Here

w

u

B_{r} = B (x_{0}, r)

, so

v = u - w \in H_{0}^{1} (B_{r})

. Subtract the weak equations and test with

v

\int_{B_{r}} ∣\nabla v ∣^{2} = \frac{1}{2} \int_{B_{r}} x_{1}^{2} u_{x_{1}} v_{x_{1}} + \int_{B_{r}} f v .

Because

B_{r} \subset B (0, 1)

∣ x_{1} ∣ \leq 1

∥ v ∥_{6} \leq C ∥\nabla v ∥_{2}

∥\nabla v ∥_{2}^{2} \leq \frac{1}{2} ∥\nabla u ∥_{2} ∥\nabla v ∥_{2} + C ∥ f ∥_{6/5} ∥\nabla v ∥_{2} .

After division and squaring,

\int_{B_{r}} ∣\nabla v ∣^{2} \leq C_{1} \int_{B_{r}} ∣\nabla u ∣^{2} + C_{2} (\int_{B_{r}} ∣ f ∣^{6/5})^{5/3} .

ii

Solution

∥ f ∥_{L^{6/5} (B_{r})} \leq ∣ B_{r} ∣^{1/3} ∥ f ∥_{L^{2} (B_{r})} .

Squaring and using

∣ B_{r} ∣ = ω_{3} r^{3}

yields

(\int_{B_{r}} ∣ f ∣^{6/5})^{5/3} \leq ω_{3}^{2/3} r^{2} \int_{B_{r}} ∣ f ∣^{2} .

Since

2 = 1 + 2 α

for

α = 1/2

, this is (6), with

C_{3} = ω_{3}^{2/3}

The Poincare-Wirtinger inequality on a ball and the assumed estimate (8) imply

\int_{B (x_{0}, ρ)} ∣ u - u_{B (x_{0}, ρ)} ∣^{2} \leq C ρ^{2} \int_{B (x_{0}, ρ)} ∣\nabla u ∣^{2} \leq C C_{4} ρ^{3 + 2 α} (∥\nabla u ∥_{L^{2} (B)}^{2} + ∥ f ∥_{L^{2} (B)}^{2}) .

This is exactly the stated Campanato space criterion with a constant

M

independent of the ball. Therefore

u \in C^{0, α} (B) .

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