I am searching for (works on) methods for solving the composite differentiable and non-differentiable convex problem:
$$ \min_{x \in B} f(x) + g(x),$$
where $B$ is a $2$-norm ball, ie: $x \in B \iff \|x\|_2 \leq C, C>0$; and the functions satisfy:
$f$ is convex and differentiable
$g$ is convex but not differentiable (possibly strongly convex)
Any literature on methods for solving that kind of problem?
If $h$ is a convex function, then the subgradient of $h$ at $x_0$ is $$ \partial h(x_0) = \{g:h(x)-h(x_0)\ge g'(x-x_0)\}. $$ This set exists and is convex on the interior of the domain of any convex function on a convex set, and if $h$ is continuous at the boundary, everywhere on the domain. For example, $\partial |0| = [-1,1]$.
The standard necessary condition ($\nabla h(x^*) = 0$ for an interior minimum) carries over to this situation, changing to set notation $0 \in \partial h(x^*)$. So for $|x|$, $0 \in [-1,1]$, and there is a local minimum at zero. If $x^*$ is on the boundary, you just get the variational inequality-type condition that for all $x' \in X$, $x' \cdot g \ge 0$ for $g \in \partial h(x^*)$, so the directional derivative is strictly positive pointed into the set.
You also have the standard results that if $h$ is convex, the set of minimizers is a convex set, and if $h$ is strictly convex, the minimizer is unique.
Authors/keywords: Clarke, Rockafellar, convex analysis, non-smooth analysis, subgradients, subderivatives
Since your functions are convex, take $h = f+g$. If $f$ is differentiable but $g$ is not, the non-differentiable points of $g$ are automatically candidate solutions since $f+g$ is also non-differentiable at those points. Since $g$ is convex, it is a.e. differentiable everywhere else, so you could compute the variational inequality $\nabla(f+g)(x^*)(x'-x^*)\ge 0$ for all $x'\neq x^*$ for the differentiable points and find the zeros there (i.e., standard FONC on the interior, and otherwise look at boundary points using KKT, since your constraint set is just the closed ball with radius $C$). Combining the two lists of candidate solutions, that might make the search much easier.
