Locally bounded harmonic families have harmonic subsequential limits

By [F2] the holomorphic atlas of $X$ is a smooth atlas, so $X$ is a smooth $2$-manifold and, under [A1], admits a compact exhaustion $K_1\subseteq K_2\subseteq \cdots$ with every $K_n$ compact, $K_n\subseteq \mathrm{int}(K_{n+1})$ and $X={\bigcup }_n{K}_n$; fix such a sequence.

For every compact $K\subseteq X$ there is a finite bound $M_K$ for all $\mathrm{\mid }{u}_n\mathrm{\mid }$ on $K$. Use the family of all open neighbourhoods furnished by local uniform boundedness, with their bounds, rather than choosing one for every point. A finite subcover exists by [F13]; the maximum of its finitely many bounds works.

Chartwise Poisson setup. Fix a holomorphic chart $\psi :U\to \mathbb{C}$ of the atlas, a point $a\in \psi (U)$, a radius $R>0$ with $\overline{B(a,R)}\subseteq \psi (U)$, and suppose $M$ satisfies $\mathrm{\mid }{u}_n\mathrm{\mid }\le M$ on ${\psi }^{-1}(\overline{B(a,R)})$ for all $n$. Put $w_n:=u_n\circ {\psi }^{-1}$ on $\psi (U)$; then each $w_n$ is plane harmonic on $\psi (U)$ by [F3] and, by [F4], satisfies $w_n(a+\rho e^{i\varphi })=\frac{1}{2\pi }{\int }_0^{2\pi }P(\rho ,\varphi -t)w_n(a+R{e}^{it})dt$ for every $\varphi$ and every $0\le \rho <R$, with $P(\rho ,\theta )=(R^2-{\rho }^2)\mathrm{/}(R^2-2R\rho \cos \theta +{\rho }^2)$ and $\mathrm{\mid }{w}_n(a+R{e}^{it})\mathrm{\mid }\le M$ on the boundary circle.

Kernel bound. For $0<\rho \le r<R$ and every $\theta$, writing $D:=R^2-2R\rho \cos \theta +{\rho }^2\ge (R-\rho {)}^2>0$, one computes ${\mathrm{\partial }}_{\rho }P=-(2\rho D+(R^2-{\rho }^2)\cdot 2(\rho -R\cos \theta ))\mathrm{/}{D}^2$, hence $\mathrm{\mid }{\mathrm{\partial }}_{\rho }P\mathrm{\mid }\le 2\rho \mathrm{/}D+2(R+\rho )(R^2-{\rho }^2)\mathrm{/}{D}^2\le 2r\mathrm{/}(R-r{)}^2+2(R+r{)}^2\mathrm{/}(R-r{)}^3=:C(R,r)<\mathrm{\infty }$, a finite constant depending only on $R$ and $r$.

Part 2. Let $\mathrm{\Omega }\subseteq \mathbb{C}$ be open, let $v_n$ be real harmonic on $\mathrm{\Omega }$ for every $n$, and let $T$ be a distribution with $T_{v_n}(\phi )\to T(\phi )$ for every $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$. By [F3] each $v_n$ is of class $C^2$ on $\mathrm{\Omega }$, hence locally integrable, so that $T_{v_n}$ is a regular distribution for every $n$ by [F9].

Differentiating in the radius. Fix $n$, $\varphi$ and $0<r<R$, and define $g(\rho ,t):=P(\rho ,\varphi -t)w_n(a+R{e}^{it})$ and $h(\rho ,t):=({\mathrm{\partial }}_{\rho }P)(\rho ,\varphi -t)w_n(a+R{e}^{it})$ on $[0,r]\times [0,2\pi ]$. On this rectangle the denominator $R^2-2R\rho \cos \theta +{\rho }^2$ is bounded below by $(R-r{)}^2>0$, so $P$ and its $\rho$-derivative are continuous there; hence $g$ and $h$ are continuous, and for each fixed $t$ the map $\rho \mapsto g(\rho ,t)$ is differentiable on $(0,r)$ with derivative $h(\rho ,t)$. By [F5], applied with the parameter $\rho$ in the first slot, the function $G(\rho ):={\int }_0^{2\pi }g(\rho ,t)dt$ is differentiable on $(0,r]$ with $G^{\mathrm{\prime }}(\rho )={\int }_0^{2\pi }h(\rho ,t)dt$. Since $G(\rho )=2\pi w_n(a+\rho e^{i\varphi })$ by step 1.3, this gives $\frac{d}{d\rho }{w}_n(a+\rho e^{i\varphi })=\frac{1}{2\pi }{\int }_0^{2\pi }({\mathrm{\partial }}_{\rho }P)(\rho ,\varphi -t)w_n(a+R{e}^{it})dt$.

For every $n$ one has $\mathrm{\Delta }{T}_{v_n}=0$. Indeed, since $v_n$ is of class $C^2$, its classical second partial derivatives are its weak second partial derivatives by [F10], and the weak test identity is equivalent to the distributional identity ${\mathrm{\partial }}_i^2{T}_{v_n}=T_{{\mathrm{\partial }}_i^2{v}_n}$; summing over $i$ gives $\mathrm{\Delta }{T}_{v_n}=T_{\mathrm{\Delta }{v}_n}$, and $\mathrm{\Delta }{v}_n=0$ pointwise on $\mathrm{\Omega }$ makes the right side the zero distribution.

Radial Lipschitz bound. Combining steps 2.1 and 1.4 with $\mathrm{\mid }{w}_n(a+R{e}^{it})\mathrm{\mid }\le M$ gives $\mid \frac{d}{d\rho }{w}_n(a+\rho e^{i\varphi })\mid \le C(R,r)M$ for every $n$, $\varphi$ and $0<\rho \le r$; integrating this radial derivative along the segment from $a$ to $z=a+\rho e^{i\varphi }$ gives $\mathrm{\mid }{w}_n(z)-w_n(a)\mathrm{\mid }\le C(R,r)M\mathrm{\mid }z-a\mathrm{\mid }$ whenever $\mathrm{\mid }z-a\mathrm{\mid }\le r$, and for $z=a$ both sides vanish.

$\mathrm{\Delta }T=0$. For every test function $\phi$, the definition of the distributional Laplacian in [F9] gives $(\mathrm{\Delta }T)(\phi )=T(\mathrm{\Delta }\phi )$, the distributional convergence in step 1.5 applied to the test function $\mathrm{\Delta }\phi$ gives $T(\mathrm{\Delta }\phi )={\lim }_n{T}_{v_n}(\mathrm{\Delta }\phi )$, and again by [F9] one has $T_{v_n}(\mathrm{\Delta }\phi )=(\mathrm{\Delta }{T}_{v_n})(\phi )=0$ for every $n$ by step 2.2; hence $(\mathrm{\Delta }T)(\phi )=0$ for every $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$.

Oscillation neighbourhoods. For every $x\in X$ and every $\epsilon >0$ there is an open neighbourhood $V$ of $x$ with compact closure such that $\mathrm{\mid }{u}_n(y)-u_n(x)\mathrm{\mid }\le \epsilon$ for all $n$ and all $y\in V$. Indeed, local uniform boundedness gives an open $W\ni x$ and $M_x$ with $\mathrm{\mid }{u}_n\mathrm{\mid }\le M_x$ on $W$ for all $n$; choose a holomorphic chart $\psi :U_0\to \mathbb{C}$ with $x\in U_0$ and replace it by its restriction to $U:=U_0\cap W$, a chart with $x\in U$ and $\mathrm{\mid }{u}_n\mathrm{\mid }\le M_x$ on $U$ for all $n$; put $a:=\psi (x)$ and choose $R>0$ with $\overline{B(a,R)}\subseteq \psi (U)$, which is possible because $\psi (U)$ is open in $\mathbb{C}$ and contains $a$. If $M_x=0$ then $u_n=0$ on $U$ for all $n$ and any open $V\ni x$ with compact closure inside $U$ works. Otherwise set $r:=R\mathrm{/}2$, $C:=C(R,r)$ from step 1.4 and $\delta :=\min \{r,\epsilon \mathrm{/}(C{M}_x)\}>0$, and put $V:={\psi }^{-1}(B(a,\delta ))$; then $V$ is open with $x\in V$, its closure lies in the compact set ${\psi }^{-1}(\overline{B(a,r)})$ by [F14] (a continuous image of a compact set is compact), and step 3.1 gives $\mathrm{\mid }{u}_n(y)-u_n(x)\mathrm{\mid }\le C{M}_x\mathrm{\mid }\psi (y)-a\mathrm{\mid }<\epsilon$ for all $n$ and all $y\in V$.

Finite oscillating covers. For every pair $j,k\ge 1$, step 4.1 and [F13] give finitely many centres $c_{j,k,i}\in K_j$ and neighbourhoods $V_{j,k,i}$ covering $K_j$ such that $\mathrm{\mid }{u}_n(y)-u_n(c_{j,k,i})\mathrm{\mid }<1\mathrm{/}k$ for all $y\in K_j\cap V_{j,k,i}$ and all $n$. Use step 4.1 with $1\mathrm{/}(2k)$ and take a finite subcover of the family of all admissible neighbourhood-centre pairs. Empty $K_j$ needs no centres.

Fixing the countable data. For each pair $(j,k)$ the set of finite cover data in step 5.1 is nonempty. Apply [A1] to this fixed countable family, and fix one finite cover with its centres for every pair. Enumerate all those centres as $p_1,p_2,\dots$ in a fixed ordering of $(j,k,i)$; pad by repetition if there are finitely many. There is at least one centre because $X$ is nonempty and the $K_j$ exhaust it. Every numerical sequence $(u_n(p_l){)}_n$ is bounded by local uniform boundedness.

Canonical nested subsequences. We give an explicit selection rule for the numerical subsequence in [F12]. A bounded real sequence $(a_l)$ has a deterministically selected convergent subsequence. Let $M$ be the least positive integer with $\mathrm{\mid }{a}_l\mathrm{\mid }\le M$ for all $l$, and start with $I_0=[-M,M]$. Bisect each closed interval $I_{r-1}$, taking its left closed half if that half contains infinitely many terms of the sequence, and its right closed half otherwise. The selected $I_r$ contains infinitely many terms and has length $2M\mathrm{/}{2}^r$. Let $\tau (r)$ be the least index greater than $\tau (r-1)$ with $a_{\tau (r)}\in I_r$, starting with $\tau (0)=0$. Nested intervals give a unique common point, to which $a_{\tau (r)}$ converges. Both recursions use uniquely specified choices. Starting with ${\sigma }_0$ the identity, apply this rule to $a_l=u_{{\sigma }_{r-1}(l)}(p_r)$ and put ${\sigma }_r={\sigma }_{r-1}\circ \tau$. Recursion on the natural numbers therefore defines all the ${\sigma }_r$ without Dependent Choice or an additional use of [A1].

Diagonal extraction. Put $n_k:={\sigma }_k(k)$. Since each ${\sigma }_{k+1}$ is a subsequence of ${\sigma }_k$ and its indexing map satisfies $\tau (l)\ge l$, one has $n_{k+1}>n_k$. For fixed $r$, the tail $k\ge r$ lies in the range of ${\sigma }_r$, so $(u_{n_k}(p_r){)}_k$ converges.

Uniform convergence on every exhaustion compact. Fix $j,k$. For every $y\in K_j$, choose one of the finitely many $V_{j,k,i}$ containing $y$. For any subsequence indices $p,q$, $\mathrm{\mid }{u}_{n_p}(y)-u_{n_q}(y)\mathrm{\mid }<2\mathrm{/}k+\mathrm{\mid }{u}_{n_p}(c_{j,k,i})-u_{n_q}(c_{j,k,i})\mathrm{\mid }\mathrm{.}$ By step 8.1, the last term is less than $1\mathrm{/}k$ for all sufficiently large $p,q$, uniformly over the finitely many centres of this cover. Thus $(u_{n_k}{\mathrm{\mid }}_{K_j}{)}_k$ is uniformly Cauchy and converges uniformly by [F15].

One global subsequence. The fixed cover data include every $K_j$, so the single subsequence constructed in step 8.1 converges uniformly on every $K_j$ by step 9.1. No second recursive extraction is needed.

Set $m_k:=n_k$. This sequence is strictly increasing and $(u_{m_k}{)}_k$ converges uniformly on every $K_j$.

Every compact $L\subseteq X$ lies in some $K_J$: the open sets $\mathrm{int}{K}_j$ cover $X$ by [F2], so a finite subcover of $L$ and nesting give such a $J$. Thus $(u_{m_k})$ converges uniformly on every compact subset of $X$, and pointwise everywhere.

Define $u(x):={\lim }_{k\to \mathrm{\infty }}{u}_{m_k}(x)$ for $x\in X$, a well-defined real function on $X$ by step 12.1. Then $(u_{m_k})$ converges to $u$ uniformly on every compact subset of $X$ and $u$ is its pointwise limit; this completes everything in part 1 except harmonicity and continuity of $u$.

$u$ is continuous. Let $x\in X$ and $\epsilon >0$; step 4.1 provides an open neighbourhood $V$ of $x$ with compact closure and $\mathrm{\mid }{u}_n(y)-u_n(x)\mathrm{\mid }\le \epsilon \mathrm{/}3$ for all $n$ and all $y\in V$. Since $(u_{m_k})$ converges uniformly on the compact set $\bar{V}$ by step 12.1, for all sufficiently large $k$ one has ${\sup }_{y\in V}\mathrm{\mid }{u}_{m_k}(y)-u(y)\mathrm{\mid }<\epsilon \mathrm{/}3$ and $\mathrm{\mid }{u}_{m_k}(x)-u(x)\mathrm{\mid }<\epsilon \mathrm{/}3$, so $\mathrm{\mid }u(y)-u(x)\mathrm{\mid }<\epsilon$ for every $y\in V$. Hence $u$ is continuous at every point of $X$.

Spherical mean property of the limit. Let $\psi :U\to \mathbb{C}$ be a holomorphic chart and put $\mathrm{\Omega }:=\psi (U)$, $w_k:=u_{m_k}\circ {\psi }^{-1}$ and $w:=u\circ {\psi }^{-1}$ on $\mathrm{\Omega }$. Each $w_k$ is plane harmonic by [F3], and $w_k\to w$ uniformly on every compact subset of $\mathrm{\Omega }$: if $L\subseteq \mathrm{\Omega }$ is compact then ${\psi }^{-1}(L)$ is a compact subset of $X$ by [F14], and ${\sup }_L\mathrm{\mid }{w}_k-w\mathrm{\mid }={\sup }_{{\psi }^{-1}(L)}\mathrm{\mid }{u}_{m_k}-u\mathrm{\mid }\to 0$ by step 12.1. Let $b\in \mathrm{\Omega }$ and $r>0$ with $\overline{B(b,r)}\subseteq \mathrm{\Omega }$; by [F6] each $w_k$ satisfies $w_k(b)=\frac{1}{2\pi }{\int }_0^{2\pi }{w}_k(b+r{e}^{is})ds$, and the integrands converge uniformly in $s\in [0,2\pi ]$ to $s\mapsto w(b+r{e}^{is})$ because the circle is a compact subset of $\mathrm{\Omega }$; [F7] and $w_k(b)\to w(b)$ therefore give $w(b)=\frac{1}{2\pi }{\int }_0^{2\pi }w(b+r{e}^{is})ds$.

Hence $w$ has the local spherical mean value property on $\mathrm{\Omega }$: given $b\in \mathrm{\Omega }$ choose $r_b>0$ with $\overline{B(b,r_b)}\subseteq \mathrm{\Omega }$; then for every $y\in B(b,r_b)$ and every $0<r<r_b-\mathrm{\mid }y-b\mathrm{\mid }$ one has $\overline{B(y,r)}\subseteq \mathrm{\Omega }$, so step 13.3 gives $w(y)=\frac{1}{2\pi }{\int }_0^{2\pi }w(y+r{e}^{is})ds$, the normalized circle average $M_w(y,r)$. Integrating this identity over the concentric radii gives the disc mean $\frac{2}{r^2}{\int }_0^r{M}_w(y,s)sds=\frac{2}{r^2}{\int }_0^{r}w(y)sds=w(y)$; the integrand extends continuously at $s=0$ because $w$ is continuous. Thus both local mean identities of [F8] hold. The chart expression $w$ is continuous by step 13.2, so [F8] and [A1] make $w$ plane harmonic on $\mathrm{\Omega }$. As the holomorphic chart was arbitrary, [F3] makes $u$ harmonic on $X$; with steps 13.1 and 13.2 this proves part 1.

By [F11] and [A1] there is a unique smooth harmonic $h$ on $\mathrm{\Omega }$ with $T=T_h$. This is exactly the representation asserted in part 2, so the proof is complete.