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Let $f(z)$ be an entire function (on $\mathbb{C}$). Assume it has a power series of the form $$\displaystyle \sum_{n=0}^\infty (-1)^nc_{2n}z^{2n},$$ where $c_{2n}\geq 0$ for all $n$.

Is there a sufficient condition on the coefficients $c_{2n}$ under which the sum is bounded for all $z=x\in \mathbb{R}?$

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    $\begingroup$ If $c_n=\frac{1}{n!}$ then the series is bounded on the real line. But I guess this is not the answer you had in mind, so you might as well be a little bit more precise in your question. $\endgroup$
    – Loïc Teyssier
    Dec 20, 2016 at 8:37
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    $\begingroup$ I am tempted to believe there is no simple criterion which would let us decide this. For example, note that if your series is bounded on the real line, then chainging a single $c_{2n}$ by any amount will lead to a function unbounded on $\mathbb R$. This tells us, for example, that any such condition has to take into account all coefficients, not just the ones with sufficiently large powers of $x$. $\endgroup$
    – Wojowu
    Dec 20, 2016 at 8:52
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    $\begingroup$ I also would say there hardly exists any suitable criterion. Notice that there are many such functions, including $\cos ax$, $\frac{\sin ax}x$, $\frac1x\int_0^x\frac{\sin at}t\,dt$, etc., and also their linear combinations with nonnegative coefficients --- possibly with infinite number of terms. $\endgroup$
    – Ilya Bogdanov
    Dec 20, 2016 at 10:30
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    $\begingroup$ @T. Amdeberhan actually the quotient $c_{2n+2}/c_{2n}$ may be unbounded; consider $\cos(x)+\cos(x^3)$ $\endgroup$
    – Pietro Majer
    Dec 20, 2016 at 20:04
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    $\begingroup$ Please see my similar question and the comments there: mathoverflow.net/questions/27100/… $\endgroup$
    – Andreas Rüdinger
    Dec 20, 2016 at 20:28

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