<<

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 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 determined by φ. The identity (1) is referred to as the . 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 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 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 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 : 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 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) [] 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

(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 .

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 , 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 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 , the space `2 (J) is a Hilbert space. We K know that this is a . 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.

Definitions. Let X be a K-vector space, equipped with an inner product   · · . Two vectors ξ, η ∈ X are said to be orthogonal, if ξ η = 0. In this case we write ξ ⊥ η. Given a set M ⊂ X, and a vector ξ ∈ X, we write ξ ⊥ M, if ξ ⊥ η, ∀ η ∈ M. Finally, two M, N ⊂ X are said to be orthogonal, and we write M ⊥ N, if ξ ⊥ η, ∀ ξ ∈ M, η ∈ N. Notation. Let X be a vector space equipped with an inner product. For a M ⊂ X, we deﬁne the set M⊥ = {ξ ∈ X : ξ ⊥ M .

Remarks 3.2. Let X be a K-vector space equipped with an inner product. A. The relation ⊥ is symmetric. B. If ξ, η ∈ X satisfy ξ ⊥ η, then one has the : kξ + ηk2 = kξk2 + kηk2. 2 2 2  This is a consequence of the equality kξ + ηk = kξk + kηk + 2Re ξ η . C. If M ⊂ X is an arbitrary subset, then M⊥ is a closed of X. This follows from the linearity of the inner product in the second variable, and from the continuity. D. For sets M ⊂ N ⊂ X, one has M⊥ ⊃ N⊥. E. For any set M ⊂ X, one has ⊥ M⊥ = Span M , where Span M denotes the norm of the of M. The inclusion ⊥ M⊥ ⊃ Span M is trivial, since we have M ⊂ Span M. Conversely, if ξ ∈ M⊥, then M ⊂ {ξ}⊥. But since {ξ}⊥ is a closed linear subspace, this gives Span M ⊂ {ξ}⊥, ⊥ i.e. ξ ∈ Span M . The following result gives a very interesting property of Hilbert spaces. Proposition 3.4. Let H be a Hilbert space, let C ⊂ H be a non-empty closed . For every ξ ∈ H, there exists a unique vector ξ0 ∈ C, such that kξ − ξ0k = dist(ξ, C). Proof. Denote dist(ξ, C) simply by d. By deﬁnition, we have δ = inf kξ − ηk. η∈C

Choose a (ηn)n≥1 ⊂ C, such that limn→∞ kξ − ηnk = δ. 84 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Claim: One has the inequality 2 2 2 2 kηm − ηnk ≤ 2kξ − ηmk + 2kξ − ηnk − 4δ , ∀ m, n ≥ 1. Use the Parallelogram Law 2 2 2 2 (5) 2kξ − ηmk + 2kξ − ηnk = k2ξ − ηm − ηnk + kηm − ηnk . 1 We notice that, since 2 (ηm + ηn) ∈ C, we have 1 kξ − 2 (ηm + ηn)k ≥ δ, so we get 2 1 2 2 k2ξ − ηm − ηnk = 4kξ − 2 (ηm + ηn)k ≥ 4δ , so if we go back to (5) we get 2 2 2 2 2 2 2kξ − ηmk + 2kξ − ηnk = k2ξ − ηm − ηnk + kηm − ηnk ≥ 4δ + kηm − ηnk , and the Claim follows. Having proven the Claim, we now notice that, since limn→∞ kξ − ηnk = δ, we immediately get the fact that the sequence (ηn)n≥1 is Cauchy. Since H is complete, it follows that the sequence is convergent to some point ξ0. Since C is closed, it follows that ξ0 ∈ C. So far we have 0 kξ − ξ k = lim kξ − ηnk = δ = dist(ξ, C), n→∞ thus proving the existence. Let us prove now the uniqueness. Assume ξ00 ∈ C is another point such that kξ − ξ00k = δ. Using the Parallelogram Law, we have 4δ2 = 2kξ − ξ0k2 + kξ − ξ00k2 = k2ξ − ξ0 − ξ00k2 + kξ0 − ξ00k2. If ξ0 6= ξ00, then we will have 2 0 00 2 1 0 00 2 4δ > k2ξ − ξ − ξ k = 4kξ − 2 (ξ + ξ )k , 1 0 00 so we have a new vector η = 2 (ξ + ξ ) ∈ C, such that kξ − ηk < δ, thus contracting the deﬁnition of δ.  Definition. Let H be a Hilbert space, and let X ⊂ H be a closed linear subspace. For every ξ ∈ H, using the above result, we let PXξ ∈ X denote the unique vector in X with the property

kξ − PXξk = dist(ξ, X).

This way we have constructed a map PX : H → H, which is called the orthogonal projection ont X. The properties of the orthogonal projection are summarized in the following result. Proposition 3.5. Let H be a Hilbert space, and let X ⊂ H be a closed linear subspace. (i) For vectors ξ ∈ H and ζ ∈ X one has the equivalence

ζ = PXξ ⇐⇒ (ξ − ζ) ⊥ X. (ii) P = Id . X X X (iii) The map PX : H → X is linear, continuous. If X 6= {0}, then kPXk = 1. §3. Hilbert spaces 85

⊥ (iv) Ran PX = X and Ker PX = X .

Proof. (i). “⇒.” Assume ζ = PXξ. Fix an arbitrary vector η ∈ X r {0}, and choose a number λ ∈ K, with |λ| = 1, such that   λ ξ − ζ η = ξ − ζ η . In particular, we have   ξ − ζ η = Re ξ − ζ λη . Deﬁne the map F : R → R by F (t) = kξ − ζ − tληk2 − kξ − ζk2.

By the deﬁnition of ζ = PXξ, we have (6) F (t) > 0, ∀ t ∈ R r {0}. 2  2 Notice that F (t) = at + bt, ∀ t ∈ R, where a = λη λη = kηk , and b =   2Re ξ − ζ λη = 2 ξ − ζ η . Of course, the property 2 at + bt > 0, ∀ t ∈ R r {0}  forces b = 0, so we indeed get ξ − ζ η = 0. “⇐.” Assume (ξ − ζ) ⊥ X. For any η ∈ X, we have (ξ − ζ) ⊥ (ζ − η), so using the Pythagorean Theorem, we get kξ − ηk2 = kξ − ζk2 + kζ − ηk2, which forces kξ − ηk ≥ kξ − ζk, ∀ η ∈ X. This proves that kξ − ζk = dist(ξ, X), i.e. ζ = PXξ. (ii). This property is pretty clear. If ξ ∈ X, then 0 = ξ − ξ is orthogonal to X, so by (i) we get ξ = PXξ. (iii). We prove the linearity of PX. Start with vectors ξ1, ξ −2 ∈ H and a λ ∈ K. Take ζ1 = PXξ1 and ζ2 = PXξ2. Consider the vector ζ = λζ1 + ζ2. For any η ∈ X, we have   λξ1 + ξ2 − ζ η = (λξ1 − λζ1) + (ξ2 − ζ2) η =     = λξ1 − λζ1 η + ξ2 − ζ2 η = λ ξ1 − ζ1 η + ξ2 − ζ2 η = 0.

By (i) we have (ξ1 − ζ1) ⊥ X and (ξ1 − ζ1) ⊥ X, so the above computation proves that (λξ1 + ξ2 − ζ) ⊥ X, so using (i) we get

PX(λξ1 + ξ2) = ζ = λζ1 + ζ2 = λPXξ1 + PXξ2, so PX is indeed linear. To prove the continuity, we start with an arbitrary vector ξ ∈ H and we use the fact that (ξ − PXξ) ⊥ PXξ. By the Pythagorean Theorem we then have 2 2 2 2 2 kξk = k(ξ − PXξ) + PXξk = kξ − PXξk + kPXξk ≥ kPXξk . In other words, we have kPXξk ≤ kξk, ∀ ξ ∈ H, 86 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS so PX is indeed continuous, and we have kPXk ≤ 1. Using (ii) we immediately get that, when X 6= {0}, we have kPXk = 1. (iv). The equality Ran PX = X is trivial by the construction of PX and by (ii). ⊥ If ξ ∈ Ker PX, then by (i), we have ξ ∈ X . Conversely, if ξ ⊥ X, then ζ = 0 satisﬁes the condition in (i), i.e. PXξ = 0.  Corollary 3.3. If H is a Hilbert space, and X ⊂ H is a closed linear subspace, then X + X⊥ = H and X ∩ X⊥. In other words the map (7) X × X⊥ 3 (η, ζ) 7−→ η + ζ ∈ H is a linear .

⊥ Proof. If ξ ∈ H then PXξ ∈ X, and ξ − PXξ ∈ X , and then the equality

ξ = PXξ + (ξ − PXξ) proves that ξ ∈ X + X⊥. The equality X ∩ X⊥ = {0} is trivial, since for ζ ∈ X ∩ X⊥, we must have ζ ⊥ ζ, which forces ζ = 0.  Exercise 3. Let H be a Hilbert space. (i) Prove that, for any closed subspace X ⊂ H, one has the equality

PX⊥ = I − PX. (ii) Prove that two closed subspaces X, Y ⊂ H, the following are equivalent: – X ⊥ Y; – PXPY = 0; – PYPX = 0. (iii) Prove that two closed subspaces X, Y ⊂ H, the following are equivalent: – X ⊂ Y; – PXPY = PX; – PYPX = PX. (iv) Let X, Y ⊂ H are closed subspaces, such that X ⊥ Y, then – X + Y is c closed linear subspace of H; – PX+Y = PX + PY. Corollary 3.4. Let H be a Hilbert space, and let X ⊂ H be a linear (not necessarily closed) subspace. Then on has the equality

⊥ X = X⊥ .

⊥ ⊥ Proof. Denote the closed subspace X⊥ by Z. Since X⊥ = X , by the previous exercise we have

P = I − P ⊥ = I − P ⊥ = I − (I − P ) = P , Z X X X X which forces

Z = Ran PZ = Ran PX = X.  §3. Hilbert spaces 87

Theorem 3.1 (Riesz’ ). Let H be a Hilbert space over K, and let φ : H → K be a linear continuous map. Then there exists a unique vector ξ ∈ H, such that  φ(η) = ξ η , ∀ η ∈ H. Moreover one has kξk = kφk. Proof. First we show the existence. If φ = 0, we simply take ξ = 0. Assume φ 6= 0. Deﬁne the subspace X = Ker φ. Notice that X is closed. Using the linear isomorphism (7) we see that the composition

map X⊥ ,→ H −−−−−−−−→ H/X is a linear isomorphism. Since

H/X = H/Ker φ ' Ran φ = K, ⊥ ⊥ it follows that dim(X ) = 1. In other words, there exists ξ0 ∈ X , ξ0 6= 0, such that ⊥ X = Kξ. Start now with some arbitrary vector η ∈ H. On the one hand, using the equality Kξ0 + X = H, there exists λ ∈ K and ζ ∈ X, such that

η = λξ0 + ζ, and since ζ ∈ X = Ker φ, we get

φ(η) = φ(λξ0) = λφ(ξ0). On the other hand, we have    2 ξ0 η = ξ0 λξ0 + ξ0 ζ = λkξ0k ,

−2 so if we deﬁne ξ = φ(ξ0)kξ0k we will have  −2  −2  ξ η = φ(ξ0)kξ0k ξ0 | η = φ(ξ0)kξ0k ξ0 η = λφ(ξ0) = φ(η). To prove uniqueness, assume ξ0 ∈ H is another vector with 0  φ(η) = ξ η , ∀ η ∈ H. In particular, we have kξ − ξ0k2 = ξ − ξ0 | ξ − ξ0  = ξ | ξ − ξ0  − ξ0 | ξ − ξ0  = φ(ξ − ξ0) − φ(ξ − ξ0) = 0, which forces ξ = ξ0. Finally, to prove the norm equality, we ﬁrst observe that when ξ = 0, the equality is trivial. If ξ 6= 0, then on the one hand, using C-B-S inequality we have  |φ(η)| = ξ η ≤ kξk · kηk, ∀ η ∈ H, so we immediately get kφk ≤ kξk. If we take the vector ζ = kξk−1ξ, then kζk = 1, and −1  φ(ζ) = ξ kξk ξ = kξk, so we also have kφk ≥ kξk.  In the remainder of this section we discuss a Hilbert space notion of . This should be thought as a “rigid” linear independence. 88 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Definition. Let X be a K-vector space, equipped with an inner product. A set F ⊂ X is said to be orthogonal, if 0 6∈ F, and ξ ⊥ η, ∀ ξ, η ∈ F, with ξ 6= η. A set F ⊂ X is said to be orthonormal, if it is orthogonal, but it also satisﬁes: kξk = 1, ∀ ξ ∈ F. Remark that, if one starts with an orthogonal set F ⊂ X, then the set F(1) = kξk−1ξ : ξ ∈ F is orthonormal.

Proposition 3.6. Let X be a K-vector space equipped with an inner product. Any orthogonal set F ⊂ X is linearly independent. Proof. Indeed, if one starts with a vanishing

λ1ξ1 + ··· + λnξn = 0, with λ1, . . . , λn ∈ K, ξ1, . . . , ξn ∈ X, such that ξk 6= ξ`, for all k, ` ∈ {1, . . . , n} with k 6= `, then for each k ∈ {1, . . . , n} we clearly have 2  λkkξkk = ξk λ − 1ξ1 + ··· + λnξn = 0, and since ξk 6= 0, we get λk = 0.  Lemma 3.2. Let X be a K-vector space equipped with an inner product, and let F ⊂ X be an orthogonal set. Then there exists a maximal (with respect to inclusion) orthogonal set G ⊂ X with F ⊂ G. Proof. Consider the sets A = G : G orthogonal subset of X ,  AF = G ∈ A : G ⊃ F , ordered with the inclusion. We are going to apply Zorn’s Lemma to AF. Let T ⊂ AF be a subcollection, which is totally ordered, i.e. for any G1, G2 ∈ T one has G1 ⊂ G2 or G1 ⊃ G2. Deﬁne the set [ M = G. G∈T

Since G ⊂ X r {0}, for all G ∈ T, it is clear that M ⊂ X r {0}. If ξ1, ξ2 ∈ M are vectors with ξ1 6= ξ2, then we can ﬁnd G1, G2 ∈ T with ξ1 ∈ G1 and ξ2 ∈ G2. Using the fact that T is totally ordered, it follows that there is k ∈ {1, 2} such that ξ1, ξ2 ∈ Gk, so we indeed get ξ1 ⊥ ξ2. It is now clear that M ∈ AF, and M ⊃ G, for all G ∈ T. In other words, we have shown that every totally ordered subset of AF has an upper bound, in AF. By Zorn’s Lemma, AF has a maximal element. Finally, it is clear that any maximal element for AF is also a maximal element in A.  Remark 3.3. Using the notations from the proof above, given an orthonormal set M ⊂ X, the following are equivalent: (i) M is maximal in A; (ii) M is maximal in A(1) = G : G orthonormal subset of X . §3. Hilbert spaces 89

The implication (i) ⇒ (ii) is trivial. Conversely, if M is maximal in A(1), we use the Lemma to ﬁnd a maximal N ∈ A with N ⊃ M. But then N(1) is orthonromal, and N(1) ⊃ M, which by the maximality of M in A(1) will force N(1) = M. Since N is linearly independent, the relations N(1) = M ⊂ N, will force N = N(1) = M. Comment. In linear we know that a linearly independent set is max- imal, if and only if it spans the whole space. In the case of orthogonal sets, this statement has a version described by the following result. Theorem 3.2. Let H be a Hilbert space, and let F be an orthogonal set in H. The following are equivalent: (i) F is maximal among all orthogonal subsets of H; (ii) Span F is dense in H in the norm . Proof. (i) ⇒ (ii). Assume F is maximal. We are going to show that Span F is dense in H, by contradiction. Denote the closure Span F simply by X, and assume X ( H. Since ⊥ X = X⊥ , we see that, the strict inclusion X ( H forces X⊥ 6= {0}. But now if we take a non-zero vector ξ ∈ X⊥, we immediately see that the set F ∪ {ξ} is still orthogonal, thus contradicting the maximality of F. (ii) ⇒ (i). Assume Span F is dense in H, and let us prove that F is maximal. We do this by contardiction. If F is not maximal, then there exists ξ ∈ H r F, such that F ∪ {ξ} is still orthogonal. This would force ξ ⊥ F, so we will also have ξ ⊥ Span F. But since Span F is dense in H, this will give ξ ⊥ H. In particular we have ξ ⊥ ξ, which would force ξ = 0, thus contradicting the fact that F ∪ {ξ} is orthigonal. (Recall that all elements of an orthigonal set are non-zero.)  Definition. Let H be a Hilbert space An orthonormal set B ⊂ H, which is maximal among all orthogonal (or orthonormal) subsets of H, is called an orthonor- mal for H. By Lemma ??, we know that given any orthonormal set F ⊂ H, there exists an B ⊃ F. By the above result, an orthonormal set B ⊂ H is an orthonormal basis for H, if and only if Span B is dense in H. Example 3.2. Let J be a non-. Consider the Hilbert space `2 (I). K For every j ∈ J, let δj : J → K be the function deﬁned by  1 if k = j δ (k) = j 0 if k 6= j

If we consider the set B = {δj : j ∈ J}, then Span B = ﬁn (J), K which is dense in `2 (J). The above result then says that B is an orthonormal basis K for `2 (J). K 90 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Lemma 3.3. Let B be an orthonormal basis for the Hilbert space H, and let F ( B be an arbitrary non-empty subset. (i) F is an orthonormal basis for the Hilbert space Span F. ⊥ (ii) Span F = Span(B r F). Proof. (i). This is clear, since F is orthonormal and has dense span. (ii). Denote for simplicity Span F = X and Span(B r F) = Y. Since ξ ⊥ η, ∀ ξ ∈ F, η ∈ B r F, it is pretty obvious that X ⊥ Y. Since X + Y clearly contains Span B, it follows that X + Y is dense in H. We know howver that X + Y is closed, so we have in fact the equality X + Y = H. This will then give

I = PH = PX + PY, so we get

PY = I − PX = PX⊥ , so ⊥ X = Ran PX⊥ = Ran PY = Y.  Theorem 3.3. Let H be a Hilbert space, and let B be an orthonormal basis 1 η for H, labelled as B = {ξj : j ∈ J}. For every vector η ∈ H, let α : J → K be the map deﬁned by  ηˆ(j) = ξj η , ∀ j ∈ J. (i) For every η ∈ H, the map ηˆ belongs to `2 (J). K (ii) The map T : H 3 η 7−→ ηˆ ∈ `2 (J) K is an isometric linear isomorphism. Proof. (i). Fix for the moment η ∈ H. One way to prove thatη ˆ belongs to `2 (J) is by showing that K  X 2 sup |ηˆ(j)| : F ∈ Pﬁn(J) < ∞. j∈F For any non-empty ﬁnite subset F ⊂ J, we deﬁne the subspace

HF = Span{ξj : j ∈ F }, and deﬁne the vector X ηF = ξj η ) · ξj. j∈F Claim: For every ﬁnite set F ⊂ J, one has the equality

ηF = PHF η.

1 This notation implicitly assumes that ξj 6= ξk, for all j, k ∈ J with j 6= k. §3. Hilbert spaces 91

It suﬃces to prove that (η − ηF ) ⊥ HF . But this is obvious, since if we start with some k ∈ F , then using the fact that  ξk ξj = 0, for all j ∈ F r {k}, together with the equality kξkk = 1, we get   X      ξk η −ηF = ξk η − ξj η · ξk ξj = ξk η − ξk η · ξk ξk = 0. j∈F Having proven the Claim, let us observe that, since the terms in the sum that deﬁnes ηF are all orthogonal, we get 2 X  2 X  2 2 X 2 kηF k = ξj η · ξj = ξj η · kξjk = |ηˆ(j)| . j∈F j∈F j∈F Combining this computation with the Claim, we now have X 2 2 2 2 |ηˆ(j)| = kηF k = kPHF ηk ≤ kηk , j∈F which proves that  X 2 2 sup |ηˆ(i)| : F ∈ Pﬁn(J) < kηk . j∈F (ii). The linearlity of T is obvious. The above inequality actually proves that kT ηk ≤ kηk, ∀ η ∈ H. We now prove that in fact T is isometric. Since T is linear and continuous, it suﬃces to prove that T is isometric. Start with some vector η ∈ Span B, Span B which means that there exists some ﬁnite set F ⊂ J, and scalars (λk)k∈F ⊂ K, P such that η = k∈F λkξk. Remark that X  λ if k ∈ F ξ | η  = λ ξ | ξ  = k j k j j 0 if k ∈ F k∈F so the elementη ˆ = T η ∈ `2 (J) is deﬁned by K  λ if k ∈ F ηˆ(k) = k 0 if k ∈ F This gives 2 X  X 2 X 2 2 kηk = λjλk ξj ξk = |λk| = |ηˆ(k)| = kηˆk , j,k∈F k∈F k∈F so we indeed get kηk = kT ηk, ∀ η ∈ Span B. Let us prove that T is surjective. Notice that, the above computation, applied to singleton sets F = {k}, k ∈ J, proves that

T ξk = δk, ∀ k ∈ J. In particular, we have Ran T ⊃ T Span B = Span T (B) = = Span{T ξ : k ∈ J} = Span{δ : k ∈ J} = ﬁn (J), k k K which proves that Ran T is dense in `2 (J). We know however that T is isometric, K so Ran T ⊂ `2 (J) is closed. This forces Ran T = `2 (J). K K  92 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Corollary 3.5 (Parseval Identity). Let H be a Hilbert space, and let B = {ξj : j ∈ J} be an orthonormal basis for H. One has:  X   ζ η = ζ ξj · ξj η , ∀ ζ, η ∈ H. j∈I   Proof. If we deﬁne α(j) = ξj ζ and ξj η , ∀ j ∈ I, then by construction we have α = T ζ and β = T η. Using the fact that T is isometric, the right hand side of the above equality is the equal to X    α(j)β(j) = α β = T ζ T η = ζ η . j∈J 

Notation. Let H be a Hilbert space, let B = {ξj : j ∈ J} be an orthonormal basis for H, and let T : H → `2 (J) be the isometric linear isomorphism deﬁned in K the previous theorem. Given an element α ∈ `2 (J), we denote the vector T −1α ∈ H K by X α(j)ξj. j∈J The summation notation is justiﬁed by the following fact.

Proposition 3.7. With the above notations, the net (ηF )F ∈Pﬁn(J) ⊂ H of partial sums, deﬁned by X ηF = α(k)ξk, ∀ F ∈ Pﬁn(J), k∈F P is convergent in the norm topology to the vector j∈J α(j)ξj. P Proof. Deﬁne the vector η = j∈I α(j)ξj. By construction we have T η = α. Likewise, if we deﬁne, for each ﬁnite set F ⊂ J, the element α ∈ `2 (J) by F K  α(k) if k ∈ F αF (k) = 0 if k ∈ J r F −1 then T αF = ηF . Using the fact that T is an , we have

kη − ηF k = kT η − T ηF k = kα − αF k, and the desired property follows from the well-known properties of `2 (J). K 

Exercise 4. Let H be a Hilbert space, let F = {ξj : j ∈ J} be an orthonormal set. Deﬁne the closed linear subspace HF = Span F. Prove that the orthogonal projection PHF is deﬁned by X  PHF η = ξj η ξj, ∀ η ∈ H. j∈J

Hints: Extend F to an orthonormal basis B. Let B be labelled as {ξi : i ∈ I} for some set

I ⊃ J. First prove that for any η ∈ H, the map βη = T η belongs to `2 (J). In particular, the J K sum X  ηF = ξj η ξj j∈J is “legitimate” and deﬁnes an element in HF (use the fact that F is an orthonormal basis for HF). Finally, prove that (η − ηF) ⊥ F, using Parseval Identity. §3. Hilbert spaces 93

The next result discusses the appropriate notion of for Hilbert spaces. Theorem 3.4. Let H be a Hilbert space. Then any two orthonormal bases of H have the same . Proof. Fix two orthonormal bases B and B0. There are two possible cases. Case I: One of the sets B or B0 is ﬁnite. In this case H is ﬁnite dimensional, since the linear span of a ﬁnite set is automatically closed. Since both B and B0 are linearly independent, it follows that both B and B0 are ﬁnite, hence their linear spans are both closed. It follows that Span B = Span B0 = H, so B and B0 are in fact linear bases for H, and then we get Card B = Card B0 = dim H. Case II: Both B and B0 are inﬁnite. The key step we need in this case is the following. Claim 1: There exists a dense subset Z ⊂ H, with Card Z = Card B0. To prove this fact, we deﬁne the set X = Span B0. Q It is clear that Card X = Card B0. Notice that X is dense in Span B0. If we work over = , then we are done. If R K R we work over K = C, we deﬁne Z = X + iX, and we will still have Card Z = Card X = Card B0. Now we are done, since clearly Z is dense in Span B0. C Choose Z as in Claim 1. For every ξ ∈ B we choose a vector ζξ ∈ Z, such that √ 2 − 1 kξ − ζ k ≤ . ξ 2

Claim 2: The map B 3 ξ 7−→ ζξ ∈ Z is injective.

Start with two vectors ξ1, ξ2 ∈ B, such that ξ1 6= ξ2. In particular, ξ1 ⊥ ξ2, so we also have ξ1 ⊥ (−ξ2), and using the Pythogorean Theorem we get 2 2 2 kξ1 − ξ2k = kξ2k + k − ξ2k = 2, which gives √ kξ1 − ξ2k = 2. Using the , we now have √ √

2 = kξ1 − ξ2k ≤ kξ1 − ζξ1 k + kξ2 − ζξ2 k + kζξ1 − ζξ2 k ≤ 2 − 1 + kζξ1 − ζξ2 k. This gives

kζξ1 − ζξ2 k ≥ 1, which forces ζξ1 6= ζξ2 . 94 CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS

Using Claim 2, we have constructed an injective map B → Z. In particular, using Claim 1 and the cardinal arithmetic rules, we get Card B ≤ Card Z = Card B0. By we also have Card B0 ≤ Card B, and then using the Cantor-Bernstein Theorem, we ﬁnally get Card B = Card B0.  Corollary 3.6 (of the proof). A Hilbert space is separable, in the norm topol- ogy, if and only of it has an orthonormal basis which is at most countable.

Proof. Use Claims 1 and 2 from the proof of the Theorem.  Definition. Let H be a Hilbert space, and let B be an orthonormal basis for H. By the above theorem, the Card B does not depend on the choice of B. This number is called the hilbertian (or orthogonal) dimension of H, and is denoted by h-dim H. Corollary 3.7. For two Hilbert spaces H and H0, the following are equivalent: (i) h-dim H = h-dim H0; (ii) There exists an isometric linear isomorphism U : H → H0. Proof. (i) ⇒ (ii). Choose a set I with h-dim H = h-dim H0 = Card I. Apply Theorem ?? to produce isometric linear T : H → `2 (I) and K T 0 : H0 → `2 (I). Then deﬁne U = T 0−1 ◦ T . K (ii) ⇒ (i). Assume one has an isometric linear isomorphism U : H → H0. Choose an orthonormal basis B for H. Then U(B) is clearly and orthonormal basis for H0, and since U : B → U(B) is bijective, we get h-dim H = Card B = Card U(B) = h-dim H0. 