LECTURE 8 Hilbert-Schmidt operators Let A : H → H be a bounded operator in a Hilbert space H ∞ and let {en }∞ n=1 and {fn }n=1 be two orthonormal basises in H. Assume that XX |(Afk , ej )|2 < ∞. (∗) k Since XX k j |(Afk , ej )| = 2 X kAfk k2 k j we obtain that (∗) is independent of {en }∞ n=1 . Besides, X XX XX kA∗ ej k2 |(A∗ ej , fk )|2 = |(Afk , ej )|2 = k j k j j and therefore (∗) is independent of {fn }∞ n=1 . This allows us to introduce the following definition: Definition. Operator A satisfying (∗) is called HilbertSchmidt. The class of such operators is denoted by S2 and we introduce X 1/2 2 kAkS2 = kAej k . j Remark 1. Any operator of finite rank is Hilbert-Schmidt. Remark 2. kAkS2 introduced above satisfies all requirements of being a norm. 1 2 Theorem 1. Let A : L2 (R) → L2 (R) be an integral operator Z Af(x) = K(x, y)f(y) dy. R A is Hilbert-Schmidt iff K ∈ L2 (R × R) and kAkS2 = kKkL2 (R×R) . Remark 3. Hilbert-Schmidt operators are compact. (For integral operators this fact has been proved before, see Lecture 7.) Fredholm-Riesz-Schauder theory let T : H → H be a compact operator. Then we know that T ∗ is also compact. We now consider Nλ := {x ∈ H : λx−Tx = 0} N∗λ := {y ∈ H : λy−T ∗ y = 0} and Rλ := {y ∈ H : y = λx−Tx} R∗λ := {x ∈ H : x = λy−T ∗ y}. Lemma 1. If T is compact, then Nλ and N∗λ are finate dimensional subspaces of H for any λ ∈ C. Lemma 2. If T is compact, then Rλ and R∗λ are closed linear subspaces of H for any λ ∈ C. Theorem 2. (Fredholm-Riesz-Schauder) Rλ = (N∗λ )⊥ which means that the equation λx − Tx = y has a solution iff y ∈ (N∗λ )⊥ . 3 Resolvent set and Spectrum Let GA = {(x, Ax), x ∈ D(A)} ⊂ H ⊕ H be the graph of the operator A. We define W : H ⊕ H → H ⊕ H such that W(x.h) = (−x, h), x, h ∈ H. Obviously W 2 = I. Theorem 3. If D(A) = H, then (WGA )⊥ = GA∗ . Proof. The fact that (h, y) ∈ H ⊕ H is orthogonal to WGA implies −(x, h) + (Ax, y) = 0. Thus (Ax, y) = (x, h) and by definition of the adjoint operator we have y ∈ D(A∗ ) and h = A∗ y. Theorem 4. The operator A∗ is always closed. Proof. Indeed, the orthogonal complement to a linear subspace is always closed. Therefore GA∗ = (WGA )⊥ , which the graph of the operator A∗ , is a closed set. Theorem 5. If D(A) = H. Then the condition D(A∗ ) = H is equivalent to A being closable. Moreover, in this case A∗∗ exists and A∗∗ = A. Proof. By Theorem 3 and since W 2 = I we obtain h i⊥ h i⊥ 2 WGA∗ = W WGA = W GA = GA . Therefore h GA∗∗ = WGA∗ i⊥ = GA = GA . 4 Theorem 6. The subspaces R(A) and N(A∗ ) are orthogonal in H and H = R(A) ⊕ N(A∗ ). Proof. The element y ∈ N(A∗ ) if and only if (Ax, y) = 0 for all x ∈ D(A). This is equivalent to y ∈ R(A). 1. S PECTRUM AND RESOLVENT OF A CLOSED OPERATOR Let A be a closed operator in a Hilbert space H. Definition. dA = defA = dim [R(A)]⊥ is called the defect of the operator A. Remark 4. By using Theorem 6 we immediately obtain that dA = dim N(A∗ ). Theorem 7. Let us assume that A is a closed operator such for some constant C > 0 kAxk ≥ Ckxk for all x ∈ D(A). Let B be an operator in H such that D(A) ⊂ D(B) and for any x ∈ D(A) kBxk ≤ akAxk, where, a < 1. Then • A + B is closed on D(A) • k(A + B)xk ≥ (1 − a)Ckxk • dA+B = dA . Proof. The graph GA is closed w.r.t. the norm |x|A = kxk + kAxk. Therefore by using the triangle inequality we have (1 − a)|x|A ≤ |x|A+B ≤ (1 + a)|x|A . Therefore the norms | · |A and | · |A+B — and thus since GA is closed then GA+B is closed. This implies that A + B is closed. 5 Now k(A + B)xk ≥ kAxk − kBxk ≥ (1 − a)kAxk ≥ (1 − a)Ckxk, which proves the second statement of the theorem Assume for a moment that dA+B < dA . Then there exists f ∈ R(A)⊥ , f 6= 0, such that f⊥R(A + B)⊥ . This implies f ∈ R(A + B) and therefore there exists y ∈ D(A) s.t. f = (A + B)y. Since f⊥R(A) we have (f, Ay) = 0. kAyk2 = (Ay, Ay) = −(By, Ay) ≤ kBykkAyk ≤ akAyk2 , which gives a contradiction. If we assume that dA + B > dA , then we can find f = Ay, f 6= s.t. f⊥R(A + B) and thus (f, (A + B)y) = 0. Finally kAyk2 = (Ay, Ay) = −(By, Ay) ≤ kBykkAyk ≤ akAyk2 . Corollary 1. Let A be an operator in H satisfying the condition kAxk ≥ Ckxk, x ∈ D(A) and let B be a bounded operator s.t. kBk < C. Then dA+B = dA . Proof. kBxk ≤ kBkkxk ≤ kBkC−1 kAxk. We now apply Theorem 7 with a = kBkC−1 . Definition. The defect of the operator A − λI is denoted by dA (λ) and called the defect of A at λ. If A − λI has a bounded inverse on its image (A − λI)(H) namely k(A − λI)xk ≥ Ckxk for some C > 0 then λ is called a quasi-regular point of A. All such points are denoted by ρ^(A). 6 Lemma 3. Let A be a closed operator in H such that (kA − λI)xk ≥ C0 kxk for some C0 > 0 and ∀x ∈ D(A). Then D := {λ ∈ C : |λ − λ0 | < C0 } ⊂ ρ^(A) and dA (λ) is constant on D. Proof. If we write A − λI = (A − λ0 ) + (λ0 − λ)I then we complete the proof by using Corollary 1. We now immediately obtain the following result: Theorem 8. The set ρ^(A) ⊂ C is open and the value of dA is constant on each connected component of ρ^(A). Definition. If dA (λ) = 0 for some λ ∈ ρ^(A), then λ is called a regular point of A. In this case the operator (A − λI)−1 is bounded. The set of all regular points of A is called a resolvent set and denoted by ρ(A). Remark 5. The set ρ(A) is open. Definitions. • The set σ(A) = C \ ρ(A) is called the spectrum of theoperator A. •σ ^ (A) = C \ ρ^(A) is called the core of the spectrum. • The set σp (A) = {λ ∈ C : N(A−λI) 6= {0}} is called the point spectrum of A and λ ∈ σp is called the eigenvalue of A. • The set σc (A) = {λ ∈ C : R(A − λI) 6= R(A − λI)} is called the continuous spectrum of A. 7 Example. R1 d 1. Let A = 1i dt defined on D(A) = {x : −1 (|x 0 (t)|2 + |x(t)|2 ) dt, } ⊂ L2 (−1, 1). D(A) is a dense in L2 (−1, 1) set. Solutions of the equation Ax = λx are x(t) = eikt and σp (A) = k, k = 0, ±1, ±2, . . . . 2. Let A be defined as Ax(t) = tx(t) in L2 (0, ∞). The operator Ais bounded and its spectrum is continuous and equal σc (A) = [0, ∞). (For the proof see Lecture 15). Definition. Let D(A) = H. An operator A is called symmetric if (Ax, y) = (x, Ay) ∀x, y ∈ D(A). It follows from the definition that A ⊂ A∗ . Therefore A can be closed and in particular A = A∗∗ . Home exercises. 1. Let A be Hilbert-Schmidt. Show that kAn kS2 = kAknS2 . 2. A is caaled of trace class (A ∈ S1 ) is X |(Aek , ek )| < ∞, where {ek } is an orthonormal basis. Show that if A, B ∈ S1 , then the product AB ∈ S2 . d defined the class of functions {x : x ∈ 3. Let A0 = 1i dt ∞ C0 (R)}. Show that A0 is symmetric and closable. R1 d 4. Let A = 1i dt defined on {x : −1 (|x 0 (t)|2 + |x(t)|2 ) dt, } ⊂ L2 (−1, 1) such that x(−1) = x(1). Show that A is self-adjoint. 8 5. Describe the closure of the class of functions {x : x ∈ C∞ 0 (R)} with respect to Z |x 0 (t)|2 dt. R