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

Question

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.

Please help me to clear this.

Thank You. :)

Have you studied any theorems related to fix point of functions. — Mhenni Benghorbal, Dec 09 '13 at 03:18
I only know that continuous bounded function on compact set has a fixed point.. — , Dec 09 '13 at 04:51

score 10 · Accepted Answer · answered Dec 09 '13 at 02:30

10

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!

answered Dec 09 '13 at 02:30

Dan

7,951

I am sorry.. I did not recognize counterexample for fourth option in what i have done... could you please explain a bit more. – Dec 09 '13 at 02:38
@PraphullaKoushik: Sure! Does $\lim_{x\to\infty}\sin(x^3)$ exist? – Dan Dec 09 '13 at 02:40
This is quite interesting... :) I do not have limit at $\infty$ for $\sin (x^3)$.. This is very beautiful.. :) – Dec 09 '13 at 02:42

score 5 · Answer 2 · answered Dec 09 '13 at 02:37

5

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

answered Dec 09 '13 at 02:37

Eric Auld

28,127

yes yes.. $\tan^{-1}x$ is increasing but i do not know if it is bounded :( – Dec 09 '13 at 02:41
3

@PraphullaKoushik It is. $-\pi/2 < \tan^{-1}(x) < \pi/2$ – Eric Auld Dec 09 '13 at 02:42
Oh my bad... I got it... I am sorry for that dumb question... I was thinking of something else... Thank you so much... – Dec 09 '13 at 02:43
@PraphullaKoushik No problem, it happens to everyone. – Eric Auld Dec 09 '13 at 02:44

score 1 · Answer 3 · answered Dec 09 '13 at 02:29

1

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

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

answered Dec 09 '13 at 02:29

Igor Rivin

25,994
1
19
40

I am sorry, I could not understand your idea... please explain a bit more. – Dec 09 '13 at 02:31
Actually, I misread your question, you had already done this part. – Igor Rivin Dec 09 '13 at 02:37
It is alright... Thank you for your interest.. :) – Dec 09 '13 at 02:44

score 0 · Answer 4 · answered Dec 09 '13 at 02:32

0

Here is an incredibly non-interesting trivial example: $f(x)=a$ for $a$ being some real number.

answered Dec 09 '13 at 02:32

Hayden

16,737

I do not really understand this is for what? please explain a bit more.. – Dec 09 '13 at 02:33
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 09 '13 at 02:35
Opps, upon reading the question further, I realize I misread. I thought you wanted an example of something that had all those properties. – Hayden Dec 09 '13 at 02:36
It is alright... Thank you :) – Dec 09 '13 at 02:37

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

4 Answers4