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This leads to aT u ≥ 0 for all u ∈ U, and bT w ≥ 0 for all w ∈ W. Since U and W are vector spaces, we conclude that a = 0 and b = 0. Therefore the inequality reduces further to uT Mw ≥ 0, ∀u ∈ U and ∀w ∈ W. Since U and W are vector spaces, we get M = 0. Now we consider the situation t > 0. Without losing generality let us scale the value of t and assume t = 1. As B(C, D) is a cone, the scaling does not change the nature of the membership checking procedure. We now rewrite (23) as (φ(u) + r)(ψ(w) + s) + (φ(u) + r)bT w + (ψ(w) + s)aT u + uT Mw = φ(u)ψ(w) + uT Mw + φ(u)bT w + ψ(w)aT u + (φ(u) + aT u)s + (ψ(w) + bT w)r + rs ≥0 (24) for all u ∈ U, w ∈ W, r ≥ 0 and s ≥ 0.

Since U and W are vector spaces, we get M = 0. Now we consider the situation t > 0. Without losing generality let us scale the value of t and assume t = 1. As B(C, D) is a cone, the scaling does not change the nature of the membership checking procedure. We now rewrite (23) as (φ(u) + r)(ψ(w) + s) + (φ(u) + r)bT w + (ψ(w) + s)aT u + uT Mw = φ(u)ψ(w) + uT Mw + φ(u)bT w + ψ(w)aT u + (φ(u) + aT u)s + (ψ(w) + bT w)r + rs ≥0 (24) for all u ∈ U, w ∈ W, r ≥ 0 and s ≥ 0. It is evident that (24) is equivalent to the following three conditions: A D-induced duality and its applications φ(u) + aT u ≥ 0 181 for all u ∈ U, (25) T ψ(w) + b w ≥ 0 for all w ∈ W, φ(u)ψ(w) + uT Mw + φ(u)bT w + ψ(w)aT u ≥ 0 for all u ∈ U, w ∈ W.

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