Parcours OJD, Ecole Polytechnique et Université Pierre et Marie Curie 05 Mai 2015

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1 Examen du cours Optimisation Stochastique Version 06/05/2014 Mastère de Mathématiques de la Modélisation F. Bonnans Parcours OJD, Ecole Polytechnique et Université Pierre et Marie Curie 05 Mai 2015 Authorized documents : none. Take care of : checking arguments; do not neglect partial answers. Problem 1 : Continuity of convex functions We recall the following results. Lemma 1. Let f be a l.s.c. (lower semicontinuous) convex function over a Banach X with values in R. Then f is continuous over the interior of its domain. Theorem 2 (Banach-Steinhaus). Let X, Y be Banach spaces and A a nonempty subset of L(X, Y ). Then {sup{ A ; A A} < } {for any x X, sup{ Ax ; A A} < }. (1) 1/ Recall the main tool of the proof of lemma 1. 2/ Give an example of a convex function X R whose subdifferential is everywhere empty, and which not l.s.c. at any point. 3/ Give an example of a nonnegative convex function over X with domain different of X, dense in X, that is not l.s.c. over its domain. 4/ Given X, Y and A as in theorem 2, show that f(x) := sup{ Ax ; A A} is l.s.c. and convex. 5/ Deduce from the previous question a proof of theorem 2. Problem 2 : Gibbs entropy Given a probability space (, F, µ), the (reverse) Gibbs entropy over X := L s (), s [1, ], is G(x) := x(ω) log(x(ω))dµ(ω), (2) where by definition g(a) := a log a has value 0 when a = 0, and g(a) = + when a < 0. Let X + := {x X; x 0 a.s.}. 1/ Prove that dom(g) = X + if s > 1. 2/ When s = 1, prove that dom(g) is a subset of X +, whose closure equals X +. Is dom(g) a convex cone? 1

2 3/ We assume in the sequel that there exists a sequence A k F of positive probability such that µ(a k ) 0. Let x dom(g). Show that G is not continuous at x. 4/ Show that, when s > 1, G is continuous over its domain, in the sense that x k x in X, with x k and x in dom(g), implies G(x k ) G( x). 5/ Show that G is not continuous over its domain, when s = 1 (it suffices to prove that it is not continuous at 0). 6/ Show that G is l.s.c. (perhaps using Fatou s lemma). 7/ Give the expression of G (x ) when s <. 8/ Give the expression of G(x) when s <. Is it a singleton? 9/ Let f be a l.s.c., convex function over a Banach space M with values at R {+ }. Let x M be such that f( x) is a singleton. Is f, possibly with a additional hypothesis, G-differentiable at x? Link this to the properties of G(x). 10/ Give the expression of G (x ) and G(x) when s =. Problem 3 : Dynamic programming approach to separable programming Consider the minimization problem Min f(x); g(x) = η (P η) x R n where f and g are real-valued functions which are separable in the sense that there exists real-valued functions over R, f 1 to f n and g 1 to g n, such that { f(x) = f1 (x 1 ) + + f n (x n ); (3) g(x) = g 1 (x 1 ) + + g n (x n ). We interpret this as a dynamic problem where the decision x k is taken at time k = 1 to n. The family of problems is Min f k(x k ) + + f n (x n ); g k (x k ) + + g n (x n ) = ν, (P k x R n ν ) 1/ Set v k (ν) := val(pν k ). Show that v k (ν) satisfies the dynamic programming principle (DPP) { v k ( (ν) = inf x k fk (x k ) + v k+1 (ν g k (x k )) ), k = 1,..., n, v n+1 (ν) = 0 if ν = 0, v n+1 (4) (ν) = if ν 0. 2/ Assuming that we have the additional constraint x k [0, 1], how could we solve the DPP in practice? 3/ When g k is for all k the identity operator, make the link with infimal convolutions. 4/ When all f k (resp. g k ) are equal to the same function ϕ (resp. ψ), and n = 2 p, how can we design an algorithm, faster than the above DPP (p steps instead of n), for computing the value of (P η ). 5/ We consider solving (P η ) by duality. How can we take advantage of separability? Is it more efficient than to use the DDP? 2

3 ANSWERS Problem 1 : Continuity of convex functions 1/ Let x int(dom(f)). Set V := {x X; f(x) f( x) + 1}. Applying Baire s lemma to the family kv, for k N, we obtain that V is a neighborhood of x, that is, f is upper bounded near x, and it is then easy to prove that f is Lipschitz near x. 2/ Assuming X to be infinite dimensional, a linear, not continuous function. 3/ Indicatrix of L (0, 1) in the space L 1 (0, 1). 4/ As a supremum of continuous convex functions, f is l.s.c. convex. 5/ Apply lemma 1 to f. If dom(f) = X then f is continuous everywhere, in particular (since f(0) = 0) has value less than 1 over some ball B(0, ε) with ε > 0. By positive homogeneity (f(αx) = αf(x) for any α > 0) we have that f(x) 1/ε over the unit ball. This proves the Banach-Steinhaus theorem. Remark The direct proof of the Banach-Steinhaus theorem (see e.g. Brézis, Anayse fonctionnelle), is also based on Baire s lemma. Problem 2 : Gibbs entropy 1/ That dom(g) X + follows from a log a = if a < 0. If s > 1 then, for a 0, a log a c(1 + a s ) for large enough c and the result follows. 2/ From the previous answer we get that dom(g) X +, and that dom(g) contains {L s () + ; s > 1} wich is a dense subset of X + (truncation argument). We next check that dom(g) is a cone (it is convex since G is). Let x dom(g), α > 0, and set y := αx. Then g(y) = αx log(αx) = αg(x) + (α log α)x (5) so that g(y) α g(x) + α log α x proving that g(y) is integrable. The result follows. 3/ For ε > 0 let x k (ω) := 1 over A k, and x k (ω) := x(ω) otherwise. Then x k dom(g), but x k x in X. The result follows. 4/ Since x k s dµ x s dµ, the result follows from the generalized dominated convergence theorem, where the pair (f k, g k ) of that theorem corresponds in our setting to (g(x k ), c(1 + x k s )) for large enough c > 0. 5/ Let a k := meas(a k ). Take x k equal to b k /a k on A k, for some b k > 0, and to 0 outside. Then x k 1 = b k, and G(x k ) := b k log(b k /a k ) = b k log b k b k log a k. (6) Take b k := log a k 1/2. Then b k 0, and therefore x k 1 0 and b k log b k 0, so that lim G(x k ) = lim b k log a k = log a k 1/2 =. Therefore, G is not continuous at 0. 3

4 6/ Let c := min{a log a, a R + }. Let x k x in X. Since g(x k ) c, Fatou s lemma applies to G(x k ) (or rather to a subsequence) and gives the result. 7/ Since g is both a Carathéodory and normal convex integrand, and g (y) = e y 1, from the lecture notes we know that G (x ) = g (x (ω))dµ(ω) = e x (ω) 1 dµ(ω), (7) with dom(g ) = {x X ; e x (ω) dµ(ω) < }. (8) 8/ Since g is both a Carathéodory and normal convex integrand, from the lecture notes we know that G (x) = {x X ; x (ω) g(ω) a.s.}. (9) Now g(z) = g (z) = 1 + log z if z > 0, and otherwise. (10) So G (x) reduces to the singleton g (x(ω)) if the latter belongs to X (this implies of course that x(ω) > 0 a.s.), and is empty otherwise. 9/ From the lecture notes we know that G-differentiability holds if in addition f is continuous at x. This is not the case for G. In addition take x dom(g) and for γ > 0, set D := {ω R + ; x(ω) γ}. Take γ large enough so that meas(d) > 0, and let y X + be unbounded over D. Then for any ε > 0, x εy dom(g), and so, G is nowhere G-differentiable. 10/ We can write x G( x) as x = x 1 + x s with x 1 L1 (), x 1 (ω) g ( x(ω)) a.s., and x s in the normal cone at x to dom(g) = L +, i.e. x s 0 and x s, x = 0. Problem 3 : Dynamic programming approach to separable programming 1/ Obvious. 2/ Space discretization, DPP formula at grid points combined with linear interpolation, as explained in the notes. 3/ When g k is for all k the identity operator, then val(p η ) = ( n f k)(η) 4/ For any nonzero q in N, set Φ q (η) := inf x { 2 q ϕ(x k ); } 2 q ψ(x k ) = η. (11) Then Φ q (η) := inf (Φ q 1(η 1 ) + Φ q 1 (η 1 )) = (Φ q 1 Φ q 1 ) (η). (12) η 1 +η 2 =η Computing Φ q 1 Φ q 1 amounts to the same operations than one step of the DPP. So we indeed can reduce the number of operations by a factor of p/2 p. 4

5 5/ Setting L k (x k ) := f k (x k ) + λg k (x k ), we can write the dual cost in the form δ(λ) = n inf L k (x k ) λη. (13) x k Its computation need n separate minimizations, which is the same cost as solving the DPP. Solving the dual problem needs to iterate over λ. By a dichotomy argument, for a precision of ε over λ, we typically need r steps with 2 r = O(ε), i.e., r = O( log ε). Therefore, solving the DPP is more effective. 5