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Ball means and sphere means are related by a radial derivative

Statement

Assume the Axiom of Countable Choice, let n≥1, let g∈C0(Rn), and for x∈Rn, r>0 let Ag(x,r):=∣Br(x)∣−1∫Br(x)g be the ball average of The average of a locally integrable function over a Euclidean ball and Mg the spherical mean of Spherical means and the weighted ball integral of space-dependent data. Then for every r>0 Ag(x,r)=nrn∫0rsn−1Mg(x,s) ds,Mg(x,r)=Ag(x,r)+rn ∂rAg(x,r).

Facts & Assumptions

Given: Countable Choice, n≥1, g∈C0(Rn), and the means Ag and Mg of the statement.

[F1]

Under Countable Choice, ∫Rnf dλn=∫0∞∫Sn−1f(sθ)sn−1 dσ(θ) ds for every Borel measurable f:Rn→[0,∞] (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F2]

Mg(x,s)=ωn−1−1∫Sn−1g(x+sω) dσ(ω) and ωn−1=σ(Sn−1) for x∈Rn, s>0 (Spherical means and the weighted ball integral of space-dependent data).

[F3]

∣Br(x)∣=ωn−1rn/n for r>0 (Sphere and ball measures scale in Rn).

[F5]

If I⊆R is order-convex with at least two elements and f:I→R is continuous on I, then for r0∈I, r↦∫r0rf is a primitive of f on I (Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a) for any primitive G).

Proof

1.1F1F2F3algebra

Polar coordinates applied to the positive and negative parts of the function u↦g(x+u)1Br(0)(u) give, using translation invariance (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation) to set y=x+u, ∣Br(x)∣Ag(x,r)=∫Br(x)g=∫0r∫Sn−1g(x+sω)sn−1 dσ(ω) ds=ωn−1∫0rsn−1Mg(x,s) ds, where the inner integral is ωn−1Mg(x,s) by the definition of the spherical mean; dividing by ∣Br(x)∣=ωn−1rn/n gives the first identity.

1.2F4algebra

The function s↦Mg(x,s) is continuous on (0,∞): for s,s0>0 with ∣s−s0∣<1 one has ∣Mg(x,s)−Mg(x,s0)∣≤sup⁡ω∈Sn−1∣g(x+sω)−g(x+s0ω)∣, and the two points x+sω and x+s0ω lie in the compact ball Bs0+1(x)‾ at distance ∣s−s0∣; since g is uniformly continuous on that ball by [F4], the supremum tends to 0 as s→s0.

1.3F5algebra

Hence s↦sn−1Mg(x,s) is continuous on (0,∞), so locally at any r>0 split the integral at a fixed r0∈(0,r); the first part is constant and [F5] differentiates the second part. Thus r↦∫0rsn−1Mg(x,s) ds is a primitive of sn−1Mg(x,s), and the first identity gives ∂r(rnAg(x,r))=n rn−1Mg(x,r) for r>0.

2.1algebra∎

Differentiating the product gives ∂r(rnAg(x,r))=nrn−1Ag(x,r)+rn∂rAg(x,r), so equating with the previous display and dividing by nrn−1>0 yields Mg(x,r)=Ag(x,r)+rn∂rAg(x,r), the second identity.

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