convexity - 1 Convexity We defined convex functions f : I R...

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Unformatted text preview: 1 Convexity We defined convex functions f : I R in the lectures. These satisfy the condition f ( x + (1 ) y ) f ( x ) + (1 ) f ( y ) for all x , y I (any real interval) and [0 , 1]. The regularity properties of convex functions can demonstrated by studying the slopes of their chords. For > 0 let + f ( u ) = f ( u + ) f ( u ) and f ( u ) = f ( u ) f ( u ) denote the slopes of chords to the right and left of u I . Good exercise: show that + f ( u ) is non-decreasing in and that similarly f ( u ) is non- increasing in . These monotonicity conditions imply that the limits of f ( u ) both exist as 0 though they may not be equal (as f could be piecewise linear with a corner at u for example). The notion of convexity can be extended to functions of more than one variable. Some further definitions are required for this generalization and to see the connection with differentiability. 1.1 Convex sets and separating hyperplanes. We have to start with a class of sets which are extremely important in optimisation and analysis. Definition 1.1 A set R n is a convex set when x , y , [0 , 1] implies that x + (1 ) y , that is, the line segment from x to y lies entirely within . The sum n 1 i x i , where each i 0 and n 1 i = 1, is the convex combination of the x i . The set = { x R n : n 1 x 2 i /a 2 i 1 } is a convex set (its an ellipsoid). To show this we must establish for any x , y , (0 , 1) , that x + (1 ) y i.e. n summationdisplay 1 1 a 2 i bracketleftbig x i + (1 ) y i bracketrightbig 2 1 . Let A = diag (1 /a 2 i ) , the n n diagonal matrix with entries 1 /a 2 i down the main diagonal. We can now write n summationdisplay 1 1 a 2 i bracketleftbig x i + (1 ) y i bracketrightbig 2 = ( x + (1 ) y ) T A ( x + (1 ) y ) (which you should check!). By expanding this product we find ( x + (1 ) y ) T A ( x + (1 ) y ) = 2 x T Ax + 2 (1 ) x T Ay + (1 ) 2 y T Ay = x T Ax + (1 ) y T Ay (1 ) bracketleftbig x T Ax 2 x T Ay + y T Ay bracketrightbig = x T Ax + (1 ) y T Ay (1 )( x y ) T A ( x y ) , (1) where the equality at * holds because x T Ay = ( x T Ay ) T = y T A T x = y T Ax . We know that for any z R n , z T Az = n 1 z 2 i /a 2 i and by assumption x T Ax 1 and y T Ay 1 . It follows immediately from (1) that ( x + (1 ) y ) T A ( x + (1 ) y ) + (1 ) 0 = 1 i.e. that x + (1 ) y and hence is convex....
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This note was uploaded on 02/04/2011 for the course MATH 3301 taught by Professor Dri.m.macphee during the Spring '10 term at Durham.

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convexity - 1 Convexity We defined convex functions f : I R...

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