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I am doing Exercise I.10.5 from textbook Analysis I by Amann/Escher.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be an increasing function. Suppose that $a, b \in \mathbb{R}$ satisfy $a<b$, $f(a)>a$ and $f(b)<b$. Prove that $f$ has at least one fixed point, that is, there is some $z \in \mathbb{R}$ such that $f(z)=z$.

Could you please verify if my attempt contains logical gaps/errors?

My attempt:

Let $A :=\sup \{x \in [a,b] \mid x \leq f(x)\}$. We have $a \in A$ and $A$ is bounded. Then $z:= \sup A$ exists. We next prove that $f(z) = z$.

If $z < f(z)$ then $f(z) \le f(f(z))$ and thus $f(z) \in A$. So $f(z) \le z$, which is a contradiction.

If $f(z) < z$ then there exists $y \in A$ such that $f(z) < y \le z$. So $f(z) < y \le f(y) \le f(z)$. It follows that $f(z) < f(z)$, which is a contradiction.

As a result, $f(z) = z$.

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    $\begingroup$ Seems like a neat proof to me. $\endgroup$ – Kavi Rama Murthy Jun 26 at 9:27
  • $\begingroup$ Thank you so much @KaviRamaMurthy ;) $\endgroup$ – MadnessFor MATH Jun 26 at 9:31
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In case $z < f(z)$, I forget to prove that $f(z) \in [a,b]$.


Since $z \le b$, $f(z) \le f(b) < b$. On ther hand, $a \le z<f(z)$. To sum up, $a< f(z) < b$.

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