Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

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 $\mathbb H$.

share|improve this answer
    
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! –  Nichole Jun 17 '12 at 18:39
    
@Nichole Each function $F_n$ is analytic in the closed upper halfplane, since it is analytic in the domain $\{z\colon \mathrm{Im}\ z>-1/n\}$. –  user31373 Jun 17 '12 at 18:55
    
@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. –  user31373 Jun 17 '12 at 19:03
    
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? –  Nichole Jun 17 '12 at 19:17
    
@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. –  user31373 Jun 17 '12 at 19:20
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.