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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.

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    You haven't transcribed the definition of the set properly. Please double-check. – 2017-02-12
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    But the thing to keep in mind is this: _this is not a set of polynomials_. After all, as its definition clearly states, it is a set of vectors in $\mathbb{R}^n$. The polynomial comes in only when deciding whether or not a particular vector $c$ is in the set; and in that definition, $c$ serves as the coefficients of a poynomial. – 2017-02-12
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    thanks for the comment. Just updated the definition. However, I am still unclear about the example. – 2017-02-12

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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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    Thanks. So the first point proves that $K$ is pointed, third one proves that $K$ is convex. I am not sure about the second one and how do we show that $K$ is solid? – 2017-02-12
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    @jhon_wick The second one proves that it is indeed a [(linear) cone](https://en.wikipedia.org/wiki/Convex_cone#Definition). What is your definition of "*solid*"? – 2017-02-12
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    here solid implies $K$ has nonempty interior. – 2017-02-12
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    @jhon_wick Note for example that all vectors with non-negative components $c_k \ge 0\,$ are in $K\,$. – 2017-02-12