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Unformatted text preview: 4. CONVEXITY AND OPTIMIZATION Convexity plays a natural role in optimization. It has tremendous intuitive appeal and provides powerful and elegant approaches to a wide variety of pure and applied problems. 4.1 Convex Sets and Functions Definition 4.1.1 (Convex sets) . We say that S R n is convex if (1 ) x + y S whenever x,y S and [0 , 1]. (See Figure 4.1.) interior of an ellipse with any subset of its boundary line segment in space quadrant any subset of the boundary of an ellipse finite set of points in space Convex sets Nonconvex sets Figure 4.1 Examples of convex and nonconvex sets Example 4.1.2 (Some convex sets) . The following sets are convex: (a) points, lines and planes; (b) boxes (as defined in Definition 1.3.3); (c) the empty set and all of R n . Proposition 4.1.3 (Intersection property) . If S 1 ,...,S k R n are all convex, then T k i =1 S i is also convex. (See Figure 4.2.) Figure 4.2 Intersection of 3 convex sets (triangular, circular, and parabolic regions) 37 Definition 4.1.4 (Epigraph of a function) . The epigraph of a function f : S R is the set of all points in S R that lie on or above the graph of f (see Figure 4.3.): epi f = { ( x, ) S R  f ( x ) } . epi f Figure 4.3 Epigraph of a (nonconvex) function Definition 4.1.5 (Convex functions) . A function on S is said to be convex on S if its epigraph is a convex set. (See Figure 4.4.) In particular, the domain of a convex function is always a convex set. Figure 4.4 Epigraphs of convex functions Definition 4.1.5 (Hypographs; concave functions) . (a) The hypograph of f : S R is defined as the set hypo f = { ( x, ) S R  f ( x ) } . (b) A function is said to be concave if its hypograph is a convex set. Equivalently, f is concave if f is convex. 38 Theorem 4.1.6 (Secant criterion for convexity) . Suppose S R n is a convex set. The function f : S R is convex if and only if f ( (1 ) x + y ) (1 ) f ( x ) + f ( y ) for all x,y S and all [0 , 1] . (See Figure 4.5.) convex nonconvex Figure 4.5 Secant criterion for convexity of a function Proof. : Suppose that f is convex on S and consider x,y S with [0 , 1] Because ( x,f ( x ) ) epi f and ( y,f ( y ) ) epi f , we have (1 ) ( x,f ( x ) ) + ( y,f ( y ) ) = ( (1 ) x + y, (1 ) f ( x ) + f ( y ) ) epi f, so f ( (1 ) x + y ) (1 ) f ( x ) + f ( y ). : Now consider ( x, ) , ( y, ) epi f and [0 , 1]. If f ( (1 ) x + y ) (1 ) f ( x ) + f ( y ) (1 ) + , then ( (1 ) x + y, (1 ) f ( x ) + f ( y ) ) = (1 )( x, ) + ( y, ) epi f....
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This note was uploaded on 03/18/2012 for the course MTH 432 taught by Professor Douglasward during the Spring '12 term at Miami University.
 Spring '12
 DouglasWard
 Sets

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