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The polar is usually defined as $A^\circ = \{ x | \langle x, a \rangle \le 1, \forall a \in A \}$. In this case, we have $A^\circ = \{ (a,b) | ax+by \le 1, \forall y \ge x^2 \}$. Suppose $(a,b) \in A^\circ$. We see that $b \le 0$, since if $b>0$, we would need to have $ax+bx^2 \le 1$ for all $x$ which is impossible. If $b = 0$, then we see that we must ...


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I would begin with the concept of support function, which is defined as follows: for a unit vector $\xi$, $h(\xi) = \sup_{a\in A}( a\cdot \xi)$. Geometrically, this is what you get by projecting $A$ onto the line with direction $\xi$, and taking the point that's furthest away in direction of $\xi$. The polar dual is formed by $0$ and the points $x\ne 0$ ...


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Just for completeness: the question is actually related with duality because can be solved using the Farkas lemma. Indeed, $P\subseteq Q$ if and only for each $i$ here is no solution to the linear system of inequalities $$Aa\geq a , B_i x<b_i,$$ otherwise said, if none of these system can be satisfied then the inclusion holds. By the Farkas lemma this ...


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Well, I've solved it and apparently it really has nothing to do with the duality theorem. The answer is as follows: let $v_i'$ be i'th row in the matrix B, $i = 1, ..., n$. for each i solve the linear program: min $v_i'\vec x$ for $A\vec x \geq \vec a$ and check if the optimum for this program is $\geq b_i $. if the optimum is smaller than $b_i$ than ...



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