$f$ is analytic if $f$ agrees with some holomorphic function on every triplet I came across the following problem asked in a prelim exam. 
Let $f$ be a function defined on the unit disk with the property that for every triplet $a, b, c$ there exists a holomorphic function $g$ such that $g$ is bounded by $1$ on the unit disk and $g(a)=f(a), g(b)=f(b)$ and $g(c)=f(c).$ Show that $f$ is holomorphic on disk and bounded by $1$. 
The fact that $f$ is bounded by $1$ is trivial. Because at every point it agrees with a function which is in turn bounded by 1. But I am trying to prove the differentiablity of $f$ and I am not able to. 
I was thinking to show that $f$ is differentiable at $0$ and showing that is enough to do so. 
I first assume that $f(0)=0.$ To prove that $f$ is differentiable at zero, I take a sequence of points $z_n\to 0$ and obtain a sequence of holomorphic functions $g_n$ such that $f(z_n)=g_n(z_n)$. I can see that $g_n$ is normal, and hence has a convergent subsequence. I choose such a subsequence, and denote the limit by $g$. I want to show that $f’(0)=g’(0).$ 
I am having problem because I can get the convergence of $g_n$ only along a subsequence, and along different subsequence I may possibly have different limits. 
Any suggestion is welcome. 
 A: You are on the right track. Using the full power of the assumption one can not only show that $\frac{f(z_{n_k})-f(0)}{z_{n_k} - 0}$ is convergent for some subsequence, but also that the limit is the same for all subsequences. It is a standard argument to conclude that the full sequence $\frac{f(z_{n})-f(0)}{z_{n} - 0}$ is convergent.

Fix $a \in \Bbb D$ (the unit disk) and define
$$
 h: \Bbb D \setminus \{ a \} \to \Bbb C, \, h(z) = \frac{f(z)-f(a)}{z-a} \, .
$$
We need to show that $\lim_{z \to a} h(z)$ exists.
First consider two arbitrary sequences $(z_n)$, $(w_n)$ in $\Bbb D$ with 
$$
 \lim_{n \to \infty} z_n = \lim_{n \to \infty} w_n = a \, .
$$
For each $n$ let $g_n$ be a holomorphic function which is bounded by one and satisfies $g_n(a) = f(a)$, $g_n(z_n) = f(z_n)$, $g_n(w_n) = f(w_n$). 
$(g_n)$ is a normal family and has a (locally uniformly) convergent subsequence $(g_{n_k})$. Let $g$ be the limit function. Then $g_{n_k}' \to g'$ uniformly in a neighborhood of $z=a$, and therefore
$$
 h(z_{n_k}) = \frac{f(z_{n_k})-f(a)}{z_{n_k} - a} = \frac{g_n(z_{n_k})-g_n(a)}{z_{n_k} - a} = \int_0^1 g_n'(a + t (z_{n_k}-a)) dt \to g'(a)
$$
and similarly
$$
 h(w_{n_k}) \to g'(a)
$$
for $k \to \infty$, i.e. $h(z_{n_k})$ and $h(w_{n_k})$ are both convergent, with the same limit.
We have therefore shown that


*

*Every sequence $z_n \to a$ has a subsequence $(z_{n_k})$ such that $h(z_{n_k})$ is convergent, and

*For any two sequences $z_n \to a$ and $w_n \to a$ such that  $h(z_n)$ and $h(w_n)$ are both convergent, the limit is the same.


It follows that $\lim_{z \to a} h(z)$ exists, i.e. $f$ is differentiable at $z=a$.
