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Dirac-delta counting formula for regular zeros

Statement

Assume the Axiom of Countable Choice. Let d≥1, let U⊆Rd be open, and let f:U→Rd be C1. Suppose every zero p∈f−1(0) is regular, meaning that the real derivative Df(p) is invertible. For every real continuous compactly supported function ψ:U→R, put

κϵ(y)=1B(0,ϵ)(y)∣B(0,ϵ)∣,y∈Rd,ϵ>0,

where B(0,ϵ) is the open Euclidean ball and ∣B(0,ϵ)∣ is its Lebesgue volume. Then the limit exists and

lim⁡ϵ↓0∫Uψ(z)κϵ(f(z)) ∣det⁡Df(z)∣ ddz=∑p∈f−1(0)ψ(p).

Only finitely many terms on the right are nonzero. This gives the precise distributional meaning, against compactly supported test functions, of δ0(f(z))∣det⁡Df(z)∣ as the counting measure on regular zeros.

Facts & Assumptions

Given: Countable choice, U,f,ψ, and the regularity of every zero as in the Statement.

[F1]

The Euclidean inverse function theorem: a C1 map between open subsets of Rd with invertible derivative at a point is a C1 diffeomorphism between suitable neighborhoods of that point and its image.

[F2]

A C1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: assuming countable choice, a C1 diffeomorphism T:V→W between open subsets of Rd satisfies ∫Wq(y) dy=∫Vq(T(z))∣det⁡DT(z)∣ dz for every nonnegative measurable q.

[F3]

The Axiom of Countable Choice (ACω): countable choice is the set-theoretic assumption used by [F2].

Proof

technique · direct
1.1givenF1algebra

Let K=supp⁡ψ, a compact subset of U. Every point of f−1(0)∩K is isolated by [F1]. This zero set is finite: if it were infinite, compactness of K would give a convergent subsequence of distinct zeros with limit p∈K; continuity gives f(p)=0, while the inverse-function neighborhood of p from [F1] contains no other zero, a contradiction.

2.1givenF1step 1.1algebra

For each of these finitely many zeros p, choose a neighborhood Vp⋐U on which f is a diffeomorphism onto an open neighborhood Wp∋0. Shrink the neighborhoods to be pairwise disjoint. The compact set K∖⋃pVp contains no zero, so continuity of f gives a positive lower bound for ∥f(z)∥ there. Choose ϵ small enough that B(0,ϵ)⊆Wp for every p and that no point of K∖⋃pVp maps into B(0,ϵ). If there are no such zeros, compactness gives the same positive lower bound on all of K, and the integral is zero for sufficiently small ϵ.

3.1givenF2F3step 2.1algebra

It is enough first to take ψ≥0; a real ψ is the difference of its nonnegative continuous compactly supported positive and negative parts. For ψ≥0 and ϵ as in step 2.1, the integrand vanishes on K∖⋃pVp. Apply [F2], under the countable-choice hypothesis [F3], to f∣Vp:Vp→Wp with the nonnegative measurable integrand q(y)=ψ(f∣Vp−1(y))κϵ(y). Change of variables makes the integral over Vp equal to ∫Wpψ(f∣Vp−1(y))κϵ(y) ddy. Summing these finitely many terms gives the original integral.

4.1givenstep 1.1step 3.1algebra

The function y↦ψ(f∣Vp−1(y)) is continuous at 0 and has value ψ(p). Its integral against κϵ is the average over B(0,ϵ); continuity implies that this average tends to ψ(p). The sum is finite by step 1.1, so taking the limit in step 3.1 gives the claimed sum for nonnegative ψ, and subtraction gives it for every real ψ.

5.1givenstep 4.1algebra∎

The limiting distribution assigns each regular zero p the unit mass ψ(p); therefore its density notation is exactly δ0(f(z))∣det⁡Df(z)∣, with no additional multiplicity or orientation sign.

Depends on

Dependency tree · 2 levels

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