1

It is the exercise 4.37 from "Introduction to Linear Optimization" by Bertsimas.

Let $\textbf{A}_1$, ..., $\textbf{A}_n$ be given vectors in $\mathbb{R}^m$. Consider the cone $C=\{\sum_{i=1}^n \textbf{A}_ix_i\mid x_i\geq 0\}$ and let $\textbf{y}^k, k=1,2,\dots$ be a sequence of elements of $C$ that converges to some $\textbf{y}$. Show that $\textbf{y}\in C$ (and hence $C$ is closed), using the following argument. With $\textbf{y}$ fixed as above, consider the problem of minimizing $\parallel\textbf{y}-\sum_{i=1}^n\textbf{A}_ix_i\parallel_\infty$, subject to the constraints $x_1,\dots,x_n\geq 0$. Here $\parallel\cdot\parallel_\infty$, stands for the $\parallel\textbf{x}\parallel_\infty=\max_i\mid x_i\mid$. Explain why the above minimization problem has an optimal solution, find the value of the optimal cost, and prove that $\textbf{y}\in C$.

In other words, I have to prove that for $A\in\mathbb{R}^{m\times n}$ then the set $E=\{Ax\mid x\in\mathbb{R}_{\geq 0}^n\}$ is closed in $\mathbb{R}^m$.

I understand why the optimal value is $0$. But I have no idea about how to prove that the minimization problem has optimal solution.

I want to use the exercise to prove the Farkas' lemma and so the Strong Duality, so I do not want to use any of them in the exercise. Is it possible to use only the Wierstrass theorem about continuous functions and at worst the particular case of the Separating Hyperplane Theorem for point versus nonempty closed convex set? Given a $y$ in the closure of $E$, I tried to use the following continuous function $x\mapsto$ $\parallel y-Ax\parallel_\infty$ and to find a compact domain that is a subset of $\mathbb{R}_{\geq 0}^n$ to find a minimum point, but I was not able to solve it.

Daniel Kawai
  • 1,027
  • 6
  • 19
  • see also this thread https://math.stackexchange.com/questions/1165286/why-are-convex-polyhedral-cones-closed – daw Nov 06 '18 at 20:40
  • and this https://math.stackexchange.com/questions/1831401/how-do-you-prove-that-ax-mid-x-geq-0-is-closed?noredirect=1&lq=1 – daw Nov 06 '18 at 20:41

1 Answers1

1

If $P$ is a polyhedron, then set set $Q=\{Ax|x \in P\}$ is also a polyhedron.

Here our $P$ is $\{x \in \mathbb{R}^n|x \ge 0\}$ and our $Q$ is $E$.

Polyhedron are intersection of halfspaces, which are closed. Hence $E$ is closed.

Siong Thye Goh
  • 153,832
  • Ok, but why is $Q$ a polyhedron? I know it looks like intuitive but a proof of it is not so trivial to me. I know a polyhedron is a set of the form $P={x\in\mathbb{R}^n\mid Ax\leq a}$, where $A\in\mathbb{R}^{r\times n}$ and $a\in\mathbb{R}^r$, but if $M\in\mathbb{R}^{m\times n}$, what are examples of new $A'$ and $a'$ such that the set $Q={Mx\mid x\in P}$ is of the form ${x\in\mathbb{R}^m\mid A'x\leq a'}$, or at least how to prove that $Q$ is a polyhedron? – Daniel Kawai Nov 06 '18 at 16:04
  • I don't have the book that you mentioned with me now but from memory, flip to the summary page of chapter $2$, look at the page just before that. It is a theorem (or perhaps corollary) proven in the book. – Siong Thye Goh Nov 06 '18 at 16:11