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A locally finite shellwise perturbation with rapidly decaying size preserves properness of a smooth exhaustion

Statement

Let $h : M \to [0, \infty)$ be a smooth proper function, and for each $n \geq 1$ let $u_{n} : M \to R$ be smooth with $supp (u_{n}) \subseteq h^{- 1} ([n - 1, n + 2]) and ∥ u_{n} ∥_{\infty} \leq 2^{- n - 2} .$ Assume the family $(supp (u_{n}))_{n \geq 1}$ is locally finite. Then the sum $u : = \sum_{n \geq 1} u_{n}$ is smooth, and $h + u$ is still proper.

Facts & Assumptions

Given: A smooth proper function $h : M \to [0, \infty)$ and a locally finite shellwise family $(u_{n})_{n \geq 1}$ 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 $\sum_{n \geq 1} 2^{- n - 2}$ converges to $1 / 4$ .

[A2]

Closed subsets of compact spaces are compact.

Proof

technique · direct

1.1

The family of supports is locally finite, so [L1] makes the sum $u = \sum_{n} u_{n}$ a smooth function.

L1given

2.1

For every $x \in M$ , the pointwise estimate gives $∣ u (x) ∣ \leq \sum_{n \geq 1} ∣ u_{n} (x) ∣ \leq \sum_{n \geq 1} 2^{- n - 2} = \frac{1}{4}$ by [A1]. Hence $(h + u) (x) \geq h (x) - \frac{1}{4}$ for all $x$ .

A1step 1.1algebra

3.1

If $(h + u) (x) \leq c$ , then step 2.1 gives $h (x) \leq c + \frac{1}{4}$ . Therefore ${x : (h + u) (x) \leq c} \subseteq {x : h (x) \leq c + \frac{1}{4}} .$ 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.

A2step 2.1givenalgebra

4.1

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

step 3.1∎

Depends on

A locally finite sum of smooth functions is smooth

Used by

Nothing in the library uses this result yet.

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Sources

Marco Gualtieri, Topology I: Smooth Manifolds, Part 11 (standard reference, not scraped)
Shintaro Fushida-Hardy, Morse theory (standard reference, not scraped)