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 '16 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 '16 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 '16 at 10:30
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    $\begingroup$ Putting together the comments above we may say, slightly more generally: If $\sum_{n=0}^\infty c_{2n}x^{2n}$ is such a series: (i) any finite subset of coefficients with positive indices is uniquely determined by all the others, because any perturbation of it changes the series by a polynomial, thus unbounded; (ii) for any $p$, the coefficients $c_{2n}$ corresponding to indices $n$ multiples of $2p+1$ are not determined by all the others, because one may add e.g. the series of $\cos x^{2p+1}$. $\endgroup$ – Pietro Majer Dec 20 '16 at 10:59
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    $\begingroup$ Please see my similar question and the comments there: mathoverflow.net/questions/27100/… $\endgroup$ – Andreas Rüdinger Dec 20 '16 at 20:28

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