**3. The Hardy space** *H*ð Þ

Let be the complex plane and ¼ f g *z*∈ :j*z*j<1 the open unit disk. The Hardy space *H*ð Þ is the set of all analytic functions *f* in the unit disk with the finite integral:

$$\int\_{0}^{2\pi} \left| f\left(\boldsymbol{\epsilon}^{i\theta}\right) \right|^{2} \,\mathrm{d}\theta. \tag{28}$$

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

$$\langle f, \mathbf{g} \rangle\_{H(\mathbb{D})} = \frac{1}{2\pi} \int\_0^{2\pi} f\left(e^{i\theta}\right) \overline{\mathbf{g}\left(e^{i\theta}\right)} \,\mathrm{d}\theta. \tag{29}$$

If *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> *<sup>H</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*bnzn*, then

$$\langle f, \mathfrak{g} \rangle\_{H(\mathbb{D})} = \sum\_{n=0}^{\infty} a\_n \overline{b\_n}. \tag{30}$$

*Analytical Applications on Some Hilbert Spaces DOI: http://dx.doi.org/10.5772/intechopen.90322*

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

**Theorem 2.5.** For any *k*∈*K* and for any *λ*>0, the problem

*λ*∥*f* ∥<sup>2</sup>

derivative *<sup>D</sup>*<sup>Φ</sup> of the quadratic and strictly convex function <sup>Φ</sup>ð Þ¼ *<sup>f</sup> <sup>λ</sup>*∥*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

*λ*∥*f* ∥<sup>2</sup>

*f* ∗ *<sup>λ</sup>*,*<sup>h</sup>* <sup>¼</sup> <sup>1</sup> *λ* þ 1

sufficient condition for the minimum at *f*. The calculation provides

and the assertion of the theorem follows at once. □

inf *f* ∈ *H*

*<sup>λ</sup>*,*<sup>k</sup>* <sup>¼</sup> *<sup>λ</sup><sup>I</sup>* <sup>þ</sup> *<sup>T</sup>*<sup>∗</sup> ð Þ *<sup>T</sup>* �<sup>1</sup>

**Proof.** The problem (23) is solved elementarily by finding the roots of the first

**Theorem 2.6.** If *T* : *H* ! *K* is an isometric isomorphism; then for any *k*∈*K* and

*<sup>H</sup>* <sup>þ</sup> <sup>∥</sup>*Tf* � *<sup>k</sup>*∥<sup>2</sup>

*T*�<sup>1</sup>

**Proof.** We have *<sup>T</sup>*<sup>∗</sup> <sup>¼</sup> *<sup>T</sup>*�<sup>1</sup> and *<sup>T</sup>* <sup>∗</sup> *<sup>T</sup>* <sup>¼</sup> *<sup>I</sup>*. Thus, by (24), we deduce the result. □

Let be the complex plane and ¼ f g *z*∈ :j*z*j<1 the open unit disk. The Hardy space *H*ð Þ is the set of all analytic functions *f* in the unit disk with the

> *f e<sup>i</sup><sup>θ</sup>* � � � � � � 2

> > 2*π* ð<sup>2</sup>*<sup>π</sup>* 0

h i *<sup>f</sup>*, *<sup>g</sup> <sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

*<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup>

*n*¼0

ð<sup>2</sup>*<sup>π</sup>* 0

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

h i *<sup>f</sup>*, *<sup>g</sup> <sup>H</sup>*ð Þ <sup>¼</sup> <sup>1</sup>

*<sup>K</sup>*. Note that for convex functions, the equation *D*Φð Þ¼ *f* 0 is a necessary and

inf *f* ∈ *H*

*f* ∗

associated with the operator *T*.

*Functional Calculus*

for any *λ*>0, the problem

**3. The Hardy space** *H*ð Þ

If *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> *<sup>H</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup>

finite integral:

**82**

has a unique minimizer given by

*k*∥<sup>2</sup>

has a unique minimizer given by

Let *λ*>0 and let *T* : *H* ! *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

*<sup>H</sup>* <sup>þ</sup> <sup>∥</sup>*Tf* � *<sup>k</sup>*∥<sup>2</sup>

*K*

*<sup>D</sup>*Φð Þ¼ *<sup>f</sup>* <sup>2</sup>*λ<sup>f</sup>* <sup>þ</sup> <sup>2</sup>*T*<sup>∗</sup> ð Þ *Tf* � *<sup>k</sup>* , (25)

*K*

� � (26)

� � (23)

*T*<sup>∗</sup> *k:* (24)

*k:* (27)

d*θ:* (28)

*f e<sup>i</sup><sup>θ</sup>* � �*g ei<sup>θ</sup>* ð Þd*θ:* (29)

*anbn:* (30)

*<sup>n</sup>*¼<sup>0</sup>*bnzn*, then

*<sup>H</sup>* þ ∥*Tf* �

The set *<sup>z</sup><sup>n</sup>* f g<sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup> forms an Hilbert's basis for the space *<sup>H</sup>*ð Þ . The Szegő kernel *Sz* given for *z*∈ , by

$$S\_{\overline{x}}(w) = \sum\_{n=0}^{\infty} \overline{z}^n w^n = \frac{1}{1 - \overline{z}w}, \quad w \in \mathbb{D}, \tag{31}$$

is a reproducing kernel for the Hardy space *H*ð Þ , meaning that *Sz* ∈ *H*ð Þ , and for all *<sup>f</sup>* <sup>∈</sup> *<sup>H</sup>*ð Þ , we have h i *<sup>f</sup>*, *Sz <sup>H</sup>*ð Þ <sup>¼</sup> *f z*ð Þ.

For *z*∈ , the function *u z*ð Þ¼ *Sz*ð Þ *w* is the unique analytic solution on of the initial problem:

$$u'(z) = w(zu'(z) + u(z)), \quad w \in \mathbb{D}, \quad u(0) = 1. \tag{32}$$

In the next of this section, we define the operators ∇, ℜ, and *L* on *H*ð Þ by

$$
\nabla f(\mathbf{z}) = f'(\mathbf{z}), \quad \Re f(\mathbf{z}) = zf'(\mathbf{z}), \quad \mathbf{L}f(\mathbf{z}) = z^2 f'(\mathbf{z}) + zf(\mathbf{z}).\tag{33}
$$

These operators satisfy the commutation rule:

$$[\nabla, L] = \nabla L - L\nabla = \mathfrak{R}\mathfrak{R} + I,\tag{34}$$

where *I* is the identity operator.

We define the Hilbert space *U*ð Þ as the space of all analytic functions *f* in the unit disk such that

$$\left\|f\right\|\_{U(\mathbb{D})}^2 = \frac{1}{2\pi} \int\_0^{2\pi} \left|\left.f'(e^{i\theta})\right|^2 \mathrm{d}\theta < \infty. \tag{35}$$

If *<sup>f</sup>* <sup>∈</sup> *<sup>U</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*anzn*, then

$$\|f\|\_{U(\mathbb{D})}^2 = \sum\_{n=1}^{\infty} n^2 |a\_n|^2. \tag{36}$$

Thus, the space *U*ð Þ is a subspace of the Hardy space *H*ð Þ . **Theorem 3.1.**


$$\|Lf\|\_{H(\mathbb{D})}^2 = \|\nabla f\|\_{H(\mathbb{D})}^2 + \|f\|\_{H(\mathbb{D})}^2 + \mathcal{Z}\langle \Re f, f \rangle\_{H(\mathbb{D})}.\tag{37}$$

#### **Proof.**

i. Let *<sup>f</sup>* <sup>∈</sup> *<sup>U</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>*. Then

$$\nabla f(z) = \sum\_{n=0}^{\infty} (n+1)a\_{n+1}z^n, \quad \Re f(z) = \sum\_{n=1}^{\infty} n a\_n z^n,\tag{38}$$

and

$$Lf(\mathbf{z}) = \sum\_{n=1}^{\infty} n a\_{n-1} \mathbf{z}^n. \tag{39}$$

Therefore

$$\|\|\nabla f\|\|\_{H(\mathbb{D})}^2 = \sum\_{n=0}^{\infty} (n+1)^2 |a\_{n+1}|^2 = \|f\|\_{U(\mathbb{D})}^2,\tag{40}$$

$$\left\|\Re f\right\|\_{H(\mathbb{D})}^2 = \sum\_{n=1}^{\infty} n^2 |a\_n|^2 = \left\|f\right\|\_{U(\mathbb{D})}^2,\tag{41}$$

**Proof.**

where *T*<sup>∗</sup>

i. If *<sup>f</sup>* <sup>∈</sup> *<sup>H</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup>

*Analytical Applications on Some Hilbert Spaces DOI: http://dx.doi.org/10.5772/intechopen.90322*

ii. If *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> *<sup>H</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup>

iii. From Theorem 2.5 we have

By (ii) we deduce that

And from this equation, *f* <sup>∗</sup>

**4. The Dirichlet space** Dð Þ

with the finite Dirichlet integral:

h i *<sup>T</sup>*1*f*, *<sup>g</sup> <sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

*T* ∗

∥*T*1*f* ∥<sup>2</sup>

*n*¼0

<sup>1</sup> *g z*ð Þ¼ *zg z*ð Þ, for *z*∈ . And therefore

*<sup>λ</sup><sup>I</sup>* <sup>þ</sup> *<sup>T</sup>*<sup>∗</sup> <sup>1</sup> *T*<sup>1</sup> � �*f* <sup>∗</sup>

ð Þ *<sup>λ</sup>* <sup>þ</sup> <sup>1</sup> *<sup>f</sup>* <sup>∗</sup>

*f* ∗ *<sup>λ</sup>*,*<sup>h</sup>*ð Þ¼ *z*

ð *f* 0 ð Þ*<sup>z</sup>* � � � � <sup>2</sup> d*x*d*y*

h i *<sup>f</sup>*, *<sup>g</sup>* Dð Þ <sup>¼</sup> *<sup>f</sup>*ð Þ <sup>0</sup> *<sup>g</sup>*ð Þþ <sup>0</sup>

If *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> Dð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup>

ffiffi *n* p n o<sup>∞</sup>

*n*¼1

The function *Kz* given for *z*∈ , by

The set 1, *<sup>z</sup><sup>n</sup>*

**85**

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

ð *f* 0 ð Þ*z g*<sup>0</sup> ð Þ*z*

*Kz*ð Þ¼ *<sup>w</sup>* <sup>1</sup> <sup>þ</sup> log <sup>1</sup>

*<sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

*n*¼1

*an*þ<sup>1</sup>*bn* <sup>¼</sup> <sup>X</sup><sup>∞</sup>

*<sup>λ</sup>*,*<sup>h</sup>*ð Þ� *<sup>z</sup> <sup>f</sup>* <sup>∗</sup>

1 *λ* þ 1

The Dirichlet space Dð Þ is the set of all analytic functions *f* in the unit disk

*<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup>

forms an Hilbert's basis for the space Dð Þ .

h i *<sup>f</sup>*, *<sup>g</sup>* Dð Þ <sup>¼</sup> *<sup>a</sup>*0*b*<sup>0</sup> <sup>þ</sup>X<sup>∞</sup>

d*x*d*y*

*n*¼1

*<sup>λ</sup>*,*<sup>h</sup>*ð Þ¼ 0 0. Hence

*<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>*, then *<sup>T</sup>*1*f z*ð Þ¼ <sup>P</sup><sup>∞</sup>

*<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup>

*<sup>λ</sup>*,*<sup>h</sup>*ð Þ¼ *<sup>z</sup> <sup>T</sup>* <sup>∗</sup>

*n*¼1

j j *an* <sup>2</sup> ≤ ∥*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

*anbn*�<sup>1</sup> <sup>¼</sup> *<sup>f</sup>*, *<sup>T</sup>* <sup>∗</sup>

<sup>1</sup> *T*1*f z*ð Þ¼ *zT*1*f z*ð Þ¼ *f z*ð Þ� *f*ð Þ 0 *:* (55)

*<sup>n</sup>*¼<sup>0</sup>*an*þ1*z<sup>n</sup>* and

*<sup>n</sup>*¼<sup>0</sup>*bnz<sup>n</sup>*, then

<sup>1</sup> *<sup>g</sup>* � �

*<sup>H</sup>*ð Þ *:* (53)

<sup>1</sup> *h z*ð Þ*:* (56)

*<sup>λ</sup>*,*<sup>h</sup>*ð Þ¼ 0 *zh z*ð Þ*:* (57)

*zh z*ð Þ*:* □ (58)

*<sup>π</sup>* , *<sup>z</sup>* <sup>¼</sup> *<sup>x</sup>* <sup>þ</sup> *iy:* (59)

*<sup>n</sup>*¼<sup>0</sup>*bnz<sup>n</sup>*, then

<sup>1</sup> � *zw* � �, *<sup>w</sup>* <sup>∈</sup> , (62)

*<sup>π</sup>* , *<sup>z</sup>* <sup>¼</sup> *<sup>x</sup>* <sup>þ</sup> *iy:* (60)

*nanbn:* (61)

*<sup>H</sup>*ð Þ , (54)

and

$$\|\|Lf\|\|\_{H(\mathbb{D})}^2 = \sum\_{n=0}^{\infty} (n+1)^2 |a\_n|^2 \le |f(\mathbb{D})|^2 + 4\|f\|\_{U(\mathbb{D})}^2. \tag{42}$$

Consequently ∇*f*, ℜ*f*, and *Lf* belong to *H*ð Þ .

ii. For *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> *<sup>U</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*anzn* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*bnz<sup>n</sup>*, one has

$$\langle \nabla f, \mathbf{g} \rangle\_{H(\mathbb{D})} = \sum\_{n=0}^{\infty} (n+1) a\_{n+1} \overline{b\_n} = \sum\_{n=1}^{\infty} n a\_n \overline{b\_{n-1}} = \langle f, \mathbf{L} \mathbf{g} \rangle\_{H(\mathbb{D})}.\tag{43}$$

Thus <sup>∇</sup><sup>∗</sup> <sup>¼</sup> *<sup>L</sup>*.

iii. Let *f* ∈ *U*ð Þ . By (ii) and (34), we deduce that

$$\left\| Lf \right\|\_{H(\mathbb{D})}^2 = \langle \nabla Lf, f \rangle\_{H(\mathbb{D})} \tag{44}$$

$$=\langle L\nabla f, f\rangle\_{H(\mathbb{D})} + \langle [\nabla, L]f, f\rangle\_{H(\mathbb{D})}\tag{45}$$

$$\mathcal{J} = \left\| \nabla f \right\|\_{H(\mathbb{D})}^2 + \left\| f \right\|\_{H(\mathbb{D})}^2 + \mathcal{Z} \langle \Re f, f \rangle\_{H(\mathbb{D})}. \quad \Box \tag{46}$$

**Theorem 3.2.** Let *f* ∈ *U*ð Þ . For all *a*, *b*∈ , one has

$$\|(\nabla + L - a)f\|\_{H(\mathbb{D})} \|(\nabla - L + ib)f\|\_{H(\mathbb{D})} \ge \|f\|\_{H(\mathbb{D})}^2 + 2\langle \Re f, f \rangle\_{H(\mathbb{D})}.\tag{47}$$

**Theorem 3.3.** Let *T*<sup>1</sup> be the difference operator defined on *H*ð Þ by

$$T\_{\mathfrak{J}}f(z) = \frac{1}{z}(\, \! \! / f(z) \, \! - \! \! / \! 0 \, \! / \! /). \tag{48}$$

i. The operator *T*<sup>1</sup> maps continuously from *H*ð Þ to *H*ð Þ , and

$$\|Tf\|\_{H(\mathbb{D})} \le \|f\|\_{H(\mathbb{D})}.\tag{49}$$

ii. For *f* ∈ *H*ð Þ and *z*∈ , we have

$$T\_1^\* f(z) = zf(z), \quad T\_1^\* T\_1 f(z) = f(z) - f(0). \tag{50}$$

iii. For any *h*∈ *H*ð Þ and for any *λ*> 0, the problem

$$\inf\_{f \in H(\mathbb{D})} \left\{ \lambda \| f \|\_{H(\mathbb{D})}^2 + \| T\_{\mathfrak{I}} f - h \|\_{H(\mathbb{D})}^2 \right\} \tag{51}$$

has a unique minimizer given by

$$f\_{\lambda,h}^\*(z) = \frac{1}{\lambda + 1} zh(z), \quad z \in \mathbb{D}.\tag{52}$$

#### **Proof.**

Therefore

*Functional Calculus*

and

∥∇*f* ∥<sup>2</sup>

∥*Lf* ∥<sup>2</sup>

ii. For *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> *<sup>U</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup>

h i <sup>∇</sup>*f*, *<sup>g</sup> <sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

Thus <sup>∇</sup><sup>∗</sup> <sup>¼</sup> *<sup>L</sup>*.

*<sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

∥ℜ*f* ∥<sup>2</sup>

*<sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

Consequently ∇*f*, ℜ*f*, and *Lf* belong to *H*ð Þ .

*n*¼0

iii. Let *f* ∈ *U*ð Þ . By (ii) and (34), we deduce that

<sup>¼</sup> ∥∇*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

ii. For *f* ∈ *H*ð Þ and *z*∈ , we have *T* ∗

has a unique minimizer given by

**84**

**Theorem 3.2.** Let *f* ∈ *U*ð Þ . For all *a*, *b*∈ , one has

∥ ∇ð Þ <sup>þ</sup> *<sup>L</sup>* � *<sup>a</sup> <sup>f</sup>* <sup>∥</sup>*<sup>H</sup>*ð Þ ∥ ∇ð Þ � *<sup>L</sup>* <sup>þ</sup> *ib <sup>f</sup>* <sup>∥</sup>*<sup>H</sup>*ð Þ ≥ ∥*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

*n*¼0

*n*¼0

*<sup>H</sup>*ð Þ <sup>¼</sup> <sup>X</sup><sup>∞</sup>

ð Þ *<sup>n</sup>* <sup>þ</sup> <sup>1</sup> <sup>2</sup>

ð Þ *<sup>n</sup>* <sup>þ</sup> <sup>1</sup> *an*þ<sup>1</sup>*bn* <sup>¼</sup> <sup>X</sup><sup>∞</sup>

∥*Lf* ∥<sup>2</sup>

*<sup>H</sup>*ð Þ <sup>þ</sup> <sup>∥</sup>*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

**Theorem 3.3.** Let *T*<sup>1</sup> be the difference operator defined on *H*ð Þ by

*z*

*<sup>T</sup>*1*f z*ð Þ¼ <sup>1</sup>

i. The operator *T*<sup>1</sup> maps continuously from *H*ð Þ to *H*ð Þ , and

<sup>1</sup> *f z*ð Þ¼ *zf z*ð Þ, *<sup>T</sup>*<sup>∗</sup>

*λ*∥*f* ∥<sup>2</sup>

1 *λ* þ 1

*<sup>H</sup>*ð Þ <sup>þ</sup> <sup>∥</sup>*T*1*<sup>f</sup>* � *<sup>h</sup>*∥<sup>2</sup>

n o

iii. For any *h*∈ *H*ð Þ and for any *λ*> 0, the problem

inf *f* ∈ *H*ð Þ

> *f* ∗ *<sup>λ</sup>*,*<sup>h</sup>*ð Þ¼ *z*

ð Þ *<sup>n</sup>* <sup>þ</sup> <sup>1</sup> <sup>2</sup>

*n*¼1 *n*2 j j *an*þ<sup>1</sup>

j j *an* <sup>2</sup> <sup>¼</sup> <sup>∥</sup>*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

*<sup>n</sup>*¼<sup>0</sup>*anzn* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup>

*n*¼1

j j *an* <sup>2</sup> <sup>≤</sup> j j *<sup>f</sup>*ð Þ <sup>0</sup> <sup>2</sup> <sup>þ</sup> <sup>4</sup>∥*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

<sup>2</sup> <sup>¼</sup> <sup>∥</sup>*<sup>f</sup>* <sup>∥</sup><sup>2</sup>

*<sup>U</sup>*ð Þ , (40)

*<sup>U</sup>*ð Þ *:* (42)

*<sup>U</sup>*ð Þ , (41)

*<sup>n</sup>*¼<sup>0</sup>*bnz<sup>n</sup>*, one has

*nanbn*�<sup>1</sup> <sup>¼</sup> h i *<sup>f</sup>*, *Lg <sup>H</sup>*ð Þ *:* (43)

*<sup>H</sup>*ð Þ <sup>¼</sup> h i <sup>∇</sup>*Lf*, *<sup>f</sup> <sup>H</sup>*ð Þ (44)

*<sup>H</sup>*ð Þ <sup>þ</sup> <sup>2</sup>h i <sup>ℜ</sup>*f*, *<sup>f</sup> <sup>H</sup>*ð Þ *:* □ (46)

ð Þ *f z*ð Þ� *f*ð Þ 0 *:* (48)

<sup>1</sup> *T*1*f z*ð Þ¼ *f z*ð Þ� *f*ð Þ 0 *:* (50)

*zh z*ð Þ, *z*∈ *:* (52)

(51)

∥*T*1*f* ∥*<sup>H</sup>*ð Þ ≤ ∥*f* ∥*<sup>H</sup>*ð Þ *:* (49)

*H*ð Þ

*<sup>H</sup>*ð Þ <sup>þ</sup> <sup>2</sup>h i <sup>ℜ</sup>*f*, *<sup>f</sup> <sup>H</sup>*ð Þ *:* (47)

<sup>¼</sup> h i *<sup>L</sup>*∇*f*, *<sup>f</sup> <sup>H</sup>*ð Þ <sup>þ</sup> h i ½ � <sup>∇</sup>, *<sup>L</sup> <sup>f</sup>*, *<sup>f</sup> <sup>H</sup>*ð Þ (45)

i.  $\|Tf\in H(\mathbb{D})\text{ with}\,f(\mathbf{z}) = \sum\_{n=0}^{\infty} a\_n \mathbf{z}^n$ , then  $T\circ f(\mathbf{z}) = \sum\_{n=0}^{\infty} a\_{n+1} \mathbf{z}^n$  and 
$$\|Tf\|\_{H(\mathbb{D})}^2 = \sum\_{n=1}^{\infty} |a\_n|^2 \le \|f\|\_{H(\mathbb{D})}^2. \tag{53}$$

ii. If *<sup>f</sup>*, *<sup>g</sup>* <sup>∈</sup> *<sup>H</sup>*ð Þ with *f z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*anz<sup>n</sup>* and *g z*ð Þ¼ <sup>P</sup><sup>∞</sup> *<sup>n</sup>*¼<sup>0</sup>*bnz<sup>n</sup>*, then

$$\langle T\_{\mathbf{j}}f, \mathbf{g} \rangle\_{H(\mathbb{D})} = \sum\_{n=0}^{\infty} a\_{n+1} \overline{b\_n} = \sum\_{n=1}^{\infty} a\_n \overline{b\_{n-1}} = \langle f, T\_1^\* \mathbf{g} \rangle\_{H(\mathbb{D})},\tag{54}$$

where *T*<sup>∗</sup> <sup>1</sup> *g z*ð Þ¼ *zg z*ð Þ, for *z*∈ . And therefore

$$T\_1^\* T\_1 f(z) = z T\_1 f(z) = f(z) - f(0). \tag{55}$$

iii. From Theorem 2.5 we have

$$(\lambda I + T\_1^\* T\_1) f\_{\lambda \mu}^\*(z) = T\_1^\* h(z). \tag{56}$$

By (ii) we deduce that

$$(\lambda + \mathbf{1})f\_{\lambda,h}^\*(z) - f\_{\lambda,h}^\*(\mathbf{0}) = zh(z). \tag{57}$$

And from this equation, *f* <sup>∗</sup> *<sup>λ</sup>*,*<sup>h</sup>*ð Þ¼ 0 0. Hence

$$f\_{\lambda,h}^\*(z) = \frac{1}{\lambda + 1} z h(z). \quad \Box \tag{58}$$
