OPERATORS SIMILAR TO PARTIAL ISOMETRIES

C. BADEA AND M. MBEKHTA

Abstract. A Hilbert space bounded linear operator is said to be regular if its range is closed and its kernel is included in the intersection of the ranges of all iterates. We prove that if T is regular, then T is similar to a partial isometry if and only if T is power bounded and there exists a power bounded operator S such that T ST = T. This is a generalization of a similarity criterion due to B. Sz.-Nagy. The regularity condition cannot be avoided. Indeed, using Pisier’s example of a polynomially bounded operator not similar to a contraction, two polynomially bounded operators T and S are constructed such that T ST = T and ST S = S, but T is not similar to a partial isometry.

1. Introduction

1A. Preamble. A Hilbert space bounded linear operator T ∈ B(H) is said to be similar to a unitary if there exists an invertible operator L ∈ B(H) such that L^{−1}T L is a unitary operator. The similarity criterion of Sz.-Nagy ([SN]) states that an invertible operator T ∈ B(H) is similar to a unitary if and only if T and its inverse T^{−1} are both power bounded, that is

sup_{n∈ℕ}∥T^{n}∥ < ∞ and sup_{n∈ℕ}∥T^{−n}∥ < ∞.

The same techniques as in [SN] (see also [CMS]) show that a left invertible operator T ∈ B(H) is similar to an isometry if and only if T is power bounded and there is a power bounded left inverse S ∈ B(H) of T. An analogous result characterizes those right invertible operators which are similar to co-isometries.

2000 Mathematics Subject Classification. 47A05, 47A10, 47B47. Key words and phrases. partial isometries, similarity problems, generalized in-

verse, generalized spectrum, the commutator equation.

The aim of this paper is to study, among other things, analogous criteria for similarity to partial isometries.

1B. Preliminaries. In general H, with or without subscripts, will denote a complex Hilbert space. Let B(H) be the C*-algebra of all bounded linear operators on H.

Two operators A and B in B(H) are similar if there is an invertible operator L ∈ B(H) such that B = L^{−1}AL. For a class C of operators on Hilbert spaces, we say that an operator A ∈ B(H) is similar to an element of C if there is B ∈ C ∩ B(H) which is similar to A.

Recall that V ∈ B(H) is an isometry if V^{∗}V = I and a co-isometry if V V^{∗} = I. Here I denotes the identity operator. The operator V is said to be a partial isometry if V V^{∗}V = V. Equivalently, V^{∗}V is a projection, or ∥V x∥ = ∥x∥ for each x in the orthogonal complement of the kernel of V. An operator P is a power partial isometry if P^{n} is a partial isometry for every integer n ≥ 1. A contraction is an operator T such that ∥T ∥ ≤ 1, equivalently, such that T^{∗}T ≤ I. It is easy to see that V is an isometry if and only if V is a contraction and V has a left inverse which is also a contraction. The Sz.-Nagy criterion stated in the Preamble shows that this characterization of isometries extends up to similarity if we replace ’contraction’ with ’power bounded’ : If V is power bounded and has a left inverse which is also power bounded, then V is similar to an isometry. An analogous result holds for operators similar to co-isometries.

It follows from [Mb3, Theorem 3.1] that T is a partial isometry if and only if T is a contraction and there is a contraction S such that T ST = T and ST S = S. It was conjectured in [Mb3] that this characterization extends up to similarity, replacing again ’contraction’ with ’power bounded’.

1C. What this paper is about. The aim of this paper is to give a counterexample to the above conjecture and to prove some positive results under the additional assumption of regularity for T (see the Abstract). Other examples are also discussed. We refer to Section 2 for a discussion of regular operators. Notice that both isometries and

SIMILARITY TO PARTIAL ISOMETRIES 3

co-isometries are regular partial isometries ; in fact, left and right invertible Hilbert space operators are regular. In all considered cases we get similarity to power partial isometries.

The counterexample shows the existence of a polynomially bounded operator with a polynomially bounded generalized inverse without being similar to a partial isometry. The proofs of the (counter)examples are based on the existence of power bounded (or even polynomially bounded) operators not similar to contractions ([Fo, Pi1]) and to the existence ([Pi2]) of a pair of commuting operators (T_{1}, T_{2}), such that each of T_{1} and T_{2} is similar to a contraction, but the pair (T_{1}, T_{2}) is not (simultaneously) similar to a pair of contractions.

1D. Organization of the paper. We start the next section with a discussion of generalized inverses and the generalized spectrum of a bounded linear operator. We discuss an equivalent definition of regularity : an operator T is regular if zero does not belong to the generalized spectrum of T. We then describe the structure of regular partial isometries (Theorem 2.3) and compare regular partial isometries with power partial isometries (Lemma 2.4). In Section 3 we give a characterization of operators similar to partial isometries in terms of existence, in a new geometry, of a contractive generalized inverse (Theorem 3.1) and we state the positive result mentioned in the Abstract (Theorem 3.2). Then the counterexample is constructed (Example 4.1) ; a related one is given in Example 4.3. Theorem 3.2 is proved in Section 5. We end the paper with some final remarks and two open questions.

2. Regular partial isometries

2A. Generalized inverses and generalized spectrum. We start by introducing some terminology and recalling known results. Generalized inverses. A bounded linear operator S ∈ B(H) is said to be a generalized inverse of T ∈ B(H) if T ST = T and ST S = S. In this case we will say that T has a generalized inverse. Notice that the first equality T ST = T is a necessary and sufficient condition for T to have a generalized inverse. Indeed, if T ST = T, then T S′T = T and S′T S′ = S′ with S′ = ST S. In particular, if T and S are contractions

4 C. BADEA AND M. MBEKHTA

and T ST = T, then S′ = ST S is a contractive generalized inverse for T. It is not hard to prove that the Hilbert space bounded linear operator T has a generalized inverse if and only if the range R(T) of T is closed. The kernel of T will be denoted by N(T). The generalized spectrum. In [Mb2] a generalization of the spectrum σ(T) of T ∈ B(H) was given by replacing the notion of invertibility which appears in the classical definition of the spectrum with the existence of an analytic generalized inverse. More precisely, we denote by reg(T) the regular set of T, defined as the set of all complex numbers λ for which there is a neighborhood U_{λ} of λ and an analytic function R : U_{λ} ⟶ B(H) such that R(μ) is a generalized inverse of T − μI for every μ ∈ U_{λ}.

The complement σ_{g}(T) = ℂ \ reg(T) of reg(T) in ℂ is called the generalized spectrum of T. It is always non-void and compact.

The Hilbert space bounded linear operator T is said to be regular if its range R(T) is closed and the kernel N(T) is included in R^{∞}(T). Here R^{∞}(T), the generalized range of T, is the intersection of the ranges of all iterates of T :

R^{∞}(T) = ⋂_{n≥0} R(T^{n}).

According to [Mb1, Théorème 2.6], T is regular if and only if 0 belongs to reg(T). Another equivalent (algebraic) definition of regularity is the following : T is regular if and only if T has a generalized inverse and we have N^{∞}(T) ⊂ R^{∞}(T). Here

N^{∞}(T) = ⋃_{n≥0} N(T^{n})

is the generalized kernel of T.

We refer to [Ap, Mb1, Mb2] for more information about the generalized spectrum. In particular, the following useful facts are true.

2.1. Remarks. (a) ∂σ(T) ⊂ σ_{g}(T) ⊂ σ(T) ;

(b) σ_{g}(LT L^{−1}) = σ_{g}(T) for any invertible L ∈ B(H);

(c) σ_{g}(T^{∗}) = \overline{σ_{g}(T)} and the generalized spectrum satisfies the spectral mapping theorem; (d) If T is regular, then R^{∞}(T) is closed and we have

T (R^{∞}(T)) = R^{∞}(T) and T (\overline{N^{∞}(T)}) = \overline{N^{∞}(T)};

(e) If T is regular, then R^{∞}(T)^{⊥} = \overline{N^{∞}(T^{∗})}.

2B. The structure of regular partial isometries. Let T be a regular operator. In the following lemma we show the invariance of R^{∞}(T) and \overline{N^{∞}(T)} under generalized inverses of T.

2.2. Lemma. Let T ∈ B(H). Suppose that T is regular and T ST = T. Then the subspaces R^{∞}(T) et \overline{N^{∞}(T)} are invariant subspaces of S.

Proof. Let u ∈ R^{∞}(T) = T (R^{∞}(T)). Then T Su = u = T v with v ∈ R^{∞}(T). We get T (Su − v) = 0 and so Su − v ∈ N(T) ⊂ R^{∞}(T). Therefore Su ∈ R^{∞}(T).

Let u ∈ N^{∞}(T) ⊂ R(T). Then there exists n ∈ ℕ and v ∈ H such that u = T v ∈ N(T^{n}). We obtain

T^{n+1}Su = T^{n+1}ST v = T^{n+1}v = T^{n}u = 0.

Hence Su ∈ N(T^{n+1}) ⊂ N^{∞}(T). Thus S(N^{∞}(T)) ⊂ N^{∞}(T) and using the continuity of S, \overline{N^{∞}(T)} is an invariant subspace of S. □

The following theorem describes the structure of all regular partial isometries. Recall that A ∈ B(H) is said to be a pure isometry (or an abstract unilateral shift) if A is an isometry such that A^{∗n} converges strongly to zero. We say that A ∈ B(H) is a pure co-isometry (or a backward shift) if A^{∗} is a pure isometry. Recall also ([RR, Chapter 1]) that an isometry A is a pure isometry if and only if R^{∞}(A) = {0}.

2.3. Theorem. Let T ∈ B(H) with 0 ∉ σ_{g}(T). Then T is a partial isometry if and only if T = T_{r} ⊕ T_{0} ⊕ T_{ℓ} with T_{r} a pure co-isometry, T_{0} unitary and T_{ℓ} a pure isometry.

Proof. We prove only the difficult implication. Let T be a regular partial isometry. Recall that T T^{∗}T = T. Denote

H_{r} = \overline{N^{∞}(T)} and H_{ℓ} = R^{∞}(T)^{⊥} = \overline{N^{∞}(T^{∗})}.

Then H_{r} and H_{ℓ} are orthogonal subspaces; indeed H_{r} ⊂ R^{∞}(T). Therefore H_{r} ⊕ H_{ℓ} is closed.

6 C. BADEA AND M. MBEKHTA

Denote

H_{0} = [H_{r} ⊕ H_{ℓ}]^{⊥}.

Then H = H_{r} ⊕ H_{0} ⊕ H_{ℓ}. Since 0 ∉ σ_{g}(T), we have T (H_{r}) = H_{r}. On the other hand, Lemma 2.2 (with S = T^{∗}) shows that H_{ℓ} is an invariant subspace of T.

The subspace H_{0} is invariant for T. Indeed, H_{0}^{⊥} = H_{r} ⊕ H_{ℓ} is an invariant subspace of T^{∗}. Denote

T_{r} = T_{|H_{r}}, T_{0} = T_{|H_{0}} and T_{ℓ} = T_{|H_{ℓ}}.

We have shown that T_{r} is right invertible. To show that T_{ℓ} is left invertible, we show that it is one-to-one and with closed range. Indeed, T_{ℓ} is one-to-one since N(T) ⊥ H_{ℓ}. On the other hand, the subspace H_{ℓ}⊕N(T) is closed, being the orthogonal sum of two closed sets. Using [Ka, Lemma 3.3.1], the range of T_{ℓ} is closed.

We show now that T_{0} is invertible. Indeed, we have

H_{0} = [H_{r} ⊕ H_{ℓ}]^{⊥} = H_{r}^{⊥} ∩ H_{ℓ}^{⊥}.

Therefore T_{0} is one-to-one and with dense range. However, H_{0}⊕N(T) is a closed subspace, again as an orthogonal sum of two closed subspaces. Using [Ka, Lemma 3.3.1], the range of T_{0} is closed.

Therefore T_{r} is a co-isometry, T_{0} is unitary and T_{ℓ} is an isometry. We show now that

R^{∞}(T_{ℓ}) = {0}.

Indeed, since H_{ℓ} is an invariant subspace of T, we have

R^{∞}(T_{ℓ}) ⊂ H_{ℓ} ∩ R^{∞}(T) = R^{∞}(T)^{⊥} ∩ R^{∞}(T) = {0}.

Thus T_{ℓ} is a pure isometry.

Since \overline{N^{∞}(T_{r})} = \overline{N^{∞}(T)} = H_{r} (or since (T_{r})^{∗} = T_{ℓ}^{∗} and T_{r} = (T_{ℓ}^{∗})^{∗}), we obtain that T_{r} is a pure co-isometry. □

We note that the above proof generalizes the well-known Wold-von Neumann decomposition of an isometry. In this case H_{r} = {0} and the isometry is decomposed as the orthogonal sum of a unitary and a pure isometry.

SIMILARITY TO PARTIAL ISOMETRIES 7

2C. Regular partial isometries as power partial isometries. A bounded linear operator T ∈ B(H) is said to be a power partial isometry if, for every n ∈ ℕ, T^{n} is a partial isometry. Equivalently, T^{n}T^{∗n}T^{n} = T^{n} (n ∈ ℕ).

According to a result of Halmos and Wallen ([HW]), a power partial isometry is a direct sum whose summands are unitary operators, pure isometries, pure co-isometries and truncated shifts. A truncated shift is an operator of the following type. Let H = H_{0} ⊕H_{0} ⊕⋯⊕H_{0} where there are n ≥ 1 summands and let J_{n}(x_{1}, ⋯, x_{n}) = (0, x_{1}, ⋯, x_{n−1}). The truncated shift J_{n} is said to be of index n. Note in particular that J_{1} = 0.

Theorem 2.3 shows that each regular partial isometry is also a power partial isometry with no truncated shift in its Halmos-Wallen decomposition. In particular, if T is a regular partial isometry, then T^{j}T^{∗j}T^{j} = T^{j} (j ∈ ℕ).

The following lemma gives a generalization of this property.

2.4. Lemma. Let T, S ∈ B(H). Suppose that 0 ∉ σ_{g}(T) and T ST = T. Let i ≥ 1 be a fixed integer. Then

T^{i}S^{i}T^{j} = \{ \begin{matrix}T^{i}S^{i−j} & if 0 ≤ j ≤ i; \\ T^{j} & if j ≥ i \end{matrix}

and

T^{j}S^{i}T^{i} = \{ \begin{matrix}S^{i−j}T^{i} & if 0 ≤ j ≤ i; \\ T^{j} & if j ≥ i. \end{matrix}

Proof. Notice first that if 1 ≤ j ≤ i, then

T^{i}S^{i}T^{j} − T^{i}S^{i−j} = ∑_{m=1}^{j} T^{i}S^{i−m}(ST − I)T^{j−m}.

Indeed, it is enough to develop the sum.

We show now that for m = 1, ⋯, j, we have T^{i}S^{i−m}(ST − I) = 0. Indeed, if x ∈ H, then (ST − I)x ∈ N(T). Using [Mb1, Corollaire

2.5], we obtain

S^{i−m}(ST − I)x ∈ N(T^{i−m+1}).

8 C. BADEA AND M. MBEKHTA

Hence

T^{i}S^{i−m}(ST − I)x = T^{m−1}T^{i−m+1}S^{i−m}(ST − I)x = 0.

Therefore, all the terms of the sum vanish and so

T^{i}S^{i}T^{j} = T^{i}S^{i−j} (1 ≤ j ≤ i).

On the other hand, if j ≥ i, then

T^{i}S^{i}T^{j} = (T^{i}S^{i}T^{i})T^{j−i} = T^{i}T^{j−i} = T^{j}.

For the second equality, we note similarly that for 1 ≤ j ≤ i we have

T^{j}S^{i}T^{i} − S^{i−j}T^{i} = ∑_{m=1}^{j} T^{j−m}(T S − I)S^{i−m}T^{i}.

We show now that for m = 1, ⋯, j, we have (T S − I)S^{i−m}T^{i} = 0. Indeed, if x ∈ H, then T^{i}x = T^{i−m+1}T^{m−1}x ∈ R(T^{i−m+1}). Using [Mb1, Corollaire 2.5], we obtain

S^{i−m}T^{i−m+1}T^{m−1}x ∈ R(T).

Therefore there is v ∈ H such that

S^{i−m}T^{i}x = S^{i−m}T^{i−m+1}T^{m−1}x = T v.

Hence (T S − I)S^{i−m}T^{i}x = (T S − I)T v = 0. Thus all the terms of the sum vanish and so

T^{j}S^{i}T^{i} = S^{i−j}T^{i} (1 ≤ j ≤ i).

If j ≥ i, then

T^{j}S^{i}T^{i} = T^{j−i}(T^{i}S^{i}T^{i}) = T^{j−i}T^{i} = T^{j}.
□

3. Main positive results

We start with the following result.

3.1. Theorem. Let T ∈ B(H). The following assertions are equivalent :

(1) T is similar to a partial isometry ;

SIMILARITY TO PARTIAL ISOMETRIES 9

(2) There exist two operators S and R such that R is positive and invertible, T ST = T, and S = RT^{∗}R^{−1}.

(3) There is an equivalent, Hilbertian norm |·| on H and an operator S ∈ B(H) such that T ST = T and |T | ≤ 1, |S| ≤ 1.

Proof. We show that (1) implies (2). If T = LV L^{−1}, with V a partial isometry and L an invertible operator, then

L^{−1}T L = V = V V^{∗}V = L^{−1}T LL^{∗}T^{∗}L^{∗−1}L^{−1}T L.

Therefore T = T LL^{∗}T^{∗}L^{∗−1}L^{−1}T. If R = LL^{∗} and S = RT^{∗}R^{−1}, then T ST = T and R is positive and invertible.

We show that (2) implies (3). Suppose that S = RT^{∗}R^{−1} for some positive invertible operator R and T ST = T. Denote

⟨⟨x, y⟩⟩ = ⟨R^{−1/2}x, R^{−1/2}y⟩ ,

where R^{1/2} denotes the positive square root of R. This is an equivalent scalar product and the corresponding Hilbertian norm is

|x| = ⟨⟨x, x⟩⟩^{1/2} = ∥R^{−1/2}x∥.

Since T S is an idempotent ((T S)^{2} = T S), the operator R^{−1/2}T SR^{1/2} is also an idempotent. Using

R^{1/2}S^{∗}R^{−1/2} = R^{−1/2}T R^{1/2} and R^{1/2}T^{∗}R^{−1/2} = R^{−1/2}SR^{1/2},

we obtain

(R^{−1/2}T SR^{1/2})^{∗} = R^{1/2}S^{∗}T^{∗}R^{−1/2}
= R^{1/2}S^{∗}R^{−1/2}R^{1/2}T^{∗}R^{−1/2}
= R^{−1/2}T R^{1/2}R^{−1/2}SR^{1/2}
= R^{−1/2}T SR^{1/2}.

Therefore R^{−1/2}T SR^{1/2} is an orthogonal projection (a self-adjoint idempotent) and so

∥R^{−1/2}T SR^{1/2}∥ = 1.

In an analogous way ∥R^{−1/2}ST R^{1/2}∥ = 1.

We have

⟨⟨T x, T x⟩⟩ = ⟨R^{−1/2}T x, R^{−1/2}T x⟩ = ⟨T^{∗}R^{−1}T x, x⟩
= ⟨R^{−1}ST x, x⟩ = ⟨R^{−1/2}ST R^{1/2} (R^{−1/2}x), R^{−1/2}x⟩
≤ ∥R^{−1/2}ST R^{1/2}∥ · ∥R^{−1/2}x∥^{2} = ∥R^{−1/2}x∥^{2}.

Therefore |T x| ≤ |x| for every x. In a similar way, S is a contraction in the new norm.

The fact that (3) implies (1) follows from [Mb3, Theorem 3.1]. □

The following is a generalization of Sz.-Nagy similarity criterion for operators similar to partial isometries.

3.2. Theorem. Let T ∈ B(H) with 0 ∉ σ_{g}(T). Then T is similar to a partial isometry if and only if T is power bounded and there exists a power bounded operator S ∈ B(H) such that T ST = T.

Theorem 3.2 implies Sz.-Nagy’s similarity criteria for unitaries (=invertible partial isometries), isometries (=left invertible partial isometries) and co-isometries (=right invertible partial isometries). A proof of Theorem 3.2 will be given in Section 5.

4. Examples

The Example 4.1 below shows that the condition 0 ∉ σ_{g}(T) cannot be removed from Theorem 3.2.

4.1. Example. There exist operators T, S ∈ B(H) such that T ST = T, ST S = S, T is polynomially bounded but not similar to a contraction and S is similar to a contraction. In particular, T and S are power bounded (even polynomially bounded) and T is not similar to a partial isometry.

Proof. According to [Pi1], there exist a Hilbert space H_{1} and an operator of the form

P = [ \begin{matrix}s^{∗} & γ \\ 0 & s \end{matrix} ] ∈ B(H_{1} ⊕ H_{1}),

SIMILARITY TO PARTIAL ISOMETRIES 11

with s an isometry, which is polynomially bounded but not similar to a contraction. Recall that P is said to be polynomially bounded if there exists a positive constant K > 0 such that

∥f(P)∥ ≤ K∥f∥_{∞}

for every polynomial f with complex coefficients. Here

∥f∥_{∞} = sup{|f(z)| : |z| = 1}

denotes the supremum norm. Consider the Hilbert space H = H_{1} ⊕ H_{1} ⊕ H_{1} and let T and S be given by the following operator matrices :

T = [ \begin{matrix}s^{∗} & γ & −γ \\ 0 & s & 0 \\ 0 & 0 & 0 \end{matrix} ] ∈ B(H),

and

S = [ \begin{matrix}s & 0 & 0 \\ 0 & s^{∗} & 0 \\ 0 & s^{∗} & 0 \end{matrix} ] ∈ B(H).

Using the fact that s^{∗}s is the identity, it can be shown that T ST = T and ST S = S.

We have

f(S) = [ \begin{matrix}f(s) & 0 & 0 \\ 0 & f(s^{∗}) & 0 \\ 0 & f(s^{∗}) & 0 \end{matrix} ]

for each polynomial f with complex coefficients satisfying f(0) = 0. Using the von Neumann inequality for the contractions s^{∗} and s, and the inequality

∥f(S)∥ ≤ ∥f(S) − f(0)I∥ + ∥f(0)I∥,

we obtain that S is polynomially bounded. In fact, using the same argument, it can be shown that S is completely polynomially bounded and thus, using Paulsen’s criterion [Pa, Chapter 10], S is similar to a contraction. The following is a more direct proof.

Let

L = [ \begin{matrix}I & 0 & 0 \\ 0 & I & 0 \\ 0 & I & I \end{matrix} ] ∈ B(H).

Then

L^{−1}SL = [ \begin{matrix}I & 0 & 0 \\ 0 & I & 0 \\ 0 & −I & I \end{matrix} ] [ \begin{matrix}s & 0 & 0 \\ 0 & s^{∗} & 0 \\ 0 & s^{∗} & 0 \end{matrix} ] [ \begin{matrix}I & 0 & 0 \\ 0 & I & 0 \\ 0 & I & I \end{matrix} ] = diag (s, s^{∗}, 0)

is a contraction.

We also have

T^{n} = [ \begin{matrix}s^{∗n} & γ_{n} & −s^{∗n−1}γ \\ 0 & s^{n} & 0 \\ 0 & 0 & 0 \end{matrix} ] ,

where γ_{n} = ∑_{j=0}^{n−1} s^{∗j}γs^{n−j−1}. Note also that

P^{n} = [ \begin{matrix}s^{∗n} & γ_{n} \\ 0 & s^{n} \end{matrix} ] .

Let f(z) = ∑_{j=0}^{d} a_{j}z^{j} be a fixed polynomial with complex coefficients and set

g(z) = \frac{f(z) − f(0)}{z} = ∑_{j=1}^{d} a_{j}z^{j−1}.

Then, using the von Neumann inequality for the contraction s^{∗}, we obtain

∥ ∑_{j=1}^{d} a_{j}s^{∗j−1}γ∥ = ∥g(s^{∗})γ∥ ≤ ∥γ∥∥g∥_{∞} ≤ 2∥γ∥∥f∥_{∞}.

Using also that P is polynomially bounded, we obtain that T is polynomially bounded.

If T were similar to a contraction, then P, the restriction of T on H_{1}⊕ H_{1}, would also be similar to a contraction, a contradiction. One way to prove that the restriction of an operator similar to a contraction to a closed invariant subspace is also similar to a contraction uses Paulsen’s criterion mentioned above. □

4.2. Remark. If one is looking only for a power bounded operator T with a power bounded generalized inverse S and not being similar to a partial isometry, then the simpler example of Foguel [Fo] of a power bounded operator not similar to a contraction can be used instead of Pisier’s example.

We present now some related examples.

4.3. Example. There exist T, S ∈ B(H) such that T ST = T, ST S = S and both T and S are similar to contractions, but T and S are not simultaneously similar to contractions.

Proof. According to [Pi2], there exist a separable Hilbert space H_{1} and two commuting operators T_{1} and T_{2} of the form

T_{j} = [ \begin{matrix}s_{j}^{∗} & δ_{j} \\ 0 & s_{j} \end{matrix} ] ∈ B(H_{1} ⊕ H_{1}), j = 1, 2,

with (s_{1}, s_{2}) a pair of commuting isometries, such that T_{1} and T_{2} are both similar to contractions but their product T_{1}T_{2} is not polynomially bounded. In particular, T_{1} and T_{2} are not jointly similar to contractions, that is, the pair (T_{1}, T_{2}) is not similar to a pair of contractions.

In order to obtain our counterexample, consider the Hilbert space H which is the direct sum of six copies of H_{1} and let T and S be given by the following operator matrices :

T = [ \begin{matrix}s_{1}^{∗} & δ_{1} & −δ_{1} \\ 0 & s_{1} & 0 \\ 0 & 0 & 0 \\ & & & s_{2} & 0 & 0 \\ & & & 0 & s_{2}^{∗} & 0 \\ & & & 0 & s_{2}^{∗} & 0 \end{matrix} ] ∈ B(H),

and

S = [ \begin{matrix}s_{1} & 0 & 0 \\ 0 & s_{1}^{∗} & 0 \\ 0 & s_{1}^{∗} & 0 \\ & & & s_{2}^{∗} & δ_{2} & −δ_{2} \\ & & & 0 & s_{2} & 0 \\ & & & 0 & 0 & 0 \end{matrix} ] ∈ B(H),

where the unwritten entries are assumed to be zeros. As in the proof of Example 4.1, we have T ST = T and ST S = S. Also, from Pisier’s result [Pi2] and the proof of Example 4.1, it follows that T and S are both similar to contractions. Since T_{1} and T_{2} are in some corners of T

14 C. BADEA AND M. MBEKHTA

and S, we obtain, using [Pi2], that T and S are not jointly similar to contractions. □

The following example answers a question asked by the (anonymous) referee.

4.4. Example. There exists a polynomial bounded Hilbert space operator P ∈ B(H) which is not similar to a contraction and with σ_{g}(P) = 𝕋. In particular, P is regular.

Proof. We use the same example by Gilles Pisier. There exist ([Pi1]) a Hilbert space H and an operator of the form

P = [ \begin{matrix}s^{∗} & γ \\ 0 & s \end{matrix} ] ∈ B(H ⊕ H),

with s an isometry having spectrum \overline{𝔻}, which is polynomially bounded but not similar to a contraction. We show now that σ_{g}(P) = 𝕋.

Since P is polynomially bounded, its spectrum and also its generalized spectrum are in the closed unit disc. In order to prove that 𝔻 ⊂ reg(P), define R : 𝔻 ↦ B(H ⊕ H) by

R(λ) = [ \begin{matrix}s(I − λs)^{−1} & −s(I − λs)^{−1}γ(I − λs^{∗})^{−1}s^{∗} \\ 0 & (I − λs^{∗})^{−1}s^{∗} \end{matrix} ] .

Then R is analytic on 𝔻 and since s is an isometry, we have

(P − λ)R(λ)(P − λI) = P − λI.

This shows that σ_{g}(P) = 𝕋. □

5. The proof of Theorem 3.2

Proof. If T is similar to a partial isometry, then T is clearly power bounded. Also, Theorem 3.1 implies the existence of an operator S which is similar to T^{∗} and T ST = T. Therefore S is similar to a contraction, in particular it is power bounded.

We prove now the converse. Let T be a power bounded regular operator such that T ST = T with a power bounded S. Denote

M = max (sup_{n∈ℕ}∥T^{n}∥; sup_{n∈ℕ}∥S^{n}∥) < ∞.

Denote H_{0} = R^{∞}(T) and H_{1} = R^{∞}(T)^{⊥}. Then H = H_{0} ⊕ H_{1} and

T = [ \begin{matrix}T_{0} & A \\ 0 & T_{1} \end{matrix} ] ∈ B(H_{0} ⊕ H_{1})

with respect to this decomposition. Since 0 ∉ σ_{g}(T), the operator T_{0} is onto and T_{1} is one-to-one.

Using Lemma 2.2, the operator S can be written with respect to the same decomposition H = H_{0} ⊕ H_{1} as

S = [ \begin{matrix}S_{0} & B \\ 0 & S_{1} \end{matrix} ] ∈ B(H_{0} ⊕ H_{1}).

Now the equality T ST = T implies

T_{0}S_{0}T_{0} = T_{0}, T_{1}S_{1}T_{1} = T_{1} and − A = T_{0}BT_{1}.

We obtain (T_{0}S_{0} − I_{0})T_{0} = 0 and T_{1}(S_{1}T_{1} − I_{1}) = 0. Since T_{0} is onto and T_{1} is one-to-one, we have T_{0}S_{0} = I ∈ B(H_{0}) and S_{1}T_{1} = I ∈ B(H_{1}).

The operators T_{0} and T_{1} are power bounded (as compressions of the power bounded operator T), as well as S_{0} and S_{1}. Applying the Sz.- Nagy similarity criterion, T_{1} is similar to an isometry and the right invertible operator T_{0} is similar to a co-isometry. A proof of the last assertion can be given using the equality S_{0}^{∗}T_{0}^{∗} = I.

We have

[ \begin{matrix}S_{0} & B \\ 0 & S_{1} \end{matrix} ]^{n} = [ \begin{matrix}S_{0}^{n} & B_{n} \\ 0 & S_{1}^{n} \end{matrix} ] ,

where

B_{n} = ∑_{k=0}^{n−1} S_{0}^{n−k−1}BS_{1}^{k}.

This implies

sup_{n∈ℕ}∥B_{n}∥ ≤ M.

Therefore

∥ ∑_{k=0}^{n−1} T_{0}^{k}BS_{1}^{k}∥ = ∥ ∑_{k=0}^{n−1} T_{0}^{k}T_{0}^{n−k}S_{0}^{n−k}BS_{1}^{k}∥

= ∥T_{0}^{n}S_{0} ∑_{k=0}^{n−1} S_{0}^{n−k−1}BS_{1}^{k}∥
≤ ∥T_{0}^{n}∥ · ∥S_{0}∥ · ∥B_{n}∥ ≤ M^{3}.

Let LIM be a Banach limit, that is a positive linear functional on ℓ^{∞}(ℂ) such that LIM(1) = 1 and LIM(x_{n+1}) = LIM(x_{n}) for all bounded sequences (x_{n}) ∈ ℓ^{∞}(ℂ). Here 1 = (1, 1, ⋯). We define the operator Z by

⟨Zx, y⟩ = LIM_{n} ⟨∑_{k=0}^{n−1} T_{0}^{k}BS_{1}^{k}x, y⟩ .

The subscript n indicates that LIM is applied to the sequence indexed by n. The preceding estimates show that Z is a well-defined bounded (linear) operator.

Using the invariance under translations of the Banach limit, we get

T_{0}ZS_{1} = Z − B.

Indeed, we have

⟨T_{0}ZS_{1}x, y⟩ = ⟨ZS_{1}x, T_{0}^{∗}y⟩ = LIM_{n} ⟨∑_{k=0}^{n−1} T_{0}^{k}BS_{1}^{k+1}x, T_{0}^{∗}y⟩

= LIM_{n} ⟨∑_{k=0}^{n−1} T_{0}^{k+1}BS_{1}^{k+1}x, y⟩
= − ⟨Bx, y⟩ + LIM_{n} ⟨∑_{k=0}^{n} T_{0}^{k}BS_{1}^{k}x, y⟩
= − ⟨Bx, y⟩ + ⟨Zx, y⟩ .

We obtain

A = T_{0}(−B)T_{1} = T_{0}(T_{0}ZS_{1} − Z)T_{1} = T_{0}^{2}Z − T_{0}ZT_{1}
= T_{0}(T_{0}Z) − (T_{0}Z)T_{1}.

Now the identity

[ \begin{matrix}I & −T_{0}Z \\ 0 & I \end{matrix} ] [ \begin{matrix}T_{0} & 0 \\ 0 & T_{1} \end{matrix} ] [ \begin{matrix}I & T_{0}Z \\ 0 & I \end{matrix} ] = [ \begin{matrix}T_{0} & A \\ 0 & T_{1} \end{matrix} ]

yields the similarity of T to the direct sum T_{0} ⊕ T_{1}. As T_{1} is similar to an isometry and T_{0} is similar to a co-isometry, we obtain that T is similar to a (regular) partial isometry. □

5.1. Remark. If T is similar to a regular partial isometry, then σ_{g}(T) ⊂ 𝕋. Indeed, the generalized spectra of co-isometries and isometries are in the unit circle. Moreover, σ_{g}(T) = 𝕋 if T is not invertible.

6. Final remarks

Remarks concerning the preceding proof. We start with the following remarks.

6.1. Remark. With the notations of the preceding proof, we have S_{1}T_{1} = I on H_{1} and S_{1} and T_{1} are both power bounded. The Sz.-Nagy criterion implies that T_{1} is similar to an isometry V. Using [CMS, Theorem 4], we also have that S_{1} is similar to V^{∗}. The similarity is not necessarily implemented by the same operator implementing the similarity of T_{1} to V (see [CMS]).

6.2. Remark. The existence of a bounded solution X = T_{0}Z for the commutator equation A = T_{0}X − XT_{1} is, in many cases, equivalent to the similarity of T with T_{0} ⊕ T_{1}. We refer to [BR] for a survey about the commutator equation and to [Ba] and its references for other applications of the commutator equation to similarity problems.

Two questions.

6.3. Question ([Mb3]). Let T, S ∈ B(H) be two operators similar to contractions such that T ST = T and ST S = S. Is it true that T is similar to a partial isometry ?

The Example 4.1 shows that it is not possible to replace ’similar to contractions’ with ’polynomially bounded’ in this question.

18 C. BADEA AND M. MBEKHTA

We do not know if the operator T from the Example 4.3 is a counterexample to Question 6.3, i.e. if it is or not similar to a partial isometry. Even if the Example 4.3 shows that the generalized inverse S of T and T itself cannot be simultaneously contractive (in a new, equivalent, Hilbertian norm), we cannot apply Theorem 3.1. Indeed, it is still possible to find another generalized inverse for T which, together with T, is jointly similar to a contraction. Also, we cannot apply Theorem 3.2 to the operator T from the Example 4.3. Indeed, it can be proven that T from the Example 4.3 is not regular.

Another remark is that each bounded linear operator T ∈ B(H) which verifies T ST = T and 0 ∉ σ_{g}(T) verifies also T^{n}S^{n}T^{n} = T^{n} for each n ∈ ℕ (see Lemma 2.4). The operators T and S from the Example 4.1 do not verify the equation T^{2}S^{2}T^{2} = T^{2}. This motivates the following revised question.

6.4. Question. Let T, S ∈ B(H) be two power bounded operators such that T^{n}S^{n}T^{n} = T^{n} and (perhaps) S^{n}T^{n}S^{n} = S^{n} for each n ∈ ℕ. Is it true that T is similar to a (power) partial isometry ?

Acknowledgment. We are grateful to Dan Timotin for useful discussions and comments and to the referee for his careful reading of our manuscript and his suggestions.

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Département de Mathématiques, UMR CNRS no. 8524, Université Lille I, F–59655 Villeneuve d’Ascq, France

  E-mail address: catalin.badea@math.univ-lille1.fr

Département de Mathématiques, UMR CNRS no. 8524, Université Lille I, F–59655 Villeneuve d’Ascq, France

  E-mail address: mostafa.mbekhta@math.univ-lille1.fr