Open access peer-reviewed chapter

# Analytical Applications on Some Hilbert Spaces

Written By

Fethi Soltani

Submitted: June 5th, 2019 Reviewed: October 30th, 2019 Published: April 23rd, 2020

DOI: 10.5772/intechopen.90322

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## Functional Calculus

Edited by Kamal Shah and Baver Okutmuştur

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## Abstract

In this paper, we establish an uncertainty inequality for a Hilbert space H. The minimizer function associated with a bounded linear operator from H into a Hilbert space K is provided. We come up with some results regarding Hardy and Dirichlet spaces on the unit disk D.

### Keywords

• Hilbert space
• Hardy space
• Dirichlet space
• uncertainty inequality
• minimizer function

## 1. Introduction

Hilbert spaces are the most important tools in the theories of partial differential equations, quantum mechanics, Fourier analysis, and ergodicity. Apart from the classical Euclidean spaces, examples of Hilbert spaces include spaces of square-integrable functions, spaces of sequences, Sobolev spaces consisting of generalized functions, and Hardy spaces of holomorphic functions. Saitoh et al. applied the theory of Hilbert spaces to the Tikhonov regularization problems [1, 2]. Matsuura et al. obtained the approximate solutions for bounded linear operator equations with the viewpoint of numerical solutions by computers [3, 4]. During the last years, the theory of Hilbert spaces has gained considerable interest in various fields of mathematical sciences [5, 6, 7, 8, 9]. We expect that the results of this paper will be useful when discussing (in Section 2) uncertainty inequality for Hilbert space H and minimizer function associated with a bounded linear operator T from H into a Hilbert space K. As applications, we consider Hardy and Dirichlet spaces as follows.

Let C be the complex plane and D=zC:z<1 the open unit disk. The Hardy space HD is the set of all analytic functions f in the unit disk D with the finite integral:

02πfe2dθ.E1

It is a Hilbert space when equipped with the inner product:

fgHD=12π02πfege¯dθ.E2

Over the years, the applications of Hardy space HD play an important role in various fields of mathematics [5, 10] and in certain parts of quantum mechanics [11, 12]. And this space is the background of some applications. For example, in Section 3, we study on HD the following two operators:

fz=fz,Lfz=z2fz+zfz,E3

and we deduce uncertainty inequality for this space. Next, we establish the minimizer function associated with the difference operator:

T1fz=1zfzf0.E4

In Section 4, we consider the Dirichlet space DD, which is the set of all analytic functions f in the unit disk D with the finite Dirichlet integral:

Dfz2dxdyπ,z=x+iy.E5

It is also a Hilbert space when equipped with the inner product:

fgDD=f0g0¯+Dfzgz¯dxdyπ,z=x+iy.E6

This space is the objective of many applicable works [5, 13, 14, 15, 16, 17] and plays a background to our contribution. For example, we study on DD the following two operators:

Λfz=fzf0,Xfz=z2fz,E7

and we deduce the uncertainty inequality for this space DD. And we establish the minimizer function associated with the difference operator:

T2fz=1zfzzf0f0.E8

## 2. Generalized results

Let H be a Hilbert space equipped with the inner product ..H. And let A and B be the two operators defined on H. We define the commutator AB by

ABABBA.E9

The adjoint of A denoted by A is defined by

AfgH=fAgH,E10

for fDomA and gDomA.

Theorem 2.1. For fDomAADomAA, one has

AfH2=AfH2+AAffH.E11

Proof. Let fDomAADomAA. Then AAf and AAf belong to H. Therefore AAfH. Hence one has

AfH2=AAffH=AAffH+AAffHE12
=AfH2+AAffH.E13

The following result is proved in [18, 19].

Theorem 2.2. Let A and B be the self-adjoint operators on a Hilbert space H. Then

AafHBbfH12ABffH,E14

for all fDomABDomBA, and all a,bR.

Theorem 2.3. Let fDomAADomAA. For all a,bR, one has

A+AafHAA+ibfHAfH2AfH2,E15

where i is the imaginary unit.

Proof. Let us consider the following two operators on DomAADomAA by

P=A+A,Q=iAA.E16

It follows that, for fDomAADomAA, we have Pf,QfH. The operators P and Q are self-adjoint and PQ=2iAA. Thus the inequality (15) follows from Theorems 2.1 and 2.2.

ΔH+fΔHffH4AfH2AfH22,E17

where

ΔH±f=fH2A±AfH2A±AffH2.E18

Proof. Let fDomAADomAA. The operator P given by (16) is self-adjoint; then for any real a, we have

PafH2=PfH2+a2fH22aPffH.E19

This shows that

minaRPafH2=PfH2PffH2fH2,E20

and the minimum is attained when a=PffHfH2. In other words, we have

minaRA+AafH2=A+AfH2A+AffH2fH2.E21

Similarly

minbRAA+ibfH2=AAfH2AAffH2fH2.E22

Then by (15), (21), and (22), we deduce the inequality (17).

Let λ>0 and let T:HK be a bounded linear operator from H into a Hilbert space K. Building on the ideas of Saitoh [2], we examine the minimizer function associated with the operator T.

Theorem 2.5. For any kK and for any λ>0, the problem

inffHλfH2+TfkK2E23

has a unique minimizer given by

fλ,k=λI+TT1Tk.E24

Proof. The problem (23) is solved elementarily by finding the roots of the first derivative DΦ of the quadratic and strictly convex function Φf=λfH2+TfkK2. Note that for convex functions, the equation DΦf=0 is a necessary and sufficient condition for the minimum at f. The calculation provides

DΦf=2λf+2TTfk,E25

and the assertion of the theorem follows at once.

Theorem 2.6. If T:HK is an isometric isomorphism; then for any kK and for any λ>0, the problem

inffHλfH2+TfkK2E26

has a unique minimizer given by

fλ,h=1λ+1T1k.E27

Proof. We have T=T1 and TT=I. Thus, by (24), we deduce the result.

## 3. The Hardy space HD

Let C be the complex plane and D=zC:z<1 the open unit disk. The Hardy space HD is the set of all analytic functions f in the unit disk D with the finite integral:

02πfe2dθ.E28

It is a Hilbert space when equipped with the inner product:

fgHD=12π02πfege¯dθ.E29

If f,gHD with fz=n=0anzn and gz=n=0bnzn, then

fgHD=n=0anbn¯.E30

The set znn=0 forms an Hilbert’s basis for the space HD.

The Szegő kernel Sz given for zD, by

Szw=n=0z¯nwn=11z¯w,wD,E31

is a reproducing kernel for the Hardy space HD, meaning that SzHD, and for all fHD, we have fSzHD=fz.

For zD, the function uz=Sz¯w is the unique analytic solution on D of the initial problem:

uz=wzuz+uz,wD,u0=1.E32

In the next of this section, we define the operators , , and L on HD by

fz=fz,ℜfz=zfz,Lfz=z2fz+zfz.E33

These operators satisfy the commutation rule:

L=LL=2+I,E34

where I is the identity operator.

We define the Hilbert space UD as the space of all analytic functions f in the unit disk D such that

fUD2=12π02πfe2dθ<.E35

If fUD with fz=n=0anzn, then

fUD2=n=1n2an2.E36

Thus, the space UD is a subspace of the Hardy space HD.

Theorem 3.1.

1. For fUD, then f, ℜf and Lf belong to HD.

2. =L.

3. For fUD, one has

LfHD2=fHD2+fHD2+2ℜffHD.E37

Proof.

1. Let fUD with fz=n=0anzn. Then

fz=n=0n+1an+1zn,ℜfz=n=1nanzn,E38

and

Lfz=n=1nan1zn.E39

Therefore

fHD2=n=0n+12an+12=fUD2,E40
ℜfHD2=n=1n2an2=fUD2,E41

and

LfHD2=n=0n+12an2f02+4fUD2.E42

Consequently f, ℜf, and Lf belong to HD.

2. For f,gUD with fz=n=0anzn and gz=n=0bnzn, one has

fgHD=n=0n+1an+1bn¯=n=1nanbn1¯=fLgHD.E43

Thus =L.

3. Let fUD. By (ii) and (34), we deduce that

LfHD2=LffHDE44
=LffHD+LffHDE45
=fHD2+fHD2+2ℜffHD.E46

Theorem 3.2. Let fUD. For all a,bR, one has

+LafHDL+ibfHDfHD2+2ℜffHD.E47

Theorem 3.3. Let T1 be the difference operator defined on HD by

T1fz=1zfzf0.E48
1. The operator T1 maps continuously from HD to HD, and

T1fHDfHD.E49

2. For fHD and zD, we have

T1fz=zfz,T1T1fz=fzf0.E50

3. For any hHD and for any λ>0, the problem

inffHDλfHD2+T1fhHD2E51

has a unique minimizer given by

fλ,hz=1λ+1zhz,zD.E52

Proof.

1. If fHD with fz=n=0anzn, then T1fz=n=0an+1zn and

T1fHD2=n=1an2fHD2.E53

2. If f,gHD with fz=n=0anzn and gz=n=0bnzn, then

T1fgHD=n=0an+1bn¯=n=1anbn1¯=fT1gHD,E54

where T1gz=zgz, for zD. And therefore

T1T1fz=zT1fz=fzf0.E55

3. From Theorem 2.5 we have

λI+T1T1fλ,hz=T1hz.E56

By (ii) we deduce that

λ+1fλ,hzfλ,h0=zhz.E57

And from this equation, fλ,h0=0. Hence

fλ,hz=1λ+1zhz.E58

## 4. The Dirichlet space DD

The Dirichlet space DD is the set of all analytic functions f in the unit disk D with the finite Dirichlet integral:

Dfz2dxdyπ,z=x+iy.E59

It is a Hilbert space when equipped with the inner product:

fgDD=f0g0¯+Dfzgz¯dxdyπ,z=x+iy.E60

If f,gDD with fz=n=0anzn and gz=n=0bnzn, then

fgDD=a0b0¯+n=1nanbn¯.E61

The set 1znnn=1 forms an Hilbert’s basis for the space DD.

The function Kz given for zD, by

Kzw=1+log11z¯w,wD,E62

is a reproducing kernel for the Dirichlet space DD, meaning that KzDD, and for all fDD, we have fKzDD=fz.

For zD, the function uz=Kz¯w is the unique analytic solution on D of the initial problem:

uzu0z=wuz,wD,u0=1.E63

In the next of this section, we define the operators Λ, , and X on DD by

Λfz=fzf0,ℜfz=zfz,Xfz=z2fz.E64

These operators satisfy the following commutation relation:

ΛX=ΛXXΛ=2.E65

We define the Hilbert space VD as the space of all analytic functions f in the unit disk D such that

fVD2=Dfz2z2dxdyπ<,z=x+iy.E66

If fVD with fz=n=0anzn, then

fVD2=n=1n3an2.E67

Thus, the space VD is a subspace of the Dirichlet space DD.

Theorem 4.1.

1. For fVD, then Λf, ℜf, and Xf belong to DD.

2. Λ=X.

3. For fVD, one has

XfDD2=ΛfDD2+2ℜffDD.E68

Proof.

1. Let fVD with fz=n=0anzn. Then

Λfz=n=1n+1an+1zn,ℜfz=n=1nanzn,E69

and

Xfz=n=2n1an1zn.E70

Therefore

ΛfDD2=n=1nn+12an+12n=2n3an2fVD2,E71
ℜfDD2=n=1n3an2=fVD2,E72

and

XfDD2=n=1n+1n2an22fVD2.E73

Consequently Λf, ℜf, and Xf belong to DD.

2. For f,gVD with fz=n=0anzn and gz=n=0bnzn, one has

ΛfgDD=n=1nn+1an+1bn¯=n=2nn1anbn1¯=fXgDD.E74

3. Let fVD. By (ii) and (65), we deduce that

XfDD2=ΛXffDDE75
=XΛffDD+ΛXffDDE76
=ΛfDD2+2ℜffDD.E77

Theorem 4.2. Let fVD. For all a,bR, one has

Λ+XafDDΛX+ibfDD2ℜffDD.E78

Theorem 4.3. Let T2 be the difference operator defined on DD by

T2fz=1zfzzf0f0.E79
1. The operator T2 maps continuously from DD to DD, and

T2fDDfDD.E80

2. For fDD with fz=n=0anzn, we have

T2fz=n=2n1nan1zn,T2T2fz=n=2n1nanzn.E81

3. For any dDD and for any λ>0, the problem

inffDDλfDD2+T2fdDD2E82

has a unique minimizer given by

fλ,dz=dΨzDD,zD,E83
Ψzw=n=1z¯n+1λn+1+nwn,wD.E84

Proof.

1. If fDD with fz=n=0anzn, then T2fz=n=1an+1zn and

T2fDD2=n=2n1an2n=2nan2fDD2.E85

2. If f,gDD with fz=n=0anzn and gz=n=0bnzn, then

T2fgDD=n=1nan+1bn¯=n=2n1anbn1¯=fT2gDD,E86

where

T2gz=n=2n1nbn1zn,zD.E87

And therefore

T2T2fz=n=2n1nanzn.E88

3. We put dz=n=0dnzn and

fλ,dz=n=0cnzn.E89

From (ii) and the equation

λI+T2T2fλ,dz=T2dz,E90

we deduce that

c1=c0=0,cn=n1λn+n1dn1,n2.E91

Thus

fλ,dz=n=1ndnλn+1+nzn+1=dΨzDD,zD.E92

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Written By

Fethi Soltani

Submitted: June 5th, 2019 Reviewed: October 30th, 2019 Published: April 23rd, 2020