v2007.09.13 - Convex Optimization

2.1. CONVEX SET 43This theorem in converse is implicitly false in so far as a convex set canbe formed by the intersection of sets that are not.Vector sum of two convex sets C 1 and C 2C 1 + C 2 = {x + y | x ∈ C 1 , y ∈ C 2 } (19)is convex.By additive inverse, we can similarly define vector difference of two convexsetsC 1 − C 2 = {x − y | x ∈ C 1 , y ∈ C 2 } (20)which is convex. Applying this definition to nonempty convex set C 1 , itsself-difference C 1 − C 1 is generally nonempty, nontrivial, and convex; e.g.,for any convex cone K , (2.7.2) the set K − K constitutes its affine hull.[228, p.15]Cartesian product of convex setsC 1 × C 2 ={[ xy]| x ∈ C 1 , y ∈ C 2}=[C1C 2](21)remains convex. The converse also holds; id est, a Cartesian product isconvex iff each set is. [147, p.23]Convex results are also obtained for scaling κ C of a convex set C ,rotation/reflection Q C , or translation C+ α ; all similarly defined.Given any operator T and convex set C , we are prone to write T(C)meaningT(C) ∆ = {T(x) | x∈ C} (22)Given linear operator T , it therefore follows from (19),T(C 1 + C 2 ) = {T(x + y) | x∈ C 1 , y ∈ C 2 }= {T(x) + T(y) | x∈ C 1 , y ∈ C 2 }= T(C 1 ) + T(C 2 )(23)