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Let $A$ be a $m\times n$ matrix and consider the cones $G_0=\{d\in\mathbb R^n:Ad<0\}$ and $G'=\{d\in\mathbb R^n:Ad\le0\}$

Prove that $G'$ is a convex closed cone.

Lets see that $G'=\overline{ G'}.$ Note that this contention $G'\subset\overline{ G'}$ is always true. Let's see the other contention.

Suppose $d\in\overline{G'}$ and $d\not\in G'. $ Thus For every open ball $B$ with $d\in B$ we have that $B\cap G'\neq\emptyset$ and $ Ad>0$

How can I reach the contradition? Is this a good way to prove it or is there a better way? I don't know

And how to prove convexity?

Please help me please thanks

Community
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user441848
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4 Answers4

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To prove that $G^\prime$ is closed use the continuity of the function $d \mapsto Ad$ and the fact that the set $\{d \in \mathbb{R}^n : d \leq 0\}$ is closed.

João Caminada
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  • How will I use continuity? Where? Could you explain with more details? – user441848 Jun 16 '18 at 00:20
  • Use the fact that the inverse image of a closed set by a continuous map is closed. – João Caminada Jun 16 '18 at 00:24
  • ? ${}{}{}{}{}{}$ – user441848 Jun 16 '18 at 00:28
  • Show that $G^\prime$ is the inverse image of the set I specified by the map I specified and use the observation I made in the last comment. – João Caminada Jun 16 '18 at 00:31
  • Note that the image of $d \to Ad$ with $d \in {d \in \mathbb{R}^n : d \leq 0}$ is $G'$ and since a continuos function takes closed sets in the domain to closed sets in the image you got that $G'$ is closed. Look here: https://math.stackexchange.com/questions/437829/the-preimage-of-continuous-function-on-a-closed-set-is-closed – Lucas Resende Jun 16 '18 at 00:31
  • @LucasResende ok I understand now, but in that case I'd need to prove first that $d \in {d \in \mathbb{R}^n : d \leq 0}$ is closed and aslo why the function is continuous? – user441848 Jun 16 '18 at 00:35
  • @LucasResende not image but inverse image. A continuous function takes closed closed sets in the codomain to closed sets in the domain. – João Caminada Jun 16 '18 at 00:36
  • right, becuase we don't know if the function is closed, so I'm lost again – user441848 Jun 16 '18 at 00:37
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And to show that $G'$ is convex just take two points $x,y \in G'$, the segment between these points must lie in $G'$. Let $0\leq \alpha\leq 1$, study the point $p = \alpha x + (1-\alpha)y$: $$ Ap = A(\alpha x + (1-\alpha)y) = \alpha Ax + (1-\alpha)Ay \leq 0 $$ the last inequality is true because $\alpha, 1-\alpha \geq 0$ and since $x,y \in G'$ than $Ax \leq 0$ and $Ay \leq 0$.

Lucas Resende
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To prove $G'$ is closed from scratch without any advanced theorems. Following your suggestion, one way $G'\subset\overline{G'}$ is trivial, let's prove the opposite inclusion by contradiction.

Let's start as you did by assuming that $\exists d\not\in G'$, $d\in\overline{G'}$. Since $d\not\in G'$, there exists one inequality among $Ad\le 0$ that is not true. In other words, there exists a row $a^T$ in the matrix $A$ such that $a^Td>0$. We denote $\rho=a^Td$ and show that for all $x$ that are close enough to $d$ we have $a^Tx>0$ too, which will mean that $x\not\in G'$ as well. Let's estimate $$ a^Tx=a^Td+a^T(x-d)\ge\rho-|a^T(x-d)|\ge\rho-\|a\|\|x-d\|. $$ From here we see that $$ \|x-d\|<\frac{\rho}{\|a\|}=\epsilon\quad\Rightarrow\quad a^Tx>0. $$ Conclusion: we have found an open ball $B$ around $d$ $$ B=\{x\colon \|x-d\|<\epsilon\} $$ such that all $x\in B$ do not belong to $G'$, that is $B\cap G'=\emptyset$. This is a contradiction with the assumption that $d\in\overline{G'}$.

A.Γ.
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  • Thank you for your answer $A.\Gamma.$ I have a question: How do you pass from $a^Tx=a^Td+a^T(x-d)\ge\rho-|a^T(x-d)|\ge\rho-|a||x-d|$ to $\Vert x-d\Vert<\frac{\rho}{||a||}?$ – user441848 Jun 17 '18 at 03:45
  • @user441848 If we want $a^Tx>0$ then it is sufficient to require $\rho-|a||x-d|>0$. The latter is equivalent to $\rho>|a||x-d|$ $\Leftrightarrow$ $\frac{\rho}{|a|}>|x-d|$. – A.Γ. Jun 17 '18 at 06:41
  • ok, got it. I have another question, you say: Since $d∉G′$, there exists one inequality among $Ad≤0$ that is not true. I thought that if $d\not\in G',$ then $Ad>0$. But I think you right if we see $G'={d\in\mathbb R^n:Ad\le0}$ equivalent to $G'={\forall d\in\mathbb R^n:Ad\le0}$.Then the negation is there exists $d$ such that.. They must be equivalent right? – user441848 Jun 17 '18 at 20:53
  • @user441848 The definition of $G'$ is the following: if $A$ is an $m\times n$ matrix then $x\in G'$ iff all $m$ inequalities are $\le$ for this $x$. Now the negaition is: $x$ is not in $G'$ iff not all $m$ inequalities are $\le$ for this $x$. Not all $\le$ is the same as to say that there is at least one $>$. – A.Γ. Jun 17 '18 at 20:57
  • got it, thank you! – user441848 Jun 17 '18 at 21:02
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$Ad \le 0$ iff $e_k^T A d \le 0$ for $i=1,...,m$.

A set of the form $\{ x | n^T x \le \alpha \}$ is quickly checked to be closed and convex from the definition.

The intersection of a collection of closed convex sets is convex.

Elaboration:

$G'=\{ x| Ax \le 0 \} = \{ x | e_1^T A x \le 0, ..., e_m^T A x \le 0 \} = \cap_{k=1}^m \{ x | e_k^T A x \le 0 \}$.

$e_k$ is the vector of zeros except for a one in the $k$th position. Note that $e_k^T A$ is a column vector.

Each set $\{ x | e_k^T A x \le 0 \}$ is a half space, and to check if it closed, suppose $x_n \to x$ and $e_k^TA x_n \le 0$. Then by continuity (of the linear function $x \mapsto e_k^T Ax$) we have , $e_k^TAx \le 0$ and so the set is closed. To check convexity, note that $x \mapsto e_k^T Ax$ is linear and so $e_k^TA (\lambda x_1+(1-\lambda)x_2) = \lambda e_k^T A x_1 + (1-\lambda)e_k^T A x_2$ and so the set is convex.

It is straightforward to check that the intersection of closed sets is closed and it is straightforward to check that the intersection of convex sets is convex.

For the $G_0$ set, the same sort of reasoning applies, the only difference is that the intersection of a finite number of open sets is open (this is the case here), not an arbitrary intersection.

copper.hat
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