TITLE: Is it possible to solve the argument maximization problem $\arg\max_x \langle x,l \rangle −f_1(x)−f_2(x)$ via convex duality?
QUESTION [8 upvotes]: I am attempting to solve the argument maximization problem
$$\arg\sup_x \{ \langle x,l \rangle − f_1(x)−f_2(x) \} \ \ \ \ \ \ \ \ \  \ (1)$$
where the functions $f_1$ and $f_2$ are concave but difficult to evaluate but their convex conjugates $f^∗_1$ and $f^∗_2$ are easy to evaluate. We can further assume that $f^∗_1$ and $f^∗_2$ are differentiable and that we can evaluate their gradients. Since the sum operation is dual to the infimal convolution (or epi-sum) operation
$$(g\#h)(x) = \inf_w \{ g(x−w) + h(w) \} $$
the standard maximization problem is easy to compute by duality using the identity
$$\sup_x\{⟨x,l⟩−f_1(x)−f_2(x)\}=\inf_w \ \{ f^*_1(l−w) + f^∗_2(w) \}.$$
Is it possible to compute the solution to problem $(1)$ is an analogous manner, making only calls to the conjugate functions $f^∗_1$ and $f^∗_2$ or their gradients?

REPLY [2 votes]: I ultimately did find a solution to this problem. First, you compute the solution $\bar{w}$ to 
$$\bar{w} \in \arg\inf_w f^*_1(\ell - w) + f_2^*(w).$$
A solution $\bar{x}$ to (1) can then be computed by choosing 
$$\bar{x} \in \partial f_1^*(\ell-\bar{w}) \cap \partial f_2^*(\bar{w})$$
where $\partial f$ denotes the subgradient of $f$. A proof of this fact appears as part of Theorem 15 in this JMLR paper: 
http://jmlr.org/papers/volume8/rifkin07a/rifkin07a.pdf