Q:

Minimization Problem: Deriving Dual Problem

Consider the following minimization problem $\min\{H(x,z) \equiv h_1(x) + h_2(z): Ax + Bz = c\}$, where $A \in \Bbb{R}^{m \times n}, B \in \Bbb{R}^{m \times p}$ and $c \in \Bbb{R}^{m}$ and $h_1, h_2$ are proper, closed and convex. 
To find the dual problem of the optimization problem, one can construct a Lagrangian:
$L(x,z;y) = h_1(x)+h_2(z) + \langle y, Ax + Bz - c \rangle$
The objective function is therefore given by
$q(y) = \min_{x, z} \{h_1(x) + h_2(z) + \langle y, Ax+Bz-c \rangle\}$
Apparently, the last line is the same $\max_{y}h_1^{*}(-A^{T}y)-h_2^{*}(-B^{T}y) - \langle c,y \rangle$
I guess that his is an application of some duality principle but I don't see how it exactly works. 

A:

Our problem is equivalent (with some domain qualification conditions) to
$$\min_{x,z}\max_{y} L(x,z,y) =\max_y \Big\{\min_{x} \{ h_1(x) + \langle y, Ax\rangle \} + \min_z \{h_2(z) + \langle y, Bz\rangle\} -\langle y, c\rangle\Big\}$$
Now use the definition of the convex conjugate to get
$$\min_{x,z}\max_y L(x,z,y) = \max_y \{-h_1^*(-A^* y) -h_2^*(-B^*y) - \langle y,c\rangle\}$$

