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By Stephen Boyd,Lieven Vandenberghe

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1)Q = tq1 q1T + qi qiT . We have n tr XY − tr X −1 = tλ1 + i=2 λi − 1/t − (n − 1), which grows unboundedly as t → ∞. Therefore Y ∈ dom f ∗ . Next, assume Y 0. If Y ≺ 0, we can find the maximum of tr XY − tr X −1 by setting the gradient equal to zero. , X = (−Y )−1/2 , and f ∗ (Y ) = −2 tr(−Y )1/2 . Finally we verify that this expression remains valid when Y 0, but Y is singular. , have closed epigraphs. 38 Young’s inequality. Let f : R → R be an increasing function, with f (0) = 0, and let g be its inverse.

Recall that the support function of a set C ⊆ R n is defined as SC (y) = sup{y T x | x ∈ C}. On page 81 we showed that SC is a convex function. (a) Show that SB = Sconv B . (b) Show that SA+B = SA + SB . (c) Show that SA∪B = max{SA , SB }. (d) Let B be closed and convex. Show that A ⊆ B if and only if SA (y) ≤ SB (y) for all y. Solution. 3 Convex functions (a) Let A = conv B. Since B ⊆ A, we obviously have SB (y) ≤ SA (y). , yT u < yT v for all u ∈ B and some v ∈ A. , v = i θi ui , with θi ≥ 0, θ = 1.

By convexity (part d), and homogeneity, MC (x + y) = 2MC ((1/2)x + (1/2)y) ≤ MC (x) + MC (y). 35 Support function calculus. Recall that the support function of a set C ⊆ R n is defined as SC (y) = sup{y T x | x ∈ C}. On page 81 we showed that SC is a convex function. (a) Show that SB = Sconv B . (b) Show that SA+B = SA + SB . (c) Show that SA∪B = max{SA , SB }. (d) Let B be closed and convex. Show that A ⊆ B if and only if SA (y) ≤ SB (y) for all y. Solution. 3 Convex functions (a) Let A = conv B.

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