PROOF OF EQUATION (5.12)
VERA ROSHCHINA
We want to show that for a closed convex cone C ⊆ Rn and a linear subspace L ⊆ Rn we have
(C ∩ L)∗ = C ∗ + L⊥ .
Recall that here the dual cone C ∗ to a closed convex cone C ⊂ Rn is defined as
C ∗ = {y ∈ Rn | hx, yi ≥ 0 ∀ x ∈ C},
and that L⊥ = L∗ is the orthogonal complement to the linear subspace L.
We will prove a more general result, namely, that for C1 , C2 ⊆ Rn , where C1 and C2 are closed
convex cones, we have
(C1 ∩ C2 )∗ = (C1∗ + C2∗ ).
We first show that C1∗ + C2∗ ⊆ (C1 ∩ C2 )∗ , and hence (C1∗ + C2∗ ) ⊆ (C1 ∩ C2 )∗ .
Observe that if y ∈ C1∗ + C2∗ , then y = u + v, where u and v are such that
hu, pi ≥ 0 ∀p ∈ C1 ,
hv, qi ≥ 0 ∀q ∈ C2 .
From here we have that
hu, xi ≥ 0, hv, xi ≥ 0 ∀x ∈ C1 ∩ C2 ,
and hence
hy, xi = hu + v, xi ≥ 0 ∀x ∈ C1 ∩ C2 ,
which means that y ∈ (C1 ∩ C2 )∗ .
It remains to show that (C1 ∩ C2 )∗ ⊆ (C1∗ + C2∗ ). For this we will assume the contrary and use
the separation theorem.
Assume there exists a point z ∈ (C1 ∩ C2 )∗ \ (C1∗ + C2∗ ). Then the point z can be separated from
the closed convex set (C1∗ + C2∗ ) by a hyperplane: there exists u ∈ Rn such that
hz, ui >
sup
hx, ui.
x∈(C1∗ +C2∗ )
Observe that the supremum on the right hand side can be either zero or +∞, since (C1∗ + C2∗ )
is a cone. It clearly can not be infinite because of the bound on the left hand side, hence
sup
hx, ui = 0,
x∈(C1∗ +C2∗ )
and
hz, ui > 0.
(1)
C1∗
Since hx, ui ≤ 0 for all x ∈
and all x ∈ C2∗ , we have −u ∈ C1∗∗ ∩ C2∗∗ = C1 ∩ C2 (the double
dual coincides with the original cone due to closedness and convexity). Since −u ∈ C1 ∩ C2 and
z ∈ (C1∗ + C2∗ ), we have
hu, zi ≥ 0,
which contradicts (1).
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