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PROOF If Y = N ( T - A), then T restricted to Y has closed range Y. If 0 € /o(T), then Ji(T) = X is closed and, hence, has to be finite dimensional. 6 Assume that — oo < a < b < oo and 1 < p < oo. Define (27)(*)= [X f(y)dy. Ja 1 p Note that T € Q5(L (a,6),L (a,6)). We want to show that T is compact if p < oo. Fix n > 1, S = (b - a)/n and define for / € Lx(a, b) (Tnf){X) = £x[rfc,rfc+1)(*) P f(s)dS, where Tk = a + k5 and XA is used to denote the characteristic function of the set A. 7. COMPACT LINEAR OPERATORS 33 a compact member of Q5(L1(a, 6), L p (a, 6)).

Let U G 55(X, M) be the extension of u as obtained in the case of real scalars. Define F{x) = U(x) - iU(ix) for all x G X. One can easily see that F is an extension of / and F G X*. If x G X and if a G C is such that aF{x) = \F(x)\, then |F(a;)| = F(ax) = U{ax) < \\U\\\\x\\. Hence \\F\\ < \\U\\ = \\u\\ < \\f\\ and, therefore, \\F\\ = \\f\\. 7 Let M be a subspace of a normed space X and assume that x G X does not belong to the closure of M. Then there exists f G X* such that ll/H = 1, f(x) = dist(rz;, M) > 0, f(y) = 0 for all y G M.

DUALS 19 Hence one can find a 6 l such that for all x G M, f(x) - c\\x - x0\\ ■ R is defined by fo(x + Xx0) = f(x) + Aa, then /o G Mo* and / 0 is an extension of / with ||/o|| = c. Let V be the collection of all pairs (AT, g) where N D M is a subspace of X and g G AT* is an extension of / with ||g|| = c. Define partial ordering -<(oii? by (AT, p) -<; (AT', g') if and only if N C N' and #' extends #.