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.
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 $y=x$.
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.
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.