Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 107 / 4 / b

Let $B⋐\Omega$ and let $h$ be the harmonic function whose boundary values are $\max (u_1,u_2)$. The harmonic extensions $(u_i{)}_B$ satisfy $(u_i{)}_B\le h$ by the maximum principle for harmonic functions, while subharmonicity gives $u_i\le (u_i{)}_B$ in $B$. Hence $\max (u_1,u_2)\le h$, proving that the maximum of two subharmonic functions is subharmonic.

The difference $u_1-u_2$ is subharmonic. If $M={\max }_{\overline{\Omega }}(u_1-u_2)>0$, its maximum set lies in $\Omega$. At any point of this set, comparison with the harmonic replacement on a small ball and the Strong maximum principle for harmonic functions show that the whole ball belongs to the maximum set. The set is therefore both open and closed in the connected domain $\Omega$, so it is all of $\Omega$, contradicting the boundary inequality. Thus $u_1\le u_2$ throughout $\Omega$. This is the comparison principle for subharmonic and superharmonic functions.

The constant $m={\min }_{\partial \Omega }\phi$ is subharmonic and lies below the boundary data, so the Perron family is nonempty. The constant $M={\max }_{\partial \Omega }\phi$ is superharmonic. Part (ii) gives $u\le M$ for every member $u$ of the family, while the member $m$ gives the lower bound. Henceso the pointwise supremum is finite and well-defined.

Fix $B⋐\Omega$ and $x_0\in B$. Choose $u_j$ in the Perron family with $u_j(x_0)\uparrow \overline{u}(x_0)$, replace successive terms by finite maxima using part (i), and take their harmonic lifts on $B$. The lifts remain in the family, are increasing, and are uniformly bounded. Interior estimates and the Arzela-Ascoli theorem give a harmonic limit $h$ on $B$ with $h\le \overline{u}$ and $h(x_0)=\overline{u}(x_0)$.

If $h(y)<\overline{u}(y)$ somewhere in $B$, take another family member larger than $h(y)$ and repeat the maximum-and-lift construction. Its harmonic limit $H$ satisfies $H\ge h$ and $H(x_0)=h(x_0)$. The strong minimum principle for elliptic operators forces $H=h$, contradicting the strict inequality at $y$. Thus $h=\overline{u}$ on $B$. Since $B$ was arbitrary, $\overline{u}$ is smooth and harmonic in $\Omega$. This is the Perron method for the Dirichlet problem.

Fix $\epsilon >0$ and choose a boundary neighbourhood $U$ of $z$ on which $|\phi -\phi (z)|<\epsilon$. The positive continuous barrier $w$ has a positive minimum $m$ on the compact set $\partial \Omega \setminus U$. Choosing $A$ large enough makeson all of $\partial \Omega$. The left function is subharmonic and belongs to the Perron family; the right function is superharmonic and dominates every family member by part (ii). ThereforeAs $x\to z$, continuity gives $w(x)\to 0$. Letting $\epsilon \downarrow 0$ proves $\overline{u}(x)\to \phi (z)$. Such a $w$ is a barrier for the Dirichlet problem, and $z$ is a regular boundary point.

Every boundary point of $D$ satisfies the exterior sphere condition. For a point on the inner spherical boundary, use a smaller ball inside the removed ball and tangent at that point; for a point on the cube, use a ball in a supporting exterior half-space. If the exterior ball has centre $y$ and radius $r$, a local positive harmonic barrier isfor $n\ge 3$, while in two dimensions use $w(x)=\log (|x-y|/r)$. Adding a sufficiently large positive multiple of a global superharmonic function extends the local barrier across the bounded domain. Hence every boundary point is regular by part (v), and the Perron method for the Dirichlet problem produces a harmonic function attaining the prescribed continuous boundary data. The maximum principle for harmonic functions gives uniqueness.