Stability for Some Operator Classes by Aluthge Transform

Gilles Cassier and Jérôme Verliat

15 juin 2009

Abstract

Our purpose is to investigate various stability properties of the Aluthge transform of

an operator acting on a Hilbert space H. We begin with the stability of partial isometries,

with some examples in Hardy spaces. Then we obtain some interesting relations about

minimum polynomial, iterated kernels and iterated ranges between an operator and its

Aluthge transform. And we finish with an analysis of stability of C0,·and C1,·

operator

classes.

Let H be a separable complex Hilbert space, and let B(H) denotes the algebra of allbounded linear operators on H. In the sequel of the paper, R(T ) and N (T ) respectively standfor the range and the kernel of the operator T . Remind that an arbitrary operator T in B(H)has a unique polar decomposition T = U |T |, where |T | denotes the modulus

√T ⋆T of T , and

U is a partial isometry such that N (U) = N (T ).

Remark 1. The previous operator U can be defined by

U : R(|T |) −→ R(T )|T |x 7−→ Tx

.

And we can also mention: R(|T | 12 ) = R(|T |) = R(T ⋆).

The purpose of the paper is to compare, regarding to various standards, the behaviour ofan arbitrary operator of B(H) and that of its Aluthge transform, defined by the following:

Definition 2. Let T = U |T | be the polar decomposition of a bounded operator T ∈ B(H). Theoperator ∆(T ) of B(H), defined by

∆(T ) = |T | 12 U |T | 12

is called the Aluthge transform of T .

Many results about this transformation have been already published. Recall in particularthe well known following claim (cf. [7], [8]) : σ(T ) = σ(∆(T )) (and σ(T )\{0} = σ(∆(T ))\{0}).In the sequel of the paper, for any subspace K of H, we denote by prK the orthogonal projectionon K.

A very useful property is the following, that is proved in [5]:

Lemma 3. Let T be an operator of B(H). For every function f which is analytic on someneighborhood of σ(T ), we can obtain the following equalities

f(∆(T ))|T | 12 = |T | 12 f(T ) and U |T | 12 f(∆(T )) = f(T )U |T | 12 .

1

1 The Class of Partial Isometries

1.a) The Aluthge Transform of a Partial Isometry

Proposition 4. Let T be a partial isometry of B(H). Then its Aluthge transform is given by

∆(T ) = T ⋆T 2.

Proof. Since T is a partial isometry, the set R(T ⋆) is closed and |T |2 = T ⋆T = prR(T ⋆). Thus

|T | 12 = prR(T ⋆). Let x = x1 + x2 be the decomposition of an arbitrary vector of H withrespect to the (orthogonal) direct sum H = N (T ) ⊕ R(T ⋆). According to Remark 1, we getTx = Tx2 = T prR(T ⋆) x, so U prR(T ⋆) = T prR(T ⋆) and N (U) = N (T ). Finally we obtainU = T and

∆(T ) = prR(T ⋆) T prR(T ⋆) = prR(T ⋆) T = T ⋆T 2,

because R(T ⋆) = [N (T )]⊥.

Example 5. Let H be the finite dimension space Cn. We denote by Sn the partial isometry

of B(H) such that its matrix in canonical basis is given by

0 0 0 · · · 01 0 0 · · · 00 1 0 0...

. . .. . .

...0 0 · · · 1 0

.

Thus ∆(Sn) is unitary equivalent to Sn−1 ⊕ 0B(H). In particular, for any nonnegative integerk :

• if k 6 n− 2, the minimum polynomial of ∆k(Sn) is Xn−k, where X denotes the positionfunction;

• if k > n − 1, the minimum polynomial of ∆k(Sn) is the null polynomial.

Proof. Let (ek)06k6n−1 be the canonical basis of H. We denote by 〈uk〉k the subspace generatedby any arbitrary family (uk)k of vectors of H. Since S⋆

nSn = pr〈ek〉k6n−2, we can form the

following matrix with respect to the blocks 〈ek〉k6n−2 and 〈en−1〉

∆(Sn) = pr〈ek〉k6n−2Sn =

(Sn−1 0

0 0

).

Minimum polynomials of each term of the sequence of the iterated of Sn can be easily computed,repeating the previous argument.

1.b) Partial Isometries Acting on the Hardy Space H2

The open unity disk {z ∈ C; |z| < 1} is denoted, as usual, by D and the torus {z ∈ C; |z| = 1}is denoted by T. We consider in this section the classical Hardy space

H2 =

f ∈ L2(T); ∀n ∈ N,

2π∫

0

f(eit)eintdm(t) = 0

.

2

Let u be an arbitrary inner function. We write Eu = H2 ⊖ uH2 and then we can define

Su : Eu −→ Eu

f 7−→ prEu(zf)

.

Since Eu is an invariant subspace of the usual shift (acting on H2), we see that S∗uf = S∗f for

any f in Eu.If u vanishes at 0, we will denote by u1 the associated function u defined by

u1 : T −→ T

z 7−→ u(z)z

.

Proposition 6. Let u be an inner function. Then, Su is a partial isometry if and only if uvanishes at 0.

Proof. Since

SuS⋆u(f) = prEu

(zf(z) − f(0)

z

)

= prEu(f − f(0))

= f − f(0)PEu1l

= f − 〈f |PEu1l〉PEu

1l

= (IdEu−PEu

1l ⊗ PEu1l)(f)

for any vector f ∈ Eu, we obtain SuS⋆u = IdEu

−PEu1l ⊗ PEu

1l. Otherwise, we can write

0 6 SuS⋆u = IdEu

−‖PEu1l‖2 PEu

1l

‖PEu1l‖ ⊗ PEu

1l

‖PEu1l‖ .

Consequently, we can observe that the spectrum of SuS⋆u is the set {1, 1−‖PEu

1l‖2}. ThereforeSuS⋆

u is an orthogonal projector if and only if ‖PEu1l‖ = 1, that is 1l ∈ Eu. It means that the

function 1l is orthogonal to vector space uH2, which is equivalent to u(0) = 0.

Proposition 7. Let u be an inner function such that u(0) = 0. Then the Aluthge transform ofthe operator Su is given by

∆(Su) = Su − S⋆u ⊗ S⋆2u.

Proof. Since u(0) = 0, Su is a partial isometry according to the previous proposition. Conse-quently, SuS⋆

u = prR(Su) and S⋆uSu = prR(S⋆

u) = Id−prN (Su). Let f be a function of Eu such

that Su(f) = 0 = prEu(zf). Since zf ∈ Eu, there exists a function g ∈ H2 such that

zf(z) = u(z)g(z) = zu1(z)g(z),

which implies that f(z) = u1(z)g(z). Besides, f ∈ Eu gives

∀h ∈ H2, 0 = 〈u1g|zu1h〉 = 〈g|zh〉 = 〈S⋆g|h〉.

It follows that S⋆g = 0 and hence there exists a complex number α such that g = α, that isf = αu1. We deduce that S⋆

uSu = Id−S⋆u ⊗ S⋆

u. As a conclusion, thanks to Proposition 4, weget ∆(Su) = Su − S⋆u ⊗ S⋆2u.

3

2 Minimum Polynomials of an Operator and its Aluthge

Transform

The aim of that section is to compare, for any arbitrary operator T acting on a finite dimensionalspace H, the minimum polynomials of T and ∆(T ).

Proposition 8. Let H be a finite dimensional vector space and T ∈ B(H). We denote by ωT

(resp. ω∆(T )) the minimum polynomial of T (resp. of ∆(T )). We decompose ωT = znω̃T (resp.ω∆(T ) = zmω̃∆(T )) where n ∈ N and ω̃T (0) 6= 0 (resp. m ∈ N and ω̃∆(T )(0) 6= 0).Then ω∆(T ) divides ωT . More precisely, we have

ω̃T = ω̃∆(T ) and m ∈ {n − 1, n}.

Proof. First of all, we prove that ω∆(T ) divides ωT . Let P be a polynomial such that P (T ) = 0.Using Lemma 3, we get

∀x ∈ R(|T | 12 ), P (∆(T ))x = 0.

According to the orthogonal decomposition H = N (T ) ⊕R(T ⋆) and Remark 1, it remains toprove that the equality P (∆(T ))x = 0 holds for any vector of

[R(|T | 12 )]⊥ = [R(T ⋆)]⊥ = N (T ) = N (|T | 12 ).

Let x be a vector of H such that Tx = 0: since ∆(T )x = |T | 12 U |T | 12 x = 0, we have P (∆(T ))x =P (0)x. Moreover, 0 belongs to the point spectrum of T . Consequently, we have P (0) = 0 and Pvanishes at ∆(T ). In short, ω∆(T ) divides ωT , which means that ω̃T divides ω̃∆(T ) and m 6 n.Thanks to Lemma 3, it follows

0 = ω∆(T )(∆(T ))|T | 12 = |T | 12 ω∆(T )(∆(T )) = |T | 12 Tmω̃∆(T )(∆(T )).

Thus, we obtain R(ω̃∆(T )(∆(T ))) ⊂ N (|T | 12 Tm) ⊂ N (Tm+1), which implies that the equalityTm+1ω̃∆(T )(∆(T )) = 0 holds. It results that znω̃T divides zm+1ω̃∆(T ). As ω̃T and zm+1 arerelatively prime, we deduce that ω̃T divides ω̃∆(T ), and we have necessarily n 6 m + 1.

This result is sharp, as it is shown in the next example. This is a consequence of the previoussection. In particular, considering the Hardy space H2 and using notations introduced before,we obtain:

Example 9. Let b be a finite Blaschke product, that is

b : T −→ T

z 7−→ λzp

n∏

j=1

z − ωj

1 − ωjz

where λ ∈ T, p is a positive integer and {ωi|1 6 i 6 n} be n distinct points in D. As before

Eb stands for the space H2 ⊖ bH2 and b1 is the function defined on torus T by z 7→ b(z)z

. Inaddition, we suppose that b vanishes at 0, so that Sb is a partial isometry. In order to simplifycomputation, we assume that the zeros of b are simple.Let us consider the minimum polynomial of ∆(Sb). We observe that

(∆(Sb))2 = (Sb − S⋆b ⊗ S⋆2b)(Sb − S⋆b ⊗ S⋆2b) = S2

b − S⋆b ⊗ S⋆3b.

4

By induction we obtain (∆(Sb))n = S2

b −S⋆b⊗ (S⋆)n+1b. Hence by linearity we prove that theequality

P (∆(Sb)) = P (Sb) − S⋆b ⊗ (P (Sb)⋆ − P (0) IdEb

)b

holds for any complex polynomial P . This formula can be generalized to every function h whichis analytic on the open unit disk and continous on the closed unit disk (notation: A(D)).Moreover, we know that the numerator of b is the minimum polynomial of Sb, so b(Sb) = 0 andhence b(∆(Sb)) = 0. We will prove that b1(∆(Sb)) = 0. If p = 1, since 0 is an eigenvalue of Sb,0 belongs also to the point spectrum of ∆(Sb), so z divides the minimum polynomial of ∆(Sb),which implies that b1(∆(Sb)) 6= 0. Now assume that p > 2, then we see that

b1(∆(Sb)) = b1(Sb) − S⋆b ⊗ b1(Sb)⋆S⋆b.

Since S⋆b b1(S

⋆b ) = 0, we get R(b1(S

⋆b )) ⊂ N (S⋆

b ) = C1l and hence we notice that b1(Sb) is arank one operator. Consequently, we can write b1(S

⋆b ) = 1l ⊗ ϕ, so b1(Sb) = ϕ ⊗ 1l. We write

eα(z) = 11−αz

for any α in the open unit disk. Make ϕ explicit:

• if 1 6 j 6 n − p, on the one hand we get b1(Sb)⋆eωj

= b1(ωj)eωj= 0; on the other hand,

it follows b1(Sb)⋆eωj

= 〈eωj|1l〉1l = ϕ(ωj)1l. Then, it implies ϕ(ωj) = 0.

• if 0 6 k 6 p − 2, on the one hand we obtain b1(Sb)⋆zk = b̃1(Sb)

⋆(S⋆b )p−1(zk) = 0,

where b̃1(z) is the Blashke product defined for any complex number z ∈ D by setting

b̃1(z) = λ

n−p∏

j=1

z − ωj

1 − ωjz; on the other hand, we can write b1(Sb)

⋆zk = 〈zk|1l〉1l = ϕ(k)(0)1l.

Thus we have ϕ(zk) = 0.

• in the same manner, we can prove that the function z 7→ ϕ(zp−1) reduces to a constantdenoted by c.

The two last assertions are true if p > 2, which is the case. Therefore, we get

ϕ(z) = λczp−1n−p∏

j=1

z − ωj

1 − ωjz= cb1(z).

Hence, we have

b1(∆(Sb)) = b1(Sb) − S⋆b ⊗ b1(Sb)⋆S⋆b

= cb1 ⊗ 1l − b1 ⊗ (c(1l ⊗ b1)S⋆b)

= cb1 ⊗ 1l − b1 ⊗ c1l

= 0.

Consequently, the minimum polynomial of ∆(Sb) is b1, while the one of Sb is b = zb1.

Remark 10. For any arbitrary complex Hilbert space H and any operator T of B(H), ifthere exists any non null function f which is analytic on a neighborhood of σ(T ) such thatf(T ) = 0, then there exists a polynomial which vanishes at T . Afterwards we can define theminimum polynomial of that operator, and the Riesz functional calculus allows us to affirm thatthe previous result holds.

5

In order to generalize the result in a more satisfactory way in the infinite dimensional case,we need a more elaborated functional calculus to define the equivalent notion of minimumpolynomial. It leads to consider the absolutely continous polynomially bounded operators(notation: PBabs(H)). In this case, we can extend naturally the definition of the class C0

given by B. Sz-Nagy and C. Foias ([11]). We can also define the minimum function of anoperator in C0 ∩ PBabs(H) as in [11]. Let us consider ∆ as an operator between B(H) anditself. We have the following stability property.

Lemma 11. If T is an operator of PBabs(H), then ∆(T ) also belongs to PBabs(H).

Proof. Assume that T ∈ PBabs(H). Lemma 3 allows us to affirm that ∆(T ) is polynomiallybounded. Let P be any arbitrary polynomial and x, y ∈ H be two vectors. Let µx,y (resp. νx,y)be a scalar quasi-spectral measure associated with the operator T (resp. ∆(T )) and the pair(x, y). We can write

〈P (T )x|y〉 =

2π∫

0

P (eiθ)dµx,y(θ) =

2π∫

0

P (eiθ)hx,y(eiθ)dm(θ)

where m denotes the Lebesgue measure on T. We decompose x = u ⊕ v with respect to the(orthogonal) direct sum H = N (T ) ⊕R(|T |).Firstly, we suppose that v ∈ R(|T |) so there exists a ∈ H such that v = |T | 12 a. Then it yields

〈P (∆(T ))x|y〉 = 〈P (∆(T ))u|y〉 + 〈P (∆(T ))|T | 12 a|y〉

Since u ∈ N (T ), we have u ∈ N (∆(T )) and 〈P (∆(T ))u|y〉 = P (0)〈u|y〉. Applying Lemma 3

on the last term, we get 〈P (∆(T ))|T | 12 a|y〉 = 〈|T | 12 P (T )a|y〉 = 〈P (T )a||T | 12 y〉. Consequently,we obtain

〈P (∆(T ))x|y〉 =

2π∫

0

P (eiθ)[〈u|y〉 + h

a,|T |12 y

(eiθ)]dm(θ).

Let us denote by γ the measure[ν

u⊕|T |12 a,y

− (〈u|y〉 + ha,|T |

12 y

(eiθ))dm]. The previous equality

shows that∀n ∈ N, γ̂(n) = 0

where γ̂(n) stands for the Fourier coefficient of index n. Applying Hoffman theorem (cf. [6]), weobtain that γ is absolutely continous with respect to Lebesgue measure m. Thus the singularpart νsing

u⊕v,y must be the zero measure.

In the general case, when v ∈ R(|T |), we can easily conclude by a density argument. Finally,νsing

x,y is null so that ∆(T ) is absolutely continous.

Now, we can give the following result concerning the minimum functions of an operator andits Aluthge transform.

Proposition 12. Let T ∈ C0 ∩PBabs(H). Then the operator ∆(T ) belongs to C0 ∩PBabs(H)and its minimum function divides the minimum function of T .

6

Proof. Let h ∈ H∞ be a function that vanishes at T . For r a real number such that 0 6 r < 1,we get h(r∆(T ))|T | 12 = |T | 12 h(rT ). All the terms are weakly convergent when r goes to 1.Letting r → 1, the previous equality becomes

h(∆(T ))|T | 12 = |T | 12 h(T ) = 0.

It follows immediately that h(∆(T )) prR(|T |) = 0.

Now, it remains to show that h(∆(T )) prN (T ) = 0. Let u ∈ N (T )\{0}, the inclusion N (T ) ⊂N (∆(T )) leads to 0 = h(T )u = h(0)u and hence h(0) = 0. Writing h(z) = zh1(z), we get

h(∆(T ))u = h1(∆(T ))∆(T )u = 0,

which completes the proof.

3 Stability for Iterated Kernels and Iterated Ranges

3.a) The Case of Iterated Kernels

In this section, we consider the classical increasing sequence (ap(T ))p∈N defined by

∀p ∈ N, ap(T ) = dimN (T p+1)

N (T p).

Recall that the ascent of an arbitrary operator T ∈ B(H) is the nonnegative integer a(T )defined by

a(T ) = inf{p > 0;N (T p) = N (T p+1)

}.

For more information, see [10]. More precisely, we are interested in studying the relationshipbetween the sequences (ap(T − αI)) and (ap(∆(T ) − αI)). Firstly, we obtain a general resultin the case α 6= 0.

Theorem 13. Let T ∈ B(H). For α ∈ C\{0} we get

∀p ∈ N, dimN ([T − αI]p+1)

N ([T − αI]p)= dim

N ([∆(T ) − αI]p+1)

N ([∆(T ) − αI]p),

that is (ap(T − αI)) = (ap(∆(T ) − αI)). In particular, the ascents of T − αI and ∆(T ) − αIcoincide.

Proof. First of all, from the equality dim(N (T − αI)) = dim(N (∆(T ) − αI)) (cf. [8]) wecan see that the case p = 0 is obvious. From now on, we suppose that p > 1. Write R =U |T | 12 . According to Lemma 3, we get RP (∆(T )) = P (T )R, for any polynomial P , andhence RN ([∆(T ) − αI]p+1) ⊂ N ([T − αI]p+1). Consequently, the application R : N ([∆(T ) −αI]p+1) → N ([T − αI]p+1) is well defined and we can consider the following mapping

f : N ([∆(T ) − αI]p+1) −→ N ([T − αI]p+1) /N ([T − αI]p)x 7−→ [Rx]

where [Rx] denotes the element associated with the vector Rx in the quotient vector space

N ([T −αI]p+1) /N ([T − αI]p) . Given any vector x ∈ N ([∆(T )−αI]p), we get Rx = U |T | 12 x ∈

7

N ([T −αI]p) (apply Lemma 3). Therefore, the function f admits a unique factorization of theform

N ([∆(T ) − αI]p+1)f

q

N ([T − αI]p+1)/N ([T − αI]p)

N ([∆(T ) − αI]p+1) /N ([∆(T ) − αI]p)f̃

where q is the canonical quotient map. Let x ∈ N ([∆(T )−αI]p+1) such that Rx ∈ N ([T−αI]p).Let us denote y = [∆(T ) − αI]px. On the one hand the equality [∆(T ) − αI]p+1x = 0 allowsus to affirm (∆(T ) − αI)y = 0. On the other hand, we have

0 = [T − αI]pRx = R[∆(T ) − αI]px = Ry

which gives ∆(T )y = 0 and, consequently, αy = 0. Since α 6= 0, we obtain y = 0 and themapping f̃ is one-to-one.In order to prove that the mapping is onto, we consider a vector y ∈ N ([T − αI]p+1) and we

resolve the equation y = [Rx] for an unknown vector x ∈ N ([∆(T )−αI]p+1). Set x = 1α|T | 12 y.

On the one hand, the equality [T − αI]p+1y = 0 leads to

0 =1

α|T | 12 [T − αI]p+1y = [∆(T ) − αI]p+1(

1

α|T | 12 y).

On the other hand, from

[T − αI]p(Rx − y) = [T − αI]p(1

αTy − y) =

1

α[T − αI]p+1y = 0,

we can see that Rx − y ∈ N ([T − αI]p). And hence f̃ is onto, and the theorem is proved.

In the case α = 0, Example 16 shows that the previous results doesn’t hold. However, aninequality between (ap(T )) and (ap(∆(T ))) is true:

Theorem 14. Let T ∈ B(H). The following inequality

∀p ∈ N\{0}, dimN (T p+1)

N (T p)> dim

N (∆(T )p+1)

N (∆(T )p)

holds.

Remark 15. Notice that dim(N (T )) 6 dim(N (∆(T ))) (cf. [8]).

Proof. We construct the application f̃ as above, and we prove that f̃ is one-to-one. Let x ∈N (∆(T )p+1) such that Rx ∈ N (T p). We get T pRx = 0 = R∆(T )px thus Ry = 0 wherey = ∆(T )px. Hence, we notice that

|T | 12 y ∈ N (U) = N (T ) = N (|T | 12 ),

which allows us to affirm that |T |y = 0. Indeed y ∈ N (|T |) = N (|T | 12 ). Since p > 1, we have

y ∈ R(|T | 12 ). Finally, we see that y ∈ N (|T | 12 ) ∩R(|T | 12 ) = {0} and f̃ is one-to-one.

8

Example 16. The partial isometry Sn defined in Example 5 provides us a counterexample forthe previous theorem in the case α = 0:

dimN (Sn

n)

N (Sn−1n )

= 1 > 0 = dimN (∆(Sn)n)

N (∆(Sn)n−1).

In particular, Sn and ∆(Sn) ascents do not coincide: a(Sn) = n > n − 1 = a(∆(Sn)).

Remark 17. If H is a finite dimensional vector space, the minimal polynomial of T can bewritten in the form

ωT (z) =∏

λ∈σ(T )

(z − λ)a(T−λI).

Thanks to the previous result, we recover that the minimum polynomial of ∆(T ) divides theminimum polynomial of T .

3.b) The Case of Iterated Ranges

As in the previous section, we consider the increasing sequence (dp(T ))p∈N by setting

∀p ∈ N, dp(T ) = dimR(T p)

R(T p+1).

Recall that the descent of an arbitrary operator T ∈ B(H) (cf. [10]) is the nonnegative integerdefined by

d(T ) = inf{p > 0;R(T p) = R(T p+1)

}.

In this section, we study the behavior of the sequence (dp(T − αI))p∈N⋆ . However, in order totake these quotient vector spaces into account, we must consider operators such that spacesR([T − αI]p) are closed: for this reason, we choose to restrict our study to the set K (H) ofcompact operators of B(H). We obtain the following result:

Theorem 18. Let T ∈ B(H) be a compact operator and α ∈ C\{0}. Then

∀p ∈ N\{0}, dimR([T − αI]p)

R([T − αI]p+1)= dim

R([∆(T ) − αI]p)

R([∆(T ) − αI]p+1).

In particular, the descents of T − αI and ∆(T ) − αI coincide.

Proof. Since T is a compact operator, we deduce successively that |T | and |T | 12 are compact.Hence ∆(T ) is also compact because K (H) is a closed ideal of B(H). For α 6= 0, the ranges ofthe operators [T − αI]m and [∆(T ) − αI]m are closed subspaces of H for any positive integerm. Consequently the quotient vector spaces R([T − αI]p)/R([T − αI]p+1) and R([∆(T ) −αI]p)/R([∆(T ) − αI]p+1) are well defined. Lemma 3 allows us to see that the mapping

g : R([T − αI]p) −→ R([∆(T ) − αI]p)/R([∆(T ) − αI]p+1)

y 7−→ [|T | 12 y],

where [|T | 12 y] denotes the element associated with |T | 12 y in the quotient vector space R([∆(T )−αI]p)/R([∆(T ) − αI]p+1), is well defined.

9

Let us prove the equality N (g) = R([T − αI]p+1). Let y ∈ N (g). Since |T | 12 y ∈ R([∆(T ) −αI]p+1), then there exists a vector x ∈ H such that

|T | 12 y = [∆(T ) − αI]p+1x =

p+1∑

k=0

(p + 1

k

)(−1)kαk∆(T )p+1−kx

= (−1)p+1αp+1x +

p∑

k=0

(p + 1

k

)(−1)kαk∆(T )p+1−kx.

Notice that α 6= 0, the previous equality implies that x ∈ R(|T | 12 ), thus there exists a vector

a ∈ H such that x = |T | 12 a. Therefore, applying Lemma 3, we obtain

|T | 12 y = [∆(T ) − αI]p+1|T | 12 a = |T | 12 [T − αI]p+1a.

Consequently, we have

y − [T − αI]p+1a ∈ N (|T | 12 ) = N (T ) ⊂ R([T − αI]p+1)

and hence y belongs to R([T −αI]p+1). Finally, we get N (g) ⊂ R([T −αI]p+1) and the reverseinclusion is obvious. Henceforth, the function g admits a unique factorization of the form

R([T − αI]p)g

q

R([∆(T ) − αI]p)/R([∆(T ) − αI]p+1)

R([T − αI]p)/R([T − αI]p+1)

g̃

.

Let us now show that g̃ is onto. Let z ∈ R([∆(T )−αI]p), we try to find a vector y ∈ R([T−αI]p)

such that |T | 12 y−z ∈ R([∆(T )−αI]p+1). But there exists x ∈ H such that z = [∆(T )−αI]px.Since [∆(T )−αI]p has a closed range, we can decompose x = (∆(T )−αI)x0 +x1 with respectto the (orthogonal) direct sum H = R(∆(T ) − αI) ⊕ N (∆(T )⋆ − αI). From the equality

(∆(T )⋆ − αI)x1 = 0, it follows that x1 = 1α∆(T )⋆x1 ∈ R(|T | 12 ). Then there exists a vector x2

such that x1 = |T | 12 x2. We can write x = (∆(T ) − αI)x0 + |T | 12 x2. Setting x′ = [T − αI]px2,we have

z − |T | 12 x′ = [∆(T ) − αI]p(x − |T | 12 x2) = [∆(T ) − αI]p+1x0 ∈ R([∆(T ) − αI]p+1).

From g̃([x′]) = [|T | 12 x′] = [z], we derive that g̃ is onto. As a conclusion, g̃ is the bijectionexpected.

Example 19. Example 5 provides a counterexample for the previous theorem in the caseα = 0:

dimR(Sn−1

n )

R(Snn)

= 1 > 0 = dimR(∆(Sn)n−1)

R(∆(Sn)n).

Remark 20. In the finite dimensional case, it can be proved that the previous theorem holdsfor α = 0 and for T being a one-to-one operator. Indeed, we can prove that the factorizationconstruction remains valid, and the equality |T | 12 R(T p) = R(∆(T )p) gives the result.

10

4 Stability of the Operator Classes C0,· and C1,·

Recall that an arbitrary power bounded operator T acting on a Hilbert space H is said to beof class C0,· (resp. of class C1,·) if Tnx → 0 (n → +∞) for any x ∈ H (resp. if the sequence(Tnx)n>0 doesn’t converge to 0 for any x ∈ H\{0}). By [11], any contraction T may be (noncanonically) triangulated in the form

T =

(T0,· ∗0 T1,·

)

where T0,· (resp. T1,·) is a contraction of class C0,· (resp. of class C1,·). This triangularizationcan be generalized in a straightforward way for power bounded operators.

4.a) The Case of Coisometries

This section is concerned with coisometries, that are operators V such that V ⋆ is an isometry.

Proposition 21. Let V be a coisometry of B(H). Then we have

∆n(V ) = V ⋆nV n+1

for any nonnegative integer n.

Proof. We proceed by induction. It is obvious for n = 0. Let n ∈ N and assume that ∆n(V ) =V ⋆nV n+1. The equality

|∆n(V )|2 = V ⋆n+1V nV ⋆nV n+1 = V ⋆n+1V n+1

shows that |∆n(V )|2 is a projection, denoted by Pn+1. Since N (V n+1) = N (Pn+1), the polardecomposition of ∆n(V ) is ∆n(V ) = ∆n(V )Pn+1. It yields to the equality

∆n+1(V ) = Pn+1∆(V )Pn+1 = V ⋆n+1V n+2

and the result is proved.

By Wold decomposition, the C0,· part of V corresponds to its pure coisometric part and theC1,· part of V is its unitary part. The following result shows that we can recapture the unitarypart of V by considering the iterated Aluthge transforms of V .

Theorem 22. Let V be a coisometry of B(H). Then the sequence (∆n(V )) strongly convergesto the unitary part of V .

Proof. From the Nagy-Foias decomposition (cf. [11]), we can write V = U ⊕W with respect tothe direct sum H = Hu⊕Hcnu where Hu (resp. Hcnu) is the reducing subspace associated withthe unitary part (resp. completly non unitary part) of V . Then we get V ⋆nV n = IHu

⊕W ⋆nWn

and I = V nV ⋆n = IHcnu⊕ WnW ⋆n. But W ⋆nWn = Qn is an orthogonal projector, so

WnW ⋆n = IHcnu. Notice that (Qn)n is a strongly decreasing sequence of positive operators and

hence its limit Q is also an orthogonal projector. Let us verify R(Q) = {0}. Since W ⋆QW = Q,we infer that QW = WQ and W ⋆Q = QW ⋆. If x ∈ R(Q), then it yields to the equalities

‖x‖2 = ‖Qx‖2 = 〈Qx|x〉 = 〈W ⋆nQWnx|x〉 = 〈QW ⋆nWnx|x〉,which implies

‖x‖2 = 〈W ⋆nWnx|Qx〉 = 〈W ⋆nWnx|x〉 = ‖Wnx‖2.

Therefore x belongs to the unitary part of W which is reduced to {0}.

11

4.b) The Case of Contractions

In this section, we restrict our study to the set of contractions of B(H) and we will showstability properties for Aluthge transform acting on C0,· and C1,· operator classes. Recall thata contraction T is of class C0,· if and only if the operator ST = lim(T ⋆nTn) is reduced to thenull application and T is of class C1,· if and only if ST is one-to-one.

Proposition 23. If T is a contraction of B(H), then the following equalities

ST = |T | 12 S∆(T )|T | 12 (1)

andS∆(T ) = |T | 12 U⋆ST U |T | 12 (2)

hold.

Proof. Let T ∈ B(H) such that ‖T‖ 6 1. On the one hand, from the inequality

∆(T )⋆n∆(T )n = |T | 12 U⋆T ⋆n−1|T | 12 |T | 12 Tn−1U |T | 12 6 ‖T‖|T | 12 U⋆T ⋆n−1Tn−1U |T | 12 ,

we get S∆(T ) 6 |T | 12 U⋆ST U |T | 12 . On the other hand, observe that

T ⋆nTn = |T | 12 ∆(T )⋆n−1|T | 12 U⋆U |T | 12 ∆(T )n−1|T | 12 6 ‖T‖|T | 12 ∆(T )⋆n−1∆(T )n−1|T | 12 ,

which implies ST 6 |T | 12 S∆(T )|T | 12 . Combining these two inequalities, we obtain successively

ST 6 |T | 12 S∆(T )|T | 12 6 |T |U⋆ST U |T | = T ⋆ST T = ST

and

S∆(T ) 6 |T | 12 U⋆ST U |T | 12 6 |T | 12 U⋆|T | 12 S∆(T )|T | 12 U |T | 12 = ∆(T )⋆S∆(T )∆(T ) = S∆(T )

and hence the formulas (1) and (2) are established.

Corollary 24. Let T be a contraction of B(H). Then T ∈ C0,· (resp. T ∈ C1,·) if and only if∆(T ) ∈ C0,· (resp. ∆(T ) ∈ C1,·).

Proof. The result concerning the class C0,· is obvious. Let T ∈ C1,·, since ST is one-to-one, |T |is also one-to-one. Let x ∈ H such that S∆(T )x = 0. From (2), we obtain

√ST U |T | 12 x = 0,

which implies successively U |T | 12 x = 0 and |T | 12 x = 0, according to Remark 1. Then it followsthat x = 0 and S∆(T ) is one-to-one which means that ∆(T ) ∈ C1,·.Conversely, assume that ∆(T ) ∈ C1,·. Let x ∈ H such that ST x = 0. From (1), we get√

ST |T | 12 x = 0 as above and hence |T | 12 x = 0. Upon applying |T | 12 U to this equality, we obtain∆(T )x = 0. Since S∆(T ) is one-to-one, ∆(T ) is also one-to-one, therefore we deduce x = 0.Finally ST is one-to-one and T ∈ C1,·.

12

4.c) Stability of the operator ST via Aluthge transform

Now we focus on the contractions T of B(H) satisfying ST = S∆(T ). We get the result:

Theorem 25. Let T ∈ B(H) be a contraction. Then the two following assertions are equiva-lent:

i) S∆(T ) = ST ;

ii) T can be written in the form T = A+B where A ∈ C0,· and B ∈ B(H) is a partial isometrysuch that BA = BA⋆ = A⋆B = 0.

Moreover, we can choose B such that the compression prN (T )⊥ B prN (T )⊥ is one-to-one.

Proof. Firstly, assume that S∆(T ) = ST . Combining the two formulas of Proposition 23, we get

|T | 12 U⋆ST U |T | 12 = |T | 12 ST |T | 12 , which can be rewritten in the following way

|T | 12 (ST − U⋆ST U)|T | 12 = 0. (3)

Let us prove now that the operator J = ST − U⋆ST U is reduced to 0. First of all, we supposethat T is one-to-one. Then |T | is a dense range operator and the previous equality leads to〈Jx|x〉 = 0 for any vector x ∈ H. Consequently, the numerical range of J is reduced to {0}and hence J = 0.In general, we consider two vectors x, y ∈ H, that we decompose with respect to the orthogonaldecomposition H = N (T ) ⊕R(|T |): x = x1 + x2 and y = y1 + y2. Then:

〈Jx|y〉 = 〈Jx1|y〉 + 〈Jx2|y1〉 + 〈Jx2|y2〉.

Since x1 ∈ N (T ), it follows that ST x1 = 0 and Ux1 = 0, so 〈Jx1|y〉 = 0. In the same manner,we obtain 〈Jx2|y1〉 = 〈x2|Jy1〉 = 0. Using (3) and a standard density argument, we easily seethat the last term is also null. Finally, the equality ST − U⋆ST U = 0 is proved.Now we decompose T with respect to the (orthogonal) direct sum H = N (ST ) ⊕R(ST ):

T =

(T1 R1

0 T2

).

Then for every nonnegative integer n, Tn =

(Tn

1 Rn

0 Tn2

)where the sequence (Rn)n∈N satisfies

Rn+1 = Tn1 R1 + RnT2. Therefore, the operators

T ⋆nTn =

(T ⋆n

1 Tn1 T ⋆n

1 Rn

R⋆nTn

1 R⋆nRn + T ⋆n

2 Tn2

)

converge to ST =

(0 00 S

)where S is a one-to-one operator that we will make precise. It forces

the sequences (T ⋆n1 Rn) and (R⋆

nTn1 ) to converge strongly to 0, and (R⋆

nRn +T ⋆n2 Tn

2 ) to convergestrongly to S. Since T2 is a contraction, (T ⋆n

2 Tn2 ) strongly converges to ST2

. Consequently,(R⋆

nRn) is strongly convergent to an operator denoted by R. From the equality

R⋆n+1Rn+1 = R⋆T ⋆n

1 Tn1 R1 + T ⋆

2 RnTn1 R1 + R⋆

1T⋆n1 RnT2 + T ⋆

2 R⋆nRnT2

13

we can deduce that R is a T2-Toeplitz operator (see [3] for a definition and related properties)because the three first terms of the right member strongly converges to 0. Finally, it impliesthat S is a T2-Toeplitz operator. Since T is a contraction and R⋆

nRn is positive operator, wecan notice that

T ⋆n2 Tn

2 6 R⋆nRn + T ⋆n

2 Tn2 6 I.

Letting n → +∞, we get ST26 S 6 I. Thus we have

T ⋆m2 ST2

Tm2 6 T ⋆m

2 STm2 6 T ⋆m

2 Tm2

for any nonnegative integer m. Letting m → +∞, we obtain S = ST2because S and ST2

areT2-Toeplitz operators.We decompose |T | with respect to the (orthogonal) direct sum H = N (ST ) ⊕ R(|T |). Fromthe equalities

|T | 12 S2T = |T | 12 ST |T | 12 ST |T | 12 = S2

T |T | 12 ,

it follows that |T | 12 ST = ST |T | 12 and |T |ST = ST |T |. As a consequence, we have |T | =(|T1| 00 L

)where T1R

⋆1 = 0 and L =

√R⋆

1R1 + T ⋆2 T2. But ST = |T | 12 ST |T | 12 = ST |T |, therefore

we deduce ST (I − |T |) = 0, which gives ST2(I − L) = 0. Since ST2

is one-to-one, we get L = Iand R⋆

1R1 + T ⋆2 T2 = I. Now we can write

T = A + B where A =

(T1 00 0

), B =

(0 R1

0 T2

).

The operator A is of class C0,·. An easy computation shows that B⋆B =

(0 00 I

)and hence B

is a partial isometry. We easily verify that BA = BA⋆ = A⋆B = 0.Conversely, decomposing A and B with respect to the orthogonal decomposition H = N (B)⊕R(B⋆), we obtain

A =

(A1 00 0

)and B =

(0 Y1

0 B2

).

Then T ⋆nTn =

(A⋆n

1 An1 A⋆n

1 Yn

Y ⋆n An

1 Y ⋆n Yn + B⋆n

2 Bn2

), where Yn satisfies Yn+1 = An

1Y1 + YnB2. We know

that T ⋆nTn strongly converges to an operator of the form

(0 00 Y

). In the same manner, we

can show that Y is a B2-Toeplitz operator and Y = SB2. Finally, we obtain that ST = 0⊕SB2

and S∆(T ) = |T | 12 ST |T | 12 = 0 ⊕ ISB2I = ST . As a conclusion, the equivalence is proved.

The extra-condition about B means B2 can be supposed one-to-one.

Corollary 26. Let T ∈ B(H) be a dense range contraction. Thus the two following assertionsare equivalent:

i) S∆(T ) = ST ;

ii) there exists an operator A ∈ C0,· acting on a subspace H1 and a unitary operator B actingon a subspace H2 such that T = A ⊕ B, with respect to the (orthogonal) direct sum H =H1 ⊕H2.

14

Proof. We only have to prove that proposition i) implies ii). Assume that S∆(T ) = ST . Ac-cording to Theorem 25, T can be decomposed as T = A + B where A and B satisfy conditionii) and we write A and B with respect to the orthogonal decomposition H = N (B) ⊕R(B⋆)

A =

(A1 00 0

)and B =

(0 Y1

0 B2

).

By Theorem 25, we may suppose that prN (T )⊥ B prN (T )⊥ is one-to-one. Let b be a vector ofR(B⋆), set y = B2B

⋆2b − b and x = Y1B

⋆2b. Having in mind that we have A⋆

1Y1 = 0 andY ⋆

1 Y1 + B⋆2B2 = I2, we obtain

A⋆1x = 0 and Y ⋆

1 x + B⋆2y = 0.

Since N (T ⋆) = {0}, it follows that x = y = 0 and, consequently, Y1B⋆2b = 0 and Y1B

⋆2 = 0. We

can see that Y1 = 0 because B⋆2 has a dense range. Finally, B2 is a dense range isometry and

hence a unitary operator.

Corollary 27. If T ∈ B(H) is a partial isometry, then S∆(T ) = ST .

4.d) The Case of Power Bounded Operators

In this section, we use tools, introduced in [3] and well adapted to the study of the asymptoticbehaviour of the sequence (T ⋆nTn)n. Given an arbitrary Banach limit L , we define the mappingET acting on B(H) associated with a bounded operator T as follows:

∀X ∈ B(H), ∀x, y ∈ H, 〈ET (X)x|y〉 = L 〈T ⋆nXTnx|y〉.

ET (I) corresponds to the Banach limit of T ⋆nTn. Firstly, we need to establish some resultsrelated to ET . If S and T are two operators of B(H), using the following inequality (provedin [3])

∀x, y ∈ H, |〈ES(|T |)x|y〉| 6√〈ES(I)x|x〉

√〈ES(|T |2)x|x〉, (4)

we obtain the two next lemmas.

Lemma 28. Let T be a power bounded operator of B(H). Then we have

ET (|T |) 6 ET (I).

Proof. Let x, y ∈ H: we can write

|〈ET (|T |)x|y〉| = |〈ET (I|T |)x|y〉| 6√〈ET (I)x|x〉

√〈ET (|T |2)y|y〉.

Since ET (|T |2) = ET (T ⋆T ) = ET (I), we can conclude that 〈ET (|T |)x|x〉 6 〈ET (I)x|x〉 and theresult is proved.

Lemma 29. For every power bounded operator T of B(H) we have

ET (I) = |T | 12 E∆(T )(|T |)|T | 12 and E∆(T )(|T |) = |T | 12 U⋆ET (I)U |T | 12 .

15

Proof. Writing T ⋆nTn = |T | 12 ∆(T )⋆n−1|T | 12 U⋆U |T | 12 ∆(T )(T )n−1|T | 12 , we can see that

ET (I) = |T | 12 E∆(T )(|T | 12 U⋆U |T | 12 )|T | 12 .

The equality U⋆U = prR(|T |) gives the first result.

The second equality derives from the following: ∆(T )⋆n|T |∆(T )n = |T | 12 U⋆(T ⋆nTn)U |T | 12 .

Thanks to these two results, we can easily obtain the stability of C0,· and C1,· classes forpower bounded operators by a different method.

Theorem 30. Let T be a power bounded operator of B(H). Then T ∈ C0,· (resp. T ∈ C1,·) ifand only if ∆(T ) ∈ C0,· (resp. ∆(T ) ∈ C1,·).

Finally, we study the sequence(E∆(T )

(|T |2n))

n∈N. Let us denote by F the spectral measure

associated with T .

Proposition 31. Let T be a power bounded operator of B(H). Then(E∆(T )

(|T |2n))

n∈Nis

increasing. Moreover, we have

E∆(T )

(|T |2n)

6 limn→+∞

E∆(T )

(|T |2n

F (]r, ‖T‖]))

for any real number r such that 0 < r < 1.

Proof. By induction, we will prove that we have

∀n ∈ N, E∆(T )(I) 6 E∆(T )(|T |) 6 E∆(T )(|T |2) 6 . . . 6 E∆(T )

(|T |2n)

,

which implies that the sequence is increasing. Writing

∆(T )⋆n∆(T )n = |T | 12 U⋆T ⋆n−1|T |Tn−1U |T | 12

and using Lemma 28, we obtain

E∆(T )(I) = |T | 12 U⋆ET (|T |)U |T | 12 6 |T | 12 U⋆ET (I)U |T | 12 .

Lemma 29 leads to E∆(T )(I) 6 E∆(T )(|T |).Given n ∈ N, assume that the inequalities E∆(T )(I) 6 E∆(T )(|T |) 6 E∆(T )(|T |2) 6 . . . 6

E∆(T )

(|T |2n)

are true. According to (4), we obtain

⟨E∆(T )

(|T |2n)

x∣∣ x

⟩6

√〈E∆(T )(I)x|x〉

√⟨E∆(T )

(|T |2n+1

)x∣∣x

⟩

6

√⟨E∆(T ) (|T |2n) x

∣∣ x⟩√⟨

E∆(T )

(|T |2n+1

)x∣∣ x

⟩.

If⟨E∆(T )

(|T |2n)

x∣∣ x

⟩6= 0, we get

⟨E∆(T )

(|T |2n)

x∣∣ x

⟩6

⟨E∆(T )

(|T |2n+1

)x∣∣∣ x

⟩; else, the

inequality is still satisfied. Consequently, we have E∆(T )

(|T |2n)

6 E∆(T )

(|T |2n+1

)in both

cases.Let us show the last part of the proposition. Set P = F ([0, r]) and Q = I − P = F (]r, ‖T‖]),where F stands for the spectral measure associated with |T |. Then, we have |T | 6 rP + |T |Q.

16

Therefore, using the classical functional calculus associated with the positive operator |T |, wecan see that

⟨|T |2n

x∣∣ x

⟩=

‖T‖∫

0

t2n

dFx,x(t)

=

∫t2

n

1l[0,r]dFx,x(t) +

∫t2

n

1l]r,‖T‖]dFx,x(t)

6⟨(

r2n

P + Q|T |2n)x∣∣ x

⟩

and hence E∆(T )

(|T |2n)

6 r2n

E∆(T )(P ) + E∆(T )

(Q|T |2n)

. Since 0 < r < 1, the sequence(r2n

E∆(T )(P ))n

converges to 0 and(E∆(T )

(Q|T |2n))

is also convergent because Q|T |2n

=

Q2|T |2n

= Q|T |2n

Q. The proposition is proved.

Remark 32. When T is a contraction, the inequality E∆(T )

(|T |2n)

6 E∆(T )(I) allows us to

affirm that the sequence(E∆(T )

(|T |2n))

n∈Nis stationnary.

References

[1] T. Ando, Aluthge Transform and the Convex Hull of the Spectrum of a Hilbert Space Oper-ator, Op. Theory Adv. and Appl. 160, 21-39, 2005.

[2] S.R. Caradus, Operators with Finite Ascent and Descent, Pacific J. of Math. 3, 437-449,1966.

[3] G. Cassier, Generalized Toeplitz Operators, Restrictions to Invariant Subspaces and Simi-larity Problems, J. Operator Theory 53:1, 49-89, 2005.

[4] J.B. Conway, A Course in Functional Analysis, Second Edition, Springer, New-York.

[5] C. Foias, I.B. Jung, E. Ko, C. Pearcy, Complete Contractivity of Maps Associated with theAluthge and Duggal Transforms, Pacific J. of Math. 209:2, 249-259, 2003.

[6] K. Hoffman, Banach Spaces of Analytic Functions, Prentice Hall Inc., 1962.

[7] I.B. Jung, E. Ko, C. Pearcy, Aluthge Transform of Operator, Integr. Equ. Oper. Theory 37,437-448, 2000.

[8] I.B. Jung, E. Ko, C. Pearcy, Spectral Pictures of Aluthge Transforms of Operators, Integr.Equ. Oper. Theory 40, 52-60, 2001.

[9] I.B. Jung, E. Ko, C. Pearcy, The Iterated Aluthge Transform of an Operator, Integr. Equ.Oper. Theory 45, 375-387, 2003.

[10] M. Mbekhta, Ascente, descente et spectre essentiel quasi-Fredholm, Rend. Circ. Mat.Palermo, 46, 175-196, 1997.

[11] B. Sz.-Nagy, C. Foias, Harmonic Analysis of Operators on Hilbert Space, North-Holland,Amsterdam 1970.

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