On Spectral Radius Algebras and
Normal Operators
A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
A BSTRACT. We associate to each normal operator N an algebra BN
that contains the commutant of N . For a subclass of normal operators
we demonstrate that BN has a nontrivial invariant subspace. Further,
we show that BN properly contains the commutant of N so that the
invariant subspace result is stronger then the existence of a hyperinvariant subspace.
1. I NTRODUCTION AND P RELIMINARIES
Let H be a complex, separable Hilbert space and let L(H ) denote the algebra of
all bounded linear operators on H . If T is an operator in L(H ), then a subspace
M ⊂ H is invariant for T if T M ⊂ M and it is hyperinvariant for T if it is
invariant for every operator in the commutant {T }0 of T . A nontrivial invariant
subspace (n.i.s.) is one that is neither H nor the zero subspace. One knows that if
N is a normal operator in L(H ) that is not a scalar multiple of identity, then {N}0
has a n.i.s. Thus, it is natural to ask whether there is a larger algebra associated to
N that also has a n.i.s.
A similar question was raised and answered in [9] for compact operators. The
algebras in question were introduced and studied under the name spectral algebras.
As we will present in Section 2, such an algebra depends, more precisely, on the
spectral radius r (A) of the operator A, so we will henceforth refer to it as a spectral
radius algebra or SRA, and denote it by BA . One of the properties of BA that is
essential for our study is that it contains {A}0 . Therefore, any subspace that is
invariant for BA is automatically hyperinvariant for A. It was shown in [9] that
such a situation arises when A is a compact operator, thus generalizing a famous
result of V. Lomonosov [11].
In this paper we concentrate on the case when N is a normal operator and we
give a complete description of BN . It follows directly from the definition of a SRA
that, if N is a scalar operator (i.e., a scalar multiple of identity), then BN = L(H ),
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Indiana University Mathematics Journal 1662
A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
an algebra with no n.i.s. In fact, as we will show in Theorem 2.7, BA = L(H )
precisely when A is similar to a constant multiple of an isometry. A normal operator of this type must be unitary so BN = L(H ) when N is a constant multiple
of a unitary operator. It is fairly easy to establish that BN contains infinitely many
compact operators and it is well known (see [14]) that an algebra that contains a
nonzero compact operator is either weakly dense or has a n.i.s. Therefore, it is of
interest to find out when BN is weakly dense.
In order to answer this question we consider the canonical decomposition of
the contraction N/kNk (cf., [12, p. 9]) given by N/kNk = U ⊕ T , where U is a
unitary operator and T is a completely nonunitary (c.n.u.) contraction. Of course,
either of the two direct summands can be absent. Our main result (Theorem 3.1)
is that, when both are present, BN has a n.i.s. and, if U is absent, then BN is a
proper, weakly dense subalgebra of L(H ). As mentioned before, if T is absent,
then BN = L(H ).
While it is not hard to show that, for any A ∈ L(H ), the algebra BA contains
{A}0 , it is much more challenging to establish that the inclusion is proper. (Of
course, if BA = {A}0 , then any n.i.s. for BA is merely hyperinvariant.) Once again,
if A is a scalar operator, then {A}0 = BA = L(H ). Our second result (Theorem
3.12) shows that this is the only obstruction and, if N is a normal operator that
is not scalar, then BN 6= {N}0 . In order to establish the existence of an operator
T ∈ BN \ {N}0 we will study an operator equation (in T ) of the form
(1.1)
NT = CT N,
C ∈ {N}0 ,
and determine condition(s) which guarantee the existence of nonzero solutions to
this equation. This is motivated by the fact that if an operator T satisfies (1.1)
with C a power bounded operator (i.e., there exists M > 0 such that kC n k ≤ M
for all n ∈ N), then T belongs to BN (Theorem 2.1 below). Moreover, if C and
N both have trivial kernels, then such an operator T cannot commute with N .
Special cases of (1.1), namely equations of the form NT = λT N where λ ∈ C
and N is not necessarily normal, were previously studied by various authors (cf.,
[3], [4], [10], [13]). This equation was also investigated in [1] when N is the
Volterra operator on L2 (0, 1) and in [2], in case N is an arbitrary C0 contraction.
In these two articles, a complex number λ for which the equation NT = λT N has
a nonzero solution T was referred to as an extended eigenvalue of N . Clearly, if
|λ| ≤ 1 is an extended eigenvalue of N , then λI is power bounded and T ∈ BN .
Moreover, if λ 6= 1, then provided N has a trivial kernel, such a (nonzero) T
belongs to BN \ {N}0 as well. Unfortunately, an extended eigenvalue λ ≠ 1 need
not always exist even when N is normal. This necessitates our study of equation
(1.1). We will show that for a normal operator N , there exist a contraction C with
the property that Ker(C − I) = (0) and a nonzero operator T satisfying (1.1).
Furthermore, such T can be chosen so that it does not commute with N . As
a consequence, when N/kNk is neither unitary nor completely nonunitary, the
invariant subspace supplied by Theorem 3.1 is “more than hyperinvariant.”
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1663
The third author would like to express gratitude to the University of North
Carolina Charlotte for its hospitality and partial support during the preparation
of the paper.
2. T HE S PECTRAL R ADIUS A LGEBRA
In this section we will briefly review some basic facts about spectral radius algebras.
Interested readers can find more details in [9].
If A ∈ L(H ) and m ≥ 1, we define
(2.1)
Rm (A) = Rm :=
∞
X
2n ∗ n n
dm
A A
1/2
where dm =
n=0
1
.
1/m + r (A)
Since dm ↑ 1/r (A) (we use the convention 10 = ∞), the sum in (2.1) is norm
convergent and the operators Rm are well defined, positive, and invertible. The
spectral radius algebra BA consists of all operators T ∈ L(H ) such that
−1
k < ∞.
sup kRm T Rm
m∈N
−1
The name comes from the fact that was proved in [8]: limm→∞ kRm
ARm k =
r (A). In general, BA can be very large. In [9] several sufficient conditions were
established for an operator T to belong to BA . The following theorem summarizes
these results (cf., [9, Proposition 2.3, Corollary 2.4]).
Theorem 2.1. If there exists a power bounded operator C in {A}0 such that
AT = CT A, then T ∈ BA . In particular, if AT = λT A with |λ| ≤ 1, then T ∈ BA .
For λ = 1, we have {A}0 ⊂ BA .
An important subset of BA is the ideal QA consisting of all operators T ∈
−1
L(H ) with the property that limm→∞ kRm T Rm
k = 0. The following theorem
(cf., [9, Theorem 3.4]) highlights the role played by QA in the search for a n.i.s.
Theorem 2.2. Let A ∈ L(H ). If QA 6= (0) and BA contains a nonzero compact
operator, then BA has n.i.s.
In particular, if A is compact one can show that QA 6= (0). Since BA clearly
contains a compact operator (namely A), it follows that BA has n.i.s.
In view of our interest in n.i.s. it is useful to know when the algebra BA is
weakly dense in L(H ). As a first step, we investigate when BA = L(H ). We start
with an easy application of the Uniform Boundedness Principle.
Proposition 2.3. T ∈ BA if and only if there exists M > 0 with the property
that
(2.2)
kRm (A)T xk ≤ MkRm (A)xk
for all m ∈ N, x ∈ H .
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A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
The following proposition illuminates the relationship between spectral radius
algebras of operators that are similar.
Theorem 2.4. Let A, B ∈ L(H ) and T be an invertible operator in L(H ) such
that T −1 BT = A. The map X , T −1 XT is an isomorphism from BB onto BA .
Proof. Clearly, the map defines an injective homomorphism from BB into
L(H ). In order to complete the proof it is enough to show that, for any X ∈
BA , the operator T XT −1 belongs to BB . Indeed, a symmetric argument can be
then used to demonstrate that Y ∈ BB implies T Y T −1 ∈ BA so the range of the
homomorphism is precisely BB . So, let X ∈ BA , f ∈ H , and m ∈ N. Notice
that r (A) = r (B) so the sequence dm defined in (2.1) is the same for both A and
B . Using (2.1) and the fact that B = T AT −1 it follows that
X
kRm (B)T XT −1 f k2 =
2n
dm
kT An XT −1 f k2
n≥0
≤ kT k2
X
2n
dm
kAn XT −1 f k2
n≥0
= kT k2 kRm (A)XT −1 f k2 .
Since X ∈ BA , Proposition 2.3 shows that there is M > 0 such that
(2.3)
kRm (B)T XT −1 f k ≤ MkT k kRm (A)T −1 f k.
Another application of (2.1) and A = T −1 BT yields
(2.4)
X
kRm (A)T −1 f k2 =
2n
dm
kT −1 B n f k2
n≥0
≤ kT −1 k2
X
2n
dm
kB n f k2
n≥0
= kT −1 k2 kRm (B)f k2 .
Combining (2.3) and (2.4) we get that
and the result now follows from Proposition 2.3.
❐
kRm (B)T XT −1 f k ≤ MkT k kT −1 k kRm (B)f k,
We will also need the following facts regarding the asymptotic behavior of the
sequence Rm .
Proposition 2.5 ([9, Proposition 3.11]). If A 6= 0, the operators Rm satisfy
limm→∞ kRm k = ∞.
Normal Operators
1665
Proposition 2.6 ([9, Proposition 3.8]). For A ∈ L(H ), we have BA = L(H )
if and only if
−1
k < ∞.
sup kRm k kRm
(2.5)
m∈N
In this paper we present a considerable improvement of Proposition 2.6.
Theorem 2.7. Let A ∈ L(H ). Then BA = L(H ) if and only if the operator A
is similar to a constant multiple of an isometry.
Proof. If A = αV where α ∈ C and V is an isometry, then for each m ∈ N,
the operator Rm (A) is scalar. Consequently, BA = L(H ). The proof of the direct
implication in the statement of the theorem now follows from Theorem 2.4.
In order to prove the converse we assume that BA = L(H ) and note that it
follows from the definition of Rm in (2.1) that
(2.6)
d2m A∗
2
2
Rm
Rm
1
A=
−
,
2
kRm k
kRm k2
kRm k2
m ∈ N.
2 k and the operators R 2 /kR k2 are positive
Also, Rm > 0 so kRm k2 = kRm
m
m
contractions for all m. Thus by passing to a subsequence {m0 } if necessary, we
2
2 → Q2 weakly as m0 → ∞, where Q is a nonmay assume that Rm
0 /kRm0 k
negative operator. Notice that, if A = 0, then the theorem is true. Therefore, we
assume that A 6= 0, in which case the definition of dm and Proposition 2.5 imply
that 1/dm → r (A) and kRm k → ∞ as m → ∞. Taking the weak limit as m0 → ∞
in (2.6), we obtain A∗ Q2 A = r 2 (A)Q2 . A result of Douglas [5, Theorem 1] now
implies that there exists an isometry Z : Ran Q → H satisfying r (A)ZQ = QA.
To complete the proof it suffices to show that Q is invertible since this implies
that A = Q−1 (r (A)Z)Q. Moreover, since Q is non-negative, it is enough to show
that Q is bounded below. By Proposition 2.6 there exists M > 0 such that for all
−1
y ∈ H and m ∈ N, kRm k kRm
yk ≤ Mkyk. In particular, for y = Rm x , we
obtain
!1/2
2
Rm
1
kRm xk
kxk ≤
kRm k
=
kRm k2
x, x
.
The result now follows by an application of the weak limit.
❐
M
3. S PECTRAL R ADIUS A LGEBRA OF A N ORMAL O PERATOR
In this section we will present the main results of this paper. First, in Theorem 3.1
we will give a complete description of BN for a normal operator N . In particular,
when N is neither unitary nor a completely nonunitary contraction, the algebra
BN has a n.i.s. Second, we will show in Theorem 3.12 that the inclusion {N}0 ⊂
BN , established by Theorem 2.1, is proper. In order to accomplish both of these
goals we will use the spectral representation of a normal operator. Namely, we will
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A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
identify a normal operator on H with a multiplication by an essentially bounded
function on L2 (X, B, µ) where X is a compact Hausdorff space, B is the σ -algebra
of Borel subsets of X and µ is a finite measure on B (cf., [7, p. 911]).
Before proceeding further, we introduce some basic notation and terminology. The symbol σ (A) denotes the spectrum of the operator A. For a measurable subset A ⊂ X of a measure space (X, B, µ), the set L2 (A) will denote
the subspace of L2 (X) that consists of functions vanishing off A. The orthogonal projection in L2 (X) with range L2 (A) will be denoted PL2 (A) . Also, for
any multiplication operator Ma on L2 (X) and any measurable subset A ⊂ X ,
the compression of Ma to the subspace L2 (A) will be denoted by MaA , that is,
MaA = MaA | L2 (A) = PL2 (A) Ma | L2 (A). As usual, when a property holds for
all x ∈ X with a possible exception of a set of µ measure 0 we will say that the
property holds almost everywhere and write a.e. [µ] or just a.e. if it is clear with
regard to what measure the statement is made. The symbol χA will stand for the
characteristic function of the set A. Recall that a point y belongs to the essential
range of a function f if, for every neighborhood U of y , f −1 (U) has positive
measure. When M is the operator of multiplication by a function a, then the
essential range of a coincides with the spectrum of Ma . A function f is essentially
bounded if its essential range is a bounded set. As usual, L∞ (X) will denote the
algebra of essentially bounded functions on X .
Now we give a description of BN when N is a normal operator. As mentioned
in the introduction, we distinguish between three basic types of normal operators
depending whether N/kNk is unitary, completely nonunitary, or a direct sum of
the first 2 types.
Theorem 3.1. Let N be a nonzero normal operator, and let A = N/kNk.
(a) If A is unitary, then BN = L(H ).
(b) If A is completely nonunitary, then BN is weakly dense in, but not equal to,
L(H ).
(c) If A is neither unitary nor completely nonunitary, then BN has a n.i.s.
Proof. Assertion (a) follows from Theorem 2.7.
We will prove assertions (b) and (c) for the case when A = N , i.e., kNk = 1.
The general case can be proved along the same lines. As mentioned before, we
represent A as the multiplication operator Ma acting on L2 (X, B, µ). Let E =
{x ∈ X : |a(x)| = 1} and G = {x ∈ X : |a(x)| < 1}. Also, let BE =
{F ∩ E : F ∈ B} and µE = µ | BE , with analogous definitions for BG and
µG . Then, as it is easy to see, the unitary part of A is the restriction of A to
L2 (E) = L2 (E, BE , µE ) and the completely nonunitary part of A is the restriction
of A to L2 (G) = L2 (G, BG , µG ). Since the remainder of the proof deals with the
situation when U is not unitary, we will have that µ(G) > 0.
First we show that BN 6= L(H ). Let f be an arbitrary function in L2 (X).
Since A is a normal operator of norm 1, its spectral radius is also 1 and, hence,
Normal Operators
1667
dm = m/(m + 1). A straightforward calculation shows that
#−1/2
ma 2
Rm f = 1 − f
1 + m
"
#1/2
ma 2
= 1−
f.
1 + m
"
and
−1
Rm
f
−1
From the definition of Rm , it follows that Rm
is a contraction for all m ∈ N. It is
not hard to see that
−1
kRm
f k2
(3.1)
for all f ∈
L2 (X).
In particular,
Z
→
X
(1 − |a|2 )|f |2 dµ,
−1
kRm
χG k
Z
→ [ (1 − |a|2 ) dµ]1/2 > 0.
G
Next, we notice that, since A∗ A is a positivep operator of norm 1, the operator
I − A∗ A fails to be invertible. Consequently, 1/(1 −p|a|2 ) does not belong to
L∞ (X) so there exists a function u ∈ L2 (X) such that 1/(1 − |a|2 ) u ∉ L2 (X).
This implies that kRm uk → ∞, as m → ∞. Indeed, suppose that {kRm uk}
has a bounded subsequence {kRmk uk}. The Lebesgue Monotone Convergence
Theorem implies that
Z
(3.2)
X
|u|2
dµ →
1 − |mk a/(1 + mk )|2
Z
X
|u|2
dµ
1 − |a|2
and the integrals
on the left side of (3.2) form a bounded sequence so it would folp
low that 1/(1 − |a|2 ) u ∈ L2 (X), contradicting our previous conclusion. Thus,
−1
kRm (u ⊗ χG )Rm
k → ∞ and u ⊗ χG ∉ BA so BA 6= L(H ).
Next, we prove assertion (b). Here, µ(E) = 0, and it follows that χG = 1 a.e. .
If δn is a positive increasing sequence that converges to 1 and Xδ = {x ∈ X :
|a(x)| ≤ δ}, then χXδ converges a.e. to χG = 1. Let u and v be any functions
n
in L2 (X). It is easy to see that (χXδn u) ⊗ v ∈ BA , n ∈ N, and that the sequence
(χXδ u ) ⊗ v converges in operator norm to u ⊗ v . Therefore, the norm closure
n
(and, hence, the weak closure) of BA contains all finite rank operators. Since the
set of finite rank operators is weakly dense in L(H ), the assertion (b) follows.
Finally, we prove (c). Again, χXδn converges a.e. to χG 6= 0 so there exists
δ ∈ (0, 1) such that µ(Xδ ) > 0. If u is any function in L2 (Xδ ), a calculation
shows that
#−1/2
mδ 2
kRm uk ≤ 1 −
kuk → [1 − δ2 ]−1/2 kuk.
1+m
"
❐
From this, we deduce that supm kRm uk < ∞, for any u ∈ L2 (Xδ ). Furthermore,
−1
vk → 0. For
if v is a nonzero function in L2 (E), formula (3.1) shows that kRm
−1
such nonzero functions u and v , kRm (u ⊗ v)Rm k → 0 and u ⊗ v ∈ QA . Since
u ⊗ v is clearly compact, the result follows from Theorem 2.2.
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A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
Theorem 3.1 shows that, unless a normal operator N is a scalar multiple of a
unitary or a completely nonunitary operator, the algebra BN has an invariant subspace. We will show that this algebra properly contains {N}0 so that the invariant
subspace in question is “more than hyperinvariant.” To that end, we consider the
operator equation (in T )
(3.3)
NT = CT N,
where N is normal and C belongs to {N}0 . The relevance of this equation in the
study of BN was pointed out in the introduction where it was noted that if T is
a solution to (3.3) and C is power bounded (in particular, a contraction), then
T ∈ BN . To study (3.3), we will need the following result which is essentially
[6, Proposition 2.4].
Theorem 3.2. Let M and N be bounded normal operators on Hilbert spaces, represented as multiplication by a and b on L2 (X, M, µ) and L2 (Y , N , ν) respectively.
The following are equivalent.
(a) There is a nonzero operator T with T M = NT .
(b) The measures ν ◦ b−1 and µ ◦ a−1 are not mutually singular.
Condition (b) of Theorem 3.2 is not always easy to verify. However, a special
case of (3.3) is relatively simple. Namely, we consider the equation MT = λT M
(λ ∈ C, λ ≠ 0, 1). The next result is an easy consequence of Theorem 3.2.
Corollary 3.3. For λ ≠ 0, 1 and with the notation of Theorem 3.2, if there is a
nonzero operator T satisfying MT = λT M , then
(3.4)
µ(a−1 (λ · σ (M)) > 0.
❐
Proof. If the measures µ ◦ a−1 and µ ◦ (a/λ)−1 are not singular, then there
exists a set S ⊂ σ (M) such that µ ◦a−1 (S) > 0 and µ ◦(a/λ)−1 (S) > 0. It follows
that µ(a−1 (λ · σ (M)) = µ ◦ (a/λ)−1 (σ (M)) > 0.
The following example shows that condition (3.4) is not sufficient to guarantee
that M has an extended eigenvalue.
Example 3.4. Let {rn } be an enumeration of the rational numbers in [0, 1]
and define a measure µ on [0, 1] as µ(rn ) = 1/2n and µ(x) = 0 √if x is irra/2. Since
tional. Let M = Ma be a multiplication by x on L2 (0, 1) and let λ = 2√
σ (M) = [0, 1] and a−1 (x) = x , we have that µ(a−1 (λ·σ (M)) = µ([0, 2/2]) >
0. Yet, µ ◦ a−1 and µ ◦ (λa)−1 are singular since µ ◦ a−1 ([0, 1] \ Q) = 0 and
µ ◦ (λa)−1 ([0, 1] ∩ Q) = 0.
Corollary 3.3 enables us to exhibit an example of a normal operator with only
trivial extended eigenvalues.
Example 3.5. Let µ be one dimensional Lebesgue measure on the set S =
{x + i : x ∈ [0, 1]}. Then the operator Mz on L2 (µ) is normal but it has no
extended eigenvalues.
Normal Operators
1669
❐
Proof. Since a(z) = z and σ (Mz ) = S , it follows that µ ◦ a−1 (λσ (Mz )) =
µ(λS) = 0 whenever λ 6= 1.
Remark 3.6. It is interesting to notice that in the preceding example, Mz is
a simple translate of a self adjoint operator Mx (multiplication by x on L2 (0, 1))
which has the entire set of positive numbers as its set of extended eigenvalues. The
latter fact is another easy consequence of Theorem 3.2.
Remark 3.7. Sometimes one can infer that an operator has no extended
eigenvalues on the basis of Rosenblum’s theorem [15]. Namely, if σ (A)∩σ (λA) =
∅, then 0 is the only operator that intertwines A and λA. This is not the case
here since, for example, with λ = (7 + 2i)/8, σ (Mz ) and σ (λMz ) intersect at
3
i + 16
= λ(i + 12 ).
We now return to the study of the operator equation (3.3). As before, we take
N = Ma on L2 (X, B, µ). Furthermore, we will take C to be of the form Mγ for
some bounded function γ on X . In case the function a is nonzero a.e., {N}0 is
a maximal Abelian von Neumann algebra. Thus, in this case, there is no loss of
generality in assuming C to be a multiplication operator as well. Therefore, we
will consider the question as to when there exists a nonzero operator T : L2 (X) →
L2 (X) satisfying
(3.5)
Ma T = Mγ T Ma .
In order to apply Theorem 3.2 it would be convenient to have Mγ invertible. The
following lemma shows how to reduce the general case to this particular one.
Lemma 3.8. Let X0 be a subset of X such that Mγ|X0 is invertible. If there exists
a nonzero operator T 0 : L2 (X) → L2 (X0 ) satisfying
(3.6)
Ma 0 T 0 = Mγ 0 T 0 Ma ,
X
X
then there exists a nonzero operator T satisfying (3.5).
❐
2
2
2
2
⊥
Proof. hDefine
i the operator T : L (X) → L (X) = L (X0 ) ⊕ (L (X0 )) by the
0
matrix T = T0 and use the fact that L2 (X0 ) is reducing for Mγ and Ma .
We now address the question whether, given a, there exists at least one contractive
function γ such that equation (3.5) has a nontrivial solution T . We would then
use this to establish the fact that for a normal operator N that is not a scalar
multiple of the identity, the algebra BN is strictly larger than {N}0 . We recall (cf.,
[16, p. 407]) that two measure spaces (X, A, µ) and (Y , B, ν) are isomorphic
if there is a one-to-one map Ψ of X onto Y such that for all A ∈ A we have
Ψ [A] ∈ B and ν(Ψ [A]) = µA, and for all B ∈ B we have Ψ −1 [B] ∈ A, and
hence µ(Ψ −1 [B]) = νB . Furthermore (cf., [16, p. 408]) a standard measure space
is one of the following:
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A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
(i) An interval [a, b] ⊂ R with m Lebesque measure or m zero measure;
(ii) A countable discrete set and m any finite measure;
(iii) The disjoint union of a measure space of type (i) and one of type (ii).
Now we state the theorem which will be instrumental in achieving our goal.
Theorem 3.9 ([16, p. 409]). Let µ be a finite Borel measure on a complete
separable metric space X . Then (X, B, µ) is isomorphic to a standard measure space.
Before we can prove that equation (3.5) has a nonzero solution, we need the
following lemma. For any function α let Dα denote the set {x : α(x) = x}.
Lemma 3.10. Let Z be a standard measure space with the property that if it is
of type (i), m is not the zero measure, and if it is of type (ii), then m is positive on at
least 3 atoms, and let Ψ : Z → C be a measure space isomorphism. Then there exist a
measurable set L ⊂ Z and a measurable function α : Z → Z such that
(a) α = α−1 ;
(b) m(L) > 0;
(c) |Ψ (α(x))| ≥ |Ψ (x)|, for x ∈ L;
(d) m(Dα ) = 0;
(e) m(α(V )) ≥ m(V ), for V ⊂ L;
(f ) m(L \ Ψ −1 (0)) > 0.
❐
Proof. Suppose that Z is a type (i) standard space, i.e., Z = [a, b] with the
Lebesgue measure, and let α(x) = a + b − x . Notice that α = α−1 so (a) holds.
Now, it is impossible to have |Ψ (α(x))| < |Ψ (x)| for a.e. x since it would
follow that |Ψ (x)| = |Ψ (α(α(x)))| < |Ψ (α(x))| < |Ψ (x)|. Therefore, there
is a set L of positive measure on which |Ψ (α(x))| ≥ |Ψ (x)|. Conditions (b)–
(d) are obvious; (e) is a consequence of the invariance of the Lebesgue measure
under rotation and translation; (f ) follows from (b) and the fact that Ψ −1 (0) is a
singleton.
Now we turn our attention to the discrete case, i.e., when Z is of type (ii).
We will identify Z with the set N, with the assumption that m{n} ≥ m{n + 1},
n ∈ N. We further assume that there are at least 3 atoms (with nonzero mass).
Since Ψ is one-to-one, at least 2 out of 3 of the complex numbers Ψ (1), Ψ (2), Ψ (3)
are not zero. Suppose that Ψ (1) and Ψ (2) are not zero. (The other possibilities can
be treated analogously.) Then we define α to be the permutation 1 ↔ 2 combined
with a permutation π of the set {3, 4, 5, . . . } such that π (n) 6= n and π 2 (n) = n
for n ≥ 3. Now, either |Ψ (1)| ≥ |Ψ (2)| > 0 or |Ψ (2)| ≥ |Ψ (1)| > 0. In the
former case we take L = {2}, in the latter L = {1}. In both cases conditions
(a)–(d) are obvious. Since L is a singleton, (e) is easy to verify. Finally, (f ) follows
from (b) and the fact that Ψ −1 (0) ∉ L.
Now we can prove the promised result.
Proposition 3.11. Let a be an essentially bounded complex function on a measure space (X, B, µ) that is not identically 0. Also, suppose that the support of µ
Normal Operators
1671
is not an atom or a pair of atoms. Then there exists a function γ : X → C with
Ker(Mγ − I) = {0} and |γ(x)| ≤ 1 a.e. [µ], such that equation (3.5) has a nonzero
solution T .
Proof. Let Ψ denote the map from a standard measure space Z with measure
m to σ (Ma ) with measure µ ◦ a−1 as set forth in Theorem 3.9. Let the map
α : Z → Z and the set L be as in Lemma 3.10. Also, let r denote the spectral
radius of Ma . We define a function ρ : σ (Ma ) → C as
(3.7)
ρ(y) =
(
Ψ αΨ −1 (y),
r + 1,
if y ∈ Ψ (L),
if y ∉ Ψ (L).
Let A = a−1 (σ (Ma )) and let γ : X → C be defined as

a(x)



,

ρ(a(x))
γ(x) =

1

 ,
2
if x ∈ A and ρ(a(x)) 6= 0,
otherwise.
First we show that |γ(x)| ≤ 1 a.e. [µ]. Indeed, if x ∉ A, then γ(x) = 12 < 1.
If x ∈ A, then a(x) ∈ σ (Ma ). Now a(x) is either in Ψ (L) or not. In the latter
case ρ(a(x)) = r + 1 ≥ |a(x)| a.e. . In the former case Ψ −1 (a(x)) ∈ L. Using
(3.7) and part (c) of Lemma 3.10, |ρ(a(x))| = |Ψ αΨ −1 a(x)| ≥ |Ψ Ψ −1 a(x)| =
|a(x)|. We conclude that |γ(x)| ≤ 1 a.e. . In particular, the operator Mγ is a
contraction. Also, this calculation shows that, if x ∈ A and ρ(a(x)) = 0 (so that
a(x) ∈ Ψ (L)), then by part (c) of Lemma 3.10, a(x) = 0. Consequently,
(3.8)
γ(x)ρ(a(x)) = a(x),
for all x ∈ A.
Next we show that γ(x) 6= 1 a.e. . Let F denote the set {x ∈ X : γ(x) = 1}.
If x ∈ F , then x ∈ A and a(x) = ρ(a(x)). From the definition of ρ and
the fact that |a(x)| ≤ r a.e., we see that, for a.e. x , a(x) ∈ Ψ (L) so a(x) =
Ψ αΨ −1 (a(x)) and Ψ −1 a(x) = αΨ −1 (a(x)). In other words, Ψ −1 a(x) ∈ Dα
a.e. [µ] and, using part (d) of Lemma 3.10, m[Ψ −1 a(F )] = 0. Since Ψ is measure
preserving, the last equality implies that µ(F ) = 0. Consequently, Ker(Mγ − I) =
{0}.
Thus it remains to show that equation (3.5) has a nonzero solution. Let
Aε = {x ∈ X : |γ(x)| ≥ ε}. We will demonstrate that µ ◦ a−1 and µ ◦ ((a/γ) |
Aε ∩ A)−1 are not mutually singular for some ε > 0 whence the result will follow
from Theorem 3.2 and Lemma 3.8. First we show that there must be ε > 0
such that µ(A ∩ Aε ) 6= 0. Otherwise, taking the union of all sets A ∩ Aε where
ε is a positive rational number, we would have that µ(A \ γ −1 (0)) = 0 and, all
the more, µ(A \ a−1 (0)) = 0. Since A = a−1 (σ (Ma )) it would follow that
µ ◦ a−1 (σ (Ma ) \ {0}) = 0 and, hence, that a = 0 a.e. [µ].
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A NIMIKH B ISWAS , A LAN L AMBERT & S RDJAN P ETROVIC
Let ε > 0 be such that µ(A ∩ Aε ) 6= 0. By (3.8),
(a/γ) | Aε ∩ A = (ρ ◦ a) | Aε ∩ A.
❐
Therefore, for any set S , µ[((a/γ) | Aε ∩A)−1 (S)] = µ[(ρ ◦a | Aε ∩A)−1 (S)] ≤
µ ◦ (ρ ◦ a)−1 (S). We claim that the restrictions of µ ◦ a−1 and µ ◦ (ρ ◦ a)−1 to
(Ψ ◦α)(L) satisfy µ◦a−1 (E) ≥ µ◦(ρ◦a)−1 (E) for all E ⊂ (Ψ ◦α)(L). To that end,
let E ⊂ (Ψ ◦ α)(L). Then there exists V ⊂ L such that E = (Ψ ◦ α)(V ). By Lemma
3.10 (e) we have that m(α(V )) ≥ m(V ). Since Ψ is measure preserving, we have
that µ ◦ a−1 [Ψ ◦ α(V )] ≥ µ ◦ a−1 [Ψ (V )]. Also, by (3.7), ρ −1 ◦ Ψ ◦ α(V ) = Ψ (V ).
So we get µ ◦a−1 [Ψ ◦α(V )] ≥ µ ◦a−1 [ρ −1 ◦Ψ ◦α(V )] = µ ◦(ρ ◦a)−1 [Ψ ◦α(V )].
Since E = Ψ ◦ α(V ), we get the desired estimate.
In order to finish the proof of the fact that µ ◦ a−1 and µ ◦ ((a/γ) | Aε ∩ A)−1
are not mutually singular, it now suffices to show there is ε > 0 such that the
restriction of µ ◦ ((a/γ) | Aε ∩ A)−1 to (Ψ ◦ α)(L) is not the zero measure.
As before, (a/γ) | Aε ∩ A = (ρ ◦ a) | Aε ∩ A and µ[(ρ ◦ a | Aε ∩ A)−1 ] =
µ[(ρ ◦ a | Aε )−1 ]. Also, µ[(ρ ◦ a | Aε )−1 ](Ψ ◦ α)(L) = µ[(ρ ◦ a)−1 (Ψ ◦
α)(L) ∩ Aε ]. We have seen that ρ −1 ◦ Ψ ◦ α(L) = Ψ (L) so the last quantity equals
µ[a−1 Ψ (L) ∩ Aε ]. If this quantity were 0 for each ε it would, once again, follow
that µ[a−1 Ψ (L) \ a−1 (0)] = 0 and by the measure invariance of Ψ we would have
that m[L \ Ψ −1 (0)] = 0 contradicting part (g) of Lemma 3.10.
Having prepared the ground, we can now prove that BN 6= {N}0 for a normal
operator N .
Theorem 3.12. Let N be a nonzero normal operator that is not a scalar multiple
of the identity, and let BN be the associated spectral radius algebra. Then there exists
an operator T ∈ BN \ {N}0 .
Proof. Once again we identify N with Ma acting on L2 (X, B, µ). Since N 6=
0, µ cannot be the zero measure. Also, µ cannot have only one atom, since that
would make N a scalar multiple of the identity. When µ is 2-atomic N can be
identified with a diagonal operator diag(αIH1 , βIH2 ) relative to the decomposition
H =H
1 ⊕H2 , with α and β distinct complex numbers, |α| ≥ |β| (so α 6= 0).
If T = 01 00 , then T N − NT = (α − β)T 6= 0 so T ∉ {N}0 . On the other hand,
NT = (β/α)T N so T ∈ BN by Theorem 2.1 with C = (β/α)I .
Thus we concentrate on the case when the assumptions of Proposition 3.11
are satisfied. Moreover, we may assume that Ker N = {0}. Indeed, one knows
that Ker N is a reducing subspace for N and, if Ker N 6= {0}, then N can be
written as N0 ⊕ 0 relative to the decomposition H = (Ker N)⊥ ⊕ Ker N and with
Ker N0 = {0}. It is not hard to see that if T0 ∈ BN0 \{N0 }0 then T0 ⊕0 ∈ BN \{N}0 .
So, suppose that Ker N = {0}. Applying Proposition 3.11 we obtain a bounded
operator Mγ and a nonzero operator T that satisfy NT = Mγ T N . Furthermore,
since |γ(x)| ≤ 1, the operator Mγ is a contraction so T ∈ BN . Thus it remains to
show that T does not commute with N . Suppose, to the contrary, that T N = NT .
Normal Operators
1673
❐
Then Mγ T N = T N and, since Ker(Mγ − I) = {0}, it follows that T N = NT = 0.
The obtained contradiction completes the proof.
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A NIMIKH B ISWAS & A LAN L AMBERT:
Department of Mathematics
UNC Charlotte
9201 University City Blvd
Charlotte, NC 28223, U. S. A.
E- MAIL: (A NIMIKH B ISWAS ) abiswas@uncc.edu
E- MAIL: (A LAN L AMBERT ) allamber@email.uncc.edu
S RDJAN P ETROVIC :
Department of Mathematics
Western Michigan University
Kalamazoo, MI 49008, U. S. A.
E- MAIL: srdjan.petrovic@wmich.edu.
K EY WORDS AND PHRASES: invariant subspaces; normal operators; spectral radius algebras.
2000 M ATHEMATICS S UBJECT C LASSIFICATION: 47A15 (47A65, 47B15)
Received : November 18th, 2005; revised: March 31st, 2006.
Article electronically published on August 9th, 2007.