Convex Collection Research Paper

Convex Optimization Solutions Manual

Stephen Boyd

Lieven Vandenberghe

January 4, 2006

Phase 2

Convex sets

Physical exercises

Exercises

Definition of convexity

2 . 1 Let C ⊆ Rn be a convex set, with x1,..., xk ∈ C, and let θ1,..., θk ∈ R gratify θi ≥ 0, θ1 + · · · + θk = 1 . Show that θ1 x1 + · · · + θk xk ∈ C. (The definition of convexity is the holds intended for k = 2; you need to show this for irrelavent k. ) Hint. Make use of induction in k. Answer. This is quickly shown by induction from your definition of convex established. We illustrate the idea for k = 3, leaving the general circumstance to the target audience. Suppose that back button 1, x2, x3 ∈ C, and θ1 & θ2 + θ3 = 1 with θ1, θ2, θ3 ≥ 0. We all will show that y = θ1 x1 + θ2 x2 + θ3 x3 ∈ C. At least one of the θi is not really equal to 1; without decrease of generality we are able to assume that θ1 = 1 . Then we can write where µ2 sama dengan θ2 /(1 − θ1 ) and µ2 sama dengan θ3 /(1 − θ1 ). Remember that µ2, µ3 ≥ 0 and µ1 + µ 2 sama dengan y = θ1 x1 + (1 − θ1 )(µ2 x2 + µ3 x3 )

1 − θ1 θ2 + θ 3 sama dengan = 1 . 1 − θ1 1 − θ1 Since C is convex and x2, x3 ∈ C, we conclude that µ2 x2 + µ3 x3 ∈ C. Since this point and x1 will be in C, y ∈ C. 2 . 2 Present that a arranged is convex if and only if the intersection with any series is convex. Show that a set is definitely affine in the event that and only if its intersection with any kind of line is affine. Solution. We confirm the first part. The intersection of two convex sets is convex. Therefore if S can be described as convex collection, the area of H with a collection is convex. Conversely, imagine the area of S i9000 with any line is definitely convex. Have any two distinct details x1 and x2 ∈ S. The intersection of S together with the line through x1 and x2 is usually convex. Consequently convex combos of x1 and x2 belong to the intersection, consequently also to S. installment payments on your 3 Midpoint convexity. A set C is midpoint convex if perhaps whenever two-points a, n are in C, the average or midpoint (a + b)/2 is within C. Naturally a convex set is usually midpoint convex. It can be demonstrated that underneath mild conditions midpoint convexity implies convexity. As a straightforward case, provide evidence that if C is closed and midpoint convex, then simply C can be convex. Solution. We have to display that θx + (1 − θ)y ∈ C for all θ ∈ [0, 1] and x, y ∈ C. Let θ(k) be the binary range of length e, i. elizabeth., a number of the contact form with ci ∈ 0, 1, closest to θ. By midpoint convexity (applied e times, recursively), θ(k) times + (1 − θ (k) )y ∈ C. Because C is shut, k→∞

θ(k) = c1 2−1 + c2 2−2 + · · · + ck 2−k

lim (θ(k) by + (1 − θ (k) )y) = θx + (1 − θ)y ∈ C.

2 . 4 Show that the convex outer skin of a arranged S is definitely the intersection of most convex pieces that contain T. (The same method can be used to show that the conic, or perhaps affine, or linear hull of a established S is definitely the intersection of conic units, or affine sets, or perhaps subspaces which contain S. ) Solution. Allow H end up being the convex hull of S and let D become the intersection of all convex sets which contain S, we. e., D= D convex, D ⊇ S.

We all will show that H = D simply by showing that H ⊆ D and D ⊆ H. 1st we demonstrate that H ⊆ G. Suppose times ∈ H, i. electronic., x is a convex mixture of some details x1,..., xn ∈ T. Now let D always be any convex set such that D ⊇ S. Evidently, we have x1,..., xn ∈ D. Since D is usually convex, and x is known as a convex mixture of x1,..., xn, it comes after that x ∈ G. We have shown that for just about any convex set D which has S, we certainly have x ∈ D. Which means that x is in the intersection of most convex sets that contain T, i. electronic., x ∈ D. Now let us show that G ⊆ L. Since H is convex (by definition) and contains H, we must have got H sama dengan D for some D in the construction of D, demonstrating the claim.

2 Examples

Convex sets

installment payments on your 5 What is the distance among two seite an seite hyperplanes x ∈ Rn and aT x = b2 ? Solution. The space between the two hyperplanes is |b1 − b2 |/ a installment payments on your To see this kind of, consider the construction in the figure below.

x2 = (b2 / a 2 )a

PSfrag substitutes

a

x1 = (b1 / a 2 )a

aT by = w 2

for x = b 1 The distance between the two hyperplanes is also the length between the two-points x1 and x2 the place that the hyperplane...



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