TITLE: Characterizing convex polynomials
QUESTION [5 upvotes]: Let $p=\sum_{i=0}^{n}a_ix^i$. Under what conditions on the coefficients $a_i$  is $p$ convex? Strictly convex?

REPLY [8 votes]: To add some more stuff to @coudy's answer. We need to check whether $p''$ is nonnegative.
To that end let us recall some basic results (copied from another MO answer of mine), discussed in these lecture notes; these results are helpful for checking if a univariate polynomial is nonnegative.
Theorem A univariate polynomial is nonnegative if and only if it is a sum of squares.
Theorem Let $\mathbf{x}=[1,x,x^2,\ldots,x^m]^T$, and let $P(x)$ have degree $2m$. Then, $P(x)$ is nonnegative if and only if there exists a $m+1 \times m+1$ positive semidefinite matrix $Q$ such that $P(x) = \mathbf{x}^TQ\mathbf{x}$.
Let $P(x)=\sum_{i=0}^{2m} p_ix^i$. It can be further shown that the matrix $Q$ must satisfy
\begin{equation*}
 p_i = \sum_{j+k=i} Q_{jk},\qquad i=0,1,\ldots,2m.
\end{equation*}
Theorem If we can find a positive semidefinite matrix $Q$ that satisfies the above linear constraints (this can be done using regular semidefinite programming software), then the polynomial $P(x)=\sum_{i=0}^{2m} p_ix^i$ is nonnegative or SOS, otherwise not.