So for the function:
$$ f(x) = \underset{i=1,\ldots,m}{max}{f_{i}(x)} $$
The set of subdifferentials at a given point $x_{k}$ is given by:
$$ \partial f(x_{k}) = conv\Big( \bigcup\limits_{i\in I_{A}(x_{k})} \partial f_{i}(x_{k}) \Big) $$
Where:
$$ I_{A}(x_{k}) = \{i | f_{i}(x_{k}) = f(x_{k})\} $$
I am not sure how can I actually compute this for given set of functions. Would it simply be:
$$ \partial f(x_{k}) = \sum\limits_{i \in I_{A}(x_{k})} \alpha_{i} \partial f_{i}(x_{k}) $$
Where $ \sum\limits_{i \in I_{A}(x_{k})} \alpha_{i} = 1$?
Basically, what I don't understand is: How is the convex hull of the union of a set of vectors different than the convex hull of those vectors?
An example for functions $\mathbb{R}\rightarrow\mathbb{R}$ is: $$ f_1(x) = |x|, \quad f_2(x) = 2x $$ Then $$ f(x) = \max[|x|, 2x] = \left\{ \begin{array}{ll} 2x &\mbox{ if $x \geq0$} \\ -x & \mbox{ if $x<0$} \end{array} \right.$$ Now at $x=0$ both functions are active and the active index set is $I=\{1, 2\}$, and \begin{align} \partial f_1 (0) &= \{m \in \mathbb{R} : -1\leq m \leq 1\} \\ \partial f_2(0) &= \{2\} \end{align} Notice that both $\partial f_1(0)$ and $\partial f_2(0)$ are indeed convex sets. So the union is: $$ U = \{2\} \cup [-1,1] $$ The convex hull of $U$ is the set of all points $z$ that can be written as $$ z = \sum_{i=1}^k \theta_i z_i $$ for some positive integer $k$, some $z_1, ..., z_k$ in $U$, and some nonnegative values $\theta_1, ..., \theta_k$ that sum to 1. But the set of all such points is just all real numbers in the closed interval from $-1$ to $2$: $$ Conv(U) = [-1, 2] $$
Claim: If $A_1, ..., A_m$ are nonempty convex subsets of $\mathbb{R}^n$ then $Conv(\cup_{i=1}^m A_i)$ can be described as the set of all points $z$ of the form: $$ z = \sum_{i=1}^m \theta_i z_i$$ for some nonnegative values $\theta_1, ..., \theta_m$ that sum to 1, and some vectors $z_1, ..., z_m$ such that $z_i \in A_i$ for all $i \in \{1, ..., m\}$.
Proof: Let $z \in Conv(\cup_{i=1}^m A_i)$. By definition, there is a positive integer $k$ and (without loss of generality, positive) values $\beta_1, ..., \beta_k$ that sum to 1 such that: $$ z = \sum_{j=1}^k \beta_j v_j $$ for some vectors $v_1, ..., v_k \in \cup_{i=1}^m A_i$. However, we can group the $v_j$ vectors into which set $A_i$ they come from, breaking ties (in cases when $v_j$ belongs to more than one $A_i$ set) in favor of the smallest $i$. For each $i \in \{1, ..., m\}$ define $B_i$ as the set of all indices $j$ such that $v_j \in A_i$, but $v_j \notin A_k$ for any $k <i$. For simplicity, assume each set $B_i$ is nonempty (this avoids distractions and the general proof without this assumption is similar). For each $i \in \{1, .., m\}$, define $\theta_i = \sum_{j \in B_i} \beta_j$ and note that $\theta_i>0$ and $\sum_{j \in B_i} \frac{\beta_j}{\theta_i} =1$. Then $$ z = \sum_{j=1}^k \beta_j v_j = \sum_{i =1}^m \sum_{j \in B_i} \beta_j v_j = \sum_{i=1}^m\theta_i \underbrace{\left[\sum_{j\in B_i} \frac{\beta_j}{\theta_i}v_j\right]}_{\in Conv(A_i)} $$ But $Conv(A_i) = A_i$ since $A_i$ is convex for each $i$. For each $i \in \{1, ..., m\}$, define $z_i = \sum_{j \in B_i} \frac{\beta_j}{\theta_i}v_j$. Then $z_i \in Conv(A_i) = A_i$ for all $i$. Also, $\theta_i$ are nonnegative values that sum to 1, and $$ z= \sum_{i=1}^m \theta_i z_i $$ The case when some $B_i$ sets are empty can be treated by repeating the above argument over only those indices $i$ for which $B_i$ is not empty, and defining $\theta_i=0$ whenever $B_i=\phi$. $\Box$