Gruber P. Convex and Discrete Geometry

3 Convex Sets, Convex Bodies and Convex Hulls 45
Proposition 3.2. Let A ⊆ E d be bounded. Then cl conv A = conv cl A.
Proof. Since conv A is convex, Proposition 3.1 shows that cl conv A is convex too.
cl conv A is a closed set which contains A. Thus it also contains cl A. Since cl conv A
is a convex set which contains cl A, it also contains conv cl A.
The set conv cl A is convex and contains A, thus it contains conv A. Since by
Corollary 3.1 conv cl A is compact and thus closed, it contains cl conv A. ⊓⊔
Convex Cones
A notion which is important in several contexts, for example in ordered topological
vector spaces and in linear optimization, is that of convex cones. A closed set C in
E d is a closed convex cone with apex o if it satisfies the following property:
λx + µy ∈ C for all x, y ∈ C, λ,µ≥ 0.
Then, in particular, C is convex and contains, with each point x, alsotheray{λx :
λ ≥ 0}. An example is the positive (non-negative) orthant {x : xi ≥ 0}. Aclosed
convex cone with apex a ∈ E d is simply the translate of a closed convex cone with
apex o by the vector a.Thelineality space L of a closed convex cone C with apex o
is the linear subspace
L = C ∩ (−C)
of E d . It is the largest linear subspace of E d which is contained in C. The convex
cone C is pointed if L ={o}. IfH is a hyperplane containing only the point o of a
pointed closed convex cone C and p ∈ C, �= o, then C ∩ (H + p) is a convex body,
sometimes called a basis of C. Itgenerates C in the sense that
C = � � λ � C ∩ (H + p) � : λ ≥ 0 � .
For later reference we prove the following simple result, where L ⊥ denotes the
orthogonal complement of L, i.e. L ⊥ ={y : x · y = 0 for all x ∈ L}. Clearly,
L ⊥ is a subspace of E d and E d = L ⊕ L ⊥ .
Proposition 3.3. Let C be a closed convex cone in E d with apex o and lineality space
L. Then
C = (C ∩ L ⊥ ) ⊕ L,
where C ∩ L ⊥ is a pointed closed convex cone with apex o.
Proof. First, the equality will be shown. Let x ∈ C. Noting that E d = L ⊥ ⊕ L, we
have x = y + z for some y ∈ L ⊥ , z ∈ L. Since C is a convex cone with apex o and
x ∈ C, −z ∈−L = L ⊆ C, it follows that y = x − z ∈ C. Hence x = y + z, where
y ∈ C ∩ L ⊥ and z ∈ L. Thus x ∈ (C ∩ L ⊥ )⊕ L. This shows that C ⊆ (C ∩ L ⊥ )⊕ L.
If, conversely, x = y + z ∈ (C ∩ L ⊥ ) ⊕ L, where y ∈ C ∩ L ⊥ , z ∈ L ⊆ C, then,
noting that C is a convex cone with apex o, it follows that x = y + z ∈ C. Thus
C ⊇ (C ∩ L ⊥ ) ⊕ L, concluding the proof of the equality.