Prove that $f$ has a fixed point.

Let $f:[0,\infty [\to[0,\infty [$ continuous such that $$\lim_{t\to\infty }\frac{f(t)}{t}=\ell\in[0,1).$$

Prove that $f$ has a fixed point, i.e. there is an $x\geq 0$ such that $f(x)=x$.

I don't really know how to solve this problem. My first intension was to use Brouwer, but it's only useable on a compact. After I tried by induction but with no success.

• This doesn't need to be true. For instance: $$f(x)=\begin{cases}3 & \text{if }x\in[0,1],\\\frac{1}{2}(x-1) & \text{else}\end{cases}$$ – Nick Peterson Aug 16 '15 at 16:03
• I guess that $f$ must be continuous, otherwise it is trivial. – Siminore Aug 16 '15 at 16:04
• yyes it's continuous, sorry :-) – idm Aug 16 '15 at 16:10
• And $\ell\in[0,1)$ is required – user261263 Aug 16 '15 at 16:11
• I know, it's already said :-) – idm Aug 16 '15 at 16:12

Consider the function $g(t)=f(t)-t$. Since $f$ is nonnegative, we must have $g(0) \ge 0$. If $g(0)=0$, we are done. Otherwise, we have $\lim_{t \to \infty} \frac{g(t)}{t}=\ell-1<0,$ so $g(t)<0$ for sufficiently large $t$. It immediately follows by the Intermediate Value Theorem that $g$ has a positive root.
I am adding the requirement that $f$ be continuous, otherwise the conclusion is trivially false.
Hint: draw $y=x$ and $y=\ell x$, where $0 \leq \ell < 1$. If $f(0)=0$, there is nothing to prove. If $f(0)>0$, then try to convince yourself that sooner or later $f(x)$ must lie below $y=x$, and conclude by continuity that $y=f(x)$ must intersect $y=x$.
Let us suppose $f(x) \neq $$x \forall x>=0. We have f(0) > 0 and because f continuous: 1) f(x) > x \forall x>0 or 2) f(x) < x \forall x>0. (Because the continuous function g(x) = f(x) - x is \neq0 \forall x so cannot change the sign) 1) Suppose f(x) > x \forall x>0. Then \frac{f(t)}{t} > 1 and$$ \lim_{t\to\infty }\frac{f(t)}{t} >= 1 $$impossible because \ell \in [0,1) 2) Suppose f(x) < x \forall x>0. Then:$$ \lim_{t\to 0 }f(t) <= 0 $$But f continuous:$$ \lim_{t\to 0 }f(t) = f(0)$$So$f(0) <= 0\$ absurd.