\overline{B_{r} (x_{0})} \subset Ω

u (x_{0}) = \frac{1}{∣ B _{r} ∣} \int_{B_{r} (x_{0})} u = \frac{1}{∣ \partial B _{r} ∣} \int_{\partial B_{r} (x_{0})} u .

u

attains its maximum

M

at an interior point, the average of the nonnegative function

M - u

over every sufficiently small centred ball is zero. Continuity makes

u = M

on each such ball, and connectedness propagates this equality throughout the domain. Applying the same argument to

- u

Solved by gpt-5.6-sol high.

b

Solution

Differentiate the ball mean-value formula and use the divergence theorem:

D u (x_{0}) = \frac{1}{∣ B _{r} ∣} \int_{B_{r} (x_{0})} D u = \frac{1}{∣ B _{r} ∣} \int_{\partial B_{r} (x_{0})} uν .

Consequently

∣ D u (x_{0}) ∣ \leq \frac{∣ \partial B _{r} ∣}{∣ B _{r} ∣} sup_{B_{r} (x_{0})} ∣ u ∣ = \frac{n}{r} sup_{B_{r} (x_{0})} ∣ u ∣.

Thus one may take

C (n) = n

Solved by gpt-5.6-sol high.

∣ D^{k + 1} u (x_{0}) ∣ \leq C (n, k) R^{- (k + 1)} sup_{B_{R} (x_{0})} ∣ u ∣.

The growth hypothesis bounds the right side by

C^{'} R^{α - 1} + o (1)

, which tends to zero as

R \to \infty

. Thus every derivative of order

k + 1

vanishes everywhere. The Taylor theorem makes

u

a polynomial of degree at most

k

Solved by gpt-5.6-sol high.

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