By Hille E.

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**Sample text**

Extending eq. 1) to tensor product of n elements, we know that {0j'=le{j : (k i ,"', kn ) E JN n } constitute a base of 0j'=lHj . Especially, if Hi = H2 = ... = H n = H, the n-fold tensor product is denoted by Higm. If {ek} kElN is a base of H, then {0j'=1 ekj : (k 1,···, kn ) E JN n } is a base of H0 n . However, in mathematical physics, one usually deals with its two subspaces: the subspaces of symmetric tensor products and antisymmetric tensor products. Here we restrict ourselves to symmetric ones.

18) is well defined. Moreover, we have This means that (',' > is the canonical bilinear form on H + X H _ which makes them mutually adjoint. • Now we are ready to prove the Minlos theorem. 6 Let

II is weaker than I . I. Let X be the completion of H with respect to II . II. Then X is a separable Banach space, and H is a linear subspace of X. If we identify the dual H* of H with H itself, then the dual X* of X can be regarded as a linear subspace of H: X* = {Y E H: xEH,iixii=l sup I(x, y)1 < oo} . ) the canonical bilinear form on X X X*. , (x,y)=(x,y), \fXEH, YEX*. r(H) the finite dimensional subspaces of X* and H, respectively. r(X*), if we denote by Cx(K) and CH(K) the a-algebras on X and on H generated by the cylinder sets based on K, respectively, then Cx(K) n H c CH(K).