HS I c Gabriel Nagy Hilbert Spaces I: Basic Properties Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we introduce an important class of Banach spaces, which carry some additional geometric structure, that enables us to use our two- or three-dimensional intuition. Convention. Throughout this note all vector spaces are over C. A. Algebraic Preliminaries: Sesquilinear Forms Definition. Given two vector spaces X and Y, a map φ : X × Y → C is said to be a sesquilinear form on X × Y, if: • for every x ∈ X , the map Y 3 y 7−→ φ(x, y) ∈ C is linear; • for every y ∈ Y, the map X 3 x 7−→ φ(x, y) ∈ C is conjugate linear, meaning that the conjugate map X 3 x 7−→ φ(x, y) ∈ C is linear. Exercise 1♥. Given a sesquilinear form φ on X × Y, prove that the map ? φ : Y × X 3 (y, x) 7−→ φ(x, y) ∈ C is a sesquilinear form on Y × X . This map is referred to as the adjoint of φ. For the remainder of this sub-section we are going to restrict our attention to the case when X = Y. In this case Lemmas 1-3 contain three easy but fundamental results. Lemma 1 (The Parallelogram Law). If X is a vector space and φ : X × X → C is a sesquilinear form, then φ(x + y, x + y) + φ(x − y, x − y) = 2φ(x, x) + φ(y, y), ∀ x, y ∈ X . (1) Proof. This follows directly, using the properties of sesquilinear forms, which yield φ(x + y, x + y) = φ(x, x) + φ(x, y) + φ(y, x) + φ(y, y), φ(x − y, x − y) = φ(x, x) − φ(x, y) − φ(y, x) + φ(y, y), for all x, y ∈ X . Lemma 2 (The Polarization Identity). If X is a vector space and φ : X × X → C is a sesquilinear form, then 3 1 X φ(x, y) = i−kφ(x + iky, x + iky), ∀ x, y ∈ X . (2) 4 k=0 1 Proof. Denote by T4 the set of roots of unity of order 4, that is, T4 = {±1, ±i}, so that the sum in the right-hand side of (2) simply reads P ζφ¯ (x + ζy, x + ζy). ζ∈T4 By the properties of sesquilinear forms, we have φ(x + ζy, x + ζy) = φ(x, x) + φ(ζy, x) + φ(x, ζy) + φ(ζy, ζy) = = φ(x, x) + ζφ¯ (y, x) + ζφ(x, y) + ζζφ¯ (y, y), ∀ x, y ∈ X , ζ ∈ C. If we specialize to the case when ζ ∈ T, i.e. |ζ| = 1(= ζζ¯ ), then the above calculation yields ζφ¯ (x + ζy, x + ζy) = ζφ¯ (x, x) + ζ¯2φ(y, x) + φ(x, y) + ζφ¯ (y, y), and then the sum in the right-hand side of (2) is: 3 X X X i−kφ(x + iky, x + iky) = ζ¯ · φ(x, x) + φ(y, y) + ζ¯2 · φ(y, x) + 4φ(x, y), k=0 ζ∈T4 ζ∈T4 so by the obvious identities P ζ¯ = P ζ¯2 = 0, the desired identity (2) follows. ζ∈T4 ζ∈T4 Definitions. Let X be a vector space, and let φ : X × X → C be a sesquilinear form. A. We say that φ is hermitian, if φ? = φ, which means: φ(x, y) = φ(y, x), ∀ x, y ∈ X . (3) B. We say that φ is positive definite, if: φ(x, x) ≥ 0, ∀ x ∈ X . (4) C. We say that φ is strictly positive definite, if: (i) φ is positive definite, and (ii) φ(x, x) = 0 ⇒ x = 0. Remarks 1-3. Let φ : X × X → C be a sesquilinear form. 1. The condition that φ is hermitian is equivalent to the condition: φ(x, x) ∈ R, ∀ x ∈ X . (5) The implication “hermitian” ⇒ (5) is trivial. Conversely, if φ satisfies (5), then (using the notations from the proof of Lemma 2), by the Polarization Identity we have 1 X 1 X φ(y, x) = ζφ(y + ζx, y + ζx) = ζφ(y + ζx, y + ζx), 4 4 ζ∈T4 ζ∈T4 so using the “change of variable” γ = ζ¯, we also get 1 X φ(y, x) = γ¯φ(y +γx, ¯ y +γx ¯ ), (6) 4 γ∈T4 2 and now the desired identity (3) follows from the properties of sesquilinear forms, which imply φ(y +γx, ¯ y +γx ¯ ) = φγ¯(x + γy), γ¯(x + γy) = = γγφ¯ (x + γy, x + γy) = φ(x + γy, x + γy), ∀ γ ∈ T4, so when we go back to (6), again by the Polarization Identity we get 1 X φ(y, x) = γφ¯ (x + γy, x + γy) = φ(x, y). 4 γ∈T4 2. If φ is hermitian, then it satisfies φ(x + y, x + y) = φ(x, x) + φ(y, y) + 2Re φ(x, y), (7) φ(x − y, x − y) = φ(x, x) + φ(y, y) − 2Re φ(x, y), (8) for all x, y ∈ X . These two identities are referred to as the Law of Cosine. 3. If φ is positive definite, then φ is hermitian. This follows immediately from Remark 1. Lemma 3 (Cauchy-Buniakowski-Schwartz Inequality). If φ : X × X → C, then |φ(x, y)|2 ≤ φ(x, x) · φ(y, y), ∀ x, y ∈ X . (9) Furthermore, if φ(x, x) 6= 0, then one has equality in (9), if and only if there exists ζ ∈ C, such that φ(ζx + y, ζx + y) = 0. (10) Proof. Fix x and y, and note that, by the Law of Cosine, one has 2 2 0 ≤ φ(αx + βy, αx + βy) = |α| φ(x, x) + 2Re αβφ¯ (x, y) + |β| φ(y, y), ∀ α, β ∈ C. (11) Fix now β ∈ C, with |β| = 1, such that βφ(x, y) = |φ(x, y)|, and define the function g : R 3 t 7−→ φ(tx + βy, tx + βy) ∈ R, which by (11) can equivalently be written as 2 g(t) = t φ(x, x) + 2t|φ(x, y)| + φ(y, y), t ∈ R. (12) By positive definiteness, we also know that g(t) ≥ 0, ∀ t ∈ R. (13) We now have two cases to consider: (i) φ(x, x) = 0; (ii) φ(x, x) > 0. In case (i) – when g is a polynomial of degree 1 – condition (13) forces g to be constant, so φ(x, y) = 0, and then (9) is trivial. In case (ii) it follows that g is a quadratic polynomial (with positive leading coefficient), so condition (13) forces the discriminant ∆ to be non-positive, i.e. 4|φ(x, y)|2 − 4φ(x, x) · φ(y, y) ≤ 0, from which (9) is trivial. Furthermore, if equality holds in (9), this means that 3 ∆ = 0, so the equation g(t) = 0 has a unique real solution t0, which means precisely that φ(t0x + βy, t0x + βy) = 0, so we also get ¯ ¯ ¯ ¯ ¯ φ(t0βx + y, tβx + y) = φ β(t0x + βy), β(t0x + βy) = ββφ(t0x + βy, t0x + βy) = 0, ¯ so (10) holds with ζ = t0β. Conversely, if (10) holds for some ζ, then by (9) it follows that 0 ≤ |φ(x, ζx + y)|2 ≤ φ(x, x) · φ(ζx + y, ζx + y) = 0, 0 ≤ |φ(y, ζx + y)|2 ≤ φ(y, y) · φ(ζx + y, ζx + y) = 0, which give φ(x, ζx + y) = φ(y, ζx + y) = 0, so we get φ(x, y) = −ζφ(x, x) and φ(y, y) = −ζφ(y, x) = −ζφ(x, y) = ζζφ¯ (x, x), thus clearly forcing |φ(x, y)|2 = |ζ|2φ(x, x)2 = φ(x, x) · φ(y, y). The final result in this sub-section establishes the characterization of positive definite sesquilinear forms in terms of convexity. Theorem 1. Let X be a C-vector space. For a map q : X → [0, ∞), the following are equivalent. (i) There exists a positive definite sesquilinear form phi : X × X → C, such that q(x) = pφ(x, x), ∀ x ∈ X . (14) (ii) q is a seminorm on X , which satisfies the Parallelogram Law: q(x + y)2 + q(x − y)2 = 2q(x)2 + 2q(y)2, ∀ x, y ∈ X . (15) Moreover, φ is unique and satisfies the Polarization Identity: 3 1 X φ(x, y) = i−kq(x + iky)2, (16) 4 k=0 Proof. (i) ⇒ (ii). Assume q is defined by (14), and let us prove first that q is a seminorm. Since φ(ζx, ζx) = ζζφ¯ (x, x) = |ζ|2φ(x, x), by taking square roots, it follows that q(ζx) = |ζ| · q(x), ∀ ζ ∈ C, x ∈ X . To prove the triangle inequality, we use the Law of Cosine and the Cauchy-Buniakowski- Schwartz Inequality, which gives: φ(x + y, x + y) = |φ(x + y, x + y)| = |φ(x, x)| + 2Re φ(x, y) + φ(y, y)| ≤ = |φ(x, x) + 2|φ(x, y)| + φ(y, y) ≤ ≤ φ(x, x) + 2pφ(x, x) · φ(y, y) + φ(y, y) = pφ(x, x) + pφ(y, y)2, so by taking square roots we get q(x + y) ≤ q(x) + q(y), ∀ x, y ∈ X . 4 The fact that q satisfies the Parallelogram Law (15), as well as the Polarization Identity (16) is immediate from Lemmas 1 and 2. (ii) ⇒ (i). Assume q be a seminorm that satisfies the Parallelogram Law (15), define φ : X × X → C by (16), and let us prove that φ is positive definite, and it also satisfies (14). Claim 1: φ(y, x) = φ(x, y), ∀ x, y ∈ X . Fix x, y ∈ X , so by definition, we have 3 1 X φ(y, x) = ikq(y + ikx)2. (17) 4 k=0 Since q is a seminorm, we have q(y + ikx) = qik(x + i−ky) = |ik| · q(x + i−ky) = q(x + i−ky), 1 P3 k −k 2 ∀ k = 0,..., 3, so if we go back to (17) we now have φ(y, x) = 4 k=0 i q(x + i y) , and using the notations from Lemma 2 we have 1 X 1 X φ(y, x) = ζq(x + ζy¯ )2 = ζq¯ (x + ζy)2 = φ(x, y).