# Convex Set

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• Topic: Convex set, Vector space, Convex geometry
• Pages : 489 (109682 words )
• Published : February 26, 2013

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Convex Optimization Solutions Manual

Stephen Boyd

Lieven Vandenberghe

January 4, 2006

Chapter 2

Convex sets

Exercises

Exercises
Deﬁnition of convexity
2.1 Let C ⊆ Rn be a convex set, with x1 , . . . , xk ∈ C, and let θ1 , . . . , θk ∈ R satisfy θi ≥ 0, θ1 + · · · + θk = 1. Show that θ1 x1 + · · · + θk xk ∈ C. (The deﬁnition of convexity is that this holds for k = 2; you must show it for arbitrary k.) Hint. Use induction on k. Solution. This is readily shown by induction from the deﬁnition of convex set. We illustrate the idea for k = 3, leaving the general case to the reader. Suppose that x 1 , x2 , x3 ∈ C, and θ1 + θ2 + θ3 = 1 with θ1 , θ2 , θ3 ≥ 0. We will show that y = θ1 x1 + θ2 x2 + θ3 x3 ∈ C. At least one of the θi is not equal to one; without loss of generality we can assume that θ1 = 1. Then we can write where µ2 = θ2 /(1 − θ1 ) and µ2 = θ3 /(1 − θ1 ). Note that µ2 , µ3 ≥ 0 and µ1 + µ 2 = y = θ1 x1 + (1 − θ1 )(µ2 x2 + µ3 x3 )

1 − θ1 θ2 + θ 3 = = 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 are in C, y ∈ C. 2.2 Show that a set is convex if and only if its intersection with any line is convex. Show that a set is aﬃne if and only if its intersection with any line is aﬃne. Solution. We prove the ﬁrst part. The intersection of two convex sets is convex. Therefore if S is a convex set, the intersection of S with a line is convex. Conversely, suppose the intersection of S with any line is convex. Take any two distinct points x1 and x2 ∈ S. The intersection of S with the line through x1 and x2 is convex. Therefore convex combinations of x1 and x2 belong to the intersection, hence also to S. 2.3 Midpoint convexity. A set C is midpoint convex if whenever two points a, b are in C, the average or midpoint (a + b)/2 is in C. Obviously a convex set is midpoint convex. It can be proved that under mild conditions midpoint convexity implies convexity. As a simple case, prove that if C is closed and midpoint convex, then C is convex. Solution. We have to show that θx + (1 − θ)y ∈ C for all θ ∈ [0, 1] and x, y ∈ C. Let θ(k) be the binary number of length k, i.e., a number of the form with ci ∈ {0, 1}, closest to θ. By midpoint convexity (applied k times, recursively), θ(k) x + (1 − θ (k) )y ∈ C. Because C is closed, k→∞

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

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

2.4 Show that the convex hull of a set S is the intersection of all convex sets that contain S. (The same method can be used to show that the conic, or aﬃne, or linear hull of a set S is the intersection of all conic sets, or aﬃne sets, or subspaces that contain S.) Solution. Let H be the convex hull of S and let D be the intersection of all convex sets that contain S, i.e., D= {D | D convex, D ⊇ S}.

We will show that H = D by showing that H ⊆ D and D ⊆ H. First we show that H ⊆ D. Suppose x ∈ H, i.e., x is a convex combination of some points x1 , . . . , xn ∈ S. Now let D be any convex set such that D ⊇ S. Evidently, we have x1 , . . . , xn ∈ D. Since D is convex, and x is a convex combination of x1 , . . . , xn , it follows that x ∈ D. We have shown that for any convex set D that contains S, we have x ∈ D. This means that x is in the intersection of all convex sets that contain S, i.e., x ∈ D. Now let us show that D ⊆ H. Since H is convex (by deﬁnition) and contains S, we must have H = D for some D in the construction of D, proving the claim.

2 Examples

Convex sets

2.5 What is the distance between two parallel hyperplanes {x ∈ Rn | aT x = b1 } and {x ∈ Rn | aT x = b2 }? Solution. The distance between the two hyperplanes is |b1 − b2 |/ a 2 . To see this, consider the construction in the ﬁgure below.

x2 = (b2 / a 2 )a

PSfrag replacements

a

x1 = (b1 / a 2 )a

aT x = b 2

aT x = b 1 The distance between the two hyperplanes is also the distance between the two points x1 and x2 where the hyperplane...