1

I've been trying to prove the Strong Duality Theorem following these slides. In those slides, we start with proving Farkas' lemma, then a corollary followed by the proof of the Strong Duality Theorem. To prove Farkas' lemma, I first proved that $\{A(x)|x \in X_n\}$ where $X_j = \{x=(x_1,x_2 \dots x_j) \in \mathbb{R}^j | x_i \geq 0 \text{ for all } 1\leq i \leq j\}$ and $A \in M(m,n)$, is a closed, convex set. This was deceptively hard and has been discussed in these answers. Using this I managed to prove the following: Let $b\in \mathbb{R}^m$. Exactly one of these two linear programs have a solution: \begin{align} Ax&=b\\ x &\in X_n \nonumber \end{align} \begin{align} (A^Ty) &\in X_n\\ \langle b,y \rangle &< 0 \nonumber\\ y &\in \mathbb{R}^m \end{align} This is Farkas' lemma. I also managed to prove this "corollary". I put it in quotes because I could not use the Farkas' lemma directly to get the result. The corollary is as follows: Let $b\in \mathbb{R}^m$. Exactly one of these two linear programs have a solution: \begin{align} Ax+s&=b\\ x &\in X_n \nonumber\\ s &\in X_m \nonumber \end{align} \begin{align} A^Ty &\in X_n \nonumber\\ \langle b,y \rangle &< 0 \\ y &\in X_m \nonumber \end{align}

The final step of this puzzle, which directly proves the Strong Duality Theorem is what I am trying to solve. This is what I am trying to prove now: For any $\alpha \in \mathbb{R}$, $b\in \mathbb{R}^m$, and $c\in \mathbb{R}^n$, prove that exactly one of these two linear programs have a solution: \begin{align} Ax+s&=b\\ \langle c,x \rangle &\leq \alpha \nonumber\\ x &\in X_n \nonumber\\ s &\in X_m \nonumber \end{align} \begin{align} \langle b,y \rangle + \alpha z&< 0 \\ A^Ty + cz&\in X_n \nonumber\\ y &\in X_m \nonumber\\ z &\in \mathbb{R}_+ \nonumber \end{align} Here are my questions:

  • Is there a way to prove Farkas' corollary using Farkas' lemma?
  • How do I prove the last step from Farkas' corollary?
KPN
  • 39
  • 1
    Here's another good thread about the fact that ${Ax \mid x \geq 0 }$ is a closed set: https://math.stackexchange.com/questions/1831401/how-do-you-prove-that-ax-mid-x-geq-0-is-closed – littleO Nov 03 '20 at 08:00

1 Answers1

1

The first system of the last alternative is $$ Ax+s = b, \ \langle c,x\rangle \le \alpha, \ x\ge 0, \ s \ge0. $$ Introducting a slack variable $t\ge 0$, this can be written as $$ \pmatrix{A& I & 0 \\ c^T&0 & 1} \pmatrix{ x \\ s \\ t} = \pmatrix {b \\ \alpha } $$ with $x\ge0,s\ge0,t\ge0$. By Farkas Lemma, either this system or the following one is solvable: $$ \pmatrix{A^T & c\\ I & 0 \\ 0 & 1} \pmatrix{y \\ z} \ge 0 $$ $$ b^Ty + \alpha z < 0, $$ which is the second alternative in the last system.

daw
  • 54,637
  • 2
  • 44
  • 85
  • From Farkas' corollary, I get that one of $(A , I) \cdot (x , s)^T = b$ or $(A^T , I)^T \cdot y \geq 0$, $b^Ty<0$ is solvable. How do I extend that to include the extra variables - $\alpha$, $z$? – KPN Nov 03 '20 at 10:44
  • 1
    I applied the Farkas Lemma (instead of the corollary). Also $\alpha$ is not a variable. You have to introduce extra variables to be able to apply it. – daw Nov 03 '20 at 11:53
  • @daw: Can you clarify how knowing that the first system in $x$ is consistent (which implies that the second in $y$ is inconsistent) shows strong duality? I thought we would like to show $"\forall\alpha:c^\top x \leq \alpha \implies b^\top y \leq \alpha" \iff c^\top x \geq b^\top y$ – jjjjjj Jun 28 '21 at 21:48