Weighted superposition operators on Fock spaces

We characterize all pairs of entire functions (u,ψ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(u,\psi )$$\end{document} for which the induced weighted superposition operator S(u,ψ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S_{(u,\psi )}$$\end{document} transforms one Fock space into another Fock space. Further analytical structures like boundedness and Lipschitz continuity of S(u,ψ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S_{(u,\psi )}$$\end{document} are described. We, in particular, show the Fock spaces support no compact weighted superposition operator.


Introduction
The theory of superposition operator has a long history in the context of real valued functions [2]. In contrast, there have been only some studies on spaces of analytic functions which includes Hardy spaces [6], Bergman spaces [7], Dirichlet type spaces [4], Bloch type spaces [1,5], and some weighted Banach spaces over the disc [3]. The goal of this note is to study the operator on the Fock spaces F p . We recall that F p is the space of entire functions f for which where d A is the usual Lebesgue area measure on the complex plane C. The space F 2 is a reproducing kernel Hilbert space with kernel function K w (z) = e wz . For each w ∈ C, a calculation shows the function k w = K w −1 2 K w ∈ F p and k w p = 1 for all p. By [10,p. B where D(z, 1) is the disc with center z and radius 1. The same conclusion with the factor 2π/ p 1/ p replaced by 1 holds whenever p = ∞.
In this note, we study the weighted superposition operator on Fock spaces where the operator S ψ is covered as a particular case. Let (u, ψ) be a pair of holomorphic functions on C. For two metric spaces H 1 and H 2 , the weighted superposition operator S (u,ψ) : and M u f = u f are respectively the superposition and multiplication operators. Recently, S (u,ψ) was studied on the Bergman and Bloch spaces [9].
The first main question now is to identify which pairs of analytic symbols (u, ψ) define weighted superposition operators S (u,ψ) from F p into F q . The questions for S (u,ψ) are in general technically more difficult than the corresponding questions for S ψ since the presence of the multiplier u can complicate proofs and arguments. We may first begin with a simple example that illustrates the problem at hand. Let f = α be a constant and g(z) = z. Then, for 0 < q ≤ ∞ and a non-zero ψ, This shows that if S (u,ψ) maps F p into F q , then both u and uψ belong to F q . On the other hand, set for example q = 2 and consider the function h(z) = sin( z 2 2 )/z 2 . Then h is an entire function which belongs to F 2 . To see this, observe that when |z| = r gets larger, then |h(z)| 2 e r 2 /r 4 and However, the function zh is not in F 2 since |zh(z)| 2 re −r 2 1/r for larger |z|. It follows that S (z,z) fails to map F 2 into itself while it is easy to see that S z does. This exhibits the existence of some degree of interplay between u and ψ in defining S (u,ψ) on Fock spaces. Our next main result provides their precise interplay. Theorem 1.1 Let ψ and u be nonzero entire functions on C, and 0 < p, q ≤ ∞.
(i) If p ≤ q, then the following statements are equivalent.
for some a, b ∈ C and u is a constant or ψ is a constant and u ∈ F q . If u is in addition non-vanishing, then u(z) = u(0)e a 1 z+a 2 z 2 , a 1 , a 2 ∈ C and |a 2 | < 1/2; (ii) If p > q, then the following statements are equivalent.
(a) S (u,ψ) maps F p into F q ; (b) ψ is a constant and u ∈ F q . If u is in addition non-vanishing, then the representation in (1.2) holds; (iii) The map S (u,ψ) : F p → F q cannot be compact for any pair of p and q.
The proof of Theorem 1.1 follows once we prove Lemma 2.2, Theorem 2.3, and Theorem 2.4 in the next section.
The operator S (u,ψ) reduces to S ψ and M u when u = 1 and ψ(z) = z respectively. Consequently, we record the following special cases of Theorem 1.1.

Corollary 1.2
Let ψ be a nonzero entire function on C and 0 < p, q ≤ ∞.
(i) If p ≤ q, then the statements: and S ψ : F p → F q is globally Lipschitz continuous, are all equivalent. (iii) S ψ : F p → F q cannot be compact for any pair of p and q.
Given the fact that F p properly contains the space F q when q < p [10,Theorem 2.10], it should be clear that a superposition from the former to the latter is possible only via constant functions.

cannot be compact for any pair of p and q.
We close this section with a word on notation. The notion U (z)

Proof of the results
Let us first consider the following necessity condition about the superposition operator S ψ . The proof of the lemma demonstrates an important test function that will be used in the proof of the main result.

Lemma 2.1 Let ψ be nonzero entire function on
Proof To prove the assertion, we need to choose test function from F p for any fixed p. Thus, for c with 1 4 < c < 1 2 , consider the function In view of the estimate, the functions f c ∈ F p . Moreover, the Fock space norm is invariant under rotations so if |λ| = |μ| = 1, then all functions f λ,μ,c (z) = λe cμz 2 ∈ F p and have the same norm as f c which can be seen by making change of variables τ = √ μz in the integration for p = ∞ and in estimating the supremum norm for p = ∞. By the same reasoning, we also observe that Now for any w = 0, we can find z, λ, μ ∈ C with |λ| = |μ| = 1 and w = λe cμz 2 and also that cμz 2 is real and positive which implies w/λ = λw. Clearly, for large w, we have Therefore, using the estimate above and in (1.1) for sufficiently large w.
Since S ψ f c q is a constant value and 1/2c < 2, the standard Cauchy estimates imply that ψ is a polynomial of degree at most one as desired.
Next, we recall the notion of order and type of an entire function and prove one more important lemma. Let f be an entire function, and If 0 < ρ < ∞, then the type of f is given by By definition, if a function f has order ρ, then for every > 0 when r → ∞. This clearly shows that any function of order less than 2 belongs to all the Fock spaces F p . Conversely, by [10,Theorem 2.12], every function in F p has order at most 2, and if its order is exactly 2, it must be of type less than or equal to 1/2. In the next lemma, we prove that if the function has no zeros, then its type can not be 1/2.

Lemma 2.2 Let 0 < p ≤ ∞ and u is a non-vanishing function in F p . Then
Proof We argue as follows. Since u ∈ F p , by [10,Theorem 2.12], ρ(u) ≤ 2 and if ρ(u) = 2, it must be of type less than or equal to 1/2. On the other hand, since u is non-vanishing, it follows from the Hadamard Factorisation Theorem that u(z) = e a 0 +a 1 z+a 2 z 2 = u(0)e a 1 z+a 2 z 2 (2.2) where a 0 , a 1 , a 2 ∈ C and |a 2 | ≤ 1/2. It remains to show that |a 2 | = 1/2 can not happen.
Assuming to the contrary, we may simply set a 2 = 1/2 (if not since the Fock space norm is invariant under rotation, we can find a μ with |μ| = 1 such that μa 2 = 1/2 ). Then for as the first integral with respect to x diverges while the second integral with respect to y converges which contradicts the assumption that u ∈ F p . If p = ∞, we replace the above integral argument with supremum to arrive at the same conclusion.
For the sake of better exposition, we split Theorem 1.1 into two theorems below.

Proof of part (i)
The sufficiency of the condition is easy to verify. It is the necessity that may require a new techniques which we present below. We shall first prove that if S (u,ψ) maps F p into F q , then ψ(z) = az + b for some a, b ∈ C. The assumption implies u ∈ F q as already seen before. To this end, if u is non-vanishing, by Lemma 2.2 u(z) = u(0)e a 1 z+a 2 z 2 , a 1 , a 2 ∈ C and |a 2 | < 1/2.

(2.4)
Suppose for the purpose of contradiction that ψ is not linear. Then there exists a sequence w n ∈ C such that |w n | → ∞ as n → ∞ and |ψ(w n )| ≥ n|w n | 2 for n ∈ N. Arguing as in the proof of Lemma 2.1, for each w n = 0, we can find z n , λ n , μ n ∈ C with |λ n | = |μ n | = 1 and w n = λ n e cμ n z 2 n (2.5) and also that cμ n z 2 n is real and positive. We may further pick a sparse subsequence of z n such that the discs D(z n , 1) are mutually disjoint. Now, for q < ∞, applying the operator to the sequence f c,λ n ,μ n (z) = λ n e cμ n z 2 and eventually invoking the estimate in (1.1), We consider two separate cases depending on whether the multiplier function u has zeros or not in C. If u is non-vanishing, by (2.4) and (2.6) |e a 0 +a 1 z n +a 2 z 2 n | q |e 2cqμ n z 2 Now, if a 2 = 0, then we may chose c in the interval 3 8 , 1 2 ) such that 2c − 1 2 > 0. Thus, the last sum in (2.6) diverges. On the other hand, if 0 < |a 2 |, then we may chose c in 1 4 and hence the sum in (2.6) still diverges. This is a contradiction and hence ψ(z) = az + b for some a, b ∈ C in this case. Next, assume that u has zeros and analyze the case when the sequence z n in the sum (2.6) could belong to the zero set of u. Let us for example assume that u is a polynomial. Then there exist positive constants C and R such that for |u(z)| ≥ C for |z| ≥ R. It follows that By Lemma 2.1, the last integral above is finite only when ψ(z) = az + b. Now if infinitely many z n s belong to the zero set of u such that last sum in (2.6) converges, then we keep changing our choice of the constant c in the interval (1/4, 1/2) above until the convergence is not possible. We explain how this can done below. From (2.5) and since c has uncountably many possible values in the said interval, we collect all these values to see that is uncountable. As known, an entire function can not have uncountable zero set. Taking this into account, we can choice the sequence z n in such a way that it does not belong to the zero set of u when n → ∞. Now, if u has order less than 2, then every choice of c in the interval (1/4, 1/2) gives that the sum in (2.6) diverges. On the other hand, if ρ(u) = 2 and its type, τ (u), is less than 1/2, then we can still choice c in the interval ( τ (u) 2 + 1 4 , 1 2 ) such that sum in (2.6) diverges. Thus, it remains to consider the extremal case when |u(z n )| e − 1 2 |z n | 2 (2.8) as n → ∞. To this end, setting the expression in (2.8) holds for all c in (1/4, 1/2). Letting c → 1 2 , we note that the sum in (2.9) converges only when the sequence {n q :∈ N} is summable, which is a contradiction again, and hence ψ(z) = az+b for some constants a, b ∈ C.
Next, we set ψ(z) = az + b and show that u is necessarily a constant function whenever a = 0. Aiming to argue in the contrary, assume that u is not a constant. Then we can pick a sparse sequence w n such that |u(w n )| → ∞ as n → ∞ and that the discs D(w n , 1) are mutually disjoint. For each positive , observe that the function g (z) = e z 2 2+ ∈ F p for all p. Then we can find a sequence μ n such that |μ n | = 1 and μ n w 2 n are real and positive for sufficiently large n. Setting h ,μ n (z) = e μn z 2 2+ − b a and eventually applying (1.1) for all > 0. Letting → 0, we observe that the sum diverges, which is again a contradiction. For q = ∞, we may simply replace the integral above by the supremum norm and argue to arrive at the same conclusion. This completes the proof of part (i).

Proof of part (ii)
The sufficiency of the condition is clear again. Let p > q. Then as already proved above in the first part, ψ(z) = az + b is a necessary condition for S (u,ψ) to map F p into F q independent of the size of p and q. Assume that a = 0 and for any f ∈ F p , consider the Then, our conclusion follows from the fact that F p \F q is non-empty [10,Theorem 2.10] and |a||α| = 0 and completes the proof of Theorem 2.3.

Theorem 2.4
Let ψ and u be nonzero entire functions on C and 0 < p, q ≤ ∞. If S (u,ψ) maps F p into F q , then it is bounded and globally Lipschitz continuous. But S (u,ψ) cannot be compact.
Proof Clearly, f ψ is a constant and u ∈ F q , then S (u,ψ) : F p → F q is bounded. Thus, lets dispose the sufficiency of the condition when ψ(z) = az + b and u = α. Consider f ∈ F p and compute where the second inequality follows from [10,Theorem 2.10].
Next, let us show that the spaces support no compact weighted superposition operators. Aiming to arrive at a contradiction, let S (u,ψ) : F p → F q be compact and hence ψ(z) = az+b for some complex number a, b and u = α or ψ = b and u ∈ F q . The sequence k w ∈ F p is bounded and converges to zero uniformly on compact subsets of C when |w| → ∞. Applying the operator to k w and eventually (1.1) S (u,ψ) k w q ≥ |ak w (w) + b||α|e −|w| 2 /2 = |a + be −|w| 2 /2 ||α| and S (u,ψ) k w q → 0 as |w| → ∞ only when a = 0. It follows that If a = 0 and u ∈ F q is non-zero, the same conclusion follows easily. It remains to show that S (u,ψ) : F p → F q is globally Lipschitz continuity. Since the case for p > q is trivial, we assume p ≤ q. An application of Theorem 2.3 gives that ψ(z) = az + b and u = α = 0 or ψ = b and u ∈ F q . If a = 0, then the conclusion follows easily. Thus, assume a = 0 and hence u = α = 0. For f , g ∈ F p where the last inequality follows from the inclusion property. This shows that S (u,ψ) is globally Lipschitz continuous with constant |aα|.

Remark
In the remainder of this section, we present alternative proofs for the necessity in Theorem 2.3 when The proof is interest of its own as it shows how the zero sets and uniqueness sets in Fock spaces play important roles in the study of weighted superposition operators. Unfortunately, a complete characterization of such sets are still an open problem as far as we know. Some necessary and sufficient conditions can be read in [10,Chapter 5].
(i) We consider the square lattice in the complex plane where Z denotes the set of all integers. As known the Weierstrass σ -function associated to is defined by Furthermore, it is known that is the zero set for σ and by [10,Lemma 5.6], σ ∈ F ∞ but not in any of the other Fock spaces F p . Aiming to argue in the contrary, suppose now that u = 0 and ψ is not a constant. Since S (u,ψ) maps F ∞ into F q , the function Now since ψ is not a constant, the function F is non-zero and vanishes on . On the other hand, by [10,Lemma 5.7], is a uniqueness set for F q for all q = ∞ which implies that F = 0. This is contradicts the fact that F = 0. Therefore, ψ is a constant.
(ii) We consider the set = − {0}. As shown in [10,p.204], is a uniqueness set for F 2 . Indeed, the function g(z) = σ (z)/z belongs to F p if and only if p > 2. Now to argue as above, assume ψ is not a constant and hence Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.