Let \begin{align*} p(x)=\sum_{i=0}^{n}a_ix^i, \end{align*} I want to restrict the coefficients of $p(x)$ to make it convex.

Here are good answers, https://mathoverflow.net/questions/28983/characterizing-convex-polynomials. The first one is @coudy's answer, a direct way. The second gives semidefinite matrix method using SOS equivalent property.

They can be used to generate convex polynomials well. However, they are not convenient to determine whether a polynomial is convex. So I am seeking for a sufficient condition.

For example, let $p(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ then $p''(x)=12a_4x^2+6a_3x+a_2.$ Thus, $p(x)$ is convex iff $(6a_3)^2-48a_2a_4\geq0.$ I think this is pretty good!

For higher degree, is there a good way like this? I just want some sufficient conditions about the coefficients of $p(x).$

I hope I've made my question clear, and any help will be appreciated.

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    $\begingroup$ Well, if $p(x)$ is convex, then $p''(x)=\sum_{i=2}^n i(i-1)a_ix^{i-2}$ is nonnegative. So any properties you can come up with about the coefficients of a nonnegative polynomial can then be mapped to the conditions for convexity. In particular, note that $a_0$ and $a_1$ are irrelevant to any consideration of convexity. $\endgroup$ – Michael Grant Jan 14 '17 at 17:26
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    $\begingroup$ I would suggest that this answer addresses your question as best as anyone is likely to, unless you can ask for more specifics. That's basically the answer I would offer. $\endgroup$ – Michael Grant Jan 14 '17 at 17:28
  • $\begingroup$ Your advice is great, and I have edit it again. I am just looking forward a sufficient condition. And is there any reference about it?@Michael Grant $\endgroup$ – VerMoriarty Jan 15 '17 at 15:08

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