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The space E does have a trivial centralizer since it has no M summands. 11. Example. Let E be the real space previous example. Here we have 1 (2) and deﬁne T as in the T ∗ ((1, 1) ◦ ψ1 ) = ψ1 and T ∗ ((−1, 1) ◦ ψ1 ) = −ψ2 and the function ϕ is not well deﬁned. Of course, the centralizer of E is not trivial in this case. There is one last piece of business we would like to attend to before closing this section. 9. The assumption that the space Y is strictly convex enables us to drop the other conditions on the range space N .

If H = hF , then H is a nonzero element of X but T H(t) = V (t)H(ϕ(t)) = 0 for all t ∈ K0 . 2, we conclude that ϕ(K0 ) is dense. If we assume that the nice operator is actually an isometry, we can prove a tiny bit more. 9. Corollary. (i) Suppose that T is an isometry deﬁned from the function module X = (Q, (Xs )s∈Q , X) onto the function module Y = (K, (Yt )t∈K , Y ) where Z(Yt ) is trivial for each t ∈ K such that Yt = {0}. Then there is a function ϕ from K0 = {t ∈ K : Yt = {0}} onto Q0 = {s ∈ Q : Xs = {0}} and a function t → V (t) from K0 into the family of nice operators from Xϕ(t) to Yt such that (10) T F (t) = V (t)F (ϕ(t)) for all t ∈ K0 and F ∈ X.

We want to relax that requirement and consider isometries deﬁned on a closed subspace M of C0 (Q, X). Once again, this parallels the approach taken in Chapter 2. The goal, as always, is to see if we can show that an isometry from such an M onto a subspace N of C0 (K, Y ) is some kind of generalized weighted composition operator. We will see that it is necessary to make some assumptions about the subspace M in order to assure the existence of enough functions of the right kind to make the arguments work.