By John B Conway

This publication is an introductory textual content in practical research. in contrast to many smooth remedies, it starts off with the actual and works its method to the extra normal. From the studies: "This publication is a wonderful textual content for a primary graduate path in practical analysis....Many fascinating and demanding functions are included....It contains an abundance of workouts, and is written within the attractive and lucid kind which we now have come to anticipate from the author." --MATHEMATICAL studies

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Since T is compact, if e > 0, there are vectors h 1 , . . , hm in ball Jf such that T(ball Jt') c U j 1 B(Thi ; e/3). So if II h II � 1, choose hi with II Th - Thi II < e/3. Thus for any integer n, II Th - Tn h II � I Th - Thj I + II Thj - Tnhj II + II Pn (Thj - Th) II � 2 11 Th - Thj II + II Thj - Tnhj II � 2ej3 + II Thi - Tnhj 11 . Using the claim we can find an integer n0 such that II Thi - Tnhi II < ej3 for 1 � j � m and n � n0• So II Th - Tnh II < e uniformly for h in ball Jt' . Therefore II T - Tn II < e for n � n 0 • (c) => (b): If { Tn } is a sequence in � 00 ( Jf, %) such that II Tn - T I -+ 0, then II T: - T* II = II Tn - T ll -+ 0.

Thus an isomorphism also preserves completeness. That is, if an inner product space is isomorphic to a Hilbert space, then it must be complete. 3. Example. Definite S: 1 2 � 1 2 by S(cx 1 , cx 2 , • • • ) = (0, cx 1 , cx 2 , • • •). Then S is an isometry that is not surjective. The preceding example shows that isometries need not be isomorphisms. A word about terminology. Many call what we call an isomorphism a unitary operator. We shall define a unitary operator as a linear transformation U: Jf � Jt that is a surjective isometry.

17. Proposition. If A e�(Jf), then the following statements are equivalent. (a) A* A = AA* = I. (b) A is unitary. ) (c) A is a normal isometry. PROOF. 2. (b) => (c): By (2. 1 7), A* A = I. But it is easy to see that the fact that A is a surjective isometry implies that A - 1 is also. 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.