Continuous function related exercise.

Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a positive and continuous function so that $\lim_{x \to +\infty} \frac{f(x)}{x}= \lambda <1.$

I must prove that: $\exists c \in \mathbb{R}, f(c)=c$.

To do so, we can consider the function $g$ defined by $\forall x \neq 0, g(x)=\frac{f(x)-x}{x}$.

Clearly, $\lim_{x \to +\infty}g(x)=\lambda - 1 <0$.

However I don't really know what to do next. My goal is to prove that $\exists a \neq 0, g(a)\leq 0 $ and $\exists b \neq 0 , g(b) \leq 0$.

Could anyone help me with that plz?

As you've currently posed it, it's not necessarily true. Maybe you're missing another condition on $f?$ — Brian Moehring, Aug 31 '19 at 23:41
Then consider $\frac{f(x)}x$. If $f(0) = 0$, then you have your fixed point. If not, how does $\frac{f(x)}x$ behave at zero? — amsmath, Aug 31 '19 at 23:44

Feng · Accepted Answer · 2019-09-01T00:48:29.070

1

We can prove it by contradiction.

Suppose that there is no $c$ such that $f(c)=c$. Since $f(x)-x$ is continuous, and $f(-1)-(-1)\geq 0+1>0$, we must have $$f(x)>x\ \text{ for all }x\in \mathbb R.$$ In this case, $\lim_{x\to+\infty}\frac{f(x)}x\geq 1$, which is a contradiction.

edited Sep 01 '19 at 00:48

answered Aug 31 '19 at 23:53

Feng

13,705

+1. You are brilliaint! – Kavi Rama Murthy Aug 31 '19 at 23:55
@KaviRamaMurthy Thanks! : ) – Feng Aug 31 '19 at 23:56
Hmmm I just don't understand why do u study f(-1)-(-1). What's the link with the continuous part? – Pablito Sep 01 '19 at 00:08
@BesoinClément Let $g(x)=f(x)-x$, then $g$ is continuous. We've assumed that $g$ can't attach $0$, so $g$ must be always positive or always negative. I studied $g(-1)$ to clarify that $g$ is always positive. – Feng Sep 01 '19 at 00:11
Oh yes nevermind really well done thanks! – Pablito Sep 01 '19 at 00:13
How do you know that $g(x)=f(x)-x$ is positive when $x>0$? – Axion004 Sep 02 '19 at 05:14
@Axion004 I’ve proven $g(-1)>0$. If there is $x_0>0$ such that $g(x_0)\leq 0$, the continuity of $g$ implies that there exists $y\in(-1,x_0]$ such that $g(y)=0$, contradicting to my hypothesis. – Feng Sep 02 '19 at 05:52

score 0 · Answer 2 · answered Sep 01 '19 at 02:36

A continuous function $f$ on $[a,b]$ must have at least one fixed point i.e. $f(c)=c$ if $f(x)\in [a,b]$.

Since $f$ must be continuous on the subset $[-M,M]$ of $\mathbb R$. It remains to prove that $f(x)\in [-M,M]$.

Let $lim_{M\rightarrow+\infty} f(M)=K$ where $K< M$ as $lim_{x\rightarrow+\infty}\frac{f(x)}{x}<1$. It proves that $f(x)\in(0,K)\subset [-M,M]$ which necessitates the existence of some $c\in \mathbb R$ s.t. $f(c)=c$.

Continuous function related exercise.

2 Answers2