Ayanna Trujillo

2022-07-01

Describe

$S=\{(u,t)\in {\mathbb{R}}^{n}\times \mathbb{R}:{u}^{\prime}x\le t\text{for all}x\in {\mathbb{R}}^{n}\text{with}Ax\le b,{u}^{\prime}x\ge t\text{for all}x\in {\mathbb{R}}^{n}\text{with}Cx\le d\}$

using finitely many linear inequalities and equations.

It's easy to show S is convex. We can describe S as

$S=\{(u,t)\in {\mathbb{R}}^{n}\times \mathbb{R}:\underset{x\in {\mathbb{R}}^{n},Ax\le b}{sup}{u}^{\prime}x\le t\le \underset{x\in {\mathbb{R}}^{n},Cx\le d}{inf}{u}^{\prime}x\}.$

Taking the supremum and infimum over infinitely many expressions is not allowed in the finitely many linear inequalities and equations. The LHS and RHS are solutions to linear programming problems and the hint is to use strong duality, but I couldn't find a way to make the number of conditions finite.

$S=\{(u,t)\in {\mathbb{R}}^{n}\times \mathbb{R}:{u}^{\prime}x\le t\text{for all}x\in {\mathbb{R}}^{n}\text{with}Ax\le b,{u}^{\prime}x\ge t\text{for all}x\in {\mathbb{R}}^{n}\text{with}Cx\le d\}$

using finitely many linear inequalities and equations.

It's easy to show S is convex. We can describe S as

$S=\{(u,t)\in {\mathbb{R}}^{n}\times \mathbb{R}:\underset{x\in {\mathbb{R}}^{n},Ax\le b}{sup}{u}^{\prime}x\le t\le \underset{x\in {\mathbb{R}}^{n},Cx\le d}{inf}{u}^{\prime}x\}.$

Taking the supremum and infimum over infinitely many expressions is not allowed in the finitely many linear inequalities and equations. The LHS and RHS are solutions to linear programming problems and the hint is to use strong duality, but I couldn't find a way to make the number of conditions finite.

plodno8n

Beginner2022-07-02Added 17 answers

Step 1

This is done with duality. Let me show one inequality:

$\underset{x}{max}\{{u}^{T}x:Ax\le b\}\le t$

$\underset{y}{min}\{{b}^{T}y:{A}^{T}y=u,y\ge 0\}\le t$

$\mathrm{\exists}y\ge 0:{b}^{T}y\le t,{A}^{T}y=u$

This is done with duality. Let me show one inequality:

$\underset{x}{max}\{{u}^{T}x:Ax\le b\}\le t$

$\underset{y}{min}\{{b}^{T}y:{A}^{T}y=u,y\ge 0\}\le t$

$\mathrm{\exists}y\ge 0:{b}^{T}y\le t,{A}^{T}y=u$

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