TITLE: Convergence of Fixed-Point Iteration of a dependent map
QUESTION [6 upvotes]: Suppose that we have two mappings $T_1(\cdot): Y \mapsto Y$ and $T_2(\cdot,\cdot) : X \times Y \mapsto X$ where both $X$ and $Y$ are compact and convex subsets of the same Euclidean space. Furthermore $T_1(\cdot)$ is nonexpansive, i.e. has Lipschitz constant equal to $1$, $T_2(\cdot, y)$ is nonexpansive for all fixed $y \in Y$ and $T_2(x,\cdot)$ is uniformly Lipschitz continuous, i.e. there exists one Lipschitz constant for all $x \in X$ .
We know that for any $\lambda \in (0,1)$ and any $y_0 \in Y$ the iteration
$y_{n+1} = (1-\lambda) y_n + \lambda T_1(y_n)$ converges to a fixed point $\bar{y} = T_(\bar{y})$.
We also know that for any $\rho \in (0,1)$ and any $x_0 \in X$ and a fixed $y \in Y$ the iteration
$x_{n+1} = (1-\rho) x_n + \rho T_2(x_n, y)$ converges to a fixed point $\bar{x} = T_(\bar{x}, y)$.
My question is: Does the iteration
$x_{n+1} = (1-\rho) x_n + \rho T_2(x_n, y_n)$
converge to a fixed point $\bar{x} = T_2(\bar{x}, \bar{y})$?
Does anybody have a hint in proving or disproving it?

REPLY [2 votes]: Take $T_1(y)=y-y^2$ with $y\in[0,1]$ and $T_2(x,y)=e^{iy}x$, $x\in\mathbb C, |x|\le 1$. Now take $x_0=1$, $y_0=1/2$, say. Then all assumptions hold, but $y_n\approx c/n$, so the rotations in the iterations sum up to infinity like a harmonic series but the contractions of absolute value of $x$ multiply to a non-zero number like the product of $e^{-n^{-2}}$, and there is no convergence.
It looks like this is the only bad scenario in the sense that if you can somehow guarantee in addition that the sum $\sum_n|y_n-\bar y|$ is finite, or that the fixed point of $T_2(\cdot,\bar y)$ is unique, or something else that would prevent this ridiculous cycling over the set of the fixed points of the limiting mapping, then the desired conclusion should follow but, since I have no idea what exactly your setup is, I haven't tried to check the details, so I may be overly optimistic here.