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3. Hilbert Spaces 3. Hilbert spaces In this section we examine a special type of Banach spaces. We start with some algebraic preliminaries. Definition. Let K be either R or C, and let Let X and Y be vector spaces over K. A map φ : X × Y → K is said to be K-sesquilinear, if • for every x ∈ X, then map φx : Y 3 y 7−→ φ(x, y) ∈ K is linear; • for every y ∈ Y, then map φy : X 3 y 7−→ φ(x, y) ∈ K is conjugate linear, i.e. the map X 3 x 7−→ φy(x) ∈ K is linear. In the case K = R the above properties are equivalent to the fact that φ is bilinear. Remark 3.1. Let X be a vector space over C, and let φ : X × X → C be a C-sesquilinear map. Then φ is completely determined by the map Qφ : X 3 x 7−→ φ(x, x) ∈ C. This can be see by computing, for k ∈ {0, 1, 2, 3} the quantity k k k k k k k Qφ(x + i y) = φ(x + i y, x + i y) = φ(x, x) + φ(x, i y) + φ(i y, x) + φ(i y, i y) = = φ(x, x) + ikφ(x, y) + i−kφ(y, x) + φ(y, y), which then gives 3 1 X (1) φ(x, y) = i−kQ (x + iky), ∀ x, y ∈ X. 4 φ k=0 The map Qφ : X → C is called the quadratic form determined by φ. The identity (1) is referred to as the Polarization Identity. Remark that the Polarization Identity can also gives 3 1 X φ(y, x) = ikQ (x + iky), ∀ x, y ∈ X. 4 φ k=0 In particular, we see that the following conditions are equivalent (i) Qφ is real-valued, i.e. Qφ(x) ∈ R, ∀ x ∈ X; (ii) φ is sesqui-symmetric, i.e. φ(y, x) = φ(x, y), ∀ x, y ∈ X. Definition. Let K be one of the fields R or C, and let X be a vector space over K.A K-sesquilinear map φ : X × X → K is said to be positive definite, if • φ is sesqui-symmetric; • φ(x, x) ≥ 0, ∀ x ∈ X. As observed before, in the complex case, the second condition actually forces the first one. With this terminology, we have the following useful technical result. 79 80 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS Proposition 3.1 (Cauchy-Bunyakowski-Schwartz Inequality). Let K be a K- vector space, and let φ : X × X → K be a positive definite K-sesquilinear map. Then (2) |φ(x, y)|2 ≤ φ(x, x) · φ(y, y), ∀ x, y ∈ X. Moreover, if equality holds then either x = y = 0, or there exists some α ∈ K with φ(x + αy, x + αy) = 0. Proof. Fix x, y ∈ X, and choose a number λ ∈ K, with |λ| = 1, such that |φ(x, y)| = λφ(x, y) = φ(x, λy). Define the map F : K → K by F (ζ) = φ(ζλy + x, ζλy + x), ∀ ζ ∈ K. A simple computation gives F (ζ) = ζλζ¯λφ¯ (y, y) + +ζλφ(x, y)ζ¯λφ¯ (y, x) + φ(x, x) = = |ζ|2|λ|2φ(y, y) + ζλφ(x, y) + ζ¯λφ(x, y) + φ(x, x) = 2 = |ζ| φ(y, y) + ζ|φ(x, y)| + ζ¯|φ(x, y)| + φ(x, x), ∀ z ∈ R. In particular, when we restrict F to R, it becomes a quadratic function: 2 F (t) = at + bt + c, ∀ t ∈ R, where a = φ(y, y) ≥ 0, b = 2|φ(x, y)|, c = φ(x, x). Notice that we have F (t) ≥ 0, ∀ t ∈ R. This forces either a = b = 0, in which case the inequality (2) is trivial, or a > 0, and b2 − 4ac ≤ 0, which reads 4|φ(x, y)|2 − 4φ(x, x)φ(y, y) ≤ 0, and again (2) follows. Let us examine now when equality holds in (2). We assume that either x 6= 0 or y 6= 0. In fact we can also assume that φ(x, x) > 0 and φ(y, y) > 0, so the fact that we have equality in (2) is equivalent to b2 − 4ac = 0, which in terms of quadratic equations says that the equation F (t) = at2 + bt + c = 0 has a (unique) solution t0. If we take α = t0λ, then the equality F (t0) = 0 is precisely φ(αy + x, αy + x) = 0. Definition. An inner product on X is a positive definite K-sesquilinear map X × X 3 (ξ, η) 7−→ ξ η ∈ K, which is non-degenerate, in the sense that • if ξ ∈ X satisfies ξ ξ = 0, then ξ = 0. Proposition 3.2. Let · · be an inner product on the K-vector space X. (i) If we define, for every ξ ∈ X the quantity kξk = p(ξ|ξ), then the map X 3 ξ 7−→ kξk ∈ [0, ∞) defines a norm on X. §3. Hilbert spaces 81 (ii) For every ξ, η ∈ X, one has the Cauchy-Bunyakovski-Schwartz inequality: (3) ξ η ≤ kξk · kηk. Moreover, if equality holds then ξ and η are proportional, in the sense that either ξ = 0, or η = 0, or ξ = λη, for some λ ∈ K. Proof. First of all, remark that by the non-degeneracy of the inner product, one has the implication (4) kξk = 0 ⇒ ξ = 0. The inequality (3) is immediate from Proposition 3.1. Moreover, if we have equality, then by Proposition 3.1, we also know that either ξ = 0, or ξ + αη ξ + αη = 0, for some α ∈ K. This is equivalent to kξ + αηk2 = 0, so by (4), we must have ξ = −αη. Finally, in order to check that k . k is a norm, we must prove (a) kλξk = |λ| · kξk, ∀ λ ∈ K, ξ ∈ X; (b) kξ + ηk ≤ kξk + kηk, ∀ ξ, η ∈ X. Property (a) is trivial, by seqsquilinearity: 2 2 2 kλξk = λξ λξ = λλ¯ λξ λξ = |λ| · kξk . Finally, for ξ, η ∈ X, we have 2 kξ + ηk = ξ + η ξ + η = ξ ξ + η η + ξ η + η ξ = 2 2 2 2 = kξk + kηk + ξ η + ξ η = kξk + kηk + 2Re ξ η . We now use the C-B-S inequality (3), which allows us to continue the above estimate to get 2 2 2 2 2 kξ + ηk = kξk + kηk + 2Re ξ η ≤ kξk + kηk + 2 ξ η ≤ 2 ≤ kξk2 + kηk2 + 2kξk · kηk = kξk + kηk , so we immediately get kξ + ηk ≤ kξk + kηk. Exercise 1. Use the above notations, and assume we have two vectors ξ, η 6= 0, such that kξ +ηk = kξk+kηk. Prove that there exists some λ > 0 such that ξ = λη. Lemma 3.1. Let X be a K-vector space, equipped with an inner product. (ii) [Parallelogram Law] kξ + ηk2 + kξ − ηk2 = 2kξk2 + 2kηk2. (i) [Polarization Identities] (a) If K = R, then 1 2 2 ξ η = kξ + ηk − kξ − ηk , ∀ ξ, η ∈ X. 4 (b) If K = R, then 3 1 X −k k 2 ξ η = i kξ + i ηk , ∀ ξ, η ∈ X. 4 k=0 Proof. (i). This is obvious, by the computations from the proof of Proposition 3.2, which give 2 2 2 kξ ± ηk = kξk + kηk ± 2Re ξ η . 82 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS (ii).(a). In the real case, the above identity gives 2 2 2 kξ ± ηk = kξk + kηk ± 2 ξ η , so we immediately get 2 2 kξ + ηk − kξ − ηk = 4 ξ η . (b). This case is a particular instance of the Polarization Identity discussed in Remark 3.1. Corollary 3.1. Let X be a K-vector space equipped with an inner product · · . Then the map X × X 3 (ξ, η) 7−→ ξ η ∈ K is continuous, with respect to the product topologies. Proof. Immediate from the polarization identities. Corollary 3.2. Let X and Y be two K-vector spaces equipped with inner products · · and · · . If T : X → is an isometric linear map, X Y K then T ξ T η = ξ η , ∀ ξ, η ∈ X. Y X Proof. Immediate from the polarization identities. Exercise 2. Let X be a normed K-vector space. Assume the norm satisfies the Parallelogram Law. Prove that there exists an inner product · · on X, such that q kξk = ξ ξ , ∀ ξ ∈ X. Hint: Define the inner product by the Polarization Identity, and then prove that it is indeed an inner product. Proposition 3.3. Let X be a K-vector space, equipped with an inner product · · . Let Z be the completion of X with respect to the norm defined by the X inner product. Then Z carries a unique inner product · · , so that the norm Z on Z is defined by · · . Moreover, this inner product extends · · , in the Z X sense that hξi hηi = ξ η , ∀ ξ, η ∈ X. Z X Proof. It is obvious that the norm on Z satisfies the Parallelogram Law. We then apply Exercise 2. Definitions. Let K be one of the fields R or C.A Hilbert space over K is a K-vector space, equipped with an inner product, which is complete with respect to the norm defined by the inner product. Some textbooks use the term Euclidean for real Hilbert spaces, and reserve the term Hilbert only for the complex case. Examples 3.1. For J a non-empty set, the space `2 (J) is a Hilbert space. We K know that this is a Banach space. The inner product defining the norm is X α β = α(j)β(j), ∀ α, β ∈ `2 (J). K j∈J §3. Hilbert spaces 83 (The fact that function αβ : J → belongs to `1 (J) is discussed in Section 1, K K Exercise 5.) More generally, a Banach space whose norm satisfies the Parallelogram Law is a Hilbert space.
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