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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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    Note that the sequence $\{F_{n}\}$ above is analytic on the UHP and has continuous extinsion the the real line, so its analytic in the closed UHP. But you are saying that your function has no analytic extinsion to the real line!2012-06-17
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    @Nichole Each function $F_n$ is analytic in the closed upper halfplane, since it is analytic in the domain $\{z\colon \mathrm{Im}\ z>-1/n\}$.2012-06-17
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    @Nichole Also, "analytic on the UHP and has continuous extinsion the the real line" is not the same as "analytic in the closed UHP". Maybe this is the source of confusion here.2012-06-17
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    Oh, ok. What I have is: in my sequence each $F_{n}$ is analytic on the open UHP and has continuous extinsion the the real line (i.e. is continuous on the real line). Anything changed?2012-06-17
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    @Nichole I added a proof that $F$ has the same properties (analytic on open halfplane, continuous on closed). It does not follow the outline you had at the beginning, because I did not see an easy way to prove that gluing $F_u$ and $F_{\mathbb R}$ gives a continuous function.2012-06-17
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    So you mean statement (*) is correct under the stated conditions?2012-06-17
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    @Nicole No, if "analytic in closed halfplane" means what I think it means. In my answer I stated exactly what I proved there.2012-06-17
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    I edited my post concerning this issue.2012-06-17
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    I have a question about step(3) and (4): So, you are saying given an integer $R$ there is a subsequence, call it $F_{n_{R}}$ which converges uniformly on $\{z\in \overline{\mathbb H}; |z|\leq R\}$, to some function say $F_{R}$. So, in this way we get a sequence $\{F_{R}\}_{R\geq 1}$ which will also converges uniformly on $\{z\in \overline{\mathbb H};|z|\leq R\}$ to some function $F$. Please correct me if I missunderstand anything! Also, I thought Arzela-Ascoli theorem works just for open subsets of $\mathbb H$!2012-06-17
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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