OPERATOR THEORY These lecture notes are based on the courses Operator Theory developed at King’s College London by G. Barbatis, E.B. Davies and J.A. Erdos, and Functional Analysis II developed at the University of Sussex by P.J. Bushell, D.E. Edmunds and D.G. Vassiliev. As usual, all errors are entirely my re- sponsibility. The same applies to the accompanying exercise sheets. 1 Introduction Spaces By IF we will always denote either the field IR of real numbers or the field C of complex numbers. 1.1. Definition A norm on a vector space X over IF is a map k · k : X → [0,∞) satisfying the conditions (i) kxk = 0 ⇐⇒ x = 0; (ii) kλxk = |λ|kxk, ∀x ∈ X, ∀λ ∈ IF; (iii) kx+yk ≤ kxk+kyk, ∀x,y ∈ X (the triangle inequality). A vector space X equipped with a norm is called a normed space. 1.2. Definition A normed space X is called a Banach space if it is complete, i.e. if every Cauchy sequence in X is convergent. 1.3. Examples 1. Any finite-dimensional normed space is a Banach space. 2. Let lp, 1 ≤ p ≤ ∞, denote the vector space of all sequences x = (x )∞ , x ∈ IF, such that k k=1 k ∞ !1/p kxk := X|x |p < ∞, 1 ≤ p < ∞, p k k=1 kxk := sup|x | < ∞. ∞ k k∈IN 1 Then k·k is a norm on lp and lp, 1 ≤ p ≤ ∞, are Banach spaces. p 3. Let C([0,1]) be the space of all continuous functions on [0,1] and (cid:18)Z 1 (cid:19)1/p kfk := |f(t)|pdt , 1 ≤ p < ∞, p 0 kfk := sup |f(t)|. ∞ 0≤t≤1 Then each k·k is a norm on C([0,1]). p The space C([0,1]) with the norm k·k is a Banach space. ∞ The space C([0,1]) with the norm k·k , 1 ≤ p < ∞, is not a Banach p space. Indeed, let us consider the sequence (f ) in C([0,1]), where n  0 if 0≤t≤ 1 − 1 ,  2 2n f (t)= nt− n−1 if 1 − 1 N, n we obtain Z 1 Z 1 |1−f(t)|pdt= |f (t)−f(t)|pdt≤kf −f kp →0 as n→∞. n n p 1/2+δ 1/2+δ Since the first integral is independent of n it has to be zero, which implies that f =1 on [1/2+δ,1] and hence on ]1/2,1] since δ was arbitrary. A similar argument shows that f =0 on [0,1/2[. Thus f is discontinuous at t=1/2 and we obtain a contradiction. 1.4. Definition Normed spaces X and Y are called isomorphic if there exists a linear isometry from X onto Y. Isometry is a map which does not change the norms of the corresponding points. 2 1.5. Theorem For any normed space (X,k · k) there exists a Banach space (X0,k·k0) and a linear isometry from X onto a dense linear subspace of X0. Two Banach spaces in which (X,k · k) can be so imbedded are isomorphic . 1.6. Definition The space (X0,k · k0) from Theorem 1.5 is called the completion of (X,k·k). The completion of the space (C([0,1]),k·k ), 1 ≤ p < ∞, is the Banach p space L ([0,1]) from the theory of Lebesgue integral. p Operators 1.7. Definition Let X and Y be vector spaces. A map B : X → Y is called a linear operator (map) if B(λx+µz) = λBx+µBz, ∀x,z ∈ X, ∀λ,µ ∈ IF. 1.8. Theorem Let X and Y be normed spaces. For a linear operator B : X → Y the following statements are equivalent: (i) B is continuous; (ii) B is continuous at 0; (iii) there exists a constant C < +∞ such that kBxk ≤ Ckxk, ∀x ∈ X. 1.9. Definition A liner operator B is bounded if it satisfies the last (and hence all) of the three conditions above. If B is bounded we define its norm by the equality kBk := inf{C : kBxk ≤ Ckxk, ∀x ∈ X}. It is easy to see that kBxk ≤ kBkkxk, ∀x ∈ X, (1.1) kBk = inf{C : kBxk ≤ C, all x s.t. kxk ≤ 1} = ( ) kBxk inf{C : kBxk ≤ C, all x s.t. kxk = 1} = sup : x 6= 0 = kxk sup{kBxk : kxk ≤ 1} = sup{kBxk : kxk = 1} 3 (prove these relations!). 1.10. Theorem Let X and Y be normed spaces. Then k · k is indeed a norm on the vector space B(X,Y) of all bounded linear operators from X into Y, and kABk ≤ kAkkBk, ∀B ∈ B(X,Y), ∀A ∈ B(Y,Z). (1.2) Moreover, if Y is complete then B(X,Y) is a Banach space. Let X be a Banach space. Theorem 1.10 says that B(X) := B(X,X) is actually what we call a Banach algebra. A vector space E is called an algebra if for any pair (x,y) ∈ E × E a unique product xy ∈E is defined with the properties (xy)z =x(yz), x(y+z)=xy+xz, (x+y)z =xz+yz, λ(xy)=(λx)y =x(λy), for all x,y,z ∈E and scalars λ. E is called an algebra with identity if it contains an element e such that for all x∈E, ex=xe=x, ∀x∈E. A normed algebra is a normed space which is an algebra such that kxyk≤kxkkyk, ∀x,y ∈E, and if E has an identity e, kek=1. A Banach algebra is a normed algebra which is complete, considered as a normed space . ˜ ˜ 1.11. Definition Let X and Y be subspaces of X and Y. An operator ˜ ˜ ˜ ˜ B : X → Y is said to be an extension of B : X → Y if Bx = Bx, ∀x ∈ X. 4 1.12. Theorem Let X and Y be Banach spaces and D be a dense linear subspace of X. Let A be a bounded linear operator from D (equipped with the X−norm) into Y. Then there exists a unique extension of A to ¯ a bounded linear operator A : X → Y defined on the whole X; moreover ¯ kAk = kAk. 1.13. Example Let X = Y be the space L ([a,b]), i.e. the completion 2 of D = (C([a,b]),k·k ), (−∞ < a < b < +∞). Consider the operator 2 A : C([a,b]) → C([a,b]) ⊂ L ([a,b]), 2 Z b (Af)(t) = k(t,τ)f(τ)dτ, where k ∈ C([a,b]2). a Theorem1.12allowsonetoextendthisoperatortoaboundedlinearoperator ¯ A : L ([a,b]) → L ([a,b]), since D is dense in its completion X = L ([a,b]). 2 2 2 We only need to prove that A : D → D is a bounded operator. Denote K = max |k(t,τ)|. (t,τ)∈[a,b]2 Using Cauchy–Schwarz inequality we obtain (cid:12) (cid:12) !1/2 !1/2 (cid:12)Z b (cid:12) Z b Z b |(Af)(t)| = (cid:12) k(t,τ)f(τ)dτ(cid:12) ≤ |k(t,τ)|2dτ |f(τ)|2dτ ≤ (cid:12) (cid:12) (cid:12) a (cid:12) a a √ K b−akfk , ∀f ∈ C([a,b]). 2 Therefore !1/2 Z b kAfk = |(Af)(t)|2dt ≤ K(b−a)kfk , ∀f ∈ C([a,b]), 2 2 a i.e. A is bounded and kAk ≤ K(b−a). (Let Y be the Banach space (C([a,b]),k·k ). Then the above proof shows ∞ ¯ that A can be extended to a bounded linear operator A : L ([a,b]) → √ 2 C([a,b]), kAk ≤ K b−a.) 1.14. Definition Let X be a normed space and x ∈ X, n ∈ IN. We n say that the series P∞ x is convergent to a vector x ∈ X if the sequence n=1 n (S ) of partial sums, S = Pj x , converges to x. We then write j j n=1 n ∞ X x = x. n n=1 5 We say that the series converges absolutely if ∞ X kx k < ∞. n n=1 1.15. Theorem Any absolutely convergent series in a Banach space is convergent. Proof: For m > k we have (cid:13) (cid:13) (cid:13) m (cid:13) m (cid:13) X (cid:13) X kSm −Skk = (cid:13) xn(cid:13) ≤ kxnk → 0, as k,m → ∞. (cid:13) (cid:13) (cid:13)n=k+1 (cid:13) n=k+1 Hencethesequence(S )isCauchyandthereforeconvergestosomex ∈ X. 2 j 2 Spectral theory of bounded linear operators Auxiliary results Let us recall that for any normed spaces X and Y we denote by B(X,Y) the space of all bounded linear operators acting from X into Y. We use also the following notation B(X) = B(X,X). 2.1. Definition Let A be a linear operator from a vector space X into a vector space Y. The kernel of A is the set Ker(A) := {x ∈ X : Ax = 0}. The range of the operator A is the set Ran(A) := {Ax : x ∈ X}. Ker(A) and Ran(A) are linear subspaces of X and Y correspondingly. (Why?) Moreover, if X and Y are normed spaces and A ∈ B(X,Y), then Ker(A) is closed (why?); this is not necessarily true for Ran(A). 6 2.2. Theorem (Banach) Let X and Y be Banach spaces and let B ∈ B(X,Y) be one-to-one and onto (i.e. Ker(B) = {0} and Ran(B) = Y). Then the inverse operator B−1 : Y → X is bounded, i.e. B−1 ∈ B(Y,X). Proof: The proof of this fundamental result can be found in any textbook on functional analysis. 2 Let X and Y be vector spaces and let an operator B : X → Y have a right inverse and a left inverse operators B−1,B−1 : Y → X, r l B−1B = I , BB−1 = I . l X r Y (By I we always denote the identity operator: Ix = x, ∀x ∈ X. Subscript (if any) indicates the space on which the identity operator acts.) Then B has a two-sided inverse operator B−1 = B−1 = B−1. Indeed, r l B−1 = I B−1 = (B−1B)B−1 = B−1(BB−1) = B−1I = B−1. r X r l r l r l Y l 2.3. Lemma Let X, Y and Z be vector spaces. (i) If operators B : X → Y and T : Y → Z are invertible, then TB : X → Z is invertible too and (TB)−1 = B−1T−1. (ii) If operators B,T : X → X commute: TB = BT, then TB : X → X is invertible if and only if both B and T are invertible. (iii) If operators B,T : X → X commute and B is invertible, then B−1 and T also commute. Proof: (i) B−1T−1TB = B−1B = I , TBB−1T−1 = TT−1 = I . X Z (ii)Accordingto(i)wehavetoproveonlythattheinvertibilityofTB implies the invertibility of B and T. Let S : X → X be the inverse of TB, i.e. STB = TBS = I. It is clear that ST is a left inverse of B. Since B and T commute, we have BTS = I, i.e. TS is a right inverse of B. Hence B is invertible and its inverse equals ST = TS. Similarly we prove that T is invertible. (iii) B−1T = B−1TBB−1 = B−1BTB−1 = TB−1. 2 7 2.4. Lemma Let X be a Banach space, B ∈ B(X) and kBk < 1. Then I −B is invertible, ∞ (I −B)−1 = XBn (2.1) n=0 and k(I −B)−1k ≤ 1/(1−kBk). Proof: It follows from (1.2) that kBnk ≤ kBn−1kkBk ≤ ··· ≤ kBkn, ∀n ∈ IN. Hence the series in the right-hand side of (2.1) is absolutely convergent and, consequently, convergent in B(X) (see Theorems 1.10 and 1.15). Let us denote its sum by R ∈ B(X). We have N (I −B)R = (I −B) lim XBn = lim (I −BN+1) = I, N→∞ N→∞ n=0 because kBN+1k ≤ kBkN+1 → 0 as N → ∞. Analogously we prove that R(I −B) = I. Thus (I −B)−1 = R ∈ B(X). Further, k(I −B)−1k = lim (cid:13)(cid:13)(cid:13)XN Bn(cid:13)(cid:13)(cid:13) ≤ lim XN kBkn = lim 1−kBkN+1 = N→∞(cid:13)(cid:13) (cid:13)(cid:13) N→∞ N→∞ 1−kBk n=0 n=0 1 . 2 1−kBk 2.5. Lemma Let X and Y be Banach spaces, A,B ∈ B(X,Y). Let A be invertible and kBk < 1/kA−1k. Then A+B is invertible, ∞ ! ∞ ! (A+B)−1 = X(−A−1B)n A−1 = A−1 X(−BA−1)n n=0 n=0 and kA−1k k(A+B)−1k ≤ 1−kBkkA−1k Proof: We have A+B = A(I +A−1B) = (I +BA−1)A and kA−1Bk ≤ kA−1kkBk < 1, kBA−1k ≤ kBkkA−1k < 1. Now it is left to apply Lemmas 2.3(i) and 2.4. 2 8 The spectrum In this subsection we will deal only with complex vector spaces, i.e. with the case IF = C, if it is not stated otherwise. 2.6. Definition Let X be a Banach space and B ∈ B(X). The resolvent set ρ(B) of the operator B is defined to be the set of all λ ∈ C such that B−λI has an inverse operator (B−λI)−1 ∈ B(X). The spectrum σ(B) of the operator B is the complement of ρ(B): σ(B) = C\ρ(B), i.e. σ(B) is the set of all λ ∈ C such that B −λI is not invertible on X. AcomplexnumberλiscalledaneigenvalueofB ifthereexistsx ∈ X\{0} such that Bx = λx. In this case x is called an eigenvector of B corresponding to the eigenvalue λ. 2.7. Lemma All eigenvalues of B belong to σ(B). Proof: Suppose λ is an eigenvalue and x 6= 0 is a corresponding eigenvector of B. Then (B −λI)x = 0. Since we have also (B −λI)0 = 0, the operator B −λI is not one–to–one. So, B −λI is not invertible on X, i.e. λ ∈ σ(B). 2 On the other hand the set of all eigenvalues may be smaller than the spectrum. 2.8. Examples 1. If the space X is finite-dimensional then the spectrum of B ∈ B(X) co- incides with the set of all its eigenvalues, i.e. with the set of zeros of the determinant of a matrix corresponding to B − λI (with respect to a given basis of X). 2. Let X be one of the Banach spaces C([0,1]), L ([0,1]), 1 ≤ p ≤ ∞, and p let B be defined by the formula Bf(t) = tf(t), t ∈ [0,1]. Then σ(B) = [0,1] but B does not have eigenvalues (exercise). 9 2.9. Lemma Let X be a Banach space and B ∈ B(X). Then σ(B) is a compact set and σ(B) ⊂ {λ ∈ C : |λ| ≤ kBk}. (2.2) Proof: Suppose |λ| > kBk. Then kBk < 1/|λ|−1 = 1/k−λ−1Ik = 1/k(−λI)−1k. Hence, according to Lemma 2.5 the operator B − λI is invertible, i.e. λ ∈/ σ(B). So, we have proved (2.2). Let us take an arbitrary λ ∈ ρ(B). Then for any λ ∈ C such that |λ−λ | < 0 0 1/k(B −λ I)−1k we conclude from Lemma 2.5 and the representation 0 B −λI = (B −λ I)+(λ −λ)I (2.3) 0 0 that the operator B−λI is invertible, i.e. λ ∈ ρ(B). Hence ρ(B) is an open set and its complement σ(B) is closed. So, σ(B) is a bounded (see (2.2)) closed set, i.e. a compact set. 2 2.10. Definition Let X be a Banach space and B ∈ B(X). The operator- valued function ρ(B) 3 λ → R(B;λ) := (B −λI)−1 ∈ B(X) is called the resolvent of the operator B. 2.11. Lemma (the resolvent equation) R(B;λ)−R(B;λ ) = (λ−λ )R(B;λ)R(B;λ ), ∀λ,λ ∈ ρ(B). 0 0 0 0 Proof: R(B;λ)−R(B;λ ) = (B −λI)−1 −(B −λ I)−1 = 0 0 (B −λI)−1((B −λ I)−(B −λI))(B −λ I)−1 = 0 0 (λ−λ )R(B;λ)R(B;λ ). 2 0 0 2.12. Lemma If operators A,B ∈ B(X) commute: AB = BA, then for any λ ∈ ρ(B) and µ ∈ ρ(A) the operators A, B, R(A;µ) and R(B;λ) all 10

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