Hilbert Spaces

By Elsevier Science

This book has evolved from the lecture course on Functional Analysis I had given several times at the ETH. The text has a strict logical order, in the style of “Definition – Theorem – Proof - Example - Exercises. The proofs are rather thorough and there many examples.

The first part of the book(the first three chapters, resp. the first two volumes) is devoted to the theory of Banach spaces in the most general sense of the term. The purpose of the first chapter (resp. first volume) is to introduce those results on Banach spaces which are used later or which are closely connected with the book. It therefore only contains a small part of the theory, and several results are stated (and proved) in a diluted form. The second chapter (which together with Chapter 3 makes the second volume) deals with Banach algebras (and involutive Banach algebras), which constitute the main topic of the first part of the book. The third chapter deals with compact operators on Banach spaces and linear (ordinary and partial) differential equations - applications of the, theory of Banach algebras.

• Language: English
• Publisher: Elsevier Science
• Release date: Jul 11, 2001
• ISBN: 9780080528359

Book preview

Spaces

5

Hilbert spaces

Corneliu Constantinescu

Most books on functional analysis treat the theory of Hilbert spaces before the theory of C*–algebras if they treat the latter at all. This is didactically justifiable since the former is simpler and is the main source of examples for the latter. But in a strictly logical sense, the theory of C*–algebras has priority. The development of this theory does not require even the definition of a Hilbert space. Conversely the theory of Hilbert spaces benefits from the theory of C*–algebras because the set of operators on a Hilbert space is a C*–algebra in a natural way. There is a didactic principle which says that in a book on mathematics the degree of difficulty should be roughly speaking an increasing function of the page number. While we do not wish to underestimate the importance of this principle, we have chosen in this instance to let logical priority be our guide and to rely on the reader not to completely ignore Hilbert spaces. This inversion of the customary order does have substantial benefits, for when dealing with the algebra of operators on a Hilbert space we may avail ourselves of all the results from the theory of C*–algebras.

5.1 Pre–Hilbert Spaces

A number of general results, constructions, and examples from the theory of Hilbert spaces are pesented in this first section.

5.1.1 General Results

Definition 5.1.1.1

(0) (von Neumann, 1928) Let E be a vector space. A scalar product on E is a positive sesquilinear form f on E (Proposition 2.3.3.3 e)), such that for every x ∈ E,

A pre-Hilbert space is a vector space endowed with a scalar product. A pre-Hilbert space is called real (complex) if the ground field in ). We shall usually denote a scalar product in a pre–Hilbert space E by

and for x ∈ E, we define

J. von Neumann defined only spearable Hilbert spaces. Non separable Hilbert spaces were considered by H. Löwing (1934) and F. Rellich (1935).

Proposition 5.1.1.2

(0) if E is a pre–Hilbert space, then the map

is a norm and

for all x, y ∈ E. The above norm is called the canonical norm on the pre–Hilbert space E. Unless otherwise stated, we always regard a pre–Hilbert space as endowed with its canonical norm. A Hilbert space is a complete pre–Hilbert space. A real (complex) Hilbert space is a complete real (complex) pre–Hilbert space.

The assertion follows immediately from Proposition 2.3.3.9.

Corollary 5.1.1.3

(0) If E is a pre–Hilbert space, then the map

is continuous.

By Schwarz’s Inequality,

for all (x,y) ∈ E × E and the assertion now follows from Proposition 1.2.9.2 c ⇒ a.

Corollary 5.1.1.4

(0) Let E, F be pre–Hilbert spaces and u : E → F a bijective linear map. u is an isornetry of the normed spaces E and F iff

for all x,y ∈ E.

The necessity follows from Proposition 2.3.3.7 a ⇒ b and Proposition 2.3.3.8 a ⇒ c & d. The sufficiency follows from

for every x ∈ E.

Corollary 5.1.1.5

Let E ≠ {0} be a pre–Hilbert space. Take , and x ∈ E#. If

then x ∈ Ker (α1 – u) and α ∈ σp(u).

Now

(Corollary 2.3.3.4). Hence

If u ≠ 0, then α ≠ 0 and x ≠ 0. Thus α ∈ σp(u). If u = 0, then

Theorem 5.1.1.6

(0) (Jordan, von Neumann) Let E be a normed space. The following are equivalent:

a) There is a scalar product on E generating the norm of E.

b) x, y ∈ E ⇒ ||x + y||² + ||x – y||² ≤ 2(||x||² + ||y||²).

c) x, y ∈ E ⇒ ||x + y||² + ||x – y||² ≥ 2(||x||² + ||y||²).

If these conditions are fulfillled, then the scalar product m a) is unique.

The uniqueness of the scalar product follows from Proposition 2.3.3.7 a ⇒ b and Proposition 2.3.3.8 a ⇒ c & d.

a ⇒ b & c follows from the parallelogram law (Proposition 2.3.3.2 b)). b ⇒ c (resp. c ⇒ b). Put

Then

so

b & c ⇒ a.

Given x, y ∈ E, put

Then, for every x,y ∈ E,

and

Thus only linearity in the first variable needs further proof.

Now

The assertion follows from Step 1, by complete induction.

By Step 1,

Hence

. By Steps 2 and 3,

Hence

by the triangle inequality and Step 3. Thus

be a sequence in Q converging to α. Then

by Steps 3, 1, and 5. Thus

by Step 4.

Given x, y ∈ E, define

Now

Hence

By Case 1,

By Steps 2 and 3,

Hence

By Steps 2, 3, and 4,

Remark. This theorem makes it possible to define pre–Hilbert (Hilbert) spaces as normed (Banach) spaces satisfying the parallelogram law.

Corollary 5.1.1.7

(0) Let E be a pre–Hilbert space and Ẽ the completion of the associated normed space. There is a unique scalar product on Ẽ generating the norm of Ẽ. Ẽ with this scalar product is called the completion of E. The scalar product of E is the restriction to E of the scalar product of Ẽ.

in E converging to x and y, respectively. By the parallelogram law (Proposition 2.3.3.2 b)),

By Theorem 5.1.1.6, there is a unique scalar product on Ẽ generating its norm. The restriction to E of this scalar product generates the norm of E. Hence it coincides with the original scalar product of E (Theorem 5.1.1.6).

Corollary 5.1.1.8

Let E,F be pre-Hilbert spaces and take . Then the following are equivalent:

a) 

b) 

a ⇒ b. The map

is an isometry and the assertion follows from Theorem 5.1.1.6.

b ⇒ a is trivial.

Proposition 5.1.1.9

(0) Let E be a real (complex) pre-Hilbert space, f a sesquilinear form on E, and take . If

for every x ∈ E, then

for all x, y ∈ E.

We may assume without loss of generality that x ≠ 0 and y ≠ 0. Consider

By Proposition 2.3.3.2 b),c),

Thus

with

Then

Proposition 5.1.1.10

Let E be a pre-Hilbert space and take x ∈ E. Let (xn)n∈ IN be a sequence in E. The following are equivalent:

a) 

.

a ⇒ b is trivial,

b ⇒ a. By Corollary 2.3.3.4,

for every n . Hence

Proposition 5.1.1.11

(0) Let E be a pre-Hilbert space and take x,y ∈ E. Put:

a) 

b) 

c) 

d) 

e) 

Then

a ⇔ b. By Corollary 2.3.3.4,

from which the assertion follows,

b ⇒ c. By Corollary 2.3.3.4,

Hence

c ⇒ a & d and d ⇒ e are trivial,

e ⇒ d. Take θ with

Then

By b ⇒ c, x and eiθy are linearly dependent. Hence x and y are lineraly dependent.

Remark,

a) It is easy to see that d) does not imply b) (take y := –x).

b) In ℓ¹,

and so a) does not imply c) in the case of arbitrary Banach spaces.

5.1.2 Examples

Example 5.1.2.1

(0) Let be a measure space. The map

is a scalar product. This scalar product renders L²(μ) a Hilbert space with norm

By taking μ to be the counting measure in the above example, we get:

Example 5.1.2.2

(0) Let T be a set. The map

is a scalar product, which renders ℓ²(Τ) a Hilbert space with norm

Example 5.1.2.3

The map

is a scalar product, ℓ² endowed with this scalar product is a Hilbert space (it is called the Hilbert space of square summable sequences). Its norm is

Example 5.1.2.4

Take n . The map

is a scalar product (it is called the canonical scalar product of n). It generates the Euclidean norm.

Example 5.1.2.5

Let T be a locally compact group, Mb(T) the involutive unital Banach algebra defined in Proposition 2.3.2.28 a), λ a left invariant Haar measure on T, and take . For x, y ∈ L²(λ), define

(Proposition 2.2.2.14 a)), and

Then f is a scalar product iff σp(u) ⊂ ]0, ∞[. This condition is fulfilled whenever μ is invertible.

By Proposition 2.3.3.10, f is a positive sesquilinear form and by Example 3.1.4.14, σp(μ+.

If 0 ∈ σp(u), then

for some x ∈ L²(λ)\{0}. Thus

and f is not a scalar product.

Assume that 0 ∉ σp(u). Take x ∈ L²(λ) with

Then

(Proposition 2.3.2.28 c), Example 2.3.2.27 b)). Thus

Hence f is a scalar product.

If μ is invertible, then μ* is invertible (Proposition 2.3.1.14), μ* * μ is invertible, and u is invertible (Proposition 2.2.2.14 e)). Hence 0 ∉ σp(u).

Example 5.1.2.6

Take p ∈ [l,∞]\{2} and let T be a set. If the norm of ℓp(T) is generated by a scalar product then CardT ≤ 1.

Assume Card T > 1. Let s, t be two distinct elements of T. Then

By the parallelogram law (Proposition 2.3.3.2 b)), the norm of ℓp(Τ) is not generated by a scalar product.

Example 5.1.2.7

Let n and take a n,n. The map

is a scalar product iff σ(a) ⊂ ]0, ∞[ and every scalar product on n is of this form.

By Example 2.3.3.6, f is a positive sesquilinear form iff σ (a+.Assume that 0 ∈ σ(a). Then there is an x n\{0} with

and so

Hence f is not a scalar product.

Now assume that 0 ∉ σ(a). Take x n with

where || · || is the Euclidean norm. Then, by Lagrange’s Theorem, there is an α with

for every i n. Thus α ∈ σ(a)