Fixed point of continuous function

Question

Given a continuous function $f : [0; +\infty) \rightarrow [0; +\infty)$ such that $$\lim_{x \to \infty}{\frac{f(x)}x}=\frac{12}{13},$$ I have to prove that this function has a fixed point.

I noticed that $\frac{f(x)}{x} \le \frac{12}{13}$ so $f(x) \le \frac{12x}{13}$, but what to do next?

Answer 1

If this $f(x)\leq {12\over 13}x $ is true then $f(x)<x$ and you can use this:

Limit of sequence $a_0=-1$, $a_{n+1}=f(a_n)$

Answer 2

• If $f(0)=0$ then $0$ is a fixed point and we are done.

• If $f(0)=a>0$ then by continuity for $\varepsilon=\frac a2,\exists \ 0<\delta<\frac a2\ \text{ st. }\ |x|\le\delta\implies |f(x)-a|<\varepsilon$

Considering both $x$ and $f(x)$ are positive quantities we can get rid of absolute values and get

$f(x)>\frac a2$ and $x<\frac a2\ $ so $\ |x|\le\delta\implies\dfrac{f(x)}{x}>1$

Similarly in infinity since $\lim\limits_{x\to+\infty}\dfrac{f(x)}{x}=\frac{12}{13}<1$ then by continuity of $f$ we can find $M>\delta$ such that $x\ge M\implies \dfrac{f(x)}{x}<1$

We can now apply IVT to the continuous function $\dfrac{f(x)}{x}$ in interval $[\delta,M]$ and there is a point $x_0$ in this interval such that $\dfrac{f(x_0)}{x_0}=1\iff f(x_0)=x_0$

Answer 3

$$\lim_{x \to \infty}{\frac{f(x)}{x}}=\frac{12}{13}$$ means that $\forall \varepsilon>0$ there exists $M_\varepsilon\in\mathbb{R}$ such that if $x>M_\varepsilon$ then $$-\varepsilon <\frac{f(x)}{x}-\frac{12}{13}<\varepsilon$$ that is $$\frac{12}{13}-\varepsilon <\frac{f(x)}{x}<\frac{12}{13}+\varepsilon$$ and then $$x\left(\frac{12}{13}-\varepsilon\right) <f(x)<x\left(\frac{12}{13}+\varepsilon\right)$$ Choose $\varepsilon=\frac{2}{3}$. We have for $x>M_{\frac23}$ $$\frac{10}{13}\,x <f(x)<\frac{14}{13}\,x$$ as $f$ is continuous, for the intermediate values theorem, there exists at least one value $\bar x$ such that $f(\bar x)=\bar x$