Fix an open, bounded $U \subset \mathbb{R}^n$ and consider a sequence $C^2(U) \cap C(\bar{U})$ of harmonic functions.
(i) Suppose that $u_n$ converges uniformly to a function $u.$ Show that u is harmonic.
(ii) Suppose instead that $u_n$ is pointwise increasing and that for some $y \in U$ that $u_n(y)$ is bounded. Show that on any $V \subset \subset U$, $u_n$ converges uniformly to a harmonic function $u$.
I am done with (i), for (ii), I am planning to prove $u_n$ is convergent pointwise. Fix $x \in U,$ suppose for the contradiction that $\lim_{n \rightarrow \infty} u_n(x)=+\infty$, but I don't know how to use condition $u_n(y)$ is bounded for some y to get the contradiction.
I also think about using Harnack's inequality to prove $u_n(x)$ is bounded in this case. But we don't have the condition $U$ is connected. Furthermore, can we apply Arzelà–Ascoli theorem to prove (ii)?
Could you please give me some ideas or other ways to prove (ii)?
The statement as written is not true. A counterexample would be $U=B_1(0)\cup B_1(3)$ and $$ u_n(x)=\begin{cases} 0,& x\in B_1(0), \\ n,& x\in B_1(3). \end{cases}$$ The $u_n$ are obviously harmonic and pontwise increasing. However, the functions do not converge uniformly as they do not converge pointwise. Indeed, $$\lim_{n\rightarrow \infty} u_n(3)=\lim_{n\rightarrow \infty} n=\infty.$$
On the other hand, if $U$ is connected, then we are fine.
First approach: The approach you suggested is doable, but hard. Fix a compact subset $A\subseteq U$ and let $d=\min \{dist(A, \partial U), 1\}$. Now pick a compact set $K\subseteq U$ such that $K$ is connected, contains $y$ and such that $$\{ x\in U \ : \ dist(x, A)\leq d/2\}\subseteq K.$$ To construct such a $K$ let $\ell:=dist(y, \partial U)$ and define $$L_n = \{ x\in U \ : \ dist(x, \partial U)\geq \ell/n\}\cap \overline{B(y, n)}.$$ Let $K_n$ be the connected component of $L_n$ which contains $y$. As connected components are closed in the subspace topology, $L_n$ is compact, we get that $K_n$ is compact, connected and contains $y$. We have to check now that there exists $n$ such that $A\subseteq K_n$. It is enough to show the statement for closed balls in $U$ (for the general case, just cover $A$ with finitely many closed balls). Let $B\subseteq U$ be a closed ball with center $x_0$. Let $\gamma:[0,1]\rightarrow U$ be a continuous function such that $\gamma(0)=y, \gamma(1)=x_0$. Then $Y=\gamma([0,1])$ is compact. Hence, $D:=dist(Y \cup B, \partial U)>0$ (as $Y\cup B$ is compact and $\partial U$ is closed). For $n$ large enough, we will have $Y\cup B\subseteq \overline{B(y,n)}$ and $D>\ell/n$. Hence, $Y\cup B\subseteq L_n$. However, $\gamma$ is a path connecting $y$ and $x_0$, thus, $B$ is in the same connected component as $y$ and hence $B\subseteq K_n$.
By assumption we know that $(u_n(y))_n$ is bounded. We can wlog assume that $u_n$ is positive on $K$ (otherwise consider $v_n=u_n-u_1+1$ which is positive and harmonic). Then by Harnack's inequality we get that the $u_n$ are uniformly bounded on $K$ as $$0\leq\sup_K u_n \leq C_K \inf_K u_n \leq C_K u_n(y).$$ As the $u_n$ are uniformly bounded on $K$, their derivatives are also uniformly bounded on $K$ and hence the family is equi-continuous on $A$. Indeed, let $x,z\in A$ with $\vert x-z\vert \leq d/2$, then we have by the mean value theorem $$ \vert u_n(x)-u_n(z)\vert \leq (\sup_m \sup_{w\in K} \vert u_m'(w)\vert) \vert x-z\vert.$$ By Arzelà-Ascoli the sequence admits a uniformly convergent subsequence on $A$. However, as the sequence is pointwise increasing, the entire sequence converges uniformly on $A$.
Alternative: Start with $A\subseteq U$ compact and $K\subseteq U$ a compact, connected set which contains $y$ and $A$. Then consider for $n>m$, $v_{n,m}=u_n-u_m+1/n$, which is harmonic and positive. By Harnack's inequality we have for $n> m$ \begin{align*} \sup_A \vert u_n -u_m\vert &\leq \sup_K \vert u_n -u_m \vert =\sup_K v_{n,m} \\ &\leq C_K \inf_K v_{n,m} \leq C_K v_{n,m}(y)\\ &=C_K (u_n(y)-u_m(y)+1/n). \end{align*} As, $(u_n(y))_n$ is a Cauchy sequence, we can make the RHS as small as we like and therefore $(u_n)_n$ converges uniformly on $A$.