Complex symmetry of weighted composition operators on the Fock space

J. Math. Anal. Appl. 433 (2016) 1757–1771

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Complex symmetry of weighted composition operators on the Fock space ✩ Pham Viet Hai, Le Hai Khoi ∗ Division of Mathematical Sciences, School of Physical and Mathematical Sciences, Nanyang Technological University (NTU), 637371 Singapore

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Article history: Received 23 July 2015 Available online 2 September 2015 Submitted by Richard M. Aron Keywords: Fock space Weighted composition operator Conjugation Complex symmetry

With the techniques of weighted composition operators, we introduce a new concept of general weighted composition conjugations on the Fock space F 2 (C) and completely solve the following problems: the description of weighted composition operators which are conjugations; the criteria for weighted composition operators to be complex symmetric; the spectrum and the point spectrum of complex symmetric operators; and the criteria for Schatten class membership as well as for normality of such operators. © 2015 Elsevier Inc. All rights reserved.

1. Introduction Many problems appearing in analysis require much research to non-Hermitian operators. Among them, the operators, known as complex symmetric operators, have become particularly important in both theoretic and application aspects (see, e.g., [4]). In a general setting, we have the following definitions. Definition 1.1. A mapping T acting on a separable complex Hilbert space H is said to be anti-linear (also conjugate-linear), if T (ax + by) = a T (x) + b T (y), ∀x, y ∈ H, ∀a, b ∈ C. Definition 1.2. An anti-linear mapping C : H → H is called a conjugation, if it is (1) involutive: C 2 = I, the identity operator; (2) isometric: Cx = x, for all x ∈ H. ✩

Supported in part by PHC Merlion Project 1.04.14.

* Corresponding author. E-mail addresses: [email protected] (P.V. Hai), [email protected] (L.H. Khoi). http://dx.doi.org/10.1016/j.jmaa.2015.08.069 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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Definition 1.3. A bounded linear operator T on H is called complex symmetric, if there exists a conjugation C on H, such that T = CT ∗ C. In this case, T is often called a C-symmetric operator (or CSO, for short). The general study of the complex symmetry was commenced by Garcia and Putinar in [5,6]. Thereafter, a number of the papers is devoted to the topic. The results show that the bounded complex symmetric operators are quite diverse. It includes the Volterra integration operators, normal operators, compressed Toeplitz operators, etc. On the other hand, given two holomorphic (or entire) functions ϕ and ψ, the weighted composition operator Wψ,ϕ is defined as Wψ,ϕ f = ψ · f ◦ ϕ. The study of weighted composition operators on function spaces has received a special attention by many authors during the past several decades (see, e.g., [1]). There is a great number of topics in the study of weighted composition operators: boundedness, compactness, compact difference, essential norm, closed range, etc. Naturally, we can ask a question of whether there exist the complex symmetric weighted composition operators. Recently, this question was analyzed in [3,9] for Hardy spaces in the unit disk. It is showed that such operators do exist. It is also worth to mention the work by Jung and colleagues [9] for the classical Hardy space and by Garcia and Hammond [3] for the weighted Hardy spaces. Independently, they discovered the complex symmetric structure of Wψ,ϕ when the conjugation is of the form Cf (z) = f (¯ z ). It should be noted that in all works mentioned above, the complex symmetry is studied mainly on the Hardy spaces. For spaces of entire functions this is still an open topic which needs addressing. Being motivated by the recent results on Hardy spaces, we investigate the complex symmetry in the Fock spaces of entire functions. 2. Preliminaries In this section, we present some basic results about the Fock space, which are used in the sequel. Technically speaking, the Fock space F 2 (C), we denote simply by F 2 , is defined as follows: ⎧ ⎫ ⎞1/2 ⎛ ⎪ ⎪  ⎨ ⎬ 1 2 2 −|z|2 ⎠ ⎝ F = f ∈ O(C) : f  = |f (z)| e dV (z) <∞ , ⎪ ⎪ π ⎩ ⎭ C

where dV is the Lebesgue measure on C and O(C) is the set of all entire functions on C. It is a reproducing kernel Hilbert space, with the inner product given by  2 1 f, g = f (z)g(z)e−|z| dV (z) π C

and the kernel Kz (u) = ezu . More generally, we have  2 1 f (m) (u) = f (z)z m ezu e−|z| dV (z), ∀f ∈ F 2 , ∀m ∈ N. π C

[m]

Here N = {0, 1, 2, . . .} is the set of non-negative integers. Setting Kz (u) = um Kz (u), the m-th derivative [m] evaluation kernel, the equality becomes f (m) (u) = f, Ku . Taking z = 0, we put [m]

Pm (u) := K0 (u) = um , m ≥ 0.

(2.1)

 z m ∞ √ m!, plays an important role in the sequel, as √ m! m=0 2 |z|2 /2 forms an orthonormal basis for F . We note that Kz  = e . This sequence of monomials (Pm ), with Pm  =

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In this paper, we also use the normalized kernel ku =

Ku . Ku 

(2.2)

Recently, T. Le [10] gave a characterization for boundedness and compactness of Wψ,ϕ on F 2 . To state these results, for convenience, we put Mz (ψ, ϕ) := |ψ(z)|2 e|ϕ(z)|

2

−|z|2

, z ∈ C, and M (ψ, ϕ) := sup Mz (ψ, ϕ). z∈C

Proposition 2.1. (See [10].) The operator Wψ,ϕ : F 2 → F 2 (i) is bounded if and only if ψ ∈ F 2 and M (ψ, ϕ) < ∞. In this case, ϕ(z) = az + b with |a| ≤ 1. (ii) is compact if and only if ψ ∈ F 2 and lim Mz (ψ, ϕ) = 0. In this case, ϕ(z) = az + b with |a| < 1. |z|→∞

It is worth to note that the key result that allows to prove Proposition 2.1 is the following. Lemma 2.2. (See [10].) Let ψ and ϕ be two entire functions with ψ ≡ 0. Suppose there is a constant C > 0 ¯ such that Mz (ψ, ϕ) ≤ C, ∀z ∈ C. Then ϕ(z) = az+b with |a| ≤ 1. In addition, if |a| = 1, then ψ(z) = ce−abz . The following well-known action on the kernel Kz plays an important role in the study of [10]. Lemma 2.3. Suppose that Wψ,ϕ is bounded on F 2 . Then, for every z ∈ C, ∗ Wψ,ϕ Kz = ψ(z)Kϕ(z) .

Furthermore, we also need the following formula of the adjoint operator acting on the sequence of polynomials (Pm ). Lemma 2.4. Suppose that Wψ,ϕ is bounded (i.e. ϕ(z) = az + b, with |a| ≤ 1). Then, ∗ Wψ,ϕ Pm

=

m    m j=0

j

[j]

aj ψ (m−j) (0)Kb , m ∈ N.

Proof. For an arbitrary g ∈ F 2 , we have ∗ g, Wψ,ϕ Pm = Wψ,ϕ g, Pm = ψ · g ◦ ϕ, K0 = (ψ · g ◦ ϕ)(m) (0) m    m j (j) = a g (b)ψ (m−j) (0) j j=0 [m]

m      m j (m−j) [j] a ψ (0)Kb , = g, j j=0

which implies the desired conclusion. 2 Remark 2.5. With a different approach than that used in [10] (where the role of adjoint operators is essential), in [8] the results in Proposition 2.1 are generalized to the case F p (C).

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In [8] we also gave some illustrative examples for boundedness and compactness of Wψ,ϕ when ψ, ϕ are specific functions. We recall here those results as they are used in the sequel. Proposition 2.6. (See [8].) Let ϕ(z) = az + b, ψ(z) = cedz , where a, b, c, and d are complex constants. Then (i) Wψ,ϕ is bounded if and only if (a) either |a| < 1 (b) or |a| = 1, d + ab = 0. In the last case, we always have Mz (ψ, ϕ) = |c|2 e|b| , ∀z ∈ C. 2

(2.3)

(ii) Wψ,ϕ is compact if and only if |a| < 1. In this case, we have Mz (ψ, ϕ) ≤ |c|2 e|b| e−(1−|a| 2

2

)|z|2 +2(|ab|+|d|)|z|

, ∀z ∈ C.

(2.4)

Finally, we note the following result that gives a useful relationship between convergence in the norm of F 2 and a point convergence. Lemma 2.7. (See [13].) For every f in F 2 , we have |f (z)| ≤ e

|z|2 2

f .

We refer the reader to the monograph [13] for more information on Fock spaces. Throughout the paper, we always assume that ψ is not identically zero. 3. Complex symmetry in F 2 In this section, we investigate the complex symmetry in F 2 . We start the section with some properties of ψ and ϕ when Wψ,ϕ is complex symmetric. 3.1. Some properties of complex symmetric weighted composition operators Recall that for a bounded linear operator T acting on a Hilbert space, σ(T ) and σp (T ) denote respectively the spectrum, point spectrum. Theorem 3.1. Let ϕ be a nonconstant entire function, C a conjugation on F 2 , and Wψ,ϕ a bounded weighted composition operator on F 2 with ϕ(z) = az + b, |a| ≤ 1. Suppose Wψ,ϕ is C-symmetric. Then the following assertions hold: (i) The function ψ(z) is nowhere vanished on C. (ii) There is the identity (CKu ) ◦ ϕ(z) ψ(u) = , ∀z, u ∈ C. ψ(z) CKϕ(u) (z)     (iii) The kernel ker Wψ,ϕ = {0} and the range Im Wψ,ϕ is dense in F 2 .     b (iv) The point spectrum σp Wψ,ϕ = ak ψ(λ) : k ∈ N , where λ = 1−a , if a = 1, the fixed point of the function ϕ.

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Proof. The arguments are, in general, similar to those of classical Hardy spaces. i) Assume in contrary that ψ(α) = 0 for some α ∈ C. Then there is a neighborhood U of α such that ∗ ψ(z) = 0 for every z ∈ U \ {α}. By Lemma 2.3, Wψ,ϕ Kα = ψ(α)Kϕ(α) = 0. But Wψ,ϕ is C-symmetric, hence ∗ Wψ,ϕ CKα = CWψ,ϕ Kα = 0. The last equation implies that for every z ∈ C ψ(z)CKα (ϕ(z)) = Wψ,ϕ CKα (z) = 0, which shows that CKα ◦ ϕ = 0 on U \ {α}. Consequently, CKα = 0: a contradiction. ii) For z, u ∈ C, again by Lemma 2.3, we have     ∗ ψ(u)CKϕ(u) (z) = C ψ(u)Kϕ(u) (z) = C Wψ,ϕ Ku (z) = Wψ,ϕ CKu (z) = ψ(z) · (CKu ◦ ϕ(z)) , from which the desired identity follows.   iii) Let f ∈ ker Wψ,ϕ . For every z ∈ C, we have ψ · f ◦ ϕ = Wψ,ϕ f = 0, which gives f (ϕ(z)) = 0. This   implies, since ϕ is linear and nonconstant, that f is necessarily the zero function. Thus ker Wψ,ϕ = {0}.      ∗  To prove Im Wψ,ϕ is dense in F 2 , since F 2 = Im Wψ,ϕ ⊕ ker Wψ,ϕ , it suffices to show that  ∗   ∗  ∗ ker Wψ,ϕ = {0}. Indeed, if f ∈ ker Wψ,ϕ , then Wψ,ϕ Cf = CWψ,ϕ f = 0. By the same arguments as above, f must necessarily be the zero function.     b iv) Let a = 1 and λ = 1−a . We first prove the inclusion σp Wψ,ϕ ⊆ ak ψ(λ) : k ∈ N . Take an arbitrary   μ ∈ σp Wψ,ϕ . Then there exists a nonzero function f ∈ F 2 , such that ψ(z)f (ϕ(z)) = Wψ,ϕ f (z) = μf (z),

for every z ∈ C.

(3.1)

If f (λ) = 0, then ψ(λ)f (λ) = ψ(λ)f (ϕ(λ)) = μf (λ), which implies that μ = ψ(λ). If f has a zero at λ of order q ∈ N, then differentiating (3.1) q times and evaluating it at the point z = λ yields μf (q) (λ) = ψ(λ)aq f (q) (λ), which gives μ = ψ(λ)aq .     Now we prove the inverse inclusion ak ψ(λ) : k ∈ N ⊆ σp Wψ,ϕ . [n] – If a is not a root of unity. By induction (along n), we can easily show, using Lemma 2.3, that Kλ has the form [n]

Kλ = vn + sn−1 vn−1 + · · · + s0 v0 , ∗ where vj is an eigenvector of Wψ,ϕ corresponding to the eigenvalue ψ(λ)aj , and sj are complex numbers. – If a is a p-th root of unity, then by a similar argument, [n]

Kλ = vn + sn−1 vn−1 + · · · + s0 v0 , ∗ whenever 0 ≤ n ≤ p − 1. Hence, ψ(λ)an is an eigenvalue of Wψ,ϕ , for n ≤ p − 1, and hence for all n. ∗ n In both cases, ψ(λ)a are eigenvalues of Wψ,ϕ . Since Wψ,ϕ is C-symmetric, ψ(λ)an are eigenvalues of Wψ,ϕ and the assertion (iv) is proved. 2

Remark 3.2. We note that for part (iv) in Theorem 3.1, the eigenvectors corresponding to the eigenvalues ψ(λ)an , in general, cannot be indicated. However, as it is seen below in Subsection 3.4, in case of symmetric weighted composition operators, we can give an explicit formula of those eigenvectors.

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3.2. Anti-linear weighted composition operators As noted in Introduction, a complex symmetric structure of Wψ,ϕ is investigated in [3,9] for the case when the conjugation is of the form Cf (z) = f (¯ z ). In this subsection, we study a more general situation of that conjugation. For given entire functions ξ and η, we define an anti-linear weighted composition operator Aξ,η acting on F 2 by (Aξ,η f ) (z) = ξ(z)f (η(z)). The first natural question to ask is: when does the operator Aξ,η act from the Fock space F 2 into itself? We note that by Lemma 2.7, a point evaluation is bounded on F 2 . Then F 2 is a functional Hilbert space, and so Aξ,η is F 2 -invariant if and only if it is bounded on F 2 . Following the arguments as in either [10] or [8] for linear weighted composition operators Wψ,ϕ , we can obtain a similar result for Aξ,η . Proposition 3.3. The anti-linear weighted composition operator Aξ,η is F 2 -invariant (or bounded on F 2 ) if and only if (1) ξ ∈ F 2 ; (2) M (ξ, η) < ∞. Moreover, in this case, η(z) = az + b with |a| ≤ 1, and  Mz (ξ, η) ≤ Aξ,η kη(z)  ≤ Aξ,η  , ∀z ∈ C.

(3.2)

Proof. • Suppose Aξ,η is bounded. The assertion (1) is true, because ξ = Aξ,η (1). Furthermore, for each u in C, since ku  = 1, by Lemma 2.7, we have Aξ,η  ≥ Aξ,η (ku ) ≥ | (Aξ,η ku ) (z)|e−

|z|2 2

, z ∈ C.

 In particular, with u = η(z), the last expression is precisely Mz (ξ, η), which gives (3.2). Then taking the supremum by z over C yields (2). • Conversely, suppose conditions (1) and (2) hold. By Lemma 2.2, η(z) = az + b with |a| ≤ 1. – If a = 0, i.e. η(z) = b, then by Lemma 2.7 and (1), for every f ∈ F 2 , Aξ,η (f ) = |f (b)|ξ ≤ ξe

|b|2 2

f .

– If a = 0, then by (2) and linearity of η, for every f ∈ F 2 , Aξ,η (f ) ≤  =  = where Cf (z) = f (z).

2



⎛ 1 M (ξ, η) ⎝ π

⎛ M (ξ, η) ⎝ 1 |a| π





⎞1/2 |Cf (η(z))|2 e−|η(z)| dV (z)⎠ 2

C

⎞1/2

|Cf (u)|2 e−|u| dV (u)⎠

C

M (ξ, η) Cf  = |a|

2

 M (ξ, η) f , |a|

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It turns out that in case η ≡ const, condition (1) in Theorem 3.3 can be removed. More precisely, if condition (2) is satisfied, then so is condition (1). We have the following result. Corollary 3.4. Let η be a nonconstant entire function. The operator Aξ,η : F 2 → F 2 is bounded if and only if M (ξ, η) < ∞. Proof. Suppose that M (ξ, η) < ∞. By Lemma 2.2, η(z) = az + b with |a| ≤ 1. Moreover, a = 0 because η is a nonconstant entire function. By Proposition 3.3, it suffices to show that ξ ∈ F 2 . Indeed, 1 ξ = π



2

Mz (ξ, η)e

−|az+b|2

C

M (ξ, η) dV (z) ≤ π

Since a = 0, the right-hand side is finite. This implies ξ ∈ F 2 .



e−|az+b| dV (z). 2

C

2

The next question is: under what conditions, Aξ,η is a conjugation on F 2 ? In what follows, we always assume that ξ and η are entire functions, η ≡ const, for which M (ξ, η) < ∞, that is, the anti-linear weighted composition operator Aξ,η is bounded on F 2 . To study an isometry, we first note that the operator Cf (z) = f (z), being a conjugation on the Hardy space H 2 (D), is also a conjugation in our case. Lemma 3.5. The operator Cf (z) = f (z) is a conjugation on F 2 . Proof. We can see that the operator C acts from F 2 to itself. Indeed, expending each f ∈ F 2 into the Taylor series, f (z) =



fn z n , z ∈ C,

n≥0

we have (Cf )(z) =



fn z n . Since (Pn ) is an orthogonal basis, we get

n≥0

Cf 2 =



|fn |2 Pn 2 = f 2 ,

n≥0

which shows that Cf ∈ F 2 . Furthermore, it is easy to verify that C has properties of anti-linearity, involution, and isometry. 2 Now for each b ∈ C, we define the operator Ab : F 2 → F 2 by Ab f (z) = k−b (z)f (z + b), where ku is defined in (2.2). We prove the following property of Ab . Lemma 3.6. For every b ∈ C, Ab is isometric on F 2 . Proof. We note the following equality: |k−β (α − β)|2 e−|α−β| = e−|α| , 2

Indeed, we have

2

for any α, β ∈ C.

(3.3)

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|k−β (α − β)|2 e−|α−β| = |e−2βα+|β| | · e−|α−β| 2

2

= |e−2βα+|β| | · e−|α| 2

2

2

−|β|2 +2 Re(βα)

= e−|α| . 2

Now take an arbitrary f ∈ F 2 . By changing the variable u = z + b, we have 1 Ab f  = π



2

=

=

1 π 1 π

|k−b (z)|2 · |Cf (z + b)|2 e−|z| dV (z) 2

C



|k−b (u − b)|2 |Cf (u)|2 e−|u−b| dV (u) 2

C



|Cf (u)|2 e−|u| dV (u) = Cf 2 = f 2 . 2

C

Here the last equality is followed by (3.3). 2 It turns out that for an isometry of Aξ,η , the functions η and ξ must be of special forms. Namely, we have the following characterization of isometry for anti-linear weighted composition operators on F 2 . Proposition 3.7. Suppose that Aξ,η : F 2 → F 2 is bounded. Then Aξ,η is isometric on F 2 if and only if the functions ξ and η are of the following forms: ξ(z) = ce−abz ,

η(z) = az + b,

(3.4)

where a, b, and c are complex numbers satisfying the conditions |a| = 1,

|c|2 e|b| = 1. 2

(3.5)

Proof. • Suppose that Aξ,η is isometric on F 2 . By Proposition 3.3, η(z) = az + b for some constants a and b with |a| ≤ 1. For every f ∈ F 2 , we have A−b f (z) = kb (z)f (z − b), and hence A−b f (η(z)) = kb (az + b)f (az). From this, it follows that Aξ,η A−b f (z) = ξ(z)A−b f (η(z)) ¯

= ξ(z)eabz+

|b|2 2

f (az) = ξ(z)kb (az + b)f (az), z ∈ C.

Obviously, Aξ,η A−b is linear. Moreover, since Aξ,η is isometric, by Lemma 3.6, we have Aξ,η A−b f  = A−b f  = f ,

for every f ∈ F 2 ,

and so Aξ,η A−b is isometric. Setting ζ(z) = ξ(z)kb (az + b) and χ(z) = az, we see that Aξ,η A−b = Wζ,χ , which is the linear weighted composition operator on F 2 , induced by symbols ζ and χ.

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Furthermore, since Wζ,χ is isometric on F 2 , by [10, Proposition 4.3], |a| = 1 and ζ must be a constant 2 of modulus 1. Consequently, denoting ς = ξ(z)kb (az + b), we have |ς| = 1 and ξ(z) = ςe−|b| /2 e−abz . Then 2 2 2 for c = ςe−|b| /2 , we get ς = ce|b| /2 . Since |ς| = 1, we have |ce|b| /2 | = 1. Thus, (3.4) and (3.5) hold. • Conversely, suppose that η, ξ satisfy (3.4) and (3.5). Take f ∈ F 2 , by (3.4), we have Aξ,η f 2 =

1 π



|ce−abz |2 |Cf (az + b)|2 e−|z| dV (z). 2

C

Furthermore, since |a| = 1, the changing the variables u = az yields 1 Aξ,η f  = π



2

=

=

1 π 1 π

|ce−bu |2 |Cf (u + b)|2 e−|u| dV (u) 2

C



|c|2 e|b| |e−bu−|b| 2

2

| |Cf (u + b)|2 e−|u| dV (u) 2

/2 2

C



|k−b (u)|2 |Cf (u + b)|2 e−|u| dV (u) = Ab f 2 = f 2 . 2

C

Here we note that the third equality holds by (3.5) and the formula k−b (u) = e−bu−|b| equality is followed by Lemma 3.6. 2

2

/2

, while the last

Now we study an involution of Aξ,η . Surprisingly, an involutive Aξ,η must necessarily be isometric. We have the following characterization of involution for anti-linear weighted composition operators on F 2 . Proposition 3.8. Suppose that Aξ,η : F 2 → F 2 is bounded. Then Aξ,η is involutive if and only if ξ and η are of the forms (3.4), i.e. η(z) = az + b,

¯

ξ(z) = ce−abz ,

where a, b, and c are complex numbers satisfying condition (3.5) with a ¯b + ¯b = 0, i.e. |a| = 1,

a ¯b + ¯b = 0,

|c|2 e|b| = 1. 2

(3.6)

Proof. Since Aξ,η is bounded on F 2 , η(z) = az + b with 0 < |a| ≤ 1, because η is nonconstant. • Suppose that Aξ,η is an involution, this means   ξ(z)ξ(η(z))f η(η(z)) = f (z), ∀f ∈ F 2 .

(3.7)

In particular, for f ≡ 1 ∈ F 2 , we get ξ(z)ξ(η(z)) = 1,

(3.8)

  f η(η(z)) = f (z), ∀f ∈ F 2 .

(3.9)

and hence equation (3.7) is reduced to

Now taking f (z) ≡ z in (3.9), we have

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η(η(z)) = z, ∀z ∈ C, which means, as η(z) = az + b, that |a|2 z + (¯ ab + ¯b) = z, ∀z ∈ C, and so |a| = 1 and a ¯b + ¯b = 0. ¯ Since |a| = 1, by Lemma 2.2, we have ξ(z) = ce−abz , c ∈ C. But a ¯b + ¯b = 0, and so we can write 2 bz 2 |b| ξ(z) = ce . Then equation (3.8) gives |c| e = 1. • Conversely, suppose that η, ξ satisfy (3.4) and (3.6). Then we have 2

ξ(z)ξ(η(z)) = |c|2 e(ab+b)z+|b| = 1, and η(η(z)) = |a|2 z + ab + b = z, from which (3.7) follows. 2 As an immediate consequence of Propositions 3.7 and 3.8, we obtain the following result. Corollary 3.9. Suppose that Aξ,η : F 2 → F 2 is bounded. If Aξ,η is involutive then it is isometric. Remark 3.10. The inverse direction of Corollary 3.9 fails to hold, as there are infinitely many triples {a, b, c}, 2 for which (3.5) holds (|a| = |c|2 e|b| = 1), but (3.6) is not satisfied (a¯b + b = 0). Combining Propositions 3.7 and 3.8 yields the following important criterion. Theorem 3.11. Let ξ and η be entire functions, with η ≡ const, such that the anti-linear weighted composition operator Aξ,η is bounded on F 2 . Then this Aξ,η represents a conjugation on F 2 if and only if η(z) = az + b,

¯

ξ(z) = ce−abz ,

where a, b, and c are complex numbers satisfying the conditions |a| = 1,

a ¯b + ¯b = 0,

|c|2 e|b| = 1. 2

Proof. Suppose that Aξ,η is a conjugation on F 2 . Since it is involutive (and isometric), by Proposition 3.8, the functions η and ξ are of the form (3.4) with conditions (3.6). Conversely, suppose η and ξ satisfy conditions (3.4) and (3.6). Again by Proposition 3.8, Aξ,η is involutive, which is also isometric, by Proposition 3.7. So Aξ,η is a conjugation. 2 ¯b + ¯b = 0, which is equivalent to a¯b + b = 0, the Remark 3.12. It is worth to mention that by the condition a formula for the symbol ξ can also be written as ¯

ξ(z) = ce−abz = cebz . In what follows, the weighted composition conjugation on F 2 is denoted by Ca,b,c , that is   Ca,b,c f (z) := cebz f az + b , where a, b, c are constants satisfying (3.6).

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Remark 3.13. It is clear that with b = 0, a = c = 1, the conjugation C1,0,1 is precisely the one considered in [3,9]. 3.3. Ca,b,c -symmetric weighted composition operators In this subsection, we consider bounded weighted composition operators and study under what conditions, these operators are Ca,b,c -symmetric. As we see below, in this case, the function ψ can be precisely computed. Note the set of all polynomials is dense in F 2 , and hence if the operator is Ca,b,c -symmetric on polynomials, then it is Ca,b,c -symmetric on the whole F 2 . To study necessary conditions for Ca,b,c -symmetry, we apply the symmetric condition to polynomials (Pm ) defined in (2.1). It turns out that the action on only P0 can already give the sufficient conditions. Proposition 3.14. Let Ca,b,c be a weighted composition conjugation on F 2 , ψ and ϕ be two entire functions with ψ ≡ 0. Suppose that Wψ,ϕ is bounded on F 2 , that is, ϕ(z) = Az + B with |A| ≤ 1. (1) If Wψ,ϕ is Ca,b,c -symmetric w.r.t. P0 , i.e. ∗ Wψ,ϕ Ca,b,c P0 = Ca,b,c Wψ,ϕ P0 ,

then the function ψ has the form ψ(z) = CeDz ,

with C = 0, D = aB − bA + b.

(2) Conversely, if the function ψ satisfies the form above, then ∗ Wψ,ϕ Ca,b,c Pm = Ca,b,c Wψ,ϕ Pm , ∀m ∈ N.

Proof. (1) Since P0 (z) ≡ 1, we have Ca,b,c P0 (z) = cebz and hence Wψ,ϕ Ca,b,c P0 (z) = ψ(z)cebϕ(z) = ψ(z)ceb(Az+B) , z ∈ C. ∗ On the other hand, by Lemma 2.4, Wψ,ϕ P0 = ψ(0)Kϕ(0) , which gives ∗ Ca,b,c Wψ,ϕ P0 (z) = cebz ψ(0)eϕ(0)(az+b) = cebz ψ(0)eB(az+b) , z ∈ C. ∗ Therefore, the equality Wψ,ϕ Ca,b,c P0 = Ca,b,c Wψ,ϕ P0 is reduced to

ψ(z) = ψ(0)e(aB−bA+b)z = CeDz , where C = ψ(0) = 0 and D = aB − bA + b. (2) Conversely, for each m ∈ N, by Lemma 2.4, we have ∗ Wψ,ϕ Pm (z)

=

m    m j=0

Then

j

Aj CDm−j z j eBz .

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∗ Ca,b,c Wψ,ϕ Pm (z) = cebz

j=0

j

Aj CDm−j (az + b)j eB(az+b)

= Cce(b+aB)z+bB

m    m j=0

(b+aB)z+bB

= Cce

j

Aj Dm−j (az + b)j

[A(az + b) + D]m

= CeDz · ceb(Az+B) [a(Az + B) + b]m = ψ(z) · cebϕ(z) [aϕ(z) + b]m = Wψ,ϕ Ca,b,c Pm (z).

2

Now with the help of illustrative examples in Section 2, we can establish the following criterion for bounded weighted composition operators to be symmetric. Theorem 3.15. Let Ca,b,c be a weighted composition conjugation on F 2 , ψ and ϕ be two entire functions with ψ ≡ 0. Let further, Wψ,ϕ be bounded on F 2 (i.e., ϕ(z) = Az + B with |A| ≤ 1). Then Wψ,ϕ is Ca,b,c -symmetric if and only if the following conditions hold: ψ(z) = CeDz , with C = 0, D = aB − bA + b,  either |A| < 1 or |A| = 1, D + AB = 0.

(3.10) (3.11)

Proof. Note that since the conjugation is involutive, the equality T = CT ∗ C (complex symmetry) can be rewritten as CT = T ∗ C. ∗ • Suppose Wψ,ϕ is Ca,b,c -symmetric on F 2 . This implies, in particular, that Ca,b,c Wψ,ϕ P0 = Wψ,ϕ Ca,b,c P0 . Hence, by Proposition 3.14 (1), ψ must be of the form (3.10). Furthermore, since Wψ,ϕ is bounded and ψ(z) = CeDz , by Proposition 2.6 (i), we get (3.11). • Conversely, suppose ψ and ϕ satisfy (3.10)–(3.11). By Proposition 3.14, we have ∗ Wψ,ϕ Ca,b,c Pm = Ca,b,c Wψ,ϕ Pm .

Since the set of all polynomials is dense in F 2 and Wψ,ϕ is bounded, we get ∗ Wψ,ϕ Ca,b,c f = Ca,b,c Wψ,ϕ f, ∀f ∈ F 2 .

That this, Wψ,ϕ is Ca,b,c -symmetric.

2

3.4. Spectral properties of Ca,b,c -symmetric operators In this subsection we provide the explicit formula for eigenvectors of Wψ,ϕ when ϕ has one fixed point B (compare with Remark 3.2). This happens if and only if A = 1, and the fixed point is λ = 1−A . Theorem 3.16. Let Wψ,ϕ be a Ca,b,c -symmetric operator on F 2 , induced by entire functions ϕ, ψ (that is ϕ(z) = Az + B, with |A| ≤ 1 and (3.10)–(3.11) hold). Then gk (z) = (z − λ)k eDz/(1−A)

(k = 0, 1, 2, . . .)

are eigenvectors of Wψ,ϕ corresponding to the eigenvalues Ak ψ(λ).

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Proof. As we have proved in Theorem 3.1, the point spectrum of Wψ,ϕ is σp (Wψ,ϕ ) = {Ak ψ(λ) : k ∈ N}. To prove the theorem, we first note that gk ∈ F 2 , for each k ≥ 0. Then by Theorem 3.15, we have Wψ,ϕ gk (z) = CeDz · (Az + B − λ)k e

D(Az+B) 1−A

= CeDz · (Az + B − λ)k e−Dz+ 1−A + 1−A Dz

DB

Dz

= ψ(λ) · (Az + B − λ)k e 1−A . Since λ = Aλ + B, the last expression is Dz

ψ(λ)(Az + B − Aλ − B)k e 1−A = Ak ψ(λ)gk (z).

2

Our next task is to determine when a Ca,b,c -symmetric Wψ,ϕ belongs to a Schatten class. Recall that a bounded linear operator T acting on a Hilbert space H is said to be in the Schatten class Sp (0 < p < ∞), if its singular value sequence {λn } belonging to p . The singular values or s-numbers of a bounded linear operator T on H are defined by λn = inf T − S,

n ≥ 1,

where the infimum is taken over all operators S with rank at most n − 1. If√T happens to be compact, then {λn } is the sequence of eigenvalues of the positive compact operator |T | = T ∗ T . Note that S1 is the space of trace class operators, while S2 is the Hilbert space of Hilbert-Schmidt operators (see, e.g., [7,11]). The following result gives a characterization for a Schatten class membership of a Ca,b,c -symmetric weighted composition operator. Theorem 3.17. Let Wψ,ϕ be a Ca,b,c -symmetric operator on F 2 , induced by entire functions ϕ, ψ (that is ϕ(z) = Az + B, with |A| ≤ 1 and (3.10)–(3.11) hold). The following assertions are equivalent: (1) Wψ,ϕ is compact. (2) Wψ,ϕ belongs to the Schatten class Sp for every p > 0. Proof. • (1) =⇒ (2). Suppose Wψ,ϕ is compact. By Proposition 2.6 (ii), |A| < 1 and for each z ∈ C, Mz (ψ, ϕ) = |ψ(z)|2 e|ϕ(z)|

2

−|z|2

≤ |C|2 e|B| e−(1−|A| 2

2

)|z|2 +2(|AB|+|D|)|z|

.

From this, it follows that  Mz (ψ, ϕ)p/2 dV (z) < ∞, ∀p > 0,

(3.12)

C

which, by [2, Theorem 2.3], is equivalent to the fact that Wψ,ϕ ∈ Sp for every p > 0. • (2) =⇒ (1). Suppose Wψ,ϕ belongs to the Schatten class Sp for every p > 0. Assume in a contrary that 2 |A| = 1. By (3.11), D + AB = 0, which in turn, by Proposition 2.6 (i), implies that Mz (ψ, ϕ) = |C|2 e|B| . But in this case, condition (3.12) does not hold: a contradiction. 2 Our next task is to compute the spectrum of the symmetric Wψ,ϕ . Remind, by Theorem 3.1, that the point spectrum of Wψ,ϕ is σp (Wψ,ϕ ) = {Ak ψ(λ) : k ∈ N}. Theorem 3.18. Let Wψ,ϕ be a Ca,b,c -symmetric operator on F 2 , induced by entire functions ϕ, ψ (that is ϕ(z) = Az + B, with |A| ≤ 1 and (3.10)–(3.11) hold). The following assertions are true:

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(1) If |A| < 1 then σ(Wψ,ϕ ) = σp (Wψ,ϕ ) ∪ {0}. (2) If |A| = 1 and A = 1, then σ(Wψ,ϕ ) = {Ak ψ(λ) : k ∈ N}. 2 (3) If A = 1, then σ(Wψ,ϕ ) = Ce|B| /2 T, where T is the unit circle. Here, λ =

B 1−A ,

with A = 1.

Proof. First note that (1) is true, because due to |A| < 1, Wψ,ϕ is compact. For (2) and (3), since |A| = 1, from (3.11) it follows that D = −AB. Then, Wψ,ϕ f (z) = Ce−ABz f (Az + B), f ∈ F 2 . Set φ(z) = e−ABz−|B|

2

/2

. Then Wψ,ϕ = Ce|B|

2

/2

Wφ,ϕ , and this gives

σ(Wψ,ϕ ) = Ce|B|

2

/2

σ(Wφ,ϕ ).

By [12, Corollary 1.4], Wφ,ϕ is unitary and ⎧  A+1 ⎨ Ak e |B|2 2 · A−1 :k∈N , σ(Wφ,ϕ ) = ⎩T,

A = 1, A = 1.

Consequently, if A = 1, then σ(Wφ,ϕ ) = Ce|B| /2 T, while if A = 1, then σ(Wφ,ϕ ) = Ce|B| /2 ×   |B|2 A+1 |B|2 A+1 |B|2 A ABB 2 Ak e 2 · A−1 : k ∈ N = {Ak ψ(λ) : k ∈ N}, because e|B| /2 e 2 · A−1 = e A−1 = e A−1 = e−ABλ = 2

2

eDλ . 2 We end up the present paper with normality problem. As mentioned in Introduction, every normal operator is complex symmetric with some conjugation. A natural question to ask is whether Ca,b,c-symmetric weighted composition operators are normal. We have the following criterion for normality of such operators. Proposition 3.19. Let Wψ,ϕ be a Ca,b,c -symmetric operator on F 2 , induced by entire functions ϕ, ψ (that is ϕ(z) = Az + B, with |A| ≤ 1 and (3.10)–(3.11) hold). Then Wψ,ϕ is normal if and only if  D=

B 1−A , 1−A

A = 1,

−B,

A = 1.

(3.13)

Proof. By definition, the operator Wψ,ϕ is normal if and only if ∗ ∗ Wψ,ϕ Wψ,ϕ f = Wψ,ϕ Wψ,ϕ f,

for all f ∈ F 2 .

Since Wψ,ϕ is bounded, and the span of the kernel functions is dense in F 2 , the equality above occurs if and only if ∗ ∗ Wψ,ϕ Wψ,ϕ Ku (z) = Wψ,ϕ Wψ,ϕ Ku (z),

On one hand, by Lemma 2.3, we have

for all u, z ∈ C.

(3.14)

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∗ Wψ,ϕ Wψ,ϕ Ku (z) = Wψ,ϕ (ψ(u)Kϕ(u) )(z)

= ψ(u)ψ(z)eϕ(u)ϕ(z) = |C|2 e|B| e(D+AB)u+|A| 2

2

uz+(D+AB)z

.

On the other hand, since Wψ,ϕ Ku (z) = CeDz eu(Az+B) = CeBu e(D+Au)z = CeBu KD+Au (z), by Lemma 2.3, we also have ∗ Wψ,ϕ Wψ,ϕ Ku (z) = CeBu ψ(D + Au)Kϕ(D+Au) (z)

= |C|2 e|D| e(B+AD)u K|A|2 u+AD+B (z) 2

= |C|2 e|D| e(DA+B)u+|A| 2

2

uz+(AD+B)z

.

So the equation (3.14) is reduced to ⎧ 2 2 ⎪ ⎪ ⎨|B| = |D| D + AB = DA + B ⎪ ⎪ ⎩D + AB = AD + B

 ⇐⇒

|B| = |D| AD + B = D + AB.

The last system, due to condition (3.11), is equivalent to (3.13).

2

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