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

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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.

One-step proximal-gradient estimate

Statement

Let $f : R^{n} \to R$ be convex and continuously differentiable with $L$ -Lipschitz gradient, $L > 0$ , and let $g$ be proper, closed and convex. Put $F = f + g$ and $x^{+} = T_{L} (y)$ . For every $z \in R^{n}$ , $F (z) \geq F (x^{+}) + \frac{L}{2} ∥ x^{+} - y ∥_{2}^{2} + L ⟨ y - z, x^{+} - y ⟩ .$

Facts & Assumptions

Given: The functions and points in the Statement.

[F1]

Three-point inequality for the proximal-gradient model: the proximal model at $y$ satisfies its three-point inequality for every comparison point.

[F2]

Descent inequality for a Lipschitz gradient: $f (x^{+}) \leq f (y) + ⟨ \nabla f (y), x^{+} - y ⟩ + (L / 2) ∥ x^{+} - y ∥_{2}^{2}$ .

[F3]

Differentiable convex functions are characterized by the gradient inequality: $f (z) \geq f (y) + ⟨ \nabla f (y), z - y ⟩$ .

[F4]

A finite continuous convex summand preserves proper closed convexity: $F$ is proper, closed and convex, and $F = f + g$ .

[F5]

Existence, resolvent characterization, and firm nonexpansiveness of the proximal map: $T_{L} (y)$ is an attained proximal minimizer; comparison with a finite-domain point shows $x^{+} \in dom g$ , so $F (x^{+})$ is finite.

Proof

1.1F1F2F3F4F5givenalgebra

Write $d = x^{+} - y$ . If $g (z) = + \infty$ , then $F (z) = + \infty$ and the asserted inequality holds because [F5] makes its right side finite. Otherwise $F (z)$ and $F (x^{+})$ are real, so ordinary subtraction is valid. By [F2] and [F3], $F (z) - F (x^{+}) \geq g (z) - g (x^{+}) + ⟨ \nabla f (y), z - x^{+} ⟩ - \frac{L}{2} ∥ d ∥_{2}^{2} .$ The three-point inequality [F1], after rearrangement, gives $g (z) - g (x^{+}) + ⟨ \nabla f (y), z - x^{+} ⟩ \geq \frac{L}{2} ∥ d ∥_{2}^{2} + \frac{L}{2} (∥ x^{+} - z ∥_{2}^{2} - ∥ z - y ∥_{2}^{2}) .$ Substitution yields $F (z) - F (x^{+}) \geq \frac{L}{2} (∥ x^{+} - z ∥_{2}^{2} - ∥ z - y ∥_{2}^{2}) .$

2.1step 1.1F5algebra∎

Since $z - y = (z - x^{+}) + d$ , the difference of squares in step 1.1 is $∥ x^{+} - z ∥_{2}^{2} - ∥ z - y ∥_{2}^{2} = - 2 ⟨ z - x^{+}, d ⟩ - ∥ d ∥_{2}^{2} = 2 ⟨ y - z, d ⟩ + ∥ d ∥_{2}^{2} .$ Multiplying by $L / 2$ gives the finite-gap inequality; adding the finite value $F (x^{+})$ yields the displayed statement.

Depends on

Used by

Dependency tree · two levels

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Sources

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