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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A locally finite shellwise perturbation with rapidly decaying size preserves properness of a smooth exhaustion

Statement

Let h:M→[0,∞) be a smooth proper function, and for each n≥1 let un:M→R be smooth with supp⁡(un)⊆h−1([n−1,n+2])and∥un∥∞≤2−n−2. Assume the family (supp⁡(un))n≥1 is locally finite. Then the sum u:=∑n≥1un is smooth, and h+u is still proper.

Facts & Assumptions

Given: A smooth proper function h:M→[0,∞) and a locally finite shellwise family (un)n≥1 as in the statement.

[L1]

A locally finite sum of smooth functions is smooth (A locally finite sum of smooth functions is smooth).

[A1]

The geometric series ∑n≥12−n−2 converges to 1/4.

[A2]

Closed subsets of compact spaces are compact.

Proof

technique · direct
1.1L1given

The family of supports is locally finite, so [L1] makes the sum u=∑nun a smooth function.

2.1A1step 1.1algebra

For every x∈M, the pointwise estimate gives ∣u(x)∣≤∑n≥1∣un(x)∣≤∑n≥12−n−2=14 by [A1]. Hence (h+u)(x)≥h(x)−14 for all x.

3.1A2step 2.1givenalgebra

If (h+u)(x)≤c, then step 2.1 gives h(x)≤c+14. Therefore {x:(h+u)(x)≤c}⊆{x:h(x)≤c+14}. The right-hand side is compact because h is proper, and the left-hand side is closed because h+u is continuous. By [A2], the left-hand side is compact.

4.1step 3.1∎

Thus h+u is proper, and the shellwise perturbation preserves properness.

Depends on

Used by

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Dependency tree · two levels

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Sources