# Hilbert modules over function algebras by R. G. Douglas, V.I. Paulsen

By R. G. Douglas, V.I. Paulsen

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Extra resources for Hilbert modules over function algebras

Example text

Since we can choose the same / throughout a neighbourhood of any given point t, and mf is continuous, the new function (j) is continuous. It is bounded by ||ra||. Given h G £2(T) of finite support, we can choose / such that / = 1 on supp/i, and then n((/))h = ji(mf)h = mfi(f)h = mh. Thus m — /x(0) is in fi{Cb{T)), as required. • In view of the above, it is natural to speculate on the multiplier algebra of Co(T,JC(H)). A naive guess might be Cb(T,B(H)), but for infinite-dimensional H this algebra is too small.

Then A n l = q(Xnl + Lan) —> q(T). Since CI is always closed, q(T) = XI for some A G C. Then (An - A)l + L a n -> T - Al G ker

We want to build a representation of A. The underlying vector space will be X 0 H^, and the inner product will be characterized by ( x 0 / i I y®k) := (7r((y , x) )h\k). (2,25) To see where this comes from, consider the case B = C. 25) is (y , x)c(h \k) = {x\ y)(h I k). 23) on the tensor product of Hilbert spaces. 64. 25). 25). 59. 25) is bilinear in (x, /i), and hence defines a linear map / y <8>/fc : XoWfr -> C. Then (y, k) i-> fy®fk is bilinear from X x Hn to (X©H*)", hence gives a linear map L on X 0 7 ^ , and we define (a \ (3) := L@(a).