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