Is the dual of a conic program $\min_{x\in K} c^T x$ subject to $Ax=b$ also a conic program

Question

Let $A$ be an $m \times n$ matrix (over $\mathbb{R}$), $b \in \mathbb{R}^m$, $c \in \mathbb{R}^n$ and $K \subseteq \mathbb{R}^n $ is a closed, convex, pointed cone with non-empty interior. We define a conic program to be $$ \min_{x\in \mathbb{R}^n} c^Tx \\ \text{subject to } Ax = b , x\in K$$

Furthermore, define the dual problem to be $$ \max_{y\in \mathbb{R}^m, z \in \mathbb{R}^n} b^Ty \\ \text{subject to } c = z + A^Ty, z \in K^*$$ where $K^*$ is the dual cone of $K$.

Question: Is the dual problem also a conic program (by the definition given)? If I try the obvious solution where $(y,z) \in \mathbb{R}^m \times K^*$, then the issue is that $\mathbb{R}^m \times K^*$ is a closed convex cone with non-empty interior but is not pointed. How can I get around this?

Thank you.

Edit: To clarify, a cone is defined as a set $K \subseteq \mathbb{R}^n$ satisfying $x \in K \implies ax \in K$ for all $a \geq 0$. A cone is pointed if it satisfies $K \cap (-K) = \{0\}.$

Answer 1

The dual program is equivalent to a conic program, by applying following transformations

1. Replace $y\in\mathbb R^m$ with $y=u - v$ for $u,v\in\mathbb R^m_+$ (cone of non-negative orthant).
2. Replace $\max ...$ with $-\min -(...)$.

Finally, we obtain the following equivalent conic program, that is we can map the feasible points of both programs into other (not necessarily bijectively) and we can map the optimal points of both programs into other:

$$ \min_{(u,v,z)} \; \begin{bmatrix}-b \\ b \\ 0 \end{bmatrix}^T \begin{bmatrix}u \\ v \\ z \end{bmatrix} $$ $$ \text{s.t. } \begin{bmatrix}A^T & -A^T & I\end{bmatrix} \begin{bmatrix}u \\ v \\ z \end{bmatrix} = c, \; \begin{bmatrix}u \\ v \\ z \end{bmatrix}\in\mathbb R^m_+ \times \mathbb R^m_+ \times K^*$$

Notes:

• If $y$ is given, then we can take $u = \max(y, 0)$ and $v = \max(-y, 0)$ component-wise.
• We can think of $u$ as the "positive" part of $y$ and $v$ as the "negative" part of $y$.