Interpolation of analytic function on unit disk 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.
Thanks
 A: 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 $f'(z_0) = L$.
A: 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?
