MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.95 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 16. Tuesday, April 7: partially reconstructed From last time Proposition 23. The invertible elements form an open subset GL(H) ⊂ B(H). Proof. Recall that we showed using the convergence of the Neumann series that if B ∈ B(H) and �B� < 1 then Id −B is invertible, meaning it has a two-sided inverse in B(H) (which we know, from the open mapping Theorem to be equivalent to it being a bijection). So, suppose G ∈ GL(H), meaning it has a two-sided (and unique) inverse G−1 ∈ B(H) : (16.1) G−1G = GG−1 = Id . Then we wish to show that B(G; �) ⊂ GL(H) for some � > 0. In fact we shall see that we can take � = �G−1�−1 . The idea is that we wish to show that G + B is a bijection, and hence invertible. To do so set (16.2) E = G−1B = G + B = G−1(Id +G−1B).⇒ This is injective if Id +G−1B is injective, and surjective if Id +G−1B is surjective, since G−1 is a bijection. From last time we know that (16.3) �G−1B� < 1 =⇒ Id +G−1B is invertible. Since �G−1B� ≤ �G−1��B� this follows if �B� < �G−1�−1 as anticipated. � Thus GL(H) ⊂ B(H), the set of invertible elements, is open. It is also a group – since the inverse of G1G2 if G1, G2 ∈ GL(H) is G−1G−1 .2 1 This group of invertible elements has a smaller subgroup, U(H), the unitary group, defined by (16.4) U(H) = {U ∈ GL(H); U−1 = U∗}. The unitary group consists of the linear isometric isomorphisms of H onto itself – thus (16.5) (Uu, U v) = (u, v), �Uu� = �u� ∀ u, v ∈ H, U ∈ U(H). This is an important object and we will use it a little bit later on. The unitary group on a separable Hilbert space may seem very similar to the familiar unitary group of n × n matrices, U(n). It is, of course it is much bigger for one thing. In fact there are some other important differences which I will describe a little later on (or get you to do some of it in the problems). On important fact that you should know, even though I will not prove it, is that U(H) is contractible as a metric space – it has no significant topology. This is to be constrasted with the U(n) which have a lot of topology, and not at all simple spaces – especially for large n. One upshot of this is that U(H) does not look much like the limit of the U(n) as n → ∞. Now, for the rest of today I will talk about the opposite of the ‘big’ operators such as the elements of GL(H). Definition 7. An operator T ∈ B(H) is of finite rank if its range has finite dimension (and that dimension is called the rank of T ); the set of finite rank operators is denoted R(B).� � � � 96 LECTURE NOTES FOR 18.102, SPRING 2009 Why not F(B)? Because we want to use this for the Fredholm operators. Clearly the sum of two operators of finite rank has finite rank, since the range is contained in the sum of the ranges (but is often smaller): (16.6) (T1 + T2u) ∈ Ran(T1) + Ran(T2). Since the range of a constant multiple of T is contained in the range of T it follows that the finite rank operators form a linear subspace of B(H). It is also clear that (16.7) B ∈ B(H) and T ∈ R(B) then BT ∈ R(B). Indeed, the range of BT is the range of B restricted to the range of T and this is certainly finite dimensional since it is spanned by the image of a basis of Ran(T ). Similalry TB ∈ R(H) since the range of TB is contained in the range of T. Thus we have in fact proved most of Proposition 24. The finite rank operators form a ∗-closed ideal in B(H), which is to say a linear subspace such that (16.8) B1, B2 ∈ B(H), T ∈ R(H) =⇒ B1T B2, T ∗ ∈ R(H). Proof. In fact it is only the fact that T ∗ is of finite rank if T is which remains to be checked. To do this let us find an explicit representation for an operator of finite rank. First, since Ran(T ) is finite dimensional, we can choose a basis, fi i = 1, . . . , N, for it. Then for any element u ∈ H, N(16.9) T u = cifi. i=1 The constants ci are determined, since the fi are a basis, and so define linear functionals u �−→ ci. These are continuous. In fact we can simply choose the fi to be orthonormal and then, pairing (16.9) with fj we see that (16.10) cj = (T u, fj ) = (u, T ∗fj ). In particular there are elements (really by Riesz’ theorem) ei = T ∗fi ∈ H sucht that N(16.11) T u = (u, ei)fi. i=1 Conversely, if T can be written in the form (16.11) then it is of finite rank, since its range is contained in the span of the fi. From (16.11) it follows that T ∗ is also of finite rank since N N(16.12) (T ∗v, u) = (v, T u) = (v, fi)(ei, u) ∀ u ∈ H = ⇒ T ∗v = (u, fi)ei. j=1 i=1 The rˆoles of the fi and ei are simply interchanged. � Next time I will show that the closure of the ideal R(H) in B(H) is the ideal of compact operators. Of course this closure is certainly closed(!) Moreover it is a ∗-closed ideal, since Tn → K in norm and B1, B2 ∈ B(H) implies (16.13) B1TnB2 → B1KB2, Tn ∗ → K∗.� �� �97 LECTURE NOTES FOR 18.102, SPRING 2009 So, once we prove that the compact operators are the closure of the finite rank operators we will know that they form a closed, ∗-ideal. Notice that the importance of the ideal condition – it is the analogue of the normal condition for a subgroup – is that the quotient B/I of the algebra by an ideal is again an algebra. The quotient by the ideal, K(H), of compact operators is a Banach space since K is closed. It is called the Calkin algebra. Lemma 11 (Row rank=Colum rank). For any finite rank operator on a Hilbert space, the dimension of the range of T is equal to the dimension of the ranfe of T ∗. Proof. We showed that a finite rank operator T always takes the form (16.11). If the fi are taken to be a basis for the range of T, so N = dim Ran(T ), then the ei must be linearly independent. Indeed, if not then one of the ei can be replaced by a linear combination ei = cj ej . Inserting this into (16.11) …