Alphabeta Math
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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.

Bregman distances and variational source rates

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Banach-space convex subgradients and dual pairings

Definition

Let X be a real Banach space and let X∗ be its continuous real dual. The dual pairing is evaluation, written ⟨ξ,x⟩X∗,X:=ξ(x) for ξ∈X∗ and x∈X. Let R:X→(−∞,+∞] be a proper convex functional: R never takes the value −∞, its effective domain dom⁡R:={x∈X:R(x)<+∞} is nonempty, and R((1−t)x+tz)≤(1−t)R(x)+tR(z) for x,z∈X and t∈[0,1], with 0⋅(+∞):=0 for this convex-combination formula. For x∈dom⁡R, its subdifferential is

∂R(x):={ξ∈X∗:R(z)≥R(x)+⟨ξ,z−x⟩X∗,X for every z∈X}.

An element of ∂R(x) is a subgradient of R at x. Set ∂R(x):=∅ when x∉dom⁡R. The subdifferential may be empty at a point of dom⁡R; no nonemptiness assertion is part of this definition.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A conjugate bound for residual and Bregman errors

Statement

Let ψ:[0,∞)→[0,∞) be continuous, nondecreasing, concave, and satisfy ψ(0)=0. Extend g=−ψ to R by g(s)=+∞ for s<0, and define H(q):=g∗(−q)=sup⁡s≥0(ψ(s)−qs) for q>0, assuming the displayed suprema below are finite. For c1,c2,C3,α>0 and d,r,e≥0, if

c1r+c2αe≤d+αψ(C3r),

then

e≤1c2(dα+H ⁣(c1C3α)),r≤2c1(d+αH ⁣(c12C3α)).

Facts & Assumptions

Given: ψ, H, c1,c2,C3,α,d,r,e satisfy the hypotheses and inequality in the Statement.

[A1]

The two values H(c1/(C3α)) and H(c1/(2C3α)) are finite.

Proof

1.1givenA1algebra

Put t=C3r. Rearranging the given inequality and dividing by c2α gives e≤dc2α+1c2(ψ(t)−c1C3αt). The parenthesized expression is bounded above by the defining supremum H(c1/(C3α)), proving the first bound.

2.1givenA1step 1.1algebra∎

Subtracting c1r/2 from the given inequality and using c2αe≥0 gives c1r2≤d+α(ψ(t)−c12C3αt)≤d+αH ⁣(c12C3α). Multiplication by 2/c1 proves the second bound.

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

Fenchel–Young equality for a Banach regularizer

Statement

Let X be a real Banach space, let R:X→(−∞,+∞] be proper and convex, and define its Fenchel conjugate R∗:X∗→(−∞,+∞] by R∗(ξ):=sup⁡z∈X(⟨ξ,z⟩X∗,X−R(z)). For every x∈dom⁡R and ξ∈X∗, the Fenchel–Young inequality and its equality characterization are

R(x)+R∗(ξ)≥⟨ξ,x⟩X∗,X,R(x)+R∗(ξ)=⟨ξ,x⟩X∗,X ⟺ ξ∈∂R(x).

Facts & Assumptions

Given: X, R, R∗, x, and ξ satisfy the hypotheses and definitions in the Statement.

[F1]

Banach-space convex subgradients and dual pairings: ξ∈∂R(x) exactly when R(z)≥R(x)+⟨ξ,z−x⟩X∗,X for every z∈X.

Proof

1.1givenalgebra

The supremum defining R∗(ξ) includes the term at z=x, so R∗(ξ)≥⟨ξ,x⟩X∗,X−R(x). Adding R(x) gives the Fenchel–Young inequality.

2.1givenF1step 1.1

If ξ∈∂R(x), [F1] gives ⟨ξ,z⟩X∗,X−R(z)≤⟨ξ,x⟩X∗,X−R(x) for all z; taking the supremum and then using the term z=x proves equality.

3.1givenF1algebra∎

Conversely, if equality holds, every term in the supremum satisfies ⟨ξ,z⟩X∗,X−R(z)≤R∗(ξ)=⟨ξ,x⟩X∗,X−R(x). Rearrangement gives [F1], hence ξ∈∂R(x).

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Bregman distances for convex regularizers

Definition

Let X be a real Banach space, let R:X→(−∞,+∞] be proper and convex, choose x†∈dom⁡R, and choose ξ†∈∂R(x†). For x∈dom⁡R, the Bregman distance generated by ξ† from the reference point x† to x is

DRξ†(x,x†):=R(x)−R(x†)−⟨ξ†,x−x†⟩X∗,X.

The subgradient inequality makes this quantity nonnegative, and DRξ†(x†,x†)=0. The definition fixes the orientation: the first argument is the comparison point and the second is the reference point carrying the chosen subgradient.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Variational source conditions on a regularizer sublevel

Definition

Let X,Y be real Banach spaces, let F:D(F)⊆X→Y, let x†∈D(F) with exact datum y=F(x†), and let R:X→(−∞,+∞] be proper and convex with x†∈dom⁡R. Fix ξ†∈∂R(x†), an exponent 1≤p<∞, a continuous nondecreasing concave function ψ:[0,∞)→[0,∞) with ψ(0)=0 and ψ(t)>0 for t>0, a coefficient 0≤β<1, and a level R0≥R(x†). The variational source condition with these data holds on the named regularizer sublevel LR0:={x∈D(F):R(x)≤R0} when

−⟨ξ†,x−x†⟩X∗,X≤βDRξ†(x,x†)+ψ ⁣(∥F(x)−F(x†)∥Yp)(x∈LR0).

The condition is an assumption on the selected subgradient and the stated sublevel; it makes no assertion that a subgradient exists at every point or that a regularized minimizer exists.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Variational source conditions yield nonlinear regularization rates

Statement

Let X,Y be real Banach spaces, D(F)⊆X, and F:D(F)→Y. Let R:X→(−∞,+∞] be proper and convex, let x†∈D(F)∩dom⁡R, set y=F(x†), and choose ξ†∈∂R(x†). Fix 1≤p<∞, 0≤β<1, and a level R0>R(x†). Let ψ:[0,∞)→[0,∞) be continuous, nondecreasing and concave, with ψ(0)=0 and ψ(t)>0 for t>0, and assume the variational source inequality

−⟨ξ†,x−x†⟩X∗,X≤βDRξ†(x,x†)+ψ ⁣(∥F(x)−F(x†)∥Yp)(x∈LR0:={x∈D(F):R(x)≤R0}),

where DRξ†(x,x†):=R(x)−R(x†)−⟨ξ†,x−x†⟩X∗,X. Extend −ψ by +∞ on negative arguments and write H(q):=(−ψ)∗(−q)=sup⁡s≥0(ψ(s)−qs); assume H(q) is finite for sufficiently large q and H(q)→0 as q→∞. For sufficiently small noise levels δ>0 and data yδ∈Y with ∥yδ−y∥Y≤δ, define Jαδ(x):=p−1∥F(x)−yδ∥Yp+αR(x). Suppose α=α(δ)>0 tends to zero, 0≤εδ≤Cεδp, and xαδ,ε∈D(F) is an εδ-minimizer, meaning Jαδ(xαδ,ε)≤inf⁡z∈D(F)Jαδ(z)+εδ. Set cp:=21−p/p and ρδ:=H(cp/α(δ)). Assume that for constants c−,c+,CH>0 independent of δ,

c−δp≤α(δ)ρδ≤c+δp,H(cp/(2α(δ)))≤CHρδ

for all sufficiently small δ. Then for all sufficiently small δ, xαδ,ε∈LR0 and

DRξ†(xαδ,ε,x†)=O(ρδ),∥F(xαδ,ε)−yδ∥Y=O(δ).

If, in addition, ψ(C0δp)=O(ρδ) for a fixed C0 such that Tδ:=∥F(xαδ,ε)−F(x†)∥Yp≤C0δp for all sufficiently small δ, then ∣R(xαδ,ε)−R(x†)∣=O(ρδ). If also DRξ†(x,x†)≥cR∥x−x†∥Xs on LR0 for constants cR,s>0, then ∥xαδ,ε−x†∥Xs=O(ρδ).

In the Hilbert special case, set p=2 and suppose X,Y are real Hilbert spaces, F:U→Y has the real open-domain interface of a nonlinear forward map, x† is an x0-minimum-norm exact solution, R(x)=12∥x−x0∥X2, and the source inequality on a regularizer sublevel contained in U is

⟨x†−x0,x†−x⟩X≤β112∥x−x†∥X2+β2∥F(x)−F(x†)∥Y

with 0≤β1<1 and β2≥0. For exact global minimizers of the nonlinear Tikhonov functional with α(δ)≍δ, the theorem gives ∥xα(δ)δ−x†∥X=O(δ) and ∥F(xα(δ)δ)−yδ∥Y=O(δ). The assertion is conditional on such a global minimizer being supplied; it does not assert existence or cover local numerical minima.

Facts & Assumptions

Given: All hypotheses in the Statement hold. Write xδ:=xαδ,ε, Tδ:=∥F(xδ)−F(x†)∥Yp, and rδ:=∥F(xδ)−yδ∥Y.

[A1]

δ>0, xδ is an εδ-minimizer of Jαδ, and ∥yδ−y∥Y≤δ, with 0≤εδ≤Cεδp.

[A2]

The VSC holds on LR0 with the chosen ξ†, β, and ψ.

[A3]

αρδ≍δp, H(cp/(2α))≤CHρδ, α→0, and ρδ→0.

[A4]

For the optional penalty conclusion, ψ(C0δp)=O(ρδ) for a fixed C0 that bounds Tδ/δp for small δ.

[A5]

For the optional norm conclusion, DRξ†(x,x†)≥cR∥x−x†∥Xs on LR0 for constants cR,s>0.

[A6]

For the Hilbert conclusion, R(x)=12∥x−x0∥X2, ξ†(h)=⟨x†−x0,h⟩X, the displayed x0-minimum-norm VSC holds with 0≤β1<1 and β2≥0, and α≍δ.

[F1]

Banach-space convex subgradients and dual pairings: for x∈dom⁡R, ∂R(x):={ξ∈X∗:R(z)≥R(x)+⟨ξ,z−x⟩X∗,X for every z∈X}.

[F2]

Bregman distances for convex regularizers: DRξ†(x,x†)=R(x)−R(x†)−⟨ξ†,x−x†⟩X∗,X and is nonnegative.

[F3]

Variational source conditions on a regularizer sublevel: −⟨ξ†,x−x†⟩X∗,X≤βDRξ†(x,x†)+ψ(∥F(x)−F(x†)∥Yp) on LR0.

[F4]

A conjugate bound for residual and Bregman errors: from c1r+c2αe≤d+αψ(C3r), the stated conjugate bounds hold for e and r.

[F5]

Nonlinear forward maps between Hilbert spaces: in the Hilbert specialization F:U→Y is a real-Hilbert forward map on an open domain.

[F6]

x₀-minimum-norm solutions of nonlinear equations: the Hilbert specialization assumes the exact solution globally minimizes distance to x0 over its exact solution set.

[F7]

The nonlinear Tikhonov functional: for a datum y∈Y, Jα,y,x0(x)=12∥F(x)−y∥Y2+α2∥x−x0∥X2.

[F9]

Real and complex inner-product spaces and their induced length: a real Hilbert norm satisfies ∥v∥X2=⟨v,v⟩X.

[F10]

Proof

1.1givenA1A3algebra

Comparing the approximate-minimum inequality with the exact solution gives Jαδ(xδ)≤δp/p+αR(x†)+εδ. Since the data term is nonnegative, R(xδ)−R(x†)≤δp/(pα)+εδ/α=O(ρδ) by [A1] and [A3]. As ρδ→0 and R0>R(x†), xδ∈LR0 for sufficiently small δ.

1.2A6F1F2F3F5F6F7F9F10algebra

By [F5]–[F7] the special case uses the open-domain Hilbert map, the stipulated global x0-minimum-norm exact solution, and the existing quadratic nonlinear Tikhonov functional. Under [A6], define ξ†(h):=⟨x†−x0,h⟩X. By [F10] this is a bounded real-linear functional; expansion using [F9] gives R(x)−R(x†)−⟨ξ†,x−x†⟩X∗,X=12∥x−x†∥X2≥0, so [F1] gives ξ†∈∂R(x†) and [F2] gives DRξ†(x,x†)=12∥x−x†∥X2. Taking ψ(s)=(β2+1)s, the stated Hilbert source inequality implies the VSC with p=2.

2.1A1F8step 1.1algebra

By [F8], Tδ1/p≤rδ+δ. For p=1, (u+v)p=up+vp; for p>1, t↦tp is continuous at zero and has nonnegative second derivative p(p−1)tp−2 on (0,∞), hence is convex on [0,∞), and its defining convexity inequality at the midpoint gives (u+v)p≤2p−1(up+vp). Taking u=rδ, v=δ yields Tδ≤2p−1(rδp+δp) and thus rδp/p≥cpTδ−δp/p. Substituting into the comparison inequality from 1.1 gives cpTδ+α(R(xδ)−R(x†))≤2δp/p+εδ.

2.2A3A6step 1.2algebra

For B:=β2+1>0, the conjugate is H(q)=sup⁡u≥0(Bu−qu2)=B2/(4q), by completing the square. With p=2, cp=1/4, so αH(cp/α)=B2α2≍δ2 and ρδ=H(cp/α)=B2α≍δ when α≍δ. The scale comparability in [A3] also holds because H(cp/(2α))=2H(cp/α).

3.1F2F3step 2.1algebra

By [F2] and [F3], R(xδ)−R(x†)=DRξ†(xδ,x†)+⟨ξ†,xδ−x†⟩X∗,X≥(1−β)DRξ†(xδ,x†)−ψ(Tδ). Combining this with the objective comparison in step 2.1 gives cpTδ+α(1−β)DRξ†(xδ,x†)≤dδ+αψ(Tδ), where dδ:=2δp/p+εδ.

4.1F4step 3.1

Apply [F4] with c1=cp, c2=1−β, C3=1, r=Tδ, e=DRξ†(xδ,x†), and d=dδ. It follows that DRξ†(xδ,x†)≤11−β(dδα+ρδ) and Tδ≤2cp(dδ+αH(cp/(2α))).

5.1A1A3F8step 4.1algebra

From [A1] and [A3], dδ=O(δp), dδ/α=O(ρδ), and dδ+αH(cp/(2α))=O(δp). Thus DRξ†(xδ,x†)=O(ρδ) and Tδ=O(δp). The triangle inequality gives rδ≤Tδ1/p+δ, so the noisy-data residual is O(δ).

6.1A4A5F2F3step 1.1step 5.1

For the penalty rate, [F2]–[F3] give R(xδ)−R(x†)≥−ψ(Tδ), while step 1.1 supplies its upper bound. If [A4] holds for a fixed C0 bounding Tδ/δp, then ∣R(xδ)−R(x†)∣=O(ρδ). If [A5] also holds, the Bregman bound in step 5.1 gives ∥xδ−x†∥Xs=O(ρδ).

7.1A6step 5.1step 1.2step 2.2algebra∎

Applying the Bregman bound from step 5.1 and the exact Bregman identity from step 1.2 gives ∥xα(δ)δ−x†∥X2=O(δ) and ∥F(xα(δ)δ)−yδ∥Y=O(δ), hence the asserted O(δ) solution error and O(δ) residual.

4 · Examples, counterexamples and false statements

None yet.

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