Given a function $f:\mathbb{R}\to\mathbb{R}$ differentiable and strictly decreasing such that $\displaystyle \lim_{x\to\infty}f(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,\infty)$. My guess is that the statement is not true but I can't find a counterexample.


Edit: added additional term to make the function actually decreasing. The idea remains the same.

Consider something like $$\frac1x + \frac{\sin(x)}{x^2}+ \frac{4 \sin^2(x/2)}{x^3}.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.

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    $\begingroup$ Is this decreasing for all reals? $\endgroup$ – marty cohen Mar 24 '19 at 15:21
  • $\begingroup$ @martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable $\endgroup$ – RiaD Mar 24 '19 at 16:37
  • $\begingroup$ @martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment. $\endgroup$ – WimC Mar 24 '19 at 18:29

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