Let $g$ and $h$ be to two partial recursive functions and let us define the following function :
$f(x,y) =\left\{ \begin{array}{ll} g(y) & \mbox{if} \ x=0,\\ f(x-1,h(x-1,y)) & \mbox{if} \ x>0. \end{array} \right.$
The goal is to show that the function $f$ is recursive (partial).
It is suggested to apply a fixed-point theorem and before to consider for all functions $\psi : \mathbb{N}^2 \to \mathbb{N}$, the function :
$\Omega(x,y) =\left\{ \begin{array}{ll} g(y) & \mbox{if} \ x=0,\\ \psi(x-1,h(x-1,y)) & \mbox{if} \ x>0. \end{array} \right.$
-The first idea could be to show that $\Omega= \psi$ only if $f = \psi$. As the function $\Omega$ is defined for all $\psi$, we can take $f=\psi$ and write : $\Omega(x,y) =\left\{
\begin{array}{ll}
g(y) & \mbox{if} \ x=0,\\
f(x-1,h(x-1,y)) & \mbox{if} \ x>0.
\end{array}
\right.$
Hence by definition of $f$, we can deduce that : $\Omega = f$. Then $\Omega = \psi$.
For the uniqueness of $f$, we can consider another function $f_1$ such that $f_1 = \psi$. Necessarily $\Omega=f_1$ and then $f_1=f$.
-The second idea as suggested is to apply a fixed-point theorem. I do not really know how to proceed. I imagine that $f$ is defined as the fixed-point of a set of equations...
Also the definition of $f$ could be a recursive one (we have cases, composition and predecessor) but the problem is that we do not know yet if $f$ or $\psi= \Omega$ are partial recursive functions.
I meet difficulties to make the link between the two hints.
Thanks in advance !