How can I prove that a continuous decreasing function on $\Bbb R$ has a fixed point?

Question

I am trying to prove that a continuous decreasing function $f: \Bbb R → \Bbb R$ has a fixed point.

I tried to use the function $g(x) = f(x) - x$, which should be a decreasing one, but I don't know how to obtain a point of $g$ that is $0$.

Answer 1

$g(x) \geq f(0)-x$ for $x<0$ so $g(x) \to \infty$ as $x \to -\infty$. Similarly, $g(x) \leq f(0)-x \to -\infty$ as $x \to \infty$. By IVP we get $g(x)=x$ for some $x$.

Answer 2

Introducing the fhe function $g:x\mapsto f(x)-x$ is not necessary. Instead, consider $a=f(0)$.

• If $a=0$, we are done.
• If $a>0$, note that $f(a)\leqslant f(0)=a$ since $a>0$ and $f$ is nonincreasing on $[0,a]$. Thus, $f(0)>0$ and $f(a)\leqslant a$. By the intermediate value theorem, there exists some $x$ in $(0,a]$ such that $f(x)=x$.
• If $a<0$, note that $f(a)\geqslant f(0)=a$ since $a<0$ and $f$ is nonincreasing on $[a,0]$. Thus, $f(0)<0$ and $f(a)\geqslant a$. By the intermediate value theorem, there exists some $x$ in $[a,0)$ such that $f(x)=x$.