Applying the Schur algorithm to finite Blaschke products A Schur function is a function which is holomorphic in the unit disk $\mathbb{D}$ satisfying $|f(z)|\leq 1$. 
The Schur algorithm is a way of producing a sequence of Schur functions starting with a given Schur fanction $f(z)$ in the following way : $$ f_{n+1}(z)=\frac{1}{z} \frac{f_n(z)-\gamma_n}{1-\overline{\gamma_n}f_n(z)} \ \ \ \ (\star) \ \ \ \ \text{with} \ \ f_0(z)=f(z) \ \ \text{and} \ \ \ \gamma_n = f_n(0) $$
One continues this algorithm as long as $|\gamma_n|<1$, if $|\gamma_m|=1$ for some $m$, we set $f_j(z) \equiv0$ for $j>m$.
I need to prove that the Schur algorithm stops ($|\gamma_m|=1$ for some $m>0$) if and only if the Schur function $f(z)=f_0(z)$ that we start with is a finite Blaschke product $\displaystyle e^{i\theta}\prod^{N}_{i=1}\frac{z-z_i}{1-\overline{z_i}z
}$; $|z_i|<1$.
This is how I approached this :
($\Leftarrow$) Proof by induction on the number of factors in the Blaschke product.
Base of induction: for $N=1$ we have $$f(z)=e^{i\theta}\frac{z-z_1}{1-\overline{z_1}z
} \ \ \ \ \ \ \ \ \ |z_1|<1 $$
We have that $$ f_1(z)= \frac{e^{i\theta}+\gamma_0\overline{z_1}}{1-|\gamma_0|^2-(\overline{z_1}+\overline{\gamma_0}e^{i\theta})z}  $$ so $$\gamma_1=\frac{e^{i\theta}+\gamma_0\overline{z_1}}{1-|\gamma_0|^2} \ \ \ \Rightarrow |\gamma_1|=|e^{-i\theta}\gamma_1|=1$$
To prove the inductive step, I need to prove that if we apply the Schur algorithm to a finite Blaschke product with $m+1$ factors, we will get another Blaschke product with $m$ factors, and then using the induction hypothesis we are done. but I had failed so far to prove this fact.
My attempt for ($\Rightarrow$): Suppose for some $m$ we have that $f_m(0)=\gamma_m \in \mathbb{T}$, where $\mathbb{T}$ is the unit circle. By a simple calculation we can show that $$f_{m-1}(z)=\frac{\gamma_{m-1}+zf_m(z)}{1+\overline{\gamma_{m-1}}zf_m(z)} \ \ \ \  \ \ (\star \star) $$ from $(\star)$. However, by maximum modulus principle we have that $$f_m(z) \equiv \gamma_m$$
so by the above equation for $f_{m-1}(z)$ we will find that $$f_{m-1}(z)=\gamma_m \frac{z-(-\overline{\gamma_m}\gamma_{m-1})}{1-\overline{\left( \overline{\gamma_m}\gamma_{m-1} \right)}z}$$ which is a Blaschke product with one factor. however, to finish the proof , I need to show that if $f_j(z)$ is a Blaschke product $\displaystyle e^{i\theta}\prod^{k}_{i=1} \frac{z-z_i}{1-\overline{z_i}z}$ then $f_{j-1}(z)$ given by $(\star \star)$ is also a Blaschke product (with $k+1$ factors?); plugging into $(\star \star)$ gives us $$f_{j-1}(z)=\frac{\gamma_{j-1}\prod_{i=1}^{k}(1-\overline{z_i}z)+z\prod_{i=1}^{k}(z-z_i)}{\prod_{i=1}^{k}(1-\overline{z_i}z)+\overline{\gamma_{j-1}}z\prod_{i=1}^{k}(z-z_i)}$$ 
The numerator is a polynomial of degree $k+1$, so I can write it as $\displaystyle \prod^{k+1}_{i=1} (z-w_i)$; but this is not at all clear to me why I can write the denominator as $\displaystyle \prod^{k+1}_{i=1} (1-\overline{w_i}z)$ for the same $w_i$ appearing in the numerator; to get a Blaschke product.
(somehow the same problem that I run into when I want to prove $(\Leftarrow)$)
Any help, hints, or suggestion of alternative ways to approach this problem is very much appreciated, Thanks !
 A: A proof of this fact doesn't need to involve any technical computations, just some simple properties of Mǒbius transfrmations.
A finite Blaschke product $B$ is a product of (a finite number of) Mőbius transformations
$$
\phi_{a,\theta}(z) = e^{i\theta} \frac {z-a}{1-\bar az},
$$
which form a group of automorphisms of the unit disc (with $a\in \mathbb D$, $\theta \in \mathbb R$). So, both $B\circ \phi_{a,\theta}$ and $\phi_{a,\theta}\circ B$ are finite Blaschke products of the same order (having the same number of zeros counted with multiplicities).
The Schur transformation (I'll call it $S$ for convenience) of a Blaschke product B of order $k>0$ can be written as
$$
S(B)(z) = \frac 1z \phi_{B(0),0} \circ B(z) =: \frac 1z B_1(z),
$$
where $B_1$ is another finite Blaschke product with $B_1(0) = 0$. Thus, one of the factors in $B_1$ is a rotation and it gets "cancelled" by $\frac 1z$. So, we've just proved that $S(B)$ is a Blaschke product of order $k-1$.
If we start with $k=1$, the we get $|S(B)| \equiv 1$ in $\mathbb D$.
The other way round, the inverse of the Schur transformation of a Blaschke product $B$ of order $k\ge 0$ is
$$
S^{-1}(B)(z) = \phi_{\gamma,0}^{-1}\circ zB(z) = \phi_{-\gamma,0}(zB(z))
$$
for some $\gamma$. And if $|\gamma| < 1$, then $S^{-1}(B)$ is a Blaschke product of order $k+1$.
Now, we can prove the theorem. If we start with a Blaschke product of order $k$, then after $k$ steps we arrive at a constant function and $|\gamma_k| = 1$, $\gamma_j = 0$, $j>k$.
For the converse, we additionally need one simple fact. If $f:\mathbb D\to \mathbb D$, then either $S(f):\mathbb D\to \mathbb D$ or $S(f)$ is constant with $|S(f)|=1$. It's easy to see that $|S(f)(z)| \le 1$ for $z\in \mathbb D$, so this fact follows from the maximum principle.
Suppose we started with some $f_0:\mathbb D\to \mathbb D$ and we reached $|\gamma_k| = 1$, $\gamma_j = 0$, $j>k$, for some $k$. (We also have $|\gamma_j|<1$ for $j<k$.) The procedure stopped for $f_k$ since $\gamma_k = f_k(0) \in \partial \mathbb D$. Function $f_k$ is constant of modulus one (by the fact above). Thus, 
$$f_{k-1}(z) = S^{-1}(f_k)(z) = \phi_{-\gamma_{k-1},0}(zf_k)$$
is a Mőbius transformation (a Blaschke product of multiplicity one). And so on.

By the way, the condition you wrote: "$\gamma_m=0$ for some $m$" is not really a stopping condition. (It should be $|\gamma_m|=1$ for some $m$.) Consider for example $f_0(z) = z^n g(z)$ with some $g$ not being a finite Blaschke product. Here $\gamma_j = 0$ for $j\le n$ but the Schur procedure never stops.
