Notes for Functional Analysis Wang Zuoqin (typed by Xiyu Zhai) December 4, 2015 1 Lecture 23 1.1 Fredholm operators Let X; Y be Banach spaces, and L 2 L(X; Y ) be continuous. Definition 1.1. L 2 L(X; Y ) is said to be a Fredholm operator if (a) dim N(L) < +1, (b) R(L) is closed in Y , (c) codimR(L) := dim Y=R(L) < 1. We will denote the space of all Fredholm operators from X to Y by F(X; Y ). This is a very important class of operators in modern mathematics. Remark. The second condition can be omitted for Banach spaces: If X; Y are Banach spaces, L 2 L(X; Y ) such that R(L) admits a closed complementary subspace, then R(L) is closed. But it is needed in defining Fredholm operators for between normed vector spaces. Definition 1.2. The index of L 2 F(X; Y ) is defined to be the integer ind(L) := dim N(L) − codimR(L): Example 1.1. If L 2 L(X; Y ) is invertible, L is a Fredholm operator of index 0. Example 1.2. If K 2 K(X), then I −K is a Fredholm operator. In fact, by PSet 12-2, 1(2), codimR(I − K) = dim X=R(I − K) = dim N(I − K∗) = dim N(I − K) < 1 This computation also shows ind(I − K) = 0: We shall see that all elements in F(X) \looks like" this form. 1 Example 1.3. Let X = l2 and consider the operator L : l2 ! l2 defined by L(a1; a2; a3; ··· ) = (a3; a4; a5; ··· ): Then obviously N(L) = f(a1; a2; 0; 0; ··· )g and thus dim N(L) = 2: Also it is easy to see R(L) = X and thus codimR(L) = 0: So L 2 F(X) with ind(L) = 2: Note that the dual operator of L, L∗ : l2 ! l2, is given by ∗ L (a1; a2; ··· ) = (0; 0; a1; a2; ··· ): One can easily see dim N(L∗) = 0 and dim l2=R(L∗) = 2. So ind(L∗) = −2: Example 1.4. Let X = C1([0; 1]);Y = C([0; 1]) and consider the map L : X ! Y; (Lf)(x) = f 0(x): Then R(L) = Y , because for any f 2 C([0; 1]), Z x L f(t)dt = f: (1) 0 It follows that codimR(L) = 0. Also it is easy to see N(L) = fconstant functionsg and thus dim N(L) = 1: So L is a Fredholm operator with index 1. Remark. Let L 2 L(X; Y ), then (a) L is injective () dim N(L) = 0. (b) L surjective () codimR(L) = 0 () ind(L) = dim N(L). 2 Proposition 1.3. If L 2 F(X; Y ), then L∗ 2 F(Y ∗;X∗) and ind(L∗) = −ind(L): Proof. Since L is Fredholm, R(L) is closed. By the closed range theorem, R(L∗) is also closed. Now we use PSet 12-2, Problem 1, to get dim N(L∗) = dim Y=R(L) = codimR(L) < 1 and codimR(L∗) = dim X∗=R(L∗) = dim N(L) < 1: So L∗ 2 F(Y ∗;X∗) and ind(L∗) = dim N(L∗) − codimR(L∗) = codimR(L) − dim N(L) = −ind(L): Proposition 1.4. Let X; Y be Banach spaces. Then (1) F(X; Y ) is an open subset in L(X; Y ), (2) The function ind : F(X; Y ) ! Z is continuous, i.e. it is locally a constant. Remark. (1) and (2) is equivalent to (3) For any L 2 F(X; Y ), there exists > 0 s.t. for any L~ 2 L(X; Y ) with kL − L~k < , we have L~ 2 F(X; Y ) and ind(L~) = ind(L): Proof. We prove (3). Suppose L 2 F(X; Y ), then dim N(L) < 1 and codimR(L) < 1. It follows that there exist closed subspaces X1 ⊂ X and Y1 ⊂ Y such that X = X1 ⊕ N(L) and Y = R(L) ⊕ Y1: Let i1 : X1 ,! X be the canonical inclusion and let p2 : Y ! R(L) be the projection associated to the second decomposition above. Then by definition, i1 is a Fredholm operator of index − dim N(L), and p2 is a Fredholm operator of index codimR(L). Let L0 = p2 ◦ L ◦ i1 : X1 ! R(L): Note that L0 is just the restriction of L onto X1 (to make it injective), and the target space is adjusted to be R(L) (to make it surjective). By definition L0 is continuous and invertible. 3 Now suppose L~ 2 L(X; Y ) is closed to L, i.e. kL~ − Lk < is small enough. We let L~0 = p2 ◦ L~ ◦ i1; then kL0 − L~0k ≤ kp2k · kL~ − Lk · ki1k is small. So L~0 is invertible for small enough. Now we apply the following lemma (whose proof is described in today's problem set): Lemma 1.5. Suppose L1 2 L(X; Y ), L2 2 L(Y; Z). Consider the three operators, L1;L2;L2 ◦ L1: If two of them are Fredholm, then the third one is also Fredholm, and ind(L2 ◦ L1) = ind(L2) + ind(L1): Since L~0; p2; i1 are all Fredholm, we conclude that L~ 2 F(X; Y ); and moreover, 0 = indL~0 = indp2 + indL~ + indi0: Comparing this with 0 = indL0 = indp2 + indL + indi0; we immediately get indL~ = indL: Next let's study the relation between Fredholm operators and compact operators. Theorem 1.6. Let X; Y be Banach spaces, and L 2 L(X; Y ). Then the followings are equivalent: (1) L 2 F(X; Y ). (2) L has a \ pseudo-inverse" (also called parametrix): there exists L~ 2 L(Y; X) and finite rank operators K1 2 L(X; X) and K2 2 L(Y; Y ) so that L~ ◦ L = IdX + K1;L ◦ L~ = IdY + K2: (2) (3) There exists L~ 2 L(Y; X) and compact operators K1 2 K(X);K2 2 K(Y ), so that L~ ◦ L = IdX + K1;L ◦ L~ = IdY + K2: (3) 4 (4) There exists L~1; L~2 2 L(Y; X) and compact operators K1 2 K(X);K2 2 K(Y ), so that L~1 ◦ L = IdX + K1;L ◦ L~2 = IdY + K2: (4) Proof. Obviously (2) ) (3) ) (4). We prove (4) ) (1), and (1) ) (2) below. (4) ) (1): Suppose such L~1; L~2;K1;K2 exists. Then since IdX + K1 is Fredholm and N(L) ⊂ N(L~1 ◦ L) = N(IdX + K1) we conclude dim N(L) < 1: Similarly from the fact that IdY + K2 is Fredholm and R(L) ⊃ R(IdY + K2) we get codimR(L) < 1: So L is Fredholm. (1) ) (2): Suppose L 2 F(X; Y ). We define X1;Y1; i1; p2 and L0 as in the proof of proposition (1.4). Since L0 is invertible (and the inverse is continuous), we can let L~ : Y ! X be the operator ~ −1 L = i1 ◦ L0 ◦ p2: Explicitly, for any y 2 Y , we can write y = y1 + y2, where y1 2 R(L) and y2 2 Y1, then ~ −1 L(y1 + y2) = L0 y1: Then L~ 2 L(Y; X) because it is the composition of three continuous operators. Moreover, if we let −K1 be the projection from X onto N(L), and let −K2 be the projection from Y to Y1, then both K1 and K2 are finite rank operators, and ~ −1 (L ◦ L − IdY )(y) = L(L0 (y1)) − y1 − y2 = −y2 = K2(y): Similarly for any x 2 X, we can write x = x1 + x2, where x1 2 X1 and x2 2 N(L), and ~ ~ −1 (L ◦ L − IdX )(x) = L(Lx1) − x1 − x2 = L0 (L0x1) − x1 − x2 = −x2 = K1(x) Remark. In the proof, we have L ◦ K1 = 0, and L~ ◦ K2 = 0. So one can choose L;~ K1;K2 in (2) so that L ◦ L~ ◦ L = L;~ L~ ◦ L ◦ L~ = L~ (5) 5 As a consequence, we have Theorem 1.7. Suppose L 2 F(X; Y ) and K 2 K(X; Y ), then L + K 2 F(X; Y ) and ind(L + K) = ind(L): Proof. Take L~, K1;K2 as in theorem 1.6 part (2). Then L~ 2 F(Y; X) and ind(L~) + ind(L) = ind(IdX + K1) = 0: So ind(L~) = −ind(L): By theorem 1.6 part (2), L~ ◦ (L + K) = IdX + K1 + L~ ◦ K and (L + K) ◦ L~ = IdY + K2 + K ◦ L:~ Apply theorem 1.6 again, we see L + K 2 F(X; Y ), and ind(L + K) = −ind(L~) = ind(L): Finally we state the following theorem whose proof is left as an exercise: Theorem 1.8. A Fredholm operator L 2 F(X; Y ) has index 0 iff there exists an invertible continuous linear operator L0 2 L(X; Y ) and a compact operator K 2 K(X; Y ) such that L = L0 + K: Proof. Left as a happy exercise. In other words, index 0 Fredholm operators are exactly compact perturbations of in- vertible operators. 6.