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If a polynomial $p(z)$ of degree $n$ with zeros $z_1,z_2,\cdots,z_n$ assumes maximum at $w$ on $|z|=1.$ Prove or disprove that the Harmonic mean of $|z_k-w|,$ $k=1,2,\cdots,n$ is greater or equal to $1.$ That is $$ \dfrac{n}{\frac{1}{|z_1-w|}+\frac{1}{|z_2-w|}+\cdots+\frac{1}{|z_n-w|}} \geq 1.$$

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  • $\begingroup$ What evidence do you have for the truth or otherwise of this claim? $\endgroup$
    – Yemon Choi
    Mar 23, 2015 at 14:10
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    $\begingroup$ I am working on a result which suggests the truth of this statement. $\endgroup$
    – Suhail
    Mar 23, 2015 at 14:17
  • $\begingroup$ So which cases can you do? Can you explain to readers who might be interested in this problem, where the difficulty lies or what is already known? $\endgroup$
    – Yemon Choi
    Mar 23, 2015 at 15:27
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    $\begingroup$ For example by Bernstein's Inequality (which states for polynomial $p(z)$ of degree $n,$ $\max_{|z|=1}|p^{\prime}(z)|\leq n\max_{|z|=1}|p(z)|$) it easily follows $\dfrac{n}{\left|\frac{1}{w-z_1}+\frac{1}{w-z_2}+\cdots+\frac{1}{w-z_n}\right|} \geq 1 .$ I am doing Bernstein's Inequality in different form, which suggests the above inequality. $\endgroup$
    – Suhail
    Mar 23, 2015 at 15:43

1 Answer 1

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This is false. Consider a polynomial whose roots are uniformly distributed on the circle of radius $r<1$. If we now send $n\to\infty$, then we obtain the elliptic integral $$ \lim_{n\to\infty} \frac{1}{n} \sum \frac{1}{|w-z_j|} = \frac{1}{2\pi} \int_0^{2\pi} \frac{dt}{|w-re^{it}|} = \frac{1}{2\pi} \int_0^{2\pi} \frac{dt}{\sqrt{ 1+r^2-2r\cos t}} . $$ In the last step, I set $w=1$ (by symmetry, it really doesn't matter any more where $w$ lies).

This would have to be $\le 1$, but for $r\ll 1$, we have that $$ \frac{1}{\sqrt{ 1+r^2-2r\cos t}} = 1 + r\cos t - \frac{r^2}{2} + \frac{3}{2}r^2\cos^2 t +O(r^3) , $$ which after averaging becomes $1+r^2/4>1$. (We really also expect things to go wrong near $r=1$, but the integral is easier to analyze for small $r$.)

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  • $\begingroup$ @ Reming; Solution looks fine, but I am little confused. Does the limiting case preserve the properties of polynomial? For example, each of the polynomial $1+\frac{z}{1}+\frac{z^2}{2!}+\cdots+\dfrac{z^n}{n!}$ have $n$ zeros, but $\lim_{n\rightarrow \infty}1+\frac{z}{1}+\frac{z^2}{2!}+\cdots+\dfrac{z^n}{n!}$ does not have any zero. $\endgroup$
    – Suhail
    Mar 24, 2015 at 2:15
  • $\begingroup$ @Suhail: I simply pass to the limit in $(1/n)\sum 1/|w-z_j|\le 1$, I don't need to make direct use of the functions the $z_j$ came from. $\endgroup$ Mar 24, 2015 at 2:20
  • $\begingroup$ Oh yes, I got it. Thanks for the answer. $\endgroup$
    – Suhail
    Mar 24, 2015 at 2:23
  • $\begingroup$ There is one detail that is a tiny little bit sloppy in my presentation: $w=w_n$ could in principle move around, but it's clear that the convergence of the Riemann sums is uniform in $w$, so this is not an issue. $\endgroup$ Mar 24, 2015 at 2:24

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