= Solution
We prove <compactness of zero-boundary Poisson solutions>. Here the printed $\partial\Omega$ must mean $\partial B_1$; otherwise the boundary condition refers to an unspecified set. Use this intended boundary condition throughout. Let $f=\Delta u$. A maximum-principle barrier gives a uniform bound on $u$: with $\psi(x)=(1-|x|^2)/(2n)$, we have $\Delta\psi=-1$ and $\psi=0$ on $\partial B_1$. Since $|f|\leq1$,
$$
\Delta(u-\psi)=f+1\geq0,\qquad\Delta(-u-\psi)=-f+1\geq0.
$$
The <weak maximum principle> gives $-\psi\leq u\leq\psi$, so $\|u\|_\infty\leq1/(2n)$.
Standard <elliptic regularity> on the smooth ball first gives $u\in C^{2,\mu}(\overline{B_1})$. The <global Schauder estimate> then gives
$$
\|u\|_{C^{2,\mu}(\overline{B_1})}\leq C\left(\|u\|_{C^0(\overline{B_1})}+\|f\|_{C^{0,\mu}(\overline{B_1})}\right)\leq C\left(1+\frac1{2n}\right).
$$
Thus the functions and all their derivatives through order two are uniformly bounded and <equicontinuous> on the closed ball. The <Arzelà-Ascoli theorem> gives a subsequence converging uniformly along with all these derivatives. Integration along line segments identifies the limiting derivatives, so the convergence is in $C^2(\overline{B_1})$.
It remains to keep the limit in the set, rather than merely proving <relative compactness>. If $u_j\to u$ in $C^2$, then the zero boundary values pass to the limit and $f_j=\Delta u_j\to f=\Delta u$ uniformly. For each pair of distinct points,
$$
\frac{|f(x)-f(y)|}{|x-y|^\mu}=\lim_j\frac{|f_j(x)-f_j(y)|}{|x-y|^\mu}.
$$
Taking suprema gives $[f]_\mu\leq\liminf_j[f_j]_\mu$, while $\|f_j\|_\infty\to\|f\|_\infty$. Hence $\|f\|_{C^{0,\mu}}\leq1$. The limit belongs to the set, proving \b[<sequential compactness> in $C^2(\overline{B_1})$]. The fixed positive Hölder exponent supplies the compactness; a bound on the <supremum norm> of $\Delta u$ alone would not supply this argument.
Back to article page
