This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 22: Subgradient Methods April 11, 2007 Lecture 22 Outline • Methods with Dynamic Stepsize Rule • Directional Derivatives and Subgradients •Subgradients andSubdifferentals Convex Optimization 1 Lecture 22 Method with Dynamic Stepsize with Known f * General Assumption • The function f ( x ) is convex and dom f = R n • The set X is nonempty closed and convex Theorem Assume that General Assumption holds and X * is nonempty. Then, { x k } generated by the subgradient method with the stepsize α k = f ( x k ) f * k s k k 2 converges to an optimal solution. Convex Optimization 2 Lecture 22 Proof We use the Basic Iterate Relation : for any y ∈ X and any k ≥ , k x k +1 y k 2 ≤ k x k y k 2 2 α k ( f ( x k ) f ( y )) + α 2 k k s k k 2 Letting y = x * for any x * ∈ X * , we obtain k x k +1 x * k 2 ≤ k x k x * k 2 2 α k ( f ( x k ) f * ) + α 2 k k s k k 2 = k x k x * k 2 α k k s k k 2 2 f ( x k ) f * k s k k 2 α k ! = k x k x * k 2 ( f ( x k ) f * ) 2 k s k k 2 The sequence { x k } is bounded, thus subgradient norms k s k k are uniformly bounded by some scalar L . Therefore, for all k and any x * ∈ X * , k x k +1 x * k 2 ≤ k x k x * k 2 ( f ( x k ) f * ) 2 L 2 (1) Convex Optimization 3 Lecture 22 Eq. (1) implies that ( f ( x k ) f * ) 2 L 2 ≤ k x k x * k 2 k x k +1 x * k 2 , and ∞ X k =0 ( f ( x k ) f * ) 2 ≤ L 2 k x x * k 2 Hence f ( x k ) → f * . Since { x k } ⊆ X is bounded, it has limit points that belong to X by closedness of X . By continuity of f and the fact f ( x k ) → f * , it follows that all limit points are optimal. We now show that { x k } has a unique limit point. To arrive at a contradic tion, suppose there are at least two: say ˜ x and ˆ x with ˜ x 6 = ˆ x . Let { x m } M and { x k } K be subsequences converging to ˜ x and ˆ x , respectively. By Eq. (1), it follows that for all m,k with m ≥ k , k x m x * k < k x k x * k for all x * ∈ X * Let x * = ˆ x , and m ∈ M and k ∈ K . Then by taking limits, we obtain lim m →∞ m ∈M k x m ˆ x k ≤ lim k →∞ k ∈K k x k ˆ x k implying that k ˜ x ˆ x k = 0 a contradiction....
View
Full
Document
This note was uploaded on 08/22/2008 for the course GE 498 AN taught by Professor Angelianedich during the Spring '07 term at University of Illinois at Urbana–Champaign.
 Spring '07
 AngeliaNedich

Click to edit the document details