I came across a problem that says:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function. If $|f'|$ is bounded, then which of the following option(s) is/are true?

(a) The function $f$ is bounded.
(b) The limit $\lim_{x\to\infty}f(x)$ exists.
(c) The function $f$ is uniformly continuous.
(d) The set $\{x \mid f(x)=0\}$ is compact.

I am stuck on this problem. Please help. Thanks in advance for your time.

  • 4
    $\begingroup$ It's been a little while since I've been active on this site. Coming back with somewhat fresh eyes, I have to say that I'm disappointed that within four minutes of posting the question the OP got two full answers. Only Asaf's answer contained the magic word "think". To me when someone asks a question like this I have to make the assumption that they want to learn how to solve such questions, not just receive a canned answer. Are those who give complete answers within four minutes really trying to help the OP or just earn points for themselves? $\endgroup$ – Pete L. Clark Dec 10 '12 at 18:31
  • $\begingroup$ (Also, I had been composing an answer which was much more socratic, asking the OP to reflect on various things and to try to test out various familiar functions to see whether they had bounded derivative and/or satisfied the multiple choice conditions. But anyone who can see the extant answers could answer these questions without really thinking, so it seems that this kind of response is mostly ruined.) $\endgroup$ – Pete L. Clark Dec 10 '12 at 18:36

You can at least eliminate a), b) and d). Let $f_1(x)=x$. Then $f'(x)=1$ for all $x$ and $|f'|$ is therefore bounded. This eliminates a) and b). Now let $f_2(x)=0$. Then $f'(x)=0$ for all $x$ and $|f'|$ is bounded. This eliminates d).



  • $f(x) = x$,

  • $f(x) = \sin(x)$,

  • $f$ is Lipschitz continuous $\Rightarrow$ $f$ is uniformly continous,

  • $f(x) = 0$.

  • $\begingroup$ thanks a lot. I have got it. $\endgroup$ – learner Dec 10 '12 at 18:34

HINT: Think about $\ln x$ and $\sin x$ as examples.

  • $\begingroup$ $\ln x$ does not have a bounded derivative on all of $\mathbb{R}$. In fact it is not even defined on all of $\mathbb{R}$. Still, this answer is very valuable in that it leaves the OP with things to think about rather than presenting a complete solution. (+1) $\endgroup$ – Pete L. Clark Dec 10 '12 at 18:22
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    $\begingroup$ @Pete: You are correct, but it's a good start (and we are interested in the positive infinity anyway). [And thank you!] $\endgroup$ – Asaf Karagila Dec 10 '12 at 18:59

Consider the function $$f(x)=x\cos(\ln (x^2+1)^{1/2}).$$

And verify (exercise) which conditions it satisfies.


(a) & (b) are false: Consider $f(x)=x$ $\forall$ $x\in\mathbb R$;

(c) is true: $|f'|$ is bounded on $\mathbb R\implies f'$ is bounded on $\mathbb R$ [See: Related result];

(d) is false: $f=0$ on $\mathbb R\implies${$x:f(x)=0$} $=\mathbb R$.


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