2 Separating Hyperplanes 3 Banach–mazur Distance

We’ll use the above result to show why the polar of the polar of a convex body is the body itself. Recall that for a convex body K, we had defined its polar K∗ to be {p|k · p ≤ 1∀k ∈ K}. Theorem 2 Let K be a convex body. Then K∗∗ = K. Proof We know that K∗ = {p|k · p ≤ 1∀k ∈ K}. Similarly K∗∗ = {y|p · y ≤ 1∀p ∈ k∗}. Let y be any point in K. Then, by the definition of the polar, for all p ∈ K∗ we have that p · y ≤ 1. The definition of the polar of K∗ implies that y ∈ K∗∗ . Since this is true for every y ∈ K, we conclude that K ⊆ K∗∗ . The other direction of the proof is the nontrivial one and we’ll have to use the convexity of the body and the separating hyperplane theorem. Suppose that we can find a y ∈ K∗∗ such that y / ∈ K. Since y ∈ K∗∗ , we have that p · y ≤ 1 for all p ∈ K∗ . Since y ∈ K, there exists a strongly separating hyperplane for y and K. Let it be H = {x|v · x = 1}. By the definition of separating hyperplane, we have v · k ≤ 1 for all k ∈ K. Hence, v ∈ K∗ . Also, v · y > 1 (since H is a separating hyperplane), and we just showed that v ∈ K∗ . This contradicts our assumption that y ∈ K∗∗ . Hence K∗∗ ⊆ K.