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oo lim ((Afn - zfn) + zfn) = g. D (iii) Assume that the operator A admits a closure A. Every point z of regular type for the operator A is also a point of regular type for A. Furthermore, R(A - z][) = (R(A - z][))~. 1). Taking into account that A ~ A, where the operator A is closed, and assertion (ii), we conclude that R(A - z][) ~ R(A - z][) => (R(A - z][))~ ~ R(A - z][).

Proof. First, we prove that A* is bounded in 1J(A*) (in this case, it certainly exists). Assume the contrary. Then there exists a sequence (gn)~=l C 1J(A*), IlgnllH = 1, such that IIA*gnIIH -+ 00 as n -+ 00. For any n E N, consider the functionals In(f) = (f, A*gn)H (f E H); Illnll = IIA*gnIIH ----n->oo 00. On the other hand, by virtue of the equality we have In(f) (Af,g)H = (f,A*g)H (f E 1J(A) = H, g E 1J(A*)) = (Af, gn)H and, therefore, Iln(f)1 ::; IIAfllH = c (f E H, n EN). 19). 3 and, hence, A* E £(H).

Taking into account that A ~ A, where the operator A is closed, and assertion (ii), we conclude that R(A - z][) ~ R(A - z][) => (R(A - z][))~ ~ R(A - z][). Let us prove the inverse inclusion. Let 9 E R(A - z][) and 9 = (A - z][)f (f E V(A)). According to the definition of A, there exists (fn)::;'=l C V(A) such that fn -> f and Afn -> Af as n -> 00. But then R(A - z][) 3 (A - z][)fn -> 9 and, D therefore, 9 E (R(A - z][))~. Thus, R(A - z][) ~ (R(A - z][)~. Let z E C be a point of regular type for the considered operator A.

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