Nuclear Toeplitz operators between Fock spaces

Tengfei Ma; Yufeng Lu; Chao Zu

doi:10.4153/S0008439526102227

Nuclear Toeplitz operators between Fock spaces

Part of: Special classes of linear operators

Published online by Cambridge University Press: 07 July 2026

Tengfei Ma ,

Yufeng Lu and

Chao Zu

Show author details

Tengfei Ma: Affiliation:
School of Mathematics Sciences, Dalian University of Technology , China e-mail: 503304283@qq.com lyfdlut@dlut.edu.cn
Yufeng Lu: Affiliation:
School of Mathematics Sciences, Dalian University of Technology , China e-mail: 503304283@qq.com lyfdlut@dlut.edu.cn
Chao Zu*: Affiliation:
School of Mathematics Sciences, Dalian University of Technology , China e-mail: 503304283@qq.com lyfdlut@dlut.edu.cn
*: e-mail: zuchao@dlut.edu.cn

Article contents

Abstract
Introduction
Preliminary
Nuclear Toeplitz operator
Trace formula and density theorem
References

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Abstract

We characterize the nuclearity of Toeplitz operators $T_\mu : F_\alpha ^p \to F_\alpha ^q$ with Borel measure symbols for $1\leq p,q\leq \infty $. For positive measures $\mu $ and $q\leq p$, we provide necessary and sufficient conditions in terms of the Berezin transform and establish a rigidity property for nuclearity across this range. In the case $p<q$, we obtain separate necessary and sufficient conditions, indicating that the Berezin transform alone is insufficient for a complete characterization. Our results extend to Fock spaces on $\mathbb {C}^n$.

Keywords

Nuclear operator Toeplitz operator Fock space

MSC classification

Primary: 47B10: Operators belonging to operator ideals (nuclear, $p$-summing, in the Schatten-von Neumann classes, etc.)

Secondary: 47B35: Toeplitz operators, Hankel operators, Wiener-Hopf operators

Information

Type: Article
Information: Canadian Mathematical Bulletin , First View , pp. 1 - 18

DOI: https://doi.org/10.4153/S0008439526102227 [Opens in a new window]
Copyright: © The Author(s), 2026. Published by Cambridge University Press on behalf of Canadian Mathematical Society

1 Introduction

Toeplitz operators on Fock spaces constitute a fundamental class of non-selfadjoint operators in complex analysis and operator theory and have attracted a great deal of interest. The theory of Toeplitz operators on Fock spaces has been extensively developed in the Hilbert space setting. For the classical Fock space $F_\alpha ^2$ , Toeplitz operators with bounded or measure-valued symbols have been studied from various perspectives, including boundedness, compactness, and Schatten class membership (see references [Reference Bauer, Coburn and Isralowitz1–Reference Berger and Coburn3, Reference Fulsche7]). In particular, Isralowitz and Zhu [Reference Isralowitz and Zhu11] obtained complete characterizations of boundedness, compactness, and Schatten class properties of Toeplitz operators with positive Borel measure symbols.

Beyond the Hilbert space situation, Hu and Lv [Reference Hu and Lv9, Reference Hu and Lv10] systematically investigated Toeplitz operators acting between different Fock spaces $F_{\alpha }^{p}$ and $F_{\alpha }^{q}$ for $0< p,q < \infty $ . They characterized bounded and compact Toeplitz operators with positive Borel measure symbols in terms of averaging functions and Berezin-type transforms. A notable feature of their results is a rigidity phenomenon specific to Fock spaces: boundedness (or compactness) from $F_{\alpha }^{p}$ to $F_{\alpha }^{q}$ for some $p \leq q$ automatically implies boundedness (or compactness) for all such q. The groundbreaking work of Hu and Lv has stimulated extensive scholarly investigations into Toeplitz and Hankel operators across diverse weighted Fock spaces [Reference Lu and Lv13, Reference Mengestie14, Reference Zhu17].

In contrast to the well-developed theory of bounded and compact Toeplitz operators, considerably less is known about Toeplitz operators belonging to finer operator ideals on Fock spaces, especially in the non-Hilbertian case. Among these ideals, the class of nuclear operators plays a distinguished role in Banach space operator theory. The aim of this article is to study nuclear Toeplitz operators acting between Fock spaces. Our first main result is presented as follows.

Theorem 1.1 Let $1\leq q\leq p<\infty , 1\leq s\leq \infty $ , and let $\mu \geq 0$ satisfy (2.1); then the following statements are equivalent:

(1) $T_\mu \in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ ;
(2) $T_\mu \in \mathcal {N}(f_\alpha ^\infty ,F_\alpha ^s)$ ;
(3) $\mu (\mathbb {C})<\infty $ .

Moreover,

(1.1) $$ \begin{align} \|T_\mu\|_{\mathcal{N}}\simeq\mu(\mathbb{C}). \end{align} $$

Since nuclear and trace-class operators coincide on Hilbert spaces, the case ${p=q=2}$ in our results recovers Zhu’s characterization of trace-class Toeplitz operators in [Reference Isralowitz and Zhu11]. Notice that Zhu’s proof relies heavily on Hilbert space structure, including orthogonal decompositions, positivity, and the classical Schatten class theory. In contrast, when $p \neq 2$ , the Fock space $F^p_\alpha $ is a Banach space lacking Hilbert geometry. The standard techniques used in the Hilbert case do not extend directly. Thus, the characterization of nuclear Toeplitz operators between Banach Fock spaces requires a different approach based on operator ideals. We emphasize that our method provides a more constructive approach to approximating $T_\mu $ with nuclear operators compared to Zhu’s proof.

Our main results provide a detailed description of nuclear Toeplitz operators in different parameter regimes. When $ 1\leq q\leq p<\infty $ and $ \mu \geq 0 $ , we obtain necessary and sufficient conditions for $ T_{\mu } $ to be nuclear in terms of the integrability of the Berezin transform of $ \mu $ . As a consequence, we show that nuclearity in this range exhibits a rigidity property analogous to that observed for boundedness: if $ T_{\mu } $ is nuclear from $ F_{\alpha }^{p} $ to $ F_{\alpha }^{q} $ for some $ q\leq p $ , then it is nuclear for all such $ q $ .

When $ p<q $ , the situation becomes more subtle. In this case, we establish separate necessary and sufficient conditions for nuclearity, showing that the Berezin transform alone does not yield a complete characterization. This reflects an inherent asymmetry between the domain and target spaces in the nuclear setting and distinguishes the nuclear theory from previously studied mapping properties of Toeplitz operators on Fock spaces. We will discuss this problem in Section 3.

Berger and Coburn [Reference Berger and Coburn3] showed that the set of all Toeplitz operators $T_\varphi $ with compact support continuous symbol is trace-norm dense in the trace ideal of $F_\alpha ^2$ and is operator-norm dense in $\mathcal {K}(F_\alpha ^2)$ . Our second main result generalizes the aforementioned finding.

Theorem 1.2 Let C denote the set of all Toeplitz operators $T_{\varphi }$ with continuous, compactly supported symbols $\varphi $ on $\mathbb {C}$ , and let $1<p,q<\infty $ . Then:

(1) C is dense in $\mathcal {N}(F^{p}_{\alpha },F^{q}_{\alpha })$ with respect to the nuclear norm.
(2) C is norm-dense in $\mathcal {K}(F^{p}_{\alpha },F^{q}_{\alpha })$ .

The proofs rely on tools from Banach space operator theory, including the duality properties of nuclear operators and the approximation and Radon–Nikodým properties of Fock spaces, which we will present in Section 2.

The article is organized as follows. In Section 2, we recall basic definitions and preliminary results concerning nuclear operators and Fock spaces. Section 3 is devoted to the characterization of nuclear Toeplitz operators between Fock spaces and contains our main results. In Section 4, we present several trace formulas and establish the density theorem.

2 Preliminary

2.1 Fock space

Let $ \alpha>0 $ and $ p>0 $ , the notation $ L^{p}_{\alpha } $ refers to the space of all Lebesgue measurable functions $ f $ on $ \mathbb {C} $ for which $ f(z)e^{-\alpha |z|^{2}/2}\in L^{p}(\mathbb {C},\mathrm {d}A) $ , where $ \mathrm {d}A $ is the Lebesgue area measure on $ \mathbb {C} $ . For $ f\in L^{p}_{\alpha } $ , its norm is given by

$$ \begin{align*}\|f\|_{\alpha,p}^{p}=\frac{p\alpha}{2\pi}\int_{\mathbb{C}}\left|f(z)e^{-\frac{\alpha}{2}|z|^{2}}\right|{}^{p}dA(z).\end{align*} $$

Correspondingly, for $ \alpha>0 $ and $ p=\infty $ , we denote by $ L_{\alpha }^{\infty } $ the space of measurable functions $ f $ such that

$$ \begin{align*}\|f\|_{\infty,\alpha}=\text{ess sup}\left\{|f(z)|e^{-\alpha|z|^{2}/2}:z\in\mathbb{C}\right\}<\infty.\end{align*} $$

The Fock space $ F_{\alpha }^{p} $ is defined as the subspace of $ L_{\alpha }^{p} $ consisting of entire functions. It is straightforward to verify that $ F_{\alpha }^{p} $ becomes a Banach space when $ 1\leq p\leq \infty $ . Let $ f_{\alpha }^{\infty } $ be the closed subspace of $ F_{\alpha }^{\infty } $ consisting of f satisfying

$$ \begin{align*}\lim_{|z|\to\infty}f(z)e^{-\frac{\alpha}{2}|z|^{2}}=0.\end{align*} $$

When $ p=2 $ , the space $ F_{\alpha }^{2} $ is a Hilbert space equipped with the inner product

$$ \begin{align*}\langle f,g\rangle_{F_{\alpha}^{2}}=\frac{\alpha}{\pi}\int_{\mathbb{C}}f(z)\overline{g(z)}e^{-\alpha|z|^{2}}\,dA(z).\end{align*} $$

As a reproducing kernel Hilbert space, its reproducing kernel is $ K_{\alpha }(z,w)=e^{\alpha z\overline {w}} $ . The normalized reproducing kernel at z is denoted by $ k_{z}(w)=e^{\alpha \overline {z}w-\frac {\alpha }{2}|z|^{2}} $ , which is a unit vector in $ F_{\alpha }^{p} $ for every $ 0<p\leq \infty $ . For $ 1\leq p<\infty $ , let $ p^{\prime } $ be the conjugate exponent satisfying $ \frac {1}{p}+\frac {1}{p^{\prime }}=1 $ . We can identify the dual of $ F_{\alpha }^{p} $ with $ F_{\alpha }^{p^{\prime }} $ , and the dual of $ f_{\alpha }^{\infty } $ with $ F_{\alpha }^{1} $ , via the integral pairing

$$ \begin{align*}\langle f,g\rangle_{F_{\alpha}^{2}}=\frac{\alpha}{\pi}\int_{\mathbb{C}}f(z)\overline{g(z)}e^{-\alpha|z|^{2}}\,dA(z),\end{align*} $$

and under this identification, the norms on the dual spaces and the Fock spaces are equivalent (see [Reference Zhu17, Corollary 2.25 and Theorem 2.26]).

Given $ \alpha>0 $ , consider the Gaussian measure

$$ \begin{align*}d\lambda_{\alpha}(z)=\frac{\alpha}{\pi}e^{-\alpha|z|^{2}}\,dA(z).\end{align*} $$

Define the integral operator $ P_{\alpha } $ on $ L_{\alpha }^{p} $ by

$$ \begin{align*}P_{\alpha}f(z)=\int_{\mathbb{C}}K_{\alpha}(z,w)f(w)\,d\lambda_{\alpha}(w),\quad f\in L_{\alpha}^{p}.\end{align*} $$

It is known that $ P_{\alpha }:L_{\alpha }^{p}\to F_{\alpha }^{p} $ is bounded for all $ p $ , and it acts as the identity on $ F_{\alpha }^{p} $ , that is, $ P_{\alpha }f=f $ for $ f\in F_{\alpha }^{p} $ (see [Reference Zhu17, Corollary 2.22]).

For a function $ \varphi \in L^{\infty }(\mathbb {C})$ , the associated Toeplitz operator $ T_{\varphi }:F_{\alpha }^{2}\to F_{\alpha }^{2} $ is defined by

$$ \begin{align*}T_{\varphi}(f)=P_{\alpha}(\varphi f),\quad f\in F_{\alpha}^{2}.\end{align*} $$

More generally, if $ \mu $ is a complex Borel measure on $ \mathbb {C} $ , the Toeplitz operator $ T_{\mu } $ is given by

$$ \begin{align*}T_{\mu}(f)(z)=\frac{\alpha}{\pi}\int_{\mathbb{C}}K_{\alpha}(z,w)f(w)e^{-\alpha|w|^{2}}\,d\mu(w),\quad z\in\mathbb{C},\end{align*} $$

provided that the integrability condition

(2.1) $$ \begin{align} \int_{\mathbb{C}}|K_{\alpha}(z,w)|e^{-\alpha|w|^{2}}\,d|\mu|(w)<\infty \end{align} $$

holds for every $ z\in \mathbb {C} $ . Let

(2.2) $$ \begin{align} K=\text{span}\{K_{z}:z\in\mathbb{C}\}. \end{align} $$

The set K is dense in $ F_{\alpha }^{p} $ for every $ 1\leq p<\infty $ and also in $ f_{\alpha }^{\infty } $ . Under assumption (2.1), the operator $ T_{\mu } $ is well-defined on K. For a measure $ \mu $ satisfying (2.1), its Berezin transform is defined as

(2.3) $$ \begin{align} \widetilde{\mu}(z)=\frac{\alpha}{\pi}\int_{\mathbb{C}}|k_{z}(w)|^{2}\mathrm{e}^{-\alpha|w|^{2}}\,d\mu(w)=\frac{\alpha}{\pi}\int_{\mathbb{C}}\mathrm{e}^{-\alpha|z-w|^{2}}\,d\mu(w). \end{align} $$

2.2 Nuclear operator

Let X and Y be complex Banach spaces. The Banach space consisting of all bounded linear operators from X to Y is represented by $\mathcal {L}(X,Y)$ . The dual space of a Banach space X is denoted by $X'$ . For $x\in X$ and $x'\in X'$ , we write $\langle x,x'\rangle $ for the dual pairing $x'(x)$ . We say that an operator $T\in \mathcal {L}(X,Y)$ is nuclear provided that it admits a representation of the form

(2.4) $$ \begin{align} T=\sum_{j=1}^{\infty}x_{j}'\otimes y_{j}, \end{align} $$ upper T equals sigma summation Underscript j equals 1 Overscript infinity Endscripts x Subscript j Superscript prime Baseline circled times y Subscript j Baseline comma

for sequences $(x_{j}')\subset X'$ and $(y_{j})\subset Y$ satisfying $\sum _{j=1}^{\infty }\|x_{j}'\|_{X'}\|y_{j}\|_{Y}<\infty .$ For $x'\in X'$ and $y\in Y$ , we denote by $x'\otimes y$ the rank-one operator from X to Y defined by

$$ \begin{align*}x'\otimes y(x)=\langle x,x'\rangle y,\quad x\in X. \end{align*} $$

The nuclear norm of T is given by

$$ \begin{align*}\|T\|_{N}:=\inf\sum_{i=1}^{\infty}\|x_{i}^{\prime}\|_{X'}\|y_{i}\|_{Y},\end{align*} $$ StartMetric upper T EndMetric Subscript upper N Baseline colon equals inf sigma summation Underscript i equals 1 Overscript infinity Endscripts StartMetric x Subscript i Superscript prime Baseline EndMetric Subscript upper X Sub Superscript prime Subscript Baseline StartMetric y Subscript i Baseline EndMetric Subscript upper Y Baseline comma

where the infimum in this definition is taken over all representations of the type (2.4). We write $\mathcal {N}(X,Y)$ for the class of all nuclear operators mapping X into Y. Equipped with the norm $\|\cdot \|_{N}$ , it becomes a Banach space. In the case where E and F coincide, we write $\mathcal {N}(E)$ for $\mathcal {N}(E,E)$ .

A Banach space E is said to have the approximation property if for every compact set $K\subset E$ and every $\varepsilon>0$ , there exists a finite-rank operator $L: E\to E$ such that

$$ \begin{align*}\|x-Lx\|<\epsilon\quad\text{for all }x\in K.\end{align*} $$

For a Banach space E with the approximation property, the nuclear trace of an operator $T\in \mathcal {N}(E)$ can be defined by

$$ \begin{align*}\mathrm{Tr}_{E}(T)=\sum_{j=1}^{\infty}\langle x_{j},x_{j}^\prime\rangle,\end{align*} $$ trace Subscript upper E Baseline left parenthesis upper T right parenthesis equals sigma summation Underscript j equals 1 Overscript infinity Endscripts left angle bracket x Subscript j Baseline comma x Subscript j Superscript prime Baseline right angle bracket comma

where $T=\sum _{j=1}^{\infty }x_{j}^{\prime }\otimes x_{j}$ is any representation as in (2.4). It follows from [Reference Gohberg, Goldberg and Krupnik8, Theorem V.1.2] that the value of the trace does not depend on the particular representation chosen. Standard references for the general theory of operator ideals include [Reference Defant and Floret4, Reference Diestel, Jarchow and Tonge5, Reference Gohberg, Goldberg and Krupnik8, Reference Pietsch15].

Definition 2.1 Let X be a Banach space, and let $(\Omega ,\Sigma ,\mu )$ be any finite measure space, if for every $\mu $ -continuous vector measure $G:\Sigma \to X$ of bounded variation, we can find a function $g\in L^{1}(\mu ,X)$ for which

$$ \begin{align*}G(E)=\int_{\Omega}g\,d\mu\quad\text{holds for all }E\in\Sigma,\end{align*} $$

we say that X has the Radon–Nikodým property.

Proposition 2.2 For $1<p<\infty $ , the Fock space $F_{\alpha }^{p}$ is a Banach space that enjoys both the approximation property and the Radon–Nikodým property.

Proof The approximation property follows from [Reference Fulsche7, Proposition 2.1]. The Radon–Nikodým property holds because $F_\alpha ^p$ is reflexive (see [Reference Diestel and Uhl6, Corollary 3.13]).

Let X be a Banach space, if $x\in X$ , define $x^{\prime \prime }: X^\prime \rightarrow \mathbb {C}$ by

$$ \begin{align*} x^{\prime\prime}(x^\prime)=x^\prime(x) \end{align*} $$

for every $x^\prime \in X^\prime $ . The map $\tau : X\rightarrow X^{\prime \prime }$ defined by $\tau (x)=x^{\prime \prime }$ for all $x\in X$ is called the canonical map. Recall that $\tau $ is an isometry from X onto $X^{\prime \prime }$ when X is reflexive.

Theorem 2.3 Let X and Y be Banach spaces such that Y is reflexive. The adjoint map $T\rightarrow T^{*}$ is an isometry from $\mathcal {N}(X,Y)$ onto $\mathcal {N}(Y^{\prime },X ^{\prime })$ .

Proof Let $T=\sum _{j=1}^{\infty }x_{j}^\prime \otimes y_{j} \in \mathcal {N}(X,Y)$ , where $(x_{j}^\prime )\in X^\prime , (y_j)\in Y$ . For any $y^\prime \in Y^{'}, x \in X$ , we have

$$ \begin{align*} T^{*}(y^\prime)(x)&=y^\prime(Tx)\\ & =y^\prime(\sum_{j=1}^{\infty}x_{j}^\prime(x)y_{j})\\ &=\sum_{j=1}^{\infty}x_{j}^\prime(x)y^\prime(y_{j}) \\ &=\sum_{j=1}^{\infty}y_{j}^{\prime\prime}(y^\prime)x_{j}^\prime(x). \end{align*} $$

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Then $T^{*}=\sum _{j=1}^{\infty }y_{j}^{\prime \prime }\otimes x_{j}^\prime $ , and $\|T^{*}\|_{\mathcal {N}}\leq \sum _{j=1}^{\infty }\|y_{j}^{\prime \prime }\|_{Y^{\prime \prime }}\|x_{j}^\prime \|_{X^{\prime }}=\sum _{j=1}^{\infty }\|y_{j}\|_{Y}\|x_{j}^\prime \|_{X^{\prime }}$ . Taking the infimum over all possible representations (2.4), we obtain $\|T^{*}\|_{\mathcal {N}}\leq \|T\|_{\mathcal {N}}$ . On the other hand, let $K\in \mathcal {N}(Y^{\prime },X^{\prime })$ . Since Y is reflexive, there exist sequences $(a_j^\prime )\in X^\prime $ and $(b_j)\in Y$ such that $K=\sum _{j=1}^{\infty }b_{j}^{\prime \prime }\otimes a_{j}^\prime $ . Let $L=\sum _{j=1}^{\infty }a_{j}^\prime \otimes b_j\in \mathcal {N}(X,Y)$ . It is easy to check that $L^*=K$ and $\|L\|_{\mathcal {N}}\leq \|K\|_{\mathcal {N}}$ . This proves the desired result.

For the dual space of compact operators and nuclear operators, we have the following characterization.

Proposition 2.4 [Reference Defant and Floret4, Proposition 16.7]

Let $ E $ and $ F $ be Banach spaces such that $ E^{\prime \prime } $ or $ F^{\prime } $ has the approximation property and $ E^{\prime \prime } $ or $ F^{\prime } $ has the Radon–Nikodým property. Then

$$ \begin{align*} (\mathcal{K}(E, F))^{\prime} &= \mathcal{N}\left(E^{\prime}, F^{\prime}\right) && \mathrm{isometrically}, \\ (\mathcal{K}(E, F))^{\prime \prime} &= \left(\mathcal{N}\left(E^{\prime}, F^{\prime}\right)\right)^{\prime} = \mathcal{L}\left(E^{\prime \prime}, F^{\prime \prime}\right) && \mathrm{isometrically}. \end{align*} $$

The duality brackets are given by

$$ \begin{align*} \langle S, T\rangle_{\mathcal{K}, \mathcal{N}} &= \sum_{n=1}^{\infty} \left\langle S^{*} y_{n}^{\prime}, x_{n}^{\prime \prime} \right\rangle_{E^{\prime}, E^{\prime \prime}} \\ \langle T, S\rangle_{\mathcal{N}, \mathcal{L}} &= \sum_{n=1}^{\infty} \left\langle y_{n}^{\prime}, S x_{n}^{\prime \prime} \right\rangle_{E^{\prime}, E^{\prime \prime}} \end{align*} $$

StartLayout 1st Row 1st Column left angle bracket upper S comma upper T right angle bracket Subscript script upper K comma script upper N 2nd Column equals sigma summation Underscript n equals 1 Overscript infinity Endscripts left angle bracket upper S Superscript asterisk Baseline y Subscript n Superscript prime Baseline comma x Subscript n Superscript double prime Baseline right angle bracket Subscript upper E prime comma upper E Sub Superscript double prime Subscript Baseline 2nd Row 1st Column left angle bracket upper T comma upper S right angle bracket Subscript script upper N comma script upper L 2nd Column equals sigma summation Underscript n equals 1 Overscript infinity Endscripts left angle bracket y prime Subscript n Baseline comma upper S x Subscript n Superscript double prime Baseline right angle bracket Subscript upper E prime comma upper E double prime EndLayout

if $ T = \sum _{n=1}^{\infty } x_{n}^{\prime \prime } \otimes y_{n}^{\prime } \in \mathcal {N}\left (E^{\prime }, F^{\prime }\right ) $ .

Proposition 2.5 Let $1<p,q<\infty $ , then

$$ \begin{align*} \mathcal{N}(F_{\alpha}^{p},F_{\alpha}^{q})^{\prime}=\mathcal{L}(F_{\alpha}^{q},F_{\alpha}^{p}),\quad \mathcal{K}(F_{\alpha}^{p},F_{\alpha}^{q})^{\prime}=\mathcal{N}(F_{\alpha}^{q},F_{\alpha}^{p}), \end{align*} $$

where the duality is induced by the trace map:

$$ \begin{align*} \langle S, T\rangle&=\mathrm{Tr}(ST)=\mathrm{Tr}(TS)\\ \langle A, B\rangle&=\mathrm{Tr}(AB)=\mathrm{Tr}(BA), \end{align*} $$

where $S\in \mathcal {N}(F_{\alpha }^p,F_{\alpha }^{q}) $ , $T\in \mathcal {L}(F_{\alpha }^{q},F_{\alpha }^{p})$ , $A\in \mathcal {K}(F_{\alpha }^{p},F_{\alpha }^{q})$ , and $B\in \mathcal {N}(F_{\alpha }^{q},F_{\alpha }^{p})$ .

Proof We use $E\cong F$ to denote that the two Banach spaces E and F are isometrically isomorphic. Then, we have

$$ \begin{align*} \mathcal{N}(F_{\alpha}^{p},F_{\alpha}^{q})^{\prime}&\cong\mathcal{N}((F_{\alpha}^{q})^\prime,(F_{\alpha}^{p})^\prime)^{\prime}\\ &\cong\mathcal{L}((F_{\alpha}^{q})^{\prime\prime},(F_{\alpha}^{p})^{\prime\prime})\\ &\cong\mathcal{L}(F_{\alpha}^{q},F_{\alpha}^{p}). \end{align*} $$

StartLayout 1st Row 1st Column script upper N left parenthesis upper F Subscript alpha Superscript p Baseline comma upper F Subscript alpha Superscript q Baseline right parenthesis prime 2nd Column approximately equals script upper N left parenthesis left parenthesis upper F Subscript alpha Superscript q Baseline right parenthesis prime comma left parenthesis upper F Subscript alpha Superscript p Baseline right parenthesis Superscript prime Baseline right parenthesis Superscript prime Baseline 2nd Row 1st Column Blank 2nd Column approximately equals script upper L left parenthesis left parenthesis upper F Subscript alpha Superscript q Baseline right parenthesis double prime comma left parenthesis upper F Subscript alpha Superscript p Baseline right parenthesis Superscript double prime Baseline right parenthesis 3rd Row 1st Column Blank 2nd Column approximately equals script upper L left parenthesis upper F Subscript alpha Superscript q Baseline comma upper F Subscript alpha Superscript p Baseline right parenthesis period EndLayout

The first step is obtained from Theorem 2.3; the second step from Proposition 2.4; and the last step follows from reflexivity.

Let $S=\sum _{j=1}^{\infty } y_{j} \otimes x_{j} \in \mathcal {N}(F_{\alpha }^{p},F_{\alpha }^{q})$ and $T\in \mathcal {L}(F_{\alpha }^{q},F_{\alpha }^{p}) $ ; applying Proposition 2.4, we obtain that the duality bracket is given by

$$ \begin{align*} \langle S, T\rangle&=\langle S^*, \tau T\tau^{-1}\rangle_{\mathcal{N}, \mathcal{L}}\\ &=\sum_{j=1}^{\infty}\langle y_j,\tau Tx_j\rangle\\ &=\sum_{j=1}^{\infty}\langle Tx_j,y_j\rangle\\ &=\mathrm{Tr}(TS)\\ &=\mathrm{Tr}(ST). \end{align*} $$

StartLayout 1st Row 1st Column left angle bracket upper S comma upper T right angle bracket 2nd Column equals left angle bracket upper S Superscript asterisk Baseline comma tau upper T tau Superscript negative 1 Baseline right angle bracket Subscript script upper N comma script upper L Baseline 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript j equals 1 Overscript infinity Endscripts left angle bracket y Subscript j Baseline comma tau upper T x Subscript j Baseline right angle bracket 3rd Row 1st Column Blank 2nd Column equals sigma summation Underscript j equals 1 Overscript infinity Endscripts left angle bracket upper T x Subscript j Baseline comma y Subscript j Baseline right angle bracket 4th Row 1st Column Blank 2nd Column equals trace left parenthesis upper T upper S right parenthesis 5th Row 1st Column Blank 2nd Column equals trace left parenthesis upper S upper T right parenthesis period EndLayout

The other proof is similar.

For convenience, in the following sections, we write $A \lesssim B$ to mean that $A \leq CB$ for some positive constant C, whose exact value is insignificant. The relation $A \simeq B$ signifies that $A \lesssim B \lesssim A$ .

3 Nuclear Toeplitz operator

In this section, given a complex Borel measure $\mu $ , we are going to characterize when the Toeplitz operator $T_\mu $ belongs to the nuclear class.

Let $1\leq p\leq \infty $ and $p^\prime $ be the conjugate exponent of p. For $f\in F_\alpha ^{p^\prime }$ , we regard f as an element of $(F_\alpha ^p)^\prime $ under the pairing $\langle \cdot ,\cdot \rangle _{F_\alpha ^2}$ , that is,

$$ \begin{align*} f(g)=\langle g,f\rangle_{F_\alpha^2} \end{align*} $$

for all $g\in F_\alpha ^p$ . Then we have $\|f\|_{(F_\alpha ^p)^\prime }\simeq \|f\|_{\alpha ,p^\prime }$ .

Definition 3.1 Let $T: F_\alpha ^p\xrightarrow {}F_\alpha ^q$ be a linear operator, where $1\leq p,q\leq \infty $ , we define the Berezin transform of T as

$$ \begin{align*} \widetilde{T}(z)=\left\langle T k_{z}, k_{z}\right\rangle _{F_\alpha^2}, \quad z \in \mathbb{C}. \end{align*} $$

Theorem 3.2 Let $1\leq p<\infty $ and $1\leq q,s\leq \infty $ . We have:

(1) If $T\in \mathcal {N}(f_\alpha ^\infty ,F_\alpha ^s)$ , then $\widetilde {T}(z)\in L^r(dA)$ for all $1\leq r\leq \infty $ .
(2) If $T\in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ and $p\leq q$ , then $\widetilde {T}(z)\in L^r(dA)$ for all $(\frac {1}{p^\prime }+\frac {1}{q})^{-1}\leq r\leq \infty $ .
(3) If $T\in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ and $p>q$ , then $\widetilde {T}(z)\in L^r(dA)$ for all $1\leq r\leq \infty $ .

Moreover,

$$ \begin{align*} \|\widetilde{T}\|_{L^r}\lesssim\|T\|_{\mathcal{N}}. \end{align*} $$

Proof To prove (1), let $T=\sum _{j=1}^{\infty } f_{j} \otimes g_{j}$ , where $\left (g_{j}\right ) \subset F_\alpha ^s, \left (f_{j}\right ) \subset F_\alpha ^1 $ be a nuclear representation. Hence,

$$ \begin{align*} \widetilde{T}(z)&=\langle T k_z,k_z\rangle_{F_\alpha^2}\\ &=\sum_{j=1}^{\infty}\langle k_z,f_j\rangle_{F_\alpha^2}\langle g_j ,k_z\rangle_{F_\alpha^2}\\ &=\sum_{j=1}^{\infty}\overline{f_j(z)}g_j(z)e^{-\alpha|z|^2}. \end{align*} $$

StartLayout 1st Row 1st Column ModifyingAbove upper T With tilde left parenthesis z right parenthesis 2nd Column equals left angle bracket upper T k Subscript z Baseline comma k Subscript z Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript j equals 1 Overscript infinity Endscripts left angle bracket k Subscript z Baseline comma f Subscript j Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline left angle bracket g Subscript j Baseline comma k Subscript z Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline 3rd Row 1st Column Blank 2nd Column equals sigma summation Underscript j equals 1 Overscript infinity Endscripts ModifyingAbove f Subscript j Baseline left parenthesis z right parenthesis With quotation dash g Subscript j Baseline left parenthesis z right parenthesis e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline period EndLayout

Let $1\leq r\leq \infty $ , we can find $m\geq 1, n\geq s$ such that $\frac {1}{m}+\frac {1}{n}=\frac {1}{r}$ . It follows from [Reference Zhu17, Theorem 2.10], we have $f_j\in F_\alpha ^1\subset F_\alpha ^m$ , $g_j\in F_\alpha ^s\subset F_\alpha ^n$ , and

$$ \begin{align*} \|f_j\|_{\alpha,m}\lesssim\|f_j\|_{\alpha,1},\|g_j\|_{\alpha,n}\lesssim\|g_j\|_{\alpha,s}. \end{align*} $$

Then, by Hölder’s inequality, we have

$$ \begin{align*} \|\widetilde{T}\|_{L^r}&\leq\sum_{j=1}^{\infty}\|{f_j(z)}e^{-\frac{\alpha}{2}|z|^2}\|_{L^m}\|g_j(z)e^{-\frac{\alpha}{2}|z|^2}\| _{L^n}\\ &\lesssim \sum_{j=1}^{\infty}\|f_j\|_{\alpha,m}\|g_j\| _{\alpha,n}\\ &\lesssim \sum_{j=1}^{\infty}\|f_j\|_{\alpha,1}\|g_j\| _{\alpha,s}\\ &\lesssim \sum_{j=1}^{\infty}\|{f_j}\|_{(f_\alpha^\infty)^\prime}\|g_j\| _{\alpha,s}. \end{align*} $$

StartLayout 1st Row 1st Column StartMetric upper T overtilde EndMetric Subscript upper L Sub Superscript r 2nd Column less than or equals sigma summation Underscript j equals 1 Overscript infinity Endscripts StartMetric f Subscript j Baseline left parenthesis z right parenthesis e Superscript minus StartFraction alpha Over 2 EndFraction StartAbsoluteValue z EndAbsoluteValue squared Baseline EndMetric Subscript upper L Sub Superscript m Baseline StartMetric g Subscript j Baseline left parenthesis z right parenthesis e Superscript minus StartFraction alpha Over 2 EndFraction StartAbsoluteValue z EndAbsoluteValue squared Baseline EndMetric Subscript upper L Sub Superscript n Baseline 2nd Row 1st Column Blank 2nd Column less than or equivalent to sigma summation Underscript j equals 1 Overscript infinity Endscripts StartMetric f Subscript j Baseline EndMetric Subscript alpha comma m Baseline StartMetric g Subscript j Baseline EndMetric Subscript alpha comma n Baseline 3rd Row 1st Column Blank 2nd Column less than or equivalent to sigma summation Underscript j equals 1 Overscript infinity Endscripts StartMetric f Subscript j Baseline EndMetric Subscript alpha comma 1 Baseline StartMetric g Subscript j Baseline EndMetric Subscript alpha comma s Baseline 4th Row 1st Column Blank 2nd Column less than or equivalent to sigma summation Underscript j equals 1 Overscript infinity Endscripts StartMetric f Subscript j Baseline EndMetric Subscript left parenthesis f Sub Subscript alpha Sub Superscript infinity Subscript right parenthesis Sub Superscript prime Subscript Baseline StartMetric g Subscript j Baseline EndMetric Subscript alpha comma s Baseline period EndLayout

Now, taking the infimum over all possible representations (2.4) yields

$$ \begin{align*} \|\widetilde{T}\|_{L^r}\lesssim\|T\|_{\mathcal{N}}<\infty. \end{align*} $$

The proof of (2) and (3) is similar.

Lemma 3.3 Let $1\leq p\leq \infty $ , $z\rightarrow k_z$ is norm-continuous on $F_\alpha ^p$ .

Proof Fix $z_0\in \mathbb {C}$ , suppose $|z-z_0|<1$ and $1\leq p<\infty $ . Then

$$ \begin{align*} \|k_z-k_{z_{0}}\|_{\alpha,p}^p=\frac{\alpha p}{2\pi}\int_{\mathbb{C}}|(k_z(w)-k_{z_0}(w))e^{-\frac{\alpha}{2}|w|^2}|^pdA(w). \end{align*} $$

Since $|z-z_0|<1$ , there exists a constant $C>0$ such that $|(k_z(w)-k_{z_0}(w))e^{-\frac {\alpha }{2}|w|^2}|\leq Ce^{-\frac {\alpha }{4}|w|^2}$ . According to the dominated convergence theorem, the conclusion holds for $1\leq p < \infty $ . When $p=\infty $ , the conclusion follows from the fact $\|f\|_{\alpha ,\infty }\leq \|f\|_{\alpha ,p}$ for any $1\leq p < \infty $ .

For a fixed positive number r, let

$$ \begin{align*}r\mathbb{Z}^{2}=\{nr+imr:n,m\in \mathbb{Z}\}.\end{align*} $$

Then, we define

$$ \begin{align*}S_{r}=\{z=x+iy:-r/2\leq x<r/2,-r/2\leq y<r/2\}.\end{align*} $$

Clearly, we have the disjoint decomposition for $\mathbb {C}$ :

$$ \begin{align*}\mathbb{C}=\bigcup\limits_{w\in r\mathbb{Z}^{2}}(S_{r}+w).\end{align*} $$

Finally, let $\{a_{n}\}$ denote any fixed enumeration of the square lattice $r\mathbb {Z}^{2}$ as a sequence.

Theorem 3.4 Let $\mu $ be a complex Borel measure on $\mathbb {C}$ with $|\mu |(\mathbb {C})<\infty $ . Then $T_\mu $ belongs to both $\mathcal {N}(f_\alpha ^\infty , F_\alpha ^s)$ and $\mathcal {N}(F_\alpha ^p, F_\alpha ^q)$ for $1\leq p<\infty $ and $1\leq q,s\leq \infty $ . Moreover,

$$\begin{align*}\|T_\mu\|_{\mathcal{N}} \leq \frac{\alpha}{\pi} \, |\mu|(\mathbb{C}). \end{align*}$$

Proof We first prove $T_\mu \in \mathcal {N}(f_\alpha ^\infty , F_\alpha ^1)$ . Let $z\in \mathbb {C}$ , we consider the operator $k_z\otimes k_z\in \mathcal {N}(f_\alpha ^\infty ,F_\alpha ^1) $ , that is,

$$ \begin{align*} (k_z\otimes k_z)f=\langle f,k_z\rangle _{F_\alpha^2}k_z. \end{align*} $$

It is obvious that $\|k_z\otimes k_z\|_{\mathcal {N}}\leq 1$ for all $z\in \mathbb {C}$ . For every $r>0$ , note that

$$ \begin{align*}\sum_{k=1}^\infty |\mu(S_r+a_k)|\leq\sum_{k=1}^\infty|\mu|(S_r+a_k)=|\mu|(\mathbb{C})<\infty,\end{align*} $$

thus we can define

$$ \begin{align*} T_r=\sum_{k=1}^\infty \mu(S_r+a_k)k_{a_k}\otimes k_{a_k} \in \mathcal{N}{(f_\alpha^\infty,F_\alpha^1)}. \end{align*} $$

Let $r_n=\frac {1}{2^n}$ , fix any $\epsilon>0$ , since $|\mu |(\mathbb {C})<\infty $ , there exists $R>0$ such that

$$ \begin{align*} |\mu|(\mathbb{C}\setminus \overline{ B(0,R)}<\epsilon. \end{align*} $$

According to Lemma 3.3, there exists a $\delta>0$ such that

$$ \begin{align*} \|k_z-k_w\|_{\alpha,1}<\epsilon \end{align*} $$

whenever $|z-w|<\delta $ , and $z,w\in \overline {B(0,R+1)}$ .

Since $ \{r_n\}$ is a monotonically decreasing sequence tending to zero, there exists an integer $N_0$ such that for all $ N> N_0 $ , the set ${\{(S_{r_N}+a_k^{(N)})}\}$ satisfies the following property: if $S_{r_N}+a_k^{(N)} \cap \overline { B(0, R)} \neq \emptyset $ , then $\overline {(S_{r_N}+a_k^{(N)})}$ is contained in $\overline {B(0, R+1)}$ .

On the one hand, there exists an integer $N_1$ such that $r_N < \sqrt {2} \delta $ for all $N> N_1$ . Fix an $N> \max (N_0, N_1)$ . For any $N'> N$ , recall that in a Banach space, any absolutely summable sequence is unconditionally summable [Reference Diestel, Jarchow and Tonge5, Proposition 1.1]. We have

$$ \begin{align*} &T_{r_N}-T_{r_{N^\prime}}=\sum_{k=1}^\infty \mu(S_{r_N}+a_k^{(N)})k_{a_k^{(N)}}\otimes k_{a_k^{(N)}}-\sum_{j=1}^\infty \mu(S_{r_{N^{\prime}}}+a_j^{(N^\prime)})k_{a_j^{(N^\prime)}}\otimes k_{a_j^{(N^\prime)}}\\&\quad =\sum_{k=1}^\infty\sum_{j=1}^\infty \mu[(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime}}}+a_j^{(N^\prime)})](k_{a_k^{(N)}}\otimes k_{a_k^{(N)}}-k_{a_j^{(N^\prime)}}\otimes k_{a_j^{(N^\prime)}})\\&\quad =I_1+I_2, \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column upper T Subscript r Sub Subscript upper N Baseline minus upper T Subscript r Sub Subscript upper N prime Baseline equals sigma summation Underscript k equals 1 Overscript infinity Endscripts mu left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Baseline circled times k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Baseline minus sigma summation Underscript j equals 1 Overscript infinity Endscripts mu left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Baseline circled times k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Baseline 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript k equals 1 Overscript infinity Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts mu left bracket left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis right bracket left parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline circled times k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline minus k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline circled times k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline right parenthesis 3rd Row 1st Column Blank 2nd Column equals upper I 1 plus upper I 2 comma EndLayout

where

$$ \begin{align*} & I_1=\sum_{k=1}^\infty\sum_{j=1}^\infty \mu[(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime}}}+a_j^{(N^\prime)})](k_{a_k^{(N)}}-k_{a_j^{(N^\prime)}})\otimes k_{a_j^{(N^\prime)}}\\ & I_2=\sum_{k=1}^\infty\sum_{j=1}^\infty \mu[(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime}}}+a_j^{(N^\prime)})]k_{a_k^{(N)}}\otimes(k_{a_k^{(N)}}-k_{a_j^{(N^\prime)}}).\\ \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column upper I 1 equals sigma summation Underscript k equals 1 Overscript infinity Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts mu left bracket left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis right bracket left parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline minus k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline right parenthesis circled times k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Baseline 2nd Row 1st Column Blank 2nd Column upper I 2 equals sigma summation Underscript k equals 1 Overscript infinity Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts mu left bracket left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis right bracket k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline circled times left parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline minus k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline right parenthesis period EndLayout

Let

$$ \begin{align*}S_1&=\{k\in \mathbb{N},(S_{r_N}+a_k^{(N)}) \cap \overline{B(0,R)}\neq\emptyset\},\\S_2&=\{k\in \mathbb{N},(S_{r_N}+a_k^{(N)}) \cap \overline{B(0,R)}=\emptyset\}.\end{align*} $$

StartLayout 1st Row 1st Column upper S 1 2nd Column equals StartSet k element of double struck upper N comma left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection ModifyingAbove upper B left parenthesis 0 comma upper R right parenthesis With quotation dash not equals normal empty set EndSet comma 2nd Row 1st Column upper S 2 2nd Column equals StartSet k element of double struck upper N comma left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection ModifyingAbove upper B left parenthesis 0 comma upper R right parenthesis With quotation dash equals normal empty set EndSet period EndLayout

It is easy to see that if $(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime }}}+a_j^{(N^\prime )})\neq \emptyset ,$ then we have $a_k^{(N)},\, a_j^{(N^\prime )}\in \overline {(S_{r_N}+a_k^{(N)})}$ , and $ |a_k^{(N)}-a_j^{(N^\prime )}|<\delta $ . Thus,

$$ \begin{align*} & \| \sum_{k\in S_1}\sum_{j=1}^\infty \mu[(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime}}}+a_j^{(N^\prime)})]k_{a_k^{(N)}}\otimes(k_{a_k^{(N)}}-k_{a_j^{(N^\prime)}})\|_{\mathcal{N}}\\&\quad \leq \sum_{k\in S_1}\sum_{j=1}^\infty |\mu|[(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime}}}+a_j^{(N^\prime)})]\|(k_{a_k^{(N)}}-k_{a_j^{(N^\prime)}})\|_{\alpha,1}\| k_{a_j^{(N^\prime)}}\|_{(f_\alpha^\infty)^\prime}\\&\quad \leq |\mu|(\mathbb{C})\epsilon. \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column StartMetric sigma summation Underscript k element of upper S 1 Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts mu left bracket left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis right bracket k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline circled times left parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline minus k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline right parenthesis EndMetric Subscript script upper N 2nd Row 1st Column Blank 2nd Column less than or equals sigma summation Underscript k element of upper S 1 Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts StartAbsoluteValue mu EndAbsoluteValue left bracket left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis right bracket StartMetric left parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline minus k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline right parenthesis EndMetric Subscript alpha comma 1 Baseline StartMetric k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline EndMetric Subscript left parenthesis f Sub Subscript alpha Sub Superscript infinity Subscript right parenthesis prime Baseline 3rd Row 1st Column Blank 2nd Column less than or equals StartAbsoluteValue mu EndAbsoluteValue left parenthesis double struck upper C right parenthesis epsilon period EndLayout

On the other hand,

$$ \begin{align*} & \| \sum_{k\in S_2}\sum_{j=1}^\infty \mu[(S_{r_N}+a_k^{(N)})\cap (S_{r_{N^{\prime}}}+a_j^{(N^\prime)})]k_{a_k^{(N)}}\otimes(k_{a_k^{(N)}}-k_{a_j^{(N^\prime)}})\|_{\mathcal{N}}\\&\quad \leq2|\mu|(\mathbb{C}\setminus \overline{ B(0,R)})\\&\quad \leq2\epsilon. \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column StartMetric sigma summation Underscript k element of upper S 2 Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts mu left bracket left parenthesis upper S Subscript r Sub Subscript upper N Subscript Baseline plus a Subscript k Superscript left parenthesis upper N right parenthesis Baseline right parenthesis intersection left parenthesis upper S Subscript r Sub Subscript upper N Sub Sub Superscript prime Sub Subscript Subscript Baseline plus a Subscript j Superscript left parenthesis upper N Super Superscript prime Superscript right parenthesis Baseline right parenthesis right bracket k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline circled times left parenthesis k Subscript a Sub Subscript k Sub Superscript left parenthesis upper N right parenthesis Subscript Baseline minus k Subscript a Sub Subscript j Sub Superscript left parenthesis upper N Sub Super Superscript prime Sub Superscript right parenthesis Subscript Baseline right parenthesis EndMetric Subscript script upper N 2nd Row 1st Column Blank 2nd Column less than or equals 2 StartAbsoluteValue mu EndAbsoluteValue left parenthesis double struck upper C minus ModifyingAbove upper B left parenthesis 0 comma upper R right parenthesis With quotation dash right parenthesis 3rd Row 1st Column Blank 2nd Column less than or equals 2 epsilon period EndLayout

Therefore, we have $\|I_2\|_{\mathcal {N}}\leq (2+|\mu |{(\mathbb {C}}))\epsilon $ . Similarly, we have $\|I_1\|_{\mathcal {N}}\leq C\epsilon $ .

Then, for any $N^\prime>N$ , we have

$$ \begin{align*} \|T_{r_N}-T_{r_N^\prime}\|_{\mathcal{N}}\leq\|I_1\|_{\mathcal{N}}+\|I_2\|_{\mathcal{N}}\leq C\epsilon. \end{align*} $$

Thus, $T_{r_n}$ is a Cauchy sequence in $\mathcal {N}({f_\alpha ^\infty ,F_\alpha ^1})$ . Then there is a $T\in \mathcal {N}(f_\alpha ^\infty , F_\alpha ^1)$ , such that $T_{r_n}\rightarrow T$ with $\|\cdot \|_{\mathcal {N}}$ . Therefore, $T_{r_n}\rightarrow T$ with $\|\cdot \|_{op}$ (operator norm), since $\|A\|_{op}\leq \|A\|_{\mathcal {N}}$ for all $A\in \mathcal {N}( f_\alpha ^\infty ,F_\alpha ^1 )$ . Then it implies that $\widetilde {T_{r_n}}\rightarrow {\widetilde {T}}$ pointwise.

Recall that $\widetilde {\mu }(z)=\frac {\alpha }{\pi }\int _{\mathbb {C}}e^{-\alpha |z-w|^2} d\mu (w)$ , we have

$$ \begin{align*} | \widetilde{T_r}(z)-\frac{\pi}{\alpha}\widetilde{\mu}(z)|&=|\sum_{k=1}^\infty \mu(S_r+a_k)e^{-\alpha|z-a_k|^2}-\int_{\mathbb{C}}e^{-\alpha|z-w|^2} d\mu(w)|\\& =|\sum_{k=1}^\infty\int_{S_r+a_k}e^{-\alpha|z-w|^2}-e^{-\alpha|z-a_k|^2}d\mu(w)|\\& \leq\sum_{k=1}^\infty\int_{S_r+a_k}|e^{-\alpha|z-w|^2}-e^{-\alpha|z-a_k|^2}|d|\mu|(w). \end{align*} $$

StartLayout 1st Row 1st Column StartAbsoluteValue ModifyingAbove upper T Subscript r Baseline With tilde left parenthesis z right parenthesis minus StartFraction pi Over alpha EndFraction ModifyingAbove mu With tilde left parenthesis z right parenthesis EndAbsoluteValue 2nd Column equals StartAbsoluteValue sigma summation Underscript k equals 1 Overscript infinity Endscripts mu left parenthesis upper S Subscript r Baseline plus a Subscript k Baseline right parenthesis e Superscript minus alpha StartAbsoluteValue z minus a Super Subscript k Superscript EndAbsoluteValue squared Baseline minus integral Underscript double struck upper C Endscripts e Superscript minus alpha StartAbsoluteValue z minus w EndAbsoluteValue squared Baseline d mu left parenthesis w right parenthesis EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column equals StartAbsoluteValue sigma summation Underscript k equals 1 Overscript infinity Endscripts integral Underscript upper S Subscript r Baseline plus a Subscript k Baseline Endscripts e Superscript minus alpha StartAbsoluteValue z minus w EndAbsoluteValue squared Baseline minus e Superscript minus alpha StartAbsoluteValue z minus a Super Subscript k Superscript EndAbsoluteValue squared Baseline d mu left parenthesis w right parenthesis EndAbsoluteValue 3rd Row 1st Column Blank 2nd Column less than or equals sigma summation Underscript k equals 1 Overscript infinity Endscripts integral Underscript upper S Subscript r Baseline plus a Subscript k Baseline Endscripts StartAbsoluteValue e Superscript minus alpha StartAbsoluteValue z minus w EndAbsoluteValue squared Baseline minus e Superscript minus alpha StartAbsoluteValue z minus a Super Subscript k Superscript EndAbsoluteValue squared Baseline EndAbsoluteValue d StartAbsoluteValue mu EndAbsoluteValue left parenthesis w right parenthesis period EndLayout

Since $e^{-\alpha |z|^2}$ is uniformly continuous on $\mathbb {C}$ , for any $\epsilon>0$ , there exists an $r_0>0$ , if $r<r_0$ , we have

$$ \begin{align*} |e^{-\alpha|z-w|^2}-e^{-\alpha|z-a_k|^2}|<\frac{\epsilon}{|\mu|(\mathbb{C})} \end{align*} $$

for every $w\in S_r+a_k$ . Hence, if $r<r_0$ , we have

$$ \begin{align*} | \widetilde{T_r}(z)-\frac{\pi}{\alpha}\widetilde{\mu}(z)|<\epsilon. \end{align*} $$

This implies $\lim \limits _{r\rightarrow 0}\widetilde {T_r}(z)=\frac {\pi }{\alpha }\widetilde {\mu }(z)$ , hence $\widetilde {T}(z)=\frac {\pi }{\alpha }\widetilde {\mu }(z)$ .

Set $F(z,w)=T_\mu K_z(w)$ and $G(z,w)=TK_z(w)$ . These functions are analytic with respect to w and are anti-holomorphic with respect to z. Furthermore, we have

$$ \begin{align*}F(z,z)=\frac{\alpha}{\pi}G(z,z)=e^{\alpha|z|^2}\widetilde{\mu}(z). \end{align*} $$

A standard polarization theorem in several complex variables (see [Reference Krantz12]) yields $F(z,w)=\frac {\alpha }{\pi }G(z,w)$ for all $z,w\in \mathbb {C}$ . Then we have $T_\mu K_z=\frac {\alpha }{\pi }TK_z$ for all $z\in \mathbb {C}$ . By the density of the finite linear span of kernel functions in $f_\alpha ^{\infty }$ , we have ${T_\mu =\frac {\alpha }{\pi }T\in \mathcal {N}(f_\alpha ^{\infty },F_\alpha ^{1})}$ .

The above proof shows that $\frac {\alpha }{\pi }T_{r_n}\stackrel {\|\cdot \|_{\mathcal {N}}}{\longrightarrow }T_\mu $ . Then we have

$$ \begin{align*} \|T_\mu\|_{\mathcal{N}}=\frac{\alpha}{\pi}\lim\limits_{n\rightarrow\infty}\|T_{r_n}\|_{\mathcal{N}}\leq\frac{\alpha}{\pi} |\mu|(\mathbb{C}). \end{align*} $$

Recall that the normalized reproducing kernel $k_z$ is a unit vector in $F_\alpha ^p$ for all $1\leq p\leq \infty $ , hence the operator $T_r$ defined above belongs to both $\mathcal {N}(f_\alpha ^\infty , F_\alpha ^s)$ and $\mathcal {N}(F_\alpha ^p, F_\alpha ^q)$ for the indices under consideration. Repeating this argument completes the proof for all cases.

From the proof of Theorem 3.4, let

$$\begin{align*}S = \{T_\mu : |\mu|(\mathbb{C}) < \infty\}, \end{align*}$$

viewed as a subset of $\mathcal {N}(F_{\alpha }^p, F_{\alpha }^q)$ . Then we have the inclusion

$$\begin{align*}S \subset \overline{\text{span}}^{\|\cdot\|_{\mathcal{N}}} \{k_z \otimes k_z : z \in \mathbb{C}\}. \end{align*}$$

In 2020, Fulsche proved that $T_{f}\in \mathcal {N}(F_{\alpha }^{p})$ if $f\in L^{1}(\mathbb {C})$ (see [Reference Fulsche7, Lemma 2.11], thus the above theorem can be viewed as a generalization of this result. Moreover, if $\mu $ is a positive measure, Theorem 1.1 shows that the condition $\mu (\mathbb {C})<\infty $ is also sufficient for $T_{\mu }\in \mathcal {N}(F_{\alpha }^{p},F_{\alpha }^{q})$ when $q\leq p$ . Next, we give the proof of Theorem 1.1.

Proof of Theorem 1.1

A direct calculation shows that $\widetilde {T_\mu }(z)=\widetilde {\mu }(z)$ . Since $\mu \geq 0$ , we can use Fubini’s theorem to obtain

$$ \begin{align*} \|\widetilde{\mu}\|_{L^1}&=\frac{\alpha}{\pi}\int_{\mathbb{C}}\int_{\mathbb{C}}e^{-\alpha|z-w|^2}d\mu(w)dA(z)\\ &=\frac{\alpha}{\pi}\int_{\mathbb{C}}\int_{\mathbb{C}}e^{-\alpha|z-w|^2}dA(z)d\mu(w)\\ &=\mu(\mathbb{C}). \end{align*} $$

StartLayout 1st Row 1st Column StartMetric ModifyingAbove mu With tilde EndMetric Subscript upper L Sub Superscript 1 2nd Column equals StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts integral Underscript double struck upper C Endscripts e Superscript minus alpha StartAbsoluteValue z minus w EndAbsoluteValue squared Baseline d mu left parenthesis w right parenthesis d upper A left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts integral Underscript double struck upper C Endscripts e Superscript minus alpha StartAbsoluteValue z minus w EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis d mu left parenthesis w right parenthesis 3rd Row 1st Column Blank 2nd Column equals mu left parenthesis double struck upper C right parenthesis period EndLayout

$ (1) \Rightarrow (3) $ : If $1\leq q\leq p<\infty $ , by Theorem 3.2(3), we have $\widetilde {T_\mu }(z)=\widetilde {\mu }(z)\in L^1(\mathbb {C},dA)$ , which implies $\mu (\mathbb {C})<\infty $ . $(3) \Rightarrow (1)$ follows from Theorem 3.4. The proof of ${(2)\Leftrightarrow (3)}$ is similar.

Estimate (1.1) follows from the equation $\|\widetilde {\mu }\|_{L^1}=\mu (\mathbb {C})$ and Theorems 3.2 and 3.4.

Remark 3.5 Suppose $1\leq p<q,$ and $\mu \geq 0$ . If $\mu (\mathbb {C})<\infty $ , by Theorem 3.4, we have $T_\mu \in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ . However, we cannot use Theorem 3.2 to prove the equivalence between $T_\mu \in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ and $\mu (\mathbb {C})<\infty $ , since in this case $(\frac {1}{p^\prime }+\frac {1}{q})^{-1}>1$ . In fact, we will show that there exists $T\in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ such that $\widetilde {T}(z)\notin L^1$ .

Theorem 3.6 Let $1\leq p<\infty $ , and let g be an entire function. Then

$$ \begin{align*}\int_{\mathbb{C}}|f(z)\overline{g(z)}|e^{-\alpha|z|^2}dA(z)<\infty\end{align*} $$

for all $f\in F_\alpha ^p$ if and only if $g\in F_\alpha ^{p^\prime }$ .

Proof $\Rightarrow :$ For any $n\in \mathbb {N}^{+}$ , define the truncated function

$$ \begin{align*} g_n(z)=\chi_{\{|g(z)|\leq n\}}\cdot g(z), \end{align*} $$

so that

$$ \begin{align*} \lim_{n\rightarrow\infty}g_n(z)=g(z) \end{align*} $$

for all $z\in \mathbb {C}$ . For any measurable function h, define $\varphi _{h}: F_\alpha ^p\rightarrow \mathbb {C}$ by

$$ \begin{align*} \varphi_{h}(f)=\int_{\mathbb{C}}f(z)\overline{h(z)}e^{-\alpha|z|^2}dA(z),\quad f\in F_\alpha^p. \end{align*} $$

Then we have $\varphi _{g_n}\in (F_\alpha ^p)^\prime $ . Fix $f\in F_\alpha ^p$ , since $|g_n(z)|\leq |g(z)|$ , we can use the dominated convergence theorem to obtain

$$ \begin{align*} \lim_{n\rightarrow\infty}\varphi_{g_n}(f)=\varphi_g(f). \end{align*} $$

Hence, $\sup _n|\varphi _{g_n}(f)|<\infty $ . It follows from the principle of uniform boundedness that we have

$$ \begin{align*} \sup_n\|\varphi_{g_n}\|=C<\infty. \end{align*} $$

Then

$$ \begin{align*} |\varphi_g(f)|=\lim\limits_{n\rightarrow\infty}|\varphi_{g_n}(f)|\leq C\|f\| \end{align*} $$

for all $f\in F_\alpha ^p$ . It implies that $\varphi _g\in (F_\alpha ^p)^\prime $ . Since $(F_\alpha ^p)^\prime $ can be identified with $F_\alpha ^{p^\prime }$ under the integral pairing $\langle \cdot ,\cdot \rangle _{F_\alpha ^2}$ , we can find $g^\prime \in F_\alpha ^{p^\prime }$ such that

$$ \begin{align*} \varphi_{g-g^\prime}=0. \end{align*} $$

Suppose $(g-g^\prime )(z)=\sum \limits _{k=0}^\infty a_kz^k$ . For any $n\in \mathbb {N}$ , we have

$$ \begin{align*} \int_{\mathbb{C}}|z^n||\overline{(g-g^\prime)(z)}|e^{-\alpha|z|^2}dA(z)<\infty. \end{align*} $$

Then we can use the dominated convergence theorem to obtain

$$ \begin{align*} \varphi_{g-g^\prime}(z^n)&= \int_{\mathbb{C}}z^n\overline{(g-g^\prime)(z)}e^{-\alpha|z|^2}dA(z)\\ &=lim_{R\rightarrow{\infty}}\int_{|z|<R}z^n\overline{(g-g^\prime)(z)}e^{-\alpha|z|^2}dA(z)\\ &=lim_{R\rightarrow{\infty}} \overline{a_n}\int_{|z|<R}|z|^{2n}e^{-\alpha|z|^2}dA(z).\\ &=\overline{a_n}\int_{\mathbb{C}}|z|^{2n}e^{-\alpha|z|^2}dA(z)\\ &=0. \end{align*} $$

StartLayout 1st Row 1st Column phi Subscript g minus g prime Baseline left parenthesis z Superscript n Baseline right parenthesis 2nd Column equals integral Underscript double struck upper C Endscripts z Superscript n Baseline ModifyingAbove left parenthesis g minus g Superscript prime Baseline right parenthesis left parenthesis z right parenthesis With quotation dash e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals l i m Subscript upper R right arrow infinity Baseline integral Underscript StartAbsoluteValue z EndAbsoluteValue less than upper R Endscripts z Superscript n Baseline ModifyingAbove left parenthesis g minus g Superscript prime Baseline right parenthesis left parenthesis z right parenthesis With quotation dash e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals l i m Subscript upper R right arrow infinity Baseline a Subscript n Baseline overbar integral Underscript StartAbsoluteValue z EndAbsoluteValue less than upper R Endscripts StartAbsoluteValue z EndAbsoluteValue Superscript 2 n Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis period 4th Row 1st Column Blank 2nd Column equals a Subscript n Baseline overbar integral Underscript double struck upper C Endscripts StartAbsoluteValue z EndAbsoluteValue Superscript 2 n Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 5th Row 1st Column Blank 2nd Column equals 0 period EndLayout

It implies that $a_n=0$ for all $n\in \mathbb {N}$ . Then we have $g=g^\prime \in F_\alpha ^{p^\prime }$ .

The other direction follows from Hölder’s inequality.

Example 3.7 It is well-known that for $1\leq p<q\leq \infty $ , we have the (proper) inclusion $F_\alpha ^p\subset F_\alpha ^q$ , and the inclusion is proper. Theorem 3.6 tells us: We can find $f\in F_\alpha ^{p^\prime }$ and $g\in F_\alpha ^q$ such that

$$ \begin{align*} \int_{\mathbb{C}}|f(z)\overline{g(z)}|e^{-\alpha|z|^2}dA(z)=\infty. \end{align*} $$

In other words, if we define $T=f\otimes g\in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ , then we have $\|\widetilde {T}\|_{L^1}=\infty .$ Subsequently, we give an explicit construction of f and g.

If $1 < p < q < \infty $ , we have $1 < q' < p' < \infty $ . Let $a = \frac {1}{q} - \frac {1}{p} = \frac {1}{p'} - \frac {1}{q'}$ and define ${\epsilon _{n} = n^{\frac {a}{2}}}$ . A sequence of indices $n_{1} < n_{2} < \cdots $ can be chosen so that $3^{-k} \leq \epsilon _{n_{k}} \leq 2^{-k}$ for all $k \in \mathbb {N}$ . Fix a constant b with $\sqrt {3} < b < 2$ . We consider the function defined by $f(z) = \sum a_{n}z^{n}$ , where its coefficients are given by

$$ \begin{align*} a_n = \begin{cases} b^k\sqrt{\frac{\alpha^{n_{k}}}{(n_k)!}} \, n_k^{\frac{1}{4} - \frac{1}{2p^\prime}} \, \epsilon_{n_k}, & \text{if}\ n = n_k \text{ for some } k, \\ 0, & \text{otherwise}. \end{cases} \end{align*} $$

a Subscript n Baseline equals StartLayout Enlarged left brace 1st Row 1st Column b Superscript k Baseline StartRoot StartFraction alpha Superscript n Super Subscript k Superscript Baseline Over left parenthesis n Subscript k Baseline right parenthesis factorial EndFraction EndRoot n Subscript k Superscript one fourth minus StartFraction 1 Over 2 p Super Superscript prime Superscript EndFraction Baseline epsilon Subscript n Sub Subscript k Subscript Baseline comma 2nd Column if n equals n Subscript k Baseline for some k comma 2nd Row 1st Column 0 comma 2nd Column otherwise period EndLayout

The membership of f in $F_{\alpha }^{p'}$ is verified by computing two sums. For $1 < p' \leq 2$ ,

$$ \begin{align*} \sum_{n=0}^{\infty} |a_{n}|^{p'} \left( \frac{n!}{\alpha^{n}} \right)^{\frac{p'}{2}} n^{-\frac{p'}{4} + \frac{1}{2}} = \sum_{k=1}^{\infty} b^{p'k} \epsilon_{n_{k}}^{p'} \leq \sum_{k=1}^{\infty} \left( \frac{b^{p'}}{2^{p'}} \right)^{k} < \infty. \end{align*} $$

If $2 \leqslant p' < \infty $ , then

$$ \begin{align*} \sum_{n=1}^{\infty} |a_{n}|^{p} \left( \frac{n!}{\alpha^{n}} \right)^{\frac{p}{2}} n^{\frac{p}{4} - \frac{1}{2}} = \sum_{k=1}^{\infty} b^{pk} n_{k}^{\frac{p}{4} - \frac{1}{2} + \frac{p}{4} - \frac{p}{2p'}} \epsilon_{n_{k}}^{p} \leq \sum_{k=1}^{\infty} \left( \frac{b^{p}}{2^{p}} \right)^{k} < \infty. \end{align*} $$

The convergence of these series, together with Theorems 1 and 4 in [Reference Tung16], implies that $ f \in F_{\alpha }^{p'} $ . Since $a_{n}\neq o\left (\sqrt {\frac {\alpha ^{n}}{n!}}n^{\frac {1}{4}-\frac {1}{2q^{\prime }}}\right )\quad (n\rightarrow \infty ), $ it follows from Corollary 5 of [Reference Tung16] that $ f\in F_{\alpha }^{p^{\prime }}\setminus F_{\alpha }^{q^{\prime }} $ .

A function g is constructed analogously. Let

$$ \begin{align*} b_{m}=\begin{cases} b^{k}\sqrt{\dfrac{\alpha^{n_{k}}}{(n_{k})!}}\,n_{k}^{\frac{1}{4}-\frac{1}{2q}}\,\epsilon_{n_{k}},&\text{if }m=n_{k},\\ 0,&\text{otherwise,} \end{cases} \end{align*} $$

b Subscript m Baseline equals StartLayout Enlarged left brace 1st Row 1st Column b Superscript k Baseline StartRoot StartFraction alpha Superscript n Super Subscript k Superscript Baseline Over left parenthesis n Subscript k Baseline right parenthesis factorial EndFraction EndRoot n Subscript k Superscript one fourth minus StartFraction 1 Over 2 q EndFraction Baseline epsilon Subscript n Sub Subscript k Subscript Baseline comma 2nd Column if m equals n Subscript k Baseline comma 2nd Row 1st Column 0 comma 2nd Column otherwise comma EndLayout

and define $ g(z)=\sum b_{m}z^{m}. $ It then follows that $ g\in F_{\alpha }^{q}\setminus F_{\alpha }^{p}. $

If $ \int _{\mathbb {C}}|f(z)g(z)|e^{-\alpha |z|^{2}}\mathrm {d}A(z)<\infty , $ then by the dominated convergence theorem, we obtain

$$ \begin{align*} \begin{aligned} \int_{\mathbb{C}}f(z)\overline{g(z)}e^{-\alpha|z|^{2}}\mathrm{d}A(z)&=\lim_{R\rightarrow\infty}\int_{|z|<R}f(z)\overline{g(z)}e^{-\alpha|z|^{2}}\mathrm{d}A(z)\\ &=\lim_{R\rightarrow\infty}\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}\overline{b_{m}}\int_{|z|<R}z^{n}\overline{z}^{m}e^{-\alpha|z|^{2}}\mathrm{d}A(z)\\ &=\lim_{R\rightarrow\infty}\sum_{n=0}^{\infty}a_{n}\overline{b_{n}}\int_{|z|<R}|z|^{2n}e^{-\alpha|z|^{2}}\mathrm{d}A(z). \end{aligned} \end{align*} $$

StartLayout 1st Row StartLayout 1st Row 1st Column integral Underscript double struck upper C Endscripts f left parenthesis z right parenthesis ModifyingAbove g left parenthesis z right parenthesis With quotation dash e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline normal d upper A left parenthesis z right parenthesis 2nd Column equals limit Underscript upper R right arrow infinity Endscripts integral Underscript StartAbsoluteValue z EndAbsoluteValue less than upper R Endscripts f left parenthesis z right parenthesis ModifyingAbove g left parenthesis z right parenthesis With quotation dash e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline normal d upper A left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals limit Underscript upper R right arrow infinity Endscripts sigma summation Underscript n equals 0 Overscript infinity Endscripts sigma summation Underscript m equals 0 Overscript infinity Endscripts a Subscript n Baseline b Subscript m Baseline overbar integral Underscript StartAbsoluteValue z EndAbsoluteValue less than upper R Endscripts z Superscript n Baseline z overbar Superscript m Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline normal d upper A left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals limit Underscript upper R right arrow infinity Endscripts sigma summation Underscript n equals 0 Overscript infinity Endscripts a Subscript n Baseline b Subscript n Baseline overbar integral Underscript StartAbsoluteValue z EndAbsoluteValue less than upper R Endscripts StartAbsoluteValue z EndAbsoluteValue Superscript 2 n Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline normal d upper A left parenthesis z right parenthesis period EndLayout EndLayout

From the construction above, we have $ a_{n}\overline {b_{n}}\geq 0 $ and $ \int _{|z|<R}|z|^{2n}e^{-\alpha |z|^{2}}\mathrm {d}A(z) $ is monotonically increasing in R. The monotone convergence theorem yields:

$$ \begin{align*} \int_{\mathbb{C}}f(z)\overline{g(z)}e^{\alpha|z|^2}dA(z)&=lim_{R\rightarrow{\infty}}\sum_{n=0}^\infty a_n\overline{b_n}\int_{|z|<R}|z|^{2n}e^{-\alpha|z|^2}dA(z)\\ &=\sum_{n=0}^\infty a_n\overline{b_n}\int_{\mathbb{C}}|z|^{2n}e^{-\alpha|z|^2}dA(z)\\ &=\frac{\pi}{\alpha}\sum_{k=1}^\infty\gamma^{2k}\delta_{n_k}\\ &\geq\frac{\pi}{\alpha}\sum_{k=1}^\infty(\frac{\gamma^2}{3})^k\\ &=\infty. \end{align*} $$

StartLayout 1st Row 1st Column integral Underscript double struck upper C Endscripts f left parenthesis z right parenthesis ModifyingAbove g left parenthesis z right parenthesis With quotation dash e Superscript alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 2nd Column equals l i m Subscript upper R right arrow infinity Baseline sigma summation Underscript n equals 0 Overscript infinity Endscripts a Subscript n Baseline b Subscript n Baseline overbar integral Underscript StartAbsoluteValue z EndAbsoluteValue less than upper R Endscripts StartAbsoluteValue z EndAbsoluteValue Superscript 2 n Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals sigma summation Underscript n equals 0 Overscript infinity Endscripts a Subscript n Baseline b Subscript n Baseline overbar integral Underscript double struck upper C Endscripts StartAbsoluteValue z EndAbsoluteValue Superscript 2 n Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals StartFraction pi Over alpha EndFraction sigma summation Underscript k equals 1 Overscript infinity Endscripts gamma Superscript 2 k Baseline delta Subscript n Sub Subscript k Baseline 4th Row 1st Column Blank 2nd Column greater than or equals StartFraction pi Over alpha EndFraction sigma summation Underscript k equals 1 Overscript infinity Endscripts left parenthesis StartFraction gamma squared Over 3 EndFraction right parenthesis Superscript k Baseline 5th Row 1st Column Blank 2nd Column equals infinity period EndLayout

But $\int _{\mathbb {C}}f(z)\overline {g(z)}\mathrm {e}^{-\alpha |z|^2}dA(z){\kern-1pt}\leq{\kern-1pt} \int _{\mathbb {C}}|f(z){g(z)}|\mathrm {e}^{-\alpha |z|^2}dA(z)$ . We obtain a contradiction.

Finally, we conclude this section with an open problem.

Question 3.8 For $1\leq p<q\leq \infty $ and $\mu \geq 0$ , what is the necessary and sufficient condition for $T_\mu \in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ ?

4 Trace formula and density theorem

In this section, for an operator $T\in \mathcal {N}(F_\alpha ^p)$ , $ (1<p<\infty )$ , we will provide some formulas for calculating its trace. Finally, we will establish a density theorem.

Lemma 4.1 Let $1<p<\infty $ and $T\in \mathcal {N}(F^p_\alpha )$ , then we have

$$ \begin{align*} \mathrm{Tr}(T)=\frac{\alpha}{\pi}\int_{\mathbb{C}}\widetilde{T}(z)dA(z). \end{align*} $$

Proof Let $T=\sum _{j=1}^{\infty } f_{j} \otimes g_{j}$ be a nuclear representation of T, where $\left (g_{j}\right ) \subset F_\alpha ^p, \left (f_{j}\right ) \subset F_\alpha ^{p^\prime }$ . Then we have $\mathrm {Tr}(T)=\sum _{j=1}^{\infty }\langle g_j,f_j\rangle =\sum _{j=1}^{\infty }\langle g_j,f_j\rangle _{F_\alpha ^2}$ . And the trace is independent of the choice of representation. Then we have

$$ \begin{align*} \frac{\alpha}{\pi}\int_{\mathbb{C}}\widetilde{T}(z)dA(z)&=\frac{\alpha}{\pi}\int_{\mathbb{C}}\langle Tk_z,k_z\rangle_{F_\alpha^2}dA(z)\\ &=\frac{\alpha}{\pi}\int_{\mathbb{C}}\sum_{j=1}^{\infty}\langle K_z,f_j\rangle_{F_\alpha^2}\langle g_j,K_z\rangle_{F_\alpha^2}e^{-\alpha|z|^2}dA(z)\\ &=\sum_{j=1}^{\infty}\int_{\mathbb{C}}\overline{f_j(z)}g_j(z)d\lambda_\alpha(z)\\ &=\mathrm{Tr}(T).\\[-37pt] \end{align*} $$

StartLayout 1st Row 1st Column StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts ModifyingAbove upper T With tilde left parenthesis z right parenthesis d upper A left parenthesis z right parenthesis 2nd Column equals StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts left angle bracket upper T k Subscript z Baseline comma k Subscript z Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline d upper A left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts sigma summation Underscript j equals 1 Overscript infinity Endscripts left angle bracket upper K Subscript z Baseline comma f Subscript j Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline left angle bracket g Subscript j Baseline comma upper K Subscript z Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline e Superscript minus alpha StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals sigma summation Underscript j equals 1 Overscript infinity Endscripts integral Underscript double struck upper C Endscripts ModifyingAbove f Subscript j Baseline left parenthesis z right parenthesis With quotation dash g Subscript j Baseline left parenthesis z right parenthesis d lamda Subscript alpha Baseline left parenthesis z right parenthesis 4th Row 1st Column Blank 2nd Column equals trace left parenthesis upper T right parenthesis period EndLayout

Notice that the case $p=q=2$ of the following theorem appears in [Reference Berger and Coburn3, Theorem 6], but we adopt a different approach here.

Theorem 4.2 Suppose $\varphi $ is Lebesgue measurable on $\mathbb {C}$ and $S\in \mathcal {L}(F_\alpha ^q,F_\alpha ^p)$ , where ${1<p,q<\infty }$ . If

(1) $\varphi K_z \in L_\alpha ^{p^{\prime }}$ for every $z\in \mathbb {C}$ ;
(2) $T_\varphi \in \mathcal {L}(F_\alpha ^p,F_\alpha ^q)$ ;
(3) $T_\varphi S \in \mathcal {N}(F_\alpha ^q)$ ;
(4) $\int _{\mathbb {C}}\int _{\mathbb {C}}|\varphi (w)||K(z,w)||SK_z(w)|d\lambda _\alpha (z)d\lambda _\alpha (w)<\infty $ .

Then we have

$$ \begin{align*} \mathrm{Tr}(T_\varphi S)=\frac{\alpha}{\pi}\int_{\mathbb{C}}\varphi(z)\widetilde{S}(z)dA(z). \end{align*} $$

Proof Let $f\in F_\alpha ^p $ , we can find a sequence $f_n\in K$ such that $f_n$ converges to f with $\|\|_{\alpha ,p}$ . The set K is defined as (2.2). By assumption (2), we have $T_\varphi f_n \stackrel {\|\cdot \|_{\alpha ,q}}{\longrightarrow }T_\varphi f$ , hence $T_\varphi f_n(z)\rightarrow T_\varphi f(z)$ for every $z\in \mathbb {C}$ .

Since $f_n\in K$ , we have

$$ \begin{align*} T_\varphi f_n(z)=\int_{\mathbb{C}}f_n(w)\varphi(w)K_\alpha(z,w)d\lambda_\alpha(w) \end{align*} $$

for all $n\in \mathbb {N}$ and $z\in \mathbb {C}$ . Then fix $z\in \mathbb {C}$ , by assumption (1), we have

$$ \begin{align*} &\left|\int_{\mathbb{C}}(f_n(w)-f(w))\varphi(w)K(z,w)d\lambda_\alpha(w)\right| \\&\quad \leq \frac{\alpha}{\pi}\int_{\mathbb{C}}|(f_n(w)-f(w))e^{-\frac{\alpha}{2}|z|^2}||\varphi(w)K(z,w)e^{-\frac{\alpha}{2}|z|^2}|dA(w)\\&\quad \leq C\|f_n-f\|_{\alpha,p}\|\varphi K_z\|_{\alpha,p^\prime}\rightarrow0(n\rightarrow\infty). \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column StartAbsoluteValue integral Underscript double struck upper C Endscripts left parenthesis f Subscript n Baseline left parenthesis w right parenthesis minus f left parenthesis w right parenthesis right parenthesis phi left parenthesis w right parenthesis upper K left parenthesis z comma w right parenthesis d lamda Subscript alpha Baseline left parenthesis w right parenthesis EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column less than or equals StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts StartAbsoluteValue left parenthesis f Subscript n Baseline left parenthesis w right parenthesis minus f left parenthesis w right parenthesis right parenthesis e Superscript minus StartFraction alpha Over 2 EndFraction StartAbsoluteValue z EndAbsoluteValue squared Baseline EndAbsoluteValue StartAbsoluteValue phi left parenthesis w right parenthesis upper K left parenthesis z comma w right parenthesis e Superscript minus StartFraction alpha Over 2 EndFraction StartAbsoluteValue z EndAbsoluteValue squared Baseline EndAbsoluteValue d upper A left parenthesis w right parenthesis 3rd Row 1st Column Blank 2nd Column less than or equals upper C StartMetric f Subscript n Baseline minus f EndMetric Subscript alpha comma p Baseline StartMetric phi upper K Subscript z Baseline EndMetric Subscript alpha comma p Sub Superscript prime Subscript Baseline right arrow 0 left parenthesis n right arrow infinity right parenthesis period EndLayout

Hence,

$$ \begin{align*} T_\varphi f(z)=\int_{\mathbb{C}}f(w)\varphi(w)K(z,w)d\lambda_\alpha(w) \end{align*} $$

for every $f\in F_\alpha ^p$ . Then we can use Fubini’s theorem to obtain

(4.1) $$ \begin{align} \begin{aligned} \mathrm{Tr}(T_\varphi S)&=\frac{\alpha}{\pi}\int_{\mathbb{C}}\widetilde{T_\varphi S}(z)dA(z)\\ &=\int_{\mathbb{C}}\langle T_\varphi SK_z,K_z\rangle_{F_\alpha^2}d\lambda_\alpha(z)\\ &=\int_{\mathbb{C}}T_\varphi SK_z(z) d\lambda_{\alpha}(z)\\ &=\int_{\mathbb{C}}d\lambda_{\alpha}(z)\int_{\mathbb{C}}\varphi(u)K_u(z)SK_z(u)d\lambda_\alpha(u)\\ &=\int_{\mathbb{C}}\varphi(u)d\lambda_\alpha(u)\int_{\mathbb{C}}K_u(z)SK_z(u)d\lambda_{\alpha}(z). \end{aligned} \end{align} $$

StartLayout 1st Row StartLayout 1st Row 1st Column trace left parenthesis upper T Subscript phi Baseline upper S right parenthesis 2nd Column equals StartFraction alpha Over pi EndFraction integral Underscript double struck upper C Endscripts ModifyingAbove upper T Subscript phi Baseline upper S With tilde left parenthesis z right parenthesis d upper A left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals integral Underscript double struck upper C Endscripts left angle bracket upper T Subscript phi Baseline upper S upper K Subscript z Baseline comma upper K Subscript z Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline d lamda Subscript alpha Baseline left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals integral Underscript double struck upper C Endscripts upper T Subscript phi Baseline upper S upper K Subscript z Baseline left parenthesis z right parenthesis d lamda Subscript alpha Baseline left parenthesis z right parenthesis 4th Row 1st Column Blank 2nd Column equals integral Underscript double struck upper C Endscripts d lamda Subscript alpha Baseline left parenthesis z right parenthesis integral Underscript double struck upper C Endscripts phi left parenthesis u right parenthesis upper K Subscript u Baseline left parenthesis z right parenthesis upper S upper K Subscript z Baseline left parenthesis u right parenthesis d lamda Subscript alpha Baseline left parenthesis u right parenthesis 5th Row 1st Column Blank 2nd Column equals integral Underscript double struck upper C Endscripts phi left parenthesis u right parenthesis d lamda Subscript alpha Baseline left parenthesis u right parenthesis integral Underscript double struck upper C Endscripts upper K Subscript u Baseline left parenthesis z right parenthesis upper S upper K Subscript z Baseline left parenthesis u right parenthesis d lamda Subscript alpha Baseline left parenthesis z right parenthesis period EndLayout EndLayout

Let $S^\prime \in \mathcal {L}(F_\alpha ^{p^\prime },F_\alpha ^{q^\prime })$ be the adjoint operator of S under the pairing $\langle \cdot ,\cdot \rangle _{F_\alpha ^2}$ . Then we have

$$ \begin{align*} \langle Sf,g\rangle_{F_\alpha^2}=\langle f,S^\prime g\rangle_{F_\alpha^2} \end{align*} $$

for all $f\in F_\alpha ^q, g\in F_\alpha ^{p^\prime }$ . Hence,

$$ \begin{align*} \int_{\mathbb{C}}K_u(z)SK_z(u)d\lambda_{\alpha}(z)&=\int_{\mathbb{C}}K_u(z)\langle SK_z,K_u\rangle_{F_\alpha^2}d\lambda_\alpha(z)\\ &=\int_{\mathbb{C}}K_u(z)\langle K_z,S^\prime K_u\rangle_{F_\alpha^2}d\lambda_\alpha(z)\\ &=\overline{\int_{\mathbb{C}}K_z(u)S^\prime K_u(z)d\lambda_{\alpha}(z)}\\ &=\overline{S^\prime K_u(u)}\\ &=e^{\alpha|u|^2}\widetilde{S}(u). \end{align*} $$

StartLayout 1st Row 1st Column integral Underscript double struck upper C Endscripts upper K Subscript u Baseline left parenthesis z right parenthesis upper S upper K Subscript z Baseline left parenthesis u right parenthesis d lamda Subscript alpha Baseline left parenthesis z right parenthesis 2nd Column equals integral Underscript double struck upper C Endscripts upper K Subscript u Baseline left parenthesis z right parenthesis left angle bracket upper S upper K Subscript z Baseline comma upper K Subscript u Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline d lamda Subscript alpha Baseline left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column equals integral Underscript double struck upper C Endscripts upper K Subscript u Baseline left parenthesis z right parenthesis left angle bracket upper K Subscript z Baseline comma upper S prime upper K Subscript u Baseline right angle bracket Subscript upper F Sub Subscript alpha Sub Superscript 2 Baseline d lamda Subscript alpha Baseline left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals ModifyingAbove integral Underscript double struck upper C Endscripts upper K Subscript z Baseline left parenthesis u right parenthesis upper S prime upper K Subscript u Baseline left parenthesis z right parenthesis d lamda Subscript alpha Baseline left parenthesis z right parenthesis With quotation dash 4th Row 1st Column Blank 2nd Column equals ModifyingAbove upper S prime upper K Subscript u Baseline left parenthesis u right parenthesis With quotation dash 5th Row 1st Column Blank 2nd Column equals e Superscript alpha StartAbsoluteValue u EndAbsoluteValue squared Baseline ModifyingAbove upper S With tilde left parenthesis u right parenthesis period EndLayout

It follows from (4.1), we have

$$ \begin{align*} \mathrm{Tr}_{F_\alpha^q}(T_\varphi S)=\frac{\alpha}{\pi}\int_{\mathbb{C}}\varphi(u)\widetilde{S}(u)dA(u).\\[-41pt] \end{align*} $$

We can now prove the following lemma.

Lemma 4.3 Let $1<p, q<\infty $ , and let $\varphi $ be a bounded, compactly supported function on $\mathbb {C} $ . Then $T_\varphi \in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ . Furthermore, for any $S\in \mathcal {L}(F_\alpha ^q,F_\alpha ^p)$ , we have

(4.2) $$ \begin{align} \mathrm{Tr}_{F_\alpha^q}(T_\varphi S)=\frac{\alpha}{\pi}\int_{\mathbb{C}}\varphi(z)\widetilde{S}(z)dA(z). \end{align} $$

Proof $T_\varphi \in \mathcal {N}(F_\alpha ^p,F_\alpha ^q)$ follows from Theorem 3.4. Applying the pointwise estimates on the Fock space, we have

$$ \begin{align*} |SK_z(u)|\leq\|SK_z\|_{\alpha,p}e^{\frac{\alpha}{2}|u|^2}\leq\|S\|e^{\frac{\alpha}{2}|z|^2}e^{\frac{\alpha}{2}|u|^2}. \end{align*} $$

Hence,

$$ \begin{align*} &\int_{\mathbb{C}}|\varphi(u)|d\lambda_\alpha(u)\int_{\mathbb{C}}|K_u(z)||SK_z(u)|d\lambda_{\alpha}(z)\\&\quad \leq C\int_{\mathbb{C}}|\varphi(u)|e^{\frac{\alpha}{2}|u|^2}d\lambda_\alpha(u)\int_{\mathbb{C}}|K_u(z)|e^{-\frac{\alpha}{2}|z|^2}dA(z)\\&\quad =C\int_{\mathbb{C}}|\varphi(u)|dA(u)\\&\quad <\infty. \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column integral Underscript double struck upper C Endscripts StartAbsoluteValue phi left parenthesis u right parenthesis EndAbsoluteValue d lamda Subscript alpha Baseline left parenthesis u right parenthesis integral Underscript double struck upper C Endscripts StartAbsoluteValue upper K Subscript u Baseline left parenthesis z right parenthesis EndAbsoluteValue StartAbsoluteValue upper S upper K Subscript z Baseline left parenthesis u right parenthesis EndAbsoluteValue d lamda Subscript alpha Baseline left parenthesis z right parenthesis 2nd Row 1st Column Blank 2nd Column less than or equals upper C integral Underscript double struck upper C Endscripts StartAbsoluteValue phi left parenthesis u right parenthesis EndAbsoluteValue e Superscript StartFraction alpha Over 2 EndFraction StartAbsoluteValue u EndAbsoluteValue squared Baseline d lamda Subscript alpha Baseline left parenthesis u right parenthesis integral Underscript double struck upper C Endscripts StartAbsoluteValue upper K Subscript u Baseline left parenthesis z right parenthesis EndAbsoluteValue e Superscript minus StartFraction alpha Over 2 EndFraction StartAbsoluteValue z EndAbsoluteValue squared Baseline d upper A left parenthesis z right parenthesis 3rd Row 1st Column Blank 2nd Column equals upper C integral Underscript double struck upper C Endscripts StartAbsoluteValue phi left parenthesis u right parenthesis EndAbsoluteValue d upper A left parenthesis u right parenthesis 4th Row 1st Column Blank 2nd Column less than infinity period EndLayout

Then the conclusion follows from Theorem 4.2.

We now present the proof of Theorem 1.2.

Proof of Theorem 1.2

We first prove (2). Assume, for contradiction, that C is not norm-dense in $\mathcal {K}(F_{\alpha }^{p},F_{\alpha }^{q})$ . An application of the Hahn–Banach theorem combined with Proposition 2.5 implies the existence of a nonzero operator $S\in \mathcal {N}(F_{\alpha }^{q},F_{\alpha }^{p})$ satisfying

$$ \begin{align*}\mathrm{Tr}(T_{\varphi}S)=0\qquad\text{for all }T_{\varphi}\in C.\end{align*} $$

By Lemma 4.3, this trace can be written as

$$ \begin{align*}\mathrm{Tr}(T_{\varphi}S)=\frac{\alpha}{\pi}\int_{\mathbb{C}}\varphi(z)\,\widetilde{S}(z)\,dA(z).\end{align*} $$

Since the equality holds for every continuous, compactly supported $\varphi $ , it follows that $\widetilde {S}(z)=0$ . Consequently, $S=0$ , which contradicts the fact that S is nonzero. This establishes statement (2).

The proof of (1) follows by a similar argument.

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Article contents

Nuclear Toeplitz operators between Fock spaces

Abstract

Keywords

MSC classification

Information

1 Introduction

2 Preliminary

2.1 Fock space

2.2 Nuclear operator

Proposition 2.4 [Reference Defant and Floret4, Proposition 16.7]

3 Nuclear Toeplitz operator

Proof of Theorem 1.1

4 Trace formula and density theorem

Proof of Theorem 1.2

References

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