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Article

Dual Toeplitz Operators on Bounded Symmetric Domains

School of Mathematics and Statistics, Tianshui Normal University, Tianshui 741000, China
Mathematics 2025, 13(10), 1611; https://doi.org/10.3390/math13101611
Submission received: 26 April 2025 / Revised: 13 May 2025 / Accepted: 13 May 2025 / Published: 14 May 2025

Abstract

:
We give some characterizations of dual Toeplitz operators acting on the orthogonal complement of the Bergman space over bounded symmetric domains. Our main result characterizes those finite sums of products of Toeplitz operators that are themselves dual Toeplitz operators. Furthermore, we obtain a necessary condition for such finite sums of dual Toeplitz products to be compact. As an application of our main result, we derive a sufficient and necessary condition for when the (semi-)commutators of dual Toeplitz operators is zero. Notably, we find that a dual Toeplitz operator is compact if and only if it is the zero operator.

1. Introduction

Let C n be an n-dimensional complex Euclidean space, and let Ω be a bounded symmetric domain in C n . We always assume that Ω is circular and in its standard (Harish–Chandra) realization so that 0 Ω . Thus, Ω is starlike; that is, z Ω implies that t z Ω for all t [ 0 , 1 ] . We can canonically define, for each w Ω , an automorphism φ w in A U T ( Ω ) , the group of all automorphisms of Ω such that
(a)
φ w φ w ( z ) = z ;
(b)
φ w ( 0 ) = w , φ w ( w ) = 0 ;
(c)
φ w has a unique fixed point in Ω .
The Harish–Chandra circularity assumption is crucial in our study as it ensures the existence of a well-behaved automorphism group that preserves the geometric and analytical properties of the bounded symmetric domain. This invariance is essential for establishing the compactness criteria of dual Toeplitz operators and deriving our main results.
It is well known that the unit disk D , the unit ball B in C n , and the unit polydisc D n can all be considered as examples of Ω .
Let d v be the normalized Lebesgue volume measure on C n . Denote by L 2 ( Ω , d v ) the space of measurable functions f on Ω such that
f 2 = Ω | f ( z ) | 2 d v ( z ) < .
Then, L 2 ( Ω , d v ) is a Hilbert space with an inner product
f , g = Ω f ( z ) g ( z ) ¯ d v ( z ) , f , g L 2 ( Ω , d v ) .
The Bergman space A 2 ( Ω ) is the closed subspace of L 2 ( Ω , d v ) consisting of all holomorphic functions. For z Ω , the Bergman reproducing kernel is the function K z A 2 ( Ω ) such that
f ( z ) = f , K z
for each f A 2 ( Ω ) . Let k z ( · ) = K z ( · ) K z be the normalized reproducing kernel. Sometimes, we use K ( · , z ) instead of K z ( · ) .
Let P be the orthogonal projection from L 2 ( Ω , d v ) onto A 2 ( Ω ) and let L ( Ω ) be the space of bounded measurable functions on Ω . Given u L 2 ( Ω ) , the multiplication operator M u , the Toeplitz operator T u , the Hankel operator H u , the dual Toeplitz operator S u , and the dual Hankel operator R u with symbol u are defined, respectively, by
M u : L 2 ( Ω , d v ) L 2 ( Ω , d v ) , M u ( f ) = u f , f L 2 ( Ω , d v ) ; T u : A 2 ( Ω ) A 2 ( Ω ) , T u ( f ) = P ( u f ) , f A 2 ( Ω ) ; H u : A 2 ( Ω ) ( A 2 ( Ω ) ) , H u ( f ) = ( I P ) ( u f ) , f A 2 ( Ω ) ; S u : ( A 2 ( Ω ) ) ( A 2 ( Ω ) ) , S u ( f ) = ( I P ) ( u f ) , f ( A 2 ( Ω ) ) ; R u : ( A 2 ( Ω ) ) A 2 ( Ω ) , R u ( f ) = P ( u f ) , f ( A 2 ( Ω ) ) .
Next, we will use Table 1 to further distinguish these operators and spaces.
Under the decomposition L 2 ( Ω , d v ) = A 2 ( Ω ) ( A 2 ( Ω ) ) , the multiplication operator M u with symbol u L ( Ω ) can be expressed as a block operator matrix relative to this decomposition. Specifically, M u can be written as
M u = T u R u H u S u ,
The identity M u M v = M u v implies the following algebraic relations between those operators:
T u v = T u T v + R u H v ; S u v = S u S v + H u R v ; a n d H u v = H u T v + S u H v .
Let Ω ¯ denote the Euclidean closure of Ω and let Ω be the topological boundary. The bounded symmetric domain Ω in its representation with normalized volume measure, the kernel function K ( · , · ) , has the following special properties:
(a)
K ( 0 , z ) = K ( z , 0 ) = 1 , z Ω ;
(b)
K ( w , z ) 0 , w Ω , z Ω ¯ ;
(c)
lim z Ω K ( z , z ) = ;
(d)
K ( · , · ) 1 is a smooth function on Ω × Ω ;
(e)
K ( w , z ) = K ( z , w ) ¯ , w , z Ω ,
For φ A U T ( Ω ) , it follows that
K ( z , w ) = K ( φ ( z ) , φ ( w ) ) · J φ ( z ) J φ ( w ) ¯ ,
where J φ ( z ) denotes the complex Jacobian. It is easy to show that
K ( φ a ( z ) , φ a ( w ) ) = K ( a , a ) K ( z , w ) K ( z , a ) K ( a , w ) .
For z Ω , define an operator U z on L 2 ( Ω , d v ) by U z f = ( f φ z ) k z ; Then U z is a unitary operator on L 2 ( Ω , d v ) and A 2 ( Ω ) . The operator U z has the following properties (see ref. [1]):
(a)
U z 2 = I ;
(b)
P U z = U z P ;
(c)
U z 1 = k z ;
(d)
If u L ( Ω ) , then U z T u U z = T u φ z .
Dual Toeplitz operators have garnered significant attention, particularly in the context of orthogonal complements of classical function spaces. For instance, Stroethoff and Zheng [2] delved into the algebraic characteristics of these operators on the orthogonal complement of the Bergman space defined on the unit disk D . Their findings revealed that for functions f and g belonging to L ( D ) , the dual Toeplitz operators S f and S g commute if and only if both functions are analytic, both are co-analytic, or they are linearly dependent modulo constants. Moreover, they determined that the product S f S g qualifies as a dual Toeplitz operator solely when either f is analytic or g is co-analytic, in which case, S f S g = S f g . Chen, Yu, and Zhao [3] explored conditions under which two dual Toeplitz operators (semi)-commute and analyzed their spectral properties in the context of the orthogonal complement of the harmonic Dirichlet space. Subsequently, their work was expanded to encompass multiple variables. He, Huang, and Lee [4] focused on dual Toeplitz operators operating on the orthogonal complement of Hardy–Sobolev spaces in a unit ball. Kong and Lu [5] characterized the algebraic properties of dual Toeplitz operators on Bergman spaces on the unit ball. Additionally, they investigated circumstances under which the summation of a finite number of dual Toeplitz product operators equates to another dual Toeplitz operator, thereby reconnecting with previously established results pertaining to commutativity or product issues. For a more comprehensive understanding of dual Toeplitz operators, one may refer to various studies [6,7,8]. Guo and Zheng [1], for instance, characterized conditions under which the sum of a finite number of small Hankel products on A 2 ( Ω ) vanishes. Ter Horst and collaborators [9] conducted spectral and compactness analyses on Toeplitz-like operators using state-space methods, providing valuable context for our study. Their work highlights the importance of understanding the behavior of these operators under various conditions and suggests potential broader implications for related operator classes. Building upon their work, this paper further explores dual Toeplitz operators on bounded symmetric domains, aiming to extend and generalize existing results to the multivariable setting. Notably, the algebraic properties of dual Toplitz operators on bounded symmetric domains remain unexplored. Consequently, this article aims to examine certain algebraic aspects of the dual Toeplitz operator on ( A 2 ( Ω ) ) , drawing inspiration from the aforementioned research findings.
Let S N be the set of all permutations on { 1 , 2 , , N } . Let H N denote the set of vectors whose components are holomorphic functions on Ω , and let H ¯ N be the set of vectors whose components are conjugate holomorphic functions. Let M N × N be the collection of N × N matrices. Now, we list our main results as follows.
Theorem 1.
Suppose that f j , g j L ( Ω ) for j = 1 , 2 , , N . Then, j = 1 N S f j S g j is a dual Toeplitz operator if and only if the following statements hold:
(i) 
j = 1 N f j g j = h for some h L ( Ω ) ;
(ii) 
There exists a matrix A M N × N and a permutation σ S N such that ( I A ) f σ H N and A ¯ g σ H ¯ N .
Here, f σ = ( f σ ( 1 ) , f σ ( 2 ) , , f σ ( N ) ) T and g σ = ( g σ ( 1 ) , g σ ( 2 ) , , g σ ( N ) ) T .
Using the methods of ([10], Theorem 3) and the above theorem, we immediately obtain the following theorem.
Theorem 2.
Suppose that f j , g j L ( Ω ) for j = 1 , 2 , , N . Then j = 1 N S f j S g j = 0 if and only if j = 1 N f j g j = 0 and one of the following equivalent conditions holds:
(i) 
j = 1 N S f j R g j = 0 .
(ii) 
There exist a matrix A M N × N and a permutation σ S N such that ( I A ) f σ H N and A ¯ g σ H ¯ N
where f σ = ( f σ ( 1 ) , f σ ( 2 ) , , f σ ( N ) ) T and g σ = ( g σ ( 1 ) , g σ ( 2 ) , , g σ ( N ) ) T .
The organization of this paper is as follows. In Section 2, we obtain a characterization of the compactness of dual Toeplitz operators and a necessary condition for the finite sum of the product of dual Toeplitz operators to be compact. Section 3 answers when the finite sum of dual Toeplitz products is another dual Toeplitz operator.

2. Bounded, Compact Dual Toeplitz Operators

Let w Ω , 0 < s < d i a m ( Ω ) 2 | w | , where d i a m ( Ω ) stands the diameter of Ω ; we now define a function on Ω by
G w , s ( z ) = ( z 1 w 1 ) ¯ χ B ( w , s ) ( z ) , z Ω ,
where B ( w , s ) is the Euclidean ball in Ω centered at w Ω with radius s and χ B ( w , s ) denotes the characteristic function of B ( w , s ) . It is easy to check that G w , s ( A 2 ( Ω ) ) . Set
g w , s ( z ) = G w , s ( z ) G w , s L 2 ( Ω , d v ) .
It follows that g w , s converges to 0 weakly in ( A 2 ( Ω ) ) as s 0 . In fact, for any f ( A 2 ( Ω ) ) and w Ω , applying the Cauchy–Schwarz inequality, then
| f , g w , s | 2 = B ( w , s ) f ( z ) g w , s ( z ) d v ( z ) 2 B ( w , s ) | f ( z ) | 2 d v ( z ) 0 , s 0 .
Lemma 1.
With the same notations as above, we have
lim s 0 + M u g w , s = | u ( w ) |
for a . e . w Ω and for each u L 2 ( Ω , d v ) .
Proof. 
There exists a positive constant c such that
G w , s L 2 ( Ω , d v ) 2 c s 2 v ( B ( w , s ) )
by ([11], Propositions 1.4.9). Hence,
| M u g w , s 2 | u ( w ) | 2 | = Ω | u ( z ) | 2 | g w , s ( z ) | 2 d v ( z ) | u ( w ) | 2 Ω | | u ( z ) | 2 | u ( w ) | 2 | | g w , s ( z ) | 2 d v ( z ) = 1 G w , s L 2 ( Ω , d v ) 2 Ω | | u ( z ) | 2 | u ( w ) | 2 | | G w , s ( z ) | 2 d v ( z ) = c G w , s L 2 ( Ω , d v ) 2 B ( w , s ) | | u ( z ) | 2 | u ( w ) | 2 | | z 1 w 1 | 2 d v ( z ) c s 2 G w , s L 2 ( Ω , d v ) 2 B ( w , s ) | | u ( z ) | 2 | u ( w ) | 2 | d v ( z ) 1 v ( B ( w , s ) ) B ( w , s ) | | u ( z ) | 2 | u ( w ) | 2 | d v ( z ) .
Let
M = w Ω : lim s 0 B ( w , s ) | | u ( z ) | 2 | u ( w ) | 2 | d v ( z ) v ( B ( w , s ) ) = 0 .
It follows that M c is a set of measure zero by Theorem 8.8 in ref. [12]. This completes the proof. □
The following lemma will be useful in our characterization of the boundedness and compactness of the dual Toeplitz operator.
Lemma 2.
For φ L 2 ( Ω , d v ) , we have
lim s 0 S φ g w , s = | φ ( w ) | , a . e . w Ω .
Proof. 
Note that for each f ( A 2 ( Ω ) ) , we have
M φ f = P ( φ f ) + ( I P ) ( φ f ) = H φ ¯ f + S φ f .
Thus,
M φ g w , s 2 = S φ g w , s 2 + H φ ¯ g w , s 2 .
Fubini’s Theorem gives
H φ ¯ g w , s 2 = Ω | H φ ¯ g w , s ( z ) | 2 d v ( z ) = Ω | P ( φ g w , s ) ( z ) | 2 d v ( z ) = Ω B ( w , s ) φ ( ζ ) g w , s ( ζ ) K ( z , ζ ) d v ( ζ ) 2 d v ( z ) Ω B ( w , s ) | φ ( ζ ) | 2 | K ( z , ζ ) | 2 d v ( ζ ) d v ( z ) B ( w , s ) | φ ( ζ ) | 2 K ( ζ , ζ ) d v ( ζ ) sup ζ B ( w , s ) K ( ζ , ζ ) B ( w , s ) | φ ( ζ ) | 2 d v ( ζ ) .
for each 0 < s < d i a m ( Ω ) 2 | w | . We, by the assumption, obtain
lim s 0 B ( w , s ) | φ ( ζ ) | 2 d v ( ζ ) = 0 .
Therefore,
lim s 0 H φ ¯ g w , s lim s 0 ¯ sup ζ B ( w , s ) K ( ζ , ζ ) lim s 0 B ( w , s ) | φ ( ζ ) | 2 d v ( ζ ) K ( w , w ) lim s 0 B ( w , s ) | φ ( ζ ) | 2 d v ( ζ ) = 0
for each w Ω , and this implies
| φ ( w ) | 2 = lim s 0 M φ g w , s 2 = lim s 0 S φ g w , s 2
for a . e . w Ω by using Lemma 1.      □
Now, we are ready to characterize the boundedness of dual Toeplitz operators on ( A 2 ( Ω ) ) .
Theorem 3.
Let f L 2 ( Ω , d v ) . Then, S f is bounded if and only if f L ( Ω ) . Furthermore, S f = f .
Proof. 
If f L ( Ω ) , it is obvious that S f f . Suppose that S f is bounded on ( A 2 ( Ω ) ) . Note that
S f g w , s S f
for all w Ω and 0 < s < d i a m ( Ω ) 2 | w | . Let s 0 in Lemma 2, we have
| f ( w ) | S f
for a . e . w Ω , so that
f S f .
This completes the proof. □
We immediately obtain the following conclusion by using the functions g w , s which converges to 0 weakly in ( A 2 ( Ω ) ) as s 0 .
Corollary 1.
Let f L ( Ω ) . Then, S f is compact if and only if S f = 0 if and only if f ( w ) = 0 a . e . w Ω .
We consider the relation between the compactness of the finite sums of finite dual Toeplitz products and their symbols. We have only the necessary conditions; unfortunately, there are no sufficient conditions here.
Theorem 4.
Let φ t , ψ t L ( Ω ) for t = 1 , 2 , , N . If t = 1 N S φ t S ψ t is compact, then t = 1 N φ t ψ t = 0 .
Proof. 
Dual Toeplitz operators are closely related to Hankel operators; we have
t = 1 N S φ t ψ t = t = 1 N S φ t S ψ t + t = 1 N H φ t H ψ t ¯ .
Lemma 2 gives
H φ t H ψ t ¯ g w , s 2 = ( I P ) ( φ t H ψ t ¯ g w , s ) 2 Ω | φ t ( z ) | 2 | H ψ t ¯ g w , s ( z ) | 2 d v ( z ) = Ω | φ t ( z ) | 2 B ( w , s ) | ψ t ( ζ ) K ( z , ζ ) ¯ | 2 d v ( ζ ) d v ( z ) = B ( w , s ) | ψ t ( ζ ) | 2 Ω | φ t ( z ) | 2 | K ( z , ζ ) | 2 d v ( z ) d v ( ζ ) ψ t φ t sup ζ B ( w , s ) K ( ζ , ζ ) B ( w , s ) d v ( ζ )
for each 0 < s < d i a m ( Ω ) 2 | w | . Hence,
lim s 0 + H φ j H ψ t ¯ g w , s = 0
for w C d , t = 1 , 2 , , N . This means that
lim s 0 S t = 1 N φ t ψ t g w , s = t = 1 N φ t ψ t = 0
by Lemma 1. This completes the proof. □
Our proofs rely on estimates of Hankel operators, particularly when characterizing the compactness of dual Toeplitz operators. Compared to the multivariable Hankel compactness results obtained by Chen and Lee [6], there are both similarities and differences. The reproducing kernels on bounded symmetric domains lack explicit forms, making the estimates relatively challenging. In contrast, the reproducing kernels on Fock spaces are more transparent, facilitating easier estimations.

3. Sums of the Dual Toeplitz Operators

In this section, we will consider the case when the finite sum of the dual Toeplitz products is a different dual Toeplitz operator. The characterization of this result requires the introduction of some basic concepts and lemmas.
For two nonzero functions f , g ( A 2 ( Ω ) ) , we use f g to denote the rank one operator on ( A 2 ( Ω ) ) defined by
( f g ) h = h , g f , h ( A 2 ( Ω ) ) .
The following lemma is important for the proof of the main result, which can be derived directly from Proposition 2.2 in ref. [1].
Lemma 3.
Let Ω be a bounded symmetric domain in its standard realization. Then, there exist polynomials p i , q i , i = 1 , 2 , , m such that
k w k w = i = 1 m T p i φ w T q i φ w ¯ .
Lemma 4.
Let f j , g j ( A 2 ( Ω ) ) for j = 1 , 2 , , N . Then,
j = 1 N f j g j = 0
if and only if there exist matrices A M N × N and σ S N such that
( I A ) f σ = 0 a n d A g σ = 0 ,
where f σ = ( f σ ( 1 ) , f σ ( 2 ) , , f σ ( N ) ) T and g σ = ( g σ ( 1 ) , g σ ( 2 ) , , g σ ( N ) ) T .
Proof. 
This is similar to the proof of Proposition 4 in ref. [13], so we omit its proof here. □
Let T be a linear operator on ( A 2 ( Ω ) ) . With fixed w Ω , we define an operator as follows:
S w ( T ) = i = 1 m S p i φ w T S q i φ w ¯ .
In particular, let T = S f for some suitable f; then, S 0 ( S f ) = i = 1 m S p i ( w ) f q i ( w ) ¯ .
Proof of Theorem 1.
Assume that j = 1 N S f j S g j = S h , and the equation
H f j H g j ¯ = S f j g j S f j S g j , j = 1 , 2 , , N .
For fixed w Ω and with u , v L ( Ω ) , it is easy to show that H u T p i φ w = S p i φ w H u , i = 1 , 2 , , m , and T q i φ w ¯ H v ¯ = H v ¯ S q i φ w ¯ , i = 1 , 2 , , m . Hence,
j = 1 N ( H f j k w ) ( H g j k w ) = j = 1 N H f j ( k w k w ) H g j ¯ = j = 1 N i = 1 m H f j T p i φ w T q i φ w ¯ H g j ¯ = i = 1 m j = 1 N S p i φ w H f j H g j ¯ S q i φ w ¯ = S w j = 1 N ( S f j g j S f j S g j ) = S w j = 1 N ( S f j g j S h ) = S w ( S j = 1 N f j g j h ) .
Note that if k 0 = 1 , then the above equation can be written as
j = 1 N ( H f j 1 ) ( H g j 1 ) = S 0 ( S j = 1 N f j g j h ) = i = 1 m S p i ( w ) q i ( w ) ¯ n = 1 N f n g n h = S i = 1 m p i ( w ) q i ( w ) ¯ n = 1 N f n g n h .
Notice that S i = 1 m p i ( w ) q i ( w ) ¯ n = 1 N f n g n h has a rank of at most N. Let w ¯ , w ¯ 2 , , w ¯ N + 1 ( A 2 ( Ω ) ) ; there exist r 1 , r 2 , , r N + 1 are not all zero, such that
S i = 1 m p i ( w ) q i ( w ) ¯ j = 1 N f j g j h l = 1 N + 1 r l w ¯ l = 0 .
Which implies that
φ ( w ) = i = 1 m p i ( w ) q i ( w ) ¯ j = 1 N f j g j h l = 1 N + 1 r l w ¯ l A 2 ( Ω ) .
By Proposition 2.2 in ref. [1], we have K ( w , w ) 1 = i = 1 m p i ( w ) q i ( w ) ¯ ; the above equation becomes
φ ( w ) = j = 1 N f j g j h l = 1 N + 1 r l w ¯ l K ( w , w ) A 2 ( Ω ) .
since K ( w , w ) as w Ω . From the maximum modulus principle, we have φ ( w ) = 0 for all w Ω . Thus,
j = 1 N f j g j = h .
It follows that
j = 1 N ( H f j 1 ) ( H g j 1 ) = 0 .
By Lemma 4, there exist A = { a i j } M N × N and σ S N such that
( I A ) ( H f σ ( 1 ) 1 , H f σ ( 2 ) 1 , , H f σ ( N ) 1 ) T = 0
and
A ( H g σ ( 1 ) 1 , H g σ ( 2 ) 1 , , H g σ ( N ) 1 ) T = 0 .
By (2), we obtain
H j = 1 N a i j f σ ( j ) 1 = H f σ ( i ) 1
for i = 1 , 2 , , N . This means that
j = 1 N a i j f σ ( j ) f σ ( i ) A 2 ( Ω ) .
Furthermore, we have ( I A ) f σ H N . By (3), we have
H i = 1 N a ¯ i j g ¯ σ ( j ) 1 = 0
for n = 1 , 2 , , N . Then,
i = 1 N a ¯ i j g ¯ σ ( j ) A 2 ( Ω ) .
So, A ¯ g σ H ¯ N .
Suppose (1) and (2) hold. Let A = { a i j } , ( I A ) f σ = ( x 1 , x 2 , , x N ) T and A ¯ g σ = ( y 1 , y 2 , , y N ) T . Then,
i = 1 N S x i S g σ ( i ) = i = 1 N j = 1 N a i j S f σ ( j ) S f i S g σ ( i ) = i = 1 N n = 1 N a i n S f σ ( n ) S g σ ( i ) i = 1 N S f i S g σ ( i ) = n = 1 N S f σ ( n ) S i = 1 N a i n g σ ( i ) i = 1 N S f i S g σ ( i ) = n = 1 N S f σ ( n ) S y n i = 1 N S f i S g i .
It follows that
i = 1 N S f i S g i = S h ,
where h = i = 1 N f σ ( i ) y i x i g σ ( i ) . Combining Theorem 4, we obtain h = i = 1 N f i g i . □
Example 1.
Consider the unit ball B n C n as a bounded symmetric domain Ω. Let f 1 ( z ) = z 1 , g 1 ( z ) = z ¯ 1 , f 2 ( z ) = z 2 , and g 2 ( z ) = z ¯ 2 be functions in L ( B n ) . We want to determine when j = 1 2 S f j S g j is a dual Toeplitz operator.
According to Theorem 1, j = 1 2 S f j S g j is a dual Toeplitz operator if and only if the following conditions hold:
1. 
j = 1 2 f j g j = h for some h L ( B n ) .
2. 
There exists a matrix A M 2 × 2 and a permutation σ S 2 such that ( I A ) f σ H N and A g σ H N , where H N is the set of vectors whose components are conjugate holomorphic functions.
Proof. 
  • Sum Condition:
    j = 1 2 f j g j = f 1 g 1 + f 2 g 2 = z 1 z ¯ 1 + z 2 z ¯ 2 = | z 1 | 2 + | z 2 | 2 = : h ( z ) .
    Clearly, h ( z ) L ( B n ) since | z 1 | 2 + | z 2 | 2 1 for z B n .
  • Matrix and Permutation Condition: Let us consider the identity matrix A = 1 0 0 1 and the trivial permutation σ = id (i.e., σ ( 1 ) = 1 , σ ( 2 ) = 2 ). Then,
    ( I A ) f σ = ( I I 2 × 2 ) ( f 1 , f 2 ) T = ( 0 , 0 ) T H N ,
    A g σ = I 2 × 2 ( g 1 , g 2 ) T = ( g 1 , g 2 ) T = ( z ¯ 1 , z ¯ 2 ) T H ¯ N .
Thus, both conditions of Theorem 1 are satisfied, and we conclude that j = 1 2 S f j S g j is a dual Toeplitz operator. □
Example 2.
Consider the following bounded functions defined on D 2 :
f 1 ( z 1 , z 2 ) = 1 , g 1 ( z 1 , z 2 ) = z 1 z 2 ¯ , f 2 ( z 1 , z 2 ) = 0 , g 2 ( z 1 , z 2 ) = 0 .
We verify the two conditions of Theorem 1:
1. 
Condition 1. Verify that j = 1 2 f j g j L ( D 2 ) .
Compute the sum:
j = 1 2 f j g j = f 1 g 1 + f 2 g 2 = 1 · z 1 z 2 ¯ + 0 · 0 = z 1 z 2 ¯ .
Since | z 1 z 2 ¯ | < 1 for all ( z 1 , z 2 ) D 2 , the function z 1 z 2 ¯ is bounded on D 2 . Thus, there exists h L ( D 2 ) (in this case, h = z 1 z 2 ¯ ) such that j = 1 2 f j g j = h .
2. 
Condition 2. Find a matrix A M 2 × 2 and a permutation σ S 2 such that ( I A ) f σ H 2 and A g σ H ¯ 2 , where H 2 denotes the subspace of functions in L 2 ( D 2 ) that are holomorphic on D 2 , and H ¯ 2 denotes the subspace of functions that are conjugate-holomorphic.
In this example, since f 2 = 0 and g 2 = 0 , the choice of A and σ is simplified. Let A = 1 0 0 1 and σ = ( 1 , 2 ) . Since A is the identity matrix, ( I A ) f σ = 0 . The zero function is trivially holomorphic, so ( I A ) f σ H 2 . Since A is the identity matrix, A = A , and g σ = ( g 1 , g 2 ) = ( z 1 z 2 ¯ , 0 ) . Thus, A g σ = g σ . The function g 1 = z 1 z 2 ¯ is conjugate-holomorphic on D 2 , so A g σ H ¯ 2 .
This example satisfies Condition 2 by choosing g 1 to be conjugate-holomorphic, aligning with the requirement that A g σ H ¯ 2 .
Let H o l ( Ω ) be the collection of all holomorphic functions on Ω . We can draw the following immediate conclusions from our main result.
Corollary 2.
Let f , g , u , v , h L ( Ω ) . Then, S h = S f S g + S u S v if and only if h = f g + u v and one of the following conditions holds:
(I) 
f , u H o l ( Ω ) ;
(II) 
g ¯ , v ¯ H o l ( Ω ) ;
(III) 
f , v ¯ H o l ( Ω ) ;
(IV) 
g ¯ , u H o l ( Ω ) ;
(V) 
f + λ u H o l ( Ω ) and v ¯ λ g ¯ H o l ( Ω ) for some constant λ 0 .
Corollary 3.
Let u , v L ( Ω ) . Then, S u and S v commute if and only if one of the following conditions holds:
(I) 
u , v H o l ( Ω ) ;
(II) 
u ¯ , v ¯ H o l ( Ω ) ;
(III) 
There exist constants λ and μ, which are not both zero, such that λ u + μ v is constant.
Corollary 4.
Let u , v L ( Ω ) . Then, S u S v is a dual Toeplitz operator if and only if u H o l ( Ω ) or v ¯ H o l ( Ω ) , in which case S u S v = S u v .
While the Harish–Chandra circularity assumption is currently necessary for our framework, it remains an open question whether this condition can be relaxed or replaced by weaker assumptions while still preserving the essential properties of dual Toeplitz operators on bounded symmetric domains. This direction of research could potentially lead to more general and applicable results.

Funding

This work was supported by the Scientific Research Project of Tianshui Normal University. (grant no. tsnu0225A01).

Data Availability Statement

This manuscript has no associated data. Data are openly available in a public repository.

Conflicts of Interest

We declare that we do not have any commercial or associative interests that represent conflicts of interest in connection with the work submitted.

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Table 1. Notation glossary.
Table 1. Notation glossary.
SymbolDescription
A 2 ( Ω ) Bergman space on bounded symmetric domain Ω
( A 2 ( Ω ) ) Orthogonal complement of A 2 ( Ω ) in L 2 ( Ω , d v )
T u Toeplitz operator with symbol u
S u Dual Toeplitz operator with symbol u
H u Hankel operator with symbol u
R u Dual Hankel operator with symbol u
ϕ w Automorphism of Ω
k w Normalized reproducing kernel at point w Ω
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Dong, J. Dual Toeplitz Operators on Bounded Symmetric Domains. Mathematics 2025, 13, 1611. https://doi.org/10.3390/math13101611

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Dong J. Dual Toeplitz Operators on Bounded Symmetric Domains. Mathematics. 2025; 13(10):1611. https://doi.org/10.3390/math13101611

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Dong, Jianxiang. 2025. "Dual Toeplitz Operators on Bounded Symmetric Domains" Mathematics 13, no. 10: 1611. https://doi.org/10.3390/math13101611

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Dong, J. (2025). Dual Toeplitz Operators on Bounded Symmetric Domains. Mathematics, 13(10), 1611. https://doi.org/10.3390/math13101611

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