Ann. Funct. Anal. 4 (2013), no. 2, 171–182
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL:www.emis.de/journals/AFA/
POSITIVE TOEPLITZ OPERATORS ON THE BERGMAN
SPACE
NAMITA DAS1 AND MADHUSMITA SAHOO2∗
Communicated by S. Barza
Abstract. In this paper we find conditions on the existence of bounded linear
operators A on the Bergman space L2a (D) such that A∗ Tφ A ≥ Sψ and A∗ Tφ A ≥
Tφ where Tφ is a positive Toeplitz operator on L2a (D) and Sψ is a self-adjoint
little Hankel operator on L2a (D) with symbols φ, ψ ∈ L∞ (D) respectively. Also
we show that if Tφ is a non-negative Toeplitz operator then there exists a rank
e
f1 (z) for some constant α ≥ 0
≥ α2 R
one operator R1 on L2a (D) such that φ(z)
f1 (z) is the
and for all z ∈ D where φe is the Berezin transform of Tφ and R
Berezin transform of R1 .
1. Introduction
Let D be the open unit disc in the complex plane C and dA(z) = π1 dxdy be the
normalized area measure on D. Let L2 (D, dA) be the space of complex-valued,
absolutely integrable, measurable functions on D with respect to the area measure dA and L2a (D) be the Bergman space consisting of all analytic functions that
are in L2 (D, dA). Here the norm k · k2 and the inner product are taken in the
space L2 (D, dA). It is [4] not difficult to see that L2a (D) is a closed subspace of
L2 (D, dA). We denote the orthogonal projection from L2 (D, dA) into L2a (D) by P .
Let L∞ (D) be the space of complex-valued, essentially bounded, Lebesgue measurable functions on D and H ∞ (D)√be the space of bounded analytic functions on
D. For n ≥ 0, n ∈ Z, let en (z) = n + 1z n . The sequence {en }∞
n=0 forms an or∞
X
1
2
thonormal basis of La (D). Let K(z, w) = Kz (w) =
=
en (z)en (w).
(1 − zw)2
n=0
Date: Received: 12 December 2012; Accepted: 25 February 2013.
∗
Corresponding author.
2010 Mathematics Subject Classification. Primary 47B15 ; Secondary 47B35.
Key words and phrases. Bergman space, Positive operators, Berezin transform, Toeplitz
operators, Little Hankel operators.
171
172
N.DAS, M.SAHOO
The function K(z, w) defined on D × D is called the Bergman kernel of D or
K(w, z̄)
1 − |z|2
Kz (w)
the reproducing kernel of L2a (D). Let kz (w) =
=
.
=
K(z, z̄)
(1 − z̄w)2
kKz k2
These functions kz are called the normalized reproducing kernels of L2a (D) for
each z ∈ D. It is clear [10] that they are unit vectors in L2a (D).
For φ ∈ L∞ (D), we define the Toeplitz operator from L2a (D) into itself by
Tφ f = P (φf ) and the Hankel operator Hφ from L2a (D) into (L2a (D))⊥ is defined
by Hφ f = (I − P )(φf ). The little Hankel operator Sφ from L2a (D) into itself
is defined as Sφ f = P (J(φf )) where J : L2 (D, dA) −→ L2 (D, dA) is defined as
Jf (z) = f (z̄). These operators [10] are all bounded.
Let L(H) denote the algebra of all bounded linear operators from the Hilbert
space H into itself. Let LC(H) denote the ideal of compact operators in L(H).
A bounded linear operator A ∈ L(H) is said to be positive if hAx, xi ≥ 0 for all
x ∈ H. The notation A ≥ 0 will mean that A is positive. We say A ≥ B when
hAx, xi ≥ hBx, xi for all x ∈ H. For arbitrary selfadjoint operators A, B ∈ L(H)
we write A ≤ B if and only if B − A ≥ 0. An operator A ∈ L(H) is called
hyponormal if A∗ A ≥ AA∗ and the operator A ∈ L(H) is called power bounded
if kAn k ≤ K for a fixed K > 0 and n = 1, 2, . . . . Let T be a bounded linear
T + T∗
T − T∗
operator on a Hilbert space H. We denote
by Re(T ) and
by
2
2i
2
Im(T ). Define the Berezin transform for operators T ∈ L(La (D)) by the formula
Te(z) = hT kz , kz i, z ∈ D.
The function Te is called the Berezin transform of T . If T ∈ L(L2a (D)) then
Te ∈ L∞ (D) and kTek∞ ≤ kT k as |Te(z)| = |hT kz , kz i| ≤ kT k for all z ∈ D. We
e
fφ = φe for φ ∈ L∞ (D). That is, φ(z)
fφ (z) for all
shall write T
= hTφ kz , kz i = T
z ∈ D.
In the set of bounded Hermitian operators from a Hilbert space H into itself,
various types of ordering by means of the cones of non-negative, positive definite
and positive invertible operators can be defined. In this paper we investigate
whether it is possible to compare the Berezin transform of non-negative Toeplitz
and little Hankel operators. In section 2, we prove a few preliminary lemmas.
In section 3, we show that if Tφ is a positive Toeplitz operator on the Bergman
space and Sψ is a self-adjoint little Hankel operator then there exist bounded
linear operators A ∈ L(L2a (D)) such that A∗ Tφ A ≥ Sψ . Similarly, we show that
there exists A ∈ L(L2a (D)) such that A∗ Tφ A ≥ Tφ . Further, one can find a
w
sequence {An } ∈ L(L2a (D)) such that An −→ 0 and A∗n Tφ An ≥ Tφ for all n. In
section 4, we prove that if Tφ is a non-negative Toeplitz operator in L(L2a (D))
e
f1 (z)
then there exists a rank one operator R1 ∈ L(L2a (D)) such that φ(z)
≥ βR
for all z ∈ D and for some constant β ≥ 0.
2. Preliminary lemmas
In this section we prove a few preliminary lemmas which will be used in proving
the main results of the paper.
POSITIVE TOEPLITZ OPERATORS
173
For finite rank operators in L(L2a (D)) one can define a trace functional tr by
n
n
X
X
tr(T ) =
hfk , gk i when T =
fk ⊗ gk .
k=1
k=1
Lemma 2.1. Let S, T ∈ L(L2a (D)). If tr(ASA) = tr(AT A) for every rank one
projection A ∈ L(L2a (D)), then S = T .
Proof. Let A = f ⊗ f , where f is a unit vector. Then A is a rank one projection
and every rank one projection takes this form. By the assumption, we have
hSf, f i = tr(Sf ⊗ f )
= tr(ASA) = tr(AT A)
= tr(T f ⊗ f )
= hT f, f i .
Thus hSf, f i = hT f, f i holds for every unit vector f ∈ L2a (D). Therefore,
hSkz , kz i = hT kz , kz i for all z ∈ D. Hence S = T .
Lemma 2.2. If Tφ is invertible and hA∗ Tφ−1 Af, gihA∗ Tφ Af, gi = hA∗ Af, gi2 for
every f, g ∈ L2a (D) and for some A ∈ L(L2a (D)) whose range is dense in L2a (D)
then φ is a constant function.
Proof. Since RangeA = L2a (D) we have Tφ−1 f, g hTφ f, gi = hf, gi2 for all f, g ∈
L2a (D). Now fix a nonzero f ∈ L2a (D). Then,
for every
g ∈ (Sp{f })⊥ ⊂ L2a (D),
we have hTφ−1 f, gi = 0 or hTφ f, gi = 0 since Tφ−1 f, g hTφ f, gi = hf, gi2 = 0. Let
Mf = g ∈ (Sp{f })⊥ : hTφ f, gi = 0 and Nf = g ∈ (Sp{f })⊥ : hTφ−1 f, gi = 0 .
Then Mf ∪ Nf = (Sp{f })⊥ . Because (Sp{f })⊥ , Mf and Nf are all closed linear
subspaces, we must have Mf ⊆ Nf = (Sp{f })⊥ or Nf ⊆ Mf = (Sp{f })⊥ .
If Nf = (Sp{f })⊥ , then Tφ−1 f ∈ Sp{f }. So there exists a λf ∈ C such that
⊥
Tφ−1 f = λf f 6= 0, that is, Tφ f = λ−1
f f . If Mf = (Sp{f }) , then Tφ f ∈ Sp{f },
that is, Tφ f = λf f for some scalar λf . Since f is arbitrary, we see that for every
f ∈ L2a (D), there is a scalar λf such that Tφ f = λf f . This implies that there
exists a λ ∈ C such that φ ≡ λ.
Corollary 2.3. Suppose that Tφ is invertible and hSψ+ Tφ−1 Sψ f, gihSψ+ Tφ Sψ f, gi =
hSψ+ Sψ f, gi2 for every f, g ∈ L2a (D) and ker Sψ = {0}. Then φ ≡ C, a constant
function.
Proof. We need only to observe that RangeSψ = L2a (D) if and only if ker Sψ =
{0}.
Lemma 2.4. Let A be a nonnegative operator in L(L2a (D)). Then ker A =
ker A1/2 and RangeA = RangeA1/2 . If RangeA is closed then RangeA1/2 is closed
and RangeA = RangeA1/2 and A = A1/2 B, for some invertible B ∈ L(L2a (D)).
Proof. Since hAf, f i = hA1/2 f, A1/2 f i, f ∈ L2a (D), it follows that ker A ⊆ ker A1/2 .
Conversely, if f ∈ ker A1/2 , we obtain Af = A1/2 A1/2 f = 0. Thus ker A =
ker A1/2 . Also, observe that RangeA = (ker A)⊥ = (ker A1/2 )⊥ = RangeA1/2 .
The lemma follows from [7].
174
N.DAS, M.SAHOO
Lemma 2.5. Let ψ ∈ C(D), the space of continuous functions on D and kψk∞ ≤
1. Let Tφ be a positive Toeplitz operator on L2a (D) such that Tφ ≤ Sψ+ Tφ Sψ where
ψ + (z) = ψ(z̄). Then Tφ = Sψ+ Tφ Sψ . Further RangeTφ reduces Sψ and Sψ |(RangeTφ )
is unitary.
1/2
Proof. Let Tφ Sψ = L. The operator L is compact [10] as ψ ∈ C(D) and Sψ is a
1/2
1/2
contraction as kψk∞ ≤ 1. Further, LL∗ = Tφ Sψ Sψ+ Tφ
≤ Tφ . This is so since
1/2
1/2
Sψ+ Tφ Sψ − Tφ Sψ Sψ+ Tφ = L∗ L − LL∗ .
Sψ∗ = Sψ+ . Hence 0 ≤ Sψ+ Tφ Sψ − Tφ ≤
Hence the operator L is hyponormal. Since L is compact, L is normal. The
1/2
1/2
normality of L implies that Tφ = Sψ+ Tφ Sψ = Tφ Sψ Sψ+ Tφ , and hence it follows
that Sψ+ is an isometry on RangeTφ and Tφ commutes with Sψ (and so also with
Sψ+ ). Consequently, Sψ+ Sψ Tφ = Sψ+ Tφ Sψ = Tφ = Tφ Sψ Sψ+ . Hence RangeTφ
reduces Sψ and Sψ |(RangeTφ ) is unitary.
3. Non-negative Toeplitz operators
In this section we show that if Tφ is a positive Toeplitz operator in L(L2a (D)) and
ψ ∈ L∞ (D) can be expressed as a linear combination of Bergman kernels and some
of its derivative then there exist bounded linear operators A ∈ L(L2a (D)) such that
A∗ Tφ A ≥ Sψ∗ Sψ . If in addition ψ(z) = ψ(z) then we can find A ∈ L(L2a (D)) such
that A∗ Tφ A ≥ Sψ . Further, we find conditions for the existence of A ∈ L(L2a (D))
such that A∗ Tφ A ≥ Tφ . It is also possible to find sequences {An } of operators in
w
L(L2a (D)) such that An −→ 0 and A∗n Tφ An ≥ Tφ for all n.
Theorem 3.1. Let Tφ be a positive Toeplitz operator in L(L2a (D)) with symbol φ ∈
L∞ (D) and Sψ be a little Hankel operator in L(L2a (D)) where
ψ(z) =
j −1
N m
X
X
j=1 γ=0
∂γ
cjγ
γ Kbj (z)
∂bj
where b = {bj }N
j=1 is a finite set of points in D, cjγ 6= 0 for all j, γ and mj is the
number of times bj appears in b. Then there exists an operator A ∈ L(L2a (D))
such that A∗ Tφ A ≥ Sψ∗ Sψ and kA∗ Tφ Ak ≥ kSψ∗ Sψ k. Further, in addition if ψ(z) =
ψ(z) then it is also possible to find A ∈ L(L2a (D)) such that A∗ Tφ A ≥ Sψ and
∗ T A)(z) ≥ S
fψ (z) where H
e denotes the Berezin transform of H ∈ L(L2 (D)),
(A^
φ
a
and kA∗ Tφ Ak ≥ kSψ k. In case A is positive, then there exists an invertible
T ∈ L(L2a (D)) such that A1/2 Tφ A1/2 ≥ (T ∗ )−1 Sψ T −1 .
Proof. From [5] it follows that Sψ is a finite rank operator on L2a (D) and therefore
Sψ∗ Sψ is a finite rank operator and Range Sψ∗ Sψ is closed in L2a (D). Notice also
that
!
[
E Tφ [λ, ∞)(L2a (D)) = ∞.
(3.1)
dim
Tφ
λ>0
2
(0, ∞)(La (D)) =
This is so as E
RangeT
φ and from [9] it follows that
RangeTφ
∗
∗
2
is infinite dimensional. Let M = Y ∈ L(La (D)) | Y Tφ Y ≥ Sψ Sψ . We first
POSITIVE TOEPLITZ OPERATORS
175
claim that 0 is in the WOT-closure of M . To show this suppose 0 is not in the
WOT-closure of M . Then there is a WOT-neighborhood
V = B ∈ L(L2a (D)) : |hBfi , gi i| ≤ , i = 1, · · · , n
of 0 (for some > 0) which does not intersect M where f1 , . . . , fn , g1 , . . . , gn ∈
L2a (D). Let K be the linear span of g1 , g2 , . . . , gn . From (3.1), it follows that there
exists λ > 0 such that dim E Tφ [λ, ∞)(L2a (D)) > n + rank(Sψ∗ Sψ ). It thus follows
T
that dim E Tφ [λ, ∞)(L2a (D)) K ⊥ ≥ rank(Sψ∗ Sψ ). Since Sψ∗ Sψ is a self adjoint
operator of finite rank, there exist real numbers {θi }ki=1 and an orthonormal basis
k
X
{δi }ki=1 for RangeSψ∗ Sψ such that Sψ∗ Sψ f =
θi hf, δi i δi and |θi | > 0 for all
i=1
i = 1, . . . , k. Let B ∈ L(L2a (D)) be such that B|(RangeS ∗ Sψ )⊥ = 0 and Bδi = ui
ψ
T
k
Tφ
2
where {ui }i=1 is an orthonormal set in E [λ, ∞)(La (D)) K ⊥ .
Now, for each g ∈ RangeSψ∗ Sψ , we have kBgk = kgk and Bg ∈ E Tφ [λ, ∞)(L2a (D)).
Thus hB ∗ Tφ Bg, gi = hTφ Bg, Bgi ≥ λkBgk2 = λkgk2 . Let f ∈ L2a (D). Then
f = g + h, where g ∈ RangeSψ∗ Sψ and h ∈ (RangeSψ∗ Sψ )⊥ . Hence
hSψ∗ Sψ f, f i =
k
X
i=1
θi |hf, δi i|2 ≤ max |θi |kgk2
i
and
hA∗ Tφ Af, f i = hTφ Af, Af i = hTφ Ag, Agi ≥ λkgk2 ≥
1 ∗
hS Sψ f, f i,
t2 ψ
λ
1
=
. Thus t2 B ∗ Tφ B ≥ Sψ∗ Sψ and tB ∈ M . Further since
2
t
max |θi |
i
T
B(L2a (D)) ⊂ K ⊥ , we have tB ∈ V . Hence V M 6= φ. This is a contradiction.
Thus there exists operator A ∈ L(L2a (D)) such that A∗ Tφ A ≥ Sψ∗ Sψ and therefore
kA∗ Tφ Ak ≥ kSψ∗ Sψ k. In case ψ(z) = ψ(z), the operator Sψ is self-adjoint. Proceeding similarly as above, one can show that there exists A ∈ L(L2a (D)) such that
∗ T A)(z) ≥ S
fψ (z) and kA∗ Tφ Ak ≥ kSψ k. If A is
A∗ Tφ A ≥ Sψ and therefore (A^
φ
positive then by Lemma 2.4 there exists an invertible operator T ∈ L(L2a (D)) such
that A = A1/2 T . Hence A∗ Tφ A ≥ Sψ implies A1/2 Tφ A1/2 ≥ (T ∗ )−1 Sψ T −1 .
where
If f (z) =
∞
X
an z n is holomorphic on D, a simple calculation shows that
n=0
∞
X
|an |2
|f (z)| dA(z) =
.
n
+
1
D
n=0
Z
2
176
N.DAS, M.SAHOO
Consequently, f ∈ L2a (D) if and only if the last expression is finite. The scalar
∞
∞
X
X
n
product of f (z) =
an z and g(z) =
bn z n , f, g ∈ L2a (D), is given by
n=0
n=0
∞
X
hf, giL2a (D) =
n=0
The truncation projections on
is defined by
L2a (D)
an b n
.
n+1
will be denoted by Pn , 0 ≤ n < ∞, and it
Pn f = Pn (a0 , a1 , a2 , · · · , an , an+1 , · · · ) = (a0 , a1 , · · · , an , 0, 0, · · · ).
These are, of course, orthogonal projections on L2a (D) which converges strongly
to the identity I on L2a (D).
Theorem 3.2. Let Tφ be a non-negative nonzero Toeplitz operator on L2a (D) with
symbol φ ∈ L∞ (D). Then
(i): For each > 0, there exists an operator A ∈ L(L2a (D)) such that
kPn APn k ≤ and A∗ Tφ A ≥ Tφ . If tr(BA∗ Tφ AB) = tr(BTφ B) for every rank one projection operator B ∈ L(L2a (D)), then A∗ Tφ A = Tφ .
(ii): If Tφ ≤ Re(A∗ Tφ ) for some A ∈ L(L2a (D)) then Tφ ≤ A∗ Tφ A. That
∗ T A(z) for all z ∈ D. Furthermore if T ≤ Re(A∗ T ) and
e
is, φ(z)
≤ A^
φ
φ
φ
Tφ = A∗ Tφ A for some A ∈ L(L2a (D)) then A∗ Tφ = Tφ .
(iii): If K = A∗ Tφ for some A ∈ L(L2a (D)) and Tφ ≤ Re(K) and A∗ is
power bounded then K = Tφ .
1/2
(iv): If for some A ∈ L(L2a (D)), kAk ≤ 1, A∗ Tφ A ≥ Tφ then Tφ A is a
hyponormal operator.
(v): Let Tφ be invertible and E be a nonzero projection and λ ∈ R, λ > 0
such that ETφ E = λE and ETφ−1 E = λ1 E. Then RangeE is a subspace of
the eigenspace of Tφ corresponding to the eigenvalue λ.
(vi): If Tφ is invertible and hA∗ Tφ−1 Af, gihA∗ Tφ Af, gi = hA∗ Af, gi2 for every f, g ∈ L2a (D) and for some A ∈ L(L2a (D)) such that RangeA = L2a (D)
then φ is a constant function.
1/2
(vii): If ψ ∈ C(D), kψk∞ ≤ 1, Sψ∗ Tφ Sψ ≥ Tφ then Tφ = Sψ∗ Tφ Sψ and Tφ Sψ
is a hyponormal operator.
Proof. We shall assume first that Tφ is one-one. For λ > 0, let Eλ be the spectral
measure of the interval [λ, ∞). Since Tφ is one-one and non-negative, hence
Eλ −→ I, the identity operator, in the strong operator topology. Thus there
exists λ = λ() > 0 such that the orthogonal projection Eλ ∈ L(L2a (D)) satisfies
√
Tφ Eλ = Eλ Tφ , k(I − Eλ )Pn k ≤ and dim(RangeEλ ) ≥ 2 dim(RangePn ). Also the spectral measure Eλ satisfies
hTφ f, f i ≥ λkf k2
(3.2)
for all f ∈ RangeEλ , From [9], it follows that RangeTφ is infinite dimensional.
Thus there exists an unitary operator U on RangeEλ such that
(RangeU Eλ Pn ) ⊥ (RangeEλ Pn ).
POSITIVE TOEPLITZ OPERATORS
177
Define A ∈ L(L2a (D)) as Af = αU Eλ f + (I − Eλ )f , where α > 0 is chosen in
such a way that A∗ Tφ A ≥ Tφ . We shall now verify that such α exists. Since Tφ
commutes with Eλ we have
hA∗ Tφ Af, f i = hTφ Af, Af i
= hαTφ U Eλ f + Tφ (I − Eλ )f, αU Eλ f + (I − Eλ )f i
= α2 hTφ U Eλ f, U Eλ f i + hTφ (I − Eλ )f, (I − Eλ )f i .
On the other hand,
hTφ f, f i = hTφ Eλ f, f i + hTφ (I − Eλ )f, f i
= hTφ Eλ f, Eλ f i + hTφ Eλ f, (I − Eλ )f i
+ hTφ (I − Eλ )f, Eλ f i + hTφ (I − Eλ )f, (I − Eλ )f i
= hTφ Eλ f, Eλ f i + hTφ (I − Eλ )f, (I − Eλ )f i.
Hence the only condition which has to be satisfied by α is
α2 hTφ U Eλ f, U Eλ f i ≥ hTφ Eλ f, Eλ f i .
The condition is satisfied by sufficiently large α because of (3.2) and because
RangeTφ is an infinite dimensional subspace of L2a (D). To show that kPn APn k ≤
, observe that kPn APn k = sup {|hPn APn f, gi| : f, g ∈ L2a (D), kf k = kgk = 1}.
Let kf k = kgk = 1. We have
|hPn APn f, gi| = |hAPn f, Pn gi|
= |hEλ APn f, Eλ Pn gi + h(I − Eλ )APn f, (I − Eλ )Pn gi|
= |hαU Eλ Pn f, Eλ Pn gi + h(I − Eλ )Pn f, (I − Eλ )Pn gi|
= |0 + h(I − Eλ )Pn f, (I − Eλ )Pn gi|
≤ k(I − Eλ )Pn f kk(I − Eλ )Pn gk
≤ k(I − Eλ )Pn kk(I − Eλ )Pn k ≤ .
To prove the general case, let M = ker Tφ . Decompose L2a (D) into an orthogonal
direct sum L2a (D) = (ker Tφ )⊥ ⊕ ker Tφ = M ⊥ ⊕ M and let Q be the orthogonal
⊥
projection onto M ⊥ . Let TφM = Tφ |M ⊥ be the restriction of T to M ⊥ . Let N =
QPn L2a (D) and let Q1 be the orthogonal projection from M ⊥ onto N . Applying
⊥
the first of the proof to the operator TφM and the projection Q1 we find an
⊥
⊥
operator A1 ∈ L(M ⊥ ) with kQ1 A1 Q1 k ≤
and A∗1 TφM A1 ≥ TφM . Let
2
kPn k
A = A1 ⊕ 0, so A1 = QAQ. Then A∗ Tφ A ≥ Tφ . It remains to show that
178
N.DAS, M.SAHOO
kPn APn k ≤ . Since Q and Q1 are self-adjoint we have
kPn APn k =
|hPn APn f, gi|
sup
kf k=kgk=1
=
|hAPn f, Pn gi|
sup
kf k=kgk=1
=
|hQAQPn f, Pn gi|
sup
kf k=kgk=1
=
|hAQPn f, QPn gi|
sup
kf k=kgk=1
≤
|hA1 f, gi|
sup
kf k≤kPn k
kgk≤kPn k
≤ kPn k2
|hA1 Q1 f, Q1 gi|
sup
kf k=kgk=1
f,g∈M ⊥
≤ kPn k2 kQ1 AQ1 k ≤ .
If further tr(BA∗ Tφ AB) = tr(BTφ B) for every rank one projection B ∈ L(L2a (D))
then from Lemma 2.1 it follows that A∗ Tφ A = Tφ . This proves (i). We shall now
prove (ii). By applying Schwarz inequality [1] to the positive semi-definite form
hf, gi −→ hTφ f, gi, f, g ∈ L2a (D) we obtain
hTφ f, f i ≤ hRe(A∗ Tφ )f, f i
= RehA∗ Tφ f, f i
≤ |hA∗ Tφ f, f i|
1
1
≤ hTφ f, f i 2 hTφ Af, Af i 2
for all f ∈ L2a (D). Hence hTφ f, f i ≤ hA∗ Tφ Af, f i for all f ∈ L2a (D). That is,
Tφ ≤ A∗ Tφ A. In addition to Tφ ≤ Re(A∗ Tφ ), if Tφ = A∗ Tφ A is assumed, then we
obtain hTφ f, f i = RehA∗ Tφ f, f i = |hA∗ Tφ f, f i| = hA∗ Tφ f, f i for all f ∈ L2a (D)
and hence Tφ = A∗ Tφ . Now we shall prove (iii). Since A∗ Tφ A − Tφ ≥ 0, it follows
2
that A∗ (A∗ Tφ A − Tφ )A ≥ 0. That is, A∗ Tφ A2 ≥ A∗ Tφ A. Repeating the process
n+1
n
n
n times, we have A∗ Tφ An+1 ≥ A∗ Tφ An . Thus, {A∗ Tφ An | n = 1, 2, . . .} is an
increasing sequence of positive operators. This sequence is bounded, since A∗ is
power bounded. Therefore, it converges to a positive operator on L2a (D), say B,
in the strong operator topology. Notice that
n
A∗ BA = A∗ ( lim A∗ Tφ An )A
n→∞
n+1
= lim A∗
n→∞
= B.
Tφ An+1
POSITIVE TOEPLITZ OPERATORS
179
(A∗ Tφ + Tφ A)
, we have
2
n
[A∗ (A∗ Tφ + Tφ A)An ]
∗n
n
A Tφ A ≤
2
n
∗
∗n
[A (A Tφ An ) + (A∗ Tφ An )A]
.
=
2
From the operator inequality Tφ ≤
∗
= Re(A∗ B). Thus B = A∗ B.
By letting n tend to ∞, we have B ≤ (A B+BA)
2
Since Tφ ≤ B, it follows that the range of Tφ is contained in the range of B,
and hence [6], we have Tφ = A∗ Tφ = K. To prove (iv) suppose kAk ≤ 1 and
A∗ Tφ A ≥ Tφ . Now
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
(Tφ A)∗ (Tφ A) − (Tφ A)(Tφ A)∗ = A∗ Tφ A − Tφ AA∗ Tφ
= A∗ Tφ A − Tφ AA∗ Tφ
1/2
1/2
≥ Tφ − Tφ AA∗ Tφ
1/2
1/2
= Tφ (I − AA∗ )Tφ
≥0
1/2
and therefore Tφ A is a hyponormal operator. To prove (v), we can assume
without loss of generality that λ = 1. Let h be any unit vector from the range
of E. Multiplying the equations ETφ E = E and ETφ−1 E = E by Fh = h ⊗ h
from the left and also from the right we obtain (h ⊗ h)T
φ (h ⊗ h)
= h ⊗ h and
(h⊗h)Tφ−1 (h⊗h) = h⊗h. These imply hTφ h, hi = 1 and Tφ−1 h, h = 1. Consider
the Cauchy-Schwarz inequality for the new inner product
(f, g) = Tφ−1 f, g , f, g ∈ L2a (D).
Insert f = Tφ h and g = h. As h is a unit vector, we see that there is equality in
the corresponding inequality
−1
T Tφ h, h 2 ≤ T −1 Tφ h, Tφ h T −1 h, h .
φ
φ
φ
This gives us that Tφ h is a nonzero scalar multiple of h. It is clear that this scalar
is necessarily 1. So we have Tφ h = h for any unit vector h from the range of E.
This proves our claim. The proof of (vi) follows from Lemma 2.2. To prove (vii),
observe that Sψ∗ = Sψ+ where ψ + (z) = ψ(z). From Lemma 2.5, it follows that
1/2
Sψ+ Tφ Sψ = Tφ and from (iv) we obtain Tφ Sψ is a hyponormal operator.
Theorem 3.3. Let Tφ be a positive Toeplitz operator on the Bergman space L2a (D)
with symbol φ ∈ L∞ (D). Then there exists a sequence {An } of operators in
L(L2a (D)) such that An −→ 0 in weak operator topology and A∗n Tφ An ≥ Tφ for all
∗ T A (z) ≥ φ(z)
e
n. Thus A^
for all z ∈ D.
n φ n
Proof. We take an index set I for the set of all pairs n = (Pn , ) where Pn is the
finite dimensional projection on L2a (D), > 0. Set (Pm , 1 ) ≺ (Pr , 2 ) if m ≤ r
and 1 > 2 . By Theorem 3.2, for each n there exists An ∈ L(L2a (D)) such that
kPn APn k ≤ and A∗n Tφ An ≥ Tφ . Let Un = {A ∈ L(L2a (D)) : kPn APn k < }. It
180
N.DAS, M.SAHOO
is not difficult to see that each n ∈ I defines a WOT-neighbourhood Un of 0.
It is also clear that in this way we obtain a basis of the weak operator topology
neighbourhoods of 0. Furthermore notice that for each n , we have Am ∈ Un for
all m > n . Hence An −→ 0 in the weak operator topology.
4. Berezin transform of positive Toeplitz operators
In this section we show that if Tφ is a non-negative Toeplitz operator in
e ≥
L(L2a (D)) then there exists a rank one operator R1 ∈ L(L2a (D)) such that φ(z)
f1 (z) for all z ∈ D and for some constant α ≥ 0. Here φe is the Berezin
α2 R
f1 is the Berezin transform of R1 .
transform of Tφ and R
Let H and K be Hilbert spaces and let T ∈ L(H, K). A maximizing vector
for T is a non-zero vector x ∈ H such that kT xk = kT kkxk. Thus a maximizing
vector for T is one at which T attains its norm. On a Banach space, even rank
one operators need not have maximizing vectors [8]. The operator (Hx)(t) =
tx(t), 0 < t < 1, is bounded on L2 (0, 1) but has no maximizing vector. However,
compact operators on Hilbert spaces do have maximizing vectors [8].
Theorem 4.1. Let Tφ be a non-negative Toeplitz operator in L(L2a (D)) with symbol φ ∈ L∞ (D) and > 0. Then there exists a non-negative operator C ∈
L(L2a (D)) such that kC − Tφ k < , R = C − Tφ = (h ⊗ h) for some h ∈ L2a (D)
e
f1 (z) for all
and the operator C has a maximizing vector. Further, φ(z)
≥ α2 R
z ∈ D and for some constant α ≥ 0.
Proof. Let Tφ be a non-negative Toeplitz operator in L(L2a (D)) and > 0. Now
kTφ k = sup hTφ g, gi = sup{hTφ g, gi : kgk = 1, g ∈ (ker Tφ )⊥ }.
g∈L2
a (D)
kgk=1
≤ hTφ h, hi.
2
Define R k = hk, hih = (h ⊗ h)k. Then R is a non-negative operator of rank
one and kR k = . Moreover,
Hence there exists a unit vector h ∈ (ker Tφ )⊥ such that kTφ k −
kTφ + R k = sup h(Tφ + R )f, f i
kf k=1
≥ h(Tφ + R )h, hi
≥ kTφ k + .
2
Now Tφ + R is non-negative, and so kTφ + R k lies in the spectrum of Tφ + R .
Since R is compact, Weyl’s theorem implies essential spectrum of Tφ +R is equal
to the essential spectrum of Tφ . But the spectrum of Tφ is bounded by kTφ k and
hence kTφ + R k must lie in the discrete spectrum of Tφ + R . In other words,
there exists a unit vector f ∈ L2a (D) such that (Tφ + R )f = kTφ + R kf . Finally,
we can assume without loss of generality that f ∈ (ker Tφ )⊥ . This is so, since
L2a (D) = ker Tφ ⊕ (ker Tφ )⊥ and if f = f1 + f2 , f1 ∈ ker Tφ , f2 ∈ (ker Tφ )⊥ then
(Tφ + R )f1 = hf1 , hih = 0.
POSITIVE TOEPLITZ OPERATORS
181
Thus if we write C = Tφ + R then C is non-negative, kC − Tφ k = kR k = and
kCf k = kCkkf k. That is, f is a maximizing vector of C. Now let = 1. Then
R1 = (h ⊗ h), khk = 1. Let
E = X ∈ L(L2a (D)) : X ≥ 0, |hXf, gi|2 ≤ hTφ f, f i hR1 g, gi for allf, g ∈ L2a (D) .
Now suppose X ∈ E. Then for f, g ∈ L2a (D),
Tφ
X
f
f
,
= hTφ f, f i + hXg, f i + hXf, gi + hR1 g, gi
X
R1
g
g
= hTφ f, f i + hR1 g, gi + 2Re hXf, gi
≥ 2 hTφ f, f i1/2 hR1 g, gi1/2 + 2Re hXf, gi
≥ 2| hXf, gi | + 2Re hXf, gi
≥ 2|hXf, gi| − 2|hXf, gi| = 0.
Tφ
X
Conversely, if X ≥ 0 and
is a positive operator in L(L2a ⊕ L2a ) then
X
R1
2 Tφ
X
f
0 Tφ
X
f
f
,
≤
,
X
R1
0
g
X
R1
0
0
Tφ
X
0
0
,
X
R1
g
g
for all f, g ∈ L2a (D). A simplification of these inner products yields
|hXf, gi|2 ≤ hTφ f, f ihR1 g, gi for all f, g ∈ L2a (D).
Hence X ∈ E. Thus
E=
X∈
L(L2a (D))
: X ≥ 0 and
Tφ
X
X
R1
is a positive operator in
L(L2a
⊕
L2a )
.
We shall now verify that max X = αR1 = α(h ⊗ h) for some constant α ≥ 0.
X∈E
Suppose Tφ is a positive invertible operator in L(L2a (D)). Then from [2], [3] it
1
1
follows that max X =
h⊗h =
R1 , a scalar multiple of R1 . If Tφ
1
1
−
−
X∈E
kTφ 2 hk
kTφ 2 hk
is an arbitrary positive operator then it follows from [2] that max X is again a
X∈E
scalar multiple of R1 , and
Tφ rR1
max X = max rR1 : r ≥ 0,
≥0 .
rR1 R1
X∈E
Tφ
rR1
f
f
The inequality
,
≥ 0 is equivalent to
rR1
R1
g
g
hTφ f, f i + rhR1 g, f i + rhR1 f, gi + hR1 g, gi ≥ 0 for all f, g ∈ L2a (D).
This can be rewritten as
hTφ f, f i + kR1 (g + rf )k2 − r2 kR1 f k2 ≥ 0
182
N.DAS, M.SAHOO
which holds for all f, g ∈ L2a (D) if and only if hTφ f, f i − r2 kR1 f k2 ≥ 0 or equivalently, r2 R1 ≤ Tφ . Thus from [3], it follows that max X = max{rR1 : r ≥
X∈E
q
0, r2 R1 ≤ Tφ } = λ(Tφ , R1 )R1 where
(
1
− 12
−2
2
kT
hk
,
if
h
∈
Range(T
φ
φ ),
λ(Tφ , R1 ) =
0, otherwise.
Tφ
αR1
Thus max X = αR1 , for some α ≥ 0. Hence
≥ 0. That is,
αR1
R1
X∈E
|hαR1 kz , kw i|2 ≤ hTφ kz , kz ihR1 kw , kw i for all z, w ∈ D. Hence
e
|α|2 |hkz , hihh, kw i|2 ≤ φ(z)|hh,
kw i|2 for all z, w ∈ D.
e ≥ |α|2 |hh, kz i|2 =
If h 6= 0 then there exists w ∈ D such that hh, kw i =
6 0. Thus φ(z)
f1 (z) for all z ∈ D.
α2 R
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1
P.G.Department of Mathematics, Utkal University, Vani Vihar, Bhubaneswar751004, Odisha, India.
E-mail address: namitadas440@yahoo.co.in
2
School of Applied Sciences (Mathematics), KIIT University, Campus-3(Kathajori
Campus), Bhubaneswar-751024, Odisha, India.
E-mail address: smita 782006@yahoo.co.in