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7. Operator Theory on Hilbert Spaces

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7. Operator Theory on Hilbert Spaces 7. Operator Theory on Hilbert spaces In this section we take a closer look at linear continuous maps between Hilbert spaces. These are often called bounded operators, and the branch of Functional Analysis that studies these objects is called “Operator Theory.” The standard notations in Operator Theory are as follows. Notations. If H1 and H2 are Hilbert spaces, the Banach space L(H1, H2) = {T : H1 → H2 : T linear continuous } will be denoted by B(H1, H2). In the case of one Hilbert space H, the space L(H, H) is simply denoted by B(H). Given T ∈ B(H1, H2) and S ∈ B(H2, H3), their composition S◦T ∈ B(H1, H3) will be simply denoted by ST . We know that B(H) is a unital Banach algebra. This Banach algebra will be our main tool used for investigating bounded operators. The general theme of this section is to study the interplay between the operator theoretic properties of a bounded linear operator, say T : H → H, and the algebraic properties of T as an element of the Banach algebra B(H). Some of the various concepts associated with bounded linear operators are actually defined using this Banach algebra. For example, for T ∈ B(H), we denote its spectrum in B(H) by SpecH(T ), that is, SpecH(T ) = {λ ∈ C : T − λI : H → H not invertible}, and we define its spectral radius radH(T ) = max |λ| : λ ∈ SpecH(T ) . We start with an important technical result, for which we need the following. Lemma 7.1. Let X and Y be normed vector spaces. Equip X×Y with the product topology. For a sesquilinear map φ : X × Y → C, the following are equivalent: (i) φ is continuous; (ii) φ is continuous at (0, 0); (iii) sup |φ(x, y)| :(x, y) ∈ X × Y, kxk, kyk ≤ 1 < ∞; (iv) there exists some constant C ≥ 0, such that |φ(x, y)| ≤ C · kxk · kyk, ∀ (x, y) ∈ X × Y. Moreover, the number in (iii) is equal to (1) min C ≥ 0 : |φ(x, y)| ≤ C · kxk · kyk, ∀ (x, y) ∈ X × Y . Proof. The implication (i) ⇒ (ii) is trivial. (ii) ⇒ (iii). Assume φ is continuous at (0, 0). We prove (iii) by contra- diction. Assume, for each integer n ≥ 1 there are vectors xn ∈ X and yn ∈ Y 2 1 with kxnk, kynk ≤ 1, but such that |φ(xn, yn)| ≥ n . If we take vn = n xn and 1 1 wn = n yn, then on the one hand we have kvnk, kwnk ≤ n , ∀ n ≥ 1, which forces limn→∞(vn, wn) = (0, 0) in X × Y, so by (iii) we have limn→∞ φ(vn, wn) = 0. On the other hand, we also have |φ(x , y )| |φ(v , w )| = n n ≥ 1, ∀ n ≥ 1, n n n2 300 §7. Operator Theory on Hilbert spaces 301 which is impossible. (iii) ⇒ (iv). Assume φ has property (iii), and denote the number sup |φ(x, y)| :(x, y) ∈ X × Y, kxk, kyk ≤ 1 simply by M. In order to prove (iv) we are going to prove the inequality (2) |φ(x, y)| ≤ M · kxk · kyk, ∀(x, y) ∈ X × Y. Fix (x, y) ∈ X×Y. If either x = 0 or y = 0, the above inequality is trivial, so we can 1 1 assume both x and y are non-zero. Consider the vectors v = kxk x and w = kyk y. We clearly have |φ(x, y)| = φ(kxkv, kykw) = kxk · kyk · |φ(v, w)|. Since kvk = kwk = 1, we have |φ(v, w)| ≤ M, so the above inequality gives (2). (iv) ⇒ (i). Assume φ has property (iv) and let us show that φ is continuous. Let C ≥ 0 is as in (iv). Let (xn)n→∞ ⊂ X and (yn)n→∞ ⊂ Y be convergent sequences with limn→∞ xn = x and limn→∞ yn = y, and let us prove that limn→∞ φ(xn, yn) = φ(x, y). Using (iv) we have |φ(xn, yn) − φ(x, y)| ≤ |φ(xn, yn) − φ(xn, y)| + |φ(xn, y) − φ(x, y)| = = |φ(xn, yn − y)| + |φ(xn − x, y)| ≤ ≤ C · kxnk · kyn − yk + C · kxn − xk · kyk, ∀ n ≥ 1, which clearly forces limn→∞ |φ(xn, yn) − φ(x, y)| = 0, and we are done. To prove the last assertion we observe first that every C ≥ 0 with |φ(x, y)| ≤ C · kxk · kyk, ∀ (x, y) ∈ X × Y, automatically satisfies the inequality C ≥ M. This is a consequence of the above inequality, restricted to those (x, y) ∈ X×Y, with kxk, kyk ≤ 1. To finish the proof, all we have to prove is the fact that C = M satisfies (iv). But this has already been obtained when we proved the implication (iii) ⇒ (iv). Notation. With the notations above, the number defined in (iii), which is also equal to the quantity (1), is denoted by kφk. This is justified by the following. Exercise 1. Let X and Y be normed vector spaces over C. Prove that the space S(X, Y) = φ : X × Y → C : φ sesquilinear continuous is a vector space, when equipped with pointwise addition and scalar multiplication. Prove that the map S(X, Y) 3 φ 7−→ kφk ∈ [0, ∞) defines a norm. With this terminology, we have the following technical result. Theorem 7.1. Let H1 and H2 be Hilbert spaces, and let φ : H1 × H2 → C be a sesquilinear map. The following are equivalent (equip H1 × H2 with the product topology): (i) φ is continuous; (ii) there exists T ∈ B(H1, H2), such that φ(ξ1, ξ2) = (T ξ1|ξ2)H2 , ∀ (ξ1, ξ2) ∈ H1 × H2, where ( . | . )H2 denotes the inner product on H2. 302 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS Moreover, the operator T ∈ B(H1, H2) is unique, and has norm kT k = kφk. Proof. (i) ⇒ (ii). Assume φ is continuous, so by Lemma 7.1 we have (3) |φ(ξ, ζ)| ≤ kφk · kξk · kζk, ∀ ξ ∈ H1, ζ ∈ H2. Fix for the moment ξ ∈ H1, and consider the map φξ : H2 3 ζ 7−→ φ(ξ, ζ) ∈ C. Using (3), it is clear that φξ : H2 → C is linear continuous, and has norm kφξk ≤ kφk·kξk. Using Riesz’ Theorem (see Section 3), it follows that there exists a unique ˜ vector ξ ∈ H2, such that ˜ φξ(ζ) = (ξ|ζ)H2 , ∀ ζ ∈ H2. Moreover, one has the equality ˜ (4) kξkH2 = kφξk ≤ kφk · kξkH1 . Remark that, if we start with two vectors ξ, η ∈ H1, then we have ˜ (ξ|ζ)H2 + (˜η|ζ)H2 = φ(ξ, ζ) + φ(η, ζ) = φ(ξ + η, ζ) = φξ+η(ζ), ∀ ζ ∈ H2, so by the uniqueness part in Riesz’ Theorem we get the equality ξ]+ η = ξ˜+η ˜. Likewise, if ξ ∈ H1, and λ ∈ C, we have ˜ ¯ ˜ ¯ (λξ|ζ)H2 = λ(ξ|ζ)H2 = λφ(ξ, ζ) = φ(λξ, ζ) = φλξ(ζ), ∀ ζ ∈ H2, which forces λξf = λξ˜. This way we have defined a linear map ˜ T : H1 3 ξ 7−→ ξ ∈ H2, with φ(ξ, ζ) = (T ξ|ζ)H2 , ∀ (ξ, ζ) ∈ H1 × H2. Using (4) we also have kT ξkH2 ≤ kφk · kξkH1 , ∀ x ∈∈ H1, so T is indeed continuous, and it has norm kT k ≤ kφk. The uniqueness of T is obvious. (ii) ⇒ (i). Assume φ has property (ii), and let us prove that φ is continuous. This is pretty clear, because if we take T ∈ B(H1, H2) as in (ii), then using the Cauchy-Bunyakovski-Schwartz inequality we have |φ(ξ1, ξ2)| = |(T ξ1|ξ2)H2 | ≤ kT ξ1k · kξ2k ≤ kT k · kξ1k · kξ2k, ∀ (ξ1, ξ2) ∈ H1 × H2, so we can apply Lemma 7.1. Notice that this also proves the inequality kφk ≤ kT k. Since by the proof of the implication (i) ⇒ (ii) we already know that kT k ≤ kφk, it follows that in fact we have equality kT k = kφk. Remark 7.1. For a sesquilinear map φ : H1 × H2 → C, the conditions (i), (ii) are also equivalent to (ii’) there exists S ∈ B(H2, H1), such that φ(ξ1, ξ2) = (ξ1|Sξ2)H1 , ∀ (ξ1, ξ2) ∈ H1 × H2. §7. Operator Theory on Hilbert spaces 303 Moreover, S ∈ B(H2, H1) is unique, and satisfies kSk = kT k. This follows from the Theorem, applied to the sesquilinear map ψ : H2 × H1 → C, defined by ψ(ξ2, ξ1) = φ(ξ1, ξ2), ∀ (ξ2, ξ1) ∈ H2 × H1. Notice that one clearly has kφk = kψk, so we also get kT k = kSk. In particular, if we start with an operator T ∈ B(H1, H2) and we consider the continuous sesquilinear map H1 × H2 3 (ξ1, ξ2) 7−→ (T ξ1|ξ2)H2 ∈ C, then there exists a unique operator S ∈ B(H2, H1), such that (5) (T ξ1|ξ2)H2 = (ξ1|Sξ2)H1 , ∀ (ξ1, ξ2) ∈ H1 × H2. This operator will satisfy kSk = kT k. Definition. Given an operator T ∈ B(H1, H2), the unique operator S ∈ ? B(H2, H1) that satisfies (5) is called the adjoint of T , and is denoted by T . By the above Remark, for any two vectors ξ1 ∈ H1, ξ2 ∈ H2, we have the identities ? (T ξ1|ξ2)H2 = (ξ1|T ξ2)H1 , ? (ξ2|T ξ1)H2 = (T ξ2|ξ1)H1 . (The second identity follows from the first one by taking complex conjugates.) The following result collects a first set of properties of the adjoint operation. Proposition 7.1. A. For two Hilbert spaces H1, H2, one has ? kT k = kT k, ∀ T ∈ B(H1, H2);(6) ? ? (T ) = T, ∀ T ∈ B(H1, H2);(7) ? ? ? (S + T ) = S + T ∀ S, T ∈ B(H1, H2);(8) ? ¯ ? (9) (λT ) = λT ∀ T ∈ B(H1, H2), λ ∈ C. B. Given three Hilbert spaces H1, H2, and H3, one has ? ? ? (10) (ST ) = T S ∀ T ∈ B(H1, H2),S ∈ B(H2, H3) Proof.
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