E extracta mathematicae Vol. 31, Núm. 1, 11 – 23 (2016)

On a ρn-Dilation of Operator in Hilbert Spaces
†

A. Salhi, H. Zerouali

PB 1014, Departement of Mathematics,
Siences Faculty, Mohamed V University in Rabat, Rabat, Morocco

radi237@gmail.com, zerouali@fsr.ac.ma

Presented by Mostafa Mbekhta Received March 21, 2016

Abstract: In this paper we define the class of ρn−dilations for operators on Hilbert spaces.
We give various properties of this new class extending several known results ρ−contractions.
Some applications are also given.

Key words: ρn-dilation, ρ-dilation.

AMS Subject Class. (2010): 47A20.

1. Introduction

Sz-Nagy and Foias introduced in [8], the subclass Cρ of the algebra L(H)
of all bounded operators on a given complex Hilbert space H. More precisely,
for each fixed ρ > 0, an operator T ∈ Cρ if there exists a Hilbert space K
containing H as a subspace and a unitary transformation U on K such that;

Tn = ρPrUn|H for all n ∈ N
∗. (1)

Where Pr : K → H is the orthogonal projection on H. The unitary op-
erator U is then called a unitary ρ-dilation of T, and the operator T is a
ρ-contraction.

Recall that T is power bounded if ∥T n∥ ≤ M for some fixed M and every
nonegative integer n. From Equation (1), it follows that every ρ-contraction
is power bounded since ∥Tn∥ ≤ ρ for all n ∈ N∗. Computing the spectral
radius of T, it comes that the spectrum of the operator T satisfies σ(T) ⊂ D,
where D = D(0, 1) is the open unit disc of the set of complex numbers C.

Operators in the class Cρ enjoy several nice properties, we list below the
most known, we refer to [7] for proofs and further information.

†This work is partially supported by Hassan II Academy of Siences and the CNRST
Project URAC 03.

11



12 a. salhi, h. zerouali

(1) The function ρ 7→ Cρ is nondecreasing, that is Cρ  Cρ′ if ρ < ρ′. We
will denote by C∞ =

∪
ρ>0

Cρ.

(2) C1 coincides whith the class of contractions (see [6]) and C2 is the class
of operators T having a numerical radius less or equal to 1 (see [1]).
The numerical radius is given by the expression, w(T) = sup{|⟨Th; h⟩| :
∥h∥ = 1}.

(3) If T ∈ Cρ so is T n. It is however not true in general that the product of
two operators in Cρ is in Cρ. Also it is not always true that ξT belongs
to Cρ when T ∈ Cρ for |ξ| ̸= 1.

(4) For any M a T−invariant subspace, the restriction of T to the subspace
M is in the class Cρ whenever T is.

(5) Any operator T such that σ(T) ⊂ D belongs to C∞.

Numerous papers where devoted the the study of differents aspects of Cρ; we
refer to [2, 4, 5] for more information.

The next theorem provides a useful characterization of the class Cρ in term
of some positivity conditions,

Theorem 1.1. Let T be a bounded operator on the Hilbert space H and
ρ be a nonnegative real. The following are equivalent

(1) The operator T belongs to the class Cρ ;

(2) for all h ∈ H; z ∈ D(0; 1)

(
2

ρ
− 1)∥zTh∥2 + (2 −

2

ρ
)Re (zTh, h) ≤ ∥h∥2; (2)

(3) for all h ∈ H; z ∈ D(0; 1)

(ρ − 2)∥h∥2 + 2Re ((I − zT)−1h, h) ≥ 0. (3)

2. Unitary ρn-dilation

We extend the notion of ρ-contractions to a more general setting. More
precisely, let (ρn)n∈N be a sequence of nonnegative numbers. We will say that
the operator T on a complex Hilbert space H belongs to the class Cρn if, there



on a ρn-dilation of operator in hilbert spaces 13

exists a Hilbert space K containing H as a subspace and a unitary operator
U such that

Tn = ρnPrU
n
|H for all n ∈ N

∗. (4)

We say in this case that the unitary operator U is a ρn-dilation for the operator
T and the operator T will be called a ρn-contraction.

Remark 2.1. .
(1) For any bounded operator T, the operator T∥T∥ is a contraction and

hence admits a unitary dilation. We deduce that,

T ∈ Cρn for ρn = ∥T∥
n for all n ∈ N.

We notice at this level that, without additional restrictive assumptions on
the sequence (ρn)n∈N, there is no hope to construct a reasonable ρn-dilation
theory. Our goal will be to extend the most usefull properties of ρ−contraction
to this more general setting.

(2) From Equation 4, for T ∈ Cρn with U a ρn-dilation, we obtain

∥Tn∥ ≤ ∥ρnPrUn|H∥ ≤ ρn.

Therefore the condition limn→∞ (ρn)
1
n ≤ 1 will ensure that σ(T) ∈ D(0; 1).

(3) In contrast with the class Cρ, the class C(ρn) is not stable by powers.
However, if T ∈ Cρn and k ≥ 1 is a given integer, we obtain Tk ∈ Cρkn. This
latter fact can be seen as a trivial extension of the case ρn = ρ0 for every n.

In the remaining part of this paper, we will assume that (ρn)n∈N is a
sequence of nonnegative numbers satisfying

lim
n→∞

(ρn)
1
n ≤ 1. (5)

We associate with the sequence (ρn)n∈N, the following function,

ρ(z) =
∑
n≥0

zn

ρn
.

It is easy to see that condition limn→∞ (ρn)
1
n ≤ 1 implies that ρ ∈ H(D).

Here H(D) is the set of holomorphic functions on the open unit disc D. Also,
the valued-operators function

ρ(zT) =
∑
n≥0

znTn

ρn



14 a. salhi, h. zerouali

is well defined and converges in norm for every |z| < 1.

We give next a necessary and sufficient condition to the membership to
the class Cρn;

Theorem 2.2. Let T be an operator on a Hilbert space H and (ρn)n∈N is
a sequence of nonnegative numbers. The operator T has a ρn-dilation if and
only if

(1 −
2

ρ0
)∥h∥2 + 2Re ⟨ρ(zT)(h); h⟩ ≥ 0 for all h ∈ H; z ∈ D(0; 1). (6)

We recall first the next well known lemma from [7, Theorem 7.1] that will be
needed in the proof of the previous theorem.

Lemma 2.3. Let H be a Hilbert space, G be a multiplicative group and
Ψ be an operator valued function s ∈ G 7→ Ψ(s) ∈ L(H) such that


Ψ(e) = I, e is the identity element of G

Ψ(s−1) = Ψ(s)∗∑
s∈G

∑
t∈G(Ψ(t

−1s)h(s); h(t)) ≥ 0

for finitely non-zero function h(s) from G.
Then, there exists a Hilbert space K containing H as a subspace and a

unitary representation U of G, such that

Ψ(s) = Pr(U(s)) (s ∈ G)

and
K =

∨
s∈G

U(s)H

Proof of Theorem 2.2. Let T be a bounded operator in the class Cρn and
U be the unitary ρn-dilation of T, given by the expression 4. We have clearly,

I + 2
∑
n≥1

znUn converges to (I + zU)(I − zU)−1

for all complex numbers z such that |z| < 1.
And

Pr(I + 2
∑
n≥1

znUn) = I + 2
∑
n≥1

zn

ρn
Tn.



on a ρn-dilation of operator in hilbert spaces 15

By writing,

I + 2
∑
n≥1

zn

ρn
Tn = (1 −

2

ρ0
)I + 2

∑
n≥0

zn

ρn
Tn

= (1 −
2

ρ0
)I + 2ρ(zT),

we get

Pr((I + zU)(I − zU)−1) = (1 −
2

ρ0
)I + 2ρ(zT).

On the other hand,

⟨(I + zU)k; (I − zU)k⟩ = ∥k∥2 + ⟨zUk; k⟩ − ⟨k; zUk⟩ − ∥zUk∥2

It follows that for every k ∈ K, we have

Re ⟨(I + zU)k; (I − zU)k⟩ = ∥k∥2 − ∥zUk∥2

= ∥k∥2 − |z|2∥k∥2

= ∥k∥2(1 − |z|2) ≥ 0 since |z| < 1.

Now if we take h = (I − zU)k we will find,

Re ⟨(I + zU)(I − zU)−1h; h⟩ = Re ⟨Pr(I + zU)(I − zU)−1h; h⟩

= Re ⟨(1 −
2

ρ0
)h + 2ρ(zT)(h); h⟩,

and hence for every h ∈ H, we obtain

Re ⟨(1 −
2

ρ0
)h + 2ρ(zT)(h); h⟩ ≥ 0

or equivalently,

(1 −
2

ρ0
)∥h∥2 + 2Re ⟨ρ(zT)(h); h⟩ ≥ 0

for every h ∈ H and all complex number z such that |z| < 1.
Conversely, let us show that condition (6) implies that the operator T

belongs to the class Cρn. To this aim, assume that (6) is satisfied and take
0 ≤ r < 1 and 0 ≤ ϕ < 2π. We introduce the next operator valued function

Q(r; ϕ) = I +
∑
n≥1

rn

ρn
(einϕTn + e−inϕT∗n).



16 a. salhi, h. zerouali

Then Q(r; ϕ) converges in the norm operator for every r and ϕ. Moreover,
from the inequality 6, we have

⟨Q(r; ϕ)l; l⟩ ≥ 0

for every l ∈ H. Therefore

J =
1

2π

∫ 2π
0

⟨Q(r; ϕ)h(ϕ); h(ϕ)⟩dϕ ≥ 0

for every h(ϕ) =
∑+∞

−∞ hne
−inϕ where (hn)n∈Z is a sequence with only finite

number of nonzero elements in H. We have

J =:
+∞∑
−∞

∥hn∥2+
∑
m

∑
n>m

rn−m

ρn−m
⟨Tn−mhn; hm⟩+

∑
m

∑
n<m

rm−n

ρm−n
⟨T∗(m−n)hn; hm⟩

for every 0 ≤ r < 1. Now taking r → 1− will imply∑
n∈Z

∑
m∈Z

⟨Ψ(ρn)(n − m)hn; hm⟩ ≥ 0,

where Ψ(ρn) : Z → L(H) is defined by Ψ(ρn)(0) = I, Ψ(ρn)(n) =
1
ρn
Tn and

Ψ(ρn)(−n) =
1
ρn
T∗n for every n > 0.

It is immediate that Ψ(ρn)(n) is nonnegative on the additive group Z of
integers. Using Lemma 2.3, there exists a unitary operator U on a Hilbert
space K containing H as a subspace and such that Ψ(ρn)(n) = PrU(n) for all
n ∈ Z.

Therefore for all n ∈ N∗

Tn = ρnPrU
n
|H.

The proof is completed.

Remark 2.4. In the case where (ρn)n∈N is a constant sequence, that is
ρn = ρ for all n ∈ N with ρ > 0, we obtain

ρ(z) =
1

1 − z
and hence, the inequality 6 becomes

(1 −
2

ρ
)∥h∥2 + 2Re ⟨(I − zT)−1h; h⟩ ≥ 0

for all h ∈ H and z ∈ D. We substitute h by l = (I − zT)−1h to retrieve
relation 3 and by Theorem (2.1) we obtain T is a ρ-contraction.



on a ρn-dilation of operator in hilbert spaces 17

The next two corollaries are immediate consequences of Equation (6) and
are related to analogous results of ρ−contraction.

Corollary 2.5. Let T ∈ Cρn and M a T−invariant subspace. Then
T|M ∈ Cρn.

Proof. It suffices to see that Equation 6 is close to restrictions.

Corollary 2.6. Let T be in the class Cρn and r ≥ 1 be a real number,
then T is in the class Crρn.

Proof. The inequality 6 is equivalent to

(ρ0 − 2)∥h∥2 + 2ρ0Re ⟨ρ(zT)(h); h⟩ ≥ 0.

Pluging rρn instead of ρn, we get

(rρ0 − 2)∥h∥2 + 2rρ0Re ⟨
1

r
ρ(zT)(h); h⟩ ≥ 0,

and thus

(1 −
2

rρ0
)∥h∥2 + 2Re ⟨

1

r
ρ(zT)(h); h⟩ ≥ 0.

Therefore T ∈ C(rρn).

We also have,

Proposition 2.7. Let T be a bounded operator on a Hilbert space H.
Then for every α > 2, there exists Γ(α) > 0 such that the operator T belongs
to C(ρn), where ρn is a sequence given by ρn = Γ(α).∥T

n∥(1 + nα).

Proof. Let Γ > 0 and ρα(z) =
∑

n≥1
zn

Γ.∥T n∥(1+nα) for all |z| ≤ 1. Then

ρα(zT) =
∑
n≥1

znTn

Γ.∥T n∥(1 + nα)
for all |z| ≤ 1.

For every vector h in H, we set

A(z) = ⟨ρα(zT)h; h⟩



18 a. salhi, h. zerouali

|A(z)| =
∣∣ ∑
n≥0

⟨
zn

Γ.∥T n∥(1 + nα)
T nh; h⟩

∣∣
≤

∑
n≥0

|
⟨T nh; h⟩

Γ.∥T n∥(1 + nα)
zn|.

Setting an =
⟨T nh;h⟩

Γ.∥T n∥(1+nα), we have

|an| ≤
∥T n∥∥h∥2

Γ.∥T n∥(1 + nα)
=

∥h∥2

Γ.(1 + nα)
< ∞.

We conclude that A(z) is holomorphic in the unit disc and continuous on
the boundary. Since the maximun is attained of the circle |z| = 1, we obtain

|A(z)| = |
∑
n≥0

anz
n|

≤
∑
n≥0

|an||z|n =
∑
n≥1

|an|

≤
∑
n≥0

∥h∥2

Γ.(1 + nα)

Now, since
∑

n≥0
1

1+(n)α
is a convergent sequence (α > 2), then choosing

Γ = 2
∑

n≥0
1

1+nα
will leads to

|A(z)| ≤
1

2
∥h∥2,

and then

∥h∥2 + 2Re ⟨ρα(zT)h; h⟩ ≥ 0 for all h ∈ H and z ∈ D.

Finally , Inequality (6) is satisfied and the operator T belongs to the class
C(ρn).

3. The Bergmann shift

We devote this section to the membership of the Bergmann shift to the
class C(ρn) for some suitable sequence ρn. Let H be a Hilbert space and (ei)i∈N∗
be an orthonormal basis of H. Recall that for a given sequence (ωn)n∈N of non
negative numbers; the weighted shift Sω associated with ωn is defined on the



on a ρn-dilation of operator in hilbert spaces 19

basis by Sω(en) = ωnen+1. A detailed study on weighted shifts can be found
in the survey [9]. On the other hand; the membership of weighted shifts to
the class Cρ is investigated in [3].

The Bergman shift is the weighted shift defined on the basis by the ex-
pression Ten = wnen+1, where

wn =
n + 1

n
for all integer n ∈ N∗.

It is easy to see that,

• ∥T∥ = sup(wn)n∈N∗ = 2.

• The weight (wn)n∈N∗ is decreasing and then

∥Tn∥ =
n∏

i=1

wi = n + 1.

In particular T is not power bounded and hence does not belong to the class
Cρ for any ρ > 0.

We have

Proposition 3.1. Let T be the Bergmann shift and ρn be the sequence
given by ρn = n

α for some α > 0. Then for every α > 2, there exists Γ(α)
such that T ∈ CΓ(α)ρn.

Proof. Let Γ > 0 and ρα(z) =
∑

n≥1
zn

Γnα
for all |z| ≤ 1 in that

ρ(zT) =
∑
n≥1

znTn

Γnα
for all |z| ≤ 1.

We set, S = ρ(zT) and let x be a vector in H. Therefore

S(x) = ρ(zT)(x) =
∑
i≥1

znTnx

Γnα
.

Writing x =
∑

i≥1 xiei, we get

S(x) =
∑
i≥1

⟨S(x); ei⟩ei =
∑
i≥1

(
∑
j≥1

xj⟨Sej; ei⟩)ei,



20 a. salhi, h. zerouali

and

⟨Sej; ei⟩ = ⟨
∑
n≥1

znTn

Γnα
(ej); ei⟩ =

∑
n≥1

zn

Γnα
⟨Tn(ej); ei⟩

=
∑
n≥1

zn

Γnα
⟨(

j+n−1∏
p=j

wp)ej+n; ei⟩.

It follows that

⟨Sej; ei⟩ =
zi−j

Γ
(i − j)α

i−1∏
p=j

wp,

and then

S(x) =
∑
i≥2

(
i−1∑
j=1

(
i−1∏
p=j

wp)xj
zi−j

Γ(i − j)α
)ei.

For the Bergman shift, we have
∏i−1

p=j wp =
i
j
and thus

ρ(zT)(x) =
∑
i≥2

(

i−1∑
j=1

i

Γj(i − j)α
xjz

i−j)ei.

Finally, we conclude that the inequality (6) is equivalent to

∑
i≥1

|xi|2 + 2Re (
∑
i≥2

i−1∑
j=1

i

Γj(i − j)α
xixjz

i−j) ≥ 0.

If we consider the function

A(z) = 2
∑
i≥2

i−1∑
j=1

i

Γj(i − j)α
xixjz

i−j

and we write n = i − j, we will obtain,

A(z) = 2
∑
i≥2

i−1∑
n=1

i

Γ(i − n)Γnα
xixi−nz

n = 2
∑
n≥1

∑
i≥n+1

i

Γ(i − n)nα
xixi−nz

n.

We denote by (Â(n))n = (an)n∈N∗ the sequence of coefficients of A,

an =
1

2

∑
i≥n+1

i

Γnα(i − n)
xixi−n



on a ρn-dilation of operator in hilbert spaces 21

Since i
n(i−n) =

1
n
+ 1

i−n ≤ 2 for every i ≥ n + 1, we obtain

|an| =
∣∣∣1
2

∑
i≥n+1

i

Γnα(i − n)
xixi−n

∣∣∣
≤

1

Γnα−1

∑
i≥n+1

|xixi−n| ≤
1

Γnα−1

∑
i≥n+1

|xixi−n|;

and by the Cauchy-Schwartz inequality, it follows,

|an| ≤
1

Γnα−1
∥x∥2 ≤ ∞.

We deduce that A(z) is holomorphic in the open unit disc and continuous on
the closed unit disc. As the maximum is attained on the circle |z| = 1, we
have

|A(z)| =
∣∣∣1
2

∑
n≥1

(
∑

i≥n+1

i
1

Γnα
(i − n)

xixi−n)z
n
∣∣∣

≤
∑
n≥1

|an||z|n =
∑
n≥1

|an|.

Now, since
∑

n≥1
1

(n)α−1
is a convergente sequence (α ≥ 2), choosing Γ =∑

n≥1
1

(n)α−1
would lead us to

|A(z)| ≤ ∥x∥2 =
∑
i≥1

|xi|2.

We derive that,

|Re (A(z))| ≤ |A(z)| ≤ ∥x∥2 =
∑
i≥1

|xi|2,

and hence ∣∣∣1
2
Re (

∑
i≥2

i−1∑
j=1

i

jΓ(i − j)α
xixjz

i−j)
∣∣∣ ≤ ∑

i≥1
|xi|2.

Therefore for all x ∈ H and a complex z such that |z| ≤ 1 we have

∑
i≥1

|xi|2 + 2Re (
∑
i≥2

i−1∑
j=1

i

jΓ(i − j)α
xixjz

i−j) ≥ 0.

We conclude that the weighted shift {wn} is a ρn-contraction with
ρn = Γn

α.



22 a. salhi, h. zerouali

Remark 3.2. We claim that for every α ≥ 1 the Bergmann shift belongs to
a class C∞,nα. A proof is not available for this claim; however it is motivated
by the incomplete computations below.

Let us set, for exemple, ρn = 4.n for all integer n ≥ 1, Let H be a Hilbert
space and (ei)i∈N∗ be a an orthonormal basis for the Hilbert space H. Consider
the Bergmann shift defined on the basis by Ten =

n+1
n

en+1 for all n ∈ N∗.
Then as in the proof of the previous proposition, we show that inequality (6)
is equivalent to the next

∑
i≥1

|xi|2 + Re (
∑
i≥2

i−1∑
j=1

i

2j(i − j)
xixjz

i−j) ≥ 0. (7)

We write

∑
i≥1

|xi|2+Re (
∑
i≥2

i−1∑
j=1

i

2j(i − j)
xixjz

i−j) ≥
∑
i≥1

|xi|2−
∑
i≥2

i−1∑
j=1

i

2j(i − j)
|xi||xj|,

and ∑
i≥1

|xi|2 +
∑
i≥2

i−1∑
j=1

i

2j(i − j)
|xi||xj| =

∑
i,j≥1

ai,j|xi||xj|,

with {
ai;i = 1 for all i ≥ 1
ai;j =

i
4j|(i−j)| for all j ̸= i

Then to show inequality (7), it suffices to prove that the infinite symmetric
matrix with the real entries M = [ai;j] is nonnegative. To this aim, we com-
pute the determinant of the first n × n-corner, to check if it is nonnegative.
An attempt on classical softwares allow to show this fact for n ≤ 150. It is
hence reasonable to conjecture that the Bergman shift belongs to C∞,n.

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(1965), 590.

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[3] G. Eckstein, A. Racz, Weighted shifts of class Cρ, Acta Sci. Math. (Szeged)
35 (1973), 187 – 194.



on a ρn-dilation of operator in hilbert spaces 23

[4] G. Eckstein, Sur les opérateurs de la classe Cρ, Acta Sci. Math.(Szeged) 33
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