Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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.

FISTA energy estimate-sequence recurrence

Statement

Let F=f+g satisfy the exact FISTA hypotheses of Proximal-gradient, FISTA, and fixed-period restarted FISTA, and suppose x∗ minimizes F. Use its iterates xk,yk,tk and set t0=0, u0=x0−x∗, and uk=tkxk−(tk−1)xk−1−x∗(k≥1). Then for every k≥1, 2tk2L(F(xk)−F(x∗))+∥uk∥22≤2tk−12L(F(xk−1)−F(x∗))+∥uk−1∥22, where the k=1 right-hand objective term is interpreted as zero.

Facts & Assumptions

Given: The exact FISTA recurrence, composite objective and minimizer in the Statement.

[F1]

Proximal-gradient, FISTA, and fixed-period restarted FISTA: t1=1, y1=x0, xk=TL(yk), tk+1=(1+1+4tk2)/2, the consequent identity tk+1(tk+1−1)=tk2, and yk=xk−1+((tk−1−1)/tk)(xk−1−xk−2) for k≥2.

[F2]

One-step proximal-gradient estimate: F(z)≥F(TL(y))+(L/2)∥TL(y)−y∥22+L⟨y−z,TL(y)−y⟩ for every z.

Proof

1.1F1giveninductionalgebra

The recurrence [F1] gives tk≥1 for every k: it starts at 1 and tk+1≥tk+1/2. Thus zk=(1−1/tk)xk−1+(1/tk)x∗ is a convex combination (and z1=x∗). The same recurrence gives tk(yk−zk)=uk−1, where at k=1 this reads y1−x∗=u0. Also uk=tk(xk−zk)=uk−1+tk(xk−yk).

2.1F1F2step 1.1givenalgebra∎

For k≥2, convexity gives F(zk)≤(1−1/tk)F(xk−1)+(1/tk)F(x∗). Apply [F2] with y=yk, z=zk, and xk=TL(yk). The upper bound for F(zk)−F(xk) and its lower bound from [F2], multiplied by 2tk2/L, imply 2tk2L(F(xk)−F(x∗))≤2tk(tk−1)L(F(xk−1)−F(x∗))−tk2∥xk−yk∥22−2tk2⟨yk−zk,xk−yk⟩. Using [F1] to replace tk(tk−1) by tk−12, and the two vector identities of step 1.1 to replace the last two terms by ∥uk−1∥22−∥uk∥22, gives the claimed recurrence for k≥2. At k=1, z1=x∗ and [F2] directly gives F(x1)−F(x∗)≤(L/2)(∥x0−x∗∥22−∥x1−x∗∥22); since t1=1 and u1=x1−x∗, this is the claimed base case without using F(x0).

Depends on

Used by

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources