LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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Compactly supported scaled Euclidean bumps

Statement

Assume ${A C}_{ω}$ . For $n \geq 1$ there is a fixed smooth $b : R^{n} \to [0, 1]$ , equal to one on ${\overline{B}}_{1} (0)$ and with support contained in $B_{2} (0)$ . For $a \in R^{n}, r > 0$ , $b_{a, r} (x) = b ((x - a) / r)$ satisfies $\int ∣ D b_{a, r} ∣ = C_{n} r^{n - 1}$ , where $C_{n} = \int ∣ D b ∣ < \infty$ . For every $0 < r < R$ one can instead obtain a smooth bump equal to one on ${\overline{B}}_{r} (a)$ and supported strictly inside $B_{R} (a)$ .

Facts & Assumptions

Given: Assume ${A C}_{ω}$ . The centre, positive radii with strict ordering, and Euclidean dimension are those of the statement. The bump and scaling constants must be constructed.

[F1]

The smooth step is zero on the negative half-line and one from one onward. (The standard smooth step function).

[F2]

The chain rule gives the derivative of a scaled bump. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F3]

Invertible linear maps scale Lebesgue measure by their determinant modulus. (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F4]

Translations preserve Lebesgue measure. (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F5]

Increasing nonnegative approximations converge in integral. (Monotone convergence for the integral).

Proof

1.1givenF1algebra

For $0 < r < R$ set $s = (r + R) / 2$ and $b_{*} (x) = σ ((s^{2} - ∣ x - a ∣^{2}) / (s^{2} - r^{2}))$ . F1 makes this smooth, between zero and one, equal to one for $∣ x - a ∣ \leq r$ , and zero for $∣ x - a ∣ \geq s$ . Its support lies in the closed s-ball, a compact subset of the open R-ball. Taking a=0,r=1,s=3/2 defines the fixed b with support inside B_2.

2.1step 1.1F2F3F4F5∎

F2 gives $D b_{a, r} (x) = r^{- 1} D b ((x - a) / r)$ . F3 and F4 imply $\int f ((x - a) / r) d x = r^{n} \int f$ for every nonnegative Borel f: this is first the set-measure identity for indicators, then a finite sum for nonnegative simple f, and finally F5 applied to $2^{- k} ⌊ 2^{k} \min (f, k) ⌋ ↑ f$ . Apply it to the continuous compactly supported $f = ∣ D b ∣$ . Its integral is finite because Db is bounded and vanishes outside a bounded ball. Multiplication by $r^{- 1}$ yields $C_{n} r^{n - 1}$ as asserted.

Source notes

Hunter, §1.9.1 Theorem 1.29 and Example 1.30, printed p. 12. The strict support margin and exact gradient scaling are computed locally.

Depends on

Used by

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Sources

Hunter, Notes on Partial Differential Equations (standard reference, not scraped)
Sung-Jin Oh, Lecture Notes for Math 222A (standard reference, not scraped)