 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-13/1/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write

ω_{n} = ∣ B_{1} ∣

. If

u

is a

C^{2}

\overline{B_{r} (x)} \subset Ω

u (x) = \frac{1}{n ω _{n} r ^{n - 1}} \int_{\partial B_{r} (x)} u d S = \frac{1}{ω _{n} r ^{n}} \int_{B_{r} (x)} u d y .

To prove the spherical property, express the normalized spherical mean on the fixed unit sphere:

M (r) = \frac{1}{n ω _{n}} \int_{S^{n - 1}} u (x + r θ) d S_{θ} .

M^{'} (r) = \frac{1}{n ω _{n} r ^{n - 1}} \int_{\partial B_{r} (x)} \partial_{ν} u d S = \frac{1}{n ω _{n} r ^{n - 1}} \int_{B_{r} (x)} Δ u d y = 0.

Since continuity gives

M (r) \to u (x)

r ↓ 0

u (x)

for every admissible radius. Integrate this identity in radial coordinates:

\int_{B_{r} (x)} u d y = n ω_{n} \int_{0}^{r} t^{n - 1} M (t) d t = ω_{n} r^{n} u (x) .

This proves the ball mean property as well. For

n = 1

, the sphere consists of the two endpoints and the same argument is the fundamental theorem of calculus.

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