EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 6, No. 3, 2013, 340-351
ISSN 1307-5543 – www.ejpam.com

A New Generalization of the Operator-Valued Poisson Kernel

Sharifa Al-Sharif1, Fatima Salem1, Basem Frasin2

1 Department of Mathematics, Yarmouk University, Irbed, Jordan
2 Department of Mathematics, Al al-Bayt University, Almafrag, Jordan

Abstract. The purpose of this paper is to give a new generalization of the operator-valued Poisson
kernel and discuss integral formulas for them.

2010 Mathematics Subject Classifications: 45P05, 47A60; 46E40, 47B38

Key Words and Phrases: Poisson kernel, Operator-valued Poisson kernel

1. Introduction

Let H be a complex Hilbert space and L (H ) denote the algebra of all bounded linear
operators from H into H . For T ∈ L (H ), its spectrum σ (T ) is the non-empty compact
subset of the complex plane C consisting of all λ ∈ C such that T − λI is non-invertible
in L (H ), where I is the identity operator on H . We write D for the open unit disk in C,
D= {z : |z|< 1}.

Let A ∈ L (H ). For a complex valued function f analytic on a domain E of the complex
plane containing the spectrum σ (A) of A, we recall Riesz-Dunford integral f (A)which is given
by

f (A) =
1

2πi

∫

C

f (z) (zI − A)−1 dz, (1)

where C is a positively oriented simple closed rectifiable contour containing σ (A).
By differentiating the integral in equation (1) with respect to A we get

f ′ (A) =
1

2πi

∫

C

f (z) (zI − A)−2 dz. (2)

Email addresses: sharifa@yu.edu.jo (S. Sharif) fatimabayui@yahoo.com (F. Salem),
bafrasin@yahoo.com (B. Frasin)

http://www.ejpam.com 340 c© 2013 EJPAM All rights reserved.



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 341

If we differentiate the integral in equation (2) with respect to A, (n− 1) times, we get

f (n) (A) =
n!

2πi

∫

C

f (z) (zI − A)−n−1 dz, (n= 0,1, 2, . . .) . (3)

Note that, expression (3) is an extension of the Riesz-Dunford integral in equation (1).
For rei t ∈ D, the (scalar) Poisson kernel Pr,t is defined by

Pr,t

�

eiθ
�

=
1− r2

�

1− rei t e−iθ
��

1− re−i t eiθ
�

=
1

1− rei t e−iθ
+

1

1− re−i t eiθ
− 1

=
∑

n≥0

rneint e−inθ +
∑

n≥0

rne−int einθ − 1. (4)

The integral formula of the (scalar) Poisson kernel

1

2π

2π
∫

0

Pr,t

�

eiθ
�

dθ = 1,

holds, where r is a real parameter satisfying |r|< 1, see[3].
For T ∈ L (H ), σ (T ) ⊂ D and rei t ∈ D, the author in [2], define the operator-valued

Poisson kernel Kr,t (T ) as follows

Kr,t (T ) =
�

I − rei t T ∗
�−1
+
�

I − re−i t T
�−1
− I , (5)

and prove the following theorem.

Theorem 1. For T ∈ L (H ) such that σ (T )⊂ D, we have

Kr,t (T ) =
�

I − rei t T ∗
�−1 �

I − r2T ∗T
��

I − re−i t T
�−1

=
∑

n≥0

rneint T ∗n+
∑

n≥0

rne−int T n− I .

Afterwards, in [1] Bulut proved the following theorem.

Theorem 2. For T ∈ L (H ) such that σ (T )⊂ D, we have

1

2π

2π
∫

0

Kr,t (T ) d t = I , (6)

where r is a real parameter satisfying |r|< 1.



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 342

A generalization of the (scalar) Poisson kernel, (4) in [3] is given by

Qa,b,t

�

eiθ
�

=
1− ab

�

1− aei t e−iθ
��

1− be−i t eiθ
� , (7)

where a and b are complex parameters satisfying |a|< 1 and |b|< 1.
In [1], Bulut introduced a generalization of the operator-valued Poisson kernel Kr,t (T ) for

T ∈ L (H ), σ (T )⊂ D and rei t ∈ D in the following way

Qa,b,t (T ) =
�

I − aei t T ∗
�−1
+
�

I − be−i t T
�−1
− I , (8)

where a and b are complex parameters satisfying |a|< 1 and |b|< 1 and prove the following
theorem.

Theorem 3 ([1]). Let T ∈ L (H ) such that σ (T )⊂ D. Then

1

2π

2π
∫

0

Qa,b,t (T ) d t = I , (9)

where a and b are complex parameters satisfying |a|< 1 and |b|< 1.

Remark 1. We note that (8) and (9) are generalizations of (5) and (6), respectively, by taking
a = b = r.

2. A New Generalization of the Operator-Valued Poisson Kernel

In this section, we set the following definition and open problem.

Definition 1. Let T ∈ L (H ) such that σ (T )⊂ D. For n= 0, 1,2, . . ., let

In = de f
1

2π

2π
∫

0

Qn+1
a,b,t (T ) d t,

where a, b, are complex parameters satisfying |a|< 1 and |b|< 1.

Open Problem: Compute In, n= 0, 1,2, . . ..
In the following theorem we give a partial answer to the open problem to certain class of

operators in L (H ).

Theorem 4. Let T ∈ L (H ) such that σ (T )⊂ D and (I − aei t T ∗) is self adjoint. Then

2π
∫

0

Qa,a,t (T )
n+1 d t =

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

(−I)l , (10)

for n= 0,1, 2, . . ., and a complex parameter a satisfying |a|< 1.



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 343

Proof. Let

In =

2π
∫

0

Qa,a,t (T )
n+1 d t =

1

2π

2π
∫

0

�

�

I − aei t T ∗
�−1
+
�

I − ae−i t T
�−1
− I
�n+1

d t

=
1

2π

2π
∫

0

n+1
∑

k=0

�

n+ 1

k

�

�

I − aei t T ∗
�−n−1+k

�

�

I − ae−i t T
�−1
+ (−I)

�k
d t

=
1

2π

2π
∫

0

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

�

I − aei t T ∗
�−n−1+k �

I − ae−i t T
�−k+l

(−I)l d t

=
1

2π

2π
∫

0

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

�

I − aei t T ∗
�−n−1+k

�

�

I − aei t T ∗
�∗�−k+l

(−I)l d t

=
1

2π

2π
∫

0

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

�

I − aei t T ∗
�−n−1+l

(−I)l d t

=
1

2π

2π
∫

0

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

e−(n+1−l)i t
�

e−i t I − aT ∗
�−n−1+l

(−I)l d t. (11)

By the change of variables, with z = e−i t , (11) becomes

In =
−1

2πi

∮

|z|=1

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

�

zI − aT ∗
�−n−1+l (−I)lzn−l dz

=
−1

2πi

n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�
∮

|z|=1

�

zI − aT ∗
�−n−1+l (−I)lzn−l dz,

where the integral along |z| = 1 is taken in the negative direction. Hence, by the Riesz-
Dunford integral (3), we have

In =
n+1
∑

k=0

k
∑

l=0

�

n+ 1

k

��

k

l

�

(−I)l , (n= 0, 1,2, . . .) .

Corollary 1. For T ∈ L (H ) such that σ (T )⊂ D and
�

I − aei t T ∗
�

is self adjoint, we have

1

2π

2π
∫

0

Qa,a,t (T )
2 d t = I , (12)



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 344

where a is complex parameter satisfying |a|< 1.

Remark 2. By taking n= 0 in (10), we obtain (9).

Definition 2. For T ∈ L (H ) such that σ (T )⊂ D, we set a generalization of the operator-valued
Poisson kernel, Qa,b,t (T ) in the following way:

Ra,b,c,d,t (T ) =
�

I − aei t T ∗
�−1
+
�

I − be−i t T
�−1
+
�

I − cei t T ∗
�−1
−
�

I − de−i t T
�−1
− I , (13)

where a, b, c, and d are complex parameters satisfying |a|< 1, |b|< 1, |c|< 1, and |d|< 1.

Remark 3. Note that Ra,b,c,d,t (T ) ∈ L (H ).

Lemma 1. For T ∈ L (H ) such that σ (T )⊂ D, we have

Ra,b,c,d,t (T ) =
∑

n≥0

aneint T ∗n+
∑

n≥0

bne−int T n+
∑

n≥0

cneint T ∗n−
∑

n≥0

dne−int T n− I . (14)

Proof. Since
�

�

�

�aei t T ∗
�

�

�

�< 1,
�

�

�

�be−i t T
�

�

�

�< 1,
�

�

�

�cei t T ∗
�

�

�

�< 1, and
�

�

�

�de−i t T
�

�

�

�< 1, we have

∑

n≥0

aneint T ∗n =
�

I − aei t T ∗
�−1

,

∑

n≥0

bne−int T n =
�

I − be−i t T
�−1

,

∑

n≥0

cneint T ∗n =
�

I − cei t T ∗
�−1

,

and
∑

n≥0

dne−int T n =
�

I − de−i t T
�−1

.

By the above four equalities and (13), we get (14). .

For an operator T ∈ L (H ) and a polynomial r (z) =
s
∑

k=0
ckzk ∈ C [z]|D , r (T ) ∈ L (H ) is

defined by

r (T ) =
s
∑

k=0

ckT k.

Lemma 2. Let T ∈ L (H ) such that σ (T )⊂ D. For r (z) ∈ C [z]|D, we have

r (bT )− r (dT ) + c0 I =
1

2π

2π
∫

0

r
�

ei t
�

Ra,b,c,d,t (T ) d t,

where a, b, c, and d are complex parameters satisfying |a|< 1, |b|< 1, |c|< 1, and |d|< 1.



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 345

Proof. Let r (z) =
s
∑

k=0
ckzk. By (14), and since

2π
∫

0
eil t d t = 0 for l ∈ Z/ {0}, we get

2π
∫

0

r
�

ei t
�

Ra,b,c,d,t (T ) d t

=

2π
∫

0

s
∑

k=0

ckeikt







∑

n≥0
aneint T ∗n+

∑

n≥0
bne−int T n

+
∑

n≥0
cneint T ∗n−

∑

n≥0
dne−int T n− I






d t

= 2πc0 I +
s
∑

k=0

2π
∫

0

ck bkT kd t + 2πc0 I −
s
∑

k=0

2π
∫

0

ckdkT kd t − 2πc0 I

= 2π
s
∑

k=0

ck bkT k − 2π
s
∑

k=0

ckdkT k + 2πc0 I

= 2πr (bT )− 2πr (dT ) + 2πc0 I .

Corollary 2. Note that, if r identically equal to 1, then

1

2π

2π
∫

0

Ra,b,c,d,t (T ) d t = I , (15)

for |a|< 1, |b|< 1, |c|< 1, |d|< 1 and T ∈ L (H ) such that σ (T )⊂ D.

In the next theorem, we give a different proof of equation (15) independent of a polyno-
mial. For this purpose we will use the Riesz-Dunford integral formula.

Theorem 5. Let T ∈ L (H ) such that σ (T )⊂ D. Then

1

2π

2π
∫

0

Ra,b,c,d,t (T ) d t = I ,

where a, b, c, and d are complex parameters satisfying |a|< 1, |b|< 1, |c|< 1, and |d|< 1.

Proof. From (13), we have

1

2π

2π
∫

0

Ra,b,c,d,t (T ) d t =
1

2π

2π
∫

0

 

�

I − aei t T ∗
�−1
+
�

I − be−i t T
�−1
+

�

I − cei t T ∗
�−1
−
�

I − de−i t T
�−1
− I

!

d t. (16)



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 346

We set

I1 =
1

2π

2π
∫

0

�

I − aei t T ∗
�−1

d t, (17)

I2 =
1

2π

2π
∫

0

�

I − be−i t T
�−1

d t, (18)

I3 =
1

2π

2π
∫

0

�

I − cei t T ∗
�−1

d t, (19)

I4 =
1

2π

2π
∫

0

�

I − de−i t T
�−1

d t, (20)

and

I5 =
1

2π

2π
∫

0

Id t. (21)

Therefore, it follows from (16)- (21) that

1

2π

2π
∫

0

Ra,b,c,d,t (T ) d t = I1+ I2+ I3− I4− I5. (22)

It is clear that
I5 = I . (23)

Next, we shall calculate I1, I2, I3 and I4. Firstly, we have

I1 =
1

2π

2π
∫

0

�

I − aei t T ∗
�−1

d t =
1

2π

2π
∫

0

e−i t
�

e−i t I − aT ∗
�−1

d t.

Making substitution z = e−i t in the last integral, we get

I1 =
−1

2πi

∫

|z|=1

�

zI − aT ∗
�−1 dz,

where the integral along |z| = 1 is taken in the negative direction. Hence, by the Riesz-
Dunford integral in the equation (1), we have

I1 = I . (24)



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 347

Similarly, we get
I3 = I . (25)

Secondly, we have

I2 =
1

2π

2π
∫

0

�

I − be−i t T
�−1

d t =
1

2π

2π
∫

0

ei t
�

ei t I − bT
�−1

d t.

If we set z = ei t , then the last integral is of the form

I2 =
1

2πi

∫

|z|=1

(zI − bT )−1 dz,

where the integral along |z|= 1 is taken in the positive direction. Hence, by the Riesz-Dunford
integral (1), we have

I2 = I . (26)

Similarly, we get
I4 = I . (27)

Therefore, from (22)-(27), we get (15).

Remark 4. By taking c = 0 and d = 0 in (13) and (15) we find that (13) and (15) are
generalizations of (8) and (9), respectively.

3. The Finite Sum of the Operator- Valued Poisson Kernel

In this section we definite a new generalization of the operator-valued Poisson kernel
M(ak ,bk)nk=0,t (T ) in 2(n+ 1) complex parameters. Let us begin by the following definition.

Definition 3. For T ∈ L (H ) such that σ (T ) ⊂ D, define the finite sum of the operator-valued
Poisson kernel in the following way.

M(ak ,bk)nk=0,t (T ) =
�

I − a0ei t T ∗
�−1
+
�

I − boe−i t T
�−1

+
n
∑

k=1

�

I − akei t T ∗
�−1
−

n
∑

k=1

�

I − bke−i t T
�−1
− I , (28)

where ak and bk are complex parameters satisfying
�

�ak

�

� < 1 and
�

�bk

�

� < 1, 0 ≤ k ≤ n, and for
n= 0, 1,2, . . ..

Remark 5. By taking n= 0 and n= 1 in (28), we obtain (8) and (13), respectively.

Remark 6. Note that M(ak ,bk)nk=0,t (T ) ∈ L (H ).



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 348

Lemma 3. For T ∈ L (H ) such that σ (T )⊂ D, we have

M(ak ,bk)nk=0,t (T ) =
∑

m≥0

am
0 eimt T ∗m+

∑

m≥0

bm
o e−imt T m

+
∑

m≥0

n
∑

k=1

am
k eimt T ∗m−

∑

m≥0

n
∑

k=1

bm
k e−imt T m− I . (29)

Proof. Since
�

�

�

�akei t T ∗
�

�

�

�< 1, and
�

�

�

�bke−i t T
�

�

�

�< 1, 0≤ k ≤ n, we have

n
∑

k=0

�

I − akei t T ∗
�−1
=
∑

m≥0

n
∑

k=0

am
k eimt T ∗m,

and
n
∑

k=0

�

I − bke−i t T
�−1
=
∑

m≥0

n
∑

k=0

bm
k e−imt T m

respectively. By the two equalities above and (28), we get (29).

For an operator T ∈ L (H ) and a polynomial r (z) =
s
∑

j=0
c jz

j ∈ C [z]|D , r (T ) ∈ L (H ) is

defined by

r (T ) =
s
∑

j=0

c j T
j .

Lemma 4. Let T ∈ L (H ) such that σ (T )⊂ D. For r (z) ∈ C [z]|D. Then

r
�

b0T
�

−
n
∑

k=1

r
�

bkT
�

+ nc0 I =
1

2π

2π
∫

0

r
�

ei t
�

M(ak ,bk)nk=0,t (T ) d t,

where ak and bk are complex parameters satisfying
�

�ak

�

� < 1 and
�

�bk

�

� < 1, 0 ≤ k ≤ n, and for
n= 0, 1,2, . . ..

Proof. From (29) and since
2π
∫

0
eil t d t = 0 for l ∈ Z/ {0}, we get

2π
∫

0

r(ei t)M(ak ,bk)nk=0,t (T ) d t =
s
∑

j=0

∑

m≥0

c ja
m
0 T ∗m

2π
∫

0

ei(m+ j)t d t

+
s
∑

j=0

∑

m≥0

c j b
m
0 T m

2π
∫

0

ei( j−m)t d t



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 349

+
s
∑

j=0

∑

m≥0

n
∑

k=1

c ja
m
k T ∗m

2π
∫

0

ei( j+m)t d t

−
s
∑

j=0

∑

m≥0

n
∑

k=1

c j b
m
k T m

2π
∫

0

ei( j−m)t d t −
s
∑

j=0

c j

2π
∫

0

ei j t d t

= 2πc0 I + 2π
s
∑

j=0

c j b
j
0T j + 2π

n
∑

k=1

c0 I

− 2π
s
∑

j=0

n
∑

k=1

c j b
j
kT j − 2πc0 I

= 2nπc0 I + 2πr
�

b0T
�

− 2π
n
∑

k=1

r
�

bkT
�

.

Corollary 3. Note that if r identically equal to 1, we have

1

2π

2π
∫

0

M(ak ,bk)nk=0,t (T ) d t = I , (30)

for complex parameters ak and bk satisfying
�

�ak

�

� < 1 and
�

�bk

�

� < 1, 0 ≤ k ≤ n, n = 0,1, 2, . . .
and T ∈ L (H ) such that σ (T )⊂ D.

Now, we give a different proof of equation (30) independent of a polynomial.

Theorem 6. Let T ∈ L (H ) such that σ (T )⊂ D. Then

1

2π

2π
∫

0

M(ak ,bk)nk=0,t (T ) d t = I ,

where ak and bk are complex parameters satisfying
�

�ak

�

�< 1 and
�

�bk

�

�< 1, 0≤ k ≤ n,
n= 0, 1,2, . . ..

Proof. From (28), we have

1

2π

2π
∫

0

M(ak ,bk)nk=0,t (T ) d t =
1

2π

2π
∫

0









�

I − a0ei t T ∗
�−1
+
�

I − boe−i t T
�−1
+

n
∑

k=1

�

I − akei t T ∗
�−1
−

n
∑

k=1

�

I − bke−i t T
�−1
− I









d t.

(31)



S. Al-Sharif, F. Salem, B Frasin / Eur. J. Pure Appl. Math, 6 (2013), 340-351 350

We set

I1 =
1

2π

2π
∫

0

�

I − a0ei t T ∗
�−1

d t, (32)

I2 =
1

2π

2π
∫

0

�

I − b0e−i t T
�−1

d t, (33)

I3 =
1

2π

2π
∫

0

n
∑

k=1

�

I − akei t T ∗
�−1

d t, (34)

I4 =
1

2π

2π
∫

0

n
∑

k=1

�

I − bke−i t T
�−1

d t, (35)

and

I5 =
1

2π

2π
∫

0

Id t. (36)

Therefore, it follows from (32)- (36) that

1

2π

2π
∫

0

M(ak ,bk)nk=0,t (T ) d t = I1+ I2+ I3− I4− I5. (37)

It is clear that
I5 = I . (38)

Following similarly the proof of Theorem 5, we get

I1 = I . (39)

I2 = I . (40)

Next, we shall calculate I3 and I4. First, we have

I3 =
1

2π

2π
∫

0

n
∑

k=1

�

I − akei t T ∗
�−1

d t =
n
∑

k=1

(
1

2π

2π
∫

0

e−i t
�

e−i t I − akT ∗
�−1

d t).

Making substitution z = e−i t in the last integral, we get

I3 =
n
∑

k=1

(
−1

2πi

∫

|z|=1

�

zI − akT ∗
�−1 dz),



REFERENCES 351

where the integral along |z| = 1 is taken in the negative direction. Hence, by the Riesz-
Dunford integral in the equation (1), we have

I3 =
n
∑

k=1

I = nI . (41)

Similarly, we get

I4 =
1

2π

2π
∫

0

n
∑

k=1

�

I − bke−i t T
�−1

d t =
n
∑

k=1

(
1

2π

2π
∫

0

ei t
�

ei t I − bkT
�−1

d t).

If we set z = ei t , then the last integral is of the form

I4 =
n
∑

k=1

(
1

2πi

∫

|z|=1

�

zI − bkT
�−1 dz),

where the integral along |z|= 1 is taken in the positive direction. Hence, by the Riesz-Dunford
integral (1), we have

I4 =
n
∑

k=1

I = nI . (42)

Therefore, from (37)-(42) we get (30).

References

[1] S Bulut. A note on the operator-valued poisson kernel. European Journal of Pure and
Applied Mathematics, 2(2):296–301, 2009.

[2] I Chalendar. The operator-valued poisson kernel and its application. Irish Mathematical
Society Bulletin, 51:21–44, 2003.

[3] H Haruki. A new generalization of the poisson kernel. Journal of Applied Analysis and
Stochastic Analysis, 10(2):191–196, 1997.