$\{u_j\}$ harmonic, then if $\max_{y\in \partial\Omega}|u_j(y)-u_k(y)|\le \frac{1}{j}+\frac{1}{k}, \ k,k=1,2,\cdots$, $\{u_j\}$ converges uniformly 
Let $\Omega$ be a bounded open of $\mathbb{R}^N$ and consider a sequence 
  $\{u_j\}$ of harmonic functions in $\Omega$, each one of them continuous 
  in $\overline{\Omega}$. Suppose that $$\max_{y\in \partial\Omega}|u_j(y)-u_k(y)|\le \frac{1}{j}+\frac{1}{k}, \ k,k=1,2,\cdots$$ Prove that $\{u_j\}$ converges uniformly in $\overline{\Omega}$ for a function $u\in C(\overline{\Omega})$, which is still harmonic in $\Omega$

I'm studying PDEs and the things that I already saw are the weak and strong maximum principle, the representation formula, the mean value property and some other minor things. I'm talking about the maximum principle because I see a maximum there, but it may have nothing to do with it. 
This condition remembers me of cauchy sequences, but I don't see anything obvious. I truly don't know how to start, so any hints will be appreciated.
 A: I assume that $\Omega$ is a bounded, connected open set and that $u_k \in C^0(\overline{\Omega})\cap C^2(\Omega)$ for all $k \in \mathbb{N}$. Assume that for all $j, k \in \mathbb{N}$ the inequality
\begin{equation}
\max_{x \in \partial \Omega} \lvert u_j(x) - u_k(x) \rvert \leq \frac{1}{j} + \frac{1}{k}\, . \quad (1)
\end{equation}
Step 1. We firts show that $(u_k)_k$ is a Cauchy sequence in $C^0(\overline{\Omega})$. Fix $k, j \in \mathbb{N}$ and set $w = u_j - u_k$. Then $w \in C^0(\overline{\Omega}) \cap C^2(\Omega)$ is harmonic and so the weak maximum principle holds, i.e., 
\begin{equation}
\max_{x \in \overline{\Omega}} \lvert w(x) \rvert = \max_{x \in \partial \Omega} \lvert w(x) \rvert \, . \quad (2)
\end{equation}
Combining $(1)$ and $(2)$, we find that
\begin{equation}
\max_{x \in \overline{\Omega}} \lvert u_j(x) - u_k(x) \rvert=\max_{x \in \partial \Omega} \lvert u_j(x) - u_k(x) \rvert \leq \frac{1}{j} + \frac{1}{k}
\end{equation}
for all $j, k \in \mathbb{N}$. From this we deduce that $(u_k)_k$ is a Cauchy sequence in $C^0(\overline{\Omega})$. As $\Omega$ is bounded, $C^0(\overline{\Omega})$ is a Banach space and hence there exists a function $u \in C^0(\overline{\Omega})$ such that $u_k \to u$ in $C^0(\overline{\Omega})$. 
Step 2. We use the mean value property to show that the limit $u$ is harmonic. That is, we show that
\begin{equation}
u(x_0) = \frac{1}{\lvert B_r(x_0) \rvert}\int_{B_r(x_0)} u(y) \, \text{d} y \quad (3)
\end{equation}
for all $B_r(x_0) \Subset \Omega$. Fix $x_0 \in \Omega$ and $r>0$ such that $B_r(x_0) \Subset \Omega$. For each $u_k$ is harmonic, we find that
\begin{align}
\bigg\lvert u(x_0) - \frac{1}{\lvert B_r(x_0) \rvert} \int_{B_r(x_0)} u\bigg \rvert &\leq \lvert u(x_0) - u_k(x_0) \rvert + \bigg\lvert u_k(x_0) - \frac{1}{\lvert B_r(x_0) \rvert} \int_{B_r(x_0)} u \bigg\rvert \\
& \leq \max_{x \in \overline{\Omega}} \lvert u(x) - u_k(x)\rvert + \frac{1}{\lvert B_r(x_0) \rvert} \int_{B_r(x_0)} \lvert u_k - u \rvert \\
& \leq 2 \max_{x \in \overline{\Omega}} \lvert u(x) - u_k(x)  \rvert \, . \quad (4)
\end{align}
As $u_k \to u$ in $C^0(\overline{\Omega})$, we can send $k \to \infty$ in $(4)$ and obtain $(3)$. This proves the claim.
