# Amann/Escher, Analysis I, Exercise I.10.5: fixed point of an increasing function

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 $$a, $$f(a)>a$$ and $$f(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, $$f(z) = z$$.

• Seems like a neat proof to me. – Kavi Rama Murthy Jun 26 at 9:27
• Thank you so much @KaviRamaMurthy ;) – MadnessFor MATH Jun 26 at 9:31

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, $$a \le z. To sum up, $$a< f(z) < b$$.