By Brewer M. J.

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For any T > 0 there is with probability 1 an N such that med xk(t) = {k{~N med {k{~n xk(t) for 0 ~ t ~ T, n ~ N Part II. If we define y(t) = lim then with probabilit~ med xk(t) for t ~ 0 , 1 : I) Yk(t) are continuous on R , Yk(O) = Xk(O) = Xk, Yk+l(t) ~ Yk(t) , for t ~ 0 and k = O, ~1, ... )'s are identical. 3) I_~f i ~ j then ~t ~ 0 : Yi(t) = yj(t)~ does not contain any interval. We omit the proof of this theorem. The equilibrium state of our ideal gas at t = 0 has been disturbed by placing at the origin the particle x o and the Poissonian random measure does not correspond to this new state and that is why we may not speak about its invariance (Doob's theorem).

E E** = E . Thus, there is linear continuous mapping R : E* ~ E with IIR[I ~ C and such that (Rf)(g) = R(f,g) for f,g E E* , and it is called the covariance operator for the LBM . 1. 8) a r Rd~ i_~s the given cont~uuous LBM, then and Rf(a) = ~((X,Z) d X(a)) , (~f, Rg) k = ~[(x,f) d (x,g) d] 49 holds for f,g E E* . Proof. 9) is dense in E* . Moreover, (Rf,~) d = (Rf)(~) = E((X,f) d (X~) d) = S ~(a) E((X,f) d X(a)) da . Rd The interchange of integration is justified by the following estimate. Since Y = (X,f) d E H x it follows that E(IYX(a) I) ~ Since @ lal I/2 (Ey2) I/2 .

Ol. 5. The covariance operator. We assume again that on (G,F,P) ~X(a) : a E Rd~ implies the continmous LBM is given. 1) P (X(a) = 0 for every p (lal ~) as lal * ~} = I 1 > ~ . ) E E~ = I . For this let be given I I < q < ~ and ~ + ~ = I . 2) p : and E such that q , I < p < ~ , E L I ( R a) . ~)p ('~ + l al ~ ~l(z~d)}. if only I1:~11, = ( J" (If(a)l(1 + lal ))q Rd and I Ilgll = (R~ (Ig(a) l I. for 1 d+ 1 )P da)~" lal x" f E E" , g r E . 4) l(~,g)dl ~< IIzl], IIg il 9 It is clear that E is reflexive.