Let $\,f : \mathbb{R} \rightarrow \mathbb{R}$ be a infinitely differentiable function that is increasing and bounded. Then is it true that $\lim_{x\to \infty}f'(x)=0$?
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No. Let $h\colon\mathbb R\to\mathbb R$ be a smooth function with $h(x)>0$ for $x\in(0,1)$ and $h(x)=0$ otherwise (a smooth "bump"). Then let $$g(x)=\sum_{n=1}^\infty h(n^2(x-n))=\begin{cases}h(n^2(x-n))&\text{if }n\le x<n+1, n\in\mathbb N\\0&\text{otherwise}\end{cases}$$ And let $f(x)=\int_0^xg(t)\,\mathrm dt$. Then
- $f$ is smooth because in each bounded interval only finitely summands are nonzero
- $\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}g(x)$ does not exist
- $f$ is increasing because $f'\ge 0$
- $f$ is bounded by $\sum_{n=1}^\infty\frac1{n^2}\int_0^1h(t)\,\mathrm dt<\infty$
answered May 10 '14 at 8:59
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Hint: One way to look at this is by the Fundamental theorem of Calculus:
$$f(b) - f(a) = \int_a^b f'(x) dx\ .$$
As the function $f$ is bounded we have
$$\int_a^\infty f'(x) dx <\infty$$
Does that imply that $f'$ tends to $0$?
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$\begingroup$ This alone does not seem that useful in answering the question (see Hagen von Eitzen's answer). $\endgroup$ – A.S May 10 '14 at 9:03
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$\begingroup$ @AndrewSalmon: Um... What do you mean by not useful? $\endgroup$ – user99914 May 10 '14 at 9:05
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$\begingroup$ How are you supposed to get a construction like Hagen von Eitzen's out of this? $\endgroup$ – A.S May 10 '14 at 19:30
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$\begingroup$ @AndrewSalmon: Note that Hagen von Eitzen's function is an explicit construction of a function I supposed in my answer. $\endgroup$ – user99914 May 10 '14 at 22:01
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Yes, if the limit at $\infty$ exists; if this is the case, then $f$ will go to $\infty$. Say the limit of $f'(x)$ as $x \rightarrow \infty=M>0$.
Then repeatedly apply the mean value theorem, so that , for a choice of an integer $n$ large-enough, $$f(n+1)-f(n)= M(n+1-n)=M$$ . This can be done for $n+1, n+2,..$ , so that $$Lim _{n\rightarrow \infty }f(n)=\Sigma_{k=n}^{\infty} [ f(k+1)-f(k)] \rightarrow \infty$$
