Corollary 3.1.21D0BED761B9E7

Proof. Let $h_ j\geq 0$ be harmonic on $D$ with $h_ j(x_0)\leq M$ for some $x_0\in D$. Let $K\subset D$ be compact and choose $\delta >0$ such that $K_\delta =\{ y:\operatorname {dist}(y,K)\leq \delta \} $ is contained in $D$. By proposition 3.1.19(3), $h_ j\leq C(K_\delta ,x_0,D)M$ on $K_\delta $ for all $j$, and (3.1.8) with radius $\delta $ then bounds each derivative $\partial ^\alpha h_ j$ on $K$ uniformly in $j$. Taking for $K$ a closed ball in $D$ and applying this bound to the first derivatives of $\partial ^\alpha h_ j$, we see that the functions $\partial ^\alpha h_ j$ are uniformly Lipschitz on every closed ball in $D$, and hence equicontinuous on compact subsets of $D$.

Let $K_ m$ be the sets of the compact exhaustion of $D$ constructed in section 2.2; every compact subset of $D$ lies in some $K_ m$. Applying proposition 2.7.4 on each $K_ m$ and making a diagonal extraction over all $m$ and all multi-indices $\alpha $, we obtain a subsequence along which every $\partial ^\alpha h_ j$ converges uniformly on compact subsets of $D$, say to $g_\alpha $. To identify the limits, let $e_ k$ be the $k$th standard basis vector. Passing to the uniform limit in

which holds whenever the segment from $y$ to $y+te_ k$ lies in $D$, we find that $\partial _ kg_\alpha =g_{\alpha +e_ k}$. By induction, $g=g_0$ is smooth with $\partial ^\alpha g=g_\alpha $, so the subsequence converges to $g$ in $C^\infty $ on compact subsets, and $\Delta g=\lim _ j\Delta h_ j=0$. This proves the first assertion.

We also need a lower Harnack bound: for compact $K\subset D$ and $x_0\in D$ there is $c>0$ such that $h\geq c\, h(x_0)$ on $K$ for every nonnegative harmonic $h$ on $D$. Indeed, if $B(a,2t)\subset D$ and $y\in B(a,t)$, then $r=|y-a|<t$, and the left inequality in (3.1.20) on $B(a,2t)$ gives $h(y)\geq (2t)^{d-2}(2t-r)(2t+r)^{1-d}h(a)\geq c_0h(a)$ with $c_0=2^{d-2}3^{1-d}$. Cover $K$ by finitely many balls $B(a_ k,t_ k)$ with $B(a_ k,2t_ k)\subset D$. Proposition 3.1.19(3), applied to the compact set $\{ x_0\} $ and the base point $a_ k$, gives $h(x_0)\leq C_ kh(a_ k)$. Hence $h\geq (c_0/\max _ kC_ k)h(x_0)$ on $K$.

Now let $(h_ j)$ be increasing and fix $x_0\in D$. The functions $h_ j-h_1$ are nonnegative and harmonic. If $h_ j(x_0)$ is bounded, the first assertion applies to every subsequence of $(h_ j-h_1)$. Since the sequence increases and, by proposition 3.1.19(3), is bounded on compact sets, it has a finite pointwise limit, and every subsequence has a further subsequence converging in $C^\infty $ on compact subsets to this limit. Hence the whole sequence converges in $C^\infty $ on compact subsets, and the limit is a finite harmonic function. If $h_ j(x_0)\to +\infty $, the lower Harnack bound gives $h_ j-h_1\geq c(h_ j(x_0)-h_1(x_0))$ on each compact $K$, and since $h_1$ is bounded on $K$, $h_ j\to +\infty $ uniformly on $K$. For a decreasing sequence, apply this to $(-h_ j)$. $\square$