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Additional info for Arguments of the Hon. Edward Stanly, of Counsel for the Receiver, and T. W. Park, Esq., of Counsel for Alvin Adams, with the Charge of the Court
Then there exists a number e> 0 such that any adjointable operator D satisfying the condition IF — DII
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 , 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 (). Let M be a Hubert B-module equipped with a strongly continuous unitary representation of C and let the group act trivially on B.