Green envelope dichotomy, logarithmic pole and leastness on a Riemann surface

Compact case and the punctured surface. If $X$ is compact, take the candidate $v_0$ of [F1]. For every $c\ge 0$, $v_0+c$ is a candidate: it is nonnegative and subharmonic, its compact support may be $X$, and $(v_0+c)+\log \mathrm{\mid }z\mathrm{\mid }=c$ near $p$. Thus $g(q)=+\mathrm{\infty }$ for every $q\mathrm{\ne }p$. In the rest of the proof assume $X$ is noncompact. A chart at $p$ is injective on a neighbourhood of $p$ and maps it onto an open subset of $\mathbb{C}$ [F2], so $X$ has more than one point and $X\setminus \{p\}\mathrm{\ne }\mathrm{\varnothing }$. Suppose $X\setminus \{p\}=A\bigsqcup B$ with $A,B$ nonempty and open in $X\setminus \{p\}$; fix open sets $U_A,U_B\subseteq X$ with $A=U_A\setminus \{p\}$ and $B=U_B\setminus \{p\}$, so that $X=U_A\cup U_B\cup \{p\}$. Choose a chart $\phi :W\to \mathbb{C}$ at $p$ [F2] and $\rho >0$ with $\overline{B(\phi (p),\rho )}\subseteq \phi (W)$, and put $V:={\phi }^{-1}(B(\phi (p),\rho ))$. Then $V\setminus \{p\}$ is homeomorphic to the punctured disc $B(\phi (p),\rho )\setminus \{\phi (p)\}$, which is connected by [F4], and it is covered by the disjoint open sets $A$ and $B$, so $V\setminus \{p\}\subseteq A$ or $V\setminus \{p\}\subseteq B$. In the first case $A\cup \{p\}=U_A\cup V$ is open in $X$: indeed $U_A\subseteq A\cup \{p\}$ and $V\subseteq A\cup \{p\}$, while $A\subseteq U_A$ and $p\in V$; moreover $p\mathrm{\notin }{U}_B$, because if $p\in U_B$ then the open neighbourhood $U_B\cap V$ of $p$ contains a point $q\mathrm{\ne }p$, giving $q\in (U_B\setminus \{p\})\cap (V\setminus \{p\})\subseteq B\cap A=\mathrm{\varnothing }$, a contradiction; hence $B=U_B$ is open in $X$ and $X=(A\cup \{p\})\bigsqcup B$ is a separation of the connected space $X$ [F2], which is impossible. The second case is symmetric, interchanging $A$ and $B$. Therefore $X\setminus \{p\}$ is connected.

Chart discs avoiding the pole. For every $x\in X\setminus \{p\}$ there are a chart $\phi :W\to \mathbb{C}$ with $x\in W$ and a radius $r>0$ such that, with $D:={\phi }^{-1}(B(\phi (x),r))$, one has $\bar{B}(\phi (x),r)\subseteq \phi (W)$ and $\bar{D}\subseteq X\setminus \{p\}$. Indeed, choose any chart $\phi$ around $x$ [F2]; shrink its domain so that $\phi (W)$ is a disc around $\phi (x)$, and choose $r>0$ so that $\bar{B}(\phi (x),r)\subseteq \phi (W)$ and, in case $p\in W$, so that $r<\mathrm{\mid }\phi (x)-\phi (p)\mathrm{\mid }$; then $p\mathrm{\notin }D$ and $\bar{D}$ is compact in the open set $X\setminus \{p\}$.

The surface Poisson modification stays in the family. Let $D={\phi }^{-1}(B(a,r))$ be a disc in a chart $\phi :W\to \mathbb{C}$ with $\bar{B}(a,r)\subseteq \phi (W)$ and $\bar{D}\subseteq X\setminus \{p\}$, and let $v\in {\mathcal{F}}_p$. Define $P_{D}v$ by $P_{D}v=v$ on $(X\setminus \{p\})\setminus D$ and by $P_{D}v:={\phi }^{-1}$-transport of the plane Poisson modification $P_{B(a,r)}(v\circ {\phi }^{-1})$ on $D$. Then $P_{D}v$ belongs to ${\mathcal{F}}_p$, is harmonic on $D$, and $P_{D}v\ge v$. To see this, put $u:=v\circ {\phi }^{-1}$ on $\phi (W)$, subharmonic on each connected component by [F3], and $h:=P_{B(a,r)}u$, which is harmonic on $B(a,r)$ and satisfies $h\ge u$ there by [F7], hence is subharmonic on $B(a,r)$ by [F16]. For $\zeta \in \mathrm{\partial }B(a,r)$ the definition [F6] writes $h={\inf }_n{h}_n$ with $h_n$ continuous on $\bar{B}(a,r)$ and $h_n={\varphi }_n\ge u$ on the boundary circle, so ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta ,z\in B(a,r)}h(z)\le {\varphi }_n(\zeta )$ for every $n$, hence at most ${\inf }_n{\varphi }_n(\zeta )=u(\zeta )$; the gluing lemma [F8], applied with $\mathrm{\Omega }$ equal to the connected component of $\phi (W)$ containing $B(a,r)$ and with $D:=B(a,r)$, therefore makes the function equal to $\max \{u,h\}=h$ on $B(a,r)$ and to $u$ outside it subharmonic on that component, while on every other component of $\phi (W)$ the same function equals $u$ and is subharmonic there as a restriction of $u$ [F3]; transporting back, $P_{D}v$ is subharmonic on $W$ [F3]. On the open set $(X\setminus \{p\})\setminus \bar{D}$ the function $P_{D}v=v$ is subharmonic as a restriction of the subharmonic function $v$ [F3], and $W\cup ((X\setminus \{p\})\setminus \bar{D})=X\setminus \{p\}$ because $D\subseteq W$, so locality [F9] makes $P_{D}v$ subharmonic on $X\setminus \{p\}$. The function $P_{D}v$ is harmonic on $D$ by [F7]; it satisfies $P_{D}v\ge v$ because $h\ge u$ on $B(a,r)$ and $P_{D}v=v$ outside $D$; it is nonnegative because $v\ge 0$ and $h\ge u\ge 0$; it vanishes outside the compact set $K\cup \bar{D}$, where $K$ is a compact set with $v=0$ off $K$ [F1], because $P_{D}v=v$ on the complement of $D$; and since $p\mathrm{\notin }\bar{D}$, the function $P_{D}v$ equals $v$ on the neighbourhood $X\setminus \bar{D}$ of $p$, so the pole condition of [F1] transfers. Hence $P_{D}v\in {\mathcal{F}}_p$.

Boundary maximum principle. Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be a bounded domain, let $u$ be subharmonic on $\mathrm{\Omega }$, and let $B\in \mathbb{R}$ satisfy ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta ,z\in \mathrm{\Omega }}u(z)\le B$ for every $\zeta \in \mathrm{\partial }\mathrm{\Omega }$. Then $u\le B$ on $\mathrm{\Omega }$. Suppose not and put $S:={\sup }_{\mathrm{\Omega }}u>B$; choose $z_j\in \mathrm{\Omega }$ with $u(z_j)\to S$ and, $\mathrm{\Omega }$ being bounded, pass to a subsequence with $z_j\to z_{\ast }\in \bar{\mathrm{\Omega }}$. If $z_{\ast }\in \mathrm{\partial }\mathrm{\Omega }$ then $S={\mathrm{lim\hspace{0.17em}sup}}_{j}u(z_j)\le {\mathrm{lim\hspace{0.17em}sup}}_{z\to z_{\ast },z\in \mathrm{\Omega }}u(z)\le B$, a contradiction; so $z_{\ast }\in \mathrm{\Omega }$. Then upper semicontinuity gives $u(z_{\ast })\ge {\mathrm{lim\hspace{0.17em}sup}}_{j}u(z_j)=S$; since subharmonic functions take values in $[-\mathrm{\infty },\mathrm{\infty })$ [F17], this forces $S<\mathrm{\infty }$ and $u(z_{\ast })=S$, so $u$ attains its finite maximum $S$ at the interior point $z_{\ast }$ and is constant $S$ on the domain $\mathrm{\Omega }$ by [F11]. But $\mathrm{\partial }\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$: otherwise $\mathrm{\Omega }$ would be a nonempty proper subset of $\mathbb{C}$ that is both open and closed, contradicting connectedness of the plane [F20]. For $\zeta \in \mathrm{\partial }\mathrm{\Omega }$ the boundary hypothesis would then give the contradiction $S={\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta }u(z)\le B$. Hence $S\le B$.

A maximizing sequence. For $n\ge 0$, let $t_n$ be an increasing sequence of real numbers with $t_n<g(x_0)$ for every $n$ and $t_n\to g(x_0)$ when $g(x_0)<+\mathrm{\infty }$, and put $t_n:=n$ when $g(x_0)=+\mathrm{\infty }$. Since $g(x_0)=\sup \{v(x_0):v\in {\mathcal{F}}_p\}$ and ${\mathcal{F}}_p\mathrm{\ne }\mathrm{\varnothing }$ [F1], every set $\{v\in {\mathcal{F}}_p:v(x_0)>t_n\}$ is nonempty, so Countable Choice [A1] gives a sequence $(v_n)\subseteq {\mathcal{F}}_p$ with $v_n(x_0)>t_n$. Setting $u_n:=\max \{v_0,\dots ,v_n\}$ gives an increasing sequence $(u_n)\subseteq {\mathcal{F}}_p$ by finite-max stability [F1]; its support is contained in the finite union of the compact supports of $v_0,\dots ,v_n$. Also $u_n(x_0)\to g(x_0)$ because $u_n(x_0)\ge v_n(x_0)>t_n$ and $u_n(x_0)\le g(x_0)$.

A subharmonic comparison function at the pole. Let $v\in {\mathcal{F}}_p$ and $\epsilon >0$, and put $w:=v+(1+\epsilon )\log \mathrm{\mid }z\mathrm{\mid }$ on $U\setminus \{p\}$. Then $w$ extends to a subharmonic function on $U$ with value $-\mathrm{\infty }$ at $p$. Indeed, in the coordinate $z$ the chart expression $v_z=v\circ z^{-1}$ is plane subharmonic on $\mathbb{D}\setminus \{0\}$ [F3] and $\log \mathrm{\mid }\cdot \mathrm{\mid }$ is harmonic on $\mathbb{D}\setminus \{0\}$ [F14], hence subharmonic [F16], so $w_z:=v_z+(1+\epsilon )\log \mathrm{\mid }\cdot \mathrm{\mid }$ is subharmonic on $\mathbb{D}\setminus \{0\}$ by [F18]. By clause 3 of [F1] there are $\delta >0$ and a constant $C_1$ with $v_z(\zeta )\le -\log \mathrm{\mid }\zeta \mathrm{\mid }+C_1$ for $0<\mathrm{\mid }\zeta \mathrm{\mid }<\delta$, so $w_z(\zeta )\le \epsilon \log \mathrm{\mid }\zeta \mathrm{\mid }+C_1\to -\mathrm{\infty }$ as $\zeta \to 0$. Setting $w_z(0):=-\mathrm{\infty }$ gives an upper semicontinuous function. For each positive integer $N$, the function $\max (w_z,-N)$ equals the constant $-N$ near $0$ and is subharmonic elsewhere by finite-max stability; locality [F9] makes it subharmonic on the full disc. These functions decrease to $w_z$. On each circle their integrals decrease to the extended integral of $w_z$ by monotone convergence after subtracting a common finite upper bound, so their submean inequalities pass to $w_z$. The latter is not identically $-\mathrm{\infty }$, hence is subharmonic by [F17], and transporting back gives the assertion.

The competitor condition is chart-independent. Let $z$ and $w$ be centred charts at $p$. The transition $\tau :=w\circ z^{-1}$ is a biholomorphism between neighbourhoods of $0$ with $\tau (0)=0$ and ${\tau }^{\mathrm{\prime }}(0)\mathrm{\ne }0$ [F1]. The quotient $\zeta \mapsto \tau (\zeta )\mathrm{/}\zeta$ is holomorphic on a punctured neighbourhood of $0$ and has the finite limit ${\tau }^{\mathrm{\prime }}(0)$ at $0$, so it extends holomorphically across $0$ by [F19]; the extension does not vanish near $0$ because its value there is ${\tau }^{\mathrm{\prime }}(0)\mathrm{\ne }0$. Hence the function $w\mathrm{/}z$, whose expression in the chart $z$ is $\zeta \mapsto \tau (\zeta )\mathrm{/}\zeta$, is holomorphic and zero-free on a punctured neighbourhood of $p$, and $\log \mathrm{\mid }w\mathrm{/}z\mathrm{\mid }$ is harmonic there by [F14] and [F15]. Therefore, if $H+\log \mathrm{\mid }z\mathrm{\mid }$ extends harmonically across $p$, then $H+\log \mathrm{\mid }w\mathrm{\mid }=(H+\log \mathrm{\mid }z\mathrm{\mid })+\log \mathrm{\mid }w\mathrm{/}z\mathrm{\mid }$ extends harmonically across $p$ too. So the hypothesis on $H$ in part 3 holds for some centred chart if and only if it holds for every centred chart.

Monotonicity of the Poisson modification. If $u,w$ are subharmonic on $X\setminus \{p\}$ with $u\le w$, and $D={\phi }^{-1}(B(a,r))$ is a chart disc as in step 1.3, then $P_{D}u\le P_{D}w$ on $D$. Work in the chart: with $U:=u\circ {\phi }^{-1}$, $W_0:=w\circ {\phi }^{-1}$ on $\phi (W)$, let $({\psi }_m)$ be a boundary approximation for $W_0$ and $k_m$ the associated harmonic functions, so that the chart expression of $P_{D}w$ on $B(a,r)$ is ${\inf }_m{k}_m$ [F6], and the chart expression of $P_{D}u$ is subharmonic on $\phi (W)$ (step 1.3). On $\mathrm{\partial }B(a,r)$ one has $k_m={\psi }_m\ge W_0\ge U$ and the chart expression of $P_{D}u$ equals $U$, while $k_m$ is continuous on $\bar{B}(a,r)$ and harmonic on $B(a,r)$; the harmonic-majorant characterization [F10], applied on the connected component of $\phi (W)$ containing $\bar{B}(a,r)$, gives $k_m\ge (P_{D}u{)}_{\phi }$ on $B(a,r)$. Taking the infimum over $m$ gives $(P_{D}u{)}_{\phi }\le (P_{D}w{)}_{\phi }$ on $B(a,r)$, that is, $P_{D}u\le P_{D}w$ on $D$.

The chart disc at the given point. By step 1.2 applied to the given point $x_0\in X\setminus \{p\}$ there is a chart disc $D={\phi }^{-1}(B(a,r))$ with $x_0\in D$ and $\bar{D}\subseteq X\setminus \{p\}$; fix such a disc and chart.

An upper bound on a smaller pole disc. Fix $0<r<1$. The circle $\mathrm{\mid }z\mathrm{\mid }=r$ lies entirely inside the chart domain, and $v$ is upper semicontinuous on its compact inverse image. Thus $B_r:={\sup }_{\mathrm{\mid }z\mathrm{\mid }=r}v+(1+\epsilon )\log r$ is finite. Apply the boundary maximum principle of step 1.4 to $w_z=v\circ z^{-1}+(1+\epsilon )\log \mathrm{\mid }\cdot \mathrm{\mid }$ on $0<\mathrm{\mid }\zeta \mathrm{\mid }<r$: at the outer circle its limsup is at most $B_r$ by upper semicontinuity inside the chart, and at $0$ it tends to $-\mathrm{\infty }$ by step 1.6. Therefore $w_z\le B_r$ for $0<\mathrm{\mid }\zeta \mathrm{\mid }<r$. No value of $z^{-1}$ on the unit circle is used.

Set $P_n:=P_D{u}_n$ for the disc $D$ of step 2.2 and the sequence $(u_n)$ of step 1.5. By steps 1.3 and 2.1 the sequence $(P_n)$ is increasing, each $P_n$ lies in ${\mathcal{F}}_p$ and is harmonic on $D$, and $P_n\ge u_n$.

The sandwich on a smaller pole disc. Fix $0<r<1$. Taking the supremum over $v$ in step 2.3, then letting $\epsilon \downarrow 0$, gives $g(q)+\log \mathrm{\mid }z(q)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }z\mathrm{\mid }=r}g+\log r$ when $0<\mathrm{\mid }z(q)\mathrm{\mid }<r$; an infinite right side is harmless. For the lower bound, the explicit candidate supported in $\mathrm{\mid }z\mathrm{\mid }<r$ has value $-\log (\mathrm{\mid }z\mathrm{\mid }\mathrm{/}r)$ there: it is a finite maximum of two harmonic functions in the larger chart and is locally zero outside that closed subdisc, hence belongs to ${\mathcal{F}}_p$ by [F1], [F9] and [F18]. Consequently $\log r\le g(q)+\log \mathrm{\mid }z(q)\mathrm{\mid }$ for $0<\mathrm{\mid }z(q)\mathrm{\mid }<r$.

Case $g(x_0)=+\mathrm{\infty }$. Then $P_n(x_0)\ge u_n(x_0)\to +\mathrm{\infty }$, so the increasing sequence $(P_n)$ of harmonic functions on the disc $D$ cannot converge locally uniformly to a finite harmonic function; by [F5] it tends to $+\mathrm{\infty }$ at every point of $D$. Since $P_n\le g$ on $D$ by [F1], it follows that $g(y)=+\mathrm{\infty }$ for every $y\in D$.

Case $g(x_0)<+\mathrm{\infty }$. Then $P_n(x_0)\le g(x_0)<+\mathrm{\infty }$, so [F5] provides a harmonic function $H$ on $D$ with $P_n\to H$ locally uniformly. Hence $H\le g$ on $D$, and $H(x_0)={\lim }_n{P}_n(x_0)\ge {\lim }_n{u}_n(x_0)=g(x_0)$ together with $H(x_0)\le g(x_0)$ gives $H(x_0)=g(x_0)$.

Comparison with an arbitrary candidate. Assume $g(x_0)<+\mathrm{\infty }$ and fix $v\in {\mathcal{F}}_p$. For every $m$ the function $W_m:=P_D\max \{v,u_m\}$ lies in ${\mathcal{F}}_p$ and is harmonic on $D$ by step 1.3, and the sequence $(W_m)$ is increasing by step 2.1 because $m\mapsto \max \{v,u_m\}$ is increasing. Also $W_m(x_0)\le g(x_0)<+\mathrm{\infty }$ by [F1], so [F5] gives a harmonic $H^{(v)}$ on $D$ with $W_m\to H^{(v)}$ locally uniformly. Then $H^{(v)}\ge v$ on $D$, since $W_m\ge \max \{v,u_m\}\ge v$; $H^{(v)}\le g$ on $D$, since every $W_m\le g$; and $H^{(v)}(x_0)={\lim }_m{W}_m(x_0)\ge {\lim }_m{u}_m(x_0)=g(x_0)$ while $H^{(v)}(x_0)\le g(x_0)$, so $H^{(v)}(x_0)=g(x_0)$. Moreover $W_m\ge P_m$ for every $m$ by monotonicity, step 2.1, so $H^{(v)}\ge H$ on $D$; the function $H^{(v)}-H$ is harmonic and nonnegative on the plane disc $\phi (D)$ and vanishes at the interior point $\phi (x_0)$, so [F12] gives $H^{(v)}=H$ on $D$. Hence $v\le H$ on $D$.

In the case $g(x_0)<+\mathrm{\infty }$, taking the supremum over $v\in {\mathcal{F}}_p$ in step 5.1 gives $g\le H$ on $D$, and comparison with step 4.2 gives $g{\mathrm{\mid }}_D=H$: the envelope is finite and harmonic on $D$.

Global dichotomy. Let $A:=\{x\in X\setminus \{p\}:g(x)=+\mathrm{\infty }\}$ and $B:=\{x\in X\setminus \{p\}:g(x)<+\mathrm{\infty }\}$. Given $x\in X\setminus \{p\}$, apply the construction of steps 2.2, 1.5, 3.1, 4.1, 4.2, 5.1 and 6.1 with $x_0:=x$; step 4.1 shows that a chart disc around $x$ lies in $A$ when $g(x)=+\mathrm{\infty }$, and step 6.1 shows that a chart disc around $x$ lies in $B$ when $g(x)<+\mathrm{\infty }$. Hence $A$ and $B$ are open; they are disjoint and cover the connected nonempty set $X\setminus \{p\}$ of step 1.1, so one of them is empty. If $B=\mathrm{\varnothing }$ then $g\equiv +\mathrm{\infty }$ on $X\setminus \{p\}$. Otherwise $g$ is finite everywhere and harmonic on a neighbourhood of every point by step 6.1, hence harmonic on $X\setminus \{p\}$ by [F3].

Strict positivity in the finite case. Assume $g<+\mathrm{\infty }$ on $X\setminus \{p\}$. By [F1] there is a centred chart $z_0:U_0\to \mathbb{D}$ at $p$ and a function $v_0\in {\mathcal{F}}_p$ equal to $-\log \mathrm{\mid }{z}_0\mathrm{\mid }$ on $U_0\setminus \{p\}$ and $0$ outside $U_0$; hence $g\ge v_0\ge 0$ everywhere on $X\setminus \{p\}$, and $g>0$ on $U_0\setminus \{p\}$ because there $v_0=-\log \mathrm{\mid }{z}_0\mathrm{\mid }>0$ (as $\mathrm{\mid }{z}_0\mathrm{\mid }<1$ on $U_0$). Let $Z:=\{q\in X\setminus \{p\}:g(q)=0\}$. Then $Z$ is closed in $X\setminus \{p\}$ because $g$ is continuous there (step 7.1), and $Z$ is open: if $q_0\in Z$, choose a chart $\psi$ whose domain $W_0$ is a connected neighbourhood of $q_0$ contained in $X\setminus \{p\}$ (possible because $g$ is defined and harmonic on the open set $X\setminus \{p\}$); the chart expression of $g$ is harmonic and nonnegative on the plane domain $\psi (W_0)$ [F3] and vanishes at $\psi (q_0)$, so it is identically $0$ by [F12] and $W_0\subseteq Z$. Since $X\setminus \{p\}$ is connected by step 1.1, the clopen set $Z$ is empty or all of $X\setminus \{p\}$; the second alternative is impossible because $g>0$ on the nonempty set $U_0\setminus \{p\}$. Hence $Z=\mathrm{\varnothing }$ and $g>0$ on $X\setminus \{p\}$.

The logarithmic pole is removable. Assume $g<+\mathrm{\infty }$ on $X\setminus \{p\}$. Step 7.1 makes $g$ continuous on $X\setminus \{p\}$, so for a fixed $0<r<1$ its supremum on the compact circle $\mathrm{\mid }z\mathrm{\mid }=r$ is finite. Step 3.2 bounds $g+\log \mathrm{\mid }z\mathrm{\mid }$ above and below on $0<\mathrm{\mid }z\mathrm{\mid }<r$. The function $g+\log \mathrm{\mid }z\mathrm{\mid }$ is harmonic on $U\setminus \{p\}$: $g$ is harmonic there by step 7.1, $\log \mathrm{\mid }z\mathrm{\mid }$ is harmonic on $U\setminus \{p\}$ because its expression in the chart $z$ is $\log \mathrm{\mid }\cdot \mathrm{\mid }$ on $\mathbb{D}\setminus \{0\}$ [F14], and sums of harmonic functions are harmonic in charts [F3, F16]. Being bounded, its chart expression in the chart $z$ has a removable singularity at $0$ by [F13] and extends harmonically over $0$ on $\mathrm{\mid }z\mathrm{\mid }<r$. This extension agrees with the original harmonic function off $0$, hence gives a harmonic function on all of $\mathbb{D}$; transporting back by [F3], $g+\log \mathrm{\mid }z\mathrm{\mid }$ extends to a harmonic function on $U$. This proves part 2 of the statement for the arbitrary centred chart $z$.

The auxiliary function for leastness. Assume $g<+\mathrm{\infty }$ on $X\setminus \{p\}$, and let $H:X\setminus \{p\}\to (0,\mathrm{\infty })$ be harmonic there with a unit logarithmic pole at $p$ in the centred chart $z$, meaning that $H+\log \mathrm{\mid }z\mathrm{\mid }$ extends to a harmonic function $h_H$ on $U$. Fix $v\in {\mathcal{F}}_p$ and $\epsilon >0$ and put $W:=v-(1+\epsilon )H$. Then: $W$ is subharmonic on $X\setminus \{p\}$, because in every chart $W_{\phi }=v_{\phi }+(1+\epsilon )(-H_{\phi })$ is a nonnegative linear combination of the subharmonic functions $v_{\phi }$ and $-H_{\phi }$ [F3, F16, F18]; $W\le 0$ on $X\setminus K$, where $K$ is a compact support of $v$ [F1], since there $W=-(1+\epsilon )H<0$; by step 1.1, $X$ is noncompact in this finite case, so $X\setminus K$ is nonempty; and $W(q)\to -\mathrm{\infty }$ as $q\to p$, because near $p$ one has $v\le -\log \mathrm{\mid }z\mathrm{\mid }+C_1$ by clause 3 of [F1] and $H=h_H-\log \mathrm{\mid }z\mathrm{\mid }$ with $h_H$ bounded near $p$, whence $W\le C_1-(1+\epsilon )h_H+\epsilon \log \mathrm{\mid }z\mathrm{\mid }\to -\mathrm{\infty }$.

Consequence: $W\le 0$. Suppose $M:={\sup }_{X\setminus \{p\}}W>0$. For every $b>0$, the superlevel set $E_b:=\{q\in X\setminus \{p\}:W(q)\ge b\}$ lies in the compact support $K$ of step 8.3 and avoids a neighbourhood of $p$ because $W(q)\to -\mathrm{\infty }$ there. Upper semicontinuity makes $E_b$ closed in $K$, hence compact. If $M=+\mathrm{\infty }$, the nested nonempty compact sets $E_n$ for positive integers $n$ have the finite-intersection property; a point in their intersection would have $W\ge n$ for every $n$, impossible since $W$ is finite on $X\setminus \{p\}$. Thus $M<+\mathrm{\infty }$. The nonempty nested sets $E_b$ for $0<b<M$ again have the finite-intersection property, so some $q_{\ast }\in K\setminus \{p\}$ satisfies $W(q_{\ast })\ge b$ for every $b<M$, hence $W(q_{\ast })=M$. In a chart around $q_{\ast }$, the subharmonic chart expression of $W$ attains its finite maximum at an interior point, so it is constant on a neighbourhood by [F11]; therefore $Z:=\{W=M\}$ is open. It is closed in the connected domain $X\setminus \{p\}$ because $W$ is upper semicontinuous and bounded above by $M$. Hence $W\equiv M>0$ there, contradicting $W<0$ on the nonempty set $X\setminus K$ from step 8.3. Therefore $M\le 0$.

Leastness. Step 9.1 gives $v\le (1+\epsilon )H$ on $X\setminus \{p\}$ for every $v\in {\mathcal{F}}_p$ and every $\epsilon >0$. Taking the supremum over $v\in {\mathcal{F}}_p$ gives $g\le (1+\epsilon )H$, and letting $\epsilon \downarrow 0$ gives $g\le H$ on $X\setminus \{p\}$. Since $H$ was an arbitrary positive harmonic unit-pole function for the centred chart $z$, this proves part 3 for that chart.

Conclusion. Part 1 is steps 7.1 and 8.1: either $g\equiv +\mathrm{\infty }$ on $X\setminus \{p\}$, or $g$ is finite, harmonic and strictly positive there. Part 2 is step 8.2. Part 3 is steps 10.1 and 1.7, which show that in the finite case $g$ is least among all positive harmonic functions on $X\setminus \{p\}$ with a unit logarithmic pole at $p$. This proves all three assertions of the statement.