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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.

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

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  • $\begingroup$ nice answer, thanks :-) $\endgroup$ – idm Aug 16 '15 at 18:28
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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 $y=x$.

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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.

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