Victoria Martín-Márquez

Transcription

1 THE METHOD OF CYCLIC PROJECTIONS ON INCONSISTENT CONVEX FEASIBILITY PROBLEMS Victoria Martín-Márquez Dpto. de Análisis Matemático Universidad de Sevilla Spain V Workshop on Fixed Point Theory and Applications Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

2 Kelowna - Canada With Heinz Bauschke and Julian Revalski Victoria Martín-Márquez (US) With Shawn Wang and Julian Revalski Inconsistent CFP November 15-16, / 18

3 Introduction ORIGIN AND MOTIVATION OF THIS WORK: Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

4 Introduction ORIGIN AND MOTIVATION OF THIS WORK: Numerous problems in mathematics and physical sciences can be recast as Convex Feasibility Problem: {C i } m i=1 H are nonempty closed convex subsets find x m i=1 C i (CFP) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

5 Introduction ORIGIN AND MOTIVATION OF THIS WORK: Numerous problems in mathematics and physical sciences can be recast as Convex Feasibility Problem: {C i } m i=1 H are nonempty closed convex subsets find x m i=1 C i Bregman (1965) proved that the method cyclic projections x n+1 = (P Cm P Cm 1 P C1 ) n x 0 (CFP) (MAP) converges weakly to x m i=1 C i if m i=1 C i /0! Bregman, The method of successive projection for finding a common point of convex sets, Soviet Math. Dokl. 6, , Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

6 Introduction if m i=1 C i = /0? Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

7 Introduction if m i=1 C i = /0? For m = 2, the MAP produces the cycle (y 1,y 2 ) that achieves the minimal distance between the two sets m = 2 Cycle: (y 1,y 2 ) C 1 C 2 if y 1 = P C1 y 2 and y 2 = P C2 y 1 Gap vectors: v 1 = y 2 y 1, v 2 = y 1 y 2 (v 1 = v 2 ) the behavior of {(P C2 P C1 ) n x 0 } n N is completely analyzed! Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

8 Introduction m = 2 T = P C2 P C1 FixT /0 if and only if there exist cycles! Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

9 Introduction m = 2 T = P C2 P C1 FixT /0 if and only if there exist cycles! The gap vectors do not depend on the cycles. v 1 = v 2 = d(c 1,C 2 ) v 1 + v 2 = 0 Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

10 Introduction m = 2 T = P C2 P C1 FixT /0 if and only if there exist cycles! The gap vectors do not depend on the cycles. v 1 = v 2 = d(c 1,C 2 ) v 1 + v 2 = 0 Even if FixT = /0, there exist the gap vectors: v = ±P C2 C 1 0 Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

11 Introduction m = 2 a 1 = P C1 x, b 1 = P C2 a 1,... a n = P C1 b n 1, b n = P C2 a n Bauschke-Borwein (1994) b n a n P C2 C 1 0, a n b n P C2 C 1 0 Moreover (dichotomy): If FixT = /0, a n, b n +. If FixT /0, a n y 1 and b n y 2, with (y 1,y 2 ) cycle. Bauschke-Borwein, Dykstra s alternating projection alforithm for two sets, Journal of Approximation Theory 79, , Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

12 Introduction if m i=1 C i = /0? For m = 3, the MAP produces the cycle (y 1,y 2,y 3 ) m 3 Cycle: (y 1,,y m ) C 1 C 2 C m if y i = P Ci y i 1, i = 1,,m, y 0 = y m Gap vectors: v i = y i+1 y 1, i = 1,,m, y m+1 = y 0 Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

13 Introduction if m i=1 C i = /0? For m = 3, the MAP produces the cycle (y 1,y 2,y 3 ) m 3 Cycle: (y 1,,y m ) C 1 C 2 C m if y i = P Ci y i 1, i = 1,,m, y 0 = y m Gap vectors: v i = y i+1 y 1, i = 1,,m, y m+1 = y 0 behavior of {(P Cm P Cm 1 P C1 ) n x 0 } n N??? Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

14 Introduction m 3 FixT /0 if and only if there exist cycles. Gap vectors do not depend of cycles. Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

15 Introduction m 3 FixT /0 if and only if there exist cycles. Gap vectors do not depend of cycles. x i n = ( P Ci P Ci 1 P Cm P C1 P Ci+1 ) n x, i = 1,,m Bauschke-Borwein-Lewis (1997) (dichotomy): If FixT = /0, x i n +, i = 1,,m. If FixT /0, (x 1 n,,x m n ) (y 1,,y m ) cycle x i+1 n x i n v i, i = 1,,m. Bauschke-Borwein-Lewis, The method of cyclic projections for closed convex sets in Hilbert space, Contemporary Mathematics 204, 1-38, Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

16 CONJECTURES m 3 Geometry: Gap vectors are described independently and they exist even if FixT = /0. Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

17 CONJECTURES m 3 Geometry: Gap vectors are described independently and they exist even if FixT = /0. Convergence: even if FixT = /0, the sequence {xn i+1 xn} i is convergent, i = 1,,m. Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

18 CONJECTURES m 3 Geometry: Gap vectors are described independently and they exist even if FixT = /0. Convergence: even if FixT = /0, the sequence {xn i+1 xn} i is convergent, i = 1,,m. Boundedness: The sequence {xn i+1 xn} i is bounded, i = 1,,m. (weaker) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

19 Geometry of the problem. Case m 3 H Hilbert space m {2, } and I = {1,,m} Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

20 Geometry of the problem. Case m 3 H Hilbert space m {2, } and I = {1,,m} Hilbert product space X = H m := {x = (x i ) i I : x i H}, x,y = x i,y i i I Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

21 Geometry of the problem. Case m 3 H Hilbert space m {2, } and I = {1,,m} Hilbert product space Displacement operator X = H m := {x = (x i ) i I : x i H}, x,y = x i,y i i I M = Id R where R is the cyclic right-shift operator R: X m X m : (x 1,x 2,...,x m ) (x m,x 1,...,x m 1 ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

22 Geometry of the problem. Case m 3 H Hilbert space m {2, } and I = {1,,m} Hilbert product space Displacement operator X = H m := {x = (x i ) i I : x i H}, x,y = x i,y i i I M = Id R where R is the cyclic right-shift operator R: X m X m : (x 1,x 2,...,x m ) (x m,x 1,...,x m 1 ) C = C 1 C m H m is closed and convex P C (x 1,,x m ) = (P 1 x 1,,P m x m ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

23 Geometry of the problem. Case m 3 H Hilbert space m {2, } and I = {1,,m} Hilbert product space Displacement operator X = H m := {x = (x i ) i I : x i H}, x,y = x i,y i i I M = Id R where R is the cyclic right-shift operator R: X m X m : (x 1,x 2,...,x m ) (x m,x 1,...,x m 1 ) C = C 1 C m H m is closed and convex P C (x 1,,x m ) = (P 1 x 1,,P m x m ) normal cone of C at x C N C (x) = {u H m : u,y x 0, y C}. Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

24 Geometry of the problem. Case m 3 x = (x 1,,x m ) is a cycle if and only if x = P C (Rx) v = (v 1,,v m ) gap vector if and only if v = Rx x Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

25 Geometry of the problem. Case m 3 x = (x 1,,x m ) is a cycle if and only if x = P C (Rx) v = (v 1,,v m ) gap vector if and only if v = Rx x x = (x 1,,x m ) is a cycle if and only if Rx x N C x 0 N C x + Mx x (N C + M) 1 (0) v = (v 1,,v m ) gap vector if and only if v (NC 1 + M 1 ) 1 (0) 0 (N C 1 + M 1 )v Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

26 Open Questions 0 ran(n 1 C + M 1 )??? Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

27 Open Questions 0 ran(n 1 C + M 1 )??? Numerical procedure to approximate a zero of (N 1 C + M 1 )??? Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

28 Open Questions 0 ran(n 1 C + M 1 )??? Numerical procedure to approximate a zero of (N 1 C + M 1 )??? N 1 C + M 1 is not maximal monotone!!! Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

29 Open Questions 1 Parallel sum of A and B [Passty (1986)] A B := (A 1 + B 1 ) 1. 0 dom(n C M)??? Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

30 Open Questions 1 Parallel sum of A and B [Passty (1986)] A B := (A 1 + B 1 ) 1. 0 dom(n C M)??? 0 ran(n C + M)??? Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

31 Open Questions 1 Parallel sum of A and B [Passty (1986)] A B := (A 1 + B 1 ) 1. 0 dom(n C M)??? 0 ran(n C + M)??? Bauschke-Martín Márquez-Moffat-Wang (2012) 0 ran(n C + M) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

32 Open Questions 1 Parallel sum of A and B [Passty (1986)] A B := (A 1 + B 1 ) 1. 0 dom(n C M)??? 0 ran(n C + M)??? Bauschke-Martín Márquez-Moffat-Wang (2012) 0 ran(n C + M) 2 Can we find an enlargement of N C + M so that there is solution??? Closure of N C M. Extended sum. Variational sum. Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

33 Asymptotic Regularity We believe that the asymptotic regularity is an important step towards a complete understanding of the sequence {(P Cm P Cm 1 P C1 ) n x 0 } n N T asymptotically regular if lim n T n x T n+1 x = 0, x H Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

34 Asymptotic Regularity We believe that the asymptotic regularity is an important step towards a complete understanding of the sequence {(P Cm P Cm 1 P C1 ) n x 0 } n N T asymptotically regular if lim n T n x T n+1 x = 0, x H (if T is SNE) 0 ran (Id T) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

35 Asymptotic Regularity We believe that the asymptotic regularity is an important step towards a complete understanding of the sequence {(P Cm P Cm 1 P C1 ) n x 0 } n N T asymptotically regular if lim n T n x T n+1 x = 0, x H (if T is SNE) 0 ran (Id T) Bauschke-Martín Márquez-Moffat-Wang (2012) If T 1,,T m FNE and asymptotically regular, then the composition T m T 1 is asymptotically regular (not necessarily FNE) Bauschke-Martín Márquez-Moffat-Wang Compositions and convex combinations of asymptotically regular firmly nonexpansive mappings are also asymptotically regular, Fixed Point Theory and Applications 2012:53, (2012) doi: / Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

36 Context H Hilbert space m {2, } and I = {1,,m} Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

37 Context H Hilbert space m {2, } and I = {1,,m} A : H 2 H maximal monotone if monotone ( x y,x y 0, (x,x ),(y,y ) gr A = {(z,az) : z H}) and there is no monotone operator whose graph contains gr A Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

38 Context H Hilbert space m {2, } and I = {1,,m} A : H 2 H maximal monotone if monotone ( x y,x y 0, (x,x ),(y,y ) gr A = {(z,az) : z H}) and there is no monotone operator whose graph contains gr A Minty s Theorem A maximal monotone J A =(I+A) 1 J A FNE with full domain Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

39 Context H Hilbert space m {2, } and I = {1,,m} A : H 2 H maximal monotone if monotone ( x y,x y 0, (x,x ),(y,y ) gr A = {(z,az) : z H}) and there is no monotone operator whose graph contains gr A Minty s Theorem A maximal monotone J A =(I+A) 1 J A FNE with full domain Hilbert product space X = H m := {x = (x i ) i I : x i H}, x,y = x i,y i i I Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

40 Context H Hilbert space m {2, } and I = {1,,m} A : H 2 H maximal monotone if monotone ( x y,x y 0, (x,x ),(y,y ) gr A = {(z,az) : z H}) and there is no monotone operator whose graph contains gr A Minty s Theorem A maximal monotone J A =(I+A) 1 J A FNE with full domain Hilbert product space X = H m := {x = (x i ) i I : x i H}, x,y = x i,y i i I diagonal subspace := { x = (x) i I x H } orthogonal complement of = { x = (x i ) i I i I x i = 0 } Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

41 Context T i : H H, i I, FNE T i = (Id + A i ) 1, A i maximal monotone, i I Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

42 Context T i : H H, i I, FNE T i = (Id + A i ) 1, A i maximal monotone, i I T = T 1 T m : X m X m : (x i ) i I (T i x i ) i I A = A 1 A m : X m X m : (x i ) i I (A i x i ) i I T = (Id + A) 1 Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

43 Context T i : H H, i I, FNE T i = (Id + A i ) 1, A i maximal monotone, i I T = T 1 T m : X m X m : (x i ) i I (T i x i ) i I A = A 1 A m : X m X m : (x i ) i I (A i x i ) i I Displacement operator T = (Id + A) 1 M = Id R where R is the cyclic right-shift operator R: X m X m : (x 1,x 2,...,x m ) (x m,x 1,...,x m 1 ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

44 Context T i : H H, i I, FNE T i = (Id + A i ) 1, A i maximal monotone, i I T = T 1 T m : X m X m : (x i ) i I (T i x i ) i I A = A 1 A m : X m X m : (x i ) i I (A i x i ) i I Displacement operator T = (Id + A) 1 M = Id R where R is the cyclic right-shift operator R: X m X m : (x 1,x 2,...,x m ) (x m,x 1,...,x m 1 ) Main properties of M M continuous, linear and maximal monotone with dom M = X m M rectangular (sup (z,z ) gr M x z,z y <, (x,y ) dom M ran M) ran M = closed Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

45 Idea of the proof (composition) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

46 Idea of the proof (composition) Brezis-Haraux (1976) A and B monotone such that A + B maximal monotone, B rectangular and doma domb. Then intran(a + B) = int(rana + ranb) and ran(a + B) = rana + ranb Brezis, Haroux Image d une somme d opérateurs monotones et applications, Israel Journal of Mathematics 23 (1976), Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

47 Idea of the proof (composition) Brezis-Haraux (1976) A and B monotone such that A + B maximal monotone, B rectangular and doma domb. Then intran(a + B) = int(rana + ranb) and ran(a + B) = rana + ranb Since A and M are maximal monotone and dom M = X m A + M is maximal monotone Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

48 Idea of the proof (composition) Brezis-Haraux (1976) A and B monotone such that A + B maximal monotone, B rectangular and doma domb. Then intran(a + B) = int(rana + ranb) and ran(a + B) = rana + ranb Since A and M are maximal monotone and dom M = X m A + M is maximal monotone Apply Brezis-Haraux, because M is rectangular, to get (ranm = ) ran(a + M) = + rana Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

49 Idea of the proof (composition) Brezis-Haraux (1976) A and B monotone such that A + B maximal monotone, B rectangular and doma domb. Then intran(a + B) = int(rana + ranb) and ran(a + B) = rana + ranb Since A and M are maximal monotone and dom M = X m A + M is maximal monotone Apply Brezis-Haraux, because M is rectangular, to get (ranm = ) ran(a + M) = + rana If 0 ran(id T i ) ( i I) 0 ran(a + M) 0 ran(id T m T m 1 T 1 ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

50 Idea of the proof (composition) Brezis-Haraux (1976) A and B monotone such that A + B maximal monotone, B rectangular and doma domb. Then intran(a + B) = int(rana + ranb) and ran(a + B) = rana + ranb Since A and M are maximal monotone and dom M = X m A + M is maximal monotone Apply Brezis-Haraux, because M is rectangular, to get (ranm = ) ran(a + M) = + rana If 0 ran(id T i ) ( i I) 0 ran(a + M) 0 ran(id T m T m 1 T 1 ) Asymptotic regularity of the composition If T i asymptotically regular ( i I) then T m T m 1 T 1 asymptotically regular Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

51 Remarks If 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

52 Remarks If 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) The converse implication fails: Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

53 Remarks If 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) The converse implication fails: Example H {0}, m = 2 and v H \ {0} T 1 (x) = x + v = 0 / ran(id T 1 ) = { v} T 2 (x) = x v = 0 / ran(id T 2 ) = {v} However T 2 T 1 = Id ran(id T 2 T 1 ) = {0} Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

54 Remarks If 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) The converse implication fails: Example H {0}, m = 2 and v H \ {0} T 1 (x) = x + v = 0 / ran(id T 1 ) = { v} T 2 (x) = x v = 0 / ran(id T 2 ) = {v} However T 2 T 1 = Id ran(id T 2 T 1 ) = {0} Optimal result: 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

55 Remarks Example If 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) The converse implication fails: H {0}, m = 2 and v H \ {0} T 1 (x) = x + v = 0 / ran(id T 1 ) = { v} T 2 (x) = x v = 0 / ran(id T 2 ) = {v} However T 2 T 1 = Id ran(id T 2 T 1 ) = {0} Optimal result: 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) In other words: If T i has fixed points ( i I) T m T m 1 T 1 has fixed points! Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18

56 Remarks If 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) The converse implication fails: Example H {0}, m = 2 and v H \ {0} T 1 (x) = x + v = 0 / ran(id T 1 ) = { v} T 2 (x) = x v = 0 / ran(id T 2 ) = {v} However T 2 T 1 = Id ran(id T 2 T 1 ) = {0} Optimal result: 0 ran(id T i ) ( i I) 0 ran(id T m T m 1 T 1 ) Example H = R 2, m = 2 T 1 = P C1, C 1 = epiexp = 0 ran(id T 1 ) T 2 = P C2, C 2 = R {0} = 0 ran(id T 2 ) However 0 ran(id T 2 T 1 ) \ ran(id T 2 T 1 ) Victoria Martín-Márquez (US) Inconsistent CFP November 15-16, / 18