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.

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Finite separable block penalties are proper, closed, and convex

Statement

Let $n \geq 1$ be finite, let $d_{i} \geq 1$ , and let each $g_{i} : R^{d_{i}} \to (- \infty, + \infty]$ be proper, closed, and convex in the epigraph convention of Proper closed convex extended-real functions on $R^{n}$ . On the product $R^{d_{1}} \times \dots \times R^{d_{n}}$ define $g (x) = \sum_{i = 1}^{n} g_{i} (x_{i})$ . Then $g$ is proper, closed, and convex. The proof below uses only finite construction and the closed epigraphs of the separate terms.

Facts & Assumptions

Given: A finite nonempty block list and the functions in the Statement.

[F1]

Proper closed convex extended-real functions on $R^{n}$ : each $g_{i}$ has nonempty effective domain, satisfies the extended-real convexity inequality, and has closed epigraph.

[F2]

The principle of mathematical induction: a property true at the base index and preserved at each successor holds for every natural index in the finite block list.

[A1]

Each block dimension $d_{i}$ is finite and positive, and the index set has size $n \geq 1$ .

Proof

1.1A1F1F2construct

For each $i$ , [F1] gives a point in $dom g_{i}$ . The empty tuple is a choice for the first zero blocks; if a tuple has been constructed for the first $k < n$ blocks, append one point from the nonempty $dom g_{k + 1}$ . Induction [F2] gives a tuple $x$ in all $n$ domains, so $g (x) < + \infty$ and $g$ is proper. Only finitely many choices are made.

1.2F1givenalgebra

If $g (x), g (y) < + \infty$ and $0 \leq θ \leq 1$ , apply the convexity inequality in [F1] to each pair $(x_{i}, y_{i})$ and add the finitely many inequalities. If either endpoint has infinite $g$ value, the extended-real inequality is automatic under the endpoint convention. Thus $g$ is convex.

1.3F1givenalgebra

Fix a block $i$ and a real $c$ . Its weak sublevel $C_{i, c} = {u_{i} : g_{i} (u_{i}) \leq c}$ is closed: the map $u_{i} \mapsto (u_{i}, c)$ preserves Euclidean distances and is continuous, so $C_{i, c}$ is its preimage of the closed set $epi g_{i}$ from [F1]. Therefore, whenever $g_{i} (x_{i}) > c$ , the open complement of $C_{i, c}$ is a neighborhood of $x_{i}$ throughout which $g_{i} > c$ .

2.1A1step 1.3givenalgebra

Take any $(x, t) \notin epi g$ . If all $g_{i} (x_{i})$ are finite, put $s = \sum_{i} g_{i} (x_{i}) > t$ and $c_{i} = g_{i} (x_{i}) - (s - t) / (2 n)$ . By step 1.3, each $x_{i}$ has an open neighborhood on which $g_{i} > c_{i}$ , so their product is a neighborhood of $x$ on which $g (u) > \sum_{i} c_{i} = (s + t) / 2$ . With $η = (s - t) / 4 > 0$ , the product of these neighborhoods and $(t - η, t + η)$ misses $epi g$ , because its height coordinate is below $t + η < (s + t) / 2$ .

2.2A1step 1.3givenalgebra

If some $g_{i} (x_{i}) = + \infty$ , choose such an index $i$ . For each $j \neq i$ , put $c_{j} = g_{j} (x_{j}) - 1$ when that value is finite, and $c_{j} = 0$ when it is infinite. Choose $B = t + 1 - \sum_{j \neq i} c_{j}$ . Since $g_{i} (x_{i}) > B$ and $g_{j} (x_{j}) > c_{j}$ , step 1.3 supplies an open neighborhood of each block on which the corresponding strict lower bound holds. Throughout their product, $g (u) > B + \sum_{j \neq i} c_{j} = t + 1$ . Restrict the epigraph-height coordinate to $(t - 1 / 2, t + 1 / 2)$ ; this neighborhood also misses $epi g$ .

3.1step 1.1step 1.2step 2.1step 2.2algebra∎

Every point outside $epi g$ is covered by one of steps 2.1 or 2.2, so its complement is open and $epi g$ is closed. Together with properness from step 1.1 and convexity from step 1.2, this proves the Statement without converting an arbitrary sequential-closure claim into topological closedness.

Depends on

Used by

Dependency tree · two levels

9 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

Stephen J. Wright, Coordinate Descent Algorithms, §3.3 (standard reference, not scraped)
Clément Royer, Regularized, Large-Scale and Distributed Optimization, Lecture 3, §3.3 (standard reference, not scraped)