By Euler, Leonhard
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Show that U is an isomorphism and find a formula for U - 1 . 5. Let (X, 0, ,u) be a a-finite measure space and let u: X -+ F be an 0-measurable function such that sup { I u(x) I : xe X } < oo. e. [,u], in which case U is surjective. 6. Let rtJ f(O) = f(2n)} and show that rtJ is dense in L2[0, 2n] . 7. Show that { ( 1/fo), ( 1/Jn ) cos nt, (1/Jn ) sin nt: 1 � n < oo} is a basis for L2[ - n, n]. 8. Let (X, 0) be a measurable space and let ,u, v be two a-finite measures defined on (X, 0). Suppose v ,u and
Yf 2 ffi . . • • . • • This is part of a more general process. Yf i for all i and 'L { II h(i) II 2 : ieJ} < oo. Yf is a Hilbert space. The main reason for considering direct sums is that they provide a way of manufacturing operators on Hilbert space. In fact, Hilbert space is a rather dull subject, except for the fact that there are numerous interesting questions about the linear operators on them that are as yet unresolved. This subject is introduced in the next chapter. EXERCISES 1 . Let { (Xh Qh Jli): ieJ} be a collection of measure spaces and define X, Q, and Jl as follows.
Hence by (2. 1 7) I = (A - 1 )*A - 1 = (A*) - 1 A - 1 = (AA*) - 1 ; this implies that A*A = AA* = I. (c) =>(a): By (2. 1 7), A* A = I. Since A is also normal, AA* = A* A = I and so A is surjective. 18. Proposition. We conclude with a very important, though easily proved, result. l. PROOF. l. l c ker A. 19. Theorem. Two facts should be noted. l. l = ran A* since ran A* may not II. Operators on Hilbert Space 36 be closed. All that can be said is that (ker A) J. L cl (ran A). = = EXERCISES 1. 5. 2.