L22_dynrule - Lecture 22 Subgradient Methods Lecture 22...

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Unformatted text preview: Lecture 22: Subgradient Methods April 11, 2007 Lecture 22 Outline • Methods with Dynamic Stepsize Rule • Directional Derivatives and Subgradients •-Subgradients and-Subdifferentals 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....
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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.

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L22_dynrule - Lecture 22 Subgradient Methods Lecture 22...

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