I have a sequence $\{F_{n}(z)\}_{n=1}^{\infty}$ of analytic functions in the open upper half plane $\mathbb H$ and continuous on $\mathbb R$, such that $|F_{n}(z)|\leq 1$ for all $n\geq 1$, and all $z$ in the closed upper half plane $\overline{\mathbb H}=\mathbb H\cup \mathbb R$. Also, restricting to the real line, $\{F_{n}(x)\}$ is continuous and uniformly Lipschitz on $\mathbb R$.
How I can get the following result:
(*) Given the sequence $\{F_{n}\}$ above,we can find a subsequence $\{F_{n'}\}$ which converges uniformly on compact subsets of $\overline{\mathbb H}$ to a function $F$, and $F$ will be analytic on $\overline{\mathbb H}$.
I tried the following: Suppose that $\{F_{n_{k}}(x)\}$ is a subsequence of $\{F_{n}(x)\}$ which converges uniformly on compact subsets of $\mathbb R$ to some continuous function, say $F_{R}(x)$ (this subsequence exists because of the uniformly Lipschitz property).
Now, consider the subsequence $\{F_{n_{k}}(z)\}$, $z\in \mathbb H$: By Montel's theorem, we can find a subsequence of $F_{n_{k}}(z)$, say $\{F_{n_{k_{j}}}\}$, which converges uniformly on compact subsets of $\mathbb H$ to an analytic function, say $F_U$. (As far as I know, the theorem doesn't say anything about convergence on the real line).
So, in this case, the new subsequence $\{F_{n_{k_{j}}}(x)\}$ will converge uniformly on compact subsets of $\mathbb R$ to $F_{R}$.
(**) Is this correct?
Now to answer my question in (*), can we get such $F$ using $F_{U}$ and $F_{R}$ above?
Edit: I changed the statement from analytic on $\overline{\mathbb H}$ to analytic on $\mathbb H$ and continuous on $\mathbb R$ to avoid confusion.