I was giving this post some thought and could not help but wonder the following. Suppose that $A_1,\dots,A_k$ are pairwise disjoint open sets and $g_1,\dots,g_k$ are smooth functions, then $1_{A_j}g_j$ (where $1_{A_j}$ is the indicator function on $A_j$) is sub-differentiable but is is true that any subgradient $G$ of $\sum_{j=1}^k I_{A_j} g_j$ can be written as $$ G = \sum_{j=1}^n 1_{A_j} \nabla g_j? $$ (since $g_j$ is continuously differentiable)
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Are you assuming that $1_{A_j}g_j$ is convex? – Angela Pretorius Apr 13 '20 at 13:34
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No, just general non-convex functions. However, $A_i$ are pairwise disjoint... if it helps.. – AB_IM Apr 14 '20 at 08:18
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$1_{A_j}g_j$ is not necessarily subdifferentiable. – Angela Pretorius Apr 14 '20 at 11:48
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I think if $A_i$ are intervals then this sould be okay? – AB_IM Apr 16 '20 at 18:41