v2009.01.01 - Convex Optimization

2.1. CONVEX SET 43
This theorem in converse is implicitly false in so far as a convex set can
be formed by the intersection of sets that are not. Unions of convex sets are
generally not convex. [173, p.22]
Vector sum of two convex sets C 1 and C 2
C 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 convex
sets
C 1 − C 2 = {x − y | x ∈ C 1 , y ∈ C 2 } (20)
which is convex. Applying this definition to nonempty convex set C 1 , its
self-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.
[266, p.15]
Cartesian product of convex sets
C 1 × C 2 =
{[ x
y
]
| x ∈ C 1 , y ∈ C 2
}
=
[
C1
C 2
]
(21)
remains convex. The converse also holds; id est, a Cartesian product is
convex iff each set is. [173, 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)
meaning
T(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)