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CSE 203B: Convex Optimization Week 4 Discuss Session

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1 CSE 203B: Convex Optimization Week 4 Discuss Session
Xinyuan Wang 01/30/2019

2 Contents Convex functions (Ref. Chap.3)
Review: definition, first order condition, second order condition, operations that preserve convexity Epigraph Conjugate function

3 Review of Convex Function
Definition ( often simplified by restricting to a line) First order condition: a global underestimator Second order condition Operations that preserve convexity Nonnegative weighted sum Composition with affine function Pointwise maximum and supremum Composition Minimization Perspective See Chap 3.2, try to prove why the convexity is preserved with those operations.

4 Epigraph ๐›ผ-sublevel set of ๐‘“: ๐‘… ๐‘› โ†’๐‘… ๐ถ ๐›ผ = ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘“ ๐‘ฅ โ‰ค๐›ผ}
๐ถ ๐›ผ = ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘“ ๐‘ฅ โ‰ค๐›ผ} sublevel sets of a convex function are convex for any value of ๐›ผ. Epigraph of ๐‘“: ๐‘… ๐‘› โ†’๐‘… is defined as ๐ž๐ฉ๐ข ๐‘“= (๐‘ฅ,๐‘ก) ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“,๐‘“ ๐‘ฅ โ‰ค๐‘ก}โŠ† ๐‘… ๐‘›+1 A function is convex iff its epigraph is a convex set.

5 Relation between convex sets and convex functions
A function is convex iff its epigraph is a convex set. Consider a convex function ๐‘“ and ๐‘ฅ, ๐‘ฆโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘กโ‰ฅ๐‘“ ๐‘ฆ โ‰ฅ๐‘“ ๐‘ฅ +๐›ป๐‘“ ๐‘ฅ ๐‘‡ (๐‘ฆโˆ’๐‘ฅ) The hyperplane supports epi ๐’‡ at (๐‘ฅ, ๐‘“ ๐‘ฅ ), for any ๐‘ฆ,๐‘ก โˆˆ๐ž๐ฉ๐ข ๐‘“โ‡’ ๐›ป๐‘“ ๐‘ฅ ๐‘‡ ๐‘ฆโˆ’๐‘ฅ +๐‘“ ๐‘ฅ โˆ’๐‘กโ‰ค0 โ‡’ ๐›ป๐‘“ ๐‘ฅ โˆ’1 ๐‘‡ ๐‘ฆ ๐‘ก โˆ’ ๐‘ฅ ๐‘“ ๐‘ฅ โ‰ค0 First order condition for convexity epi ๐’‡ ๐‘ก ๐‘ฅ Supporting hyperplane, derived from first order condition

6 Pointwise Supremum If for each ๐‘ฆโˆˆ๐‘ˆ, ๐‘“(๐‘ฅ, ๐‘ฆ): ๐‘… ๐‘› โ†’๐‘… is convex in ๐‘ฅ, then function ๐‘”(๐‘ฅ)= sup ๐‘ฆโˆˆ๐‘ˆ ๐‘“(๐‘ฅ,๐‘ฆ) is convex in ๐‘ฅ. Epigraph of ๐‘” ๐‘ฅ is the intersection of epigraphs with ๐‘“ and set ๐‘ˆ ๐ž๐ฉ๐ข ๐‘”= โˆฉ ๐‘ฆโˆˆ๐‘ˆ ๐ž๐ฉ๐ข ๐‘“(โˆ™ ,๐‘ฆ) knowing ๐ž๐ฉ๐ข ๐‘“(โˆ™ ,๐‘ฆ) is a convex set (๐‘“ is a convex function in ๐‘ฅ and regard ๐‘ฆ as a const), so ๐ž๐ฉ๐ข ๐‘” is convex. An interesting method to establish convexity of a function: the pointwise supremum of a family of affine functions. (ref. Chap 3.2.4)

7 Conjugate Function Given function ๐‘“: ๐‘… ๐‘› โ†’๐‘… , the conjugate function
๐‘“ โˆ— ๐‘ฆ = sup ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘ฆ ๐‘‡ ๐‘ฅ โˆ’๐‘“(๐‘ฅ) The dom ๐‘“ โˆ— consists ๐‘ฆโˆˆ ๐‘… ๐‘› for which sup ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘ฆ ๐‘‡ ๐‘ฅ โˆ’๐‘“(๐‘ฅ) is bounded Theorem: ๐‘“ โˆ— (๐‘ฆ) is convex even ๐‘“ ๐‘ฅ is not convex. Supporting hyperplane: if ๐‘“ is convex in that domain Slope ๐‘ฆ=โˆ’1 ( ๐‘ฅ , ๐‘“( ๐‘ฅ )) where ๐›ป ๐‘“ ๐‘‡ ๐‘ฅ =๐‘ฆ if ๐‘“ is differentiable Slope ๐‘ฆ=0 (0, โˆ’ ๐‘“ โˆ— (0)) (Proof with pointwise supremum) ๐’š ๐‘ป ๐’™ โˆ’๐’‡(๐’™) is affine function in ๐’š

8 Examples of Conjugates
Derive the conjugates of ๐‘“:๐‘…โ†’๐‘… ๐‘“ โˆ— ๐‘ฆ = sup ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘ฆ๐‘ฅ โˆ’๐‘“(๐‘ฅ) Affine quadratic Norm See the extension to domain ๐‘น ๐’ in Example 3.21

9 Examples of Conjugates
Log-sum-exp: ๐‘“ ๐‘ฅ = log ( ฮฃ ๐‘–=1 n ๐‘’ ๐‘ฅ ๐‘– ) , ๐‘ฅโˆˆ ๐‘… ๐‘› . The conjugate is ๐‘“ โˆ— ๐‘ฆ = sup ๐‘ฅโˆˆ๐‘‘๐‘œ๐‘š ๐‘“ ๐‘ฆ ๐‘‡ ๐‘ฅ โˆ’๐‘“(๐‘ฅ) Since ๐‘“(๐‘ฅ) is differentiable, first calculate the gradient of ๐‘” ๐‘ฅ = ๐‘ฆ ๐‘‡ ๐‘ฅ โˆ’๐‘“(๐‘ฅ) ๐›ป๐‘” ๐‘ฅ =๐‘ฆโˆ’ ๐‘ ๐Ÿ ๐‘‡ ๐‘ where ๐‘= ๐‘’ ๐‘ฅ 1 , โ‹ฏ ๐‘’ ๐‘ฅ ๐‘› ๐‘‡ . The maximum of ๐‘” ๐‘ฅ is attained at ๐›ป๐‘” ๐‘ฅ =0, so that ๐‘ฆ= ๐‘ ๐Ÿ ๐‘‡ ๐‘ . ๐‘ฆ ๐‘– = ๐‘’ ๐‘ฅ ๐‘– ฮฃ ๐‘–=1 n ๐‘’ ๐‘ฅ ๐‘– , ๐‘–=1,โ€ฆ, ๐‘› Does the bounded supremum (not โ†’โˆž) exist for all ๐’šโˆˆ ๐‘น ๐’ ?

10 Examples of Conjugates
Does the bounded supremum (not โ†’โˆž) exist for all ๐’šโˆˆ ๐‘น ๐’ ? If ๐‘ฆ ๐‘– <0, then the gradient on ๐›ป ๐‘– ๐‘”= ๐‘ฆ ๐‘– โˆ’ ๐‘’ ๐‘ฅ ๐‘– ฮฃ ๐‘–=1 n ๐‘’ ๐‘ฅ ๐‘– <0 for all ๐‘ฅ ๐‘– โˆˆ๐‘…. Then ๐‘” ๐‘ฅ reaches maximum at โˆž at ๐‘ฅโ†’โˆ’โˆž. So ๐‘ฆโ‰ฝ0. To have ๐‘ฆ= ๐‘ 1 ๐‘‡ ๐‘ solvable, we notice that ๐Ÿ ๐‘‡ ๐‘ฆ= ๐Ÿ ๐‘‡ ๐‘ 1 ๐‘‡ ๐‘ =1. If ๐Ÿ ๐‘‡ ๐‘ฆโ‰ 1 and ๐‘ฆโ‰ฝ0, we choose ๐‘ฅ=๐‘ก๐Ÿ s.t. ๐‘” ๐‘ฅ = ๐‘ฆ ๐‘‡ ๐‘ฅ โˆ’๐‘“ ๐‘ฅ =๐‘ก ๐‘ฆ ๐‘‡ ๐Ÿโˆ’ log ๐‘› ๐‘’ ๐‘ก =๐‘ก ๐‘ฆ ๐‘‡ ๐Ÿโˆ’1 โˆ’ log ๐‘› ๐‘” ๐‘ฅ reaches maximum at โˆž when we let ๐‘กโ†’โˆž(โˆ’โˆž), which is not bounded. The conjugate function ๐‘“ โˆ— ๐‘ฆ = ฮฃ ๐‘–=1 n ๐‘ฆ ๐‘– log ๐‘ฆ ๐‘– if ๐Ÿ ๐‘‡ ๐‘ฆ=1 and ๐‘ฆโ‰ฝ0 โˆž otherwise


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