LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 convex function is bounded above and below on a smaller interior cube

Statement

Let $n \geq 1$ , let $C \subseteq R^{n}$ be convex, let $f : C \to R$ be convex, and suppose the closed sup-norm cube

$Q (a, r) = {x : ∥ x - a ∥_{\infty} \leq r}$

with $r > 0$ lies in $C$ (Axis-parallel rectangles in $R^{m}$ and their volume, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ). Then $f$ is bounded above on the full cube and bounded above and below on the concentric half-sized cube.

Facts & Assumptions

Given: The data in the Statement. Finite maxima exist by Every nonempty finite set of reals has a maximum and a minimum and have the convention of Maximum and minimum of a set.

[L1]

For a positive finite family of points in $C$ and nonnegative weights summing to one, $f$ of their weighted Euclidean sum is at most the weighted sum of their $f$ -values (Finite Jensen inequality for convex functions on $R^{n}$ ).

Proof

technique · direct

1.1

Every point of $Q (a, r)$ is an explicit convex combination of its finite vertex set. By [L1], its value is at most the corresponding weighted average of the vertex values, hence at most their maximum $M$ .

L1givenalgebra

2.1

If $x \in Q (a, r / 2)$ , then the reflection $2 a - x$ lies in $Q (a, r)$ and $a = (x + (2 a - x)) / 2$ . Convexity gives $f (a) \leq (f (x) + f (2 a - x)) / 2$ , so step 1.1 yields $2 f (a) - M \leq f (x) \leq M .$ Thus the half-sized cube has both bounds.

step 1.1givenalgebra∎

Depends on

Used by

A convex function on an open convex set is locally Lipschitz Theorem

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

D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lecture 4 (standard reference, not scraped)