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Chapter 4 Hilbert Space 4.1 Inner Product Space Let E be a complex vector space, a mapping (·, ·) : E × E → is called an inner product on E if i) (x, x) ≥ 0 ∀x ∈ E and (x, x) = 0 if and only if x = 0; ii) (·, x) is linear on E for each x ∈ E; and iii) (x, y) = (y, x) for all x, y in E. With such an inner product E is called an inner product space. If we write x for (x, x), then · is a norm on E and hence E is a normed vector space. For this fact, we show ﬁrst Schwarz Inequality |(x, y)| ≤ x · y , for x, y ∈ E. Proof. For t ∈ Ê we have 0 ≤ (x + ty, x + ty) = (x, x + ty) + (ty, x + ty) 2 = x + 2t Re(x, y) + t2 y 2 , from which we necessarily have | Re(x, y)|2 ≤ x 2 · y 2 , and hence | Re(x, y)| ≤ x · y . In the above, replace y by θy for some θ ∈ with |θ| = 1 and Re(x, θy) = |(x, y)|, then |(x, y)| = Re(x, θy) ≤ x · θy = x · y . ¿From Schwarz inequality, it follows that 2 2 2 2 2 x+y = x + 2 Re(x, y) + y ≤ x +2 x · y + y = ( x + y )2 , 41 42 CHAPTER 4. HILBERT SPACE or x+y ≤ x + y , i.e. triangular inequality holds for · . Hence · is a norm on E. For an inner product space E, the norm on E is the norm so deﬁned unless stated otherwise. Examples. n i) Let E = ′ ′ . For z = (z1 , · · · , zn ) and z ′ = (z1 , · · · , zn ) in n let n (z, z ′ ) = zj z ′ j . j=1 Æ ii) Let E = l2 ( ) = {(z1 , z2 , · · · ) : ∞ j=1 |zj |2 < ∞}. For z = (z1 , z2 , · · · ), ′ ′ Æ and z ′ = (z1 , z2 , · · · ) in l2 ( ) let ∞ (z, z ′ ) = zj z ′ j . j=1 Æ The space E = l2 ( ) will be hereafter simply denoted by l2 . iii) Let E = L2 (Ω, Σ, µ). For f , and g in L2 (Ω, Σ, µ), deﬁne (f, g) := f gdµ. Ω Æ Exercise 4.1.1. i) For z, z ′ ∈ l2 ( ), s how that {|zj z ′ j |}j∈Æ is summable and hence (z, z ′ ) deﬁned above is absolutely convergent. Æ ii) Show that l2 ( ) is complete. An inner product space is called a Hilbert space if it is complete. Both n Æ and l2 ( ) are Hilbert spaces. Since L2 (Ω, Σ, µ) is complete with respect to the metric deﬁned by L2 -norm and since L2 -norm is given by the inner product deﬁned in Example iii), L2 (Ω, Σ, µ) is a Hilbert space of which both n and Æ l2 ( ) are special cases. Exercise 4.1.2. Deﬁne real inner product space and real Hilbert space. 4.2 Geometry in Hilbert Space Theorem 4.2.1. Let E be an inner product space, and M a complete convex subset of E. Suppose x ∈ E, then the following are equivalent: 1) y ∈ M satisﬁes x − y = minz∈M x − z ; 2) y ∈ M satisﬁes Re(y − x, y − z) ≤ 0 ∀ z ∈ M . Furthermore, there is unique y ∈ M satisfying 1) and 2). 4.2. GEOMETRY IN HILBERT SPACE 43 Proof. 1) ⇒ 2): For z ∈ M and 0 < θ ≤ 1, let 2 2 f (θ) = x − {(1 − θ)y + θz} = x − y + θ(y − z) 2 = x−y + θ2 y − z 2 + 2θ Re(x − y, y − z). 2 Since f (θ) ≥ f (0) = x − y for 0 < θ ≤ 1, we have f (θ) − f (0) lim = 2 Re(x − y, y − z) ≥ 0. θ↓0 θ 2) ⇒ 1): For z ∈ M we have Re(y − x, y − z) = − Re(x − y, y − x + x − z) 2 = x−y − Re(x − y, x − z) ≤ 0, hence 2 x−y ≤ Re(x − y, x − z) ≤ x − y · x − z , and consequently x−y ≤ x−z for all z ∈ M . That there exists at most one such y follows from 2), for if both y1 and y2 satisfy 2) for all z ∈ M , then 2 0 ≤ y1 − y2 = (y1 − y2 , y1 − y2 ) = (y1 − x, y1 − y2 ) + (y2 − x, y2 − y1 ) = Re(y1 − x, y1 − y2 ) + Re(y2 − x, y2 − y1 ) ≤ 0, so y1 = y2 . To show that there exists y satisfying 1), let α = inf x−z . z∈M Consider then a sequence {zn } ⊂ M satisfying 1 α2 ≤ x − zn 2 ≤ α2 + . n We claim that {zn } is a Cauchy sequence. We have 2 2 zn − zm = (zn − x) − (zm − x) 2 2 = zn − x + zm − x − 2 Re(zn − x, zm − x) ; 2 zn + zm 2 2 4 −x = zn − x + zm − x + 2 Re(zn − x, zm − x) , 2 44 CHAPTER 4. HILBERT SPACE consequently 2 2 2 2 zn + zm zn − zm = 2 zn − x + 2 zm − x −4 −x 2 1 1 1 1 ≤ 2 α2 + + 2 α2 + − 4α2 = 2 + , n m n m which shows that {zn } is a Cauchy sequence. Now since M is complete, there is y ∈ M with y = limn→∞ zn . Obviously x − y = limn→∞ x − zn = α. The map t : E → M deﬁned by tx = y, where y is the unique element in M which satisﬁes 1) and 2) of Theorem 1 is called the projection from E onto M and is more precisely denoted by tM if necessary. Theorem 2.1 is usually applied in the special case when M is a closed convex subset of a Hilbert space. Corollary 4.2.1. Let M be a closed convex subset of a Hilbert space E, then t = tM has the following properties: i) t2 = t (t is idempotent); ii) tx − ty ≤ x − y (t is contractive); and iii) Re(tx − ty, x − y) ≥ 0 (t is monotone). Proof. i) is obvious. ii): From Re(tx − x, tx − ty) ≤ 0 and Re(ty − y, ty − tx) ≤ 0 we obtain Re(x − y − (tx − ty), tx − ty) ≥ 0 , hence tx−ty 2 ≤ Re(x−y, tx−ty) ≤ x−y · tx−ty , from which tx−ty ≤ x − y follows. iii) : Again from Re(x − y − (tx − ty), tx − ty) ≥ 0 we have 2 0 ≤ tx − ty ≤ Re(x − y, tx − ty). Exercise 4.2.1. If M is a closed convex cone of a Hilbert space E and x ∈ E, then y = tx if and only if Re(x − y, y) = 0 and Re(x − y, z) ≤ 0 for all z ∈ M . Note : A convex set M in a vector space is called a convex cone if αx ∈ M for x ∈ M and α > 0. Exercise 4.2.2. Let M be a closed convex cone in a Hilbert space E and let N = {y ∈ E : Re(y, x) ≤ 0 ∀ x ∈ M }. Put t = tM and s = tN . Show that i) s = 1 − t, 1 being the identity map of E. ii) t(λx) = λtx if λ ≥ 0 (t is positively homogeneous), iii) x 2 = tx 2 + sx 2 , x ∈ E, 4.3. LINEAR TRANSFORMATION 45 iv) N = {x ∈ E : tx = 0}, M = {x ∈ E : sx = 0}. v) Re(tx, sx) = 0 and x = tx + sx; conversely if x = y + z, y ∈ M , z ∈ N and Re(y, z) = 0, then y = tx, z = sx. In the remaining part of the exercise, suppose that M is a closed vector subspace of E. Show that vi) N = M ⊥ := {y ∈ E : (y, x) = 0 ∀ x ∈ M }. vii) both t and s are continuous and linear. viii) M = tE = ker s; N = ker t = sE. ix) (tx, y) = (x, ty) ∀ x, y ∈ E. x) tx and sx are the unique elements y ∈ M and z ∈ M ⊥ such that x = y +z. 4.3 Linear transformation In this section we consider a linear transformation T from a normed vector space X into a normed vector space Y over the same ﬁeld Ê or . Exercise 4.3.1. Show that T is continuous on X if and only if it is continuous at one point. Theorem 4.3.1. T is continuous if and only if there is C ≥ 0 such that Tx ≤ C x (4.1) for all x ∈ X. Proof. If there is C ≥ 0 such that (3.1) holds for all x ∈ X, then T is obviously continuous at x = 0 and hence by Exercise 3.1 it is continuous on X. Conversely, suppose that T is continuous on X, and is hence continuous at x = 0. There is then δ > 0 such that if x ≤ δ, then T x ≤ 1. Let now δ δ 1 x ∈ X and x = 0, then x x = δ, so T ( x x) ≤ 1. Thus T x ≤ δ x . 1 If we choose C = δ , then (3.1) holds for x = 0. But when x = 0, (3.1) holds always. ¿From this theorem it follows that if T is a continuous linear transformation from X into Y , then Tx T := sup < +∞ (4.2) x∈X , x=0 x and is the smallest C for which (3.1) holds. T is called the norm of T . Of course, T can be deﬁned for any linear transformation T from X into Y , and T is continuous if and only if T < +∞. Hence a continuous linear transformation is also called a bounded linear transformation. 46 CHAPTER 4. HILBERT SPACE Exercise 4.3.2. Show that T = supx∈X, x =1 Tx . Exercise 4.3.3. Let L(X, Y ) be the space of all bounded linear transformations from X into Y . Show that it is a normed vector space with norm given by (3.2). Theorem 4.3.2. If Y is a Banach space, then L(X, Y ) is a Banach space. Proof. We will show that L(X, Y ) is complete. Let {Tn } be a Cauchy se- quence in L(X, Y ). Since Tn x − Tm x = (Tn − Tm )x ≤ Tn − Tm · x , {Tn x} is a Cauchy sequence in Y for each x ∈ X. Put T x = limn→∞ Tn x. T is obviously a linear transformation from X into Y . We claim now T ∈ L(X, Y ). Since {Tn } is Cauchy Tn ≤ C for some C > 0, and for all n. Now T x = lim Tn x ≤ lim inf Tn · x n→∞ n→∞ ≤ sup Tn x ≤C x n for each x ∈ X. Hence T is a bounded linear transformation. We show next limn→∞ Tn − T = 0. Given ε > 0, there is n0 such that Tn − Tm < ε if n, m ≥ n0 . Let n ≥ n0 , we have Tn − T = sup Tn x − T x x∈X, x =1 = sup lim Tn x − Tm x x∈X, x =1 m→∞ ≤ sup lim inf Tn − Tm · x x∈X, x =1 m→∞ ≤ sup ε x = ε, x∈X, x =1 this shows that limn→∞ Tn − T = 0, or limn→∞ Tn = T . Thus the sequence {Tn } has a limit in L(X, Y ). Ê L(X, ), or L(X, ), depending on whether X is a complex or a real vector space, is called the topological dual of X and is denoted by X ′ . X ′ is a Banach space. Theorem 4.3.3. (Riesz Representation Theorem) Let X be a Hilbert space and ℓ ∈ X ′ , then there is y0 ∈ X such that ℓ(x) = (x, y0 ) for x ∈ X. Furthermore, the mapping ℓ → y0 is conjugate linear and ℓ = y0 . 4.4. LEBESGUE-NIKODYM THEOREM 47 Proof. We may assume ℓ = 0. Let M = ker ℓ, then M ⊥ is one dimensional. For x ∈ X, x can be uniquely expressed as x = v + λx0 , where x0 is a ﬁxed non-zero element of M ⊥ , v ∈ M and λ a scalar. We have ℓ(x) = ℓ(v) + λℓ(x0 ) = λℓ(x0 ) and (x, x0 ) = (v + λx0 , x0 ) = λ x0 2 . ℓ(x0 ) Hence if we let y0 = x0 2 x0 , then (x, y0 ) = λℓ(x0 ) = ℓ(x). All the other assertions are obvious. 4.4 Lebesgue-Nikodym Theorem Let (Ω, Σ, µ) be a measure space and f a Σ-measurable function on Ω. Suppose that Ω f dµ has a meaning, then the set function ν deﬁned by ν(A) = f dµ, A ∈ Σ A is called the indeﬁnite integral of f . Then ν(∅) = 0 and ν is σ-additive i.e. if {An } ⊂ Σ is a disjointed sequence, then ν An = ν(An ). n n It enjoys also the following property: ν(A) = 0 whenever A ∈ Σ and µ(A) = 0. This fact suggests the following deﬁnition of absolute continuity of a measure with respect to another measure. Let (Ω, Σ, µ) and (Ω, Σ, ν) be measure spaces, ν is said to be absolutely continuous w.r.t. µ if ν(A) = 0 whenever A ∈ Σ and µ(A) = 0. Theorem 4.4.1. (Lebesgue-Nikodym Theorem) Let (Ω, Σ, µ) and (Ω, Σ, ν) be measure spaces with µ(Ω) < +∞ and ν(Ω) < +∞. Suppose that ν is absolutely continuous w.r.t. to µ, then there is a unique h ∈ L1 (Ω, Σ, µ) such that ν(A) = hdµ, A ∈ Σ. A Furthermore, h ≥ 0 µ − a.e. Proof. Let ρ = µ + ν, then ρ is a ﬁnite measure on Σ. Consider the real Hilbert space L2 (Ω, Σ, ρ) and consider the linear functional ℓ on L2 (Ω, Σ, ρ) deﬁned by ℓ(f ) = f dν. 48 CHAPTER 4. HILBERT SPACE Since 1/2 1/2 1/2 |ℓ(f )| ≤ |f |2 dν 1dν ≤ ν(Ω)1/2 |f |2 dρ = ν(Ω)1/2 f L2(ρ) , ℓ is a bounded linear functional on L2 (Ω, Σ, ρ). By Riesz Representation Theo- rem there is unique g ∈ L2 (Ω, Σ, ρ) such that f dν = f gdρ = f gdµ + f gdν for all f ∈ L2 (Ω, Σ, ρ), or f (1 − g)dν = f gdµ (4.3) for all f ∈ L2 (Ω, Σ, ρ). Claim 1. 0 ≤ g(x) < 1 for ρ − a.e. x on Ω. Let A1 = {x ∈ Ω : g(x) < 0} and A2 = {x ∈ Ω : g(x) ≥ 1}. If we let f = χA1 in (4.1), then 0 ≤ ν(A1 ) ≤ A1 (1 − g)dν = A1 gdµ, from which follows that µ(A1 ) = 0 and hence ν(A1 ) = 0. Thus ρ(A1 ) = 0. Now in (4.1) choose f = χA2 , we have 0 ≥ A2 (1 − g)dν = A2 gdµ ≥ µ(A2 ). This implies µ(A2 ) = 0, hence ν(A2 ) = 0. Consequently, ρ(A2 ) = 0. Thus Claim 1 is established. Claim 2. (4.1) holds for all Σ-measurable and ρ − a.e. non-negative functions f. For each positive integer n, let fn = f ∧n. Since 1−g > 0 and g ≥ 0 ρ−a.e., 0 ≤ fn (1 − g) ր f (1 − g), and 0 ≤ fn g ր f g, then from Monotone Convergence Theorem and (4.1) it follows that f (1 − g)dν = lim fn (1 − g)dν = lim fn gdµ = f gdµ, n→∞ n→∞ which proves the claim. z For a Σ-measurable and ρ − a.e. ≥ 0 function z, choose f = 1−g in (4.1), then g zdν = z dµ = zhdµ, (4.4) 1−g g where h = 1−g . If for A ∈ Σ we take z = χA in (4.2), then ν(A) = IA hdν = hdµ. A Since ν(Ω) < +∞, we know hdµ < +∞, hence h ∈ L1 (Ω, Σ, µ). That such an h is unique is obvious. That h ≥ 0 µ − a.e. is also obvious. 4.5. THE LAX-MILGRAM THEOREM 49 Exercise 4.4.1. A measure space (Ω, Σ, µ) is said to be σ-ﬁnite if there are A1 , A2 , · · · in Σ such that An = Ω and µ(An ) < +∞, n = 1, 2, · · · . Show that Lebesgue-Nikodym Theorem holds if both (Ω, Σ, µ) and (Ω, Σ, ν) are σ- ﬁnite. But in this case h may not be µ-integrable. 4.5 The Lax-Milgram Theorem Let X be a Hilbert space. For deﬁniteness, let X be a complex Hilbert space. B(·, ·) : X × X → is called sesquilinear if for x, x1 , x2 in X and λ1 , λ2 ∈ the following equalities hold B(λ1 x1 + λ2 x2 , x) = λ1 B(x1 , x) + λ2 B(x2 , x), ¯ ¯ B(x, λ1 x1 + λ2 x2 ) = λ1 B(x, x1 ) + λ2 B(x, x2 ). B is said to be bounded if there is r > 0 such that |B(x, y)| ≤ r x · y for all x and y in X; and B is said to be positive deﬁnite if there exists ρ > 0 such that B(x, x) ≥ ρ x 2 for all x in X. Exercise 4.5.1. Suppose that B is a bounded, positive deﬁnite and sesquilinear function on X × X and assume that B(x, y) = B(y, x) for all x and y in X. let ((·, ·)) = B(·, ·), then X, ((·, ·)) is a Hilbert space which is equivalent to X, (·, ·) as Banach space. Theorem 4.5.1. (The Lax-Milgram Theorem) Let X be a Hilbert space and B a bounded, positive deﬁnite and sesquilinear functional on X × X. Then there is a unique bounded linear operator S : X → X such that (x, y) = B(Sx, y) for all x and y in X and S ≤ ρ−1 . Furthermore S −1 exists and is bounded with S −1 ≤ r. Proof. Let D = {y ∈ X : ∃ y ∗ ∈ X such that (x, y) = B(x, y ∗ ) ∀ x ∈ X}. D is not empty, for 0 ∈ D. Also y ∗ is uniquely determined by y. For, if ∗ ∗ ∗ ∗ B(x, y1 ) = B(x, y2 ) = (x, y) ∀ x ∈ X, then B(x, y1 − y2 ) = 0 ∀ x ∈ X, and ∗ ∗ ∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗ hence 0 = B(y1 − y2 , y1 − y2 ) ≥ ρ y1 − y2 , implying y1 − y2 =0, or y1 = y2 . For y ∈ D, let Sy = y ∗ . Since B is sesquilinear, D is a vector subspace of X and S is linear on D. Furthermore, from ρ Sy 2 ≤ B(Sy, Sy) = (Sy, y) ≤ y · Sy , it follows that Sy ≤ ρ−1 y for y ∈ D. Thus S is bounded on D with S ≤ ρ−1 . We proceed to show that D = X. For this we show ﬁrst that D is closed. Let {yn }∞ ⊂ D with limn→∞ yn = y for some y ∈ X, then n=1 (x, y) = lim (x, yn ) = lim B(x, Syn ), n→∞ n→∞ 50 CHAPTER 4. HILBERT SPACE for all x ∈ X. Since S is bounded on D, Syn is Cauchy in X, and hence has a limit z ∈ X. ¿From this and the fact that B is bounded, it follows that (x, y) = lim B(x, Syn ) = B(x, z), n→∞ for all x ∈ X. Hence y ∈ D and z = Sy. So D is closed. Now if D = X, there is y0 ∈ D⊥ , y0 = 0. Consider the linear functional ℓ deﬁned on X by ℓ(x) = B(x, y0 ), x ∈ X. As B is bounded, ℓ is a bounded linear functional on X, and hence by Riesz Representation Theorem there is x0 ∈ X such that B(x, y0 ) = (x, x0 ), x ∈ X. Thus x0 ∈ D and ρ y0 2 ≤ B(y0 , y0 ) = (x0 , y0 ) = 0. Hence y0 = 0, this contradicts the fact that y0 = 0. Therefore D = X. Thus S is a bounded linear operator on X and S ≤ ρ−1 . As Sy = 0 implies (x, y) = B(x, Sy) = 0 ∀ x ∈ X and hence y = 0, S is an one-to-one map. Applying Riesz Representation Theorem again, as in the last paragraph, for each y ∗ in X, there is y ∈ X such that (x, y) = B(x, y ∗ ) ∀ x ∈ X, i.e. y ∗ = Sy. Thus S is an onto map. Hence S −1 exists. But from S −1 y 2 = |(S −1 y, S −1 y)| = |B(S −1 y, y)| ≤ r S −1 y · y , it follows that S −1 ≤ r. 4.6 Gram-Schmidt Orthogonalization Procedure A family {xα }α∈I of non-zero elements of a Hilbert space X is said to be or- thogonal if (xα , xβ ) = 0 whenever α = β. An orthogonal family is obviously linearly independent. A ﬁnite or countable orthogonal family is usually referred to as an orthogonal system in X. ¿From a linearly independent system {xn } in X, one can construct an orthogonal system {yn } in the following way. Let Ek = < x1 , · · · , xk >, the vector subspace of X generated by {x1 , · · · , xk }, and let tk = tEk be the orthogonal projection of X onto Ek . {yn } is deﬁned inductively as follows. Let y1 = x1 ; suppose y1 , · · · , yk have been deﬁned, let yk+1 = xk+1 − tk xk+1 . Exercise 4.6.1. Show that (i) < y1 , · · · , yn > = < x1 , · · · , xn >, n = 1, 2, 3, · · · ; (ii) {yn } is an orthogonal system in X. 4.7. BESSEL INEQUALITY AND PARSEVAL RELATION 51 This procedure of obtaining an orthogonal system from a linearly indepen- dent system is called the Gram-Schmidt orthogonalization procedure. An orthogonal family {xα }α∈I is called an orthonormal family if (xα , xβ ) = δαβ , where δαβ = 1 or 0 according as α = β or α = β. If {yn } is the orthogonal- ized system of {xn } through Gram-Schmidt procedure, the system {en } , en = yn yn , n = 1, 2, 3, · · · , is called the Gram-Schmidt orthonormalization of {xn }. 4.7 Bessel Inequality and Parseval Relation Let {en } be an orthonormal system in a Hilbert space X, U be the closed vector subspace of X generated by {en }, and tU be the orthogonal projection of X onto U . For each positive integer k, let Ek = < e1 , · · · , ek > and tk = tEk , the orthogonal projection of X onto Ek . The following propositions hold: 1) tk x = k (x, ej )ej , x ∈ X. j=1 k Proof. Let tk x = j=1 λj ej , λj ∈ . Then k k (tk x, ei ) = λj ej , ei = λj (ej , ei ) j=1 j=1 = λi , 1 ≤ i ≤ k. But (x, ei ) = tk x + (1 − tk )x, ei = (tk x, ei ) = λi , 1 ≤ i ≤ k, ⊥ where the fact that 1 − tk is the orthogonal projection of X onto Ek (See Exercise 2.2) has been used, hence k tk x = (x, ej )ej . j=1 2) For x ∈ X, limk→∞ tk x = tU x. Proof. For y ∈ U and ε > 0, there is a ﬁnite linear combination z = N such that z − y < ε. For k ≥ N , z ∈ Ek , and hence tk z = z, j=1 λj ej then tk y − y = tk y − tk z + z − y ≤ tk (y − z) + z − y ≤ 2 y − z < 2ε. Consequently, (limk→∞ tk y = y.) Now for x ∈ X, tU x ∈ U , thus from what is proved above (limk→∞ tk (tU x) = tU x.) But tk ◦ tU = tk . 52 CHAPTER 4. HILBERT SPACE k 3) For each k and x, y in X, (tk x, tk y) = j=1 (x, ej )(y, ej ). Proof. This follows easily from 1). ∞ 4) For x, y in X, (tU x, tU y) = j=1 (x, ej )(y, ej ). Proof. Since (tU x, tU y) − (tk x, tk y) = (tU x − tk x, tU y) + (tk x, tU y − tk y), we have |(tU x, tU y) − (tk x, tk y)| ≤ tU y · tU x − tk x + tk x · tU y − tk y ≤ y · tU x − tk x + x · tU y − tk y , and hence by 2) and 3) k ∞ (tU x, tU y) = lim (tk x, tk y) = lim (x, ej )(y, ej ) = (x, ej )(y, ej ). k→∞ k→∞ j=1 j=1 ∞ 5) ( j=1 |(x, ej )|2 ≤ x 2 , x ∈ X.) (Bessel inequality) ∞ Proof. ¿From 4), ( j=1 |(x, ej )|2 = tU x 2 ≤ x 2. ) ∞ 6) ( j=1 |(x, ej )|2 = x 2 ) for all x ∈ X) if and only if U = X. Proof. If U = X, then tU = 1 and hence from 4) ∞ |(x, ej )|2 = tU x 2 = x 2 j=1 for all x ∈ X. On the other hand, if U = X, there is x ∈ X such that x = tU x, hence ∞ x 2 = tU x 2 + (1 − tU )x 2 > tU x 2 = |(x, ej )|2 . j=1 An orthonormal system {en } in X is called complete if U = X. Exercise 4.7.1. Show that an orthonormal system {en } is complete if and only if from the fact that (en , x) = 0 for all n it follows that x = 0. A Hilbert space is called separable if it contains a countable dense subset. Theorem 4.7.1. A separable Hilbert X is isometrically isomorphic either to n for some n or to l2 . 4.7. BESSEL INEQUALITY AND PARSEVAL RELATION 53 Proof. Let {zk }∞ be a sequence of elements which is dense in X. By a k=1 well-known elementary selection procedure in linear algebra, one can extract from {zk } an independent subsequence {xk } such that {xk } = {zk } . If {xk } is ﬁnite, then the proof that X is isometrically isomorphic to n , where n is the cardinality of {xk }, is an easy imitation of that of the case when {xk } is inﬁnite. Hence we assume that {xk } is inﬁnite. From Gram-Schmidt orthonormalization procedure we construct from {xk } an orthonormal system {ek }∞ such that k=1 < {xk } > = < {ek } >. As before, let U be the closure of {ek } , then U = X; thus for x, y in X we have from 4) ∞ (x, y) = (x, ek )(y, ek ), (4.5) k=1 because tU = I. Deﬁne a map τ : X → l2 by letting, for x ∈ X, τ x = (αk )∞ ,k=1 ∞ where αk = (x, ek ). Since x 2 = k=1 |(x, ek )|2 by 6), τ x is in l2 and x = τ x l2 , so τ is an isometry. That τ is linear is obvious. We show now that τ is onto l2 . Let (αk )∞ ∈ l2 . For each positive integer n, let k=1 n xn = αk ek . k=1 We claim that {xn } is a Cauchy sequence in X. For n > m we have n 2 xn − xm = |αk |2 , k=m+1 which tends to 0 as m → ∞. Hence {xn } is a Cauchy sequence. Let (x = limn→∞ xn ). Then n (x, ek ) = lim , αj ej , ek = αk n→∞ j=1 and hence τ x = (αk )∞ . Therefore τ is onto. That τ is an isomorphism follows k=1 from (7.1). ∞ 2 2 The equality j=1 |(x, ej )| = x for x ∈ X and (7.1) are called Parseval relations.

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