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Let $\mathscr X$ be a complete separable metric space and $\mathbb B$ be the Banach space of all real-valued bounded measurable functions on $\mathscr X$. The partial order on this space is introduced by $ f\leq g \text{ iff }f(x)\leq g(x)\text{ for all }x\in \mathscr X. $ The operator $\mathscr A:\mathbb B\to\mathbb B$ is called monotone if $f\leq g$ implies $\mathscr Af\leq \mathscr Ag$, such operator is not necessary linear. Let us consider the function $f_0\in \mathbb B$ such that $\mathscr Af_0\geq f_0$ and construct the sequence $f_{n+1} = \mathscr A f_n$. Clearly, for any fixed $x\in \mathscr X$ the limit $\lim\limits_{n}f_n(x)$ exists (though it may be infinite) and the convergence is monotone.

Let us assume that for any $x\in\mathscr X$ the limit is finite and denote it by $f(x)$. Is it true that $ f = \mathscr Af\quad? $

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    @BrianM.Scott $Y$ou are right. Failed attempt...2012-01-13

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Let the metric space have one point, and identify $\mathbb B$ with $\mathbb R$. Let $\mathscr A(x)=\sqrt[3]{x}$ if $x<1$, $\mathscr A(x)=2$ if $x\geq 1$. Let $f_0=\frac{1}{2}$. Then $\mathscr Af_0\geq f_0$, and $\lim\limits_{n\to\infty}\mathscr A^nf_0=1=f$, but $\mathscr Af=2$.

In general $\mathscr A f\geq f$ is true, but this example shows that the equality need not hold.

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    Oh, I mixed it up again. You're right, sorry2012-01-15