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.

A quadratic regularization has quadratic growth from its minimizer

Statement

Let $F : R^{n} \to (- \infty, + \infty]$ be convex, let $t > 0$ , $v \in R^{n}$ , and suppose $p$ minimizes $Q (y) = F (y) + ∥ y - v ∥_{2}^{2} / (2 t)$ with $Q (p) < + \infty$ . Then for every $z$ with $F (z) < + \infty$ , $Q (z) \geq Q (p) + \frac{∥ z - p ∥_{2}^{2}}{4 t} .$

Facts & Assumptions

Given: The extended-real convexity convention Proper closed convex extended-real functions on $R^{n}$ and Euclidean quadratic identity The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ .

[F1]

Convexity gives $F ((p + z) / 2) \leq (F (p) + F (z)) / 2$ for finite-valued endpoints (Proper closed convex extended-real functions on $R^{n}$ ).

[F2]

The midpoint identity is $∥ (p + z) / 2 - v ∥_{2}^{2} = (∥ p - v ∥_{2}^{2} + ∥ z - v ∥_{2}^{2}) / 2 - ∥ p - z ∥_{2}^{2} / 4$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

Since $Q (p) < + \infty$ and the quadratic term is finite, $F (p) < + \infty$ ; together with the stated $F (z) < + \infty$ , both endpoints are in the finite domain required for [F1].

Proof

1.1F1F2F3algebra

[F3] puts both endpoints in the finite domain. Apply [F1] to $p, z$ and combine with [F2], after multiplying the latter by $1 / (2 t)$ , to obtain $Q ((p + z) / 2) \leq (Q (p) + Q (z)) / 2 - ∥ z - p ∥_{2}^{2} / (8 t)$ .

2.1givenstep 1.1algebra∎

Minimality gives $Q (p) \leq Q ((p + z) / 2)$ . Combining with step 1.1 and rearranging yields $Q (z) \geq Q (p) + ∥ z - p ∥_{2}^{2} / (4 t)$ , including $z = p$ with equality.

Depends on

Used by

Vanishing aggregate-model error controls the proximal candidate Lemma

Dependency tree · two levels

18 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

Mateo Díaz and Benjamin Grimmer, Optimal Convergence Rates for the Proximal Bundle Method, §§2–3 (standard reference, not scraped)
Dominikus Noll, Olivier Prot, and Aude Rondepierre, A proximity control algorithm to minimize nonsmooth and nonconvex functions, §§2–4 (standard reference, not scraped)