Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and that its image $f(\mathbb{R})$ is bounded. Prove there is a solution to the equation $f(x)=x$ for some $x\in \mathbb{R}$.
We want to find a solution to the equation $f(x)-x=0$ which we can do using the intermediate value theorem, if we can show
$\hspace{150pt} f(a)-a<0<f(b)-b$.
Since $f(\mathbb{R})$ is bounded, then $\exists M\in \mathbb{R}$ such that $|f(x)|\le M$ for all $x\in \mathbb{R}$.
$\hspace{160pt} -M\le f(x) \le M$
$\hspace{140pt} -M-x\le f(x) -x\le M-x$
I'm not sure what I can do after this though.
Best Answer
Choose $a < -M$; then
$$-M - a \le f(a) - a$$
But by our choice of $a$, it's necessary that $-M - a > 0$, so we have
$$0 < f(a) - a$$
Likewise choose a $b > M$, so that
$$f(b) - b \le M - b < 0$$
Now apply the intermediate value theorem.