Dual cone of nonnegative continuous functions in $C([a,b])$ This is an example in Luenberger's "Optimization by Vector Space Methods" (Example $2$ in section $8.2$).
Let $P$ be the set of all nonnegative functions in $C([a,b])$, then the dual cone $P^*$ consists all linear functionals on $C([a, b])$ represented by $v$ of bounded variation and nondecreasing on $[a,b]$. In a normed vector space $X$, the dual cone of a convex cone $P$ is defined as
\begin{align*}
P^* = \{ x^* \in X^* \colon \langle x, x^* \rangle \ge 0 \qquad \text{for all } x \in P \}.
\end{align*}
I am struggling to understand why $v$ has to be nondecreasing. By classical Riesz Representation Theorem, $v$ is certainly of bounded variation. Thanks.
 A: Every function $v$ of bounded variation defines a linear functional on $C([a,b])$ via the Stieltjes integral 
$$f\mapsto \int_a^b f\,dv\tag1$$
Actually, we can assume $v$ to be continuous from the right at every point, because redefining $v(t)$ as $v(t+)$ does not change the integral $(1)$ for continuous $f$. (At $b$, we can make $v$ continuous from the left by letting $v(b) = v(b-)$.)
We want those $v$ that are in the dual cone of $P$, which means
$$f\ge 0 \implies \int_a^b f\,dv \ge 0 \tag2$$
A nondecreasing $v$ satisfies $(2)$ by the definition of the integral. To prove the converse, suppose $v$ fails to be nondecreasing; that is there exist $x_1<x_2$ such that $v(x_1)>v(x_2)$. The one-sided continuity discussed above implies there exist intervals $I_1$ and $I_2$, containing $x_1$ and $x_2$, such that $v(t_1)>v(t_2)$ for all $t_1\in I_1$ and $t_2\in I_2$. 
Let $f$ be a continuous function that is $0$ to the left of $I_1$, increases from $0$ to $1$ within $I_1$, stays constant $1$ between $I_1$ and $I_2$, decreases to $0$ within $I_2$ and stays there. For such $f$ we have  $\int_a^b f\,dv<0$, which shows that $v$ is not in the dual cone. 

represented by $v$ of bounded variation and nondecreasing on $[a,b]$

I would add "continuous from the right" (or from the left if you prefer) to this description because otherwise representation is not unique: for example, $v _1 = \chi_{(c,b]}$ and $v_2 = \chi_{[c,b]}$, where $a<c<b$, define the same functional.
