Fenchel dual vs Lagrange dual Consider the Fenchel dual and the Lagrangian dual. 
Are these duals equivalent? In other words, is using one of the these duals (say for solving an optimization), would give the same answer as using the other one? 
I think the answer is no, but I am not sure. One reason for saying that, is that, in the Lagrange dual, we have a relatively straightforward way to add the constraints into the objective function. But what about the Fenchel? I have not seen any. 
But I have seen some problems in which both of these give same answers. So, I would assume that, on a subset of problems, these two dualities, are exactly the same. 
And also, if they are different, how would you choose which one to use on your problem? 

Let the fenchel dual of a function $f : \mathbb{R}^n \to \mathbb{R}$ be $D_f$. Consider the convex program $P^{+} = \min f(x)$ s.t. $x \ge \boldsymbol{0}$ and let $D_l$ be this program's lagrangian dual then for $\boldsymbol{\lambda} \ge \boldsymbol{0}$ $D_f(\lambda) = -D_l(\lambda)$. So on the positive orthant the fenchel dual agrees with the lagrangian dual of $P^{+}$.
Similarly on the negative orthant $D_f$ agrees with the dual of $P^{-} =  \min f(x)$ s.t. $x \le \boldsymbol{0}$. 
The general case is let $P^{A,b} = \min f(x)$ s.t. $Ax \le b$ and let $D_l^{A,b}$ be its dual, then for $\boldsymbol{\lambda} \ge \boldsymbol{0}$ $D_f(-A^T\boldsymbol{\lambda}) + D_l^{A,b}(\boldsymbol{\lambda}) + \boldsymbol{\lambda}^Tb=0$. 
So the Fenchel dual is stitched together from lots of lagrange duals.
Or conversely the lagrange dual of convex problems on polyhedrons can be constructed from the fenchel dual.
