TITLE: Interpolation of analytic function on unit disk
QUESTION [8 upvotes]: 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

REPLY [9 votes]: 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 $f'(z_0) = L$.