LECTURE 20, 18.155, 22 NOVEMBER 2011One of the main differences between finite dimensional and infinitedimensional Hilbert spaces is that the latter can have non-closed linearsubspaces. Similarly for operators one of the new phenomena in infinitedimensions is that the range of a bounded operator need not be closed.Let us consider the set of operators with closed range – it is not a linearspace:-(1) CR(H) = {A ∈ B(H); A(H) = A(H)}.Let Nul(A) ⊂ H be the null space – always closed for any A ∈ B(H)and Ran(A) = A(H) be the range. Purely algebraically, any operatorA defines a bounded isomorphism, i.e. a linear bijection, from Nul(A)⊥to Ran(A) and this is bounded with respect to the induced norm onRan(A). So, assuming A ∈ CR(H) the range is closed, hence is itself aHilbert space and by the Open Mapping theorem, this restricted maphas a bounded inverse. Denote the extension of this inverse by G,specfically(2)G(Au) = u if u ∈ Nul(A)⊥, Gv = 0 if v ∈ Ran(A)⊥=⇒ G ∈ B(H).This is the ‘generalized inverse’ of A ∈ CR(H), and clearly satisfies(3) GA = Id −ΠNul(A), AG = Id −ΠRan(A)⊥.These two projections can also be written as ΠNul(A)⊥and ΠRan(A)butthe form with ‘Id −Π’ emphasises that the projections measure thefailure of A to be an isomorphism.Now, conversely the identities in (3) imply that A has closed range(although G does not have to be the generalized inverse to satisfy (3)since adding a map from Ran(A)⊥to Nul(A) would not change theidentity).Consider then the two smaller sets of bounded operators(4)eF(H) = {A ∈ CR(H); either Nul(A) or Ran(A)⊥is finite dimensional},F(H) = {A ∈ CR(H); both Nul(A) and Ran(A)⊥are finite dimensional}.The first is the set of semi-Fredholm operators and the second the set ofFredholm operators. Thus an operator is semi-Fredholm if it has closedrange and one (at least) of the two projections ΠNul(A)and ΠRan(A)⊥has finite rank. It is Fredholm if they both have finite rank.12 LECTURE 20, 18.155, 22 NOVEMBER 2011In all three cases F(H) ⊂eF(H) ⊂ CR(H), composition on the rightor left by a bounded invertible operator maps these sets into themselves.Note that if A ∈ B(H) has range with a closed complement – mean-ing that there is a closed subspace F ⊂ H such that F ∩ Ran(A) = {0}but H = F + Ran(A) – then Ran(A) is in fact closed and A ∈ CR(H).Indeed, the map F ⊕ H 3 (f, u) 7−→ f + Au is clearly bounded andsurjective, so from the discussion above (meaning the open mappingtheorem) has a bounded generalized inverse G. If vn∈ Ran(A) con-verges to v ∈ H then Gvn= (fn, un) → (f, u) but fn= 0 sinceF ∩ Ran(A) = {0} and hence Au = v and Ran(A) is closed. In par-ticular if a bounded linear map has range with a finite dimensionalcomplement then it is semi-Fredholm. Note that it is not true (maybeyou need to worry about the truth of the Axiom of Choice here) thata general subspace with a finite dimensional complement is closed. Anon-trivial, but non-continuous linear functional has null space whichis not closed but does have a 1-dimensional complement (spanned byany vector not in the null space). So it is important here that we areconsidering the range of a bounded operator – which is not an arbitrarylinear subspace.It is very convenient in the subsequent discussion to have a somewhatcoarser characterization of Fredholm operators.Lemma 1. An operator A ∈ B(H) is Fredholm if and only if it hasboth a left and a right parameterix, in the sense that there are operatorsBRand BL∈ B(H) such that(5)ABR= Id −KR, KR∈ K(H),BLA = Id −KL, KL∈ K(H).Here a parametrix (modulo an ideal, in this case the compact operators)is a (one-sided in this case) inverse modulo the ideal, i.e. with a ‘error’term which is in the ideal.Proof. That (5) can be arranged for a Fredholm operator is immediate,since the generalized inverse gives (5) in the much stronger form (3)with KRand KLthe finite rank projections onto the null space andthe orthocomplement of the range.The converse, that (5) holds follows from special case that A =Id −K where K ∈ K(H) is Fredholm. Clearly this satisfies (5) withBR= BL= Id . Conversely, once we know that each operator Id −Kis Fredholm where K ∈ K(H) we see that the operators on the rightin (5) are Fredholm. In particular Id −KRhas finite dimensional nullspace and Id −KLhas closed range of finite codimension. However theLECTURE 20, 18.155, 22 NOVEMBER 2011 3two identities in (5) imply that(6) Nul(A) ⊂ Nul(Id −KR), Ran(A) ⊃ Ran(Id −KL).Thus the null space of A must be finite dimensional, since it is con-tained in a finite-dimensional space and Ran(A) must have a finitedimensional complement since it contains a subspace with finite di-mensional complement. In any case, Ran(A) contains a closed linearsubspace of finite codimension which ensures that it too is closed.Thus we are reduced to showing that Id −K is Fredholm where K ∈K(H). This follows from the decomposition discussed earlier that(7) Id −K = (Id −B1)(Id −D)(Id −B2), kB1k < 1, kB2k < 1and where ΠNDΠN= D for the projection onto the span of the first Nelements of an orthonormal basis. Thus Id −D has finite dimensionalnull space, since it is contained in the range of ΠNand finite dimen-sional complement to the range, since the range contains the range ofId −ΠN. Then the general case follows from (7) since the outer factorsare invertible. Although this result is stated in terms of the existence of a separateright and left parametrix, as in (5), it follows directly that, once bothexist, each is actually a two-sided parametrix in the same sense. Itis useful to have this a little more abstractly so that it coverse otherideals that K(H).Lemma 2. If A is an associative algebra, I ⊂ A is a two-sidedideal and A ∈ A has left and right parametrices, BL, BR∈ A, soId −ABR∈ I and Id −BLA ∈ I then BR− BL∈ I and each is atwo-sided parametrix modulo I.This is just the uniqueness of inverses in the associative algebra A/I.Proof. By assumptionBL= BL(Id −ABR+ABR) = BLABR+I1= (BLA−Id)BR+BR+I1= BR+I1−I2where I1= BL(Id −BR) ∈ I and I2= (Id −BLA)BR∈ I. Addingan element of I to a left or right parametrix gives another, so theconclusions follow. Proposition 1. The Fredholm operators form an open set in B(H)which is closed under composition and adjoints; the image under aFredholm operator of a closed subspace is closed. The semi-Fredholmoperators for an open and dense set