# Does my fixed point proof hold?

I have am looking for existence of a fixed point for an operator that I have. I already looked at some related fixed point theorems such as Schrauder's and Rothe's. But most of them seem to require that the space that my operator maps to (and from) is compact. A property which I unfortunately don't have. Now I tried working around this, please let me know if my result is correct or if there is a fixed point theorem that I can apply. Any feedback is welcome.

The givens: Let $L = \{ f : 0 \leq f(x) \leq B \}$ for some $B>0$. Here $f : X \mapsto X$ and $X$ is some compact interval in $\mathbb{R}$. The operator $T$ maps $L \mapsto L$ and furthermore it is monotone, ie. if $f(x) \leq g(x) \ \forall x$, then $Tf(x) \leq Tg(x) \ \forall x$.

Attempt at proof of existence fixed point: I iterate on this operator by starting the with "lowest" function in $L$ ($f = 0$) and argue with sequences in $\mathbb{R}$, as follows:

Start with $f_0 = 0$ and define $f_{n+1} = Tf_n$ recursively. For every fixed $x_0$, we have by monotonicity of $T$ that the sequence $(f_n(x_0))_{n\geq 0}$ is monotone. Furthermore, since the monotone sequence $(f_n(x_0)) \in L$ for all $n$ (closed and bounded), we know it converges to some $f(x_0)^*$. We can do this for every $x \in X$ and define our fixed point $f^*$ to be such that $f^*(x) = f(x)^*$ for all $x$.

Two questions: (1) Is this result correct? (2) Did I overlook a fixed point theorem that I could have easily applied here?

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I see several problems, but can start at the beginning: Your use of strict inequalities in the definition of “monotone” does not allow you to conclude that the sequence $(f_n)$ is monotone. – Harald Hanche-Olsen Sep 19 '13 at 13:57
I also don't see how you can get $Tf^*=f^*$. But here is a possible way to get a fixed point, provided monotonicity is defined using non-strict inequalities: Consider all functions $f$ with $Tf\ge f$. This class is non-empty, since the zero function belongs. Use Zorn's lemma to show it has a maximal element (for any totally ordered set of such functions, the supremum should be one too). I strongly suspect this will be a fixed point. – Harald Hanche-Olsen Sep 19 '13 at 14:15
Thank you, I see the problem. I'm making a small edit, my operator satisfies the "monotone" condition also for weak inequalities. – Bas van Heiningen Sep 19 '13 at 14:22
Thank you for your suggestion. I will try to work with that a bit. – Bas van Heiningen Sep 19 '13 at 14:25
Thanks Harald. It worked like a charm! – Bas van Heiningen Sep 19 '13 at 15:20