We have p u p 0 t u for any u where maximizes g

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Unformatted text preview: (6), then f (x) = inf x L(˜, λ) = g (λ), where L is the Lagrangian x ˜ of the problem and g is the dual function. Therefore strong duality holds, and x is globally optimal. (b) Next we show that the conditions (6) are also necessary. Assume that x is globally optimal for (5). We distinguish two cases. (i) (ii) x 2 < 1. Show that (6) holds with λ = 0. x 2 = 1. First prove that (A + λI )x = −b for some λ ≥ 0. (In other words, the negative gradient −(Ax + b) of the objective function is normal to the unit sphere at x, and point away from the origin.) You can show this by contradiction: if the condition does not hold, then there exists a direction v with v T x < 0 and v T (Ax + b) < 0. Show that f (x + tv ) < f (x) for small positive t. It remains to show that A + λI 0. If not, there exists a w with wT (A + λI )w < 0, and without loss of generality we can assume that wT x = 0. Show that the point y = x + tw with t = −2wT x/wT w satisfies y 2 = 1 and f (y ) < f (x). (c) The optimality conditi...
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This note was uploaded on 09/10/2013 for the course C 231 taught by Professor F.borrelli during the Fall '13 term at University of California, Berkeley.

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