Analytical Applications on Some Hilbert Spaces

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.

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 $\mathbb{C}$ be the complex plane and $\mathbb{D}=\left\{z\in \mathbb{C}:|z|<1\right\}$ the open unit disk. The Hardy space $H\left(\mathbb{D}\right)$ is the set of all analytic functions $f$ in the unit disk $\mathbb{D}$ with the finite integral:

$${\int }_0^{2\pi }{\left|f\left(e^{\mathit{i\theta }}\right)\right|}^2\mathrm{d}\theta .$$E1

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

$${⟨f,g⟩}_{H\left(\mathbb{D}\right)}=\frac{1}{2\pi }{\int }_0^{2\pi }f\left(e^{\mathit{i\theta }}\right)\overline{g\left(e^{\mathit{i\theta }}\right)}\mathrm{d}\theta .$$E2

Over the years, the applications of Hardy space $H\left(\mathbb{D}\right)$ 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 $H\left(\mathbb{D}\right)$ the following two operators:

$$\nabla f\left(z\right)=f^{\prime }\left(z\right),\mathit{Lf}\left(z\right)=z^2{f}^{\prime }\left(z\right)+\mathit{zf}\left(z\right),$$E3

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

$$T_{1}f\left(z\right)=\frac{1}{z}\left(f\left(z\right)-f\left(0\right)\right).$$E4

In Section 4, we consider the Dirichlet space $\mathcal{D}\left(\mathbb{D}\right)$, which is the set of all analytic functions $f$ in the unit disk $\mathbb{D}$ with the finite Dirichlet integral:

$${\int }_{\mathbb{D}}{\left|f^{\prime }\left(z\right)\right|}^2\frac{\mathrm{d}x\mathrm{d}y}{\pi },z=x+\mathit{iy}.$$E5

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

$${⟨f,g⟩}_{\mathcal{D}\left(\mathbb{D}\right)}=f\left(0\right)\overline{g\left(0\right)}+{\int }_{\mathbb{D}}{f}^{\prime }\left(z\right)\overline{g^{\prime }\left(z\right)}\frac{\mathrm{d}x\mathrm{d}y}{\pi },z=x+\mathit{iy}.$$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 $\mathcal{D}\left(\mathbb{D}\right)$ the following two operators:

$$\mathrm{\Lambda }f\left(z\right)=f^{\prime }\left(z\right)-f^{\prime }\left(0\right),\mathit{Xf}\left(z\right)=z^2{f}^{\prime }\left(z\right),$$E7

and we deduce the uncertainty inequality for this space $\mathcal{D}\left(\mathbb{D}\right)$. And we establish the minimizer function associated with the difference operator:

$$T_{2}f\left(z\right)=\frac{1}{z}\left(f\left(z\right)-z{f}^{\prime }\left(0\right)-f\left(0\right)\right).$$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 $\left[A,B\right]$ by

$$\left[A,B\right]≔\mathit{AB}-\mathit{BA}.$$E9

The adjoint of $A$ denoted by $A^{\ast }$ is defined by

$${⟨\mathit{Af},g⟩}_H={⟨f,A^{\ast }g⟩}_H,$$E10

for $f\in \mathrm{Dom}\left(A\right)$ and $g\in \mathrm{Dom}\left(A^{\ast }\right)$.

Theorem 2.1. For $f\in \mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$, one has

$$\parallel A^{\ast }f{\parallel }_H^2=\parallel \mathit{Af}{\parallel }_H^2+{⟨\left[A,A^{\ast }\right]f,f⟩}_H.$$E11

Proof. Let $f\in \mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$. Then ${\mathit{AA}}^{\ast }f$ and $A^{\ast }\mathit{Af}$ belong to $H$. Therefore $\left[A,A^{\ast }\right]f\in H$. Hence one has

$$\begin{array}{l}\parallel A^{\ast }f{\parallel }_H^2={⟨{\mathit{AA}}^{\ast }f,f⟩}_H={⟨A^{\ast }\mathit{Af},f⟩}_H+{⟨\left[A,A^{\ast }\right]f,f⟩}_H\end{array}$$E12

$$\begin{array}{l}=\parallel \mathit{Af}{\parallel }_H^2+{⟨\left[A,A^{\ast }\right]f,f⟩}_H.\square \end{array}$$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

$$\parallel \left(A-a\right)f{\parallel }_H\parallel \left(B-b\right)f{\parallel }_H\ge \frac{1}{2}\mid {⟨\left[A,B\right]f,f⟩}_H\mid ,$$E14

for all $f\in \mathrm{Dom}\left(\mathit{AB}\right)\cap \mathrm{Dom}\left(\mathit{BA}\right)$, and all $a,b\in \mathbb{R}$.

Theorem 2.3. Let $f\in \mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$. For all $a,b\in \mathbb{R}$, one has

$$\parallel \left(A+A^{\ast }-a\right)f{\parallel }_H\parallel \left(A-A^{\ast }+\mathit{ib}\right)f{\parallel }_H\ge \mid \parallel \mathit{Af}{\parallel }_H^2-\parallel A^{\ast }f{\parallel }_H^2\mid ,$$E15

where $i$ is the imaginary unit.

Proof. Let us consider the following two operators on $\mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$ by

$$P=A+A^{\ast },Q=i\left(A-A^{\ast }\right).$$E16

It follows that, for $f\in \mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$, we have $\mathit{Pf},\mathit{Qf}\in H$. The operators $P$ and $Q$ are self-adjoint and $\left[P,Q\right]=-2i\left[A,A^{\ast }\right]$. Thus the inequality (15) follows from Theorems 2.1 and 2.2. $\square$

Theorem 2.4. Let $f\in \mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$. Then

$${\mathrm{\Delta }}_H^{+}\left(f\right){\mathrm{\Delta }}_H^{-}\left(f\right)\ge \parallel f{\parallel }_H^4{\left(\parallel \mathit{Af}{\parallel }_H^2-\parallel A^{\ast }f{\parallel }_H^2\right)}^2,$$E17

where

$${\mathrm{\Delta }}_H^{\pm }\left(f\right)=\parallel f{\parallel }_H^2\parallel \left(A\pm A^{\ast }\right)f{\parallel }_H^2-{\left|⟨\left(A\pm A^{\ast }\right)f,f⟩{}_H\right|}^2.$$E18

Proof. Let $f\in \mathrm{Dom}\left({\mathit{AA}}^{\ast }\right)\cap \mathrm{Dom}\left(A^{\ast }A\right)$. The operator $P$ given by (16) is self-adjoint; then for any real $a$, we have

$$\parallel \left(P-a\right)f{\parallel }_H^2=\parallel \mathit{Pf}{\parallel }_H^2+a^2\parallel f{\parallel }_H^2-2a{⟨\mathit{Pf},f⟩}_H.$$E19

This shows that

$$\underset{a\in \mathbb{R}}{min}\parallel \left(P-a\right)f{\parallel }_H^2=\parallel \mathit{Pf}{\parallel }_H^2-\frac{{\left|{⟨\mathit{Pf},f⟩}_H\right|}^2}{\parallel f{\parallel }_H^2},$$E20

and the minimum is attained when $a=\frac{{⟨\mathit{Pf},f⟩}_H}{\parallel f{\parallel }_H^2}$. In other words, we have

$$\underset{a\in \mathbb{R}}{min}\parallel \left(A+A^{\ast }-a\right)f{\parallel }_H^2=\parallel \left(A+A^{\ast }\right)f{\parallel }_H^2-\frac{{\left|⟨\left(A+A^{\ast }\right)f,f⟩{}_H\right|}^2}{\parallel f{\parallel }_H^2}.$$E21

Similarly

$$\underset{b\in \mathbb{R}}{min}\parallel \left(A-A^{\ast }+\mathit{ib}\right)f{\parallel }_H^2=\parallel \left(A-A^{\ast }\right)f{\parallel }_H^2-\frac{{\left|⟨\left(A-A^{\ast }\right)f,f⟩{}_H\right|}^2}{\parallel f{\parallel }_H^2}.$$E22

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

Let $\lambda >0$ and let $T:H\to K$ 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 $k\in K$ and for any $\lambda >0$, the problem

$$\underset{f\in H}{inf}\left\{\lambda \parallel f{\parallel }_H^2+\parallel \mathit{Tf}-k{\parallel }_K^2\right\}$$E23

has a unique minimizer given by

$$f_{\lambda ,k}^{\ast }={\left(\mathit{\lambda I}+T^{\ast }T\right)}^{-1}{T}^{\ast }k.$$E24

Proof. The problem (23) is solved elementarily by finding the roots of the first derivative $D\mathrm{\Phi }$ of the quadratic and strictly convex function $\mathrm{\Phi }\left(f\right)=\lambda \parallel f{\parallel }_H^2+\parallel \mathit{Tf}-k{\parallel }_K^2$. Note that for convex functions, the equation $D\mathrm{\Phi }\left(f\right)=0$ is a necessary and sufficient condition for the minimum at $f$. The calculation provides

$$D\mathrm{\Phi }\left(f\right)=2\mathit{\lambda f}+2T^{\ast }\left(\mathit{Tf}-k\right),$$E25

and the assertion of the theorem follows at once. $\square$

Theorem 2.6. If $T:H\to K$ is an isometric isomorphism; then for any $k\in K$ and for any $\lambda >0$, the problem

$$\underset{f\in H}{inf}\left\{\lambda \parallel f{\parallel }_H^2+\parallel \mathit{Tf}-k{\parallel }_K^2\right\}$$E26

has a unique minimizer given by

$$f_{\lambda ,h}^{\ast }=\frac{1}{\lambda +1}{T}^{-1}k.$$E27

Proof. We have $T^{\ast }=T^{-1}$ and $T^{\ast }T=I$. Thus, by (24), we deduce the result. $\square$

3. The Hardy space $H\left(\mathbb{D}\right)$

Let $\mathbb{C}$ be the complex plane and $\mathbb{D}=\left\{z\in \mathbb{C}:|z|<1\right\}$ the open unit disk. The Hardy space $H\left(\mathbb{D}\right)$ is the set of all analytic functions $f$ in the unit disk $\mathbb{D}$ with the finite integral:

$${\int }_0^{2\pi }{\left|f\left(e^{\mathit{i\theta }}\right)\right|}^2\mathrm{d}\theta .$$E28

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

$${⟨f,g⟩}_{H\left(\mathbb{D}\right)}=\frac{1}{2\pi }{\int }_0^{2\pi }f\left(e^{\mathit{i\theta }}\right)\overline{g\left(e^{\mathit{i\theta }}\right)}\mathrm{d}\theta .$$E29

If $f,g\in H\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and $g\left(z\right)={\sum }_{n=0}^{\infty }{b}_n{z}^n$, then

$${⟨f,g⟩}_{H\left(\mathbb{D}\right)}=\sum _{n=0}^{\infty }{a}_n\overline{b_n}.$$E30

The set ${\left\{z^n\right\}}_{n=0}^{\infty }$ forms an Hilbert’s basis for the space $H\left(\mathbb{D}\right)$.

The Szegő kernel $S_z$ given for $z\in \mathbb{D}$, by

$$S_z\left(w\right)=\sum _{n=0}^{\infty }{\overline{z}}^n{w}^n=\frac{1}{1-\overline{z}w},w\in \mathbb{D},$$E31

is a reproducing kernel for the Hardy space $H\left(\mathbb{D}\right)$, meaning that $S_z\in H\left(\mathbb{D}\right)$, and for all $f\in H\left(\mathbb{D}\right)$, we have ${⟨f,S_z⟩}_{H\left(\mathbb{D}\right)}=f\left(z\right)$.

For $z\in \mathbb{D}$, the function $u\left(z\right)=S_{\overline{z}}\left(w\right)$ is the unique analytic solution on $\mathbb{D}$ of the initial problem:

$$u^{\prime }\left(z\right)=w\left(z{u}^{\prime }\left(z\right)+u\left(z\right)\right),w\in \mathbb{D},u\left(0\right)=1.$$E32

In the next of this section, we define the operators $\nabla$, $\Re$, and $L$ on $H\left(\mathbb{D}\right)$ by

$$\nabla f\left(z\right)=f^{\prime }\left(z\right),\mathit{\Re f}\left(z\right)=z{f}^{\prime }\left(z\right),\mathit{Lf}\left(z\right)=z^2{f}^{\prime }\left(z\right)+\mathit{zf}\left(z\right).$$E33

These operators satisfy the commutation rule:

$$\left[\nabla ,L\right]=\nabla L-L\nabla =2\Re +I,$$E34

where $I$ is the identity operator.

We define the Hilbert space $U\left(\mathbb{D}\right)$ as the space of all analytic functions $f$ in the unit disk $\mathbb{D}$ such that

$$\parallel f{\parallel }_{U\left(\mathbb{D}\right)}^2=\frac{1}{2\pi }{\int }_0^{2\pi }{\left|f^{\prime }\left(e^{\mathit{i\theta }}\right)\right|}^2\mathrm{d}\theta <\infty .$$E35

If $f\in U\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$, then

$$\parallel f{\parallel }_{U\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }{n}^2{\left|a_n\right|}^2.$$E36

Thus, the space $U\left(\mathbb{D}\right)$ is a subspace of the Hardy space $H\left(\mathbb{D}\right)$.

Theorem 3.1.

1. For $f\in U\left(\mathbb{D}\right)$, then $\nabla f$, $\mathit{\Re f}$ and $\mathit{Lf}$ belong to $H\left(\mathbb{D}\right)$.

2. ${\nabla }^{\ast }=L$.

3. For $f\in U\left(\mathbb{D}\right)$, one has

$$\parallel \mathit{Lf}{\parallel }_{H\left(\mathbb{D}\right)}^2=\parallel \nabla f{\parallel }_{H\left(\mathbb{D}\right)}^2+\parallel f{\parallel }_{H\left(\mathbb{D}\right)}^2+2{⟨\mathit{\Re f},f⟩}_{H\left(\mathbb{D}\right)}.$$E37

Proof.

1. Let $f\in U\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$. Then

   $$\nabla f\left(z\right)=\sum _{n=0}^{\infty }\left(n+1\right)a_{n+1}{z}^n,\mathit{\Re f}\left(z\right)=\sum _{n=1}^{\infty }{\mathit{na}}_n{z}^n,$$E38

   and

   $$\mathit{Lf}\left(z\right)=\sum _{n=1}^{\infty }{\mathit{na}}_{n-1}{z}^n.$$E39

   Therefore

   $$\parallel \nabla f{\parallel }_{H\left(\mathbb{D}\right)}^2=\sum _{n=0}^{\infty }{\left(n+1\right)}^2{\left|a_{n+1}\right|}^2=\parallel f{\parallel }_{U\left(\mathbb{D}\right)}^2,$$E40
   $$\parallel \mathit{\Re f}{\parallel }_{H\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }{n}^2{\left|a_n\right|}^2=\parallel f{\parallel }_{U_{}\left(\mathbb{D}\right)}^2,$$E41

   and

   $$\parallel \mathit{Lf}{\parallel }_{H\left(\mathbb{D}\right)}^2=\sum _{n=0}^{\infty }{\left(n+1\right)}^2{\left|a_n\right|}^2\le {\left|f\left(0\right)\right|}^2+4\parallel f{\parallel }_{U\left(\mathbb{D}\right)}^2.$$E42

   Consequently $\nabla f$, $\mathit{\Re f}$, and $\mathit{Lf}$ belong to $H\left(\mathbb{D}\right)$.

2. For $f,g\in U\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and $g\left(z\right)={\sum }_{n=0}^{\infty }{b}_n{z}^n$, one has

   $${⟨\nabla f,g⟩}_{H\left(\mathbb{D}\right)}=\sum _{n=0}^{\infty }\left(n+1\right)a_{n+1}\overline{b_n}=\sum _{n=1}^{\infty }{\mathit{na}}_n\overline{b_{n-1}}={⟨f,\mathit{Lg}⟩}_{H\left(\mathbb{D}\right)}.$$E43

   Thus ${\nabla }^{\ast }=L$.

3. Let $f\in U\left(\mathbb{D}\right)$. By (ii) and (34), we deduce that

   $$\begin{array}{l}\parallel \mathit{Lf}{\parallel }_{H\left(\mathbb{D}\right)}^2={⟨\nabla \mathit{Lf},f⟩}_{H\left(\mathbb{D}\right)}\end{array}$$E44
   $$\begin{array}{l}={⟨L\nabla f,f⟩}_{H\left(\mathbb{D}\right)}+{⟨\left[\nabla ,L\right]f,f⟩}_{H\left(\mathbb{D}\right)}\end{array}$$E45
   $$\begin{array}{l}=\parallel \nabla f{\parallel }_{H\left(\mathbb{D}\right)}^2+\parallel f{\parallel }_{H\left(\mathbb{D}\right)}^2+2{⟨\mathit{\Re f},f⟩}_{H\left(\mathbb{D}\right)}.\square \end{array}$$E46

Theorem 3.2. Let $f\in U\left(\mathbb{D}\right)$. For all $a,b\in \mathbb{R}$, one has

$$\parallel \left(\nabla +L-a\right)f{\parallel }_{H\left(\mathbb{D}\right)}\parallel \left(\nabla -L+\mathit{ib}\right)f{\parallel }_{H\left(\mathbb{D}\right)}\ge \parallel f{\parallel }_{H\left(\mathbb{D}\right)}^2+2{⟨\mathit{\Re f},f⟩}_{H\left(\mathbb{D}\right)}.$$E47

Theorem 3.3. Let $T_1$ be the difference operator defined on $H\left(\mathbb{D}\right)$ by

$$T_{1}f\left(z\right)=\frac{1}{z}\left(f\left(z\right)-f\left(0\right)\right).$$E48

1. The operator $T_1$ maps continuously from $H\left(\mathbb{D}\right)$ to $H\left(\mathbb{D}\right)$, and

   $$\parallel T_{1}f{\parallel }_{H\left(\mathbb{D}\right)}\le \parallel f{\parallel }_{H\left(\mathbb{D}\right)}.$$E49

2. For $f\in H\left(\mathbb{D}\right)$ and $z\in \mathbb{D}$, we have

   $$T_1^{\ast }f\left(z\right)=\mathit{zf}\left(z\right),T_1^{\ast }{T}_{1}f\left(z\right)=f\left(z\right)-f\left(0\right).$$E50

3. For any $h\in H\left(\mathbb{D}\right)$ and for any $\lambda >0$, the problem

   $$\underset{f\in H\left(\mathbb{D}\right)}{inf}\left\{\lambda \parallel f{\parallel }_{H\left(\mathbb{D}\right)}^2+\parallel T_{1}f-h{\parallel }_{H\left(\mathbb{D}\right)}^2\right\}$$E51

   has a unique minimizer given by

   $$f_{\lambda ,h}^{\ast }\left(z\right)=\frac{1}{\lambda +1}\mathit{zh}\left(z\right),z\in \mathbb{D}.$$E52

Proof.

1. If $f\in H\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$, then $T_{1}f\left(z\right)={\sum }_{n=0}^{\infty }{a}_{n+1}{z}^n$ and

   $$\parallel T_{1}f{\parallel }_{H\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }{\left|a_n\right|}^2\le \parallel f{\parallel }_{H\left(\mathbb{D}\right)}^2.$$E53

2. If $f,g\in H\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and $g\left(z\right)={\sum }_{n=0}^{\infty }{b}_n{z}^n$, then

   $${⟨T_{1}f,g⟩}_{H\left(\mathbb{D}\right)}=\sum _{n=0}^{\infty }{a}_{n+1}\overline{b_n}=\sum _{n=1}^{\infty }{a}_n\overline{b_{n-1}}={⟨f,T_1^{\ast }g⟩}_{H\left(\mathbb{D}\right)},$$E54

   where $T_1^{\ast }g\left(z\right)=\mathit{zg}\left(z\right)$, for $z\in \mathbb{D}$. And therefore

   $$T_1^{\ast }{T}_{1}f\left(z\right)={\mathit{zT}}_{1}f\left(z\right)=f\left(z\right)-f\left(0\right).$$E55

3. From Theorem 2.5 we have

   $$\left(\mathit{\lambda I}+T_1^{\ast }{T}_1\right)f_{\lambda ,h}^{\ast }\left(z\right)=T_1^{\ast }h\left(z\right).$$E56

By (ii) we deduce that

$$\left(\lambda +1\right)f_{\lambda ,h}^{\ast }\left(z\right)-f_{\lambda ,h}^{\ast }\left(0\right)=\mathit{zh}\left(z\right).$$E57

And from this equation, $f_{\lambda ,h}^{\ast }\left(0\right)=0$. Hence

$$f_{\lambda ,h}^{\ast }\left(z\right)=\frac{1}{\lambda +1}\mathit{zh}\left(z\right).\square$$E58

4. The Dirichlet space $\mathcal{D}\left(\mathbb{D}\right)$

The Dirichlet space $\mathcal{D}\left(\mathbb{D}\right)$ is the set of all analytic functions $f$ in the unit disk $\mathbb{D}$ with the finite Dirichlet integral:

$${\int }_{\mathbb{D}}{\left|f^{\prime }\left(z\right)\right|}^2\frac{\mathrm{d}x\mathrm{d}y}{\pi },z=x+\mathit{iy}.$$E59

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

$${⟨f,g⟩}_{\mathcal{D}\left(\mathbb{D}\right)}=f\left(0\right)\overline{g\left(0\right)}+{\int }_{\mathbb{D}}{f}^{\prime }\left(z\right)\overline{g^{\prime }\left(z\right)}\frac{\mathrm{d}x\mathrm{d}y}{\pi },z=x+\mathit{iy}.$$E60

If $f,g\in \mathcal{D}\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and $g\left(z\right)={\sum }_{n=0}^{\infty }{b}_n{z}^n$, then

$${⟨f,g⟩}_{\mathcal{D}\left(\mathbb{D}\right)}=a_0\overline{b_0}+\sum _{n=1}^{\infty }{\mathit{na}}_n\overline{b_n}.$$E61

The set ${\left\{1,\frac{z^n}{\sqrt{n}}\right\}}_{n=1}^{\infty }$ forms an Hilbert’s basis for the space $\mathcal{D}\left(\mathbb{D}\right)$.

The function $K_z$ given for $z\in \mathbb{D}$, by

$$K_z\left(w\right)=1+log\left(\frac{1}{1-\overline{z}w}\right),w\in \mathbb{D},$$E62

is a reproducing kernel for the Dirichlet space $\mathcal{D}\left(\mathbb{D}\right)$, meaning that $K_z\in \mathcal{D}\left(\mathbb{D}\right)$, and for all $f\in \mathcal{D}\left(\mathbb{D}\right)$, we have ${⟨f,K_z⟩}_{\mathcal{D}\left(\mathbb{D}\right)}=f\left(z\right)$.

For $z\in \mathbb{D}$, the function $u\left(z\right)=K_{\overline{z}}\left(w\right)$ is the unique analytic solution on $\mathbb{D}$ of the initial problem:

$$\frac{u^{\prime }\left(z\right)-u^{\prime }\left(0\right)}{z}=w{u}^{\prime }\left(z\right),w\in \mathbb{D},u\left(0\right)=1.$$E63

In the next of this section, we define the operators $\mathrm{\Lambda }$, $\Re$, and $X$ on $\mathcal{D}\left(\mathbb{D}\right)$ by

$$\mathrm{\Lambda }f\left(z\right)=f^{\prime }\left(z\right)-f^{\prime }\left(0\right),\mathit{\Re f}\left(z\right)=z{f}^{\prime }\left(z\right),\mathit{Xf}\left(z\right)=z^2{f}^{\prime }\left(z\right).$$E64

These operators satisfy the following commutation relation:

$$\left[\mathrm{\Lambda },X\right]=\mathrm{\Lambda }X-X\mathrm{\Lambda }=2\Re .$$E65

We define the Hilbert space $V\left(\mathbb{D}\right)$ as the space of all analytic functions $f$ in the unit disk $\mathbb{D}$ such that

$$\parallel f{\parallel }_{V\left(\mathbb{D}\right)}^2={\int }_{\mathbb{D}}{\left|f^{\prime }\left(z\right)\right|}^2{\left|z\right|}^2\frac{\mathrm{d}x\mathrm{d}y}{\pi }<\infty ,z=x+\mathit{iy}.$$E66

If $f\in V\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$, then

$$\parallel f{\parallel }_{V\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }{n}^3{\left|a_n\right|}^2.$$E67

Thus, the space $V\left(\mathbb{D}\right)$ is a subspace of the Dirichlet space $\mathcal{D}\left(\mathbb{D}\right)$.

Theorem 4.1.

1. For $f\in V\left(\mathbb{D}\right)$, then $\mathrm{\Lambda }f$, $\mathit{\Re f}$, and $\mathit{Xf}$ belong to $\mathcal{D}\left(\mathbb{D}\right)$.

2. ${\mathrm{\Lambda }}^{\ast }=X$.

3. For $f\in V\left(\mathbb{D}\right)$, one has

   $$\parallel \mathit{Xf}{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2=\parallel \mathrm{\Lambda }f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2+2{⟨\mathit{\Re f},f⟩}_{\mathcal{D}\left(\mathbb{D}\right)}.$$E68

Proof.

1. Let $f\in V\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$. Then

   $$\mathrm{\Lambda }f\left(z\right)=\sum _{n=1}^{\infty }\left(n+1\right)a_{n+1}{z}^n,\mathit{\Re f}\left(z\right)=\sum _{n=1}^{\infty }{\mathit{na}}_n{z}^n,$$E69

   and

   $$\mathit{Xf}\left(z\right)=\sum _{n=2}^{\infty }\left(n-1\right)a_{n-1}{z}^n.$$E70

   Therefore

   $$\parallel \mathrm{\Lambda }f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }n{\left(n+1\right)}^2{\left|a_{n+1}\right|}^2\le \sum _{n=2}^{\infty }{n}^3{\left|a_n\right|}^2\le \parallel f{\parallel }_{V\left(\mathbb{D}\right)}^2,$$E71
   $$\parallel \mathit{\Re f}{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }{n}^3{\left|a_n\right|}^2=\parallel f{\parallel }_{V\left(\mathbb{D}\right)}^2,$$E72

   and

   $$\parallel \mathit{Xf}{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2=\sum _{n=1}^{\infty }\left(n+1\right)n^2{\left|a_n\right|}^2\le 2\parallel f{\parallel }_{V\left(\mathbb{D}\right)}^2.$$E73

   Consequently $\mathrm{\Lambda }f$, $\mathit{\Re f}$, and $\mathit{Xf}$ belong to $\mathcal{D}\left(\mathbb{D}\right)$.

2. For $f,g\in V\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and $g\left(z\right)={\sum }_{n=0}^{\infty }{b}_n{z}^n$, one has

   $${⟨\mathrm{\Lambda }f,g⟩}_{\mathcal{D}\left(\mathbb{D}\right)}=\sum _{n=1}^{\infty }n\left(n+1\right)a_{n+1}\overline{b_n}=\sum _{n=2}^{\infty }n\left(n-1\right)a_n\overline{b_{n-1}}={⟨f,\mathit{Xg}⟩}_{\mathcal{D}\left(\mathbb{D}\right)}.$$E74

3. Let $f\in V\left(\mathbb{D}\right)$. By (ii) and (65), we deduce that

   $$\begin{array}{l}\parallel \mathit{Xf}{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2={⟨\mathrm{\Lambda }\mathit{Xf},f⟩}_{\mathcal{D}\left(\mathbb{D}\right)}\end{array}$$E75
   $$\begin{array}{l}={⟨X\mathrm{\Lambda }f,f⟩}_{\mathcal{D}\left(\mathbb{D}\right)}+{⟨\left[\mathrm{\Lambda },X\right]f,f⟩}_{\mathcal{D}\left(\mathbb{D}\right)}\end{array}$$E76
   $$\begin{array}{l}=\parallel \mathrm{\Lambda }f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2+2{⟨\mathit{\Re f},f⟩}_{\mathcal{D}\left(\mathbb{D}\right)}.\square \end{array}$$E77

Theorem 4.2. Let $f\in V\left(\mathbb{D}\right)$. For all $a,b\in \mathbb{R}$, one has

$$\parallel \left(\mathrm{\Lambda }+X-a\right)f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}\parallel \left(\mathrm{\Lambda }-X+\mathit{ib}\right)f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}\ge 2{⟨\mathit{\Re f},f⟩}_{\mathcal{D}\left(\mathbb{D}\right)}.$$E78

Theorem 4.3. Let $T_2$ be the difference operator defined on $\mathcal{D}\left(\mathbb{D}\right)$ by

$$T_{2}f\left(z\right)=\frac{1}{z}\left(f\left(z\right)-z{f}^{\prime }\left(0\right)-f\left(0\right)\right).$$E79

1. The operator $T_2$ maps continuously from $\mathcal{D}\left(\mathbb{D}\right)$ to $\mathcal{D}\left(\mathbb{D}\right)$, and

   $$\parallel T_{2}f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}\le \parallel f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}.$$E80

2. For $f\in \mathcal{D}\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$, we have

   $$T_2^{\ast }f\left(z\right)=\sum _{n=2}^{\infty }\frac{n-1}{n}{a}_{n-1}{z}^n,T_2^{\ast }{T}_{2}f\left(z\right)=\sum _{n=2}^{\infty }\frac{n-1}{n}{a}_n{z}^n.$$E81

3. For any $d\in \mathcal{D}\left(\mathbb{D}\right)$ and for any $\lambda >0$, the problem

   $$\underset{f\in \mathcal{D}\left(\mathbb{D}\right)}{inf}\left\{\lambda \parallel f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2+\parallel T_{2}f-d{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2\right\}$$E82

   has a unique minimizer given by

   $$f_{\lambda ,d}^{\ast }\left(z\right)={⟨d,{\mathrm{\Psi }}_z⟩}_{\mathcal{D}\left(\mathbb{D}\right)},z\in \mathbb{D},$$E83
   $${\mathrm{\Psi }}_z\left(w\right)=\sum _{n=1}^{\infty }\frac{{\overline{z}}^{n+1}}{\lambda \left(n+1\right)+n}{w}^n,w\in \mathbb{D}.$$E84

Proof.

1. If $f\in \mathcal{D}\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$, then $T_{2}f\left(z\right)={\sum }_{n=1}^{\infty }{a}_{n+1}{z}^n$ and

   $$\parallel T_{2}f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2=\sum _{n=2}^{\infty }\left(n-1\right){\left|a_n\right|}^2\le \sum _{n=2}^{\infty }n{\left|a_n\right|}^2\le \parallel f{\parallel }_{\mathcal{D}\left(\mathbb{D}\right)}^2.$$E85

2. If $f,g\in \mathcal{D}\left(\mathbb{D}\right)$ with $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ and $g\left(z\right)={\sum }_{n=0}^{\infty }{b}_n{z}^n$, then

   $${⟨T_{2}f,g⟩}_{\mathcal{D}\left(\mathbb{D}\right)}=\sum _{n=1}^{\infty }{\mathit{na}}_{n+1}\overline{b_n}=\sum _{n=2}^{\infty }\left(n-1\right)a_n\overline{b_{n-1}}={⟨f,T_2^{\ast }g⟩}_{\mathcal{D}\left(\mathbb{D}\right)},$$E86

   where

   $$T_2^{\ast }g\left(z\right)=\sum _{n=2}^{\infty }\frac{n-1}{n}{b}_{n-1}{z}^n,z\in \mathbb{D}.$$E87

   And therefore

   $$T_2^{\ast }{T}_{2}f\left(z\right)=\sum _{n=2}^{\infty }\frac{n-1}{n}{a}_n{z}^n.$$E88

3. We put $d\left(z\right)={\sum }_{n=0}^{\infty }{d}_n{z}^n$ and

   $$f_{\lambda ,d}^{\ast }\left(z\right)=\sum _{n=0}^{\infty }{c}_n{z}^n.$$E89

From (ii) and the equation

$$\left(\mathit{\lambda I}+T_2^{\ast }{T}_2\right)f_{\lambda ,d}^{\ast }\left(z\right)=T_2^{\ast }d\left(z\right),$$E90

we deduce that

$$c_1=c_0=0,c_n=\frac{n-1}{\mathit{\lambda n}+n-1}{d}_{n-1},n\ge 2.$$E91

Thus

$$f_{\lambda ,d}^{\ast }\left(z\right)=\sum _{n=1}^{\infty }\frac{{\mathit{nd}}_n}{\lambda \left(n+1\right)+n}{z}^{n+1}={⟨d,{\mathrm{\Psi }}_z⟩}_{\mathcal{D}\left(\mathbb{D}\right)},z\in \mathbb{D}.\square$$E92