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Been thinking about this problem for a long time without any progress, can someone help?

Consider a bounded function $f: \mathbb{D} \rightarrow \mathbb{D}$ with the following property : for every finite sequence $z_1, z_2, \dots z_n \in \mathbb{D}$, there exists an analytic function $g: \mathbb{D}\rightarrow \mathbb{D}$ such that $g(z_j)=f(z_j)$ for $j=1,2,\dots, n$. Show that $f$ itself is analytic.


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Your function $f$ is already bounded, as it takes valuen on $\mathbb D$. – Matemáticos Chibchas Jan 16 '13 at 1:20
Maybe he mean open disc, which I believe is indeed the case.. – lee Jan 16 '13 at 1:31
It is open unit disk – KWO Jan 16 '13 at 1:31
I think we can use the Morera's theorem. But I don't know how to do this. – lee Jan 16 '13 at 1:35
@lee But in this case you need to know in advance that $f$ is continuous. – Matemáticos Chibchas Jan 16 '13 at 1:49
up vote 8 down vote accepted

This is my idea: let $\{z_n: n\geq1\}$ be any countable dense subset of $\mathbb D$. For each $n$ let $g_n$ be a holomorphic function from $\mathbb D$ into $\mathbb D$ such that $g_n(z_k)=f(z_k)$ for $k=1,\dots,n$. Then the sequence $(g_n)_{n\geq1}$ is uniformly bounded, so by Montel's theorem there is a holomorphic function $h:\mathbb D\to\mathbb C$ and a subsequence $(g_{n_k})_{k\geq1}$ such that $g_{n_k}\xrightarrow{k\to\infty}h$ uniformly on compact subsets of $\mathbb D$, and $h$ is holomorphic. If $m\geq1$ then for all $k$ such that $n_k>m$ we have $g_{n_k}(z_m)=f(z_m)$. Taking $k\to\infty$ we obtain $h(z_m)=f(z_m)$, and so $h$ and $f$ agree on a dense subset of $\mathbb D$.

Let $z_0\in\mathbb D$ such that $z_0\ne z_m$ for all $m\geq1$. We repeat the previous construction, now including $z_0$ into the dense subset of $\mathbb D$. We obtain a holomorphic function $\tilde h$ defined on $\mathbb D$ such that $\tilde h(z_m)=h(z_m)=f(z_m)$ for all $m\geq1$, and $\tilde h(z_0)=f(z_0)$. By continuity of both $h$ and $\tilde h$ we conclude that $h=\tilde h$, and so $h(z_0)=\tilde h(z_0)=f(z_0)$, which shows that $h=f$. Am I wrong?

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Looks good! A natural follow-up: would the conclusion hold if $n$ was a fixed number instead of arbitrary integer? For example, $n=3$? – user53153 Jan 16 '13 at 3:12
Thanks for your help. – KWO Jan 16 '13 at 11:50

Let $z_n$ be a sequence with $z_n \to z_0 \in \mathbb D$. Using the Cauchy integral formula, there are constants $K$ and $\epsilon > 0$ such that any analytic function $g: {\mathbb D} \to {\mathbb D}$ has $$\left| \dfrac{g(z) - g(z_0)}{z - z_0} - g'(z_0) \right| \le K |z - z_0| \text{ for } |z - z_0| < \epsilon $$ and thus for $n$ and $m$ large enough, $$ \left|\dfrac{g(z_n) - g(z_0)}{z_n -z_0} - \dfrac{g(z_m) - g(z_0)}{z_m - z_0}\right| \le K (|z_n - z_0| + |z_m - z_0|)$$ Since there is such a $g$ with $g(z_0) = f(z_0)$, $g(z_n) = f(z_n)$ and $g(z_m) = f(z_m)$, the same is true with $g$ replaced by $f$. But that says $\dfrac{f(z_n) - f(z_0)}{z_n - z_0}$ forms a Cauchy sequence. If the limit is $L$, we must have $$\lim_{z \to z_0} \dfrac{f(z) - f(z_0)}{z - z_0} = L$$ i.e. $f$ is differentiable at $z_0$ with $f'(z_0) = L$.

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It's worth mentioning that your solution uses $n=3$ only. – user53153 Jan 16 '13 at 3:54
Yes, and the number 3 is the best we can do! – lee Jan 16 '13 at 8:20
Thanks for the solution. – KWO Jan 16 '13 at 11:50
@RobertIsrael Great solution, better than mine, you deserve the "chosen answer". – Matemáticos Chibchas Jan 17 '13 at 22:28

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