By P. M. Cohn

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Suppose that X is a right Hilbert A-module, and that T is a linear operator from X to X. Then T is a positive element of C(X) if and only if (T(x) , x) > 0 for all x G X. Hilbert C*-Modules 20 Proof. If T > 0 in £(X), then T = S*S for some S £ £(X) and (T(x) , X)A = (S(x) , S(X))A > 0. Now assume (T(x) , x) > 0 for all x G X. 16) and (T(z) , 2;) = (z , T(z)) for all 2 G X, it follows that (T(x) , y) = (x , T(y)) for all x, y G X. Thus T is adjointable with T* = T. Now the functional calculus allows us to write T = S - R with S, i?

57. a) and / G C 0 (T,/C). We first show that Lm(f) G Co(T, K). Since 11-» m(t)h is continuous for each ft, a standard compactness argument shows that for every compact set K in H, t — i > m(t)h is uniformly continuous for ft G if. Thus £ — i > m(t)S is continuous from T into /C for each S e JC. Fix s e T and e > 0. Choose a neighbourhood (7 of 5 such that ||m(t)/(s) - m(s)f(s)\\ < e/2 and \\f(t) - f(s)\\ < 6/(2117711100) for all t G U. Then t G C/ implies \\m(t)f(t)-m(s)f(s)\\ < e. Since ||m(*)/(*)|| < ||m|UI/WII, this proves that L m ( / ) G Co(T,/C).

So the standard procedure is to construct such an algebra M(A), and to prove that this concretely defined object is a maximal unitization, but to remember how individual elements were constructed. 2]). Since we are interested in the interaction between multipliers and Hilbert modules, it is more helpful for us to define M(A) to be the C*-algebra £(AA), SO that individual multipliers are adjointable operators on A A- This elegant approach comes from [94]. For those who have seen the traditional definition of M(A), the next Theorem says that M(A) = C{AA)] for others, it will motivate our definition of M(A).