Brown measure and iterates of the Aluthge transform for some operators arising from measurable actions

We consider the Aluthge transform $|T|^{1/2}U|T|^{1/2}$ of a Hilbert space operator $T$, where $T=U|T|$ is the polar decomposition of $T$. We prove that the map that sends $T$ to its Aluthge transform is continuous with respect to the norm topology and with respect to the $*$--SOT topology on bounded sets. We consider the special case in a tracial von Neumann algebra when $U$ implements an automorphism of the von Neumann algebra generated by the positive part $|T|$ of $T$, and we prove that the iterated Aluthge transform converges to a normal operator whose Brown measure agrees with that of $T$ (and we compute this Brown measure). This proof relies on a theorem that is an analogue of von Neumann's mean ergodic theorem, but for sums weighted by binomial coefficients.


Introduction
For a Hilbert space H, we let B(H) denote the set of bounded linear operators on H. Let T ∈ B(H) and let T = U|T | be its polar decomposition. The Aluthge transform of T is the operator T = |T | 1/2 U|T | 1/2 . This was first studied in [1] and has received much attention in recent years; (see, for example, the papers cited below). One reason the Aluthge transform is interesting is in relation to the invariant subspace problem. Jung, Ko and Pearcy prove in [7] that T has a nontrivial invariant subspace if and only if T does. They also note that when T is a quasiaffinity, then T has a nontrivial hyperinvariant subspace if and only if T does. (A quasiaffinity is a operator with zero kernel and dense range; the invariant and hyperinvariant subspace problems are interesting only for quasiaffinities.) Clearly, the spectrum of T equals that of T . Jung, Ko and Pearcy prove in [7] that other spectral data are also preserved by the Aluthge transform; see also [8] for more and related results.
The iterated Aluthge transforms (or Aluthge iterates) of T are the operators T (k) , k ≥ 0, defined by setting T (0) = T and letting T (k+1) be the Aluthge transform of T (k) . An interesting result, due to Yamazaki [14] (see also [13]), is that the spectral radius of T is equal to the limit as k → ∞ of the norm of T (k) . Recently, It was conjectured in [9] that T (k) converges in strong operator topology to a normal operator as k → ∞. Antezana, Pujals and Stojanoff [2] proved this conjecture for operators on finite dimensional Hilbert spaces. While this conjecture has been shown in [4] not to hold in general, we believe, based partly on the evidence of examples described below, that it will hold for elements of finite von Neumann algebras.
The examples we consider are when T = U|T |, where U is unitary and where the positive part |T | belongs to an abelian algebra that is normalized by U, (i.e. on which U acts by conjugation). In this situation, we describe the Aluthge iterates of T and show that the conjecture mentioned above holds. We also caculate the Brown measure for these operators. Our proof relies on an analogue of von Neumann's mean ergodic theorem for sums weighted with binomial coefficients, and on a theorem of Haagerup and Schultz.
The authors thank the referee for identifying some needed improvements in the paper.

Continuity properties in B(H)
The results of this section are applications of the continuous functional calculus for positive operators.
Lemma 2.1. For n ∈ N consider the function f n (t) = max( 1 n , t) defined for real, non-negative t and let A n = f n (|T |). Then Proof. The norm bounds (1)-(4) follow easily from the continuous functional calculus. For (5), we use Proof. Take ǫ < 1, for convenience. By Lemma 2.1, we may choose n so large that , with f n as in Lemma 2.1. Let p and q be real polynomials of one variable such that Then we have A n − p(T * T ) < ǫ, A −1 n − q(T * T ) < ǫ. Using the estimates in Lemma 2.1, we get . Now replace ǫ in the above argument by ǫ/(3R 3/2 ).
In the statements below, · refers to the operator norm topology and * -SOT to the * -strong operator topology on B(H). Proof. For (a), let T ∈ B(H) and take ǫ ∈ (0, 1]. Let R = T + 1 and let p and q be polynomials as found in Lemma 2.2 for these values of R and ǫ. Let δ ∈ (0, 1] be such that T − S < δ implies p(T * T )T q(T * T ) − p(S * S)Sq(S * S) < ǫ.
Then T − S < δ implies where the last inequality is by choice of p and q. This proves part (a). For (b), let R > 1, ǫ > 0 and let p and q be the polynomials found in Lemma 2.2 for these values. Let x ∈ H. If S, T ∈ B(H), each with norm ≤ R, then Since multiplication is * -SOT continuous on bounded subsets of B(H), by taking S in some neighborhood of T in this topology, both quantities (p( x can be forced to be arbitrarily small. This proves (b).

The Aluthge transform in finite von Neumann algebras
In this section, we consider a von Neumann algebra M equipped with a normal, faithful, tracial state τ , acting on the Hilbert space H := L 2 (M, τ ), which is the completion of M with respect to the norm x 2 = τ (x * x) 1/2 . For x ∈ M, we denote the corresponding element of L 2 (M, τ ) byx. Clearly, the Aluthge transform T of any T ∈ M also lies in M. In M, convergence in SOT in B(H) implies convergence in · 2 , because x 2 = x1 H . On the other hand, on bounded subsets of M, convergence in · 2 implies convergence in * -SOT. Indeed, let a n be a sequence in the unit ball of M such that a n 2 → 0 as n → ∞, let η ∈ L 2 (M, τ ), and let us show a n η H → 0 as n → ∞. Let ǫ > 0. Then there is x ∈ M such that η −x H < ǫ. Letting ρ denote the right action of M op on H, we have a n η H ≤ 2ǫ + a nx H = 2ǫ + ρ(x)a n1 H ≤ 2ǫ + x a n 2 , and from this we conclude a n η H → 0 as n → ∞. Since a * n 2 = a 2 2 , we conclude a n → 0 in * -SOT.
Therefore, we have the following immediate corollary of Theorem 2.3(b).
The proof of Theorem 2.2 of [9] carries over to the setting of finite von Neumann algebras to prove the following result.
Moreover, equality holds if and only if T is normal.
of the sequence of Aluthge interates of T converges in · 2 to a limit L, then L is a normal operator.
is a sequence of projections decreasing to zero as n → ∞, we get (7). The Brown measure [3] of an operator T ∈ M is defined as 1 2π times the Laplacian of the function C ∋ λ → ∆(λ − T ), where ∆ is the Fuglede-Kadison determinant [6], defined by ∆(X) = exp(τ (log |X|)). By Theorem 4.3 of [3], the Brown measures µ T and µ e T are the same. In light of this, one would like in Proposition 3.3 to have that L has the same Brown measure as T . This is at present unknown. However, we will conjecture even more than this, namely the actual convergence of Aluthge iterates.
n=1 of Aluthge iterates of T converges in · 2 to a normal operator N whose Brown measure is equal to the Brown measure of T .
This conjecture is the analogue for finite von Neumann algebras of Conjecture 5.6 of [9], which is about SOT-convergence of Aluthge iterates of T ∈ B(H). Although Cho, Jung and Lee [4] solve Conjecture 5.6 of [9] in the negative by giving an example of a weighted shift operator T whose Aluthge iterates fail to converge even in weak operator topology, that example clearly generates an infinite von Neumann algebra. As remarked earlier, this conjecture has recently been proved for matrices in [2].

An Ergodic theorem for sums weighted with binomial coefficients
In this section, we prove a result that resembles von Neumann's mean ergodic theorem, but for sums that are weighted with binomial coefficients. Our proof is based on the proof of von Neumann's theorem found in [12]. This theorem is covered in the case of a measure-preserving transformation by Theorem 1 of [11]. However, check of the relevant conditions in our case seems to be just as much work as proving the theorem directly. After proving the theorem, we also draw some consequences that will be used in some examples in Section 5.
It is clear that if v ∈ ker(T − I), then H n (v) = v for all n. We will show that if v ∈ ker(T − I) ⊥ , then H n (v) converges to zero as n → ∞. By linearity, this will imply that for general v ∈ H, H n (v) converges to P v, In order to show that H n (v) converges to zero for v ∈ ker(T − I) ⊥ , it will suffice to show it for v ∈ ran(T − I), because this latter space is dense in ker(T − I) ⊥ and each H n is a linear contraction. So we may assume v = T y − y, some y ∈ H. Then Since It will suffice to show lim n→∞ 1 2 n n k=1 |2k − n + 1| k n k = 0.
Let us now consider a probability space (X, µ) and a µ-preserving, invertible transformation α : X → X. Let E α : L 1 (µ) → L 1 (µ) α denote the conditional expectation onto the subspace L 1 (µ) α of α-invariant functions in L 1 (µ). In analogous notation, we will also write E α : L 2 (µ) → L 2 (µ) α and E α : L ∞ (µ) → L ∞ (µ) α for the restrictions of E α to the indicated subspaces. For a random variable b and n ≥ 0, we will consider the random variable Applying Theorem 4.1 in the case H = L 2 (µ) and T f = f • α, we have the following result. Let Since we're working over a probability space, from Corollary 4. We now consider a random variable b :  Proof. Fix K > 0 large and ǫ > 0 and δ > 0 small. Consider the α-invariant set . Thus, for n sufficiently large we have For integers m with m > |R| and with g m as defined in (13), This implies H = E α (b) a.e. By Egoroff's Theorem, exp(E α (g m • b)) converges in measure to exp(E α (b)), so E α (g m • b) converges in measure to E α (b). Therefore, for some m we have Again by Corollary 4.3, lim n→∞ h n (g m • b) − E α (g m • b) 1 = 0, so for n sufficiently large we have Finally, since h n (b) ≤ h n (g m • b), for n sufficiently large we have Combining (14) and (15) finishes the proof.  converges in probability to E α (b) as n → ∞.

Some examples in finite von Neumann algebras
Let T ∈ B(H) and let T = U|T | be its polar decomposition. The Aluthge transform of T is by definition T = |T | 1/2 U|T | 1/2 . It is easily seen that if V ∈ B(H) is any partial isometry whose restriction to U * UH equals U, then T = |T | 1/2 V |T | 1/2 . In particular, if T belongs to a finite von Neumann algebra, then we may take V to be a unitary extension of U.
We consider a probability space (X, µ) and we let α be an invertible, measurepreserving transformation of X. Let M = L ∞ (X) ⋊ α Z be the crossed product von Neumann algebra, which is, therefore, generated by a copy of L ∞ (X) = L ∞ (X, µ) and a unitary U such that for f ∈ L ∞ (X), we have Uf U * = f • α. For convenience, we will writeα is then strongly dense in M, and there is a normal tracial state τ on M uniquely determined by We will investigate the Aluthge iterates of operators of the form T = U|T | ∈ M with U as above and with |T | ∈ L ∞ (X).
Lemma 5.1. The n'th Aluthge iterate of T is where Proof. The proof proceeds by induction over n. Clearly, (17) and (18) hold for n = 0 (with the convention that T (0) = T ). Let N ≥ 1, and assume that (18) holds for n = N − 1. Then for the Nth Aluthge iterate we have: and this shows that (17) and (18) hold for n = N as well.
Given a random variable b : X → [−∞, R] for some real number R, we let E α (b) denote the conditional expectation of b onto the space of such α-invariant random variables, as described immediately before Corollary 4.4.
Proof. In light of (17), it will suffice to show | T (n) | · 2 −→ H as n → ∞. Since the Aluthge iterates ( T (n) ) ∞ n=1 form a norm-bounded sequence in L ∞ (X), it clearly suffices to show that | T (n) | converges in probability to H. This is equivalent to convergence in probability of log | T (n) | to E α (log(|T |)). with h n as defined in (12). Now the required convergence follows from Corollary 4.4.
If α is ergodic, then E α (b) = τ (b) and we have the following.
Corollary 5.3. If T = U|T | as above and if α is ergodic, then the Aluthge iterates We will now verify Conjecture 3.5 for our operators T by proving that the Brown measure of T is the same as that of UH, with H as in Theorem 5.2. This is an application of very general results from [10]. and Moreover, UH is a normal operator and the Brown measures of T and UH agree.
Proof. We will prove (19). Note that ) m∈N is uniformly bounded, convergence in probability implies converges in · 2 and, hence, also in SOT to H. The SOTconvergence (20) follows analogously.
Since H is α-invariant, H commutes with U and UH is normal. We will now show that the Brown measures µ T and µ U H agree. First note that both measures are invariant under rotations. Indeed, for θ ∈ R we have an isomorphism π θ : M → M given by π θ (U) = e iθ U, π θ (f ) = f, f ∈ L ∞ (X).