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Then there exists a unique nonempty, closed and convex set A ⊂ Rn such that hA = h. Proof. If h ≡ ∞, then h = hRn and, by the support theorem, Rn is the only closed convex set A with hA ≡ ∞. Hence, we may now assume that h is proper. We consider h∗ . For α > 0, we obtain from the positive homogeneity h∗ (x) = sup ( x, y − h(y)) = sup ( x, αy − h(αy)) y∈Rn y∈Rn = α sup ( x, y − h(y)) = αh∗ (x). 3. THE SUPPORT FUNCTION 55 Therefore, h∗ can only obtain the values 0 and ∞. We put A := dom h∗ . 4(a), A is nonempty, closed and convex, and h∗ = δA .

Remark. As a consequence, we obtain that K(u) consists of one point, if and only if hK (u; ·) is linear. 6, the latter is equivalent to the differentiability of hK at u. If all the support sets K(u), u ∈ S n−1 , of a nonempty, compact convex set K consist of points, the boundary bd K does not contain any segments. Such sets K are called strictly convex. Hence, K is strictly convex, if and only if hK is differentiable on Rn \ {0}. We finally consider the support functions of polytopes. We call a function h on Rn piecewise linear, if there are finitely many convex cones A1 , .

G. we may assume that dim K = n, hence 0 ∈ int K. If we copy the proof of (b) with B(ε) = εB(1) replaced by εK, we obtain a polytope P with K ⊂ P ⊂ (1 + ε)K.

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