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We define the monotone nonnegative cone as

$$K_{m+} = \{ x \in \mathbb{R}^n \mid x_1 \ge x_2 \ge \dots \ge x_n \ge 0\}$$

i.e. all nonnegative vectors with components sorted in nonincreasing order.

Find the dual cone $K^*_{m+}$. Hint: use the identity

\begin{align*} \sum^n_{i=1} x_iy_i = & (x_1 - x_2)y_1 + (x_2 - x_3)(y_1+y_2) + (x_3 - x_4)(y_1+y_2+y_3) \\ &+ \dots + (x_{n-1} -x_n)(y_1+ \dots + y_{n-1}) + x_n(y_1+ \dots + y_n) \end{align*}

I proved the cone is a proper cone if that knowledge helps somehow.

The same problem was posted here but the solution does not make sense to me.

Rodrigo de Azevedo
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CEP
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1 Answers1

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The hint can be easily proven by induction on $n$, so I will skip the details for this. Assume $y\in K_{m+}^*$. For any given $k$ between $1$ and $n$, define $$ z_k = \begin{bmatrix} 1\\\vdots\\1\\0\\\vdots\\0 \end{bmatrix}, $$ where the last $1$ is in row $k$. Since $z_k\in K_{m+}$, we have $$ 0 \le y^Tz_k = y_1 + \cdots + y_k. $$ Reciprocally, if $y_1 +\cdots+ y_k \ge0$ for all $k$, the hint shows that $$ y^Tx = \sum_{i=1}^nx_iy_i $$ can be expressed as a sum of non-negative terms $(x_{i}-x_{i-1})(y_1+\cdots+y_i)$, which makes it non-negative, showing that $y\in K^*_{m+}$.

Leandro Caniglia
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