Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 107 / 5 / c / Solution

Take ${\Omega }^{\prime }$ nonempty and compactly contained in $\Omega$. The argument is a compactness proof of a nonnegative elliptic solution estimate, with the global Schauder estimate retaining control all the way to the boundary. The operator and the Hölder space exponent are fixed throughout.

Suppose the estimate fails. For each integer $k$ there is an admissible nonnegative solution $u_k$, with forcing $f_k$, such thatPut $v_k=u_k/M_k$ and $g_k=f_k/M_k$. Then $0\le v_k\le 1$, ${\max }_{\overline{\Omega }}{v}_k=1$, $v_k=0$ on the boundary, $L{v}_k=g_k$, andGlobal elliptic regularity and the global Schauder estimate for zero Dirichlet boundary condition giveThis estimate includes the $C^0$ term, so it does not require invertibility of $L$ or a sign restriction on $c$. The compact embedding of Hölder spaces, or the Arzelà-Ascoli theorem applied through second partial derivatives, gives a subsequence converging in $C^2(\overline{\Omega })$ to $v$. Thus $v\ge 0$, $Lv=0$, $v=0$ on the boundary, and $\max v=1$.

Choose $x_k\in {\Omega }^{\prime }$ with $v_k(x_k)\le {\inf }_{{\Omega }^{\prime }}{v}_k+1/k$. A further subsequence has $x_k\to x_{\ast }\in \overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence yields $v(x_{\ast })=0$. Part (b) says that a nonnegative homogeneous solution on the connected domain is either zero everywhere or strictly positive inside. Both alternatives contradict, respectively, $\max v=1$ and $v(x_{\ast })=0$. ThereforeThe interior infimum and the forcing control the global supremum. Global boundary control is crucial here: an interior-only compactness argument could lose the normalized maximum at the boundary.