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These are the definitions I'm using (cf Rockaffeller):

  • normal cone to a convex set $C$: $$\mathcal{N}_C(x)=\{d\ | <d,y-x>\leq 0,\ \forall y\in C\}$$
  • tangent cone to a convex set $C$: $$\mathcal{T}_C(x)=\{u\ | <d,u>\leq 0,\ \forall d\in\mathcal{N}_C(x)\}$$

Let's take $C=\mathbb{R}_+$ (which is convex): if I am not mistaken, $$\begin{equation} \mathcal{N}_{\mathbb{R}_+}(x)=\begin{cases} \{0\} & \text{if $x>0$} \\ \mathbb{R}_- & \text{if $x=0$} \end{cases} \quad\text{ so }\quad\mathcal{T}_{\mathbb{R}^+}(x)=\begin{cases} \mathbb{R} & \text{if $x>0$} \\ \{0\} & \text{if $x=0$}\end{cases} \end{equation}$$

the normal cone of which I found to be: $$\mathcal{N}_{\mathcal T_{\mathbb{R}_+}(x)}(y)=\begin{cases} \{0\}&\text{if $x>0$} \\ \begin{cases} \mathbb{R}^+ & \text{if $y>0$} \\ \mathbb{R} & \text{if $y=0$} \\ \mathbb{R}_-&\text{if $y<0$} \end{cases}&\text{if $x=0$} \end{cases}$$

Did I make a mistake in the calculation? For some reason, I was excepting not to see $\mathbb{R}$ (maybe $\{0\}$) in the latter expression.

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I indeed made a mistake; I give the actual results after verification, in case it helps someone (feel free to delete the question though): $$\begin{equation} \mathcal{N}_{\mathbb{R}_+}(x)=\begin{cases} \{0\} & \text{if $x>0$} \\ \mathbb{R}_- & \text{if $x=0$} \end{cases} \quad\text{ so }\quad\mathcal{T}_{\mathbb{R}^+}(x)=\begin{cases} \mathbb{R} & \text{if $x>0$} \\ \mathbb{R}_+ & \text{if $x=0$}\end{cases} \end{equation}$$

the normal cone of which I found to be: $$\mathcal{N}_{\mathcal T_{\mathbb{R}_+}(x)}(y)=\begin{cases} \{0\}&\text{if $x>0$} \\ \begin{cases} \{0\} & \text{if $y>0$} \\ \mathbb{R}_- & \text{if $y=0$} \\ \end{cases}&\text{if $x=0$} \end{cases}$$

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  • $\begingroup$ I think, this is not true. It should be $\mathbb{R}^+$ in the tangent cone of $\mathbb{R}^+$. $\endgroup$ – gerw Feb 5 '15 at 8:10
  • $\begingroup$ @gerw: You are absolutely right (the final result is correct). Edited. $\endgroup$ – anderstood Feb 5 '15 at 16:08

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