6

New! Save questions or answers and organize your favorite content.
Learn more.

$\begingroup$

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. :)

  • real-analysis

asked Dec 9, 2013 at 2:25

$\endgroup$

2

  • $\begingroup$ Have you studied any theorems related to fix point of functions. $\endgroup$

    Dec 9, 2013 at 3:18

  • $\begingroup$ I only know that continuous bounded function on compact set has a fixed point.. $\endgroup$

    –user87543

    Dec 9, 2013 at 4:51

4 Answers 4

$\begingroup$

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 9, 2013 at 2:30

Dan's user avatar

$\endgroup$

3

  • $\begingroup$ I am sorry.. I did not recognize counterexample for fourth option in what i have done... could you please explain a bit more. $\endgroup$

    –user87543

    Dec 9, 2013 at 2:38

  • $\begingroup$ @PraphullaKoushik: Sure! Does $\lim_{x\to\infty}\sin(x^3)$ exist? $\endgroup$

    Dec 9, 2013 at 2:40

  • $\begingroup$ This is quite interesting... :) I do not have limit at $\infty$ for $\sin (x^3)$.. This is very beautiful.. :) $\endgroup$

    –user87543

    Dec 9, 2013 at 2:42

$\begingroup$

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

answered Dec 9, 2013 at 2:37

Eric Auld's user avatar

$\endgroup$

4

  • $\begingroup$ yes yes.. $\tan^{-1}x$ is increasing but i do not know if it is bounded :( $\endgroup$

    –user87543

    Dec 9, 2013 at 2:41

  • $\begingroup$ @PraphullaKoushik It is. $-\pi/2 < \tan^{-1}(x) < \pi/2$ $\endgroup$

    Dec 9, 2013 at 2:42

  • $\begingroup$ Oh my bad... I got it... I am sorry for that dumb question... I was thinking of something else... Thank you so much... $\endgroup$

    –user87543

    Dec 9, 2013 at 2:43

  • $\begingroup$ @PraphullaKoushik No problem, it happens to everyone. $\endgroup$

    Dec 9, 2013 at 2:44

$\begingroup$

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 9, 2013 at 2:29

Igor Rivin's user avatar

$\endgroup$

3

  • $\begingroup$ I am sorry, I could not understand your idea... please explain a bit more. $\endgroup$

    –user87543

    Dec 9, 2013 at 2:31

  • $\begingroup$ Actually, I misread your question, you had already done this part. $\endgroup$

    Dec 9, 2013 at 2:37

  • $\begingroup$ It is alright... Thank you for your interest.. :) $\endgroup$

    –user87543

    Dec 9, 2013 at 2:44

$\begingroup$

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

answered Dec 9, 2013 at 2:32

Hayden's user avatar

$\endgroup$

4

  • $\begingroup$ I do not really understand this is for what? please explain a bit more.. $\endgroup$

    –user87543

    Dec 9, 2013 at 2:33

  • $\begingroup$ 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. $\endgroup$

    Dec 9, 2013 at 2:35

  • $\begingroup$ Opps, upon reading the question further, I realize I misread. I thought you wanted an example of something that had all those properties. $\endgroup$

    Dec 9, 2013 at 2:36

  • $\begingroup$ It is alright... Thank you :) $\endgroup$

    –user87543

    Dec 9, 2013 at 2:37