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Let $p$ be a prime number, $\mathbb C_p$ be the completion of an algebraic closure of $\mathbb F_q\left(\left(\frac1T\right)\right)$ for the valution $v(x)=-\deg(x)$. Let $\sum_{n\ge0}a_nz^n$ be a power series that converge for all $z\in\mathbb C_p$. Schnirelmann's famous result assert that for every $r\in\mathbb R^+$, one has $\sup_{\substack{z\in\mathbb C_p\\\deg(z)\le r}}\limits\deg(f(z))=\sup_{n\in\mathbb N}\limits\{deg(a_n)+nr\}$. From that, we can deduce that for every sequence $(b_n)_n$ of $\mathbb C_p$ with $\deg(b_n)\le 0$, for the power series $$g(z)=\sum_{n\ge0}a_nb_nz^n,$$ one has $$\sup_{\substack{z\in\mathbb C_p\\\deg(z)\le r}}\limits\deg(g(z))\le \sup_{\substack{z\in\mathbb C_p\\\deg(z)\le r}}\limits\deg(f(z))$$

I wonder whether there exists such a result in characteristic $0$ case. Let $f(x)=\sum_{n\ge0}a_nx^n$ be an entire function on $\mathbb C$. Consider a sequence $(b_n)_n$ of complex numbers such that $0<|b_n|\le1$ for all $n\in\mathbb N$. Does one have $$\sup_{\substack{z\in\mathbb C\\|z|\le r}}|\sum_{n\ge0}a_nb_nz^n|\le \sup_{\substack{z\in\mathbb C\\|z|\le r}}|\sum_{n\ge0}a_nz^n|+K$$ where $K$ would be a constant depending on the sequence $(a_n)_n$ only.

Thanks in advance for any answers or hints.

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No, over the complex field the result is much weaker. Even this is not true: $$\max_{|z|\leq r}\left|\sum_{n=0}^\infty a_nb_nz^n\right|\leq K\max_{|z|\leq r}\left|\sum_{n=0}^\infty a_nz^n\right|,$$ with $K$ independent of $z$. I suppose that the simplest counter-example is $a_n=e^{i\alpha_n}/n!$ and $b_n=e^{-i\alpha_n}$ where $\alpha_n=2\pi in^2\alpha$, and $\alpha=(\sqrt{5}-1)/2$. This can be obtained by combining the results of P. Levy, Sur la croissance des fonctions entières, Bull. Soc. Math. France, 58 (1930) 29–59 et 127–149, and Theorem 4 in my paper with I. Ostrovskii, On the "pits effect" of Littlewood and Offord, Bull. London Math. Soc., 39 (2007) 929-939.

http://www.math.purdue.edu/~eremenko/dvi/ostr.pdf

Combination of these results gives $e^r$ in the LHS and $o(e^r)$ in the RHS.

Alternatively, one can use independent random arguments $\alpha_n$. In this case, the LHS is $e^r$ while the RHS is $$\max_{|z|\leq r}\left|\sum_{n=0}^\infty e^{2\pi i\alpha_n}z^n/n!\right|\approx r^{-1/4}e^r\sqrt{\log r},$$ P. Erdos and A. Renyi, On random entire functions, Zastosowania Matematyki (Applicationes mathematicae) 10 (1969) 47-55. Here $\approx$ means that the ratio is between positive constants.

EDIT: For recent results on the subject see Filevich, On Influence of the Arguments of Coefficients of a Power Series Expansion of an Entire Function on the Growth of the Maximum of Its Modulus, Siberian math. J., 44 (2003), 3, 529-538.

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  • $\begingroup$ Thanks for this accurate answer. Can you know what kind of results we can expect in this way? $\endgroup$ – joaopa Mar 12 '16 at 23:01
  • $\begingroup$ you mean $\alpha_n = \pi n^2 (\sqrt{5}-1)$ such that $|e^{- i \alpha_n}| = 1$. that's interesting as I always wondered if $\max_{|z| \le r} |\sum_n a_n z^n| \sim \max_{|z| \le r} |\sum_n |a_n| r^n$, hence you say it is not true (what about $\max_{|z| \le r} \log |\sum_n a_n z^n| \sim \max_{|z| \le r} \log(\sum_n |a_n| r^n)$, is it false too ?) $\endgroup$ – reuns Mar 13 '16 at 14:49
  • $\begingroup$ I am not sure about log. Probably it's true if you allow some small exceptional set in r. But without an exceptional set I am not sure. $\endgroup$ – Alexandre Eremenko Mar 13 '16 at 22:05
  • $\begingroup$ There is nothing special about $(\sqrt{5}-1)/2$. Any number which is badly approximable is fine. $\endgroup$ – Alexandre Eremenko Mar 13 '16 at 22:07

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