= Solution
First replace $u$ by $u+\varepsilon$ and later let $\varepsilon\downarrow0$. In the weak subsolution inequality use the admissible truncations approximating $\eta^2u^{\alpha-1}$. Uniform ellipticity, the coefficient bound, <Cauchy-Schwarz inequality>, and <Young inequality> give
$$
(\alpha-1)\lambda\int\eta^2u^{\alpha-2}|Du|^2
\leq2\Lambda\int\eta u^{\alpha-1}|Du||D\eta|
$$
and therefore
$$
\int|Du|^2u^{\alpha-2}\eta^2
\leq\frac{C(\lambda,\Lambda)}{(\alpha-1)^2}\int u^\alpha|D\eta|^2.
$$
Apply the <Sobolev embedding theorem> to $w=\eta u^{\alpha/2}$. The preceding estimate yields, for concentric balls $B_r\subset B_R\subset B_1$,
$$
\|u\|_{L^{\alpha\sigma}(B_r)}
\leq\left(\frac{C\alpha^2}{(\alpha-1)^2(R-r)^2}\right)^{1/\alpha}
\|u\|_{L^\alpha(B_R)}.
$$
Starting with $\alpha=p>1$, taking $\alpha_k=p\sigma^k$, and choosing radii decreasing to $1/2$, the product of constants converges because $\sum_k\alpha_k^{-1}<\infty$. Letting $k\to\infty$ proves
$$
\sup_{B_{1/2}}u\leq C(n,\lambda,\Lambda,p)\|u\|_{L^p(B_1)}.
$$
This exponent-raising argument is <Moser iteration>.
Back to article page
