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By Hari Kishan

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Hence, x∗ ◦ψϕ(t) ∈ ext(C0 (Q, X1 )∗ ), and since T is an isometry, y ∗ ◦ ψt = (T ∗ )−1 (x∗ ◦ ψϕ(t) ) is an extreme point for the unit ball of N ∗ . 5. 3 can differ from ch(N ). 9. Example. Let E be two-dimensional real space with norm √ |a| + |b|( 2 − 1) if |b| ≤ |a|; √ (a, b) o = |b| + |a|( 2 − 1) if |a| ≤ |b|. The dual norm is given by   |α| (α, β) d = |α|+|β| 2   |β| if if if √ |β| ≤ ( 2 − 1); √ √ ( 2 − 1) ≤ |β| ≤ ( 2 + 1)|α|; √ ( 2 + 1)|α| ≤ |β|. Let N = sp{G1 , G2 }, where G1 , G2 are elements of C0 ({1, 2, 3}, E) defined by G1 (1) = e1 , G2 (1) = e2 G1 (2) = −e1 , G2 (2) = e1 G1 (3) = −e1 , G2 (3) = −e1 .

We begin with a theorem that will reveal the major change in the description of into isometries from what we have seen in the previous work of this chapter. As usual, it will be helpful to introduce some new notation. We assume now that T is an isometry from C0 (Q, X) onto a subspace N of C0 (K, Y ), where, as before, Q, K are locally compact Hausdorff spaces and X, Y are Banach spaces. Given x ∈ S(X) and s ∈ Q, let F(x, s) = {F ∈ C0 (Q, X) : F (s) = F x}, and B(x, s) = {t ∈ K : T F (t) = F for F ∈ F(x, s)}.

8, T F (tα ) = 0 for such α and we are left to conclude that T F (t) = 0 as well. Furthermore, because A is normal, there exists g ∈ C0 (Q) such that g ≡ 1 on U . For G = g(·)x, T G(tα ) = 1 for every α ≥ α0 . To see this last statement, recall that G(sα ) = x, and since sα ∈ σ(A), there exists some f ∈ A such that f (sα ) = 1 = f . Let G1 = f (·)x and observe that T G1 (tα ) = 1 since tα ∈ B(x, sα ). 8 it follows that T H(tα ) = 0. Therefore, T G(tα ) = T G1 (tα ) = 1 and this holds for all α ≥ α0 .

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