0=R and take the same source and test law μ0=μ=(δae1+δb0e1)/2. Compare the two allowed operators E1=0 and E2=−r(1+r)−1P1, whose norms are at most r. In the first world, the test truth ae1 has observation mean ae1; in the second, truth b0e1 has the same mean because (I+E2)b0e1=ae1. The joint law of the full training sample and this conditional test observation is identical in the two cases, including every noisy coordinate. Calling that common law H, the corresponding two full test risks satisfy R1≥12EH∥G−ae1∥2 and R2≥12EH∥G−b0e1∥2. Therefore [A2] gives max⁡(R1,R2)≥(b0−a)2/8=R2r2/[8(1+r)2]≥R2r2/18. This holds for every Borel learner and all training sizes. If r=0, the asserted zero operator lower bound is immediate.

2.1A1step 1.1constructalgebra

If σ<R, use t from step 1.1 and k=⌈t⌉≤2t. On the coordinate block k≤j<2k, put a=2−s−1Rk−s−1/2 and let μ0=μ be the uniform law on the 2k vectors having independent coordinates θj=±a on this block and zero elsewhere. Each such vector lies in the ellipsoid because ∑jj2sθj2≤k(2k)2sa2=R2/4. Again choose E=0. For every block coordinate, its test sign is independent of the whole training sample and of all other test coordinates. Its signal amplitude is j−pa≤k−pa≤σ, by the definition of t and k≥t. Applying the density-crossing argument in [A1], conditional on all ancillary data, gives sign error at least b. Summing the coordinate squared errors gives risk at least bka2=b2−2s−2R2k−2s≥b2−4s−2Qp,s. Finite averaging over the sign prior and integration over the independent ancillary observations suffice; no choice of an unknown prior by the learner is being assumed.

3.1step 1.1step 1.2step 2.1step 1.3algebra∎

Steps 1.2 and 2.1 give RN≥c0Qp,s with c0=bmin⁡{1,2−4s−2}>0, and step 1.3 gives RN≥R2ρA2/18. The maximum of these bounds is at least 12min⁡{c0,1/18}(Qp,s+R2ρA2), proving the lower bound with that explicit positive cp,s. For fixed p,s,R>0, the upper bound tends to zero when σ,ρA→0, and the lower bound forces both limits whenever the risk tends to zero. Both lower constructions have μ=μ0, so they remain admissible even at zero truth-law shift; the upper construction ignores the training law entirely. This explains why N and ρX do not enter the matching-order rate for this worst-law formulation.

Scope of the resolution

This gives matching-order upper and lower bounds for the exact original worst-law problem, including simultaneous unknown relative-operator shift and truth-law Wasserstein shift. It neither identifies the exact minimax constant nor says that finite-sample distribution learning is unimportant for a fixed restricted prior family. The source paper's conditional-law DRO bound is adjacent context; the two shift mechanisms and finite sign-prior proofs above are local and do not substitute its finite-dimensional hypotheses.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Ideal-marginal score error allows unbounded finite-Euler Wasserstein error

Example

For the separated-mixture question reproduced below, there are admissible fixed one-dimensional parameters with L=2, fixed positive score-error allowance εs, and every fixed integer J≥2, for which the supremum of the displayed expected fidelity loss over the original finite-valued Borel score class is +∞, for every finite M≥1. Each individual field in the construction has a finite-first-moment Euler chain and obeys the required score bound at every time. Thus the original ideal-marginal L2 hypothesis supplies no finite uniform bound for the requested loss, however large the fixed budget JM is. The original question is retained verbatim under Original question.

Original question

Fix dimension d≥1, component count L≥2, noise scale σ>0, diffusion horizon T>0, and a continuous schedule β:[0,T]→(0,∞). Assume the parameter choices below admit at least one component configuration and mixture weight vector. Consider a posterior density on Rd of the form π(x)=∑ℓ=1Lwℓ N(x;μℓ,Σℓ),∑ℓ=1Lwℓ=1,wℓ≥wmin⁡>0, Here N(⋅;μℓ,Σℓ) denotes the normal density with mean μℓ and covariance matrix Σℓ. The component means satisfy ∥μℓ∥2≤M0 and ∥μℓ−μr∥2≥Δ>0 for ℓ≠r, and every eigenvalue of each positive-definite Σℓ lies in [vmin⁡,vmax⁡] for fixed 0<vmin⁡≤vmax⁡<∞. Assume 0<wmin⁡≤1/L and Δ≥8dvmax⁡. For t∈[0,T], let pt be the density of Xt under dXt=−12β(t)Xt dt+β(t) dWt,X0∼π, where Wt is a standard d-dimensional Brownian motion, and let st=∇log⁡pt be its exact score. Let s^t be a supplied jointly Borel, finite-valued score field satisfying the integrated error bound sup⁡0≤t≤T∫Rd∥s^t(x)−st(x)∥22pt(x) dx≤εs2. Consider fields whose Euler chain has finite first moment. For h=T/J, use the reverse-time Euler convention Z0∼pT and Zk+1=Zk+h[12β(T−kh)Zk+β(T−kh)s^T−kh(Zk)]+hβ(T−kh) ζk,k=0,…,J−1, with ζk∼N(0,Id) independent of Z0 and of one another. Generate M≥1 independent terminal values X(1),…,X(M) and let π^J,M=M−1∑i=1MδX(i); one score evaluation per reverse step gives online budget B=JM. Define disjoint mode regions Rℓ={x:∥x−μℓ∥2<Δ/4}, target masses qℓ=π(Rℓ), and empirical masses q^ℓ=π^J,M(Rℓ). For probability measures on Rd, let W1 be the infimum of E∥X−Y∥2 over all couplings with the specified marginals. Determine whether the integrated score-error assumption alone yields finite minimax-sharp nonasymptotic upper and lower bounds, uniformly over the stated score-field class, for E W1(π^J,M,π)+Δ∑ℓ=1LE∣q^ℓ−qℓ∣+Δ P{min⁡1≤ℓ≤Lq^ℓ=0} as a function of d,L,wmin⁡,Δ,vmin⁡,vmax⁡,M0,T,β,εs,J,M. In particular, characterize the compute regimes B=JM that guarantee simultaneous control of distributional error and missed-mode probability uniformly over this family. This asks about score-based reverse diffusion with finite computation; it does not assume that the posterior is unimodal or that a good posterior mean certifies its mode weights.

Facts & Assumptions

Given: Choose d=1, π=(N(−a,v)+N(a,v))/2 with a,v>0 and 2a≥8v. Choose L=2, wmin⁡=1/2, M0=a, Δ=2a, vmin⁡=vmax⁡=v, T>0, β=1, εs>0, J≥2 and h=T/J. The extraneous positive parameter called σ in the original formulation may be arbitrary; it is not used by its displayed sampler.

[A1]

The integrating factor et/2 in the scalar forward equation gives Xt=e−t/2X0+e−t/2∫0teu/2 dWu. The stochastic integral is the mean-square limit of deterministic dyadic sums of independent Gaussian increments; these sums are centered Gaussian, with variances tending to ∫0teu du=et−1, and are independent of X0. The limit is therefore Gaussian with that variance and independent of X0, as follows by taking limits of its Gaussian characteristic functions. At time t, the noised target is the symmetric Gaussian mixture with means ±e−t/2a and common variance vt=e−tv+1−e−t>0. Its exact score is a posterior-weighted average of −(x−e−t/2a)/vt and −(x+e−t/2a)/vt, so ∣st(x)∣≤(∣x∣+a)/min⁡{v,1} and is continuous. These formulas follow directly by differentiating the sum of the two Gaussian densities.

[A2]

A convolution with a nondegenerate Gaussian has a strictly positive continuous density: continuity follows from the bounded Gaussian derivative, and positivity from its positive kernel. Gaussian densities have finite first moments and exponentially decaying tails.

Verification

Given: The fixed parameters and Gaussian identities in [A1]–[A2].

1.1A1A2construct

Use the exact score at all grid times before the last two updates. All those states have finite first moment because each exact-score drift has linear growth by [A1]. Let U be the state before the penultimate update, corresponding to ideal time 2h. Its law has a strictly positive continuous density ρ: for J=2 it is exactly pT, and otherwise the last preceding nondegenerate Euler noise gives [A2]. Since p2h is also positive and continuous, the density ratio ρ/p2h has positive minimum c and finite maximum C on [−1,1].

2.1A1step 1.1constructalgebra

For every sufficiently small q>0, choose the centered interval Aq⊆[−1,1] with ∫Aqp2h(x) dx=q. Such an interval exists by continuity and positivity of that density. Then cq≤P{U∈Aq}≤Cq. Put bq=εs/(2q) and add bq1Aq to the score at time 2h. Its squared ideal-marginal error is exactly εs2/4. Set f(x)=(1+h/2)x+hs2h(x) and H=sup⁡∣x∣≤1∣f(x)∣<∞. After this update the state is V=f(U)+hbq1Aq(U)+h G.

3.1A1A2step 2.1algebra

Let Iq=[hbq−H−h,hbq+H+h], and let rq=∫Iqph(x) dx>0. On {U∈Aq,∣G∣≤1} the state V lies in Iq, so P{V∈Iq}≥cqP{∣G∣≤1}. The ideal density ph is a fixed Gaussian mixture. Since the interval width is constant and its center hbq tends to infinity, its Gaussian density formula gives constants C1,c1>0, independent of small q, such that rq≤C1e−c1bq2. For example, once hbq≥2(H+h+a), every point of the interval is at distance at least hbq/2 from either component mean; use the largest fixed variance in its Gaussian exponent and the constant interval length.

4.1A1step 2.1step 3.1constructalgebra

At the final evaluation time h add dq1Iq to the exact score, where dq=εs/(2rq). This second squared ideal-marginal error is also εs2/4. At all other times use the exact score. These modifications at two distinct times define a jointly Borel finite-valued field with the original supremum-in-time score bound. Every individual field has finite first moments at all Euler steps: it is an affine-growth exact score plus two bounded, finite-amplitude additions.

5.1A1step 2.1step 3.1step 4.1algebra

The first moment of V is bounded uniformly as q↓0: the added term has expectation at most hbqCq=hCεsq/2, while E∣f(U)∣ and E∣hG∣ are fixed and finite. The terminal state is X=(1+h/2)V+hsh(V)+hdq1Iq(V)+hG′. By [A1] the absolute expectation of the first two terms is uniformly bounded; EG′=0. Therefore EX≥hεscqP{∣G∣≤1}/(2rq)−C2→+∞, because bq2=εs2/(4q) and the exponential bound in step 3.1 dominates every power of q.

6.1givenstep 4.1step 5.1algebra∎

Every coupling of two finite-first-moment probabilities satisfies ∣EX−EY∣≤E∣X−Y∣ by the integral triangle inequality. Thus W1(π^J,M,π)≥∣M−1∑iX(i)∣ because the target mean is zero. Taking expectations gives EW1≥∣EX∣, which tends to infinity by step 5.1 for every finite M. The other two terms in the requested fidelity loss are nonnegative. Its supremum is consequently infinite despite fixed positive score-error allowance and fixed admissible target parameters.

Scope of the resolution

The negative answer concerns every positive error allowance and every finite J>=2. It does not assert the same obstruction at J=1 or at exactly zero error. The mechanism is mismatch between ideal marginal mass and actual chain mass, not merely missing a mode by iid sampling. No numerical observation or source-specific failure experiment is used as the proof.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Coordinate-free Fourier inverse maps have an odd-mesh risk floor

Example

Consider the complete original Fourier architecture and diagonal inverse question reproduced below. For every odd integer m≥3, every N≥1, and every admissible bounded parameter class with a measurable empirical minimizer, its risk satisfies RN,mERM(S,B)≥R2(3p−2p)24 32(p+s)>0. The same bound holds for every measurable learner whose realized predictor belongs to the stated coordinate-free Fourier class, whether or not it minimizes empirical risk. On the other hand, the unrestricted risk tends to zero as m→∞ for fixed p,s,R,σ>0, independently of N. Thus no growth of N,S,B can make this original architecture attain the vanishing unrestricted risk along an odd-mesh subsequence. This is a partial resolution: no complete joint bounded-class rate, or corresponding fixed positive even-mesh floor, is asserted. The original question is retained under Original question with explicit positive coordinate indices and empty-class conventions.

Original question

Let T=R/(2πZ) carry normalized Lebesgue measure and fix the real orthonormal Fourier basis ϕ1(x)=1,ϕ2k(x)=2cos⁡(kx),ϕ2k+1(x)=2sin⁡(kx)(k≥1). Use this basis to identify the real Hilbert space ℓ2(N>0) with L2(T;R). Fix integers N,m≥1 and parameters s,p,R,σ>0, and define Θs(R)={θ∈ℓ2:∑j≥1j2sθj2≤R2},(Apθ)j=j−pθj,Xp(R)=ApΘs(R). The inverse map G†:Xp(R)→ℓ2 is G†(Apθ)=θ. For m≥1, let Pm retain the first m Fourier coefficients and let Im:Rm→ℓ2 append zero coordinates.

Fix a globally Lipschitz nonpolynomial function σ0:R→R, called the activation. For x∈Rm, form the scalar input field uˉ0(x;z)=∑j=1mxjϕj(z). A pointwise affine lift u0(x;z)=Vinuˉ0(x;z)+bin, with Vin∈Rc0 and bin∈Rc0, creates c0 channels. For fields uℓ(z)∈Rcℓ with positive integer channel counts cℓ, apply an integer number D≥1 of layers indexed by ℓ=0,…,D−1: uℓ+1(z)=σ0 ⁣(Wℓuℓ(z)+bℓ+(Kℓuℓ)(z)). Here σ0 acts coordinatewise, Wℓ∈Rcℓ+1×cℓ and bℓ∈Rcℓ+1 are pointwise channel weights, and Kℓ is a Fourier multiplier meaning (Kℓu)^(k)=Rℓ,ku^(k), where u^(k)=(2π)−1∫02πu(z)e−ikz dz is computed componentwise, with matrices Rℓ,k∈Ccℓ+1×cℓ satisfying Rℓ,−k=Rℓ,k‾ and Rℓ,k=0 for ∣k∣>q for some integer cutoff q≥0. A pointwise affine output map v(x;z)=aout⊤uD(z)+bout gives a real scalar field, and the network output is Ψm(x)=(⟨v(x;⋅),ϕj⟩L2)j=1m. Let FS,B(m) be the union of these maps over architectures whose depth, channel counts, Fourier cutoff, and number of real scalar parameters are each at most the integer S≥1, with every trainable parameter bounded in absolute value by B>0. A parameter array used at mesh size m is the same array at every other mesh size. For fixed S, there are finitely many allowed architecture indices; equip their disjoint union of bounded real parameter boxes with its Borel structure.

Write P(K) for the Borel probability laws on a Borel set K. For a Borel probability law μ on Θs(R), draw independent training truths θi∼μ and observations Xi,m=PmApθi+ηi,m,Yi,m=Pmθi,ηi,m∼N(0,σ2m−1Im),i=1,…,N, with all noises independent, and write DN,m=((Xi,m,Yi,m))i=1N. A selector is Borel into the finite disjoint union of architecture indices and their bounded parameter boxes, and its realized map minimizes the empirical loss. For each N,m,S,B with FS,B(m)≠∅, assume there is at least one such selector τ that, from DN,m, returns a minimizer of 1N∑i=1N∥Ψ(Xi,m)−Yi,m∥22 over Ψ∈FS,B(m); let TN,m(S,B) be the set of all such selectors, and set it to ∅ when FS,B(m)=∅. At test time, draw an independent θ∼μ and ηm∼N(0,σ2m−1Im), and define RN,mERM(S,B)=inf⁡τ∈TN,m(S,B)sup⁡μ∈P(Θs(R))E∥Imτ(DN,m)(PmApθ+ηm)−θ∥ℓ22. Use the convention inf⁡∅=+∞, so an empty architecture class has infinite ERM risk. Also define RN,mall by the same infimum over all Borel predictors from (DN,m,Xm) to Rm, with Xm=PmApθ+ηm. Determine matching minimax upper and lower rates for RN,mERM(S,B) jointly in N,m,S,B,p,s,R,σ. In particular, characterize the joint growth regimes of m=mN and S=SN that separate statistical estimation, neural-operator approximation, and continuum truncation error, and determine when the bounded shared-parameter Fourier architecture attains the unrestricted rate RN,mall.

Facts & Assumptions

Given: The original real Fourier basis ϕ1=1, ϕ2=2cos⁡z, ϕ3=2sin⁡z and all higher complete cosine/sine pairs; the architecture has only constant pointwise channel weights and biases, Fourier multipliers and a scalar pointwise Lipschitz activation. Its parameters may depend on all training data but are fixed during test evaluation.

[A1]

Translation Tαu(z)=u(z−α) is an isometry of L2(T), sends ϕ2 to ϕ3 for α=π/2, and preserves each full two-dimensional cosine/sine frequency pair. The first-m-coordinate subspace is translation-invariant when m is odd.

[A2]

A vector with distribution N(0,τ2Im) is invariant under every orthogonal matrix: its density is a constant times e−∥x∥2/(2τ2), and orthogonal changes of variables preserve norm and volume.

[A3]

∥x−a∥2+∥x−b∥2≥∥a−b∥2/2 by completing the square.

Verification

Given: The original architecture, training and test model, and [A1]–[A3].

1.1A1givenconstructalgebra

Each Fourier multiplier commutes with translation because the kth translated Fourier coefficient is multiplied by the scalar e−ikα, which commutes with its channel matrix. Every constant channel matrix, constant bias and coordinatewise scalar activation also commutes with translation. By induction through the lift, all layers and output map, the realized scalar output field is translation-equivariant. The layer fields are in L2: a globally Lipschitz activation has linear growth, the pointwise weights are finite and the finite Fourier multipliers are bounded. For odd m, projection onto the first m coefficients commutes with translation by [A1]. Hence the zero-extended output map F=ImΨm satisfies F(Uαx)=TαF(x), where Uα is the orthogonal translation matrix on the retained coefficients. This holds for every individual parameter array, also after data-dependent selection.

1.2A1A2givenconstructalgebra

Put a=R3−(p+s), choose θc=a2pe2 and θs=a3pe3, and let μ=(δθc+δθs)/2. These truths lie in the ellipsoid because their weighted squared norms are at most a232(p+s)=R2. Their forward coefficient vectors are respectively ae2 and ae3, related by U=Uπ/2. Conditional on any training sample, the test truth is still an independent equal-probability draw from these two values. Couple the corresponding test noises as η and Uη; by [A2] both have the required N(0,σ2m−1Im) law.

1.3givenconstructalgebra

To see unrestricted vanishing without training, put km=min⁡{m,max⁡{1,⌊m1/(2(2p+1))⌋}}. Output jpXm,j for j≤km and zero otherwise. Its risk for every truth is at most σ2m−1∑j≤kmj2p+R2(km+1)−2s≤σ2m−1km2p+1+R2(km+1)−2s→0. It is a Borel predictor and is allowed in the unrestricted class. This proves the contrast with the positive odd-mesh architecture floor.

2.1A1A3step 1.1step 1.2algebra

For fixed training data, let F be its selected predictor. By step 1.1, F(ae3+Uη)=Tπ/2F(ae2+η). Translating the first output and truth preserves their norm. Consequently the conditional average of the two squared risks is 12E[∥Tπ/2F(ae2+η)−Tπ/2θc∥2+∥Tπ/2F(ae2+η)−θs∥2], which is at least ∥Tπ/2θc−θs∥2/4=a2(3p−2p)2/4 by [A3]. This lower bound is independent of the training sample. Integrating over it, taking the worst law and then the infimum over selectors proves the stated floor.

3.1step 1.1step 2.1step 1.3algebra∎

The mismatch arises because the diagonal inverse rescales the first cosine and sine by different factors 2p and 3p, while the coordinate-free architecture must respect translation between them. It is independent of finite training or parameter-size limitations. The proof deliberately uses odd m so both the input noise law and retained output subspace are translation-invariant; it supplies no identical assertion for even m, whose last frequency pair is incomplete.

Scope of the resolution

The odd-mesh consistency subquestion is answered negatively, and the unrestricted vanishing comparison is proved. The full requested joint N,m,S,B minimax rate is not resolved here; in particular even meshes break the complete-pair symmetry used by the lower bound, and different activations/parameter budgets may have distinct approximation behavior. This is not the EIT question: the generalized EIT architecture has coordinate/chart lifting, so this coordinate-free translation obstruction is not transferred to it.

Sources