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v2009.01.01 - Convex Optimization

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2.13. DUAL CONE & GENERALIZED INEQUALITY 191<br />

To the i th extreme direction v = Γ i ∈ R n of cone K , ascribe a coordinate<br />

t ⋆ i(x)∈ R of x from definition (432). On domain K , the mapping<br />

⎡ ⎤<br />

t ⋆ 1(x)<br />

t ⋆ ⎢ ⎥<br />

(x) = ⎣ . ⎦: R n → R N (433)<br />

t ⋆ N (x)<br />

has no nontrivial nullspace. Because x −t ⋆ v must belong to ∂K by definition,<br />

the mapping t ⋆ (x) is equivalent to a convex problem (separable in index i)<br />

whose objective (by (298)) is tightly bounded below by 0 :<br />

t ⋆ (x) ≡ arg minimize<br />

t∈R N<br />

N∑<br />

i=1<br />

Γ ∗T<br />

j(i) (x − t iΓ i )<br />

subject to x − t i Γ i ∈ K ,<br />

i=1... N<br />

(434)<br />

where index j ∈ I is dependent on i and where (by (331)) λ = Γj ∗ ∈ R n is an<br />

extreme direction of dual cone K ∗ that is normal to a hyperplane supporting<br />

K and containing x − t ⋆ iΓ i . Because extreme-direction cardinality N for<br />

cone K is not necessarily the same as for dual cone K ∗ , index j must be<br />

judiciously selected from a set I .<br />

To prove injectivity when extreme-direction cardinality N > n exceeds<br />

spatial dimension, we need only show mapping t ⋆ (x) to be invertible;<br />

[106, thm.9.2.3] id est, x is recoverable given t ⋆ (x) :<br />

N∑<br />

x = arg minimize Γ<br />

x∈R n j(i) ∗T(x<br />

− t⋆ iΓ i )<br />

i=1<br />

(435)<br />

subject to x − t ⋆ iΓ i ∈ K , i=1... N<br />

The feasible set of this nonseparable convex problem is an intersection of<br />

translated full-dimensional pointed closed convex cones ⋂ i K + t⋆ iΓ i . The<br />

objective function’s linear part describes movement in normal-direction −Γj<br />

∗<br />

for each of N hyperplanes. The optimal point of hyperplane intersection is<br />

the unique solution x when {Γj ∗ } comprises n linearly independent normals<br />

that come from the dual cone and make the objective vanish. Because the<br />

dual cone K ∗ is full-dimensional, pointed, closed, and convex by assumption,<br />

there exist N extreme directions {Γj ∗ } from K ∗ ⊂ R n that span R n . So<br />

we need simply choose N spanning dual extreme directions that make the<br />

optimal objective vanish. Because such dual extreme directions preexist<br />

by (298), t ⋆ (x) is invertible.<br />

Otherwise, in the case N ≤ n , t ⋆ (x) holds coordinates of biorthogonal<br />

expansion. Reconstruction of x is therefore unique.

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