Let $f$ be a continuous function such that $f((a,b))\subset [a,b]$ show that there exists $c \in [a,b]$ such that $f(c)=c$

Question

Let $f$ be a continuous function on the interval $(a,b)$ such that $$f((a,b))\subset [a,b].$$ Show that, there exists $c \in [a,b]$ such that $f(c)=c$.

I thought about using the Intermediate Value Theorem in the function
$$\psi(x)=f(x)-x,\: \forall \: x\in [a,b],$$ but I stopped at the problem of $ f $ not being set to $ a $ and $ b $. I don't know if it's a problem to work around or not.

Answer 1

The statement is not true, here are some counterexamples for functions $ f:(a, b) \to [a, b]$ without fixed point:

• $f(x) = a$,
• $f(x) = a + \frac 12 (x-a)$,
• $f(x) = a + \frac{(x-a)^2}{b-a}$.

If $f$ has no fixed point in $(a, b)$ then either $$ a \le f(x) < x \quad \text{ for all } x \in (a, b) $$ or $$ x < f(x) \le b \quad \text{ for all } x \in (a, b) \, . $$ so that $$ \lim_{x\to a}f(x) = a \quad \text{ or } \quad \lim_{x\to b}f(x) = b \, . $$

It follows that $f$ has a fixed point, or $f$ can be continuously extended to (at least) one of the boundary points of the interval, and the extended function has a fixed point.