On Infinite-Dimensional Integration in Weighted Hilbert Spaces

Transcription

1 On Infinite-Dimensional Integration in Weighted Hilbert Spaces Sebastian Mayer Universität Bonn Joint work with M. Gnewuch (UNSW Sydney) and K. Ritter (TU Kaiserslautern). HDA 2013, Canberra, Australia. 1/22

2 Introduction Finite-dimensional integration: f(x 1,...,x s ) = u {1,...,s} f u(x u ), x u = (x j ) j u f = u {1,...,s} reproducing kernel Hilbert spaces γu 1/2 f u 2/22

3 Introduction Finite-dimensional integration: f(x 1,...,x s ) = u {1,...,s} f u(x u ), x u = (x j ) j u f = u {1,...,s} reproducing kernel Hilbert spaces Infinite-dimensional integration: f(x) = u f u(x u ), finite subsets u N γu 1/2 f u How to define integration for infinitely many variables? 2/22

4 1 The Setting 2 The Function Spaces 3 Finite-dimensional Integration 4 Infinite-Dimensional Integration 3/22

5 Basic assumptions, domain (A1) k 0 reproducing kernel on D D where D (A2) H(k) H(1) = {0} (A3) (γ u ) u family of non-negative weights such that γ u m u <, m := inf k(x,x) x D u 4/22

6 Basic assumptions, domain (A1) k 0 reproducing kernel on D D where D (A2) H(k) H(1) = {0} (A3) (γ u ) u family of non-negative weights such that γ u m u <, m := inf k(x,x) x D u The domain for functions of infinitely-many variables: X = x DN : γ u k(x j,x j ) < (A3) X u j u 4/22

7 The kernel Notation: 1 : s = {1,...,s} Define reproducing kernels on X X by k u (x,y) := j uk(x j,y j ) K 1:s (x,y) := K(x,y) := u u 1:s γ u k u (x,y) γ u k u (x,y) = lim s K 1:s(x,y) 5/22

8 The kernel Notation: 1 : s = {1,...,s} Define reproducing kernels on X X by k u (x,y) := j u k(x j,y j ) D u D u K 1:s (x,y) := K(x,y) := u u 1:s γ u k u (x,y) D 1:s D 1:s γ u k u (x,y) = lim s K 1:s(x,y) 5/22

9 1 The Setting 2 The Function Spaces 3 Finite-dimensional Integration 4 Infinite-Dimensional Integration 6/22

10 The function spaces Proposition Wasilkowski, Woźniakowski (2004) Every f H(K 1:s ) has a unique decomposition f = f u, f u H(γ u k u ). Moreover, u 1:s f K1:s = u 1:s proof based on inductive argument γ 1/2 u f u ku. 7/22

11 The function spaces Proposition Wasilkowski, Woźniakowski (2004) Every f H(K 1:s ) has a unique decomposition f = f u, f u H(γ u k u ). Moreover, u 1:s f K1:s = u 1:s proof based on inductive argument γ 1/2 u f u ku. Proposition Gnewuch, Mayer, Ritter (2012) Given (A1) - (A3): H(K) = H(γ u k u ). u 7/22

12 1 The Setting 2 The Function Spaces 3 Finite-dimensional Integration 4 Infinite-Dimensional Integration 8/22

13 Further basic assumptions In addition to (A1) - (A3): (A4) ρ a probability measure on D (A5) k measurable and ρ( { x D : k(x,x) = 0 } ) = 0 (A6) H(k) L 1 (ρ) 9/22

14 Further basic assumptions In addition to (A1) - (A3): (A4) ρ a probability measure on D (A5) k measurable and ρ( { x D : k(x,x) = 0 } ) = 0 (A6) H(k) L 1 (ρ) H(k) L 1 (ρ) holds true if e.g. k square-root integrable along the diagonal k(x,x)ρ(dx) <. D 9/22

15 { } H(k) L 1 (ρ), = J : H(k) L closed graph theorem 1 (ρ) bounded i.e. H(k) L 1 (ρ) H(K) L 1 (ρ). 10/22

16 { } H(k) L 1 (ρ), = J : H(k) L closed graph theorem 1 (ρ) bounded i.e. H(k) L 1 (ρ) H(K) L 1 (ρ). representer h H(k): h 2 k = D D g dρ = g,h k D k(x, y)ρ(dx)ρ(dy) 10/22

17 { } H(k) L 1 (ρ), = J : H(k) L closed graph theorem 1 (ρ) bounded i.e. H(k) L 1 (ρ) H(K) L 1 (ρ). representer h H(k): h 2 k = D D g dρ = g,h k D k(x, y)ρ(dx)ρ(dy) if k 0 then J = h k. (see e.g. Hinrichs (2010), Gnewuch, Mayer, Ritter (2012)) 10/22

18 Integration on H(k u )? For finite u N consider h u (x) = j uh(x j ), x D u. h u H(k u ), h u ku = h u k. 11/22

19 Integration on H(k u )? For finite u N consider h u (x) = j uh(x j ), x D u. h u H(k u ), h u ku = h u k. Straight-forward calculation: for f span { k u (,x) : x D u} D u f dρ u = f,h u ku 11/22

20 Integration on H(k u )? For finite u N consider h u (x) = j uh(x j ), x D u. h u H(k u ), h u ku = h u k. Straight-forward calculation: for f span { k u (,x) : x D u} What about the closure? I.e. is D u f dρ u = f,h u ku H(k u ) L 1 (ρ u )? 11/22

21 Continuous embedding H(k u ) L 1 (ρ) Proposition (Gnewuch, Mayer, Ritter (2012)) Suppose u. Then H(k u ) L 1 (ρ u ) and h u k J u ( π/2 J ) u. Furthermore, D u f dρ u = f,h u ku, f H(k u ). Proof: Little Grothendieck Theorem,change of measure argument; cf. Hinrichs (2010). 12/22

22 Continuous embedding H(k u ) L 1 (ρ) Proposition (Gnewuch, Mayer, Ritter (2012)) Suppose u. Then H(k u ) L 1 (ρ u ) and h u k J u ( π/2 J ) u. Furthermore, D u f dρ u = f,h u ku, f H(k u ). Proof: Little Grothendieck Theorem,change of measure argument; cf. Hinrichs (2010). k square-root integrable along the diagonal proof elementary, J u ( D k(x,x)ρ(dx) ) u. 12/22

23 Continuous embedding H(k u ) L 1 (ρ) Proposition (Gnewuch, Mayer, Ritter (2012)) Suppose u. Then H(k u ) L 1 (ρ u ) and h u k J u ( π/2 J ) u. Furthermore, D u f dρ u = f,h u ku, f H(k u ). Proof: Little Grothendieck Theorem,change of measure argument; cf. Hinrichs (2010). k square-root integrable along the diagonal proof elementary, J u ( D k(x,x)ρ(dx) ) u. k 0 = J u = J u = h u k. 12/22

24 1 The Setting 2 The Function Spaces 3 Finite-dimensional Integration 4 Infinite-Dimensional Integration 13/22

25 Infinite-dimensional integration Put µ = ρ N. Perspective 1: Integration with respect to µ, I 1 (f) = f(x)µ(dx) if µ(x) = 1. X 14/22

26 Infinite-dimensional integration Put µ = ρ N. Perspective 1: Integration with respect to µ, I 1 (f) = f(x)µ(dx) if µ(x) = 1. µ(x) {0,1} X 14/22

27 Infinite-dimensional integration Put µ = ρ N. Perspective 1: Integration with respect to µ, I 1 (f) = f(x)µ(dx) if µ(x) = 1. µ(x) {0,1} Perspective 2: Integral as limit of representers. If X h = u γ u h u exists in H(K), define I 2 (f) = f,h K. See Kuo, Sloan, Wasilkowski, Woźniakowski (2010), Plaskota, Wasilkowski (2011). 14/22

28 Integration with respect to the product measure For f = u f u H(K) let f (s) : D N R, f (s) (y) = f u (y u ). u 1:s 15/22

29 Integration with respect to the product measure For f = u f u H(K) let f (s) : D N R, f (s) (y) = f u (y u ). u 1:s Definition: We say H(K) L 1 (µ) if the sequence (f (s) ) s converges in L 1 (µ) for every f H(K). 15/22

30 Integration with respect to the product measure For f = u f u H(K) let Definition: We say f (s) : D N R, f (s) (y) = H(K) L 1 (µ) u 1:s f u (y u ). if the sequence (f (s) ) s converges in L 1 (µ) for every f H(K). Definition: We say if H(K) L 1 (µ) and H(K) L 1 (µ) T : H(K) L 1 (µ), Tf = lim s f (s) defines a bounded linear mapping. 15/22

31 Integration with respect to the product measure Lemma If µ(x) = 1 and H(K) L 1 (µ) then H(K) L 1 (µ) and Tf dµ = f dµ X. D N X µ(x) = 1 and H(K) L 1 (µ) are satisfied if e.g. ( γ u u D k(x,x) dρ(x)) u <. 16/22

32 Proposition (Gnewuch, Mayer, Ritter (2012)) Let f = u f u H(K), f u H(γ u k u ). If u γ u h 2 u k < then h = u γ uh u H(K) and f,h K = f u (x u )ρ u (dx u ). u D u 17/22

33 Proposition (Gnewuch, Mayer, Ritter (2012)) Let f = u f u H(K), f u H(γ u k u ). If u γ u h 2 u k < then h = u γ uh u H(K) and f,h K = f u (x u )ρ u (dx u ). u D u γ u J u 2 < H(K) L 1 (µ) u u γ u h 2 u k <. 17/22

34 Proposition (Gnewuch, Mayer, Ritter (2012)) Let f = u f u H(K), f u H(γ u k u ). If u γ u h 2 u k < then h = u γ uh u H(K) and f,h K = f u (x u )ρ u (dx u ). u D u γ u J u 2 < H(K) L 1 (µ) u u γ u h 2 u k <. If H(K) L 1 (µ) then f,h = Tf(x)µ(dx). D N 17/22

35 Example 1: Non-negative Kernel ρ the uniform distribution on D = [0,1] k(x,y) = min(x,y) Fix α > 0; γ 1:s = α s ; γ u = 0 otherwise. 18/22

36 Example 1: Non-negative Kernel ρ the uniform distribution on D = [0,1] k(x,y) = min(x,y) Fix α > 0; γ 1:s = α s ; γ u = 0 otherwise. u γ u J u 2 < H(K) L 1 (µ) α < 3, u γ u h 2 u < k µ(x) = 1 α < exp(1). 18/22

37 Example 1: Non-negative Kernel ρ the uniform distribution on D = [0,1] k(x,y) = min(x,y) Fix α > 0; γ 1:s = α s ; γ u = 0 otherwise. u γ u J u 2 < H(K) L 1 (µ) α < 3, u γ u h 2 u < k µ(x) = 1 α < exp(1). For α [ exp(1),3 ), H(K) L 1 (µ) holds true, but µ(x) = 0. 18/22

38 Example 1: Non-negative Kernel ρ the uniform distribution on D = [0,1] k(x,y) = min(x,y) Fix α > 0; γ 1:s = α s ; γ u = 0 otherwise. { }... α < 3, µ(x) = 1 α < exp(1), 19/22

39 Example 1: Non-negative Kernel ρ the uniform distribution on D = [0,1] k(x,y) = min(x,y) Fix α > 0; γ 1:s = α s ; γ u = 0 otherwise. ( γ u u D { }... α < 3, µ(x) = 1 α < exp(1), k(x,x)dρ(x)) u < α < 2, X = D N α < 1. 19/22

40 Example 1: Non-negative Kernel ρ the uniform distribution on D = [0,1] k(x,y) = min(x,y) Fix α > 0; γ 1:s = α s ; γ u = 0 otherwise. ( γ u u D { }... α < 3, µ(x) = 1 α < exp(1), k(x,x)dρ(x)) u < α < 2, X = D N α < 1. H(K) L 1 (µ) u γ u( D k(x,x)dρ(x) ) u < 19/22

41 Example 2: ANOVA-type kernel ρ uniform distribution on D = [ 1,1] k(x,y) = x y γ u as before α < 4 u u γ u h 2 u k < α > 0 (trivial), γ u J u 2 < H(K) L 1 (µ) α 4, ( γ u u D µ(x) = 1 α < exp(2), u k(x,x)dρ(x)) < α < 3, X = D N α < 1. 20/22

42 Example 2: ANOVA-type Kernel α < 4 u u γ u h 2 u k < α > 0 (trivial), γ u J u 2 < H(K) L 1 (µ) α 4, µ(x) = 1 α < exp(2), For α ( 4,exp(2) ) f,h K well-defined and µ(x) = 1. H(K) L 1 (µ) not satisfied. 21/22

43 Summary, open questions Sometimes the number I 2 (f) = f,h = lim s u 1:s may exist, but can not be interpreted as I 1 (f) = Tf dµ. D N D u f u dρ u I 1 (f) can make sense non-trivially even if µ(x) = 0. Open questions: How to characterize H(k) L 1 (ρ)? A necessary condition for H(K) L 1 (µ)? 22/22