I've come across rather general statement about convergence of iterates of an analytic function $U(x,y)$ to a solution of an equation $U(x,y)x=y,$ but I can't understand how to prove it. The statement is the following:
Let $A,B\subset \mathbb{C}^p$ be open connected neighborhoods of $0\in\mathbb{C}^p$ and $U:A\times B\to\mathbb{C}^{p\times p}$ be an analytic function (i.e. a $p\times p$ matrix whose entries are analytic in $A\times B$). It is assumed that the sequence of iterates $u_0(x)=0,$ $u_n(x)=U(x,u_{n-1}(x))\cdot x,$ $n\geq 1,$ are well-defined in the sense that for all $n\geq 0,$ $u_n(A)\subset B.$ Further, it is assumed that restrictions of $u_n$ onto every compact $L\subset A$ are relatively compact in $C(L;\mathbb{C}^p).$ Then it is stated that iterates $u_n$ converge (convergence is uniform on every compact in $A$) to an analytic function $v:A\to B$ which is a unique analytic solution of the equation $U(x,v(x))\cdot x=v(x).$
Update. I've made certain progress on this question and proved it for $p=1$ in the following way:
1) As $A, B$ are open neighborhoods of $0\in \mathbb{C}$ there exists $\gamma>0$ such that the closed ball $B'(0,\gamma)\subset A\cap B.$ Let $C>0$ be such that $$ \sup_{|x|,|y|\leq \gamma}\|D_y U(x,y)\|<C, $$ and $\alpha\in(0,\min(\gamma,\frac{1}{2C}))$ be such that $$ \sup_{n\geq 0}\sup_{|x|\leq \alpha}|u_n(x)|<\frac{1}{2}\gamma $$ (one can choose $\alpha$ because restrictions $u_n|_{B'(0,\gamma)}$ form a relatively compact sequence, and $u_n(0)=0$ by definition). Next we estimate for all $x\in B'(0,\alpha)$ and $n\geq 1$ $$ |u_{n+1}(x)-u_n(x)|\leq \|U(x,u_{n}(x))-U(x,u_{n-1}(x))\||x|\leq $$ $$ \leq C\alpha|u_n(x)-u_{n-1}(x)|\leq \frac{1}{2}|u_n(x)-u_{n-1}(x)| $$ It follows that $$ |u_n(x)-u_{n-1}(x)|\leq \frac{1}{2^{n-1}}|u_1(x)| $$ and for each $x\in B'(0,\alpha)$ sequence $u_n(x)$ converges.
2) $A$ is connected, restrictions $u_n|_L$ to any compact subset $L\subset A$ form a relatively compact sequence, and $u_n(x)$ converges on a set of uniqueness. Consequently, for all $x\in A$ there is a limit $\lim_{n\to\infty }u_n(x)=v(x),$ $v:A\to \mathbb{C}$ is analytic and convergence is uniform on each compact $L\subset A.$
3) To pass to the limit in the equation $u_n(x)=U(x,u_{n-1}(x))\cdot x$ it remains to show that $v(A)\subset B.$
If $v$ is constant, then $v(x)=v(0)=0\in B.$ Assume $v$ is non-constant.
Let $a\in A$. As $v$ is non-constant, there exists $r>0$ such that $v(z)\ne v(a)$ for $|z-a|=r$. It remains to apply the Rouche's theorem to functions $$ f_n(z)=u_n(z)-v(a), \ g(z)=v(z)-v(a) $$ to deduce that for some $n$ and some $z$ in a neighbourhood of zero one has $$ v(a)=u_n(z)\in u_n(A)\subset B. $$
So, iterates $u_n$ converge uniformly on compact sets to an analytic function $v:A\to B$ which is a unique analytic solution of the equation $U(x,v(x))x=v(x)$ (uniqueness follows from estimates of step 1).
Question. In the case $p>1$ I can't show inclusion $v(A)\subset \bar{B}.$ Is it possible that a limit of a sequence $u_n:A\to B$ take values in $\bar{B}\setminus B?$