Regular exhaustion and Dirichlet solutions on relatively compact surface domains

The holomorphic charts of $X$ have holomorphic transition maps, and holomorphic functions of one variable are smooth; hence these charts form a $C^{\mathrm{\infty }}$ atlas on the connected Hausdorff second countable space $X$. So $X$ is a smooth $2$-manifold, and the hypotheses of [F6] are met.

Locality of plane subharmonicity. Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be a plane domain and let $f:\mathrm{\Omega }\to [-\mathrm{\infty },\mathrm{\infty })$ be upper semicontinuous, not identically $-\mathrm{\infty }$ on any component of $\mathrm{\Omega }$, and subharmonic near each of its points. To use the harmonic-majorant characterization, fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ and a continuous $k$ on it, harmonic on $D(a,r)$, with $k\ge f$ on the boundary. If $k<f$ somewhere, set $s:=f-k$. Then $s$ is extended-valued upper semicontinuous, is positive somewhere, and satisfies $s\le 0$ on the boundary; it may equal $-\mathrm{\infty }$. It is bounded above on the compact disc: the open sets $\{s<n\}$ for positive integers $n$ cover it, so a finite subcover and nesting give a common upper bound. Thus $M:=\sup s$ is finite and positive. For each real $b<M$, the set $E_b:=\{s\ge b\}$ is closed by [F10] and nonempty by the definition of $M$; these sets have the finite-intersection property, so compactness gives a point $z_1\in {\bigcap }_{b<M}{E}_b$, where $s(z_1)=M$. Since $s\le 0$ on the boundary, $z_1$ is interior. Let $Z:=\{z\in D(a,r):s(z)=M\}$; it is nonempty and closed in the open disc by upper semicontinuity.

By [F6] and [A1] there is a smooth proper function $h:X\to [0,\mathrm{\infty })$ whose sublevel sets $\{h\le c\}$ are compact for every $c\in \mathbb{R}$. Fix a point $x_0\in X$ and put $M_0:=h(x_0)$, a nonnegative real number.

$Z$ is open in $D(a,r)$: given $z\in Z$, choose $\rho >0$ with $\overline{D(z,\rho )}\subseteq D(a,r)$ and with $D(z,\rho )$ contained in a neighbourhood of $z$ on which $f$ is subharmonic. On the disc $D(z,\rho )$ the function $f$ is subharmonic and $-k$ is subharmonic (it is harmonic), so $s=f+(-k)$ is subharmonic on $D(z,\rho )$; it attains the maximum $M$ at the interior point $z$ and therefore is constant equal to $M$ on $D(z,\rho )$ by the strong maximum principle, that is, $D(z,\rho )\subseteq Z$. Since $D(a,r)$ is connected and $Z\mathrm{\ne }\mathrm{\varnothing }$ is both open and closed in it, $Z=D(a,r)$ and $s\equiv M$ on the disc.

For each $n\ge 1$ the open interval $(M_0+n,M_0+n+1)$ is nonempty and, by [F7], it contains a regular value of $h$, since the regular values of $h$ are dense in $\mathbb{R}$. By [A1] fix such a regular value $c_n$ for every $n$. Then $c_n<M_0+n+1\le M_0+n+2$ and $c_{n+1}>M_0+n+1>c_n$, so $c_1<c_2<\cdots$ and $c_n\to \mathrm{\infty }$.

Step 2.2 leads to a contradiction, completing step 1.2: for $b\in \mathrm{\partial }D(a,r)$ the upper semicontinuity of $s$ at $b$ (as a function defined on a neighbourhood of $\overline{D(a,r)}$) gives $s(b)\ge {\mathrm{lim\hspace{0.17em}sup}}_{D(a,r)\ni z\to b}s(z)=M>0$, contradicting $s\le 0$ on the boundary circle. Hence no such counterexample exists, the harmonic comparison of step 1.2 always holds, and the harmonic-majorant characterization makes $f$ subharmonic on $\mathrm{\Omega }$.

For each $n$ let $K_n$ be the connected component of the closed set $\{h\le c_n\}$ containing $x_0$ and set $D_n:=\mathrm{int}{K}_n$. Since $\{h\le c_n\}$ is compact and $K_n$ is a component of it, $K_n$ is closed in $X$ and compact. Also $D_n=K_n\cap \{h<c_n\}$: a point $p\in K_n$ with $h(p)<c_n$ has a ball $B$ with $p\in B\subseteq \{h<c_n\}\subseteq \{h\le c_n\}$; this ball is connected and contains $p\in K_n$, hence lies in the component $K_n$, so $p$ is interior to $K_n$; conversely an interior point $p$ of $K_n$ cannot have $h(p)=c_n$, because $c_n$ is a regular value, so $h$ takes values larger than $c_n$ arbitrarily close to $p$ and no neighbourhood of $p$ is contained in $\{h\le c_n\}$. In particular $x_0\in D_n$ for every $n$.

At every point $p$ of $h^{-1}(c_n)$ the differential of $h$ is nonzero, so $h$ is a submersion at $p$ and by [F8] there are smooth coordinates $(u,v)$ around $p$ with $u=h-c_n$. Hence $h^{-1}(c_n)$ is a smooth embedded $1$-submanifold of $X$, and near each of its points the set $\{h\le c_n\}$ is the closed half-space $\{u\le 0\}$ in those coordinates, a set which is path-connected after shrinking the chart.

Gluing subharmonic functions on a surface. Let $\mathrm{\Omega }\subseteq X$ be open, let $u$ be subharmonic on $\mathrm{\Omega }$, let $B\subseteq \mathrm{\Omega }$ be any open subset, let $v$ be subharmonic on $B$, and assume ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \eta ,z\in B}v(z)\le u(\eta )$ for every $\eta \in \mathrm{\partial }B\cap \mathrm{\Omega }$. Define $w:=u$ on $\mathrm{\Omega }\setminus B$ and $w:=\max \{u,v\}$ on $B$. Then $w$ is upper semicontinuous on $\mathrm{\Omega }$: at points of $B$, $w$ is a finite maximum of upper semicontinuous functions; at points of $\mathrm{\Omega }\setminus \bar{B}$ it equals $u$; and at $\eta \in \mathrm{\partial }B\cap \mathrm{\Omega }$ the limsup of $w$ is the larger of the limsups of $u$ and of $v$ along $B$, at most $\max \{u(\eta ),u(\eta )\}=u(\eta )=w(\eta )$ by the seam hypothesis. Also $w\ge u$, and $u$ is not identically $-\mathrm{\infty }$ on any component of $\mathrm{\Omega }$, so neither is $w$.

$K_n=\overline{D_n}$, and $D_n$ is connected. Indeed $D_n\subseteq K_n$ and $K_n$ is closed, so $\overline{D_n}\subseteq K_n$; conversely, for $p\in K_n$ with $h(p)=c_n$ the local half-space from step 4.2 contains points $q$ with $h(q)<c_n$ arbitrarily close to $p$, and the path-connectedness of the local half-space puts each such $q$ in the same component of $\{h\le c_n\}$ as $p$, that is, in $K_n\cap \{h<c_n\}=D_n$; a point of $K_n$ with $h(p)<c_n$ already lies in $D_n$. So $K_n\subseteq \overline{D_n}$ and equality holds. If $D_n=A\cup B$ were a disjoint union of nonempty open subsets, then $K_n=\bar{A}\cup \bar{B}$ would be a union of two nonempty sets; if $\bar{A}\cap \bar{B}=\mathrm{\varnothing }$ this contradicts connectedness of $K_n$, so pick $p\in \bar{A}\cap \bar{B}$. If $p\in D_n$ then $p$ lies in the open set $A$ (say) which is disjoint from $B$, contradicting $p\in \bar{B}$; so $p\in \mathrm{\partial }{K}_n$, and near $p$ the set $D_n$ contains the path-connected local half-space $\{h<c_n\}$ intersected with the chart, which must lie entirely in $A$ or entirely in $B$, say in $A$; then all points of $D_n$ near $p$ lie in $A$, contradicting $p\in \bar{B}$. Hence $D_n$ is connected.

For every point $p\in \mathrm{\Omega }$ there is a chart $\theta$ with $p$ in its domain such that the chart expression $w_{\theta }$ is plane subharmonic on a neighbourhood of $\theta (p)$. If $p\mathrm{\notin }\bar{B}$, take a chart whose domain is contained in $\mathrm{\Omega }\setminus \bar{B}$; then $w_{\theta }=u_{\theta }$ is plane subharmonic by [F2]. If $p\in B$, take a chart with domain contained in $B$; then $w_{\theta }=\max \{u_{\theta },v_{\theta }\}$ is a finite maximum of plane subharmonic functions, hence plane subharmonic by [F3]. If $p\in \mathrm{\partial }B\cap \mathrm{\Omega }$, take a coordinate-disc neighbourhood $T$ of $p$ contained in $\mathrm{\Omega }$, with chart $\theta :T\to D(0,r)$. Apply the plane gluing lemma [F3] with ambient domain $\theta (T)$ and open inside set $\theta (T\cap B)$. The chart expression of $u$ is subharmonic on $\theta (T)$, that of $v$ is subharmonic on each component of $\theta (T\cap B)$, and every seam point inside $\theta (T)$ corresponds to a point of $\mathrm{\partial }B\cap T$, where step 4.3 supplies the limsup bound. Thus $w_{\theta }$ is subharmonic near $\theta (p)$.

For every $n$, $\mathrm{\partial }{D}_n$ is a nonempty compact smooth embedded $1$-submanifold of $X$ and $D_n=\mathrm{int}\overline{D_n}$. Indeed $\mathrm{\partial }{D}_n=\overline{D_n}\setminus D_n=K_n\setminus D_n=K_n\cap h^{-1}(c_n)$ by step 5.1, which is closed in the compact set $K_n$, hence compact, and is a smooth $1$-submanifold by step 4.2. If $\mathrm{\partial }{D}_n$ were empty then $D_n=K_n$ would be open and closed in the connected space $X$, so $D_n=X$ and $X$ would be compact, contradicting the hypothesis that $X$ is noncompact; thus $\mathrm{\partial }{D}_n\mathrm{\ne }\mathrm{\varnothing }$.

For every $n$ one has $K_n\subseteq D_{n+1}$. The set $K_n$ is connected, contains $x_0$, and satisfies $K_n\subseteq \{h\le c_n\}\subseteq \{h<c_{n+1}\}$ because $c_n<c_{n+1}$; hence $K_n$ lies in the component $K_{n+1}$ of $\{h\le c_{n+1}\}$ containing $x_0$, and since $h<c_{n+1}$ on all of $K_n$ no point of $K_n$ lies in $h^{-1}(c_{n+1})$; by step 4.1 applied at level $n+1$, $K_n\subseteq K_{n+1}\cap \{h<c_{n+1}\}=D_{n+1}$. In particular $\overline{D_n}=K_n\subseteq D_{n+1}$ and $D_n\subseteq D_{n+1}$.

$w$ is subharmonic on $\mathrm{\Omega }$. Let $\theta$ be any chart with $\mathrm{dom}\theta \cap \mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. By step 5.2 and the biholomorphic invariance of subharmonicity [F3], every point of $\theta (\mathrm{dom}\theta \cap \mathrm{\Omega })$ has a plane neighbourhood on which $w_{\theta }$ is subharmonic: for a point $\theta (p)$, use the chart ${\theta }_p$ of step 5.2 and write $w_{\theta }=w_{{\theta }_p}\circ ({\theta }_p\circ {\theta }^{-1})$ on the overlap, a composition of the plane subharmonic function $w_{{\theta }_p}$ with the biholomorphism ${\theta }_p\circ {\theta }^{-1}$ between plane domains. Moreover $w_{\theta }$ is upper semicontinuous and is not identically $-\mathrm{\infty }$ on any component, because these properties hold for $w$ by step 4.3 and are read in charts. By the locality of plane subharmonicity, step 3.2, applied separately to each component, $w_{\theta }$ is plane subharmonic on every component of $\theta (\mathrm{dom}\theta \cap \mathrm{\Omega })$. As $\theta$ was an arbitrary chart, $w$ is subharmonic on $\mathrm{\Omega }$.

$X={\bigcup }_{n\ge 1}{D}_n$. Let $x\in X$. By [F9] the connected manifold $X$ is path-connected and locally path-connected, so choose a path $\gamma$ from $x_0$ to $x$; its image $C$ is compact, so $h(C)$ is a compact subset of $[0,\mathrm{\infty })$ and is bounded, say by $N$. Choose $n$ with $c_n\ge N$, possible because $c_n\to \mathrm{\infty }$. Then $C\subseteq \{h\le c_n\}$ is connected and contains $x_0$ and $x$, so $x\in K_n\subseteq D_{n+1}$ by step 6.2. Hence every point of $X$ lies in some $D_n$.

Steps 6.1, 6.2 and 7.1 prove part 1: $(D_n)$ is a sequence of connected relatively compact domains with $\overline{D_n}\subseteq D_{n+1}$, each with nonempty smooth boundary and $D_n=\mathrm{int}\overline{D_n}$, whose union is $X$.

Maximum principle. Let $v$ be subharmonic on $D$ and let $B\in \mathbb{R}$ satisfy ${\mathrm{lim\hspace{0.17em}sup}}_{q\to \eta ,q\in D}v(q)\le B$ for every $\eta \in \mathrm{\partial }D$. Then $v\le B$ on $D$. For any real $b>B$, the set $E_b:=\{q\in D:v(q)\ge b\}$ is closed in $\bar{D}$: it is closed inside $D$ by upper semicontinuity, and its closure cannot meet $\mathrm{\partial }D$ by the boundary limsup bound. Thus $E_b$ is compact. If $s:={\sup }_{D}v=+\mathrm{\infty }$, the nonempty nested compact sets $E_n$ for integers $n>B$ have the finite-intersection property, so some point satisfies $v(q)\ge n$ for every such $n$, impossible because subharmonic functions have no $+\mathrm{\infty }$ values. Hence $s<+\mathrm{\infty }$. If $s>B$, the nested nonempty compact sets $E_b$ for $B<b<s$ again have the finite-intersection property; a point in their intersection has $v(q)\ge b$ for every $b<s$, so $v(q)=s$. Since $s>B$, the point is interior, and the strong maximum principle forces $v\equiv s$ on connected $D$. Taking any boundary point then gives $s={\mathrm{lim\hspace{0.17em}sup}}_{q\to \eta }v(q)\le B$, a contradiction. Therefore $s\le B$.

In the situation of step 9.1 the maximum set $\{q\in D:v(q)=s\}$ is open: for $q$ in it, take a chart $\theta$ around $q$ with $\theta (q)=0$ and $\theta (\mathrm{dom}\theta \cap D)$ a plane domain; the chart expression $v_{\theta }$ is plane subharmonic and attains the finite maximum $s$ at the interior point $0$, so by the strong maximum principle $v_{\theta }$ is constant on $\theta (\mathrm{dom}\theta \cap D)$ and $v$ is constant on a neighbourhood of $q$. The set is also closed in $D$, because it is $D\cap \{v\ge s\}$ and superlevel sets of the upper semicontinuous function $v$ are closed. Since $D$ is connected the set is all of $D$, so $v\equiv s>B$ on $D$. But $\mathrm{\partial }D\mathrm{\ne }\mathrm{\varnothing }$: otherwise $D$ would be open and closed in the connected space $X$, forcing $D=X$ and contradicting compactness of $\bar{D}$. At any $\eta \in \mathrm{\partial }D$ the boundary hypothesis then gives the contradiction $s={\mathrm{lim\hspace{0.17em}sup}}_{q\to \eta }v(q)\le B$, so $s\le B$ and the maximum principle holds for all $v$.

The Perron family. Fix a connected relatively compact domain $D\subseteq X$ with smooth boundary and a continuous datum $\phi :\mathrm{\partial }D\to \mathbb{R}$, and define $\mathcal{P}:=\{v\text{ subharmonic on }D:{\mathrm{lim\hspace{0.17em}sup}}_{q\to \eta ,q\in D}v(q)\le \phi (\eta )\text{ for every }\eta \in \mathrm{\partial }D\}\mathrm{.}$ The boundary $\mathrm{\partial }D$ is compact (it is closed in the compact set $\bar{D}$), so its continuous image $\phi (\mathrm{\partial }D)$ is a compact subset of $\mathbb{R}$ [F9]. The real-valued extreme-value theorem [F10] gives $m:={\min }_{\mathrm{\partial }D}\phi$ and $M:={\max }_{\mathrm{\partial }D}\phi$. The constant function $m$ is harmonic, hence subharmonic, on the nonempty open set $D$, and its limsup at each $\eta$ equals $m\le \phi (\eta )$; so $m\in \mathcal{P}$ and $\mathcal{P}$ is nonempty. By the maximum principle step 10.1 every $v\in \mathcal{P}$ satisfies $v\le M$ on $D$. Hence the pointwise supremum $U(q):=\sup \{v(q):v\in \mathcal{P}\}$ satisfies $m\le U\le M$ on $D$.

The envelope. Let $H$ be the upper-semicontinuous regularization of $U$, $H(x):={\mathrm{lim\hspace{0.17em}sup}}_{y\to x,y\in D}U(y)$, equivalently the least upper semicontinuous majorant of $U$ [F5]. Then $U\le H$ and, since $m\le U\le M$, also $m\le H\le M$. Moreover $H$ is subharmonic on $D$: for every chart $\theta$ the family of plane subharmonic functions $\{v_{\theta }:v\in \mathcal{P}\}$ on the plane domain $\theta (\mathrm{dom}\theta \cap D)$ is nonempty and bounded above by $M$, its pointwise supremum is $U_{\theta }$ and its upper-semicontinuous regularization is $H_{\theta }$ (regularization is a local, hence chart-invariant, operation); so the upper envelope theorem [F3] makes $H_{\theta }$ plane subharmonic. As $\theta$ was arbitrary, $H$ is subharmonic on $D$.

Poisson modification on the surface. Let $B\subseteq D$ be a coordinate disc with $\bar{B}\subseteq D$, written $B={\theta }^{-1}(D(w_0,r))$ for a chart $\theta$ and an open disc $D(w_0,r)$ whose closure lies in $\theta (\mathrm{dom}\theta \cap D)$. For $v\in \mathcal{P}$ let $P_{\theta }v$ denote the plane Poisson modification of $v_{\theta }$ on the disc $D(w_0,r)$, which is well defined by [F4], and define the surface modification $\tilde{v}:=\{\begin{array}{ll}{P}_{\theta }v\circ \theta & \text{on }B,\\ v& \text{on }D\setminus B\mathrm{.}\end{array}$ Then $\tilde{v}$ is harmonic on $B$ and $\tilde{v}\ge v$ on $B$, because the plane modification has these properties and harmonicity and the inequality are read in the chart $\theta$; and ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \eta ,z\in B}\tilde{v}(z)\le v(\eta )$ for every $\eta \in \mathrm{\partial }B$: if ${\phi }_n\downarrow v_{\theta }$ is the decreasing sequence of continuous boundary approximants used to define the modification and $h_n$ are their harmonic Poisson extensions, then $P_{\theta }v={\inf }_n{h}_n\le h_n$ on the disc, so ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \eta ,z\in B}\tilde{v}(z)\le \mathrm{lim\hspace{0.17em}sup}{h}_n(\theta (z))={\phi }_n(\theta (\eta ))$ for every $n$, and letting $n\to \mathrm{\infty }$ gives the bound $v(\eta )$.

$\tilde{v}$ is subharmonic on $D$ and belongs to $\mathcal{P}$. Indeed $\tilde{v}=v$ on the open set $D\setminus \bar{B}$, so its boundary limsups at points of $\mathrm{\partial }D$ equal those of $v$ and are at most $\phi$; and subharmonicity follows from the gluing lemma step 6.3 applied with $\mathrm{\Omega }:=D$, the subharmonic function $v$, the coordinate disc $B$, and the inside function $\tilde{v}{\mathrm{\mid }}_B$, whose seam limsups are at most $v$ by step 13.1: the glued function is $\max \{v,\tilde{v}\}=\tilde{v}$ on $B$ and $v$ outside $B$, that is, $\tilde{v}$ itself. Hence $\tilde{v}\in \mathcal{P}$, and in particular $\tilde{v}\le M$.

Monotonicity and directedness. If $v,w\in \mathcal{P}$ satisfy $v\le w$ on $D$, then $\tilde{v}\le \tilde{w}$ on $B$: choose decreasing continuous approximants ${\alpha }_n\downarrow v_{\theta }$ and ${\beta }_n\downarrow w_{\theta }$ on the circle $\mathrm{\partial }D(w_0,r)$ and set ${\gamma }_n:=\min \{{\alpha }_n,{\beta }_n\}$; then the ${\gamma }_n$ are continuous and decrease to $\min \{v_{\theta },w_{\theta }\}=v_{\theta }$, while ${\gamma }_n\le {\beta }_n$, so positivity of the Poisson kernel [F4] gives $P[{\gamma }_n]\le P[{\beta }_n]$ on $D(w_0,r)$ and hence $P_{\theta }v={\inf }_{n}P[{\gamma }_n]\le {\inf }_{n}P[{\beta }_n]=P_{\theta }w$, that is, $\tilde{v}\le \tilde{w}$. Also, if $v,w\in \mathcal{P}$ then $t:=\max \{v,w\}\in \mathcal{P}$: it is subharmonic on $D$ since both are, and at each $\eta \in \mathrm{\partial }D$ its limsup is at most $\max \{\mathrm{lim\hspace{0.17em}sup}v,\mathrm{lim\hspace{0.17em}sup}w\}\le \phi (\eta )$. By monotonicity $\tilde{t}\ge \tilde{v},\tilde{w}$ on $B$.

Harnack control. Define $h_{\ast }(q):=\sup \{\tilde{v}(q):v\in \mathcal{P}\}$ on $B$. Every $\tilde{v}$ is harmonic on $B$ and satisfies $\tilde{v}\le U\le H$ on $B$, so $h_{\ast }\le H$ on $B$; and $h_{\ast }\ge m$ since $m\in \mathcal{P}$ and $\tilde{m}\ge m$. Fix $a\in B$ and $\epsilon >0$. Since $h_{\ast }(a)$ is a real number and is the supremum of the values $\tilde{v}(a)$, there is $v\in \mathcal{P}$ with $\tilde{v}(a)>h_{\ast }(a)-\epsilon$. For any $w\in \mathcal{P}$ put $t:=\max \{v,w\}$; then $\tilde{t}\ge \tilde{v},\tilde{w}$ and $0\le (\tilde{t}-\tilde{v})(a)\le h_{\ast }(a)-\tilde{v}(a)<\epsilon ,$ while $\tilde{t}-\tilde{v}$ is harmonic and nonnegative on $B$ by step 15.1.

$h_{\ast }$ is harmonic on $B$. Choose a chart subdisc ${\theta }^{-1}(D(a,\rho ))\subseteq B$ centred at the point $a$ of step 16.1, and fix $r<\rho$. For $\epsilon >0$ choose $v$ as in step 16.1 and let $w\in \mathcal{P}$ be arbitrary. The nonnegative harmonic function $u:=\tilde{t}-\tilde{v}$ on $B$ satisfies $u(a)<\epsilon$, so Harnack's inequality [F4] applied on the disc $D(a,\rho )$ gives $0\le \tilde{t}(z)-\tilde{v}(z)=u(z)\le \frac{\rho +r}{\rho -r}u(a)<C\epsilon (z\in \overline{D(a,r)}),$ with $C:=(\rho +r)\mathrm{/}(\rho -r)$ independent of $w$. Hence $\tilde{w}\le \tilde{v}+C\epsilon$ on $\overline{D(a,r)}$ for every $w\in \mathcal{P}$, and taking the supremum over $w$ gives $h_{\ast }-\tilde{v}\le C\epsilon$ on $\overline{D(a,r)}$, while $h_{\ast }\ge \tilde{v}$. So for every $\epsilon >0$ there is a function harmonic on $B$ that approximates $h_{\ast }$ uniformly on $\overline{D(a,r)}$ within $C\epsilon$; in particular $h_{\ast }(b)={\lim }_{\epsilon \downarrow 0}{k}_{\epsilon }(b)$ where $k_{\epsilon }$ is such an approximant, and the circle average of $h_{\ast }$ over $\mathrm{\partial }D(b,s)$ differs from $h_{\ast }(b)$ by at most $2C\epsilon$ for every $\epsilon >0$, hence equals it. Thus $h_{\ast }$ satisfies the local mean-value property on $B$ and is continuous, so it is harmonic on $B$ by the converse of the mean-value property [F4].

$h_{\ast }=H$ on $B$. Fix $a\in B$ and $\epsilon >0$, and choose $D(a,\rho )\subseteq \theta (B)$ centred at $\theta (a)$, denoting this coordinate again by $a$. Choose $0<r<\rho$ so small that $(\frac{\rho +r}{\rho -r}-1)(M-H(a)+2\epsilon )<\epsilon \mathrm{.}$ By the definition of the regularization in step 12.1, there is $z\in D(a,r)$ with $U(z)>H(a)-\epsilon$, and then some $v\in \mathcal{P}$ has $v(z)>U(z)-\epsilon >H(a)-2\epsilon$. Thus $\tilde{v}(z)\ge v(z)>H(a)-2\epsilon$. The nonnegative function $k:=M-\tilde{v}$ is harmonic on $B$ by step 14.1. Harnack's inequality [F4], applied at the actual distance $\mathrm{\mid }z-a\mathrm{\mid }<r$, gives $k(a)\le \frac{\rho +\mathrm{\mid }z-a\mathrm{\mid }}{\rho -\mathrm{\mid }z-a\mathrm{\mid }}k(z)\le \frac{\rho +r}{\rho -r}(M-H(a)+2\epsilon )<M-H(a)+3\epsilon \mathrm{.}$ Hence $h_{\ast }(a)\ge \tilde{v}(a)>H(a)-3\epsilon$. Letting $\epsilon \downarrow 0$ gives $h_{\ast }(a)\ge H(a)$; the reverse inequality is step 16.1.

$H$ is harmonic on $D$: every point of the open set $D$ has, by local compactness of the manifold $X$ [F9], a coordinate disc $B$ around it with $\bar{B}\subseteq D$, and $H=h_{\ast }$ is harmonic on $B$ by steps 17.1 and 17.2.

A local peak function. Fix $\zeta \in \mathrm{\partial }D$ and a chart $\theta$ with $\theta (\zeta )=0$ whose domain $U$ is a coordinate disc $\theta (U)=D(0,2{\rho }_0)$. Since $\mathrm{\partial }D$ is a smooth embedded curve, after composing $\theta$ with a rotation we may assume that near $0$ the image $\theta (\mathrm{\partial }D\cap U)$ is the graph of a smooth function $\gamma$ with $\gamma (0)={\gamma }^{\mathrm{\prime }}(0)=0$, and that $\theta (D\cap U)\cap D(0,2{\rho }_0)=\{(x,y):x^2+y^2<(2{\rho }_0{)}^2,y>\gamma (x)\};$ shrinking ${\rho }_0$ we may also assume $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}(x)\mathrm{\mid }\le \frac{1}{4}$ for $\mathrm{\mid }x\mathrm{\mid }\le 2{\rho }_0$. Then every point $w=(x,y)$ of $\theta (D\cap U^{\mathrm{\prime }})$, where $U^{\mathrm{\prime }}:={\theta }^{-1}(D(0,2{\rho }_0))$, satisfies $y>\gamma (x)\ge -\frac{1}{4}\mathrm{\mid }x\mathrm{\mid }>-\frac{1}{2}\mathrm{\mid }x\mathrm{\mid }$; that is, $\theta (D\cap U^{\mathrm{\prime }})$ is contained in the sector $S:=\{y>-\frac{1}{2}\mathrm{\mid }x\mathrm{\mid }\}$, whose half-angle at the origin, measured from the positive vertical axis, is $\frac{\pi }{2}+\arctan \frac{1}{2}$. Put $\mu :=\pi /(\pi +2\arctan \frac{1}{2})$ and define, for $w\in S$, $Q(w):=\mathrm{Re}((-iw{)}^{\mu }),$ where $(-iw{)}^{\mu }:=\exp (\mu \mathrm{Log}(-iw))$. Since $\mathrm{\mid }\arg (-iw)\mathrm{\mid }<\frac{\pi }{2}+\arctan \frac{1}{2}<\pi$ on $S$, this is holomorphic by [F11]; writing $w=r{e}^{i(\pi \mathrm{/}2+\vartheta )}$ gives $Q(w)=r^{\mu }\cos (\mu \vartheta )$. Finally define $q(z):=-Q(\theta (z))(z\in D\cap U^{\mathrm{\prime }})\mathrm{.}$

The function $q$ has the following properties: it is subharmonic on $D\cap U^{\mathrm{\prime }}$; it is negative there; $q(z)\to 0$ as $z\to \zeta$; and with $c_0:=\cos (\mu (\frac{\pi }{2}+\arctan \frac{1}{4}))>0$ one has $q(z)\le -c_0\mathrm{\mid }\theta (z){\mathrm{\mid }}^{\mu }(z\in D\cap U^{\mathrm{\prime }})\mathrm{.}$ Indeed $Q$ is harmonic on $S\setminus \{0\}$ (it is the real part of a holomorphic function) and $0\mathrm{\notin }\theta (D\cap U^{\mathrm{\prime }})$, so $q$ is harmonic, hence subharmonic, on $D\cap U^{\mathrm{\prime }}$; and for $w\in \theta (D\cap U^{\mathrm{\prime }})$ the angle $\vartheta$ of $w$ relative to the positive vertical axis satisfies $\mathrm{\mid }\vartheta \mathrm{\mid }\le \frac{\pi }{2}+\arctan \frac{1}{4}$, because $y>-\frac{1}{4}\mathrm{\mid }x\mathrm{\mid }$; since $\mu (\frac{\pi }{2}+\arctan \frac{1}{4})<\mu (\frac{\pi }{2}+\arctan \frac{1}{2})=\frac{\pi }{2}$, we get $\cos (\mu \vartheta )\ge c_0>0$, whence $q(w)=-\mathrm{\mid }w{\mathrm{\mid }}^{\mu }\cos (\mu \vartheta )\le -c_0\mathrm{\mid }w{\mathrm{\mid }}^{\mu }<0$ and $q\to 0$ at $0$.

Boundary limit from above. Fix $\zeta \in \mathrm{\partial }D$ and $\epsilon >0$, and take the chart, ${\rho }_0$, $U^{\mathrm{\prime }}$ and $q$ of steps 19.1 and 20.1. Shrinking $U^{\mathrm{\prime }}$ further, using continuity of $\phi$ at $\zeta$, we may assume $\mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }<\epsilon$ for every $\eta \in \mathrm{\partial }D\cap U^{\mathrm{\prime }}$. Let $B:={\theta }^{-1}(D(0,{\rho }_0))$, so that $q\le -c_0{\rho }_0^{\mu }$ on $\mathrm{\partial }B\cap D$ by step 20.1. Fix $v\in \mathcal{P}$ and choose an integer $A\ge 0$ with $A{c}_0{\rho }_0^{\mu }\ge M-\phi (\zeta )-\epsilon$. Define $G:=\max \{\phi (\zeta )+\epsilon ,v+Aq\}\text{ on }D\cap B,G:=\phi (\zeta )+\epsilon \text{ on }D\setminus B\mathrm{.}$ At every $\eta \in \mathrm{\partial }B\cap D$ the limsup of $v+Aq$ is at most $v(\eta )+Aq(\eta )\le M-A{c}_0{\rho }_0^{\mu }\le \phi (\zeta )+\epsilon$, and the constant $\phi (\zeta )+\epsilon$ is subharmonic, so the gluing step 6.3, with open inside set $D\cap B\subseteq D$, shows that $G$ is subharmonic on $D$. Its boundary limsup at every $\eta \in \mathrm{\partial }D$ is at most $\phi (\zeta )+\epsilon$: for $\eta \in \mathrm{\partial }D\setminus B$ this is the definition, and for $\eta \in \mathrm{\partial }D\cap B$ it is $\max \{\phi (\zeta )+\epsilon ,\mathrm{lim\hspace{0.17em}sup}(v+Aq)\}\le \max \{\phi (\zeta )+\epsilon ,\phi (\eta )\}=\phi (\zeta )+\epsilon$, because $q<0$ on $D\cap B$ and $\phi (\eta )<\phi (\zeta )+\epsilon$. By the maximum principle step 10.1, $G\le \phi (\zeta )+\epsilon$ on $D$. Because the chosen $A$ depends only on $M,\phi (\zeta ),\epsilon$ and the fixed chart, it is independent of $v$. Thus $U\le \phi (\zeta )+\epsilon -Aq$ on $D\cap B$. The right-hand side is continuous there, so its upper-semicontinuous regularization satisfies $H=U^{\ast }\le \phi (\zeta )+\epsilon -Aq$; since $q(z)\to 0$ as $z\to \zeta$, ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta }H(z)\le \phi (\zeta )+\epsilon$.