Fixed points for increasing function

Question

Let $f(x):[a,b] \to \mathbb{R}$ a real-valued function that is strictly increasing in $x$. Further, $f(b)>b$. I would like to show that the function has at most two fixed points; is that true? Graphing the function, I find three cases:

1. A single fixed point if $f(a)<a$
2. Either zero or two fixed points if $f(a)>a$.

Any ideas of how I can prove this formally? Thanks!

Answer 1

This is clearly not true. As an example, take $f$ over $[0,2]$ to be given by $$ f(x) = \begin{cases} x & x<1,\\ 2x-1 & x\geq 1. \end{cases} $$ The function $f$ satisfies the necessary conditions but has infinitely many fixed points.

If you prefer a smooth example or an example with finitely many fixed points, you could also take a function of the form $f(x) = x + a\sin(bx)$ for suitable values of $a$ and $b$. For example, here is the graph of one such function over $[0,2\pi]$.

Answer 2

Building on the other answers. Here is a differentiable one: $$ f(x) = x + \sin \left[\frac{k \pi(2x - a - b)}{2(b-a)} \right]. $$ Let $k$ vary as much as you want to get any number of zeros you'd like (ensure that $\sin(\dots)$ remains positive though!).

Answer 3

As other already said the proposition is not true. Under which assumptions could this be true? Consider if $x,y$ are two fixed points, so $f(x)=x$, $f(y)=y$. Then this proposition is essentially equivalent to $f$ not intersecting the diagonal in a third point.

This would for example be true if $f$ was a strictly convex or concave function, which is probably what you had in mind when graphing your functions.

In this case your cases are almost correct assuming $f$ is continuous:

If $f(a)<a$ and $f(b) > b$ then by continuity you get at least one intersection $x$ such that $f(x) = x$. Assume $f$ is strictly convex. Then no point in $[a,x]$ can be a fixed point. But if you get another fixed point $y > x$ then no point in $[a,y]$ can be a fixed point, so $x$ is the only one (essentially any fixed point must be the smallest fixed point). Strictly concave goes the same way, just the other direction.

If on the other hand $f(a) > a$ and $f$ is strictly concave then there cannot be such a point. If $f$ is strictly concave all three $0,1$ or $2$ solutions are possible. If $f(a) = a$ $f$ can have 1 or 2 fixed points.