# Normal cone to the tangent cone of $\mathbb{R}_+$

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, $$$$\mathcal{N}_{\mathbb{R}_+}(x)=\begin{cases} \{0\} & \text{if x>0} \\ \mathbb{R}_- & \text{if x=0} \end{cases} \quad\text{ so }\quad\mathcal{T}_{\mathbb{R}^+}(x)=\begin{cases} \mathbb{R} & \text{if x>0} \\ \{0\} & \text{if x=0}\end{cases}$$$$

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.

I indeed made a mistake; I give the actual results after verification, in case it helps someone (feel free to delete the question though): $$$$\mathcal{N}_{\mathbb{R}_+}(x)=\begin{cases} \{0\} & \text{if x>0} \\ \mathbb{R}_- & \text{if x=0} \end{cases} \quad\text{ so }\quad\mathcal{T}_{\mathbb{R}^+}(x)=\begin{cases} \mathbb{R} & \text{if x>0} \\ \mathbb{R}_+ & \text{if x=0}\end{cases}$$$$
the normal cone of which I found to be: $$\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}$$
• I think, this is not true. It should be $\mathbb{R}^+$ in the tangent cone of $\mathbb{R}^+$. – gerw Feb 5 '15 at 8:10