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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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Is $\mathcal{A}$ assumed to be linear? – William Jan 13 '12 at 18:04
@WNY no, $\mathscr A$ only assumed to be monotone – Ilya Jan 13 '12 at 18:10
Maybe I misunderstand the problem, but what about $X=\mathbb R_+$, $Af(x)=\sup_{0\leq t\leq x}f(t)$. $A$ is monotone, $Af\geq f$ for all $f$ and $A^2f=Af$ so $f_n=A^nf=Af$ is converging, but we don't need to have $Af=f$. – Davide Giraudo Jan 13 '12 at 18:36
@Davide: You’ve misunderstood: Ilya’s asking whether the limit function has to be a fixed point. In your example the limit is $Af$, and indeed it’s true that $Af=A(Af)$. – Brian M. Scott Jan 13 '12 at 18:54
@BrianM.Scott You are right. Failed attempt... – Davide Giraudo Jan 13 '12 at 18:56
up vote 5 down vote accepted

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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Perfect answer! Thank you very much, Jonas. By the way, to make the operator monotone on the whole $\mathbb R$ I guess we should add something like $\mathscr A(x) = -1$ for $x< 0$. – Ilya Jan 14 '12 at 11:47
@Ilya: It already is monotone on $\mathbb R$. That's why I used $\sqrt[3]{x}$ instead of $\sqrt x$. – Jonas Meyer Jan 14 '12 at 17:45
isn't it that $\sqrt[3]{-1/3}<-1/3$? – Ilya Jan 15 '12 at 18:04
@Ilya: Increasing means $x\leq y\implies Ax\leq Ay$, not $\forall x,Ax\geq x$. It is true that $x\leq y\Longleftrightarrow x^3\leq y^3$. – Jonas Meyer Jan 15 '12 at 18:12
Oh, I mixed it up again. You're right, sorry – Ilya Jan 15 '12 at 18:57

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