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 ∈ L(X,Y ) be continuous. Deﬁnition 1.1. L ∈ L(X,Y ) is said to be a Fredholm operator if (a) dim N(L) < +∞,

(b) R(L) is closed in Y ,

(c) codimR(L) := dim Y/R(L) < ∞. 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 ∈ 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 ∈ F(X,Y ) is defined to be the integer

ind(L) := dim N(L) − codimR(L).

Example 1.1. If L ∈ L(X,Y ) is invertible, L is a Fredholm operator of index 0. Example 1.2. If K ∈ 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) < ∞

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) = {(a1, a2, 0, 0, ··· )} and thus dim N(L) = 2. Also it is easy to see R(L) = X and thus

codimR(L) = 0.

So L ∈ 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 ∈ 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) = {constant functions} and thus dim N(L) = 1. So L is a Fredholm operator with index 1. Remark. Let L ∈ 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 ∈ F(X,Y ), then L∗ ∈ 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) < ∞ and codimR(L∗) = dim X∗/R(L∗) = dim N(L) < ∞. So L∗ ∈ 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 ∈ F(X,Y ), there exists > 0 s.t. for any L˜ ∈ L(X,Y ) with kL − L˜k < , we have L˜ ∈ F(X,Y ) and ind(L˜) = ind(L).

Proof. We prove (3). Suppose L ∈ F(X,Y ), then dim N(L) < ∞ and codimR(L) < ∞. 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˜ ∈ 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 ∈ L(X,Y ), L2 ∈ 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˜ ∈ 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 ∈ L(X,Y ). Then the followings are equivalent: (1) L ∈ F(X,Y ).

(2) L has a “ pseudo-inverse” (also called parametrix): there exists L˜ ∈ L(Y,X) and ﬁnite rank operators K1 ∈ L(X,X) and K2 ∈ L(Y,Y ) so that

L˜ ◦ L = IdX + K1,L ◦ L˜ = IdY + K2. (2)

(3) There exists L˜ ∈ L(Y,X) and compact operators K1 ∈ K(X),K2 ∈ K(Y ), so that

L˜ ◦ L = IdX + K1,L ◦ L˜ = IdY + K2. (3)

4 (4) There exists L˜1, L˜2 ∈ L(Y,X) and compact operators K1 ∈ K(X),K2 ∈ 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) < ∞.

Similarly from the fact that IdY + K2 is Fredholm and

R(L) ⊃ R(IdY + K2) we get codimR(L) < ∞. So L is Fredholm. (1) ⇒ (2): Suppose L ∈ 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 ∈ Y , we can write y = y1 + y2, where y1 ∈ R(L) and y2 ∈ Y1, then ˜ −1 L(y1 + y2) = L0 y1. Then L˜ ∈ 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 ∈ X, we can write x = x1 + x2, where x1 ∈ X1 and x2 ∈ 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 ∈ F(X,Y ) and K ∈ K(X,Y ), then L + K ∈ F(X,Y ) and

ind(L + K) = ind(L).

Proof. Take L˜, K1,K2 as in theorem 1.6 part (2). Then L˜ ∈ 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 ∈ 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 ∈ F(X,Y ) has index 0 iﬀ there exists an invertible continuous linear operator L0 ∈ L(X,Y ) and a compact operator K ∈ 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 invertible operators.