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For simplicity, let $X=\mathbb{R}$ and $K=[0,1]$. Consider $I_K:\mathbb{R}\to \{0,+\infty\}$ given by $I_K(x)=0$ if $x\in K$ and $+\infty$ otherwise. What is the subdifferential (in the sense of convex analysis) of $I_{K}$?

Since $R\simeq R^{*}$ it is easy to see that for every $x\in K$ we have $\partial I_{K}(x)=0$. Let's take $x_{0}\notin K$. Then

$$\partial I_{K}(x_{0})=\{x\in \mathbb{R}:x\cdot(v-x_{0})\le I_K(v)-I_K(x_0)\,\mbox{for all}\, v\in \mathbb{R}\}=\emptyset.$$

Will the same result hold if I consider $X$ being abstract Banach space with $K$ being convex and closed?

zorro47
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  • You have $\partial I_K(1) = [0,\infty)$. The subdifferential is empty for $x \notin K$. – copper.hat May 31 '17 at 16:13
  • The issue is that if $I(x_0) = \infty $ and $I(x) < \infty$ there can be no $v \in X^*$ such that $I(x) \ge I(x_0) + v(x-x_0)$. – copper.hat May 31 '17 at 16:20
  • You're right $\partial I_K(1)=[0,\infty)$. – zorro47 May 31 '17 at 16:27
  • My second comment is an answer to your question. As long as there is some $x \in K$, the subdifferential is empty when $x_0 \notin K$. This is not specific to $\mathbb{R}^n$. – copper.hat May 31 '17 at 17:20

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