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This one should be easy, however for some reason I can't find an easy way to solve it. So if f is a $C^1$ function over $\mathbb{R}$ that is decreasing and positive (so converging to some value, let say 0 since it does not really matter). Can we know for sure that $f'$ will be converging to 0? I know this is not true if f is not decreasing or increasing.

Thank for any hint

All the best T.

Tochoka
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    The answer is no, you might even assume that $f$ is smooth. See this http://math.stackexchange.com/questions/788813/the-limit-of-the-derivative-of-an-increasing-and-bounded-function-is-always-0/788818#788818 – Gabriel Romon Feb 28 '16 at 18:17
  • Thanks, what if $xf(x)$ also converges to 0? – Tochoka Feb 28 '16 at 18:22
  • Is your limit point $\infty$? If so, then if $f(x)$ does not converge to zero what makes you think that $xf(x)$ can converge to zero? – Beni Bogosel Feb 28 '16 at 18:23
  • No of course, $f$ must converge to 0 then – Tochoka Feb 28 '16 at 18:24

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It is not true, here is counterexample. The idea is that a function can have a derivative of any size for a short enough interval that the net contribution to the value is small. Drawing a little picture would help.

Let $f_n(x) = {2 \over \pi}\arctan (n^3x)$ and note that $|f_n(x)| \le 1$ for all $x $,$f_n$ is increasing, and $f'_n(0) = {2 \over \pi} n^3$.

Define $\phi(x) = \sum_{n=1}^\infty {1 \over n^2} f_n(x-n)$. Then $\phi$ is smooth, increasing, $\phi(0) = 0$ and $\bar{\phi}=\lim_{x \to \infty} \phi(x) = \sum_{n=1}^\infty {1 \over n^2}$. Furthermore, $\phi'(n) > {2 \over \pi} n$.

Now let $f(x) = \bar{\phi}-\phi(x)$ to get the desired function.

copper.hat
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  • I see, that was I was affraid of. But if we impose that $f$ converge to 0 "rather fast", for example $f(x)x$ also converge to 0, can we rule out all the counter example? – Tochoka Feb 29 '16 at 11:04