Dual of the relative entropy cone I've been trying to calculate the dual cone of the relative entropy cone, which is given by: 
$$\mathbb{H}_n = \Bigg\{\,(\theta\oplus \kappa\oplus x)\in\mathbb{R}^n_+\oplus\mathbb{R}_+^n\oplus\mathbb{R}^n\,\colon \theta_i\log\bigg(\frac{\theta_i}{\kappa_i}\bigg)\leq x_i,\text{ for each } i\in [n]\Bigg\}.$$
I know that this cone is created from direct sums of its 1-dimensional version $\mathbb{H}_1$ and that for any sets $K$ and $L$ we have $(K\oplus L)^\ast=K^\ast\oplus L^\ast$. Moreover, I know that 
$$(x\oplus \theta \oplus \kappa)\in\mathbb{G}_1\iff (\theta \oplus \kappa\oplus x)\in\mathbb{H}_1.$$
Where $\mathbb{G}_1$ is the 1-dimensional version of 
$$\mathbb{G}_n=\Bigg\{\,(x\oplus\theta\oplus\kappa) \in\mathbb{R}^n\oplus\mathbb{R}_+\oplus\mathbb{R}_+\,\colon \theta\sum_{i\in [n]}\exp\bigg(\frac{-x_i}{\theta}\bigg)\leq\kappa \Bigg\}.$$
And also  $A(K)^\ast=(A^\ast)^{-1}(K^\ast).$ Thus, I am trying to calculate the dual of $\mathbb{G}_n$. A similar result is presented in https://tel.archives-ouvertes.fr/tel-00006861/document Sec 6.3 and my attempt is to adapt it to my case. However, I found the following problems:
1- While considering the case where $\theta=0$, we still have that $(x,0,\kappa)\in\mathbb{G}_n$ for each $(x,\kappa)\in\mathbb{R}^n\oplus\mathbb{R}_+$. Because $x$ is not necessairly nonnegative, I couldn't obtain any corresponding sign constraints on $x^\ast$ and $\kappa^\ast$ (besides $x^\ast=0$ and $\kappa^\ast\geq0$). 
2- While minimizing a single term of the summation, I would obtain that $t_i=−\log(\frac{|x_i^\ast|}{\kappa^\ast})$ because there is no $\log$ for negative numbers, is this right? 
3- Why is the minimum equal to zero when $\kappa^\ast=0$?
4- The two final cases colapse and then we obtain that the minimum is $−x_i^\ast\log(\frac{x_i^\ast}{\kappa^\ast})$ either way? 
5- Finally, we conclude that $x^\ast\in$?, $\kappa^\ast\in$? and $\theta^\ast\geq\sum_i x_i^\ast\log(\frac{x_i^\ast}{\kappa^\ast})−x_i^\ast$. Right? 
It seems like the absence of the plus sign in my definition of $\mathbb{G}_n$ (i.e. allowing negative values of $x$, which is not usually done but gives me the above bijection) would force the first coordinate of the dual variables of $\mathbb{G}_n$ to be zero. In other words, I would not be able to use the knowledge of $\mathbb{G}_1^\ast$ to deduce $\mathbb{H}_1^\ast$. Thanks
 A: The dual cone is simply given by:
$$\mathbb{H}_1^* = \{ (a,b,c) : ax+by+cz \geq 0 \; \forall (x,y,z) : x \log(x / y) \leq z, x \geq 0, y \geq 0 \},$$
but you probably want to simplify this expression :) 
The dual cone is equivalent to:
$$\mathbb{H}_1^* = \{ (a,b,c) : \min_{x,y,z}\{ ax+by+cz : x \log(x / y) \leq z, x \geq 0, y \geq 0 \} \geq 0 \},$$
We formulate the dual of the minimization problem using this convenient method (observe that strong duality holds). Note that $x \log(x / y)$ is the perspective of $x \log x$.
The objective function is $f_0(x,y,z) = ax+by+cz$. The constraint function is $f_1(x,y,z) = x \log(x / y) - z$ if $x \geq 0$, $y\geq 0$ ($\infty$ otherwise).
We have $f_0^*(u_0,v_0,w_0) = 0$ if $u_0=a$, $v_0=b$, $w_0=c$ ($\infty$ otherwise), and $f_1^*(u_1,v_1,w_1)= 0$ if $e^{u_1-1} \leq v_1$ and $w_1 = -1$ ($\infty$ otherwise). The latter can be derived as follows:
$$\begin{align}
& \sup_{x \geq 0, y \geq 0, z} \{ xu+vy+zw - x\log(x/y) + z \} \\
= \; & \sup_{y \geq 0} \{ vy + y \sup_{x \geq 0} \{ (x/y)u - (x/y)\log(x/y) \} \} + \sup_z \{ zw + z \} \\
= \; & \sup_{y \geq 0} \{ vy + y \sup_{t \geq 0} \{ tu - t\log t  \} \} + \sup_z \{ z(w+1) \} \\
= \; & \sup_{y \geq 0} \{ vy + y \exp(u-1) \} + \sup_z \{ z(w+1) \} \\
= \; & 0 \text{ if } v+\exp(u-1) \leq 0 \text{ and } w+1=0 \text{ (}\infty \text{ otherwise).}
\end{align}$$
The dual problem is therefore
$$\begin{align}
\max_{u,v,w,\lambda} \quad & 0 \\
\text{s.t.} \quad & u_0=a, v_0=b, w_0 = c \\
& \lambda e^{u_1/\lambda - 1} \leq -v_1, w_1=-\lambda \\
& u_0+u_1=0, v_0+v_1=0, w_0+w_1=0 \\
& \lambda \geq 0 
\end{align}$$
which can be simplified to $\max\{0 : c e^{-a/c-1} \leq b \}$ if $c \geq 0$. Since strong duality holds, we can replace the minimization problem in the definition of the dual cone:
$$\mathbb{H}_1^* = \{ (a,b,c) : \max\{0 : c e^{-a/c-1} \leq b \} \geq 0, c \geq 0 \}.$$
This is just:
$$\mathbb{H}_1^* = \{ (a,b,c) : c e^{-a/c-1} \leq b, c \geq 0 \}.$$
