Here is the text of the problem I'm attempting:
Suppose that $(u_j)$ is a sequence of non-negative harmonic functions on $B_1(0)$ such that the sequence of numbers $u_j(0)$ converges. Show that there is a non-negative harmonic function u on $B_1(0)$ and a subsequence $(j')$ such that $u_{j'}\to u$ in $C^k(B_\rho(0))$ for every $k \in \mathbb{N}$ and every $\rho\in(0;1)$. Must the whole sequence $(u_j)$ converge to $u$? What if we drop the non-negativity assumtpion on $u_j$
Now my attempt is as follows:
Let $\rho\in(0;1)$, then by Harnack's inequality in $B_\rho(0)$ we have that:$$sup_{B_\rho(0)}u_j\leq Cu_j(0)$$ for every $x\in B_\rho(0)$. Furthermore we have derivative estimates:$$sup_{ B_\rho(0)}|D^\alpha u_j| \leq C\int_ {B_\rho(0)}u_j =Cu_j(0) $$ where in the last equality we used the mean value property for harmonic functions and this hold for every multiindex $\alpha$. Thus we get by Arzela-Ascoli that there exists $u\in C^k( B_\rho(0))$ and a subsequence $(j')$ such that $u_j\to u$ in $C^k( B_\rho(0))$ for every $k\in\mathbb{N}$ and every $\rho\in(0;1)$. Now $u$ is non-negative since all the $u_j$ are and harmonic since it satisfies the mean value property by passing to the limit in the mean value equality for $u_j$.
Now for the last two questions, I'd say that the whole sequence need not converge since we could consider the sequence to be such that $u_{2j}=v$ and $u_{2j+1}=w$, where both $v$ and $w$ are non-negative harmonic functions with $v(0)=w(0)$ but $v(x)\neq w(x)$ for $x\neq 0$.
The last question however I'm not sure. I'd guess that dropping non-negativity means that this won't work anymore as we won't have any way to control the supremum norm, but I really don't know nor am I sure about what it might mean.
Is this proof correct? Have I missed something or got something blatantly wrong?
