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Dec 9, 2016 at 15:06 comment added anonymous @IlyaBogdanov Oh, indeed. The combination of negation and changes of direction was just a bit too much for me, it seems. What we want is that with decreasing distance, $\alpha$ also decreases. Which means that $\alpha$ is increasing; in other words (since we don't need it to be strictly increasing), non-decreasing. And we cannot generally make it so if it converges to 1 when distances go to zero.
Dec 9, 2016 at 14:58 comment added Ilya Bogdanov @anonymous If $\alpha$ is non-increasing, then $\alpha(|x_{n+1}-x_n|)\geq \alpha(|x_n-x_{n-1}|)\geq \dots\geq \alpha(|x_1-x_0|)$. So $|x_{n+1}-x_n|/|x_1-x_0|$ can be estimated just as a product of these values --- what next?
Dec 9, 2016 at 14:56 comment added anonymous @IlyaBogdanov I think the argument does work the way it is. Can you be more precise about where you see a problem?
Dec 9, 2016 at 14:46 comment added Ilya Bogdanov @AlexandreEremenko: I do not think this relaxation makes much difference. If $\alpha(d)$ is close to 1, AND if for an extremal pair of points $x$, $y$ there exists "almose" midpoint $z$, then $\alpha$ of around $d/2$ is also close to 1. If there is no such midpoint, then we will never come to such distant pairs of points.
Dec 9, 2016 at 14:10 comment added Alexandre Eremenko @Ilya Bogdanov: Yes. I propose to relax his condition: $\alpha(t_n)\to 1$ implies $t_j\to 0$ or $t_n\to\infty$.
Dec 9, 2016 at 11:54 comment added anonymous @IlyaBogdanov you're right, that's what I meant. We can make it non-increasing by assumption.
Dec 9, 2016 at 7:22 comment added Ilya Bogdanov @anonymous: we can make $\alpha$ to be non-increasing, not non-decreasing.
Dec 9, 2016 at 7:21 comment added Ilya Bogdanov I suppose this is also prohibited: the only sequences mapped to convergent to 1 are those convergent to 0. In this case, $\alpha$ may be done decreasing a in my answer.
Dec 9, 2016 at 4:27 comment added Alexandre Eremenko @Ilya Bogdanov: I finally got the point:-) But what if $\alpha$ is neither increasing nor decreasing but $\alpha(x)\to 1$ both for $x\to+\infty$ and for $x\to 0$?
Dec 9, 2016 at 3:27 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 8, 2016 at 21:37 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 8, 2016 at 20:27 comment added Ilya Bogdanov But $\alpha$ may be monotonically decreasing, not increasing...
Dec 8, 2016 at 20:03 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 8, 2016 at 18:27 comment added Isra El Thank you Alexandre Eremenko ,but i don't agree with you ,because $k$ depent to $n$ .
Dec 8, 2016 at 6:51 comment added Ilya Bogdanov The crucial inequality (with $k^n$) does not seem to be correct --- or at least proved.
Dec 8, 2016 at 2:14 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 7, 2016 at 22:44 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 7, 2016 at 22:11 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 7, 2016 at 22:06 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 7, 2016 at 21:57 history undeleted Alexandre Eremenko
Dec 7, 2016 at 21:53 history deleted Alexandre Eremenko via Vote
Dec 7, 2016 at 21:53 history edited Alexandre Eremenko CC BY-SA 3.0
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Dec 7, 2016 at 21:48 comment added Isra El I think we can sove this probelm without continuity of $\alpha$ .Thank you
Dec 7, 2016 at 21:42 history answered Alexandre Eremenko CC BY-SA 3.0
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