GAUSSIAN MEASURES ON 1.1 BOREL MEASURES ON HILBERT SPACES CHAPTER 1

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1 CAPTE GAUSSIAN MEASUES ON ILBET SPACES The aim of this chapter is to show the Minlos-Sazanov theorem and deduce a characterization of Gaussian measures on separable ilbert spaces by its Fourier transform. By using the notion of the ellinger integral we prove the Kakutani theorem on infinite product measures. As a consequence we obtain the Cameron-Martin theorem. For Gaussian measures on Banach spaces and their relationship with parabolic equations with many infinitely variables we refer to [] and [] and the references therein.. BOEL MEASUES ON ILBET SPACES Let be a real separable ilbert space, B the Borel σ-algebra on. Then B is a separable σ-algebra. A measure on the measurable space, B is called a Borel measure on. ere we only investigate finite Borel measures. Definition.. Let µ be a finite Borel measure on. The Fourier transform of µ is defined by µx : e i<x,y> µdy, x. Clearly µ possesses the following properties. Proposition.. The Fourier transform of a finite Borel measure satisfies the following properties

2 Gaussian measures on ilbert spaces µ0 µ. µ is continuous on. 3 µ is positive definite in the sense that l, µx l x k α l α k 0.. for any n, x, x,, x n, and α, α,, α n C. Proof: We have only to prove the third assertion. For n, x, x,..., x n, and α, α,..., α n C we have µx l x k α l α k e i<xl,y> e i<xk,y> α l α k µdy l, l, l, l e i<x l,y> α l e i<x k,y> α k µdy e i<xl, > α l, e i<xk, > α k L,µ e i<xk,y> α k µdy 0. ere L, µ denotes the space of all measurable functions f : satisfying fx µdx <. A natural question arises. Is any positive definite continuous functional on the Fourier transform of some finite Borel measure? The answer is affirmative if dim <. This is exactly the classical Bochner theorem see Theorem A..3. But in the infinite dimensional case the answer is negative. Take, for example, φx : exp x, x. Then it is easy to see that φ is a positive definite functional on. But φ is not the Fourier transform of any finite Borel measure on as we will see later see Proposition... To this end let us prove some auxiliary results. Lemma..3 Let φ be a positive definite functional on. x, y, Then, for any

3 . Borel measures on ilbert spaces 3 φx φ0, φx φ x. φx φy φ0 φ0 φx y. 3 φ0 φx φ0φ0 φx. Proof: For x, y, set φ0 φx A : φ x φ0 B : φ0 φx φy φ x φ0 φy x φ y φx y φ0 Since φ is positive definite, one can see that both A and B are positive definite matrices. In particular A t A. ence, φx φ x for all x. From deta 0, it follows that φx φ0. On the other hand, we have detb φ0 3 φ0 φx y φx[φ0φx φx yφy] + φy[φxφx y φ0φy] φ0 3 φ0 φx y φ0 φx φy + [φyφxφx y φ0]. Using the inequality a 3 ab a a b for b < a, we find Therefore, φ0 3 φ0 φx y φ0 φ0 φx y. 0 detb 4φ0 φ0 φx y φ0 φx φy This proves. Finally 3 follows from φ0 φx φ0 φx φ0 φx φ0 φ0φx + φx φ0 φ0φx. The following lemma will be useful for the proof of the Minlos-Sazanov theorem. Lemma..4 Let µ be a finite Borel measure on. Then the following assertions are equivalent.

4 4 Gaussian measures on ilbert spaces i x µdx <. ii There exists a positive, symmetric, trace class operator Q such that for x, y Qx, y x, z y, z µdz.. If ii holds, then TrQ x µdx. Proof: Suppose that ii holds. Let e n n N be an orthonormal basis of. Then x µdx x, e n µdx Qe n, e n TrQ <..3 n n Conversely, assume that i is satisfied. Thus, x, z y, z µdz x y z µdz. By the iesz representation theorem there exists Q L such that. is satisfied. Obviously, Q is positive and symmetric. Furthermore, by.3, TrQ x µdx <. ence Q is of trace class. Let show now the Minlos-Sazanov theorem. Theorem..5 Let φ be a positive definite functional on a separable real ilbert space. Then the following assertions are equivalent. φ is the Fourier transform of a finite Borel measure on. For every ε > 0 there is a symmetric positive operator of trace class Q ε such that Q ε x, x < φ0 φx < ε. 3 There exists a positive symmetric operator of trace class Q on such that φ is continuous or, equivalently, continuous at x 0 with respect to the semi-norm Q, where x Q : Qx, x Q / x, x. Proof: : Let φ µ. By applying the inequality cos ϑ ϑ, ϑ,

5 . Borel measures on ilbert spaces 5 we obtain, for any γ > 0, φ0 φx cos x, z µdz x, z µdz + µ {z : z > γ}. { z γ} Set µ A : µa { z γ} for A B, and apply Lemma..4 to µ. Thus there is a positive symmetric operator of trace class B γ such that B γ z, z { z γ} z, z z, z µdz. On the other hand, for every ε > 0 there is γ > 0 such that µ{z : z > γ} ε 4. Put Q ε : ε B γ, then φ0 φx ε Q εx, x + ε. : Assume that holds. Then φx is continuous at x 0. So, by Lemma..3-, φ is continuous on. Now, take any orthonormal basis e n n N of and for n put f i,,i n ω,, ω n : φω e + + ω n e n, ω j, j n. Then f i,,i n is a positive definite function on n. By the classical Bochner theorem see Theorem A..3 there exists a finite Borel measure µ i,,i n on n such that f i,,i n µ i,,i n. The family {µ i,,i n } satisfies the consistency conditions of Kolmogorov s extension theorem for measures cf. [30], p. 44. ence there is a unique finite Borel measure γ on, B such that µ i,,i n γ X i,, X in, where γ X i,, X in is defined by γ X i,, X in A γx i,, X in A for A B, and X j ω ω j, ω ω,, ω n,, j N. Claim: X k < γ-a.e.. Let P n be the standard Gaussian measure on n. Then e iay+ +anyn P n dy exp n a j.

6 6 Gaussian measures on ilbert spaces By assumption, we know that for every ε > 0 there is a positive symmetric operator Q ε of trace class such that ence, by Lemma..3-, Q ε x, x < φ0 φx < ε. φ0 φx ε + φ0 Q ε x, x for x. By Fubini s theorem we obtain ence, φ0 exp X k+j γdω φ0 γdω exp i y j X k+j P n dy n φ0 P n dy exp i y j X k+j γdω n φ0 P n dyφ y j e k+j n [φ0 φ y j e k+j ]P n dy n ε + φ0 Q ε y j e k+j, y l e k+l P n dy n l ε + φ0 Q ε e k+j, e l+j y j y l P n dy l, n ε + φ0 Q ε e k+j, e k+j yj P n dy } n {{} ε + φ0 Q ε e k+j, e k+j. φ0 exp X k+j γdω ε + φ0 Q ε e j, e j. jk+

7 . Borel measures on ilbert spaces 7 Now, let k, and ε 0, so we get lim exp X j γdω φ0 γ 0. k + jk+ This means that the function exp jk+ X j converges in L, γ to the constant function. Thus there is a subsequence of exp jk+ converging to γ a.e., which implies that and the claim is proved. Finally, let X j Xj < γ a.e., Xω : X j ωe j, ω. Then X is defined on γ-a.e., and X is an -valued measurable function. Put µ : γ X. Then µ is a finite Borel measure on and since µ i,,i n γ X i,, X in we obtain µ x, e j e j f,,n x, e,, x, e n φ x, e j e j. By letting n we obtain µ φ and the equivalence is proved. 3: Assume that holds. In take ε k for k N and λ k > 0 such that λ ktrq <. Set Q : k λ kq. It is obvious that Q is k a positive symmetric operator of trace class on. Moreover Q satisfies Qx, x < λ k Q x, x < k φ0 φx < k.

8 8 Gaussian measures on ilbert spaces So, by Lemma..3, φ is continuous on with respect to Q and hence 3 is proved. 3 : Conversely, suppose 3 is satisfied. So for every ε > 0 there is δ > 0 such that x Q < δ φ0 φx < ε. Set Q ε : δ Q. Then, Q ε x, x < φ0 φx < ε and Q ε satisfies.. GAUSSIAN MEASUES ON ILBET SPACES We will study a special class of Borel probability measures on. We first introduce the notions of mean vectors and covariance operators for general Borel probability measures. Definition.. Let µ be a Borel probability measure on. If for any x the function z x, z is integrable with respect to µ, and there exists an element m such that m, x x, z µdz, x, then m is called the mean vector of µ. If furthermore there is a positive symmetric linear operator B on such that Bx, y z m, x z m, y µdz, x, y, then B is called the covariance operator of µ. Mean vectors and covariance operators do not necessarily exist in general. But if x µdx <, then by iesz representation theorem, the mean vector m exists, and m x µdx. If furthermore, x µdx <, then by Lemma..4, there is a positive symmetric trace class operator Q such that Qx, y x, z y, z µdz x, y. Set Bx Qx m, x m, x. Then it is easy to verify that B is the covariance operator of µ. Note that B is also a positive symmetric trace class operator. We introduce now Gaussian measures.

9 . Gaussian measures on ilbert spaces 9 Definition.. Let µ be a Borel probability measure on. If for any x the random variable x, has a Gaussian distribution, then µ is called a Gaussian measure. emark..3 The scalar function x, has a Gaussian distribution means that there exists a real number m x and a positive number σ x such that µx e i x,z µdz exp im x σ x, x. In the sequel we will characterize Gaussian measures by means of Fourier transform. Lemma..4 Let α j j N such that α j. Then there exists a sequence of real numbers β j such that α j β j 0 for all j, βj < and α j β j. Proof: Set n 0 0 and define n k inductively as follows Then, n k. Put β j : α j k + n k : inf{l : n k+ jn k + α j l jn k + Then, for all j, α j β j 0, and βj n k+ α j }, k., n k + j n k+, k 0,,.... βj k0 jn k + k0 k + <, α j β j n k+ k0 jn k + k0 k0 k + α j β j k+ jn k + k +. α j The following result gives a characterization of Gaussian measures on separable ilbert spaces.

10 0 Gaussian measures on ilbert spaces Theorem..5 A Borel probability measure µ on is a Gaussian measure if and only if its Fourier transform is given by µx exp i < m, x > < Bx, x >, where m, B is a positive symmetric trace class operator on. In this case, m and B are the mean vector and covariance operator of µ respectively. Moreover, x µdx TrB + m. Proof: Let µ be a Gaussian measure on. Claim: x µdx <. By assumption, for any x, x, has a Gaussian distribution. Thus there are m x, and σ x > 0 such that µx e i<x,z> µdz exp im x σ x..4 Let e j be an orthonormal basis of. Since ξ m N m, σ dξ σ and ξn m, σ dξ m, we have x µdx x, e j µdx x jµdx j σe j + m e j. Let β j such that β j m ej 0 and β j <. Set ξx : β j e j, x By Schwarz inequality, the above series converges absolutely and ξx βj x, x. Moreover, ξ is linear. So by iesz representation theorem there is z such that ξx z, x, x. By assumption ξ z, is a Gaussian variable with a finite mean, i.e., β jm ej <. Now, by Lemma..4,

11 . Gaussian measures on ilbert spaces m e j <. Thus, in order to prove x µdx <, it suffices to check σ j <. By Theorem..5, there is a positive, symmetric, trace class operator Q such that Qx, x < µx < 6. ence, exp σ x µx Qx, x +, x..5 6 Without loss of generality we may assume that the kernel of Q is {0}. For x \ {0}, set y : 3<Qx,x> x. Then σ y 3 Qx, x σ x, and Qy, y 3. eplacing x by y in.5, we obtain exp σ x 6 Qx, x This implies that Thus, σ x 6 log 6 Qx, x, x. σe j 6 log 6TrQ <. ence, x µdx < and the claim is proved. So by the remark following Definition.. the mean vector m and the covariance operator B of µ exist. The above notation gives m x x, z µdz m, x and σ x From.4 we obtain µx exp x, z µdz m x [ x, z m, x ]µdz x, z m µdz Bx, x. i m, x Bx, x, x.

12 Gaussian measures on ilbert spaces Moreover, x µdx σe j + m e j TrB + m which proves the first implication. Conversely, let m and B be a positive, symmetric, trace class operator, and consider the positive definite functional φx exp i m, x Bx, x, x. Set Qx : Bx + m, x m, x. Then Q is a positive, symmetric, trace class operator on. Define Q on as follows x Q Q / x Qx, x / Bx, x + m, x /. Then φx is continuous at x 0 with respect to Q. So by Theorem..5, φ is the Fourier transform of some Borel probability measure µ on. Clearly for any x, x, is a Gaussian random variable with mean m, x and covariance Bx, x under µ. Thus, µ is a Gaussian measure. A Gaussian measure with mean vector m and covariance operator B will be denoted by N m, B. We propose now to compute some Gaussian integrals. Proposition..6 Let N 0, B be a Gaussian measure on. Then there is an orthonormal basis e n of such that Be n λ n e n, λ n 0, n N. Moreover, for any α < α 0 : inf n λ n, we have e α x N 0, Bdx αλ k deti αb. Proof: The first assertion follows from the fact that B is symmetric and positive. Since TrB λ k <, it follows that 0 αλ k < for α < α0. Furthermore, e α <x,e> N 0, Bdx e α ξ N 0, λ dξ πλ αλ. e α ξ e ξ λ dξ

13 . Gaussian measures on ilbert spaces 3 In similar way we have e α P n n <x,e k> N 0, Bdx αλ k and the result follows from the monotone convergence theorem. Before proving a more general result we propose first to study the transformation of a Gaussian measure by an affine mapping. Lemma..7 Let and be two separable ilbert spaces. Consider the affine transformation F : defined by F x Qx + z, where Q L, and z. If we set µ F : N m, B F, the measure defined by µ F A N m, BF A, A B, then µ F N Qm + z, QBQ. Proof: obtain Let compute the Fourier transform of µ F. From Theorem..5 we µ F x e i x,ey µ F dỹ e e i x,qy+z µdy e i x,z e i Q x,y µdy e i x,qm+z e QBQ x,x N Qm + z, QBQ x for x. So the lemma follows from Theorem..5. From the above lemma follows the following result. Proposition..8 Let α 0 : inf k λ k. Then, for any α < α 0, e α x N m, Bdx deti αb α exp I αb m, m.

14 4 Gaussian measures on ilbert spaces Proof: From Lemma..7 we have e α x N m, Bdx e α x+m N 0, Bdx e α m e α m πλk πλk e α m e πλk e α λ k α m k e πλk e α ξ +αm k ξ e ξ λ k dξ h i e αλk λ ξ αm k ξ k dξ λ k α m k αλ k m k αλ k αλ k αm k e αλ k deti αb e α I αb m,m. e αλ k λ ξ λ k αm k k αλ k dξ λk e ξ dξ αλ k Example..9 Let compute the integrals a x m N 0, Bdx, m N, b My N 0, Bdy, where M L. a For the integral in a we consider the function fα : e α x N 0, Bdx deti αb for α < α 0. Now, it is easy to see that, α 0 α deti αb is C and d dα deti αb TrBI αb deti αb, α < α 0. Furthermore we can differentiate under the integral sign. ence, x m N 0, Bdx m dm dα m deti αb α0.

15 . Gaussian measures on ilbert spaces 5 This implies that and b It follows from Lemma..7 that My N 0, Bdy x N 0, Bdx TrB x 4 N 0, Bdx TrB + TrB. y N 0, MBM dy. So by Theorem..5 we have My N 0, Bdy TrMBM TrM MB..6 By a same computation as above one has Proposition..0 For any α, m, we have e α,x N m, Bdx e α,m e Bα,α. We end this section by proving that the positive definite functional on defined by ϕx e x, x, is not the Fourier transform of any Borel measures provided that dim. Proposition.. Let Q be a positive, symmetric operator on a separable ilbert space. Then the functional φx exp < Qx, x >, x, is the Fourier transform of a probability measure on if and only if TrQ <. Proof: Suppose that TrQ <. Then φ0 and φ is Q -continuous positive functional on. So by Theorem..5 there exists a probability measure µ such that µx φx, x. To show the converse, assume that there is a probability measure µ such that e i<x,y> µdy exp < Qx, x >. Then by Theorem..5, for any ε 0, 3, there exists a positive, symmetric operator Q ε of trace class such that < Q ε x, x > < φ0 eφx < ε < Qx, x > < 3ε.

16 6 Gaussian measures on ilbert spaces Let now y 0 and < Q ε y 0, y 0 >: c, with c > 0. Let d > c arbitrary. Then y < Q 0 ε d, y0 c d > d <. ence, < Q y0 d, y0 d > < ε, i.e. < Qy 0, y 0 > < εd. Letting d c, we have < Qy 0, y 0 > ε < Q ε y 0, y 0 >. Since y 0 is arbitrary, we obtain < Qy, y > ε < Q ε y, y > for all y. In particular, for an orthonormal basis e n n N of, we obtain TrQ n < Qe n, e n > ε n < Q ε e n, e n > εtrq ε <. As an immediate consequence we obtain that the functional φx exp x, x, is not the Fourier transform of any probability measure on if dim..3 TE ELLINGE INTEGAL AND TE CAMEON-MATIN TEOEM The Cameron-Martin formula permits us to differentiate under the integral sign with respect to Gaussian measures in infinite dimensional ilbert spaces. This allows us to obtain some regularity results for parabolic equations with many infinitely variables. First we need some preparations. We denote by L + the space of all positive, symmetric operators of trace class on a separable ilbert space. Let B L + and consider an orthonormal basis e n n N of and a sequence λ n n N + such that Be n λ n e n, n N. Suppose also that ker B {0}. If we denote by x n :< x, e n >, then We set also Bx λ n x n e n and B x λ xn e n, x. n B n x : n λ k x k e k and B n x : Let consider, for a and n N, the function g a,n x : a, B n x λ k x ka k. λ k x ke k.

17 .3 The ellinger integral and the Cameron-Martin theorem 7 If a B then one can define the function g a x : λ k x ka k, x. The following proposition shows that it is always possible to define g a as an L, µ-function even if a B. Proposition.3. Let B L + with ker B {0} and µ : N 0, B its corresponding Gaussian measure on. Then the limit exists in L, µ. Moreover, for a given a. lim n + g a,n : g a g a x µdx a Proof: We have g a,n+p x g a,n x µdx n+p n+p kn+ h,kn+ n+p kn+ n+p kn+ λ k x ka k µdx λ h λ k ah a k λ k a k a k. x kµdx x h x k µdx ence g a,n n N is a Cauchy sequence in L, µ. Moreover, g a,n µdx a k λ k x kµdx a k and the theorem is proved by letting n. emark.3. Suppose that ker B {0} and take x such that B a, x 0 for all a. ence, B x 0 and so Bx 0, which implies that x 0. This proves that B is dense in. For the converse, let x with Bx 0. Thus, B x 0 and hence, B x, y x, B y 0 for

18 8 Gaussian measures on ilbert spaces all y. Since B, it follows that x 0. By the same arguments as in the proof of Proposition.3. one can show that g a is well defined as an L, µ function and g a L,µ a for a B. In the sequel we will use the notation g a x : a, B x, x. Proposition.3.3 Let B L + with ker B {0} and µ : N 0, B its corresponding Gaussian measure on. Then the limit lim n ega,n : e ga exists in L, µ for a given a. Moreover, for any a, e a,b x N 0, Bdx e a. Proof: By applying Proposition..0 we obtain e ga,n e ga,m µdx e B n a,x e B n a,x + B m a,x + e B m a,x µdx e P n a k + e P m a k e B n +B m a,x µdx e P n a k + e P m a k e P n a k + P m kn+ a k e P n a k + e P m kn+ a k e P m kn+ a k 0 n, m. This proves that e ga,n is a Cauchy sequence in L, µ and one can see that e a,b x N 0, Bdx e a is satisfied for every a. We propose now to recall the definition of the ellinger integral. Let µ, ν be two probability measures on a measurable space Ω, Σ. We say that µ and ν are singular notation: µ ν if there is a set B Σ such that µb 0 and νω \ B 0. It is easy to see that two probability measures µ and ν are singular if and only if for any ε > 0 there is B Σ such that µb < ε and νω \ B < ε. Further, µ is called ν-absolutely continuous notation: µ ν if νb 0 implies µb 0 for any B Σ. So by the theorem of adon-nikodym we know that if µ is ν-absolutely continuous,

19 .3 The ellinger integral and the Cameron-Martin theorem 9 then there is a non-negative measurable function ϕ defined on Ω, called the density function of µ, such that µb ϕωνdω for any B Σ. The density ϕ is denoted by B ϕω : dµ ω, ω Ω. dν If µ ν and ν µ are satisfied then µ and ν are called equivalent notation: µ ν. If µ ν, then the two density functions ϕ dµ dν dν and ψ dµ satisfy ϕωψω, a.e. ω Ω. ence, ϕω > 0 µ-a.e. ω Ω. Let now µ and ν two arbitrary probability measures on Ω, Σ. Let γ be a probability measure on Ω, Σ such that µ γ and ν γ. Such a measure exists, we have to take for example γ µ + ν. Thus, the following integral is well-defined µ, ν : Ω dµ dν ω ω γdω. dγ dγ This integral will be called the ellinger integral. Let now consider the measurable space, B, where B is the Borel field of subsets B of. On, B we consider two sequences of measures µ n and ν n with Then one has µ n, ν n Let us consider two infinite product measures µ n ν n, n N..7 n dν n dµ n x n µ n dx n. µ : µ n and ν : defined on, B. The following result is du to S. Kakutani [] and gives a condition under which these two measures µ and ν are equivalent. Theorem.3.4 Assume that.7 is satisfied. Then the following assertions hold. i If n µ n, ν n > 0 then µ ν and n ν n dν dµ x dν k x, a.e. x. dµ k

20 0 Gaussian measures on ilbert spaces ii If n µ n, ν n 0 then µ ν. Moreover, µ, ν µ n, ν n..8 n Proof: If we set ψ n x : n dν k dµ k x k for x and n N, then ψ n L,µ n dν k dµ k x k µdx n ν k dx k and ψ n ψ m L,µ n n dν k dµ k x k dν k dµ k x k m kn+ m kn+ dν k µ k, ν k m dν k x k µdx dµ k m dν k x k µdx dµ k kn+ x k µ k dx k dµ k.9 for any positive integers n and m with n < m. i If n µ n, ν n > 0 then lim m n,m kn+ µ k, ν k. ence, by.9, ψ n is a Cauchy sequence in L, µ and so there is ψ L, µ such that lim n ψ n ψ L,µ 0. Let prove now that ν µ and dν dµ x ψx, x, i.e. νb ψx µdx B

21 .3 The ellinger integral and the Cameron-Martin theorem for any B B. To this purpose it follows from ölder s inequality and.9 that ψ m x ψ n x µdx ψ m x + ψ n x µdx ψ m x ψ n x µdx 4 ψ m x ψ n x µdx m 8 µ k, ν k for n < m. Thus, kn+ lim n ψ n ψ L,µ 0. Finally let B B and set χ n x : χ B P n x, x, where χ B denotes the characteristic function of the measurable set B and P n x : x,..., x n, 0,.... So we have χ n x νdx χ B x,..., x n, 0,... ν dx... ν n dx n n n χ n x n dν k dµ k x k χ n xψ n x µdx. n µ k dx k Since ψn ψ in L, µ and by letting n we obtain νb ψx µdx. In a similar way one can see that µ ν. So we obtain µ ν. Finally, since µ ν, we have µ, ν ψx µdx lim ψ n x µdx n n dν k lim x k µ k dx k n dµ k lim n n µ k, ν k.

22 Gaussian measures on ilbert spaces So we obtain.8. ii If µ k, ν k 0 then for any ε > 0 there is n N such that n µ k, ν k < ε. Let B n B n with B n : {x,..., x n n : ψ n x,..., x n, 0,... Then, n µ k B n < B n n µ k dx n dν k dµ k x k > }. n ψ n x,..., x n, 0,... µ k dx B n n n dν k x k µ k dx B n dµ k n µ k, ν k < ε. By the same computation we obtain n ν k n \ B n n µ k, ν k < ε. Therefore, if we set B : B n kn+, then µb < ε and ν \ B < ε. This proves that µ ν. Suppose now that µ ν. Then there exists B B such that µb 0 and ν \ B 0. So by ölder s inequality, it follows that µ, ν B B dµ dγ dν x x γdx + dγ \B dµ x γdx dγ \B B dµ x γdx dγ dµ dγ dν x γdx dγ \B dν x x γdx dγ + dν x γdx dγ µb νb + µ \ B ν \ B 0. Therefore,.8 holds. This end the proof of the theorem. Let prove now the Cameron-Martin formula. We note here that the measure space, B can be identified with, B.

23 .3 The ellinger integral and the Cameron-Martin theorem 3 Corollary.3.5 Let B L + such that ker B {0} and µ : N 0, B and ν : N m, B be two Gaussian measures on, B. Then the following assertions hold. i The Gaussian measures µ and ν are equivalent if and only if m B. Moreover the adon-nikodym derivative is given by dν x exp dµ B m + < B x, B m >. ii The measures µ and ν are singular if and only if m B. Proof: We will apply Theorem.3.4 to the Gaussian measures µ and ν. To this purpose let compute the associated ellinger integral using.8. It follows from Proposition..0 that dν k µ k, ν k x k µ k dx k dµ k So by.8 we obtain This implies that k 4λ k e m e m k 8λ k. µ, ν µ, ν > 0 e m k x k λ k N 0, λ k dx k e m k 8λ k. m k λ k < m B. Moreover, in this case, it follows from Theorem.3.4 that dν dµ x dν k x dµ k exp e m k λ k e x k m k λ k B m + B x, B m, where x x ke k with x k : x, e k for an orthonormal basis e n of such that Be n λ n e n for n N. ere we used Proposition.3.3. Finally it is clear that the measures µ and ν are singular if and only if m B.

24 4 Gaussian measures on ilbert spaces Exercise.3.6 The Feldman-ajek theorem Let consider two linear operators B, B L + with ker B ker B {0} and an orthonormal basis e n of such that B e n λ n e n, n N, where λ n > 0 for all n N. On, B we consider the Gaussian measures µ : N 0, B and µ : N 0, B.. The commutative case: Suppose that B B B B. By using Theorem.3.4 show that a. if λ n α n n λ n+α n <, then µ µ. In this case dµ dµ x n b. if λ n α n n λ n+α n, then µ µ. exp λ n α n x, e n, λ n α n ere α n > 0, n N, are such that B e n α n e n, n N.. The General case: a Assume that there is S L + such that Show that µ µ. B B Id SB. b Assume that S L + and S <. Show that dµ x [deti S] exp dµ SI S B x, B x, x. ere L + is the set of positive ilbert-schmidt bounded linear operators on. That is, B L + if and only if B L, B positive and n Be n <.