I'm trying to find a factorization for $f(z) = z - c$ in inner and outer functions.
So, first I try to find for what $c$, $f$ is cyclic for the unilateral shift $M_z$. There is a theorem that states that if $A$ is a bounded operator on a Hilbert space $H$ then $x \in H$ is cyclic for $A$ iff zero is the only element that is orthogonal to all $A^n x$ for $n = 0,1,2,\ldots$.
Let $M_z$ be the multiplication operator by $z$ ("the unilateral shift on $H^2$").
So, if $g_n(z) = M_z^n (z - c) = -cz^n + z^{n + 1}$ I find that the power series coefficients of $g_n$ (which is given by $g_n(z) = \sum_m a_{n,m} z^m$) are $a_{n,n} = -c$ and $a_{n,n+1} = 1$. If I insert this in the inner product (where $h(z) = \sum \overline{b_n} z^n$), $(g_n, h) = 0$ for all $n = 0, 1, 2, \ldots$. I obtain that $b_{n + 1} - c b_n = 0$ for $n \geq 0$. This gives that $b_n = C \cdot c^n$. But $\sum c^n z^n = 0$ only if $c = 0$. This would imply that $f(z) = z$ is cyclic but a cyclic vector cannot have a zero in the unit disk! What is wrong here?
If I would find this $c$, then I would find a factorization if the $c$ makes $f$ cyclic because then the inner function is $1$ and the outer function $f$. How to do it if $f$ is not cyclic?