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?

  • $\begingroup$ "finite dimensional comapct and covex [sic] hilbert spaces", i.e. you're working on Euclidean space, right? $\endgroup$ – Jaap Eldering Jun 29 '15 at 15:55
  • $\begingroup$ oh sorry, i edited the corresponding part of the question $\endgroup$ – Rufio Jun 29 '15 at 17:22
  • $\begingroup$ Have you looked at the properties of the corresponding iteration the product space? $\endgroup$ – Dirk Aug 29 '15 at 11:45

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.


If I understand you correctly, this follows straightforwardly from the Fiber Contraction Theorem due to Hirsch and Pugh, "Stable manifolds and hyperbolic sets" (1970). The original article is not available to me, but its statement and proof can also be found in Vanderbauwhede "Centre manifolds, normal forms and elementary bifurcations" (1989) page 105, or in appendix D of my book on normal hyperbolicity.

  • 2
    $\begingroup$ As far as i have understood, the fiber contraction theorem only works for contraction mappings and not for nonexpansive mappings. But the hint that such mappings are denoted fiber mappings is very useful, thanks. $\endgroup$ – Rufio Jun 29 '15 at 17:24

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