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I would like help with exercise:

Be $S = \{ d\in\mathbb{R}^2\mid d \ge 0,d_1d_2=0 \}$.

$(a)$ Show that $S$ is a non-convex cone;

$(b)$ To determine $P(s) = \{p\in\mathbb{R}^2\mid p^{T}d\le 0, \forall d \in S\}$

$(c)$ Geometrically represent the sets $S$ e $P(S)$

What did I do:

$(a)$ Just take $u = (1,0)$ and $v = (0,1)$ and choose $t = \frac{1}{2}$. To have, $(1-t)u + tv = (1-t)(1,0) + t(0,1) = (1-t, t) = (\frac{1}{2},\frac{1}{2})$ that is not in the set.

$(b)$ Be $d \in S$ e $P=(x,y)\in\mathbb{R}$ we have to $p^{T}d = (x \ y){d_1 \choose d_2} = xd_1 + yd_2 \le 0$ I can't finish.

$(c)$ I do not know how to do it.

if anyone can help me, I appreciate it. Sorry for the English, because it is not my native language.

Amissadai ferreira
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1 Answers1

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For a point $d=(d_1,d_2)$ to be in $S$ you need at least one of $d_1$ or $d_2$ to be zero. Both of them have to be non-negative also. So, $S$ is the union of the positive $X$ semi-axis and the positive $Y$ semi-axis, including the origin.

$P(S)$ is the set of points (vectors) that form an obtuse angle with all the vectors in $S$. Clearly, these are the points in the third quadrant (just draw the picture), that is $P(S)=\{(d_1,d_2):\, d_1\le 0, d_2\le 0\}$.

Hope this helps.

GReyes
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