Malliavin's Calculus and Applications in Stochastic

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    Malliavins calculus and applications in stochasticcontrol and finance

    Warsaw, March, April 2008

    Peter Imkellerversion of 5. April 2008

    Malliavins calculus has been developed for the study of the smooth-ness of measures on infinite dimensional spaces. It provides a stochasticaccess to the analytic problem of smoothness of solutions of parabo-lic partial differential equations. The mathematical framework for thisaccess is given by measures on spaces of trajectories.

    In the one-dimensional framework it is clear what is meant by smooth-ness of measures. We look for a direct analogy to the smoothness problemin infinite-dimensional spaces. For this purpose we start interpreting theWiener space as a sequence space, to which the theory of differentiationand integration in Euclidean spaces is generalized by extension to infinitefamilies of real numbers instead of finite ones.

    The calculus possesses applications to many areas of stochastics, inparticular finance stochastics, as is underpinned by the recently publis-

    hed book by Malliavin and Thalmayer. In this course I will report onrecent applications to the theory of backward stochastic differential equa-tions (BSDE), and their application to problems of the fine structure ofoption pricing and hedging in incomplete finance or insurance markets.

    At first we want to present an access to the Wiener space as sequencespace.

    1 The Wiener space as sequence space

    Definition 1.1 A probability space (, F, P) is called Gaussian if thereis a family (Xk)1kn or a sequence (Xk)kN of independent Gaussianunit random variables such that

    F= (Xk : 1 k n) resp. (Xk : k N)(completed by sets of P-measure 0).

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    Example 1:Let = C(R+, Rm), Fthe Borel sets on generated by the topologyof uniform convergence on compact sets of R+, P the m-dimensional

    canonical Wiener measure on F. Let further W = (W1

    , , Wm

    ) be thecanonical m-dimensional Wiener process defined by the projections onthe coordinates.

    Claim: (, F, P) is Gaussian.Proof:

    Let (gi)iN be an orthonormal basis of L2(R+),

    Wj(gi) =

    gi(s)dWjs , i

    N, 1

    j

    m,

    in the sense of L2-limits of Ito integrals. Then (modulo completion) wehave

    F= (Wt : t 0).Let t 0, (ai)iN a sequence in l2 such that

    1[0,t] =iN

    ai gi.

    Then we have for 1

    j

    m

    Wjt = limn

    ni=1

    ai Wj(gi) =

    i=1

    aiWj(gi),

    hence Wjt is (modulo completion) measurable with respect to (Wj(gi) :

    i N). Therefore (modulo completion)F= (Wj(gi) : i N, 1 j m).

    Moreover, due to

    E(Wj(gi)Wk(gl)) = jkgi, gl = jk il, i, l N, 1 j, k m,

    hence the Wj(gi) are independent Gaussian unit variables. In the following we shall construct an abstract isomorphism between

    the canonical Wiener space and a sequence space. Since we are finallyinterested in infinite dimensional spaces, we assume from now on

    Assumption: the Gaussian space considered is generated by infinitelymany independent Gaussian unit variables.

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    Let RN = {(xi)iN : xi R, i N} be the set of all real-valuedsequences, and for n N denote by

    n : RN

    Rn, (xi)i

    N

    (xi)1

    i

    n,

    the projection on the first n coordinates. Let Bn be the -algebra ofBorel sets in Rn,

    BN = (nN1n [Bn]).Let for n N

    1(dx) =12

    exp(x2

    2) dx, = P(Xn)nN = iN1, n = 1n .

    This notation is consistent for n = 1.

    We want to construct an isomorphism between the spaces of integrablefunctions on (, F, P) and (RN, BN, ). For this purpose, it is necessaryto know how functions on the two spaces are mapped to each other. Itis clear that for BN-measurable f on RN we have

    F = f ((Xn)nN)is F-measurable on .

    Lemma 1.1Let F be

    F-measurable on . Then there exists a

    B

    N-measurable function f on RN such that

    F = f ((Xn)nN).Proof

    1. Let F = 1A with A = ((Xi)1in)1[B], B Bn. Then set f = 11n [B].f is by definition BN-measurable and we have

    f((Xn)nN) = 1B((Xi)1in) = 1A = F.

    Hence the asserted equation is verified by indicators of a generating setof Fwhich is stable for intersections. Hence by Dynkins theorem it isvalid for all indicators of sets in F.

    2. By part 1. and by linearity the claim is then verified for linearcombinations of indicator functions ofF-measurable sets. The assertionis stable for monotone limits in the set of functions for which it is verified.Hence it is valid for all F-measurable functions by the monotone classtheorem.

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    Theorem 1.1 Let p 1. Then the mappingLp(RN, BN, )) f F = f ((Xn)nN) Lp(, F, P))

    defines a linear isomorphism.

    ProofThe mapping is well defined due to

    ||F||pp = E(|f((Xn)nN)|p)=

    |f(x)|p(dx) (transformation theorem)

    = ||f||pp,

    and bijective due to Lemma 1.1. Linearity is trivial. Theorem 1.1 allows us to develop a differential calculus on the se-

    quence space (RN, BN, ), and then to transfer it to the canonical space(, F, P). For this purpose we are stimulated by the treatment of theone-dimensional situation.

    Questions of smoothness of probability measures are prevalent. Westart considering them in the setting of R.

    2 Absolute continuity of measures on R

    Our aim is to study laws of random variables defined on (, F, P), i.e. theprobability measures PX for random variables X. By means of Theorem1.1 these measures correspond to the measures f1 for BN-measurablefunctions f on RN. The one-dimensional version of these measures isgiven by 1 f1 for B1-measurable functions f defined on R.

    We first discuss a simple analytic criterion for absolute continuity ofmeasures of this type.

    Lemma 2.1 Let be a finite measure on B1. Suppose there exists c Rsuch that for all C1(R) we have

    |

    (x)(dx)| c||||.Then < < , i.e. is absolutely continuous with respect to , theLebesgue measure on R.

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    ProofLet 0 f be continuous with compact support, and define

    (x) =x

    f(y)dy,

    Then f d =

    (x)(dx) c|||| = c

    fd.

    By a measure theoretic standard argument this inequality follows forbounded measurable f. Therefore we conclude for A B1

    (A) c (A),which clearly implies

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    For the analysis on Gaussian spaces things are different. We sketch theanalogue of a differential calculus with respect to duality on Gaussianspaces. For g, h L2(R, 1) denote

    g|h = g(x)h(x) 1(dx).To apply Lemma 1.1 formally to the measure = 1 f1 for some B1-measurable f, we have to write, assuming all operations are justified,

    (x)1 f1(dx) =

    f(x)1(dx) = f|1 = ( f)| 1

    f.

    Now, as in the classical setting, we want to transfer the derivation tothe other argument. For this purpose we continue calculating for g, h

    C0 (R)

    g|h = 12

    g(x) h(x) exp(x

    2

    2) dx

    = 12

    g(x)

    d

    dx[h(x) exp(x

    2

    2)] dx

    =

    g(x) exp(x2

    2)

    d

    dx[h(x) exp(x

    2

    2)] 1(dx)

    = g| h + xh.So in the setting of Gaussian spaces, if we define as before dg as distribu-tional derivative in the generalized sense, its dual operator on a suitablespace of functions (to be described later) has to be defined by

    h = h + xh.In this sense we have the following duality relationship, completely ana-logously to the classical formula

    dg|h = g|h.For the combination of the derivative operator and its dual we obtainthis time the following operator

    L = d = d2

    dx2+ x

    d

    dx,

    in the suitable distributional sense.

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    d will be called Malliavin derivative, Skorokhod integral, and LOrnstein-Uhlenbeck operator. The domains of these operators will bemore precisely defined in the higher dimensional setting. The present

    exposition is given for motivating the notions to be studied.Let us return to the problem of smoothness of the measure 1 f1.

    Lemma 2.2 Let g, h L2(R, 1) be such that dg,h L2(R, 1). Thenwe have

    dg|h = g|h.Moreover for f L2(R, 1) such that ( 1df) L2(R, 1) we have

    1

    f1

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    C0 (R), even, due to the integrability properties of the Gaussian density,on the space of polynomials in one real variable.

    Definition 3.1 For n

    0 let

    Hn = n 1.

    Hn is called Hermite polynomial of degree n.

    By definition we have for x RH0(x) = 1,

    H1(x) = 1 = x,

    H2(x) = x =

    1 + x2,

    H3(x) = (1 + x2) = x 2x + x3 = x3 3x.Theorem 3.1 Hn is a polynomial of degree n, with leading coefficient1. Moreover for n N

    (i) Hn = Hn+1,

    (ii) dHn = nHn1,(iii) LHn = n Hn.

    In particular, Hn is an eigenvector of L with eigenvalue n.

    Proof(i): follows by definition.

    (ii): We first briefly investigate the commutator of d and . In fact, wehave for h C0 (R)(dd) h = d(h+xh)(h+xh) = h+xh+h(h+xh) = h.

    This means that(d d) = id.

    With this in mind we proceed by induction on the degree n. The claimis clear for n = 1. Assume it holds for n 1. Then

    dHn = dHn1 = dHn1 + Hn1= (n 1)Hn2 + Hn1 = nHn1.

    (iii): LHn = dHn = nHn1 = nHn.

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    Corollary 3.1 For g L2(R) define the Fourier transform by

    g(u) =12

    R

    eiuxg(x) dx, u R.

    Then (Hn e

    x2

    2 )(u) = (iu)n eu2

    2 .

    ProofChoose u R. Then

    (Hn e

    x22 )(u) =

    (n1 e

    x2

    2 )(u)

    = eiu|n1

    = dn

    e

    iu

    |1= (iu)neiu|1= (iu)ne

    u2

    2 .

    With these preliminaries, we can show that the Hermite polynomials

    constitute an orthonormal basis of our Gaussian space in one dimension.

    Theorem 3.2 ( 1n!

    Hn)n0 is an orthonormal basis of L2(R, 1).

    Proof1. Let n, k N, and suppose that n < k. Then by Theorem 3.1

    Hn|Hk = n1|k1 = dkn1|1 = 0,while

    Hn|Hn = dnn1|1 = n!1|1 = n!.2. It remains to show that (Hn)n0 is complete in L2(R, 1), i.e. the

    set of linear combinations of Hermite polynomials is dense in L2(R, 1).

    For this purpose, it sufficesto show: if L2(R, 1) satisfies for all n 0 we have Hn| = 0,

    then = 0.For z C let

    F(z) =R

    (v)eivz12v

    2

    dv.

    Then we have for k N, t R by Cauchy-Schwarz

    |

    R

    |vk(v)

    |evt

    12v2dv

    [R

    2(v)e12v2dv

    R

    v2ke2vt12v2dv]

    12 0 by translational invariance of Lebesgue measure

    1

    [f(x + a) f(x)](x)dx = 1

    f(x)[(x a) (x)]dx.Now use (i), let 0, to identify the limit as daf(x)(x)dx on theleft hand side, and as f(x)da(x)dx on the right hand side. Hencefor any C0 (Rk)

    daf, = f, da.This means by definition that f possesses the weak derivative daf whichbelongs to Lp(Rk).

    2. Let us now prove that (ii) implies (i). Fix C0 (Rk), a =(a1, , ak) Rk. Then by Taylors formula with integral remainderterm and Fubinis theorem we have for any > 0

    Rk

    1

    [f(x + a) f(x)](x)dx

    =Rk

    1

    f(x)[(x a) (x)]dx

    =

    Rk

    f(x)[1

    0

    k

    i=1ai

    xi(x

    a)d]dx

    = 1

    0

    [Rk

    ki=1

    ai

    xi(x a)f(x)dx]d

    =1

    0

    [Rk

    (x a)ua(x)dx]d

    =1

    0

    [Rk

    (x)ua(x + a)dx]d

    =Rk

    [1

    0

    ua(x + a)d](x)dx.

    It remains to prove that 1

    0 ua(+ a)d converges to ua in Lp(Rk). This

    is certainly true provided ua C0 (Rk). But for any f, g Lp(Rk) wehave uniformly in > 0

    ||1

    0

    f( + a)d 1

    0

    g( + a)d||p 1

    0

    ||f( + a) g( + a)||pd=

    ||g

    f

    ||p.

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    By means of this observation we can transfer the desired result fromC0 (R

    k) to Lp(Rk), since C0 (Rk) is dense in Lp(Rk).

    Corollary 4.1 Lete1,

    , ek denote the canonical basis ofR

    k, let(fn)nN

    be a sequence in Wp1 such that(i) ||fn f||p 0 as n ,(ii) for any 1 i k the sequence (difn)nN converges in Lp(Rk).Then f Wp1 and ||fn f||1,p 0 as n .Proof

    We have to show that f is weakly differentiable in direction ei for 1 i k, and dif = limn difn Lp(Rk). For this purpose let

    ui = limn difn,

    which exists due to assumption (ii). Then by (i) for any C0 (Rk), 1 i k

    f(x)di(x)dx = limn

    fn(x)di(x)dx

    = limn

    difn(x)(x)dx =

    ui(x)(x)dx.

    This means that f possesses weak directional derivatives in direction ei

    and dif = ui Lp(Rk). Now Theorem 4.1 is applicable and finishes theproof. Corollary 4.2 Let p 1. Then Wp1 is a Banach space with respect tothe norm | | | |1,p, and for any a Rk the mapping da : Wp1 Lp(Rk) iscontinuous.

    ProofWe have to prove that Wp1 is complete with respect to ||||1,p. Let therefore(fn)nN be a Cauchy sequence in W

    p

    1 . Then setting f = limn fn inLp(Rk), we see that the hypotheses (i) and (ii) of Corollary 4.1 aresatisfied, and it suffices to apply this Corollary.

    We finally need a local version of Sobolev spaces.

    Definition 4.4 For p 1, s N letWps,loc = {f : f : Rk R measurable f Wps for C0 (Rk).}

    (local Sobolev space of order (s, p)).

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    Theorem 4.2 Let p 1, s N. Then f Wps,loc iff for any x0 Rkthere exists an open neighborhoodVx0 ofx0 such that for any C0 (Rk)with support in Vx0 we have f Wps .

    ProofWe only need to prove the only if part of the claim. For any x0 Rklet therefore Vx0 be given according to the statement of the assertion.Then (Vx0)x0Rk is an open covering of R

    k. Then there exists a locallyfinite partition of the unit (k)kN C0 (Rk) which is subordinate tothe covering, i.e. such that

    (i) 0 n 1, for any n N,(ii) for any n

    N there exists x0(n) such that supp(n)

    Vx0(n),

    (iii) nN n = 1,(iv) for any compact set K Rk the intersection of K and supp(n)

    is non-empty for at most finitely many n.Now let C0 (Rk). Then for any k N (ii) gives supp(k) Vx0(k)

    and thus by assumption

    kf Wps , k N.Since by (iv) the support of k is non-trivial for at most finitely many

    k, (iii) and linearity yield the desired

    f Wps .

    We now turn to Gaussian Sobolev spaces. Our analysis will againbe based on the differential operator we know from the above sket-ched classical calculus. Only the measure with respect to which we

    consider duality changes from the Lebesgue to the Gaussian measure.Since we thereby pass from an infinite to a finite measure, integra-bility properties for functions and therefore the domains of the dualoperators change. This is why the notion of local Sobolev spaces isimportant. On these spaces, we can define our operators locally, wi-thout reference to integrability first. In fact, using Theorem 4.2, and fors N, p 1, 1 j1, , js k, f Wps,loc we can define

    dj1dj2

    djs f

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    locally on an open neighborhood Vx0 of an arbitrary point x0 Rk by thecorresponding generalized derivative of f with C0 (Rk) such that = 1 on an open neighborhood Ux0 Vx0 of x0. This gives a globallyunique notion, since x0 is arbitrary.

    Definition 4.5 Let s N, p 1, 1 j k, f Wps,loc, and denoteby dj the directional derivative in direction of the jth unit vector in R

    k

    according to the preceding remark. Let then

    f = (d1f, , dkf),jf = djf + xj f,

    Lf =k

    j=1jdjf =

    kj=1

    [djdjf + xjdjf].

    For any 1 r s we define more generallyrf = (dj1dj2 djrf : 1 j1, j2, , jr k).

    This definition gives rise to the following notion of Gaussian Sobolevspaces.

    Definition 4.6 Let p 1, s N. Then letDps(R

    k) = {f Wps,loc :s

    r=0

    |||rf| ||p < },

    ||f||s,p =s

    r=0

    |||rf| ||p(k-dimensional Gaussian Sobolev space of order (s, p)).

    RemarkDps(Rk) is a Banach space. This is seen by arguments as for the proof of

    Corollary 4.2.

    Since our calculus will be based mostly on the Hilbert case p = 2, weshall restrict our attention to this case whenever convenient. In this case,our ONB composed of k-dimensional Hermite polynomials as investiga-ted in the previous chapter will play a central role, and adds structureto the setting. To get acquaintance with Gaussian Sobolev spaces, let us

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    compute the operators defined on the series expansions with respect tothis ONB.

    For f L2(Rk, k) we can write

    f =

    pEk

    cp(f)

    p!Hp

    with coefficients cp(f) R, p Ek. Due to orthogonality, the Gaussiannorm is given by

    ||f||2 =

    pEk

    cp(f)2

    p!2Hp|Hp =

    pEk

    cp(f)2

    p!.

    We also write f (cp(f)) to denote this series expansion. Denote byP the linear hull of the k-dimensional Hermite polynomials. Plainly,P Wps,loc for any s N, p 1. According to chapter 3, P is densein L2(Rk, k). And for functions in P, the generalized derivatives dj are

    just identical to the usual partial derivatives in direction j, 1 j k.We first calculate the operators on Hermite polynomials. In fact, for

    p Ek, 1 j k we have in the non-trivial cases

    djHp = pj i=j HpiHpj1, jHp = i=j HpiHpj+1, LHp = |p|Hp.Hence for f (cp(f)) P, 1 j k we may write

    djf =

    pEk

    cp(f)

    p!pji=j

    HpiHpj1,

    jf =

    pEk

    cp(f)

    p!

    i=j

    HpiHpj+1,

    Lf = pEk

    cp(

    f)p! |p|Hp.

    According to Corollary 4.2 and the calculations just sketched, thenatural domains of the operators extending , j and L beyond Pmustbe those distributions in Rk for which the formulas just given generateconvergent series in the L2-norm with respect to k. The most importantdomain is the one of, the Sobolev space D21(Rk). For f (cp(f)) P

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    we have

    |||f| ||22 =Rk

    |f|2(x)k(dx)

    =k

    j=1Rk

    |djf|2(x)k(dx)

    =k

    j=1

    pEk

    p2jcp(f)

    2

    p!2i=j

    pi!(pj 1)!

    =k

    j=1

    pEk

    pjcp(f)

    2

    p!

    =

    p

    Ek

    |p|cp(f)2

    p!.

    If in addition f L2(Rk, k), we may write f (cp(f)) and approximateit by fn =

    pEk,|p|n

    cp(f)p!

    P, n N. Hence, according to corollary 4.2,f belongs to D21(R

    k) if the following series converges

    |||f| ||22 = limn |||fn| ||22= lim

    n

    pEk,|p|n|p|cp(f)

    2

    p!

    =

    pEk|p|cp(f)2

    p!< .

    Along these lines, we now turn to describing Gaussian Sobolev spacesand the domains of our principal operators for p = 2 by means of Hermiteexpansions. We start with the case k = 1.

    Theorem 4.3 Let r N, f (cp(f)) L2(R, 1) W2r,loc. Denotefp =

    cp(f)

    p!H

    p, p

    0.

    Then the following are equivalent:(i) rf L2(R, 1),(ii)

    p0pr||fp||22 < ,

    (iii) f D2r(R),(iv) rf L2(R, 1).In particular, D2r(R) is the domain of r, r in L2(R, 1). For f, g

    D21(R) we havef|g = f|g.

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    Proof1. We prove equivalence of (i) and (ii). We have

    f

    = p1p cp(f)

    p!H

    p1 = p0cp+1(f)

    p!H

    p,

    and therefore by iteration

    rf = p0

    cp+r(f)

    p!Hp1.

    Therefore

    ||rf||22 =p0

    cp+r(f)2

    p!, ||fp||22 =

    cp(f)2

    p!,

    and hence||rf||22 =

    p0

    (p + r)!

    p!||fp+r||22 <

    if and only if p0

    (p + r)r||fp+r||22 < ,

    and this is the case if and only if

    p0pr

    ||fp

    ||22 1, f Lp(RN), (fn)nN the corresponding sequenceaccording to the above remarks. Suppose that supnN ||fn||1,p < . Then

    for any j N the sequence (djfn n)nN converges in Lp(RN), toa limit that we denote by djf. Corresponding statements hold true forhigher order derivatives.

    ProofLet n N, j N. Then for n j we have

    E(djfn+1 n+1|Cn) = djfn n.This means that (djfn n)nj is a martingale with respect to (Cn)njwhich, due to

    supn

    j

    ||djfn n||p supn

    N

    ||fn||1,p < ,

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    is bounded in Lp(RN) and hence converges in Lp(RN), due to p > 1. The preceding Lemmas give rise to the following definition of Sobolev

    spaces.

    Definition 5.1 Let p 1, s N. ThenDps(R

    N) = {f Lp(RN, ) : fn Dps(Rn), n N, supnN

    ||fn||s,p < },

    (infinite dimensional Sobolev space of order (s, p)), endowed with thenorm

    ||f||s,p = supnN

    ||fn||s,p, f Dps(RN).

    This definition makes sense, for the following reasons.

    Theorem 5.1 Let p > 1, s N. Then Dps(RN) is a Banach space withthe norm | | | |s,p.

    ProofWe prove the claim for s = 1. Let (fm)mN be a Cauchy sequence inD

    p1(R

    N), and (fmn )n,mN the corresponding finite dimensional functionsaccording to the remarks above. Then for m, l N, n N Jensensinequality and the martingale statement in the preceding proof give thefollowing estimate

    lim supm,l

    ||fmn fln||1,p limm,l

    ||fm fl||1,p = 0.

    Dp1(R

    n) being a Banach space for n N, we know thatfn = limm f

    mn Dp1(Rn)

    exists. Let fn = fn n. Now let f = limm fm in Lp(RN). Then byuniform integrability

    E(f|Cn) = E( limm fm|Cn) = limm E(fm|Cn) = limm fmn = fn.Moreover

    supnN

    ||fn||1,p supm,nN

    ||fmn ||1,p supmN

    ||fm||1,p < .

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    Hence by definition f Dp1(RN), and by Fatous lemma||f fm||1,p liminf

    l||fm fl||1,p 0

    as m . According to Lemma 5.2, the gradient on the infinite dimensional

    Gaussian Sobolev spaces is defined as follows.

    Definition 5.2 Let p > 1, f Dp1(RN). Then letf = (djf)jN

    (Malliavin gradient or Malliavin derivative), where for any j N ac-cording to Lemma 5.2

    djf = limn djfn n.

    Accordingly, for s N we define rf, 1 r s, for f Dps(RN).Remark

    The gradient being a continuous mapping from Dp1(Rn) to Lp(Rn, n)for any finite dimension n, Lemma 5.2 and the definition of the Mal-

    liavin gradient imply, that is a continuous mapping from Dp

    1(RN

    ) toLp(RN, ).Let us now again restrict our attention to p = 2 and describe Gaussian

    Sobolev spaces by means of the generalized Hermite polynomials. Firstof all, suppose f =

    pE

    cp(f)p!

    L2(RN, ). We shall continue to use thenotation f (cp(f)). Then for n N, we have fn = pEn c(p,0)(f)p! Hp,where we put (p, 0) = (p1, , pn, 0, 0, ) for p = (p1, , pn) En.Therefore, we also have fn =

    pEn

    c(p,0)(f)

    p!H(p,0). Let again P be the

    linear hull generated by all generalized Hermite polynomials.As in the preceding chapter, we may calculate the gradient norms forf (cp(f)) D21(RN). In fact, we have for j N

    djf = limn djfn n = limn

    pEn

    c(p,0)(f)

    p!pji=j

    HpiHpj1

    =pE

    cp(f)

    p!pji=j

    HpiHpj1.

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    Furthermore, for f D21(RN) let us compute the norm of |f| =[jN(djf)2]

    12 in L2(RN, ). In fact, we have, using the calculation of

    gradient norms in the preceding chapter,

    > supnN

    |||fn n|||22 = |||f|||22

    = supnN

    pEn

    |p|c(p,0)(f)2

    p!=pE

    |p|cp(f)2

    p!.

    We therefore obtain the following main result about the descriptionof the infinite dimensional Gaussian Sobolev spaces or order (1, 2).

    Theorem 5.2 For f

    L2(RN, ) the following are equivalent:

    (i) f D21(RN),(ii)

    pE |p|cp(f)

    2

    p!< ,

    (iii) |fn| n = [jN(djfn)2 n] 12 converges in L2(RN, ) to |f|.Moreover, D21(R

    N) is a Hilbert space with respect to the scalar product

    (f, g)1,2 = f|g +

    jNdjf|djg, f, g D21(RN).

    For p 2 P is dense in Dp1(RN). Analogous results hold for Sobolevspaces of order (s, 2) with s

    N.

    6 Absolute continuity in infinite dimensional Gaus-sian space

    We are now in a position to discuss the main result of Malliavins calculusin the framework of infinite dimensional Gaussian sequence spaces. Theresult is about the smoothness of laws of random variables defined on theGaussian space. We start with a generalization of Lemma 1.1 to finitemeasures on Bd for d N.Lemma 6.1 Let |Bd be a finite measure. Assume there exists c Rsuch that for all C1(Rd) with bounded partial derivatives, and any1 j d we have

    |

    xj(x) (dx)| c ||||.

    Then

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    ProofFor simplicity, we argue for d = 2, and omit the superscript denoting2-dimensional Lebesgue measure.

    1. Assume that C1

    (R

    2

    ) possesses compact support. We show:

    [

    ||2d] 12 12

    [

    | x1

    |2d +

    | x2

    |2d].

    In fact, we have

    [

    ||2d] 12 [

    supx1R

    |(x1, x2)|dx2

    supx2R

    |(x1, x2)|dx1]12

    [

    | x1

    (x1, x2)|dx1 dx2

    | x2

    (x1, x2)|dx2 dx1]12

    12

    [|

    x1|2d + |

    x2|2d].

    2. Let 0 u be continuous with compact support, and such thatud = 1, define for > 0 u =

    12

    u(1). Moreover, let

    =

    u( y)(dy)be a smoothed version of . Then we obtain for h continuous with com-

    pact support, using Fubinis theorem,(x)h(x)dx =

    [

    u(x y)(dy)]h(x)dx=

    [

    u(x y)h(x)dx](dy)=

    [

    u(x)h(x + y)dx](dy)

    h(y)(dy).

    3. We show:

    L2

    (R2

    ) g gd Ris a continuous linear functional.

    In fact, let C1(R2) have compact support, and let > 0. Then byhypothesis and smoothness of with a calculation as in 2.

    |

    xi(x)(x)dx| = |

    (x)

    xi(x)dx|

    = |

    [

    u(x y) xi

    (x)dx](dy)|

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    = |

    [

    xiu(x y)(x)dx](dy)|

    =

    |

    [

    yi

    u(x

    y)(x)dx](dy)

    | c|| u(x )(x)dx|| c||||.

    Generalizing this inequality to bounded measurable , and then taking = sgn() yields the inequality

    | xi

    |d c

    for any > 0. Now let > 0, g L2(R2) be given. Then, using 1. andthe estimate above

    |

    (x)g(x)dx| [

    |(x)|2dx

    |g(x)|2dx] 12

    12

    [

    | x1

    |d +

    | x2

    |d]12 ||g||2 c||g||2.

    Applying this inequality in the special case, in which g is continuouswith compact support, and using 2. we get

    |

    g(x)(dx)| c||g||2.

    Finally extend this inequality to g L2(R2) by approximating it withcontinuous functions of compact support. This yields the desired conti-nuity of the linear functional.

    4. It remains to apply Riesz representation theorem to find a squareintegrable density for .

    We now consider a vector f = (f1, , fd) with components in

    L2(RN, BN, ). Our aim is to study the absolute continuity with respectto d of the law off under , i.e. of the probability measure f1. Forthis purpose we plan to apply the criterion of Lemma 6.1. Let C1(Rd)possess bounded partial derivatives. Then, the integral transformationtheorem gives

    xid f1 =

    xi

    fd.

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    In case d = 1 at this place we use integration by parts hidden in therepresentations

    d(

    f) = (f)df,

    (f) = d( f) 1df

    .

    Our infinite dimensional analogue of d is the Malliavin gradient .Hence, we need a chain rule for .Theorem 6.1 Let p 2, f Dp1(RN)d, C1(Rd) with bounded par-tial derivatives. Then

    f Dp

    1(RN

    ) and [ f] =d

    i=1

    xi(f) fi

    .

    ProofUse Theorem 5.2 to choose a sequence (fn)nN Pd such that for any1 i d

    ||fin fi||1,p 0.For each n N we have

    [ fn] =d

    i=1

    xi(fn) fin.

    Since is continuous on Dp1(RN), and since the partial derivatives of are bounded, we furthermore obtain that

    [ f] = limn [ fn] =

    di=1

    xi(f) fi

    in L2(RN, ). This completes the proof.We next present a calculation leading to the verification of the abso-

    lute continuity criterion of Lemma 6.1. We concentrate on the algebraicsteps, and remark that their analytic background can be easily providedwith the theory of chapter 5. The first aim of the calculations must beto isolate, for a given test function C1(Rd) with bounded partialderivatives, the expression

    xi(f), 1 i d. Recall the notation

    (x, y) =i=1

    xiyi, x, y l2.

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    For 1 i, k d letik = (fi, fk).

    Then we have for 1 k d

    (( f), fk) = j=1

    dj( f)djfk

    =

    j=1

    1id

    xi(f)djf

    idjfk

    =

    1id

    xi(f)ik.

    We now assume that the matrix is (almost everywhere) invertible.

    Then, denoting its inverse by 1

    we may write

    xi(f) =

    1kd

    (( f), fk1ki )

    =

    1kd

    j=1

    dj( f)1ki djfk.

    We next assume, that the dual operator j of dj, which is defined in theusual way on P, is well defined and the series appearing is summable.Then we have

    xi

    (f)d =

    1kd

    j=1

    dj( f)1ki djfkd

    =

    f[ 1kd

    j=1

    j(djfk1ki )]d.

    The right hand side can be estimated by c|||| with

    c = || 1kd

    j=1

    j(djfk1ki )||2

    in L2(RN, ). It can be seen (in analogy to Theorem 4.3) that this seriesmakes sense under the hypotheses of the following main theorem.

    Theorem 6.2 Suppose that f = (f1, , fd) L2(RN, ) satisfies(i) fi D42(RN) for 1 i d,(ii) ik = (fi, fk), 1 i, k d, is -a.s. invertible and 1ki

    D41(RN) for 1 i, k d.

    Then we have

    f1

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    ProofApproximate f by polynomials, use continuity properties of the operators.

    7 The canonical Wiener space: multiple integrals

    We now return to the canonical Wiener space. The transfer betweensequence and canonical space is provided by the isomorphism of chapter1. We briefly recall it. Let (gi)iN be an orthonormal sequence in L2(R+).Then it is given by

    T : Lp(RN, BN, ) Lp(, F, P), f f ((W(gi)iN),where W(gi) is the Gaussian stochastic integral ofgi for i

    N. It will be

    constructed in the following chapter. For simplicity we confine our atten-tion to the canonical Wiener space in one dimension, i.e. = C(R+, R),F the (completed) Borel -algebra on generated by the topology ofuniform convergence on compact sets in R+, P Wiener measure on F.

    In the approach of differential calculus on Gaussian sequence spaces,in the Hilbert space setting the most important tool proved to be theHermite expansions of functions in L2(RN, ). In the setting on the ca-nonical space, they can be given a different interpretation which we shall

    now develop.According to our isomorphism, the objects corresponding to generali-

    zed Hermite polynomials on the canonical space are given byi=1

    Hpi(W(gi)),

    p E. We shall interpret these objects as iterated Ito integrals. To dothis, we use the abbreviation B1+ for the Borel sets of R+.

    Definition 7.1 For m N letEm = {f|f : Rm+ R, f =

    ni1,,im=1

    ai1im1Ai1Aim ,

    (Ai)1in B1+ p.d., ai1im = 0 in case ik = il for some k = l}.Remark

    For f L2(R+) of the formf =

    ni=1

    ai1Ji, (Ji)1in p.d. intervals in R+

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    let

    W(f) =n

    i=1

    aiW(Ji) =n

    i=1

    ai(Wti Wsi),

    if Ji =]si, ti], 1

    i

    n. Then by Itos isometry we have

    ||W(f)||22 = ||f||22.Since E1 is dense in L2(R+), we can extend the linear mapping f W(f)to L2(R+). Therefore, in particular for A B1+ with finite Lebesguemeasure, W(A) = W(1A) is defined. It will be used for the definitionof the following multiple stochastic integrals. In this chapter, the scalarproduct , will be with respect to Lebesgue measure on Rm+ with

    unspecified integerm

    .Definition 7.2 For f =

    ni1,,im=1 ai1im1Ai1Aim Em let

    Im(f) =n

    i1,,im=1ai1imW(Ai1) W(Aim).

    The additivity of B1+ A W(A) R implies that Im is well defined.We state some elementary properties of Im. Denote by

    Sm the set of

    all permutations of the numbers 1, , m.Lemma 7.1 Let m, q N, f Em, g Eq.

    (i) Im|Em is linear,(ii) if f(t1, , tm) = 1m!

    Sm f(t(1), , t(m)) (symmetrization of

    f), thenIm(f) = Im(f),

    (iii) E(Im(f)Iq(g)) = m!f , g, if m = q, and 0 otherwise.

    Proof

    1. (i) follows from the additivity of the map A W(A).2. (ii) is a direct consequence of the fact that in the definition of Im

    the product W(Ai1) W(Aim) is invariant under permutations of thefactors.

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    3. By (ii), we may assume that f, g are symmetric. By choosing com-mon subdivisions, we may further assume that

    f =n

    i1,,im=1 ai1im1Ai1Aim ,

    g =n

    i1,,iq=1bi1iq1Ai1Aiq .

    Now if m = q, by the assumptions that (Ai)1in consists of p.d. Borelsets, and that coefficients vanish if two of the indices coincide,E(Im(f)Iq(g)) = 0 is evident. Assume m = q. Then, again by thesetwo assumptions and symmetry

    E(Im(f)Im(g)) =

    n

    i1,,im=1

    n

    j1,,jm=1 a

    i1imbj1jmE(m

    p=1

    W(Aip)W(Ajp))

    = m!2

    i1

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    where bi1im = 1 if Bi1 Bim A1 Am, and 0, if not. LetI = {(i1, , im) : ik = il for k = l}, and J = {1, , n}m \ I. Then bydefinition

    f = (i1,,im)Ibi1im1Bi1Bim Emand we have

    ||1A1Am f||22 =

    (i1,,im)Jb2i1im

    mp=1

    (Bip)

    m(m 1)2

    ni=1

    (Bi)2(

    ni=1

    (Bi))m2

    m(m 1)

    2(

    n

    i=1 (Bi))m1

    Finally, we have to choose small enough. Using Lemma 7.2, we may now extend Im to L

    2(Rm+).

    Definition 7.3 The linear and continuous extension ofIm|Em to L2(Rm+)which exists according to Lemma 7.2 is called multiple Wiener-Ito inte-gral of degree m and also denoted by Im.

    Properties of the elementary integral are transferred in a straightfor-ward way.

    Theorem 7.1 Let m, q N, f L2(Rm+), g L2(Rq+). Then(i) Im|L2(Rm+) is linear,(ii) we have

    Im(f) = Im(f),

    (iii) E(Im(f)Iq(g)) = m!f , g, if m = q, and 0 otherwise,(iv) I1(f) = W(f), f L

    2

    (R+).

    NotationWe write

    Im(f) =Rm+

    f(t1, , tm)dWt1 dWtm=

    R

    m+

    f(t1, , tm)W(dt1) W(dtm).

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    We next aim at explaining the relationship between generalized Her-mite polynomials and multiple Wiener-Ito integrals. For this purposewe will need a recursive relationship between Hermite polynomials of

    different degrees.RemarkRecall the definition of Hermite polynomials in one variable, given by

    Hn = n1.

    Moreover, we may compute for n NxHn = (d + )Hn = nHn1 + Hn+1, or Hn+1 = xHn nHn1.

    For technical reasons, we need the following operation of contraction.

    Definition 7.4 Let m N. Suppose f L2(Rm+), g L2(R+). Thenfor t1, , tm, t R+

    f g(t1, , tm, t) = f(t1, , tm) g(t) (tensor product),

    f1 g(t1, , tm1) =R+

    f(t1, , tm)g(tm) dtm (contraction).

    The recursion relation for Hermite polynomials will emerge from therecursion relation between Wiener-Ito integrals stated in the followingLemma.

    Lemma 7.3 Let m N, f L2(Rm+), g L2(R+). Then we haveIm(f)I1(g) = Im+1(f g) + mIm1(f1 g).

    Proof

    1. By linearity and density ofEm in L2

    (R

    m

    +) we may assume thatf = 1A1Am, g = 1A0 or g = 1A1,

    where (Ai)0im B1+ is a collection of p.d. Borel sets with finite Le-besgue measure.

    2. The case g = 1A0 is trivial. Then the second term on the righthand side of the claimed formula vanishes, and the other two terms areobviously identical by definition of the elementary integral.

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    3. Let now g = 1A1. For > 0 choose a collection of p.d. setsB1, , Bn B1+ such that A1 = ni=1Bi, and for any 1 i n wehave (Bi) < . Then

    Im(f)I1(g) = W(A1)2W(A2) W(Am)=

    i=j

    W(Bi)W(Bj)W(A2) W(Am)

    +

    1in[W(Bi)

    2 (Bi)]W(A2) W(Am)+ (A1)W(A2) W(Am).

    a) We now prove that the first term on the right hand side of ourformula is close to Im+1(f

    g). In fact, let

    h =i=j

    1BiBjA2Am Em+1.

    Then

    ||h f g||22 n

    i=1

    (Bi)2(A2) (Am)

    (A1) (Am).

    b) Let us next prove that the second term on the right hand side isnegligeable in the limit 0. In fact, denote it by R. Then, since for1 i n the variance of W2(Bi) (Bi) is given by c(Bi)2 with someconstant c, we obtain

    E(R2) cn

    i=1

    (Bi)2(A2) (Am) c(A1)(A2) (Am).

    c) To evaluate the last term, note that

    1A1Am 1 1A1 = 1m 1A2Am (A1).Therefore

    (A1)W(A2) W(Am) = mIm1( 1A1Am 1 1A1),and we obtain the desired recursion formula.

    This finally puts us in a position to derive the relationship betweenHermite polynomials and iterated stochastic integrals.

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    Theorem 7.2 Let m N, h L2(R+) be such that ||h||2 = 1. Denoteby hm the m-fold tensor product of h with itself. Then we have

    Hm(W(h)) = Im(hm).

    Let H0 = R, Hm = Im(L2(Rm+)), m N. Then: (Hm)mN is a sequenceof pairwise orthogonal closed linear subspaces ofL2(, F, P) and we have

    L2(, F, P) = m=0Hm.In particular, for any F L2(, F, P) there exists a sequence (fm)m0of functions fm L2(Rm+) such that

    F = m=0

    Im(fm).

    The representation with symmetric fm is m-a.e. unique, m N.

    Proof1. We first have to prove:

    Hm(W(h)) = Im(hm).

    This is done by induction on m. For m = 1, the formula is clear fromH1 = x, I1(h) = W(h). Now assume it is known for m. Then Lemma 7.3and the recursion formula for Hermite polynomials given above combineto yield, remembering that ||h||2 = 1,

    Im+1(hm+1) = Im(hm)I1(h) mIm1(hm 1 h)

    = Im(hm)I1(h) mIm1(hm1)

    = Hm(W(h))H1(W(h)) mHm1(W(h))

    = Hm+1(W(h)).

    2. Let L2s(Rm+) be the linear space of symmetric functions in L

    2(Rm+).Then by Theorem 7.1

    ||Im(f)||22 = m!||f||22,hence Hm = Im(L2s(Rm+)) is closed. Orthogonality is also a consequenceof Theorem 7.1.

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    3. Let (gi)iN be an orthonormal basis of L2(R+), F L2(, F, P).Let f L2(RN, BN, ) be such that T(f) = F. Assume f (cp(f)),according to the notation of chapter 3. For m 0, define

    fm =

    pE,|p|=mcp(f)

    p!iN

    gpii ,

    considered as a function of m variables. Then fm L2(Rm+) and withthe help of Lemma 7.3 we see

    Im(fm) =

    pE,|p|=m

    cp(f)

    p!Im(

    iN

    gpii )

    = pE,|p|=mcp(f)

    p! iN Ipi(gpi

    i )

    =

    pE,|p|=m

    cp(f)

    p!

    iN

    Hpi(W(gi)).

    Summing this expression over m yields the desired

    F =

    m=0

    Im(fm).

    The remaining claims are obvious.8 The canonical Wiener space: Malliavins deriva-

    tive

    In this chapter we shall investigate the analogue of the gradient we en-countered in the differential calculus on the sequence space. Fix againan orthonormal basis (gi)iN of L2(R+), and recall the isomorphism

    T : L2(RN, BN, ) L2(, F, P), f f((W(gi)iN).Of course, every permutation of the orthonormal basis functions givesanother orthonormal basis. So here we encounter a problem of coordinatedependence of our objects of study. How can we define Malliavins deri-vative on the canonical space in a both consistent and basis independentway? According to Theorem 5.2

    f = (djf)j

    N

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    takes values in l2. The corresponding object on the side of the canonicalspace is L2(R+). It is therefore plausible if we set

    Definition 8.1 For n

    N let Cp (Rn) denote the set of smooth functi-

    ons the partial derivatives of which possess polynomial growth. Let

    S = {F|F = f(W(h1), , W(hn)), h1, , hn L2(R+),f Cp (Rn), n N}.

    For F = f(W(h1), , W(hn)) S, t 0 let

    DtF =n

    i=1

    xif(W(h1), , W(hn)) hi(t).

    To see if this is a good candidate for the definition of the Mallia-vin gradient in the setting of the canonical space, let us verify in de-tail the independence on the specific representation of functionals. Leth1, , hn L2(R+), and g1, , gm L2(R+) orthonormal, such thatthe linear hulls of the two systems are identical, and such that withf Cp (Rn), g Cp (Rm) we have

    f(W(h1),

    , W(hn)) = g(W(g1),

    , W(gm)).

    For 1 i n writehi =

    mj=1

    hi, gjgj.

    Then, denoting = (hi, gj)1in,1jm,

    we obviously have f = g. Thereforem

    j=1

    xj (f )(W(g1), , W(gm))gj =m

    j=1

    n

    i=1

    xif(W(h1), , W(hn))hi, gjgj

    =n

    i=1

    xif(W(h1), , W(hn))hi.

    This proves that the definition ofD is independent of the representationof functionals in S.

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    If h L2(R+) is another function, we have by definition

    DF, h =mi=1

    xif(W(g1), , W(gm))gi, h,

    in particular for i N

    DF, gi = xi

    f(W(g1), , W(gm)) = dif(W(g1), , W(gm)).

    We therefore may interpret DF, gi as directional derivative in directionof gi, and we have by Parsevals identity

    DF,DF =mi=1

    DF,gi2 =mi=1

    (dif)2(W(g1), , W(gm))

    = |f|2(W(g1), , W(gm)).Analogously, higher derivatives are related to each other. So we see thatthe isomorphism T also maps DF,DF to |f|2. Consequently, we can

    just transfer the definitions of Gaussian Sobolev spaces to the setting ofthe canonical Wiener space.

    Definition 8.2 Let p 2, s N. For F L2(, F, P) denote by f L2(RN, BN, ) the function for which we have F = f((W(gi)iN). Then

    letDps = {F|f Dps(RN)}

    (canonical Gaussian Sobolev space of order (s, p)), with the norm

    ||F||s,p = ||f||s,p.For F = T(f) Dps, 1 r s, let

    DrF =

    j1,

    ,jr=1

    dj1 djrf(W(gi)iN)gj1 gjr

    (canonical Malliavin derivative of order r).

    RemarkFrom our knowledge of sequence spaces we can easily derive that Dps isa Banach space with respect to the norm | | | |s,p for p 2, s N, andthat for F Dps we have

    ||F||s,p =s

    r=0

    ||DrF, DrF||p.

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    We know that Dps is the closure ofSwith respect to the norm | | | |s,p.Turning to p = 2, we know that D21 is a Hilbert space with respect to

    the scalar product

    (F, G)1,2 = E(F G) + E(DF,DG), F, G D21.Moreover, we know that D is a closed operator, defined on D21, which iscontinuous as a mapping from D21 to L

    2(, F, P).Let us now investigate how D operates on the decomposition into

    Wiener-Ito integrals.

    Theorem 8.1 Let F =

    m=0 Im(fm) L2(, F, P) be given, fm sym-metric for any m

    0. Then we have

    F D21 if and only if

    m=1m m! ||fm||22 < .

    In this case we have

    DtF =

    m=1m Im1(fm(, t))

    (for P -a.e. (, t) R+).

    Proof1. Suppose that with respect to an orthonormal basis (gi)iN of L2(R+)we have F = T(f) with f (cp(f)). As before, for m 0 let

    fm =

    pE,|p|=m

    cp(f)

    p!

    iN

    gpii .

    We interpretiN g

    pii as

    kj=1 g

    pijij , ifi1, , ik are precisely those indices

    for which pi1

    ,

    , pik

    > 0.2. Let now p E such that |p| = m, and let t 0. Then

    DtIm(iN

    gpii ) = Dt

    iN

    Ipi(gpii )

    = DtHp((W(gi)iN)=

    iN

    pij=i

    Hpj(W(gj))Hpi1(W(gi))gi(t)

    = Im1(i

    N

    pij

    =i

    gpjj g

    pi1i gi(t)).

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    Hence by closedness of D, symmetry of fm and |p| = m, the desiredformula

    DtIm(fm) = mIm1(fm(, t))follows.

    3. For n N let now

    Fn =n

    m=0

    Im(fm).

    By the closedness of the operator D and the remarks above, we knowthat

    F D21 if and only if (Fn)nN is Cauchy in D21.

    Now we know from the first part of the proof that

    DFn =n

    m=1m Im1(fm(, )).

    Let n, m N, n m be given. Then

    E(D(Fn Fm), D(Fn Fm)) =n

    k=m+1

    k2R+

    (k 1)!fk(, t), fk(, t)dt

    =

    n

    k=m+1 k2

    (k 1)!||fk||2

    2.

    Hence (DFn)nN is a Cauchy sequence in D21 if and only if

    k=0 kk!||fk||22 0and X H2(Rm) let

    ||X||22, = E(T0

    et|Xt|2dt),and H2,(Rm) the space H2(Rm) endowed with the norm | | | |2,.

    We next describe the general hypotheses we want to require for the

    parameters of our BSDE. The terminal condition will be supposed tobelong to L2(Rm). The generator will be a function

    f : R+ Rm Rnm Rm,which is product measurable, adapted in the time parameter, and whichfulfills

    (H1) f(, 0, 0) H2(Rm),f is uniformly Lipschitz, i.e. there exists C R such that for any(y1, z1), (y2, z2) R

    m

    Rnm

    , P -a.e. (, t) R+(H2) |f(, t, y1, z1) f(, t, y2, z2)| C[|y1 y2| + |z1 z2|].

    Here for z Rnm we denote |z| = (tr(zz))12 .Definition 10.1 A pair of functions (f, ) fulfilling, besides the mentio-ned measurement requirements, hypotheses (H1), (H2), is said to be astandard parameter.

    Given standard parameters, we shall solve the problem of finding apair of (Ft)0tT-adapted processes (Yt, Zt)0tT such that the backwardstochastic differential equation (BSDE)

    () dYt = Zt dWt f(, t , Y t, Zt)dt, Y T = ,is satisfied.

    In order to construct a solution, a contraction argument on suitableBanach spaces will be used. For its derivation we shall need the followinga priori inequalities.

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    Lemma 10.1 For i = 1, 2 let (fi, i) be standard parameters, (Yi, Zi) H2(Rm) H2(Rnm) solutions of (*) with corresponding standard para-meters. Let C be a Lipschitz constant for f1. Define for 0 t T

    Yt = Y1t Y2t ,2ft = f

    1(, t , Y 2t , Z2t ) f2(, t , Y 2t , Z2t ).Then for any triple (,,) with > 0, 2 > C, C(2 + 2) + 2 wehave

    ||Y||22, T[eTE(|YT|2) +1

    2||2f||22,],

    ||Z||22, 2

    2

    C

    [eTE(|YT|2) + 12

    ||2f||22,].Proof

    1. Let (Y, Z) H2(Rm) H2(Rnm) be a solution of (*) with standardparameters (f, ). This means that we may write for 0 t T

    () Yt = Tt

    ZsdWs +Tt

    f(, s , Y s, Zs)ds.We show:

    sup0tT

    |Yt| L2(Rm).

    In fact, due to (*) we havesup

    0tT|Yt| || +

    T0

    |f[, s , Y s, Zs)|ds + sup0tT

    |Tt

    Zs dWs|,and, with the help of Doobs inequality

    E( sup0tT

    |Tt

    ZsdWs|2) 4E( sup0tT

    |t0

    Zs dWs|2) 8E(T0

    |Zs|2ds).

    Since in addition (H1) and (H2) guarantee that ||+T0 |f(, s , Y s, Zs)|ds L2(R), we obtain the desired

    E( sup0tT

    |Yt|2) < .2. Now we derive a preliminary bound. Apply Itos formula to the

    semimartingale (es|Ys|2)0sT to obtain for 0 t TeT|YT|2 et|Yt|2

    = Tt

    es|Ys|2ds + 2Tt

    esYs, f1(, s , Y 1s , Z1s )

    f2(

    , s , Y 2s , Z

    2s )

    ds

    2Tt

    es

    Ys, Z

    sdWs

    +Tt

    es

    |Zs

    |2ds.

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    By reordering the terms in the equation we obtain

    et|Yt|2 + Tt

    es|Ys|2ds +Tt

    es|Zs|2ds

    = eT|YT|2 + 2 T

    t esYs, ZsdWs

    2Tt

    esYs, f1(, s , Y 1s , Z1s ) f2(, s , Y 2s , Z2s )ds.3. We prove for 0 t T:

    E(et|Yt|2) E(eT|YT|2) + 12

    E(Tt

    es|2fs|2ds).

    To prove this, first take expectations on both sides of the inequality

    obtained in 2., with the result

    E(et|Yt|2) + E(Tt

    es|Ys|2ds) + ETt

    es|Zs|2ds) E(eT|YT|2)

    +2E(Tt

    esYs, f1(, s , Y 1s , Z1s ) f2(, s , Y 2s , Z2s )ds).Now by our assumptions for 0 s T

    |f1

    (, s , Y 1

    s , Z

    1

    s ) f2

    (, s , Y 2

    s , Z

    2

    s )| |f1

    (, s , Y 1

    s , Z

    1

    s ) f1

    (, s , Y 2

    s , Z

    2

    s )|+|2fs|

    C[|sY| + |sZ|] + |2fs|.The latter implies

    Tt

    E(2es|Ys, f1(, s , Y 1s , Z1s ) f2(, s , Y 2s , Z2s )|dsTt

    2esE(|Ys|[C(|sY| + |sZ|) + |2fs|]ds= T

    t2es[CE(|Ys|2) + E(|sY|(C|sZ|) + |2fs|)]ds.

    Now for C, y, z, t > 0 with , > 0

    2y(Cz + t) = 2Cyz + 2yt

    C[(y)2 + ( z

    )2] + (y)2 + (t

    )2

    = C(z

    )2 + (

    t

    )2 + y2(2 + C2).

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    With this we can estimate the last term in our inequality further:Tt

    2es[CE(|Ys|2) + E(|sY|(C|sZ| + |2fs|))]ds

    T

    t es[2CE(|Ys|2) +

    C

    2E(|sZ|2)+

    1

    2E(|2fs|2) + (2 + C2)E(|sY|2]ds

    =Tt

    es[(2 + C(2 + 2))E(|Ys|2)

    +C

    2E(|sZ|2) + 1

    2E(|2fs|2)]ds.

    Summarizing, we obtain, using our assumptions on the parameters

    () E(et|Yt|2) E(Tt

    es|Ys|2ds)[+ C(2 + 2) + 2]

    +E(Tt

    es|Zs|2ds)[ C2

    1] + E(eT|YT|2)

    +1

    2E(Tt

    es|2fs|2ds) + E(eT|YT|2)

    E(eT|YT|2) + 12

    E(Tt

    es|2fs|2ds).

    This is the claimed inequality.4. In order to obtain the first inequality in the assertion, it remains to

    integrate the inequality resulting from 3. in t [0, T].5. The second inequality in the assertion follows from (**) by taking

    the second term from the right hand side to the left. This completes theproof.

    We are in a position to state existence and uniqueness results for ourBSDE (*).

    Theorem 10.1 Let (, f) be standard parameters. Then there exists auniquely determined pair(Y, Z) H2(Rm)H2(Rnm) with the property

    (BSDE) Yt = Tt

    ZsdWs +Tt

    f(, s , Y s, Zs)ds, 0 t T.Proof

    Consider

    : H2,(Rm)

    H2,(Rnm)

    H2,(Rm)

    H2,(Rnm), (y, z)

    (Y, Z),

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    where (Y, Z) is a solution of the BSDE

    () Yt = Tt

    Zs dWs +Tt

    f(, s , ys, zs)ds, 0 t T.

    1. We prove: (Y, Z) is well defined. First of all, our assumptions yield

    +Tt

    f(, s , ys, zs)ds L2(), 0 t T.Therefore

    Mt = E( +T0

    f(, s , ys, zs)ds|Ft), 0 t T,is a well defined martingale. M possesses a continuous version, due to

    the fact that we are working in a Wiener filtration. M is square integra-ble. Hence we may apply the martingale representation theorem, whichprovides (a unique) Z H2(Rnm) such that

    Mt = M0 +t0

    ZsdWs, 0 t T.Let now

    Yt = Mt t0

    f(, s , ys, zs)ds.

    Then Y is square integrable, and we have

    Yt = E( +Tt

    f(, s , ys, zs)ds|Ft), 0 t T.Hence

    YT = = M0 +T0

    ZsdWs T0

    f(, s , ys, zs)ds,and thus for 0 t T

    Yt =

    M0

    T0

    ZsdWs +T0

    f(

    , s , ys, zs)ds

    + M0 +t0

    ZsdWs t0

    f(, s , ys, zs)ds=

    Tt

    Zs dWs +Tt

    f(, s , ys, zs)ds.

    2. We prove: For > 2(1 + T)C the mapping is a contracti-on. For this purpose, let (y1, z1), (y2, z2) H2,(Rm) H2,(Rnm),(Y1, Z1), (Y2, Z2) corresponding solutions of (*) according to 1. We ap-ply Lemma 10.1 with C = 0, = 2, and fi = f(, yi, zi). With this

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    choice we obtain

    ||Y||2, T

    [E(T0

    es|f(, s , y1s , z1s) f(, s , y2s , z2s)|2ds)]12 ,

    ||Z||2, 1

    [E(T0

    es|f(, s , y1s , z1s) f(, s , y2s , z2s)|2ds)]12 .

    Since f is Lipschitz continuous, we further obtain

    ||Y||2, 2T C

    [||y||2, + ||z||2,],

    ||Z||2, 2C

    [||y||2, + ||z||2,].

    We summarize to obtain

    () ||Y||2, + ||Z||2, 2C(T + 1)

    [||y||2, + ||z||2,].

    By choice of , is a contraction.3. Now let (Y , Z) be the fixed point of , which exists due to 2. Let

    Yt = E( +Tt

    f(, s , Y s, Zs)ds|Ft), 0 t T.

    Then Y is continuous and P-a.s. identical to Y. Then (Y, Z) is a solutionof our BSDE.

    4. Uniqueness follows from the contraction property of and the un-iqueness of the fixed point.

    The construction of solutions in the preceding proof rests upon a recur-sive algorithm. The algorithm converges, as we shall note in the followingCorollary.

    Corollary 10.1 Let > 2(1 + T)C, ((Yk, Zk))k0 the sequence of pro-cesses, given by Y0 = Z0 = 0,

    Yk+1t = Tt

    (Zk+1s )dWs +

    Tt

    f(, s , Y ks , Zks )ds

    according to the proof of the preceding Theorem. Then((Yk, Zk))k0 con-verges in H2,(Rm) H2,(Rnm) to the uniquely determined solution(Y, Z) of (BSDE).

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    ProofThe inequality (**) in the proof of Theorem 10.1 recursively yields

    ||Yk+1

    Yk

    ||2, +

    ||Zk+1

    Zk

    ||2,

    k[

    ||Y1

    Y0

    ||2, +

    ||Z1

    Z0

    ||2,],

    with = 2C(T+1)

    < 1. This implies

    kN

    [||Yk+1 Yk||2, + ||Zk+1 Zk||2,] < .

    Now a standard argument applies.

    11 Interpretation of backward stochastic differen-

    tial equations in Malliavins calculus

    In this chapter we shall establish the vital connection between Mallia-vins calculus and the structure of solutions of BSDE. We shall see that,provided the standard parameters are sufficiently smooth, the processZ can be interpreted as the Malliavin trace of the process Y. For doingthis, we first have to introduce the version of the space L21 which corre-sponds to integrability in some arbitrary power p 2. For simplicity, welet the dimension of our underlying Wiener process be one, i.e. for thischapter we set n = 1.

    Definition 11.1 Let p 2, and

    Lp1(Rm) = {u|u adapted, [

    T0

    |ut|2dt] 12 Lp(),ut (Dp1)m for a.a. t 0,

    for some measurable version

    of (s, t) Dsut we have E([T

    0T0 |Dsut|2dsdt]

    p

    2) < }.For u Lp1(Rm) define

    ||u||1,p = E([T0

    |ut|2dt]p

    2 ) + E([T0

    T0

    |Dsut|2dsdt]p

    2 ).

    To abbreviate, for k N, v L2( [0, T]k) denote||v|| = [

    [0,T]k

    |vt|2dt]12 .

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    In these terms, Jensens inequality gives

    E(||Du||p) Tp21T0

    ||Dsu||ppds.

    We next prove that solutions of BSDE for regular standard parametersare Malliavin differentiable, and that Z allows an interpretation as a Mal-liavin trace ofY. We need some versions of the process spaces consideredin the previous chapter that correspond to p-integrable random varia-bles. For p 2 denote by Sp(Rm) the linear space of all measurable (Ft)-adapted continuous processes X : [0, T] Rm, endowed with thenorm ||X||Sp = E(sup0tT |Xt|p)

    1p . Let further Hp(Rm) denote the linear

    space of measurable (

    Ft)-adapted processes X :

    [0, T]

    Rm endo-

    wed with the norm ||X||p = E(||X||p)1p . To abbreviate, let Bp(Rm) =

    Sp(Rm) Hp(Rm), with the norm ||(Y, Z)||p = [||Y||pSp + ||Z||pp]1p .

    We are ready to state our main result.

    Theorem 11.1 Let (f, ) be standard parameters such that D21 L4(Rm), f : [0, T] Rm Rm Rm continuously differentiablein (y, z), with uniformly bounded and continuous partial derivatives, andsuch that for (y, z) Rm Rm we have

    (H3) f(, y , z) L21, f(, 0, 0) H4(Rm),for t [0, T] and (y1, z1, y2, z2) (Rm Rm)2 we have

    (H4) |Dsf(, t , y1, z1) Dsf(, t , y2, z2)| Ks(t)[|y1 y2| + |z1 z2|](in L2( [0, t]2)), with a real-valued measurable process (Ks(t))0stwhich is (Ft)-adapted in t, and satisfies

    T0 ||Ks||44ds < .For the unique solution (Y, Z) of the BSDE (*) we moreover suppose

    T0

    ||Dsf(, Y , Z )||2ds < P-a.s..

    Then we have:(Y, Z)

    L21(R

    m)

    L21(R

    m),

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    and a (measurable) version of (DsYt, DsZt)0s,tT) possesses the proper-ties

    DsYt = DsZt = 0, 0

    t < s

    T,

    DsYt = Ds Tt

    DsZudWu

    +Tt

    [

    yf(, u , Y u, Zu) DsYu +

    zf(, u , Y u, Zu) DsZu

    +Dsf(, u , Y u, Zu)]du, 0 s t T,(DsYs)0sT is a version of (Zs)0sT.

    Proof1. For further simplifying notation, we assume m = 1. As in chapter 10,our arguments are mainly based upon several a priori estimates. Thefirst one is an analogue of Lemma 10.1 and investigates the propertiesof the contraction map on B2(R).

    Lemma 11.1 Let p 2, assume f(, 0, 0) Hp(R), and define : Bp(R) Bp(R), (y, z) (Y, Z),

    where (Y, Z) is the solution of the BSDE

    (+) Yt = Tt

    ZsdWs +Tt

    f(, s , ys, zs)ds.

    Let further for i = 1, 2 (Yi, Zi) be the solutions corresponding to (yi, zi)in (+), and let

    Y = Y1 Y2, Z = Z1 Z2, y = y1 y2, z = z1 z2.Then there exists a constant Cp not depending on (y , z , Y , Z ) such that

    (i) ||Y||pSp CpE([||p) + Tp

    2(T0

    |f(, s , ys, zs)|2ds)p

    2 ]),

    (ii) ||Z||pp CpE([||p) + Tp

    2(T0

    |f(, s , ys, zs)|2ds)p

    2 ]),

    (iii) ||(Y,Z)||pp CpTp

    2 ||(y,z)||pp.Proof

    a) Taking up the notation of the proof of Lemma 10.1, we show: iswell defined.

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    To do this, recall for 0 t TYt = E( +

    Tt

    f(, s , ys, zs)ds|Ft).

    We have |Yt| E(|| + Tt

    |f(, s , ys, zs)|ds|Ft),hence Doobs inequality provides a universal constant C1p such that

    ||Y||pSp C1pE([|| +T0

    |f(, s , ys, zs)|ds]p).Moreover, by Cauchy-Schwarz inequalityT

    0|f(, s , ys, zs)|ds T12 [

    T0

    |f(, s , ys, zs)|2ds]12 ,

    hence with another universal constant C2p we have

    () ||Y||pSp C2pE([||p + [T0

    |f(, s , ys, zs)|2ds]p

    2 ]).

    Invoke f(, 0, 0) Hp(R), that f is Lipschitz continuous, and that(y, z) Bp(R), to see that the right hand side of the preceding ine-quality is finite.

    We next prove that Z Hp(R). For this purpose we shall use theinequality of Burkholder. It yields further universal constants C3p ,

    , C5p

    such that

    () E(||Z||p) C3pE(|T0

    ZsdWs|p) C4pE(| +

    T0

    f(, s , ys, zs)ds Y0|p) C5pE(||p + T

    p

    2 [T0

    |f(, s , ys, zs)|2ds]p

    2).

    Hence we obtain Z Hp(R), and therefore (Y, Z) Bp(R). The ine-qualities (i) and (ii) have also been established.

    b) We prove: (iii). The solution (Y,Z) belongs to the generatorf(, t , y1t , z1t ) f(, t , y2t , z2t ), and = 0. Therefore (i) and (ii), as wellas an appeal to the Lipschitz condition, give, with universal constantsC6p , C

    7p

    ||(Y,Z)||pp C6pTp

    2 E([T0

    |f(, t , y1t , z1t ) f(, t , y2t , z2t )|2dt]p

    2)

    C7pTp

    2 [||y||pSp + ||z ||pp]= C7pT

    p

    2

    ||(y,z)

    ||pp.

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    This completes the proof. Let us return to the proof of Theorem 11.1. We define approximations

    of the solution of the BSDE recursively. Let for k 0, 0 t TY0 = Z0 = 0,

    Yk+1t = Tt

    Zk+1s dWs +Tt

    f(, s , Y ks , Zks )ds.We show:

    ||(Yk, Zk) (Y, Z)||4 0 (k ).Recall the universal constant C4 from Lemma 11.1, (iii). We may (mo-

    dulo repeating the argument finitely often on successive subintervals of[0, T]) assume that [C4T2]

    14 < 1. With this condition, Lemma 11.1 im-

    plies that is a contraction, and the solution (Y, Z) of the BSDE itsunique fixed point in B4(R). From this observation, we obtain our as-sertion via the Cauchy sequence property of the approximate solutionswhich follows from

    ||(Yk, Zk)(Yl, Zl)||4 l

    r=k+1

    ||(Yr, Zr)(Yr1, Zr1)||4 0 (k, l ).

    2. We prove by recursion on k:

    (Yk, Zk) L21(R) L21(R).This is trivial for k = 0. Let it be guaranteed for k. According to thechain rule for the Malliavin derivative and our hypotheses concerningthe standard parameters we know for 0 t T

    + T

    t f(, s , Y k

    s , Zk

    s )ds D2

    1,

    with Malliavin derivative

    Ds +Tt

    [

    yf(, u , Y ku , Zku) DsYku +

    zf(, u , Y ku , Zku) DsZku

    + Dsf(, u , Y ku , Zku)]du.This is seen by discretizing the Lebesgue integral, using the chain ru-le, and then approximating by means of the boundedness properties of

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    the partial derivatives, the Lipschitz continuity properties of f and clo-sedness of the operator D. Consequently, Lemma 9.2 yields for fixed0 t T

    Yk+1

    t = E( + T

    t f(, s , Y k

    s , Zk

    s )ds|Ft) D2

    1

    as well. Consequently, also

    Tt

    Zk+1s dWs = +Tt

    f(, s , Y ks , Zks )ds Yk+1t D21.

    Now an appeal to Theorem 9.4 implies that Zk+1 L21(R) and in L2([0, T]2) we have the equation

    Ds

    T

    tZk+1

    udW

    u=

    T

    tD

    sZk+1

    udW

    u, s

    t,

    DsTt

    Zk+1u dWu = Zk+1s +

    Ts

    DsZk+1u dWu, s > t.

    All stated differentiabilities go along with square integrability of theMalliavin derivatives in all variables. This means that

    (Yk+1, Zk+1) L21(R) L21(R),and the recursion step is completed. We also can identify the Malliavin

    derivative by the formula valid for 0 s t T in the usual sense( ) DsXk+1t = Ds

    Tt

    DsZk+1u dWu

    +Tt

    [

    yf(, u , Y ku , Zku) DsYku +

    zf(, u , Y ku , Zku) DsZku

    +Dsf(, u , Y ku , Zku)]du.

    3. In this step we show:

    (DYk, DZk) (Y, Z) in L2( [0, T]2),where for 0 s T (Ys, Zs) is the solution of the BSDE

    ()Yst = Ds Tt

    ZsudWu +Tt

    [

    yf(, u , Y u, Zu) Ysu +

    zf(, u , Y u, Zu) Zsu

    +Dsf(, u , Y u, Zu)]du, 0 s t T,Yst = Z

    st = 0, 0

    t < s

    T.

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    We first consult our hypotheses to verify that, at least for -a.e. 0 s T the parameters (Fs, Ds) with

    F

    s

    (, t , y , z) =

    y f(, t , Y t, Zt) y +

    z f(, t , Y t, Zt) z + Dsf(, t , Y t, Zt),0 t T , y , z R, are standard. Hence (Ys, Zs) is well defined (andset trivial on the set of s where the parameters eventually fail to bestandard). Also in this case our arguments will be based on a prioriinequalities.

    Lemma 11.2 Let (fi, i), i = 1, 2, be standard parameters of a BSDE,p

    2. Suppose

    i Lp(), fi(, 0, 0) Hp(R), i = 1, 2.Let (Yi, Zi) Bp(R) be the corresponding solutions, C a Lipschitz con-stant for f1. Put

    Y = Y1Y2, Z = Z1Z2, 2ft = f1(, t , Y 2t , Z2t )f2(, t , Y 2t , Z2t ),0 t T. Then for T small enough there exists a constant Cp,T suchthat

    ||(Y,Z)||pp Cp,T[E(|YT|p) + E((T0

    |2fs|ds)p)] Cp,T[E(|YT|p) + T

    p

    2 ||2fs||p].Proof

    With a calculation analogous to the one used to prove (i) and (ii) inLemma 11.1 we arrive at the following inequality which is valid with uni-versal constants C1p , , C3p , and for which we also use Doobs and Cauchy-Schwarz inequalities,||Y||pSp + ||Z||pp C1pE(|YT|p

    +(T0

    |f1(, t , Y 1t , Z1t ) f2(, t , Y 2t , Z2t )|dt)p) C2pE(|YT|p + (

    T0

    [|Ys| + |Zs| + |2fs|]ds)p) C3p [E(|YT|p

    +(T0

    |2fs

    |ds)p) + (Tp

    ||Y

    ||pSp + T

    p

    2

    ||Z

    ||pp)].

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    Now choose T small enough to ensure C3p(Tp+T

    p

    2) < 1. This being done,we may take the last two expressions in the previous inequality from theright to the left hand side, to obtain the desired estimate.

    Let us now apply Lemma 11.2 to prove that for -a.a. 0 s T wehave (Ys, Zs) B2(R). For this purpose, we apply the Lemma with

    Y1 = Ys, Y2 = 0,

    1 = Ds, 2 = 0,

    f1(, t , y , z) = ( y

    f(, t , Y t, Zt) y + z

    f(, t , Y t, Zt) z + Dsf(, t , Y t, Zt),f2 = 0,

    0 t T , y , z R. Then we have2ft = Dsf(, t , Y t, Zt), 0 t T.

    We obtain with some universal constant C the inequality

    ||(Ys, Zs)||22 CE(|Ds|2 + ||Dsf(, Y , Z )||2),and therefore

    T

    0 ||(Ys

    , Z

    s

    )||2

    2ds < .This implies the desired integrability.

    To obtain estimates for differences of (DYk, DZk) and (Y, Z), let usnext, fixing k N, apply Lemma 11.2 to the following parameters

    1 = 2 = Ds,

    f1(t) =

    yf(, t , Y kt , Zkt ) DsYkt +

    zf(, t , Y kt , Zkt ) DsZkt +Dsf(, t , Y kt , Zkt ),

    f2(t) =

    y f(, t , Y t, Zt) Yst +

    z f(, t , Y t, Zt) Zst + Dsf(, t , Y t, Zt),s t T. Set for abbreviation

    kt = f1(t) f2(t), s t T.

    The Lemma yields the inequality

    ||(DsYk+1 Ys, DsZk+1 Zs)||22 C1E((Ts

    |kt |dt)2)

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    with a universal constant C1. Let us now further estimate the right handside of this inequality. We have, fixing 0 s T

    E((T

    s |kt

    |dt)2)

    C2[A

    sk(T) + B

    sk(T) + C

    sk(T)],

    where

    Ask(T) = E([Ts

    |Dsf(, t , Y t, Zt) Dsf(, t , Y kt , Zkt )|dt]2),

    Bsk(T) = E([Ts

    | y

    f(, t , Y kt , Zkt )(Yst DsYkt )|dt]2)

    + E([Ts

    | z

    f(, t , Y kt , Zkt )(Zst DsZkt )|dt]2),

    Csk(T) = E([T

    s |

    y f(, t , Y t, Zt)

    y f(, t , Y kt , Zkt )| |Yst |dt]2)

    + E([Ts

    | z

    f(, t , Y t, Zt) z

    f(, t , Y kt , Zkt )| |Zst |dt]2).

    With further universal constants C3, C4 we deduce, using (H4)

    Ask(T) E([Ts

    Ks(t)[|Yt Ykt | + |Zt Zkt |]dt]2)

    C3E(

    Ts

    Ks(t)2dt[

    Ts

    |Yt

    Ykt

    |2dt +

    Ts

    |Zt

    Zkt

    |2dt])

    C4[E(Ts

    Ks(t)4dt]

    12 [E(

    Ts

    |YtYkt |4dt)12 + E(

    Ts

    |ZtZkt |4dt)12 ].

    Hence by part 1. of the proof

    limk

    T0

    Ask(T)ds = 0.

    Furthermore, since the partial derivatives off with respect to y, z arebounded and continuous, and since E(

    T0 ||(Ys, Zs)||2ds < , dominated

    convergence allows to conclude

    limk

    T0

    Csk(T)ds = 0.

    Let us finally discuss the convergence of the Bsk(T) as k . Againby boundedness of the partial derivatives of f we obtain with a universalconstant C5

    Bsk(T)

    C5T

    2

    ||DsY

    k

    Ys, DsZ

    k

    Zs)

    ||22.

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    Now choose T small enough to ensure = C5T2 < 1. Let > 0. Then

    by what has been shown there exists N N large enough so that fork N we have

    E(T0

    ||(DsYk+1 Ys, DsZk+1 Zs)||2ds) + E(

    T0

    ||(DsYk Ys, DsZk Zs)||2ds).By recursion we obtain for k N

    E(T0

    ||(DsYk Ys, DsZk Zs)||2ds) (1 + + 2 + + kN1)+kNE(

    T0

    ||(DsYN Ys, DsZN Zs)||2ds)

    1 + kNE(T

    0||(DsYN Ys, DsZN Zs)||2ds).

    Now let k . Since is arbitrary, we conclude

    limk

    T0

    Bsk(T)ds = 0.

    3. Since L21

    (R) is a Hilbert space, and D is a closed operator, weobtain that (Y, Z)

    L21(R)

    L21(R), and that (Y

    s, Zs)0s

    T is a version

    of (DsY, DsZ)0sT in the usual sense.

    4. We show:

    (DtYt)0tT is a version of (Zt)0tT.

    For t s we haveYs = Yt +

    st

    ZrdWr st

    f(, r , Y r, Zr)dr.

    Hence by Theorem 9.4 for t < u sDuYs = Zu +

    su

    DuZrdWr

    su

    [

    yf(, r , Y r, Zr) DuYr +

    zf(, r , Y r, Zr) DuZr

    +Duf(, r , Y r, Zr)]dr.By continuity in t of (Ys, Zs) we may choose u = s, to obtain the desiredidentity.

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