Question

Let i_c(x) be the indicator function of the closed convex set C. Show that the sub-differential of the function i_c at a point c in C is the normal cone to C at the point c.

Answer 1

Answer on Question#37090 - Math- Other

Let $I_{\mathcal{C}}(x)$ be the indicator function of the closed convex set $\mathcal{C}$ . Show that the sub-differential of the function $I_{\mathcal{C}}$ at a point $\mathcal{C}$ in $\mathcal{C}$ is the normal cone to $\mathcal{C}$ at the point $\mathcal{C}$ .

Solution.

We have the convex set $\mathcal{C}$ and the indicator function $I_{\mathcal{C}}(x)$ . We know that:

1. For $c \notin C$ , $\partial I_{\mathcal{C}}(c) = \emptyset$ (by convention).

2. For $c \in C$ , we have $g \in \partial I_{\mathcal{C}}(c)$ if $I_{\mathcal{C}}(z) \geq I_{\mathcal{C}}(c) + g'(z - c), \forall z \in C$ , or equivalently $g'(z - c) \leq 0$ for all $z \in C$ .

Thus $I_{\mathcal{C}}(c)$ is the normal cone of $\mathcal{C}$ at $c$ , denoted $N_{\mathcal{C}}(c)$ :

$N_{\mathcal{C}}(c) = \{g\mid g^{\prime}(z - c)\leq 0,\forall z\in \mathcal{C}\} .$