PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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 pointwise supremum of convex functions is convex wherever it is finite

Statement

Let $(f_{i})_{i \in I}$ be a nonempty family of convex real-valued functions on a common convex set $C$ , put $D = {x \in C : \sup_{i \in I} f_{i} (x) < \infty}$ , and define $g (x) = \sup_{i \in I} f_{i} (x)$ on $D$ . Then $D$ is convex and $g$ is convex on $D$ . Here $x \in D$ means precisely that the nonempty set ${f_{i} (x) : i \in I}$ is bounded above, so its real supremum exists.

Facts & Assumptions

Given: The family, domain, and finite-valued set in the Statement.

[F1]

The function $f : C \to R$ is convex when $f ((1 - t) x + t y) \leq (1 - t) f (x) + t f (y)$ for all $x, y \in C$ and $t \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

[L1]

Every nonempty subset of $R$ that is bounded above has a least upper bound in $R$ (Dedekind completeness: the least-upper-bound property).

Proof

technique · direct

1.1F1L1givenalgebra

Let $x, y \in D$ and $t \in [0, 1]$ . For every $i$ , [F1] gives $f_{i} ((1 - t) x + t y) \leq (1 - t) f_{i} (x) + t f_{i} (y) \leq (1 - t) g (x) + t g (y) .$ This common finite upper bound and [L1] show that the supremum exists at the combined point, so that point lies in $D$ and $D$ is convex.

2.1step 1.1F1L1algebra∎

The common upper bound from step 1.1 also bounds the least upper bound supplied by [L1]. Hence $g ((1 - t) x + t y) \leq (1 - t) g (x) + t g (y),$ which is [F1] for $g$ .

Depends on

Used by

Nothing in the library uses this result yet.

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

S. Boyd and L. Vandenberghe, Convex Optimization, §3.2.3 (standard reference, not scraped)