Math 273a: Optimization Convex Conjugacy

Transcription

1 Math 273a: Optimization Convex Conjugacy Instructor: Wotao Yin Department of Mathematics, UCLA Fall 2015 online discussions on piazza.com

2 Convex conjugate (the Legendre transform) Let f be a closed proper convex function. The convex conjugate of f is f is convex (in y). f (y) = sup {y T x f(x)} x domf Reason: for each fixed x, (y T x f(x)) is linear in y. Hence, f is point-wise maximum of linear functions, that is, point y and over x. As long as f is proper, f is proper closed convex.

3 Geometry For fixed y and z, consider linear function: g(x) = y T x z the corresponding hyperplane is H = {(x, g(x)) : x R n } R n+1 [y; 1] is the downward normal direction of H H crosses the (n + 1)th axis at g(0) = z y (tilt) and z (height) uniquely define the hyperplane H and function g Our task: for each y, determine z using the function f, thus determining g

4 If we set two rules 1. g(x) = f(x) at some point x, i.e., H intersects epi(f), 2. H is as low as possible, i.e., z is as small as possible, then H will be the supporting hyperplane of f By rule #1, point x domf, y T x z = f(x) or z = f(x) y T x By rule #2, z = inf domf {f(x) y T x} = z = sup x domf {y T x f(x)} Therefore, z = f (x) g(y) = y T x f (x) H is the supporting hyperplane

5 Geometry 1. H intersects epi(f) 2. H is as low as possible = f (y) = inf x domf {f(x) y T x} epi(f ) f (y) f(x) x f almost completely characterizes f. Almost is because covers only up to closure. This is a result of the Hahn-Banach Separation Theorem.

6 Relation to Lagrange duality Consider convex problem minimize f(x) subject to Ax = b. x Lagrangian: L(x; y) = f(x) y T (Ax b). Lagrange dual function: d(y) = inf x domf L(x; y) = sup {y T (Ax b) f(x)} x domf = f (A T y) b T y. Lagrange dual problem (given in terms of convex conjugate f ): minimize y f (A T y) b T y or maximize b T y f (A T y). y

7 Exercise Derive a Lagrange dual problem for minimize x R n f(x) + g(ax)

8 Primal and dual subdifferentials Suppose that [y; 1] is the downward normal of the hyperplane touching epi(f) at x; therefore, y = f(x ) In general, for proper closed convex f, y f(x ). Therefore, domf = { f(x) : x domf}. Theorem (biconjugation) Let x domf and y domf. Then, y f(x) x f (y). If the relation holds, then f(x) + f (y) = y T x. The result is very useful in deriving optimality conditions.

9 Fenchel s inequality Theorem (Fenchel s inequality) For arbitrary x domf and y domf, we have f(x) + f (y) y T x. Proof. Since x is not necessarily the maximizing point for f(y) = sup x { }, we have f (y) y T x f(x).

10 Theorem If f is proper, closed, convex, then (f ) = f, i.e., f(x) = sup y domf {y T x f (y)}. Proof. Consider linear function g y,z, defined as g y,z(x) = y T x z. Step 1. (cont.) g y,z f y T x z f(x), x y T x f(x) z, x sup{y T x f(x)} z x f (y) z (y, z) epi(f )

11 Proof. Step 2. From the Hahn-Banach Separation Theorem, Step 3. sup y,z f(x) = sup{g y,z(x) : g y,z f}, x domf. y,z {g y,z (x) : g y,z f} = sup{g y,z (x) : f (y) z} by Step 1 y,z = sup{y T x z : f (y) z} y,z = sup{y T x f (y)} y = (f ) (x) Combining Steps 2 and 3, we get f = (f ).

12 Examples f(x) = ι C(x), indicator function of nonempty closed convex set C; then is the support function of C σc := f (y) = sup y T x x C Applying the theorem, we get (ι C) = ι C and (σ C) = σ C.

13 Examples f(x) = ι { 1 x 1}, indicator function of the unit hypercube; then f (y) = sup y T x = y 1 1 x 1 f(x) = ι { x 2 1}, then f(x) = 1 p x p p, 1 < p <, then f (y) = y 2 f (y) = 1 q x q q 1 p + 1 q = 1 lots of smooth examples...

14 Alternative representation Previously, we can represent f by f via f(x) = sup y domf {x T y f (y)} We can introduce a more general representation: f(x) = sup {(Ax b) T y h (y)} = h(ax b) y domh so that h might be simpler than f (or h is simpler than f) in form.

15 Example f(x) = x 1 Let C = {y = [y 1; y 2] : y 1 + y 2 = 1, y 1, y 2 0} and [ ] 1 A =. 1 We have x 1 = sup y {(y 1 y 2) T x ι C(y)} = sup{y T Ax ι C(y)}. Since ι C([x 1; x 2]) = 1 T (max{x 1, x 2}), where max is taken entry-wise, and we have sup {(Ax b) T y h (y)} = h(ax b) y domh x 1 = ι C(Ax) = 1 T (max{x, x}). y

16 Application: dual smoothing Idea: strongly convexify h = f becomes Lipschitz-differentiable Suppose that f is represented in terms of h as f(x) = sup {y T (Ax b) h (y)} y domh Let us strongly convexify h by adding strongly convex function d: (a simple choice is d(y) = 1 2 y 2 ) ĥ (y) = h (y) + μd(y) Obtain a Lipschitz-differentiable approximation: f μ(x) = sup {y T (Ax b) ĥ (y)} y domh f μ(x) is differentiable since h (y) + μd(y) is strongly convex.

17 Example: augmented l 1 primal problem: min{ x 1 : Ax = b} dual problem: max{b T y + ι [ 1,1] n(a T y)} f(y) = ι [ 1,1] n(y) is non-differentiable plan: strongly convexify the primal so that dual becomes smooth and can be quickly solved let f (x) = x 1 and f(y) = ι [ 1,1] n(y) = sup x {y T x f (x)}, add μ 2 x 2 to f (x) and obtain f μ(y) = sup{y T x ( x 1 + μ x 2 x 2 2)} = 1 2μ y Proj [ 1,1] n(y) 2 2 f μ(y) is differentiable; f μ(y) = 1 μ shrink(y). On the other hand, we can also directly smooth f (x) = x 1 and obtain differentiable f μ(x) by adding d(y) to f(y). (see the next slide...)

18 Example: smoothed absolute value Let x R. Recall f(x) = x = sup{yx ι [ 1,1] (y)} y add d(y) = y 2 /2 to ι [ 1,1] (y) and obtain: f μ = sup{yx (ι [ 1,1] (y) + μy 2 /2)} = y which is the Huber function. { x 2 /(2μ), x μ, x μ/2, x > μ, If x μ, the maximizing y stays within [ 1, 1] even if ι [ 1,1] (y) vanishes, so it leaves with μy 2 /2 only. If x > μ, the maximizing y occurs at ±1, so μy 2 /2 = μ/2. The Huber function is used in robust least squares.

19 add d(y) = 1 1 y 2, which is well defined and strongly convex in [ 1, 1]: fμ = sup{yx (ι [ 1,1] (y) μ 1 y 2 )} μ = x 2 + μ 2 μ, y which is used in reweighted least-squares methods Recall for C = {y : y 1 + y 2 = 1, y 1, y 2 0}. x = sup{(y 1 y 2 )x ι C (y)} y Add negative entropy d(y) = y 1 log y 1 + y 2 log y 2 + log 2: f μ(x) = sup y {(y 1 y 2)x (ι C(y) + μd(y))} = μ log ex/μ + e x/μ. 2

20 x log(x) 1.5 x log(x) x x log(x) is strongly convex between [0, C] for any finite C > 0

21 Compare three smoothed functions Courtesy of L. Vandenberghe

22 Example: smoothed maximum eigenvalue Let X S n. Consider f(x) = λ max(x), which is the l norm on the matrix spectrum. Recall the dual of l is l 1, and we have x = sup{x T y ι {1 T y=1,y 0}(y)} y Let C = {Y S n : try = 1, Y 0}. We have Next, strongly convexify ι C(Y): f(x) = λ max(x) = sup{y X ι C(Y)} Y

23 Negative entropy of {λ i(y)}: n d(y) = λ i(y) log λ i(y) + log n i=1 (Courtesy of L. Vandenberghe) Smoothed function ( 1 f μ(x) = sup{y X (ι C(Y) + μd(y))} = μ log Y n n i=1 e λ i(x)/μ )

24 Application: smoothed minimization 1 Instead of solving solve min f(x), x min x f μ(x) = sup y domh { y T (Ax + b) [h (y) + μd(y)] } by gradient descent, with acceleration, line search, etc... 1?

25 Since h (y) + μd(y) is strongly convex, f μ(x) is given by: f μ(x) = A T ȳ, where ȳ = arg max y domh {y T (Ax + b) [h (y) + μd(y)]}. Theorem If d(y) is strongly convex with modulus ν > 0, then h (y) + μd(y) is strongly convex with modulus at least μν f μ(x) is Lipschitz continuous with constant no more than A 2 /μν.

26 Nonsmooth optimization Examples: min Ax b 1 min TV(x) s.t. Ax b 2 σ Worst-case complexity for ɛ-approximation: If f is convex and f is L-Lipschitz, accelerated gradient method takes O( L/ɛ) iterations. If f is convex and nonsmooth, f is G-Lipschitz, subgradient method takes O(G 2 /ɛ 2 ) iterations. Smooth optimization has much better complexities.

27 Nesterov s complexity analysis 1. Construct smooth approximate satisfying f μ f f μ + μd and consequently f(x) f f μ (x) fμ + μd 2. Choose μ such that μd ɛ/2 = 1 μ 2D ɛ 3. Minimize f μ such that f μ (x) f μ ɛ/2 Step 3 has complexity 1 D O( μɛ ) = O( ɛ ), which can be much better than the subgradient method s O(G 2 /ɛ 2 ).

28 Theorem Consider h(x) = h μ(x) = sup y T x h (y), y domh sup y T x (h (y) + μd(y)), y domh where d(y) 0 is strongly convex with modulus ν > 0. Then, 1. h μ is (μν) 1 -Lipschitz; 2. if d(y) D for y domh, then h μ(x) h(x) h μ(x) + μd, x domh.

29 Example: Huber function Recall h μ(x) = sup{yx (ι [ 1,1] (y) + μy 2 /2)} = y { x 2 /(2μ), x μ, x μ/2, x > μ, We have and h μ(x) x h μ(x) + μ/2 h μ(x) = { x/μ, x μ, sign(x), x > μ, which is Lipschitz with constant μ 1. Apply to l 1-norm: h μ(x i) x 1 h μ(x i) + nμ/2. i i

30 Robust least squares Consider min x f(x) = Ax b 1 Representation where C = {y : y 1}. x 1 = sup{y T (Ax b) ι C(y)} y Add μd(y) = μ 2 y 2 2 to ι C(y) and obtain min x f μ(x) = m h μ(ai T x b i). i=1

31 Other examples and questions Total variation / analysis l 1 minimization examples Nuclear norm examples More other one nonsmooth terms... Stopping criteria