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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.

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I think the statement is false. Let $\mathbb D$ be the unit disk. Take a function $f\colon \mathbb D\to\mathbb D$ that is Lipschitz in $\overline{\mathbb D}$ but is not analytic in the closed disk (i.e., does not have an analytic extension to a larger disk). For example, $\displaystyle f(z)=\frac{1}{10}\sum_{n=1}^{\infty}\frac{z^n}{n^3}$. Transplant this function to the upper half-plane via substitution $F(z)=f((i-z)/(i+z))$. The composition $F$ is also Lipschitz, but does not extend analytically to a neighborhood of $0$, since $f$ does not extend to a neighborhood of $1$. Finally, let $F_n(z)=F(z+i/n)$: this sequence has all the properties stated, and converges to $F$.


[added] Let $\mathbb H=\{z : \mathrm{Im}\ z>0\}$. Here is a proof of the following statement: "Given the sequence $\{F_n\}$ above, we can find a subsequence $\{F_{n_k}\}$ which converges uniformly on compact subsets of $\overline{\mathbb H}$ to a function $F$, where $F$ is analytic on $\mathbb H$ and continuous on $\overline{\mathbb H}$."

Step 1: the functions $F_n$ are actually uniformly Lipschitz on $\overline{\mathbb H}$. Indeed, let $L$ be the Lipschitz constant of ${F_n}\big|_{\mathbb R}$. For any $h\in \mathbb R$ the difference $g(z)=F_n(z+h)-F_n(z)$ is bounded by $2$ in $\mathbb H$ and bounded by $Lh$ on $\mathbb R$. By the maximum principle, $|g(z)|\le Lh$ on $\mathbb H$. Passing to the limit $h\to 0$, we obtain $|F_n'|\le L$ on $\mathbb H$. It follows that $F_n$ is $L$-Lipschitz on $\overline{\mathbb H}$.

Step 2: for each integer $R$ there is a subsequence that converges uniformly on $\{z\in \overline{\mathbb H} : |z|\le R\}$. This follows from 1 and the Arzela-Ascoli theorem.

Step 3: The usual diagonal argument gives a subsequences that converges uniformly on $\{z\in \overline{\mathbb H} : |z|\le R\}$ for every $R$. Let $F$ be its limit.

Step 4, conclusion: The function $F$ is continuous on $\overline{\mathbb H}$, and is analytic on $\mathbb H$.

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    @Nicole A-A is normally stated for compact sets... Yes, we can simply take the diagonal subsequence $F_{n,n}$.2012-06-17