TITLE: Cone of polynomials nonnegative on [0,1] is a proper cone.
QUESTION [1 upvotes]: Example 2.16 in Boyd's book: 
Cone of polynomials non-negative on $[0,1]$ can be defined as: 
$K =\{c \in R^n | c_1 + c_2 t + \dots + c_n t^{n-1} \geq 0 \text{ for } t \in [0,1] \}$. 
Its mentioned that $K$ is the cone of (coefficients of) degree $n-1$ that are non-negative on the interval $[0,1]$ and $K$ a proper cone.
How polynomials satisfy that? Any explanation is appreciated. Thanks.

REPLY [1 votes]: Hint: write the polynomial inequality associated with $c \in K$ and show that:

$0_n \in K$ where $0_n$ is the vector with all components equal to $0\,$;
$c \in K \implies \lambda c \in K$ for any real $\forall \lambda \ge 0\,$;
$c,c' \in K \implies \lambda c + (1-\lambda)c' \in K$ for any real $\forall \lambda \in [0,1]\,$.
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