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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.

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    (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.)2012-12-10

5 Answers 5

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(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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Hints:

  • $f(x) = x$,

  • $f(x) = \sin(x)$,

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

  • $f(x) = 0$.

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    th$a$nks a lot. I have got it.2012-12-10
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HINT: Think about $\ln x$ and $\sin x$ as examples.

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    @Pete: You are correct, but it's a good start (and we are interested in the positive infinity anyway). [And thank you!]2012-12-10
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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).

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

And verify (exercise) which conditions it satisfies.