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.� � � � 108 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 19. Thursday, April 16 I am heading towards the spectral theory of self-adjoint compact operators. This is rather similar to the spectral theory of self-adjoint matrices and has many useful applications. There is a very effective spectral theory of general bounded but self-adjoint operators but I do not expect to have time to do this. There is also a pretty satisfactory spectral theory of non-selfadjoint compact operators, which it is more likely I will get to. There is no satisfactory spectral theory for general non-compact and non-self-adjoint operators as you can easily see from examples (such as the shift operator). In some sense compact operators are ‘small’ and rather like finite rank operators. If you accept this, then you will want to say that an operator such as (19.1) Id −K, K ∈ K(H) is ‘big’. We are quite interested in this operator because of spectral theory. To say that λ ∈ C is an eigenvalue of K is to say that there is a non-trivial solution of (19.2) Ku − λu = 0 where non-trivial means other than than the solution u = 0 which always exists. If λ = 0 we can divide by � λ and we are looking for solutions of (19.3) (Id −λ−1K)u = 0 which is just (19.1) for another compact operator, namely λ−1K. What are properties of Id −K which migh show it to be ‘big? Here are three: Proposition 26. If K ∈ K(H) is a compact operator on a separable Hilbert space then null(Id −K) = {u ∈ H; (IdK )u = 0} is finite dimensional (19.4) Ran(Id −K) = {v ∈ H; ∃u ∈ H, v = (Id −K)u} is closed and Ran(Id −K)⊥ = {w ∈ H; (w, Ku) = 0 ∀ u ∈ H} is finite dimensional and moreover (19.5) dim (null(Id −K)) = dim Ran(Id −K)⊥ . Definition 9. A bounded operator F ∈ B(H) on a Hilbert space is said to be Fredholm if it has the three properties in (19.4) – its null space is finite dimensional, its range is closed and the orthocomplement of its range is finite dimensional. For general Fredholm operators the row-rank=colum-rank result (19.5) does not hold. Indeed the difference of these two integers (19.6) ind(F ) = dim (null(Id −K)) − dim Ran(Id −K)⊥ is a very important number with lots of interesting properties and uses. Notice that the last two conditions are really independent since the orthocom-plement of a subspace is the same as the orthocomplement of its closure. There are for instance bounded opertors on a separable Hilbert space with trivial null space and dense range which is not closed. How could this be? Think for instance of the operator on L2(0, 1) which is multiplication by the function x. This is assuredly bounded and an element of the null space would have to satisfy xu(x) = 0 almost everywhere, and hence vanish almost everywhere. Moreover the density of the L2� � � 109 LECTURE NOTES FOR 18.102, SPRING 2009 functions vanishing in x < � for some (non-fixed) � > 0 shows that the range is dense. However it is clearly not invertible. Before proving this result let’s check that the third condition in (19.4) really follows from the first. This is a general fact which I mentioned, at least, earlier but let me pause to prove it. Proposition 27. If B ∈ B(H) is a bounded operator on a Hilbert space and B∗ is its adjoint then (19.7) Ran(B)⊥ = (Ran(B))⊥ = {v ∈ H; (v, w) = 0 ∀ w ∈ Ran(B)} = Nul(B∗). Proof. The definition of the orthocomplement of Ran(B) shows immediately that (19.8) v ∈ (Ran(B))⊥ ⇐⇒ (v, w) = 0 ∀ w ∈ Ran(B) ←→ (v, Bu) = 0 ∀ u ∈ H ⇐⇒ (B∗v, u) = 0 ∀ u ∈ H ⇐⇒ B∗v = 0 ⇐⇒ v ∈ Nul(B∗). On the other hand we have already observed that V ⊥ = (B)⊥ for any subspace – since the right side is certainly contained in the left and (u, v) = 0 for all v ∈ V implies that (u, w) = 0 for all w ∈ V by using the continuity of the inner product to pass to the limit of a sequence vn w. �→ Thus as a corrollary we see that if Nul(Id −K) is always finite dimensional for K compact (i.e. we check it for all compact operators) then Nul(Id −K∗) is finite dimensional and hence so is Ran(Id −K)⊥. Proof of Proposition 26. First let’s check this in the case of a finite rank operator K = T. Then (19.9) Nul(Id −T ) = {u ∈ H; u = T u} ⊂ Ran(T ). A subspace of a finite dimensional space is certainly finite dimensional, so this proves the first condition in the finite rank case. Similarly, still assuming that T is finite rank consider the range (19.10) Ran(Id −T ) = {v ∈ H; v = (Id −T )u for some u ∈ H}. Consider the subspace {u ∈ H; T u = 0}. We know that this this is closed, since T is certainly continuous. On the other hand from (19.10), (19.11) Ran(Id −T ) ⊃ Nul(T ). Remember that a finite rank operator can be written out as a finite sum N(19.12) T u = (u, ei)fi i=1 where we can take the fi to be a basis of the range of T. We also know in this case that the ei must be linearly independent – if they weren’t then we could write one of them, say the last since we can renumber, out as a sum, eN = ciej , of j<N multiples of the others and then find N−1(19.13) T u = (u, ei)(fi + cj fN ) i=1 showing that the range of T has dimension at most N − 1, contradicting the fact that the fi span it.� 110 LECTURE NOTES FOR 18.102, SPRING 2009 So, going back to (19.12) we know that Nul(T ) has finite codimension – every element of H is of the form N(19.14) u = u� + diei, u� ∈ Nul(T ). i=1 So, going back to (19.11), if Ran(Id −T ) =� Nul(T ), and it need not be equal, we can choose – using the fact that Nul(T ) is closed – an element g ∈ Ran(Id −T ) \Nul(T ) which is orthogonal to Nul(T ). To do this, start with any a vector g� in Ran(Id −T ) which is not in Nul(T ). It can be split as g� = u�� + g where g ⊥Nul(T ) (being a closed subspace) and u�� ∈ Nul(T ), then g =� 0 is in Ran(Id −T ) and orthongonal to Nul(T ). Now, the new space Nul(T ) ⊕ Cg is again closed and contained in Ran(Id −T ). But we can continue this process