Arguments of the Hon. Edward Stanly, of Counsel for the by Anonymous

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Then there exists a number e> 0 such that any adjointable operator D satisfying the condition IF — DII 0 such that if liP'1 — D1 <5 and D1 is an operator, then D1 is also an isomorphism.

Since Ildil < k 1/2 k 1/2 — (i d)(1 < thus completing the induction step. So we obtain that as k —* oc. Since the strictly positive element h generates the whole A 0 [104], the lemma is proved. Let us continue the proof of the theorem. 3, we can choose a form an sequence (x2) in the module M such that the partial sums of > (xi, approximate unit in A. Define the map T: A 12(M) by the formula a E A. T(a) = (xia,. ), = a*a is convergent in A, one has Since the series (Ta,Ta) = T(a) E 12(M). Moreover, the adjoint operator is well defined, T* 12(M) —+ A, T* (y2) = (Xj,Yj) E A for (yr) E 12(M).

9. 27) M' := P8(M), where the sum is supposed to be completed either as a Hubert sum. or as a closure in M of the algebraic sum (which is the same). Then M' = M. 28) PROOF. Assume that a C-linear functional f on M vanishes on M and that x E M is an arbitrary vector. Then, for any set of indices, we have E M', so that 0= f dg. 5, f(T9 (x)) = 0 holds almost everywhere = f(x) = and, by continuity, it vanishes everywhere. In particular, Hence, by the Hahn—Banach theorem, M = M. 0. 10 ([91]). Let M be a Hubert B-module equipped with a strongly continuous unitary representation of C and let the group act trivially on B.

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