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.

Three-point inequality for the proximal-gradient model

Statement

Let $g$ be proper, closed and convex, let $L > 0$ , let $f$ be differentiable at $y$ , and set $x^{+} = {prox}_{g / L} (y - \nabla f (y) / L)$ . For every $z \in R^{n}$ , $g (x^{+}) + ⟨ \nabla f (y), x^{+} - y ⟩ + \frac{L}{2} ∥ x^{+} - y ∥_{2}^{2} + \frac{L}{2} ∥ x^{+} - z ∥_{2}^{2} \leq g (z) + ⟨ \nabla f (y), z - y ⟩ + \frac{L}{2} ∥ z - y ∥_{2}^{2} .$

Facts & Assumptions

Given: The function, point, and scalar in the Statement.

[F1]

Existence, resolvent characterization, and firm nonexpansiveness of the proximal map: $x^{+} = {prox}_{g / L} (w)$ iff $L (w - x^{+}) \in \partial g (x^{+})$ .

[F2]

Subdifferentials of proper extended-real convex functions: $p \in \partial g (x)$ implies $g (z) \geq g (x) + ⟨ p, z - x ⟩$ for every $z$ .

[F3]

Proximal mapping: the proximal step uses $w = y - \nabla f (y) / L$ and parameter $1 / L$ .

Proof

1.1F1F2F3givenalgebra

By [F1] and [F3], $L (y - x^{+}) - \nabla f (y) \in \partial g (x^{+})$ . For finite $g (z)$ , [F2] gives $g (z) - g (x^{+}) \geq ⟨ L (y - x^{+}) - \nabla f (y), z - x^{+} ⟩ .$ Add $⟨ \nabla f (y), z - x^{+} ⟩$ and $\frac{L}{2} (∥ z - y ∥_{2}^{2} - ∥ x^{+} - y ∥_{2}^{2})$ to both sides. Using $∥ z - y ∥_{2}^{2} - ∥ x^{+} - y ∥_{2}^{2} = ∥ z - x^{+} ∥_{2}^{2} + 2 ⟨ z - x^{+}, x^{+} - y ⟩$ and $y - x^{+} = - (x^{+} - y)$ cancels the cross terms and leaves $L ∥ z - x^{+} ∥_{2}^{2} / 2$ . Rearranging is the claimed inequality.

2.1step 1.1algebra∎

If $g (z) = + \infty$ , the right side is $+ \infty$ and the inequality holds automatically, so the finite- $g (z)$ calculation proves the statement for every $z$ .

Depends on

Used by

One-step proximal-gradient estimate Lemma

Dependency tree · two levels

28 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

Neal Parikh and Stephen Boyd, Proximal Algorithms, §4.2 (standard reference, not scraped)
Amir Beck and Marc Teboulle, A Fast Iterative Shrinkage-Thresholding Algorithm for Linear Inverse Problems, Lemma 2.3 (standard reference, not scraped)