Let $D$ be the unit disc in $\mathcal{R}^2$ and let $f : \mathcal{R}^2 → \mathcal{R}$ be the function $f(x) = d(x, D)$ = the distance from $x$ to $D$. We are looking for the sub differential of $f$ at $\bar x$ that is we are looking for $\xi \in \mathcal{R}^2$ such that $\forall x \in \mathcal{R}^2$ : $$f(x) \geq f(\bar x)+\xi^T (x-\bar x)$$ First, I considered the case : $\|\bar x\| \leq 1$, that is we are looking for $\xi \in \mathcal{R}^2$ such that $\forall x \in \mathcal{R}^2$ : $$f(x) \geq \xi^T (x-\bar x)$$ Since $f(\bar x)=0$ because $\|\bar x\| \leq 1$. In this case I find the sub diffential $\xi=0$. $$ $$ But for the case $\|\bar x\| > 1$, I'm stuck and I don't know how to start...
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$f$ is a radial version of a very simple function – Calvin Khor Dec 06 '19 at 14:08
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$f$ such that : $$ f(x)=| x | -1~~ if~~ | x | >1 ~~and ~~f(x) = 0 ~~otherwise $$ right ? – Sylvain Lhermite Dec 06 '19 at 14:15
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SylvainLhermite yes – Calvin Khor Dec 06 '19 at 17:37