Lecture 2

Gaussian measures in infinite dimen- sion

In this lecture, after recalling a few notions about -algebras in Banach spaces, we intro- duce Gaussian measures in separable Banach spaces and we prove the Fernique Theorem. This is a powerful tool to get precise estimates on the Gaussian integrals. In most of the course, our framework is an infinite dimensional separable real Banach space that we denote by X, with norm or when there is risk of confusion. The open (resp. k · k k · kX closed) ball with centre x X and radius r>0 will be denoted by B(x, r)(resp. B(x, r)). 2 We denote by X , with norm , the topological dual of X consisting of all linear ⇤ k · kX⇤ continuous functions f : X R. Sometimes, we shall discuss the case of a separable ! Hilbert space in order to highlight some special features. The only example that is not a Banach space is RI (see Lecture 3), but we prefer to describe this as a particular case rather than to present a more general abstract theory.

2.1 -algebras in infinite dimensional spaces and characteristic functions

In order to present further properties of measures in X, and in particular approximation of measures and functions, it is useful to start with a discussion on the relevant underly- ing -algebras. Besides the Borel -algebra, to take advantage of the finite dimensional reductions we shall frequently encounter in the sequel, we introduce the -algebra E (X) generated by the cylindrical sets, i.e, the sets of the form

C = x X :(f (x),...,f (x)) C , 2 1 n 2 0 n o n where f1,...,fn X⇤ and C0 B(R ), called a base of C. According to Definition 2 2 1.1.8, E (X)=E (X, X⇤). The following important result holds. We do not present its most general version, but we do not even confine ourselves to Banach spaces because

13 14 we shall apply it to R1, which is a Fr´echet space. We recall that a Fr´echet space is a complete metrisable locally convex topological vector space, i.e., a vector space endowed with a sequence of seminorms that generate a metrisable topology such that the space is complete.

Theorem 2.1.1. If X is a separable Fr´echet space, then E (X)=B(X). Moreover, there is a countable family F X separating the points in X (i.e., such that for every pair of ⇢ ⇤ points x = y X there is f F such that f(x) = f(y)) such that E (X)=E (X, F). 6 2 2 6

Proof. Let (xn)beasequencedenseinX, and denote by (pk) a family of seminorms which defines the topology of X. By the Hahn-Banach theorem for every n and k there is ` X such that p (x )=` (x ) and sup ` (x): p (x) 1 = 1. As a n,k 2 ⇤ k n n,k n { n,k k  } consequence, for every x X and k N we have pk(x)=sup `n,k(x) . Therefore, for 2 2 n{ } every r>0

B (x, r):= y X : p (y x) r = y X : ` (y x) r E (X). k { 2 k  } { 2 n,k  } 2 n N \2 As X is separable, there is a countable base of its topology. For instance,we may take the sets Bk(xn,r) with rational r, see [DS1, I.6.2 and I.6.12]. These sets are in E (X), hence the inclusion B(X) E (X) holds. The converse inclusion is trivial. ⇢ To prove the last statement, just take F = `n,k,n,k N . It is obviously a countable { 2 } family; let us show that it separates points. If x = y,thereisk N such that pk(x y)= 6 2 sup `n,k(x y) > 0 and therefore there isn ¯ N such that `n,k¯ (x y) > 0. n 2 In the discussion of the properties of Gaussian measures in Banach spaces, as in Rd, the characteristic functions, defined by

µˆ(f):= exp if(x) µ(dx),fX⇤, { } 2 ZX play an important role. The properties of characteristic functions seen in Lecture 1 can be extended to the present context. We discuss in detail only the extension of property (iii) (the injectivity), which is the most important for our purposes. In the following proposi- tion, we use the coincidence criterion for measures agreeing on a system of generators of the -algebra, see e.g. [D, Theorem 3.1.10].

Proposition 2.1.2. Let µ1,µ2 be two probability measures on (X, B(X)).Ifµ1 = µ2 then µ1 = µ2. b b Proof. It is enough to show that ifµ ˆ =0thenµ = 0 and in particular, by Theorem 2.1.1, d that µ(C)=0whenC is a cylinder with base C0 B(R ). Let beµ ˆ = 0, consider 21 d F = span f1,...,fd X⇤ and define µF = µ PF ,wherePF : X R is given by { } ⇢ d ! PF (x)=(f1(x),...,fd(x)). Then for any ⇠ R 2

µ (⇠)= exp i⇠ y µ (dy)= exp i⇠ P (x) µ(dx)= exp iP ⇤ ⇠(x) µ(dx)=0, F { · } F { · F } { F } ZF ZX ZX b 15

d where P ⇤ : R X⇤ is the adjoint map F ! d

PF⇤ (⇠)= ⇠ifi. Xi=1

It follows that µF = 0 and therefore the restriction of µ to the -algebra E (X, F)isthe null measure.

2.2 Gaussian measures in infinite dimensional spaces

Measure theory in infinite dimensional spaces is far from being a trivial issue, because there is no equivalent of the Lebesgue measure, i.e., there is no nontrivial measure invariant by translations.

Proposition 2.2.1. Let X be a separable Hilbert space. If µ : B(X) [0, + ] is a ! 1 -additive set function such that:

(i) µ(x + B)=µ(B) for every x X, B B(X), 2 2 (ii) µ(B(0,r)) > 0 for every r>0, then µ(A)=+ for every open set A. 1 Proof. Assume that µ satisfies (i) and (ii), and let e be an orthonormal basis in X. { n} For any n N consider the balls Bn with centre 2ren and radius r>0; they are pairwise 2 disjoint and by assumption they have the same measure, say µ(Bn)=m>0 for all n N. 2 Then, 1 1 1 B B(0, 3r)= µ(B(0, 3r)) µ(B )= m =+ , n ⇢ ) n 1 n=1 n=1 n=1 [ X X hence µ(A)=+ for every open set A. 1 Definition 2.2.2 (Gaussian measures on X). Let X be a Banach space. A probability 1 measure on (X, B(X)) is said to be Gaussian if f is a Gaussian measure in R for every f X .Themeasure is called centred (or symmetric) if all the measures 2 ⇤ f 1 are centred and it is called nondegenerate if for any f =0the measure f 1 is 6 nondegenerate.

Our first task is to give characterisations of Gaussian measures in infinite dimensions in terms of characteristic functions analogous to those seen in Rd, Proposition 1.2.4. Notice that if f X then f Lp(X, ) for every p 1: indeed, the integral 2 ⇤ 2

p p 1 f(x) (dx)= t ( f )(dt) | | | | ZX ZR 1 is finite because f is Gaussian in R. Therefore, we can give the following definition. 16

Definition 2.2.3. We define the mean a and the covariance B of by

a (f):= f(x) (dx),fX⇤, (2.2.1) 2 ZX B (f,g):= [f(x) a (f)] [g(x) a (g)] (dx),f,gX⇤. (2.2.2) 2 ZX Observe that f a (f) is linear and (f,g) B (f,g) is bilinear in X . Moreover, 7! 7! ⇤ B (f,f)= f a (f) 2 0 for every f X . k kL2(X,) 2 ⇤ Theorem 2.2.4. A Borel probability measure on X is Gaussian if and only if its char- acteristic function is given by

1 ˆ(f)=exp ia(f) B(f,f) ,fX⇤, (2.2.3) 2 2 n o where a is a linear functional on X⇤ and B is a nonnegative symmetric bilinear form on X⇤.

Proof. Assume that is Gaussian. Let us show that ˆ is given by (2.2.3) with a = a and B = B.Indeed,wehave:

1 1 2 (f)= exp i⇠ ( f )(d⇠)=exp im , R { } 2 Z n o where m and 2 areb the mean and the covariance of f 1, given by

1 m = ⇠( f )(d⇠)= f(x)(dx)=a (f), ZR ZX and 2 2 1 2 = (⇠ m) ( f )(d⇠)= (f(x) a (f)) (dx)=B (f,f). ZR ZX Conversely, let be a Borel probability measure on X and assume that (2.2.3) holds. Since a is linear and B is bilinear, we can compute the Fourier transform of f 1, for f X , as follows: 2 ⇤

1 1 \f (⌧)= exp i⌧t ( f )(dt)= exp i⌧f(x) (dx) { } { } ZR ZX 1 =exp i⌧a(f) ⌧ 2B(f,f) . 2 n o According to Remark 1.2.2, f 1 = N (a(f),B(f,f)) is Gaussian and we are done. Remark 2.2.5. We point out that at the moment we have proved that a (X ) 2 ⇤ 0 and B (X X ) , the algebraic duals consisting of linear functions (not necessarily 2 ⇤ ⇥ ⇤ 0 continuous). We shall see that a and B are in fact continuous. 17

As in the finite dimensional case, we say that is centred if a = 0; in this case, the 2 bilinear form B is nothing but the restriction of the inner product in L (X, )toX⇤,

2 B (f,g)= f(x)g(x) (dx),B(f,f)= f 2 . (2.2.4) k kL (X,) ZX In the sequel we shall frequently consider centred Gaussian measures; this requirement is equivalent to the following symmetry property.

Proposition 2.2.6. Let be a Gaussian measure on a Banach space X and define the measure µ by µ(B):=( B), B B(X). 8 2 Then, is centred if and only if = µ.

Proof. We know that (f)=exp ia (f) 1 f a (f) 2 . On the other hand, since { 2 k kL2(X,)} µ = R 1 with R : X X given by R(x)= x,then ! b if(x) if(x) 1 2 µ(f)= e µ(dx)= e (dx)=exp ia (f) f a (f) 2 . { 2k kL (X,)} ZX ZX Thenb µ = if and only if a (f) = 0 for any f X , whence the statement follows by 2 ⇤ Proposition 2.1.2. b b Let us draw some interesting (and useful) consequences from the above result.

Proposition 2.2.7. Let be a centred Gaussian measure on X.

(i) If µ is a centred Gaussian measure on a separable Banach space Y , then µ is a ⌦ centred Gaussian measure on X Y . ⇥ (ii) If µ is another centred Gaussian measure on X, then the convolution measure µ, ⇤ defined as the image measure in X of µ on X X under the map (x, y) x + y ⌦ ⇥ 7! is a centred Gaussian measure and is given by

µ(B)= µ(B x)(dx)= (B x)µ(dx). (2.2.5) ⇤ ZX ZX

1 (iii) For every ✓ R the image measure ( ) R in X X under the map R✓ : 2 ⌦ ✓ ⇥ X X X X, R (x, y):=(x cos ✓ + y sin ✓, x sin ✓ + y cos ✓) is again . ⇥ ! ⇥ ✓ ⌦ 1 (iv) For every ✓ R the image measures ( ) , i =1, 2 in X under the maps 2 ⌦ i : X X X, i ⇥ ! (x, y):=x cos ✓ + y sin ✓, (x, y):= x sin ✓ + y cos ✓ 1 2 are again . 18

Proof. All these results follow quite easily by computing the relevant characteristic func- tions. We start with (i) taking into account that for f X and g Y we have 2 ⇤ 2 ⇤ 2 2 B (f,f)= f 2 ,B(g, g)= g 2 . k kL (X,) µ k kL (Y,µ) For every ` (X Y ) ,wedefinef and g by 2 ⇥ ⇤ `(x, y)=`(x, 0) + `(0,y)=:f(x)+g(y).

Then

\µ(`)= exp i`(x, y) ( µ)(d(x, y)) ⌦ X Y { } ⌦ Z ⇥ = exp if(x) (dx) exp ig(y) µ(dy) { } { } ZX ZY 1 2 1 2 =exp f 2 g 2 2k kL (X,) 2k kL (Y,µ) ⇢ 1 2 2 =exp f 2 + g 2 . 2 k kL (X,) k kL (Y,µ) ⇢ ⇣ ⌘ On the other hand, since and µ are centred

f(x)g(y)( µ)(d(x, y)) = f(x)(dx) g(y)µ(dy)=0, X Y ⌦ X Y Z ⇥ Z Z and therefore

2 2 2 2 f 2 + g 2 = f(x) (dx)+ g(y) µ(dy) k kL (X,) k kL (Y,µ) ZX ZY = `(x, 0)2(dx)+ `(0,y)2µ(dy) ZX ZY = `(x, 0)2( µ)(d(x, y)) + `(0,y)2( µ)(d(x, y)) X Y ⌦ X Y ⌦ Z ⇥ Z ⇥ = (`(x, 0)2 + `(0,y)2)( µ)(d(x, y)) X Y ⌦ Z ⇥ = (`(x, 0) + `(0,y))2( µ)(d(x, y)) X Y ⌦ Z ⇥ = `(x, y)2( µ)(d(x, y)). X Y ⌦ Z ⇥ So we have

B µ(`, `)=B(f,f)+Bµ(g, g) ⌦ if we decompose ` by `(x, y)=f(x)+g(y), f(x)=`(x, 0), g(y)=`(0,y) as before. 19

The proof of statement (ii) is similar; indeed if h : X X X is given by h(x, y)= ⇥ ! x + y,then

1 1 ( \µ) h (`)= exp i`(x) ( µ) h (dx) ⌦ { } ⌦ ZX = exp i`(h(x, y)) ( µ)(d(x, y)) X Y { } ⌦ Z ⇥ = exp i`(x) (dx) exp i`(y) µ(dy) { } { } ZX ZY 1 2 2 =exp ` 2 + ` 2 . 2 k kL (X,) k kL (X,µ) ⇢ for every ` X . Using the notation of Remark 1.1.14, for every B B(X)wehave 2 ⇤ 2 (x, y) X X : h(x, y) B = B x, { 2 ⇥ 2 }x (x, y) X X : h(x, y) B y = B y. { 2 ⇥ 2 } 1 Applying the Fubini Theorem to the characteristic functions of h (B) we deduce that the convolution measure is given by (2.2.5). 1 To show (iii), set µ := ( ) R ; taking into account that for any ` (X X) ⌦ ✓ 2 ⇥ ⇤ we have

`(R (x, y)) =`(x cos ✓ + y sin ✓, x sin ✓ + y cos ✓) ✓ = `(x, 0) cos ✓ `(0,x)sin✓ + `(y, 0) sin ✓ + `(0,y) cos ✓ =:f✓(x) =:g✓(y) we find | {z } | {z }

µˆ(`)= exp i`(R✓(x, y)) ( )(d(x, y)) X X { } ⌦ Z ⇥ = exp if (x) (dx) exp ig (y) (dy) { ✓ } { ✓ } ZX ZX 1 =exp (B (f ,f )+B (g ,g )) . 2 ✓ ✓ ✓ ✓ ⇢ since B is bilinear,

B(f✓,f✓)+B(g✓,g✓)=B(f0,f0)+B(g0,g0), where f0(x)=`(x, 0), g0(y)=`(0,y) and

B(f0,f0)+B(g0,g0)=B (`, `) ⌦ by the proof of statement (i). To prove (iv), we notice that the laws of under the projection maps p ,p : ⌦ 1 2 X X X, ⇥ ! p1(x, y):=x, p2(x, y):=y 20 are by definition of product measure. Since = p R , i =1, 2 we deduce i i ✓ 1 1 1 1 1 ( ) =( ) (p R ) =(( ) R ) p =( ) p = . ⌦ i ⌦ i ✓ ⌦ ✓ i ⌦ i

2.3 The Fernique Theorem

In this section we prove the Fernique Theorem; we start by proving it in the case of centred Gaussian measure and then we extend the result to any Gaussian measure.

Theorem 2.3.1 (Fernique). Let be a centred Gaussian measure on a separable Banach space X. Then there exists ↵ > 0 such that

exp ↵ x 2 (dx) < . { k k } 1 ZX Proof. If is a Dirac measure the result is trivial, therefore we may assume that this is not the case. The idea of the proof is to show that the measures of suitable annuli decay fast enough to compensate the growth of the exponential function under the integral. Let us fix t>⌧ > 0 and let us estimate ( x ⌧ )( x >t ). Using property (iii) of {k k } {k k } Proposition 2.2.7 with ✓ = ⇡ , we obtain 4 ( x X : x ⌧ )( x X : x >t ) { 2 k k } { 2 k k } =( )( (x, y) X X : x ⌧ (x, y) X X : y >t ) ⌦ { 2 ⇥ k k } ⇥ { 2 ⇥ k k } =( )( x ⌧ y >t ) ⌦ {k k } \ {k k } x y x + y =( ) k k ⌧ k k >t . ⌦ p2  \ p2 ✓n o n o◆ x+y x y The triangle inequality yields x , y k k k k , which implies the inclusion k k k k 2 2 x y x + y t ⌧ t ⌧ k k ⌧ k k >t x > y > . p2  \ p2 ⇢ k k p2 \ k k p2 n o n o n o n o As a consequence, we have the estimate

t ⌧ 2 ( x ⌧ )( x >t ) X B 0, . (2.3.1) {k k } {k k }  \ p2 ✓ ⇣ ⌘◆ We leave as an exercise, Exercise 2.1, the fact that if is not a Dirac measure, then (B(0, ⌧)) < 1 for any ⌧ > 0. Let us fix ⌧ > 0 such that c := B(0, ⌧) (1/2, 1) and set 2 1 c ↵ := log , 24⌧ 2 1 c ⇣ ⌘ n+1 t0 := ⌧,tn := ⌧ + p2tn 1 = ⌧(1 + p2) p2 1 ,n1. 21

tn ⌧ Applying estimate (2.3.1) with t = tn and recalling that = tn 1, we obtain p2 2 (X B¯(0,tn 1)) (X B¯(0,tn 1)) 2 (X B¯(0,t )) \ = \ c \ n  (B¯(0, ⌧)) c ⇣ ⌘ and iterating n 1 c 2 (X B¯(0,t )) c . \ n  c ✓ ◆ Therefore

exp ↵ x 2 (dx)= exp ↵ x 2 (dx)+ { k k } { k k } ZX ZB(0,⌧) 1 + exp ↵ x 2 (dx) { k k } n=0 B(0,tn+1) B(0,tn) X Z \ 1 c exp ↵⌧ 2 + exp ↵t2 (X B(0,t )).  { } { n+1} \ n n=0 X n+2 Since (p2 1)2 2n+2 for every n N,  2 1 1 c 2n exp ↵ x 2 (dx) c exp ↵⌧ 2 + exp 4↵⌧ 2(1 + p2)22n { k k }  { } { } c X n=0 Z ⇣ X ⇣ ⌘ ⌘ 1 1 c = c exp ↵⌧ 2 + exp 2n log +4↵⌧ 2(1 + p2)2 { } c n=0 ⇣ X n ⇣ ⌘o⌘ 1 1 p2 1 c = c exp ↵⌧ 2 + exp 2n log . { } 2 3 c n=0 ⇣ X n ⇣ ⌘ ⇣ ⌘o⌘ 1 c The last series is convergent because c>1/2 and hence log c < 0. The validity of the Fernique Theorem can be extended to any Gaussian measure, not necessarily centred. Corollary 2.3.2. Let be a Gaussian measure on a separable Banach space X. Then there exists ↵ > 0 such that

exp ↵ x 2 (dx) < + . { k k } 1 ZX Proof. Let us set µ(B)=( B) for any B B(X). According to (2.2.5), the measure 2 = µ is given by ( µ) h 1, h(x, y)=x + y, and is a centred Gaussian measure. 1 ⇤ ⌦ Therefore, there exists ↵1 > 0 such that

2 2 + > exp ↵1 x 1(dx)= exp ↵1 x + y ( µ)(d(x, y)) 1 X { k k } X X { k k } ⌦ Z Z ⇥ 2 2 = exp ↵1 x y ( )(d(x, y)) = (dy) exp ↵1 x y (dx). X X { k k } ⌦ X X { k k } Z ⇥ Z Z 22

Then, for –a.e. y X, 2 exp ↵ x y 2 (dx) < + . { 1k k } 1 ZX Using the inequality 2ab "a2 + 1 b2, which holds for any a, b and " > 0, we get  " 1 x 2 x y 2 + y 2 +2 x y y (1 + ") x y 2 + 1+ y 2. k k k k k k k kk k k k " k k ⇣ ⌘ For any ↵ (0, ↵ ), by setting " = ↵1 1, we obtain 2 1 ↵ ↵↵ exp ↵ x 2 (dx) exp 1 y 2 exp ↵ x y 2 (dx). { k k }  ↵ ↵k k { 1k k } ZX ⇢ 1 ZX Then for any ↵ < ↵1, exp ↵ x 2 (dx) < + . { k k } 1 ZX

As a first application of the Fernique theorem, we notice that for every 1 p< we  1 have x p (dx) < + (2.3.2) k k 1 ZX since x p c exp ↵ x 2 for all x X and for some constant c depending on ↵ and k k  ↵,p { k k } 2 ↵,p p only. We already know, through the definition of Gaussian measure, that the functions f X belong to all Lp(X, ) spaces, for 1 p< . The Fernique Theorem tells us much 2 ⇤  1 more, since it gives a rather precise description of the allowed growth of the functions in p L (X, ). Moreover, estimate (2.3.2) has important consequences on the functions a and B. Proposition 2.3.3. If is a Gaussian measure on a separable Banach space X, then a : X⇤ R and B : X⇤ X⇤ R are continuous. In addition, there exists a X ! ⇥ ! 2 representing a, i.e., such that

a (f)=f(a), f X⇤. 8 2 Proof. Let us define

c := x (dx),c:= x 2 (dx). (2.3.3) 1 k k 2 k k ZX ZX Then, if f,g X 2 ⇤ a (f) f x (dx)=c f , | | k kX⇤ k k 1k kX⇤ ZX B (f,g) f(x) a (f) g(x) a (g) (dx) | |  | || | ZX f g ( x + c )2 (dx)=(c +3c2) f g . k kX⇤ k kX⇤ k k 1 2 1 k kX⇤ k kX⇤ ZX 23

To show that a can be represented by an element a X, by the general duality theory 2 (see e.g. [Br, Proposition 3.14]), it is enough to show that the map f a (f) is weakly 7! ⇤ continuous on X⇤, i.e., continuous with respect to the duality (X⇤,X). Moreover, since X is separable, by [Br, Theorem 3.28] weak⇤ continuity is equivalent to continuity along weak⇤ convergent sequences. Let (fj) be a sequence weakly⇤ convergent to f,i.e.,

f (x) f(x), x X. j ! 8 2 By the Uniform Boundedness Principle,

sup fj X⇤ < + , j N k k 1 2 and by the Lebesgue Dominated Convergence Theorem, we deduce

lim a(fj)= lim fj(x) (dx)= f(x) (dx)=a(f). j + j + ! 1 ! 1 ZX ZX

The space X is continuously embedded in L2(X, ) by the map j : X L2(X, ), ⇤ ⇤ !

j(f)=f a (f),fX⇤. 2

The operator j is continuous because j(f) 2 (1 + c ) f ,wherec is defined k kL (X,)  1 k kX⇤ 1 in (2.3.3). We define the reproducing kernel as the closure of the range of this embedding,

Definition 2.3.4 (Reproducing kernel). The reproducing kernel is defined by

2 X⇤ := the closure of j(X⇤) in L (X, ), (2.3.4) i.e. X consists of all limits in L2(X, ) of sequences of functions j(f )=f a (f ) ⇤ h h h with (f ) X . h ⇢ ⇤ The definitions of characteristic function, mean and covariance can be extended to the reproducing kernel.

Proposition 2.3.5. Let be a Gaussian measure on a separable Banach space. Then, the functions , a and B admit extensions defined on X⇤ that are continuous with respect 2 to the L (X, ) topology. In particular, a =0on X⇤ and b 1 2 (f)=exp f 2 , f X⇤. 2k kL (X,) 8 2 n o Proof. The extensions ofb a and B are obvious since the maps

f a (f)= f(x) (dx), 7! ZX 24 and

(f,g) B(f,g)= f(x) a(f) g(x) a(g) (dx) 7! X Z h ih i are continuous in the L2(X, ) topology. So, for f,g X we have a (f) = 0, 2 ⇤

B (f,g)= f(x)g(x)(dx)= f,g 2 , h iL (X,) ZX and ˆ(f)= exp if(x) (dx). { } ZX 2 Let gh = j(fh), fh X⇤, be a sequence of functions converging to f in L (X, ). Then, 2 it limh a(gh) = 0. Using the fact that the map t e is 1-Lipschitz, we have !1 7!

exp igh(x) exp if(x) (dx) exp igh(x) exp if(x) (dx) X { } { }  X | { } { }| Z Z gh(x) f(x) (dx)  | | ZX 1/2 2 gh(x) f(x) (dx) 0  X | | ! ⇣Z ⌘ Therefore, 1 1 ˆ(f)= lim ˆ(gh)= lim exp a(gh) B(gh,gh) =exp f L2(X,) . h h 2 2k k !1 !1 n o n o

Notice that if is nondegenerate then two di↵erent elements of X⇤ define two di↵erent elements of X⇤,butif is degenerate two di↵erent elements of X⇤ may define elements coinciding -a.e. By Proposition 2.3.5, we may define the operator R : X (X ) by ⇤ ! ⇤ 0

R f(g):= f(x)[g(x) a (g)] (dx),fX⇤,g X⇤. (2.3.5) 2 2 ZX Observe that R f(g)= f,g a (g) 2 . (2.3.6) h iL (X,) It is important to notice that indeed R maps X⇤ into X. Proposition 2.3.6. The range of R is contained in X, i.e., for every f X there is 2 ⇤ y X such that R f(g)=g(y) for all g X . 2 2 ⇤ Proof. As in the proof of Proposition 2.3.3, we show that for every f X the map 2 ⇤ g R f(g) is weakly continuous on X , i.e., continuous with respect to the duality 7! ⇤ ⇤ (X⇤,X). By the general duality theory (see e.g. [Br, Proposition 3.14]) we deduce that R f X. Recall that, since X is separable, weak continuity is equivalent to continuity 2 ⇤ 25 along weak convergent sequences. Let then (g ) X be weakly convergent to g,i.e., ⇤ k ⇢ ⇤ ⇤ g (x) g(x) for every x X. Then, by the Uniform Boundedness Principle the sequence k ! 2 (g ) is bounded in X and by the Dominated Convergence Theorem a (g ) a (g) and k ⇤ k ! R f(g )= f(x)[g (x) a (g )] (dx) f(x)[g(x) a (g)] (dx)=R f(g). k k k ! ZX ZX

Remark 2.3.7. Thanks to Proposition 2.3.6, we can identify R f with the element y X 2 representing it, i.e. we shall write

R f(g)=g(R f), g X⇤. 8 2 2.4 Exercises

Exercise 2.1. Prove that if is a Gaussian measure and is not a Dirac measure, then for any r>0 and x X, 2 (B(x, r)) < 1. Exercise 2.2. Let X be an infinite dimensional Banach space. Prove that there is no nontrivial measure µ on X invariant under translations and such that µ(B) > 0 for any ball B. Hint: modify the construction described in the Hilbert case using a sequence of elements in the unit ball having mutual distance 1/2. Exercise 2.3. Prove that a Gaussian measure on a Banach space is degenerate i↵ there exists X f = 0 such that ˆ(f) = 1 and hence i↵ there exists a proper closed subspace ⇤ 3 6 V X with (V ) = 1. ⇢ Exercise 2.4. Let be centred. Prove that for any choice f ,...,f X ,setting 1 d 2 ⇤

P (x)=(f1(x),...,fd(x)),

1 d n P is the Gaussian measure N (0,Q), with Qi,j = fi,fj L2(X,).IfL : R R is h i ! another linear map, compute the covariance matrix of (L P ) 1. Exercise 2.5. Let = N (a, Q) be a nondegenerate Gaussian probability measure on Rd. Show that x 4 (dx)=(TrQ)2 + 2Tr(Q2). d | | ZR 2 Hint: Consider the function F (")= e" x (dx) and compute F (0). Rd | | 00 Exercise 2.6. Let be a centredR Gaussian measure on a separable Banach space X. Compute the integrals

f(x) k e (dx), f(x) (dx),k N 2 ZX ZX for every f X . 2 ⇤ 26

Bibliography

[Br] H. Brezis: Functional Analysis, Sobolev spaces and partial di↵erential equations, Springer, 2011.

[D] R. M. Dudley: Real Analysis and Probabillity, Cambridge University Press, 2004.

[DS1] N. Dunford, J. T. Schwartz: Linear operators I, Wiley, 1958.