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Far-field asymptotics of compact-source Newtonian potentials

Statement

Assume the Axiom of Countable Choice and let n≥2. Fix R>0, and let f:Rn→R be Lebesgue measurable, integrable, and compactly supported, with ∥f∥1:=∫Rn∣f(y)∣ dy<∞, and zero outside the open Euclidean ball BR(0). Put M:=∫Rnf(y) dy and let ωn−1 be the sphere-area normalization in the definition of Φ.

For every x∈Rn with r:=∣x∣>2R, the Newtonian integral Nf(x)=∫RnΦ(x−y)f(y) dy is absolutely finite. If n≥3, then

∣Nf(x)−Mr2−n(n−2)ωn−1∣≤2n−1Rωn−1∥f∥1r1−n.

If n=2, then

∣Nf(x)+M2πlog⁡r∣≤Rπ∥f∥1r−1.

The constants are independent of the direction of x. In particular, when M=0 these are the respective improved orders Nf(x)=O(r1−n) for n≥3 and Nf(x)=O(r−1) for n=2.

Facts & Assumptions

Given: Assume ACω, let n≥2, let R>0, and let f:Rn→R be Lebesgue measurable with ∫Rn∣f∣<∞ and f=0 off BR(0).

[A1]

The Axiom of Countable Choice, written ACω, says every sequence of nonempty sets has a choice function. (The Axiom of Countable Choice (ACω)).

[F2]

Away from its pole the kernel is Φ(z)=∣z∣2−n/((n−2)ωn−1) for n≥3 and Φ(z)=−(2π)−1log⁡∣z∣ for n=2; the value at zero is assigned as in the kernel convention. (Fundamental solution for the positive operator minus Laplacian).

[F3]

A real measurable function is integrable exactly when ∫∣f∣<∞, and its integral is the difference of the finite integrals of its positive and negative parts. (Integrable real and complex functions, and their integrals).

[F4]

The open ball is B(0,R)={y:∥y∥2<R} for R>0. (Open ball, closed ball and sphere in a metric space).

[F5]

The Borel sigma-algebra is generated by the open sets. (The Borel sigma-algebra of a topological space).

[F6]

Continuous maps have Borel preimages of Borel sets. (A continuous map has Borel preimages of Borel sets).

[F7]

Under ACω, every Borel subset of Rn is Lebesgue measurable. (Assuming countable choice, every Borel subset of Rn is Lebesgue measurable).

[F8]

Sums, differences, products and absolute values of measurable real functions are measurable. (Arithmetic and lattice operations preserve measurability whenever they are defined).

[F9]

The nonnegative integral is monotone and homogeneous for nonnegative scalars. (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F10]

The integral is linear on L1. (The Lebesgue integral is linear on L1(μ)).

[F11]

For an integrable function, ∣∫h∣≤∫∣h∣. (The modulus of an integral is bounded by the integral of the modulus).

[F12]

If a real function is continuous on a closed interval and differentiable on its interior, the mean value theorem gives its difference as an interior derivative times the interval length. (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

[F13]

For α∈R, (tα)′=αtα−1 on t>0, and tα is continuous there. (Continuity and derivatives of positive-base real powers).

[F15]

Differentiability of a real function at every point of an open interval implies continuity there. (A function differentiable at c is continuous at c).

[F16]

The Newtonian potential is the pointwise integral wherever it is absolutely finite. (Newtonian potential of compactly supported data).

Proof

technique · direct
1.1givenF1F2

Fix x with r:=∣x∣>2R and y∈BR(0), and set s:=∣x−y∣. By [F1], s≥r−∣y∣>r−R>r/2>0 and ∣s−r∣≤∣y∣<R. Every number between r and s is therefore greater than r/2, so neither kernel evaluation below is at the pole.

1.2F2F13F14F15algebra

For t>0, write qn(t)=t2−n/((n−2)ωn−1) when n≥3 and q2(t)=−(2π)−1log⁡t when n=2. By [F2], [F13], and [F14], qn′(t)=−1/(ωn−1tn−1) for n≥3 and q2′(t)=−1/(2πt). These profiles are continuous on every closed interval in (0,∞) and differentiable in its interior; for the logarithm, continuity follows from [F15].

2.1step 1.1step 1.2F12algebracases

If s=r, then Φ(x−y)−Φ(x)=0. Otherwise apply [F12] between r and s; the derivative bounds in step 1.2 and the lower bound in step 1.1 give ∣Φ(x−y)−Φ(x)∣≤Cn∣y∣r1−n, where Cn=2n−1/ωn−1 for n≥3 and C2=1/π. The same bound holds when s=r.

2.2A1F1F2F4F5F6F7F13F14F15F17step 1.1

For this fixed x, y↦Φ(x−y) is continuous on BR(0) by step 1.1 and [F2], [F13]–[F15], and the Euclidean norm is continuous by [F1]. Extend this function by zero outside the open ball to obtain gx, and extend y↦Φ(x−y)−Φ(x) the same way to obtain hx. For any open O⊆R, continuity makes each preimage inside BR(0) relatively open and [F6] makes it Borel; since BR(0) is open by [F17], this relative preimage is open in Rn. The extension's preimage is that set, with BR(0)c adjoined when 0∈O, so [F4], [F5], and [F17] show that gx and hx are Borel. Thus [F7] makes both functions Lebesgue measurable under [A1].

3.1F3F8F9F16step 2.1step 2.2

By [F8], fgx=fΦ(x−⋅) and fhx are measurable. Step 2.1 gives ∣f(y)hx(y)∣≤CnRr1−n∣f(y)∣, while ∣gx(y)∣≤∣Φ(x)∣+CnRr1−n gives ∣f(y)Φ(x−y)∣=∣f(y)gx(y)∣≤(∣Φ(x)∣+CnRr1−n)∣f(y)∣. Since f∈L1, [F3] and [F9] make both products integrable. In particular the Newtonian integral is absolutely finite at x, so [F16] defines Nf(x).

4.1F2F9F10F11step 3.1

Since f vanishes outside BR(0), f(y)Φ(x−y)=Φ(x)f(y)+f(y)hx(y) for every y. Integrability from step 3.1 and [F10] give Nf(x)=MΦ(x)+∫f(y)hx(y) dy. By [F11], then the pointwise bound of step 3.1 and [F9], ∣Nf(x)−MΦ(x)∣≤∫∣f(y)hx(y)∣ dy≤CnR∥f∥1r1−n. Substituting the profiles [F2] yields the stated two estimates, with constants independent of the direction of x.

5.1A1F2F7F16step 4.1cases∎

If M=0, the leading term in step 4.1 vanishes, giving the two improved orders; if f=0, then M=Nf=0 and both bounds hold with equality. The assumptions n≥2 and r>2R exclude dimensions zero and one and the endpoint r=2R. Countable Choice is used only through the kernel and potential conventions [F2], [F16] and the Borel-to-Lebesgue interface [F7]; no full Axiom of Choice is invoked.

Source notes

Hunter, §2.7 equation (2.24) and the following exterior-asymptotic passage, printed pp.36–37, defines the Newtonian potential and, for n≥3, rewrites it as the total-charge leading term times a kernel ratio, then uses dominated convergence to obtain the leading asymptotic. For n=2 Hunter states only that the potential generally grows logarithmically. That passage does not prove the explicit O(r1−n) and O(r−1) remainders or the zero-mass improvements; steps 1.1–4.1 derive those quantitatively from the radial derivatives. Oh, §4.2 Theorem 4.4 and Corollary 4.7, printed pp.59–60, give the harmonic-derivative and growth-class context for applications of these estimates; they are not used to prove the far-field bounds here.

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