Skew-symmetric and essentially unitary operators via Berezin symbols

Definition 1.1

An operator T ∈ ℬ ( ℋ ) is said to be skew-symmetric, if T = − T ∗ .

Proposition 1

If { e j } j ≥ 1 is an orthonormal basis for the RKHS ℋ = ℋ ( Ω ) , then

where the convergence is in ℋ . In particular,

It follows from the aforementioned proposition that k ℋ , z ( λ ) = k ℋ , λ ( z ) ¯ . Writing k ℋ ( z , λ ) = k ℋ , λ ( z ) , we have k ℋ ( λ , z ) = k ℋ ( z , λ ) ¯ for all z , λ ∈ Ω . The norm of k ℋ , λ is easily determined: ∥ k ℋ , λ ∥ 2 = 〈 k ℋ , λ , k ℋ , λ 〉 = k ℋ , λ ( λ ) .

The function

is called the normalized reproducing kernel at λ .

Definition 2.1

Let T be a bounded linear operator on ℋ , the Berezin symbol of T is defined by T ˜ ( λ ) = 〈 T k ^ ℋ , λ , k ^ ℋ , λ 〉 for λ ∈ Ω .

Note that the Berezin symbol T ˜ is a complex-valued bounded function, because T ˜ ( λ ) ≤ ∥ T ∥ for all λ ∈ Ω . Let ber ( T ) denote the Berezin number of T defined by

For any T ∈ ℬ ℋ , we can write

Thus, T is uniquely determined by the function T ∗ k ℋ , z ( λ ) on Ω × Ω , also by the function

defined at all points, where k ( z , λ ) ≠ 0 . It follows from the definition that the mapping T → T ( z , λ ) is linear, T ∗ ( z , λ ) = T ( λ , z ) ¯ , and I ( z , λ ) = 1 , where I is the identity operator and 1 the constant function one. Also, if the elements of ℋ are continuous (smooth, holomorphic), then so is T ( z , λ ¯ ) in each variable, at all points where k ℋ ( z , λ ¯ ) ≠ 0 .

Assume now that:

*the functions T ( z , λ ) are uniquely determined by their restrictions T ( z , z ) to the diagonal.

This is the case, for example, whenever the functions T ( z , λ ) are holomorphic in z and λ ¯ , by a well-known classical theorem in complex analysis (see, for example, Folland [11] and Stroethoff [12, Lemma 2.3]). The following uniqueness lemma is due to the second author (see [13, Lemma 2]).

Lemma 2

Let ℋ = ℋ ( Ω ) be an RKHS whose elements are functions on some set Ω . If ℋ satisfies condition (*), then the correspondence T ↔ T ˜ is one-to-one, i.e., T = 0 if and only if T ˜ ( λ ¯ ) = 0 for all λ ∈ Ω .

Definition 2.2

The RKHS ℋ = ℋ ( Ω ) is said to be standard (see Nordgren and Rosenthal [14]) if the underlying set Ω is a subset of a topological space and the boundary ∂ Ω is nonempty and has the property that k ˆ ℋ , λ n converges weakly to 0 whenever { λ n } is a sequence in Ω that converges to a point in ∂ Ω .

Lemma 3

If T φ is a Toeplitz operator on H 2 , then T ˜ φ ( λ ) = φ ˜ ( λ ) for all λ ∈ D , i.e., T ˜ φ = φ ˜ .

Lemma 4

Let φ ∈ L ∞ ( T ) and let φ ˜ be its harmonic extension into D (which is the Berezin symbol of T φ ). Then T φ k ˆ H 2 , λ − T ˜ φ ( λ ) k ˆ λ → 0 radially, i.e.,

for almost all t ∈ 0, 2 π ) .

The following set is defined by Engliš in [17, (6) in Section 3]

(note that | T ˜ ( λ ) | = | T ∗ ˜ ( λ ) | for all λ ∈ D ). Engliš proved (see [17, Section 3]) that ε H 2 is a C ∗ -algebra. We call ε H 2 the Engliš algebra of operators on the Hardy space H 2 . It follows from Lemma 4 that for any φ ∈ L ∞ ( T ) , T φ ∈ ε H 2 . Here we study the solution of the operator equations T X = K + Y and X T = K + Y in the set

and in the Engliš algebra ε ℋ , where K is compact and Y is a skew-symmetric operator, which is defined on an RKHS ℋ = ℋ ( Ω ) , as follows:

As in the case ℋ = H 2 , it can be shown that actually the set ε ℋ is a C ∗ - algebra (see the proof of (A1) in Section 3 of Engliš’s paper [17]). We also characterize the skew-symmetric operators in terms of Berezin symbols.

Proposition 5

Let ℋ ( Ω ) be an RKHS such that the functions T ( z , λ ) satisfy condition (*) whenever T is a bounded linear operator on ℋ = ℋ ( Ω ) . Then the operator T is skew-symmetric if and only if Re T ˜ = 0, i.e., if its Berezin symbol T ˜ is a purely imaginary complex-valued function.

Proof

According to Lemma 2, T ∗ = − T if and only if T ∗ ˜ ( λ ) = − T ˜ ( λ ) for all λ ∈ Ω , or equivalently T ∗ + T ˜ ( λ ) = 0  for all λ ∈ Ω . This means that

(because T ∗ ˜ = T ˜ ¯ ), and hence

for all λ ∈ Ω . Consequently, T ∗ = − T if and only if Re T ˜ = 0 for all λ ∈ Ω , as desired.□

Theorem 6

Let T , K , Y ∈ ℬ ( ℋ ( Ω ) ) be operators such that Y is skew-symmetric.

1. If K ˜ ( λ ) → 0 as λ → ∂ Ω and X ∈ A ℋ is a solution of the equation

   (2) T X = K + Y ,

   then

   lim λ → ∂ Ω Re ( X ˜ ( λ ) T ˜ ( λ ) ) = 0 .

2. If T ∗ ∈ A ℋ and X ∈ B ( ℋ ( Ω ) ) is a solution of equation (2) with K ˜ ( λ ) → 0 as λ → ∂ Ω , then

   lim λ → ∂ Ω Re ( X ˜ ( λ ) T ˜ ( λ ) ) = 0 .

3. Let ℋ ( Ω ) be an RKHS such that the functions T ( z , λ ) satisfy condition (*) whenever T is a bounded linear operator on ℋ = ℋ ( Ω ) . If X ∈ B ( ℋ ( Ω ) ) is a solution of the equation T X = K + Y such that R a n g e ( X ) ⊥ R a n g e ( T ∗ ) , then K is skew-symmetric.

Proof

1. Let X ∈ A ℋ satisfy (2), that is, T X = K + Y . Then X ∗ T ∗ = K ∗ + Y ∗ = K ∗ − Y . Since T X ˜ ( λ ) = K ˜ ( λ ) + Y ˜ ( λ ) and X ∗ T ∗ ˜ ( λ ) = K ∗ ˜ ( λ ) − Y ˜ ( λ ) , we get T X ˜ λ + X ∗ T ∗ ˜ λ = K ∗ + K ˜ λ for all λ ∈ Ω . Since T X ∗ ˜ λ = T X ˜ ¯ λ , the latter means that

   2 Re T X ˜ λ = K ∗ + K ˜ λ ,

   and hence

   (3) Re T X k ˆ ℋ , λ , k ˆ ℋ , λ = 1 2 ( K ∗ + K ) ˜ ( λ )

   for all λ ∈ Ω . Then we have

   1 2 ( K ∗ + K ) ˜ ( λ ) = Re X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ , T ∗ k ˆ ℋ , λ + X ˜ ( λ ) k ˆ ℋ , λ , T ∗ k ˆ ℋ , λ = Re X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ , T ∗ k ˆ ℋ , λ + Re X ˜ ( λ ) T ˜ ( λ ) ,

   from which we have that

   (4) Re ( X ˜ ( λ ) T ˜ ( λ ) ) ≤ 1 2 K ∗ + K ˜ ( λ ) + X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ , T ∗ k ˆ ℋ , λ ≤ 1 2 K ∗ + K ˜ ( λ ) + ∥ T ∥ X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ ℋ

   for all λ ∈ Ω . On the other hand, by considering that X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ ⊥ X ˜ ( λ ) k ˆ ℋ , λ , we have

   X k ˆ ℋ , λ ℋ 2 = X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ ℋ 2 + | X ˜ ( λ ) | 2 ,

   and hence

   X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ ℋ 2 = X k ˆ ℋ , λ ℋ 2 − | X ˜ ( λ ) | 2

   for all λ ∈ Ω . This shows that

   lim λ → ζ ∈ ∂ Ω X k ˆ ℋ , λ ℋ 2 − | X ˜ ( λ ) | 2 = 0

   if and only if

   lim λ → ζ ∈ ∂ Ω X k ˆ ℋ , λ − X ˜ ( λ ) k ˆ ℋ , λ ℋ = 0 .

   So, since

   K ∗ + K ˜ ( λ ) → 0 as λ → ζ ∈ ∂ Ω ,

   we have from the inequality (4) that

   lim λ → ζ ∈ ∂ Ω ( Re ( X ˜ ( λ ) T ˜ ( λ ) ) ) = 0 ,

   which implies the desired result.

2. It follows from (3) that

   Re X k ˆ ℋ , λ , T ∗ k ˆ ℋ , λ − T ∗ ˜ ( λ ) k ˆ ℋ , λ + Re ( X ˜ ( λ ) T ˜ ( λ ) ) = 1 2 ( K ∗ + K ) ˜ ( λ ) ,

   and therefore,

   Re ( X ˜ ( λ ) T ˜ ( λ ) ) ≤ ∥ X ∥ T ∗ k ˆ ℋ , λ − T ∗ ˜ ( λ ) k ˆ ℋ , λ + 1 2 ( K ∗ + K ) ˜ ( λ )

   for all λ ∈ Ω . Now the result follows immediately from this inequality. This proves (ii).

3. It follows from (3) that

for all λ ∈ Ω , and since X ℋ ⊥ T ∗ ℋ , we have

for all λ ∈ Ω . This implies by Lemma 2 that

as desired. This proves the theorem.□

Remark 1

Assertion (ii) in Theorem 3 shows that the necessary condition

in (i) is not in general a sufficient condition for a solution X of (2) being from the class A ℋ .

Corollary 7

Let ℋ ( Ω ) be an RKHS such that the functions T ( z , λ ) satisfy condition (*) whenever T is a bounded linear operator on ℋ . Let T , K , Y ∈ ℬ ( ℋ ) be operators such that Y is skew-symmetric and K is not skew-symmetric. Then the operator equation T X = K + Y has no solution with the property that R a n g e ( X ) ⊥ R a n g e ( T ∗ ) .

Remark 2

It is easy to see from the proof of Theorem 6 that the same results can be obtained for the operator equation X T = K + Y ; we omit them.

Proposition 8

Let ℋ = ℋ ( Ω ) be an RKHS, and let T 1 , T 2 , K 1 , K 2 , Y 1 , Y 2 ∈ ℬ ( ℋ ) be operators such that K ˜ 1 ( λ ) → 0 , K ˜ 2 ( λ ) → 0 as λ → ∂ Ω , Y 1 and Y 2 are the skew-symmetric operators. If X ∈ ε ℋ satisfies the equations

and

then

and

Proof

The proof is similar to the proof of (i) in Theorem 6. Indeed, if X ∈ ε ℋ is a common solution of equations (5) and (6), then as in the proof of (i) in Theorem 6, we obtain that

and

Since X ∈ ε ℋ and K i ˜ ( λ ) , K i ∗ ˜ ( λ ) → 0 i = 1 , 2 as λ → ∂ Ω , the desired results are immediate from these inequalities. This proves the proposition.□

Corollary 9

Let ℋ = ℋ ( Ω ) be a standard RKHS, and let T 1 , T 2 , K 1 , K 2 , Y 1 , Y 2 ∈ ℬ ( ℋ ) be operators such that K 1 , K 2 are compact and Y 1 , Y 2 are skew-symmetric. If X ∈ ε ℋ satisfies equations ( 5 ) and ( 6 ), then

The proof of the following corollary is immediate from Lemmas 3 and 4 and Proposition 8.

Corollary 10

Let ℋ = H 2 in Proposition 8. If the Toeplitz operator T φ with the symbol φ ∈ L ∞ ( T ) is a common solution of (5) and (6), then

Remark 3

Note that the Berezin symbol of φ ˜ of a function φ ∈ L ∞ ( D , d A ) is defined to be the Berezin symbol of the Toeplitz operator T φ on the Bergman space L a 2 = L a 2 ( D ) with the normalized reproducing kernel k ˆ a , λ ( z ) ≔ 1 − | λ | 2 ( 1 − λ ¯ z ) 2 . In other words, φ ˜ = T ˜ φ . Because T φ k ˆ a , λ , k ˆ a , λ = P ( φ k ˆ a , λ ) , k ˆ a , λ = φ k ˆ a , λ , k ˆ a , λ , we obtain the formula

where d A ( z ) = d x d y π is the normalized Lebesgue area measure and P is the Bergman projection.

The Berezin symbol of a function in L ∞ ( D , d A ) often plays the same important role in the theory of Bergman spaces as the harmonic extension of a function in L ∞ ( ∂ D ) plays in the theory of Hardy spaces.

The Toeplitz algebra T is the C ∗ -subalgebra of ℬ ( L a 2 ) generated by { T g : g ∈ H ∞ } . Let U denote the C ∗ -subalgebra of L ∞ ( D , d A ) generated by H ∞ . As it is well known (see [20, Proposition 4.5]), U equals the closed subalgebra of L ∞ ( D , d A ) generated by the set of bounded harmonic functions on D . Although the map u ↦ T u is not multiplicative on L ∞ ( D , d A ) , the identities T u ∗ = T u ¯ , T u T v = T u v , and T v ¯ T u = T v ¯ u hold for all u ∈ L ∞ ( D , d A ) , and v ∈ H ∞ . This implies that T equals the closed subalgebra of ℬ ( L a 2 ) generated by the Toeplitz operators with bounded harmonic symbol, and that T also equals the closed subalgebra of ℬ ( L a 2 ) generated by { T u : u ∈ U } . The main goal of the paper [20] is to study the boundary behavior of the Berezin symbols of the operators in T and of the functions in U . Namely, the author’s study shows (see [20, Theorem 2.11]) that if S ∈ T , then S ˜ ∈ U . Also, they prove (see [20, Corollary 3.4]) that if u ∈ U , then u ˜ − u has nontangential limit 0 at almost every point of ∂ D . Using similar techniques, they prove (see [20, Corollary 3.7]) that if u ∈ U , then the function λ ↦ T u − u ˜ ( λ ) k ˆ a , λ 2 has nontangential limit 0 at almost every point of ∂ D . Theorem 3.10 in [20] describes the functions u ∈ U such that u ˜ ( λ ) − u ( λ ) → 0 as λ ↦ ∂ D . The aforementioned assertions show that the function λ ↦ T u − u ˜ ( λ ) k ˆ a , λ 2 has nontangential limit 0 at almost every point of ∂ D . Since this property is mainly used in the proof of Corollary 10, the same results also can be proved for the Bergman space Toeplitz operators T u with u ∈ U , which we omit.

Remark 4

As an application of Proposition 5 and also for its usefulness note the following: if φ ∈ H ∞ is a non-constant function and θ is a nontrivial analytic map of D onto itself (i.e., θ ( D ) ⊂ D and θ ≠ constant ), then we can construct a weighted composition operator with symbols φ , θ ,

for f in the Hardy space H 2 , where T φ is an analytic Toeplitz operator and C θ is a composition operator on H 2 .

Example 1

Let φ ∈ H ∞ be as in Remark 4. Then W φ , θ f is skew-symmetric if and only if

Proof

Let λ ∈ D be arbitrary. Then by using formula (7), we have:

hence, Re W φ , θ ˜ ( λ ) = 0 if and only if Re φ ( λ ) 1 − λ θ ( λ ) ¯ = 0 . Since λ ∈ D is arbitrary, Proposition 5 works.□

Example 2

Let φ ∈ H ∞ be a non-constant function and θ be a non-trivial analytic self-map of the unit disc D . Then C θ ∗ T φ is a skew-symmetric operator on H 2 if and only if Re Ψ λ ˜ ( λ ) = 0 for all λ ∈ D , where Ψ λ denotes the function φ ( ζ ) 1 − λ ζ ¯ 1 − λ θ ( ζ ) ¯ on T .

Proof

Indeed, we have:

It is easy to see that Ψ λ ∈ L ∞ since | λ θ ( ζ ) ¯ | < 1 for all λ ∈ D . Hence, the last formula means that C θ ∗ T φ ˜ ( λ ) = Ψ λ ˜ ( λ ) , where Ψ λ ˜ denotes the harmonic extension of Ψ λ into D . Then, by Proposition 5, we deduce that C θ ∗ T φ is a skew-symmetric if and only if Re Ψ λ ˜ ( λ ) = 0 for all λ ∈ D , as required.□