# Continuous bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$

Question is to check which of the following holds (only one option is correct) for a continuous bounded function $f:\mathbb{R}\rightarrow \mathbb{R}$.

• $f$ has to be uniformly continuous.
• there exists a $x\in \mathbb{R}$ such that $f(x)=x$.
• $f$ can not be increasing.
• $\lim_{x\rightarrow \infty}f(x)$ exists.

What all i have done is :

• $f(x)=\sin(x^3)$ is a continuous function which is bounded by $1$ which is not uniformly continuous.
• suppose $f$ is bounded by $M>0$ then restrict $f: [-M,M]\rightarrow [-M,M]$ this function is bounded ad continuous so has fixed point.
• I could not say much about the third option "$f$ can not be increasing". I think this is also true as for an increasing function $f$ can not be bounded but i am not sure.
• I also believe that $\lim_{x\rightarrow \infty}f(x)$ exists as $f$ is bounded it should have limit at infinity.But then I feel the function can be so fluctuating so limit need not exists. I am not so sure.

So, I am sure second option is correct and fourth option may probably wrong but i am not so sure about third option.

Thank You. :)

• Have you studied any theorems related to fix point of functions. – Mhenni Benghorbal Dec 9 '13 at 3:18
• I only know that continuous bounded function on compact set has a fixed point.. – user87543 Dec 9 '13 at 4:51

For the third point, consider $f(x) = \arctan(x)$. For the fourth point, you've already found a counterexample in one of your other points!

• I am sorry.. I did not recognize counterexample for fourth option in what i have done... could you please explain a bit more. – user87543 Dec 9 '13 at 2:38
• @PraphullaKoushik: Sure! Does $\lim_{x\to\infty}\sin(x^3)$ exist? – Dan Dec 9 '13 at 2:40
• This is quite interesting... :) I do not have limit at $\infty$ for $\sin (x^3)$.. This is very beautiful.. :) – user87543 Dec 9 '13 at 2:42

$\tan^{-1}x$ is increasing. $\sin (x^3)$ has no limit at infinity.

• yes yes.. $\tan^{-1}x$ is increasing but i do not know if it is bounded :( – user87543 Dec 9 '13 at 2:41
• @PraphullaKoushik It is. $-\pi/2 < \tan^{-1}(x) < \pi/2$ – Eric Auld Dec 9 '13 at 2:42
• Oh my bad... I got it... I am sorry for that dumb question... I was thinking of something else... Thank you so much... – user87543 Dec 9 '13 at 2:43
• @PraphullaKoushik No problem, it happens to everyone. – Eric Auld Dec 9 '13 at 2:44

Well, for $x$ really, really large, what can you say about $f(x) - x?$

For $x$ really, really small, what can you say about $f(x) - x?$

• I am sorry, I could not understand your idea... please explain a bit more. – user87543 Dec 9 '13 at 2:31
Here is an incredibly non-interesting trivial example: $f(x)=a$ for $a$ being some real number.
• This is a uniformly continuous, non-increasing function that has a fixed point (namely, $x=a$) with the property that the limit at infinity exists. – Hayden Dec 9 '13 at 2:35