subgradient trajectories : the convex case

Transcription

1 trajectories : Université de Tours ley Joint work with : Jérôme Bolte (Paris vi) Aris Daniilidis (U. Autonoma Barcelona & Tours) and Laurent Mazet (Paris xii)

2 inequality inequality [ 1963] f : R N R is analytic. Let a be a critical point of f. Then there exists a neighborhood U of a, C > 0 and θ (0, 1) such that f (x) f (x) f (a) θ for all x U. Finite length of the gradient trajectories, Every critical point is limit of a gradient trajectory, etc.

3 Basic assumptions To simplify : H is a real Hilbert space f : H [0, + ) is smooth (f 0) For all r > 0, C r := {f r} (A0) (0 is a critical point and a global minimum) 0 C 0 (A1) (0 is an isolated critical value) There exits r 0 > 0 such that : x C r0 and f (x) > 0 f (x) 0 (A2) (Sublevel compactness) There exits r 0 > such that : C r0 = {f r 0 } is compact.

4 Generalization : K L-inequality We say that f satisfies inequality [Kurdyka 1998] if : There exists ϕ KL(0, r 0 ) such that : where : (ϕ f )(x) 1 for all x C r0 \ C 0. KL(0, r 0 ) = { ϕ : [0, r 0 ] R + continuous, ϕ(0) = 0, ϕ C 1 (0, r 0 ), ϕ > 0 }. Lojasiewicz inequality is a particular with ϕ(r) = 1 C(1 θ) r 1 θ.

5 The convex From now on, we assume that f is convex Issues : 1 Characterizations of the K L-inequality in 2 Does a convex function satisfy the K L-inequality?

6 Piecewise gradient curves Gradient dynamical system : { Ẋx (t) = f (X x (t)), t 0 X x (0) = x A piecewise gradient curve γ is a countable family of gradient curves X xi ([0, t i )) with f (X xi (0)) = f (x i ) = r i, f (X xi (t i )) = r i+1 r i C r0 x 0 X x0 (t) i + 0 C r2 C r1 C r3 x 1 x 2 C 0 0

7 Classical properties of the convex gradient curves Lemma. For all x 0 C r0 \ C 0, 1 t f (X x0 (t)) is convex, L 1 (0, + ) and decreasing with limit 0. 2 Each trajectory goes closer to all minima at the same time, i.e., for each a C 0, 3 For all T > 0, d dt X x 0 (t) a 2 2f (X x0 (t)) < 0. T 0 Ẋ x0 (t) dt 1 2 x 0 log T.

8 Characterizations of the K L-inequality (f convex) Theorem. The following statements are equivalent : 1 f satisfies the K L-inequality in C r0 : (ϕ f )(x) 1 with ϕ KL(0, r 0 ). 2 f satisfies the K L-inequality globally in H with ϕ KL(0, + ) which is concave. 1 3 r (0, r 0 ] is integrable. f (x) inf f (x)=r 4 For all piecewise gradient curves γ in C r0 we have length(γ) = ti i=0 0 Ẋ xi (t) dt <.

9 Length of the convex gradient curves [Brézis 1973] Given x 0 C r0, do we have length(x x0 ) = 0 Ẋ x0 (t) dt <? Theorem. [Brézis 1973] Yes if int(argmin(f )). finite length finite length Xx0 (t) Ẋx0 (t) C0 C0 unknown length Theorem. [Baillon 1978] No in general (counter-example in infinite dimension)

10 A sufficient condition for a convex function to satisfy K L-inequality Theorem. Assume that there exists m : [0, + ) [0, + ) continuous increasing with m(0) = 0 such that f m(dist(, C 0 )) on C r0 and r0 0 m 1 (r) dr < + (growth condition). (1) r Then K L-inequality holds for f. Proof : f (x) f (x), x p C0 (x) f (x) dist(x, C 0 ) f (x) m 1 (f (x)). non analytic examples : m(r) = exp( 1/r α ), α (0, 1) satisfies (1).

11 A smooth convex counterexample to K L-inequality Theorem. There exists a C 2 convex function f : R 2 R + with {f = 0} = D(0, 1) for which K L-inequality fails. Note that the gradient trajectories have uniform finite length.

12 An auxiliary problem A farmer rakes its (convex) field in several steps in the following way : l0 l1 l2 If he is unlucky, is it possible that he walks an infinite path? (i.e. l i = + ) i 0

13 An auxiliary problem A farmer rakes its (convex) field in several steps in the following way : l0 l1 l2 If he is unlucky, is it possible that he walks an infinite path? (i.e. l i = + ) i 0 Answer : Yes!

14 Hausdorff distance between nested convex sets Lemma. There exists a decreasing sequence of compact convex subsets {T k } k in R 2 such that : (i) T 0 is the disk D := D(0, 2) ; (ii) T k+1 int T k for every k N ; (iii) k N T k is the unit disk D(0, 1) ; (iv) + k=0 dist Hausdorff (T k, T k+1 ) = +.

15 Picture of the sequence of convex sets 1 l k := dist Hausdorff (T k, T k+1 ) R i R i+1 and N i Ri R i+1 It suffices to take R i R i+1 = 1 in order that i 2 i R i R i+1 < and k l k i N i(r i R i+1 ) i Ri R i+1 = +.

16 End of the construction Construction of a convex function with prescribed sublevel sets T k : [Torralba 1996]. Smoothing of the function.