TITLE: Show tangent cone and linearized fesible directions of this set
QUESTION [2 upvotes]: Consider a feasible set $\Omega = \{(x_1,x_2) \in \mathbb{R}^2: -x_2 \leq 0\ and \ x_2-{x_1}^2\leq 0 \} $. For $x^*={(0,0)}^T$. Determine if the tangent cone and linearized feasible solutions are equal.

I am having difficulty in finding the tangent cones. I built sequences $(w_1/k,w_2/k)$, where $w_1 \neq0 \ and \ w_2 \geq0$ and $(w_1/k,{w_1}^2/k^2)$, where $w_1 \in \mathbb{R}$. By definition of tangent cone, I have to find all vectors tangent to $\Omega$ at $x^*$. So I am not sure if I have covered all vectors.

REPLY [0 votes]: Just found this question while looking for something else, and I'm not familiar with the syntax on this site.
You can rewrite the functions to:
$x_2\ge0$ and $x_2\le x_1^2$. Since $(0,0)$ is your only feasible point, your tangent cone is the point $(0,0)$. For the linearized feasible directions there is only the $x$-axis. Therefore the tangent cone and the linearized tangent cone are not the same.
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