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**Sample text**

We begin in this section with the diﬀerentiable, inequality constrained problem ⎧ ⎪ f (x) ⎨ inf subject to gi (x) ≤ 0 for i = 1, 2, . . 5) ⎪ ⎩ x ∈ C. For a feasible point x¯ we deﬁne the active set I(¯ x) = {i | gi (¯ x) = 0}. For this m problem, assuming x¯ ∈ int C, we call a vector λ ∈ R+ a Lagrange multiplier vector for x¯ if x¯ is a critical point of the Lagrangian m λi gi(x) L(x; λ) = f (x) + i=1 x) = 0) and complementary slackness (in other words, ∇f (¯ x) + λi ∇gi (¯ holds: λi = 0 for indices i not in I(¯ x).

2) L(x; λ) = f (x) + λT g(x), is called the Lagrangian. A feasible solution is a point x in dom f satisfying the constraints. We should emphasize that the term ‘Lagrange multiplier’ has diﬀerent ¯ ∈ Rm meanings in diﬀerent contexts. In the present context we say a vector λ + is a Lagrange multiplier vector for a feasible solution x¯ if x¯ minimizes the func¯ over E and λ ¯ satisﬁes the complementary slackness conditions: tion L(·; λ) ¯ x) < 0. λi = 0 whenever gi (¯ We can often use the following principle to solve simple optimization problems.

Proof. 6, we simply have to show that for any ﬁxed d in E there is a subgradient φ satisfying φ, d = f (¯ x; d). Choose a basis {e1 , e2 , . . , en } for E with e1 = d if d is nonzero. Now deﬁne a x; ·), and pk (·) = sequence of functions p0 , p1 , . . , pn recursively by p0 (·) = f (¯ pk−1 (ek ; ·), for k = 1, 2, . . , n. We essentially show that pn (·) is the required subgradient. 2, each pk is everywhere ﬁnite and sublinear. 7 we know lin pk ⊃ lin pk−1 + span {ek }, for k = 1, 2, . . , n, so pn is linear.