# Fixpoint of monotone operators

Let $X$ be some set and let $F$ be the set of all functions with a domain $X$ and a range $[0,1]$. We consider $F$ to be a partially ordered set with $f\leq g$ if and only if $f(x)\leq g(x)$ for all $x\in X$. $\newcommand{\i}{\mathscr{I}}$ $\newcommand{\k}{\mathscr{K}}$ $\newcommand{\i}{\mathscr{I}}$

Let $A$ be another set - to be thought of as a set of optimization parameters. Consider a family of operators $(\mathscr I_a)_{a\in A}$ acting on $F$ and assume that $\mathscr I_a$ is monotone for any $a\in A$, i.e. $$\mathscr I_a f\leq \mathscr I_a g$$ for all pairs $f\leq g$ and for all parameters $a$. Assume in addition that $\mathscr I_a$ commutes with limits of sequences, i.e. if $f = \lim\limits_n f_n$ then $\mathscr I_a f = \lim\limits_n \mathscr I_a f_n$.

Introduce two new operators: $$\mathscr K^+f:=\sup\limits_a \mathscr I_a f,\quad \mathscr K^-f:=\inf\limits_a \mathscr I_a f.$$ Clearly, both of them are monotone operators. I am given some function $f_0 = g_0\in F$ which initialize two sequences: $$f_n := (\k^+)^n f_0, \quad g_n := (\k^-)^n g_0.$$ In both cases it holds that $f_0\leq f_1$ and $g_0\leq g_1$. Clearly, by monotonicity arguments we have $$f_{n}\leq f_{n+1}, \quad g_{n}\leq g_{n+1}$$ for all $n \in \Bbb N$, so that the correspondent limits do exist: $$f^* := \lim\limits_n f_n\in F,\quad g^* := \lim\limits_n g_n\in F.$$ As a result, we have that $f^* = \sup\limits_n f_n$ and $g^* = \sup\limits_n g_n$. I want to characterize them as fixpoints of some operators.

1. For the first sequence it holds that: \begin{align} \k^+ f^* &= \sup\limits_{a\in A}\i_a \left[\lim_n f_n\right] \\ \\ &=\sup\limits_{a\in A}\lim\limits_n\i_a\left[f_n\right] \\ \\ &=\sup\limits_{a\in A}\sup\limits_n\i_a\left[f_n\right] \\ \\ &=\sup\limits_n\sup\limits_{a\in A}\i_a\left[f_n\right] \\ \\ &=\sup\limits_n (\k^+)^{n+1}f = \lim\limits_n f_{n+1} = f^*. \end{align} Thus, $f^*$ is the (least) fixpoint of the operator $\k^+$.

2. For the second sequence I would expect that $g^*$ is a fixpoint of $\k^-$, but I'm only able to derive till the following point: \begin{align} \k^- g^* &= \inf_{a\in A}\i_a \left[\lim_n g_n\right] \\ \\ &=\inf_{a\in A}\lim_n\i_a\left[g_n\right] \\ \\ &=\inf_{a\in A}\sup_n\i_a\left[g_n\right] \end{align} where as far as I know it's not possible to chance $\lim$ and $\sup$, or $\inf$ and $\sup$. So, I can only show that $\k^+ g^* = \lim\limits_n \k^+ g_n$ which does not seem to lead to some fixpoint characterization.

Would you help me to understand, is it possible to derive a fixpoint characterization for $g^*$? I will also be very grateful for the comments whether it is true what I've written, or there are mistakes.

P.S. I was not sure which tags should I put, so please fix them if needed. I put [measure-theory] since $\i_a$ are often integral operators.

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Why do $f^*$ and $g^*$ exist? – Mercy King Jul 11 '12 at 17:50
@Mercy: for each fixed $x$, the sequence $f_n(x)$ is non-decreasing and bounded from above. Thus, the point-wise limit always exists and coincides with a supremum – Ilya Jul 11 '12 at 19:24
This exactly what I do not understand! Why is $f_n(x)$ bounded from above? – Mercy King Jul 11 '12 at 20:05
@Mercy: because the class $F$ only contains function with a range $[0,1]$ – Ilya Jul 12 '12 at 9:14
Thanks, it's clear now! – Mercy King Jul 12 '12 at 9:21