A Quantum Harmonic Analysis Approach to Segal Algebras

Abstract

In this article, we study a commutative Banach algebra structure on the space $L^1(\mathbb {R}^{2n})\oplus {\mathcal {T}}^1$, where the ${\mathcal {T}}^1$ denotes the trace class operators on $L^2(\mathbb {R}^{n})$. The product of this space is given by the convolutions in quantum harmonic analysis. Towards this goal, we study the closed ideals of this space, and in particular its Gelfand theory. We additionally develop the concept of quantum Segal algebras as an analogue of Segal algebras. We prove that many of the properties of Segal algebras have transfers to quantum Segal algebras. However, it should be noted that in contrast to Segal algebras, quantum Segal algebras are not ideals of the ambient space. We also give examples of different constructions that yield quantum Segal algebras.

1 Introduction

Classical mechanics of a particle on ${\mathbb {R}}^n$ is often described by endowing a symplectic structure on its phase space ${\mathbb {R}}^{2n}$. Given a point $(p,q)\in {\mathbb {R}}^{2n}$, the point p describes the position and q describes the momentum of a particle. In quantum physics, functions on ${\mathbb {R}}^{2n}$ are typically referred to as classical observables, while (not necessarily bounded) operators on ${\mathcal {H}}=L^2({\mathbb {R}}^n)$ are referred to as quantum observables. Weyl’s canonical commutation relations on ${\mathcal {H}}$ play the same role as the symplectic phase space structure in classical mechanics. Usually, the theory for the classical observables and the quantum observables are considered on different spaces, with quantization acting as a way to go between classical and quantum observables. In his 1984 paper [25], R. Werner introduced a framework for the simultaneous treatment of both the classical and the quantum observables, which he called quantum harmonic analysis. Here, he extends notions in harmonic analysis for the classical observables, such as convolutions and Fourier transforms, to the quantum observables. In doing so, one achieves interesting interactions between the classical and the quantum side of the theory. Besides being an object of interest for mathematical physics and also of intrinsic interest, this framework also allows for important applications in several fields, e.g., localization operators, Wigner distributions and Toeplitz operators. Reformulating problems in terms of quantum harmonic analysis offers a new framework for tackling problems, which often proves fruitful. For more information, see e.g., [1, 10, 11, 13, 19, 20, 24, 26].

Being a bit more precise, let $W_z$ denote the Weyl operators on the Hilbert space ${\mathcal {H}}$ defined by

$$\begin{aligned} W_{(x,\,\xi )}\phi (y)=e^{i\xi y-ix\xi /2}\phi (y-x), \quad z=(x,\,\xi )\in {\mathbb {R}}^{2n}. \end{aligned}$$

As is common, we have set Planck’s constant to $\hbar =1$. The Weyl operators are unitary operators on ${\mathcal {H}}$, and satisfies the Weyl commutation relations

$$\begin{aligned} W_z W_{z'}=e^{-i\sigma (z,\,z')/2}W_{z+z'} = e^{-i\sigma (z,\,z')} W_{z'} W_z, \quad z, z' \in {\mathbb {R}}^{2n}, \end{aligned}$$

where $\sigma$ denotes the standard symplectic form on ${\mathbb {R}}^{2n}$. Given two trace class operators S and T, their convolution is defined by

$$\begin{aligned} S *T(z) :={{\,\textrm{tr}\,}}(S W_z P T PW_{-z}), \quad z \in {\mathbb {R}}^{2n}, \end{aligned}$$

where P is the parity operator acting on ${\mathcal {H}}$ by $P\phi (x) = \phi (-x)$. The expression $\alpha _z(A) :=W_z A W_{-z}$ plays the role of translating the operator A by z, analogous to the translation $\alpha _z(f) = f(\cdot -z )$ used in the definition of function convolution. Defining the Fourier-Weyl transform (also called Fourier-Wigner transform) by

$$\begin{aligned} {\mathcal {F}}_W(S)(z) :={{\,\textrm{tr}\,}}(S W_z), \quad z \in {\mathbb {R}}^{2n}, \end{aligned}$$

we have that

$$\begin{aligned} {\mathcal {F}}_\sigma (S*T) = {\mathcal {F}}_W(S){\mathcal {F}}_W(T), \end{aligned}$$

where ${\mathcal {F}}_\sigma$ is the symplectic Fourier transform. Additionally, given a function $f\in L^1({\mathbb {R}}^{2n})$ and a trace class operator $A\in {{\mathcal {T}}}^1({\mathcal {H}})$, one defines the operator $f *A$ by

$$\begin{aligned} (f*A)\phi = \frac{1}{(2\pi )^n}\int _{{\mathbb {R}}^{2n}}f(z)W_z AW_{-z}\phi \,\textrm{d}z, \quad z \in {\mathbb {R}}^{2n}. \end{aligned}$$

Again, we have a relation with the Fourier transforms given by

$$\begin{aligned} {\mathcal {F}}_W(f*A)={\mathcal {F}}_\sigma (f){\mathcal {F}}_W(A). \end{aligned}$$

Using the convolutions, one can define a product on $L^1({\mathbb {R}}^{2n})\oplus {{\mathcal {T}}}^1({\mathcal {H}})$ by

$$\begin{aligned}(f,S)*(g,T)=(f*g+S*T, f*T+g*S),\end{aligned}$$

making the space into a commutative Banach algebra. Hence, this new space contains both the classical and the quantum observables.

It is not hard to find reasons why $L^2({\mathbb {R}}^{2n})$ and the Hilbert-Schmidt operators ${{\mathcal {T}}}^2({\mathcal {H}})$ play similar roles for classical and quantum observables. For example, the Fourier-Weyl transform can be extended to an isomorphism between these two spaces by the Plancherel theorem, see e.g., [9, Sec. 7.5]. For other function spaces on ${\mathbb {R}}^{2n}$, the paper [25, Sec. IV] gives a definition of corresponding spaces. We say that a space of classical observables ${\mathcal {D}}_0$ and a space of quantum observables ${\mathcal {D}}_1$ are corresponding spaces if

$$\begin{aligned} (L^1({\mathbb {R}}^{2n})\oplus {{\mathcal {T}}}^1({\mathcal {H}}))*({\mathcal {D}}_0\oplus {\mathcal {D}}_1)\subset {\mathcal {D}}_0\oplus {\mathcal {D}}_1. \end{aligned}$$

(1)

Using this definition $L^2({\mathbb {R}}^{2n})$ and ${{\mathcal {T}}}^2({\mathcal {H}})$ are corresponding spaces, after extending the definition of convolution appropriately, the same is true for $L^\infty ({\mathbb {R}}^{2n})$ and the space of bounded operators ${\mathcal {L}}({\mathcal {H}})$. It should be noted that this correspondence only becomes unique, in the sense that given a space ${\mathcal {D}}_0$ there is only one corresponding space ${\mathcal {D}}_1$, upon assuming additional topological properties on ${\mathcal {D}}_0$ and ${\mathcal {D}}_1$. Notice that when ${\mathcal {D}}_0\oplus {\mathcal {D}}_1\subset L^1({\mathbb {R}}^{2n})\oplus {{\mathcal {T}}}^1({\mathcal {H}})$, equation (1) implies that ${\mathcal {D}}_0\oplus {\mathcal {D}}_1$ is an ideal.

This paper has the dual purpose of both furthering our understanding of the space $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$ as a commutative Banach algebra and introducing a quantum version of Segal algebras. For the first goal, we completely classify the closed ideals of $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$. As is well-known, closed ideals of $L^1({\mathbb {R}}^n)$ are identical to closed translation-invariant subspaces. This analogy breaks down for closed ideals of $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$, where translation-invariance is merely a necessary condition for being a closed ideal, but not sufficient. Nevertheless, there is still a rich connection between translation-invariant subspaces and closed ideals, which we investigate. As a special case, we describe the Gelfand theory of $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$. From the Gelfand theory of $L^1({\mathbb {R}}^{2n})$ it should come as no surprise that this is related to the Fourier transforms.

The second purpose of the paper is to introduce a notion of quantum Segal algebras. The quantum Segal algebras play an analogous role in quantum harmonic analysis to the Segal algebras in classical harmonic analysis. The definition of a quantum Segal algebra is in spirit the same as Segal algebras: A quantum Segal algebra is a dense subalgebra of $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$ that comes with its own Banach algebra norm such that translations act isometrically and continuously on it. We show that quantum Segal algebras are, in analogy with Segal algebras, the same as essential $L^1$ modules. Further, we prove that under the technical assumption of the space being graded (i.e., the algebra has the structure of a direct sum of its classical and quantum part), they always have the same Gelfand theory as $L^1( {\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$. As a last general result on quantum Segal algebras, we give a description of the intersection of all graded quantum Segal algebras.

After having discussed the general theory of quantum Segal algebras, we turn towards examples of such algebras. We give two different ways of constructing such algebras: By convolving a Segal algebra with a regular trace class operator or by using the Weyl quantization. As a special case of a quantum Segal algebra, we discuss the quantum Feichtinger algebra. It is obtained as the direct sum of the Feichtinger algebra ${\mathcal {S}}_0({\mathbb {R}}^{2n})$ and the subspace of ${{\mathcal {T}}}^1({\mathcal {H}})$ obtained by the Weyl quantization of symbols in ${\mathcal {S}}_0({\mathbb {R}}^{2n})$. Due to the importance of the classical Feichtinger algebra, we find it appropriate to look into this example in some more detail. In particular, we give some conditions on when an operator belongs to the quantum Feichtinger algebra in terms of the operations of quantum harmonic analysis. Not by chance, this turns out to be entirely analogous to the characterization of the membership of functions in the classical Feichtinger algebra.

Our paper is structured as follows: In Sect. 2, we give a more detailed account of the foundations for our work. We start by recalling basics of Segal algebras and thereafter review some results in quantum harmonic analysis. In particular, we introduce the notion of modulation of an operator, which when applying the Fourier Weyl transform is turned into function translations. This notion of modulation seems not to be present in the literature thus far. In Sect. 3, we conduct the investigation of the closed ideals of $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$ as well as the Gelfand theory. Section 4 is then devoted to the introduction and the basic properties of quantum Segal algebras. Finally, Sect. 5 will discuss the above mentioned constructions of quantum Segal algebras, in particular the quantum Feichtinger algebra.

2 Preliminaries

2.1 Classical Segal Algebras

Before introducing Segal algebras, let us quickly recall some basic facts about the space of integrable functions $L^1({\mathbb {R}}^n)$. We will let $\alpha _x:L^1({\mathbb {R}}^n)\rightarrow L^1({\mathbb {R}}^n)$ denote the translation operator

$$\begin{aligned}\alpha _xf(y):=f(y-x), \qquad f \in L^{1}({\mathbb {R}}^{n}), \ x\in {\mathbb {R}}^n.\end{aligned}$$

For elements $f, g \in L^{1}({\mathbb {R}}^{n})$ their convolution product is

$$\begin{aligned}(f *g)(x) :=\frac{1}{(2\pi )^{n/2}}\int _{{\mathbb {R}}^{n}}f(y)\alpha _{y}g(x) \, \textrm{d}y = \frac{1}{(2\pi )^{n/2}}\int _{{\mathbb {R}}^{n}}f(y)g(x - y) \, \textrm{d}y,\end{aligned}$$

making $L^1({\mathbb {R}}^n)$ into a Banach algebra. We want to note that the factors $\frac{1}{(2\pi )^{n/2}}$ serve as appropriate normalizations for the Haar measure, which makes the connection between classical and quantum harmonic analysis most natural.

The algebra $L^{1}({\mathbb {R}}^{n})$ is equipped with the involution

$$\begin{aligned}f^{*}(x) :=\overline{f(-x)}.\end{aligned}$$

We use the notation ${\mathcal {F}}f$ for the standard Fourier transform of $f \in L^{1}({\mathbb {R}}^{n})$ given by

$$\begin{aligned}{\mathcal {F}}f(\xi ) :=\frac{1}{(2\pi )^{n/2}}\int _{{\mathbb {R}}^{n}}f(t)e^{- i t \xi } \, \textrm{d}t.\end{aligned}$$

We emphasize at this point that the above standard Fourier transform will not play a role later in this paper. It is usually replaced by the symplectic Fourier transform, and we stress already at this point that ${\widehat{f}}$ will not denote the standard Fourier transform of f as above, but its symplectic Fourier transform ${\mathcal {F}}_\sigma (f)$, see (10).

Remark 1

During the preparation of this paper, we of course also encountered the “eternal struggle” of choosing the “right” conventions regarding the Fourier transform. In the end, we have chosen to use the normalization of the Fourier transform used in Werner’s initial work on quantum harmonic analysis [25]. This gives us a slightly different, but equivalent framework from the one that can be found in, e.g., [24].

Let us recall the definition of a Segal algebra, as in [21].

Definition 2.1

A Segal algebra S is a linear subspace of $L^{1}({\mathbb {R}}^{n})$ satisfying the following four properties:

• (S1) The space S is dense in $L^1({\mathbb {R}}^n)$.

• (S2) The space S is a Banach algebra with respect to a norm $\Vert \cdot \Vert _S$ and convolution as the product.

• (S3) The space S is isometrically translation-invariant, i.e., for $f\in S$ and $x \in {\mathbb {R}}^n$ it holds true that $\alpha _x(f)\in S$ with $\Vert \alpha _x(f)\Vert _S = \Vert f\Vert _S$.

• (S4) The translations $\alpha _x$ act strongly continuous, i.e., for each $f \in S$ the map ${\mathbb {R}}^n \ni x \mapsto \alpha _x(f) \in S$ is continuous.

To check the continuity condition in property (S4) one only needs to check that given an arbitrary $f \in S$ and $\epsilon > 0$ there exists a neighborhood U of the origin in ${\mathbb {R}}^{n}$ such that

$$\begin{aligned}\Vert \alpha _{x}f - f\Vert _{S} < \epsilon \quad \text { for } \quad x \in U.\end{aligned}$$

Remark 2

Segal algebras are often defined on a locally compact group G as a dense subspace of $L^{1}(G)$ equipped with the left Haar measure and left translations. In this more general case, the existence and variety of Segal algebras depend heavily on the group G in question. For discrete groups the only Segal algebra is the whole space $l^{1}(G) = L^{1}(G)$. On the other hand, for G compact the space $L^{p}(G)$ is a Segal algebra for every $1\le p< \infty$. In particular, the Segal algebra $L^2(G)$ is heavily used in Peter-Weyl theory.

A well known result is that Segal algebras are (two-sided) ideals of $L^1({\mathbb {R}}^n)$, i.e., for a Segal algebra S, we have that $f *g \in S$ whenever $f \in L^1({\mathbb {R}}^n)$ and $g \in S$. Moreover, Segal algebras always satisfy by [21, Prop. 4.1] the estimate

$$\begin{aligned} \Vert f *g\Vert _{S} \le \Vert f\Vert _{L^{1}({\mathbb {R}}^{n})} \Vert g\Vert _{S} \end{aligned}$$

(2)

for $f \in L^{1}({\mathbb {R}}^{n})$ and $g \in S$. More generally, one has that if $T\subset S$ are both Segal algebras then

$$\begin{aligned} \Vert f *g\Vert _{T} \le C \Vert f\Vert _{S} \Vert g\Vert _{T}\qquad \ f\in S, g\in T. \end{aligned}$$

(3)

We refer the reader to [21, Prop. 4.2] for a generalization of (2) involving bounded measures.

Definition 2.2

We say that a Segal algebra S is star-symmetric if the involution is an isometry in the S-norm, meaning that for $f\in S$ we have that $f^*\in S$ and $\Vert f^*\Vert _S = \Vert f\Vert _S.$

Clearly $L^{1}({\mathbb {R}}^{n})$ is a star-symmetric Segal algebra. In the following example, we will review a few classical examples of Segal algebras of central importance.

Example 2.3

• (E1) Let $C_{0}({\mathbb {R}}^{n})$ denote the continuous function on ${\mathbb {R}}^{n}$ that vanish towards infinity. Then the space $S_{\infty } :=L^{1}({\mathbb {R}}^{n}) \cap C_{0}({\mathbb {R}}^{n})$ is a star-symmetric Segal algebra with the norm

  $$\begin{aligned}\Vert f\Vert _{S_{\infty }} :=\Vert f\Vert _{L^{1}({\mathbb {R}}^{n})} + \Vert f\Vert _{L^\infty ({\mathbb {R}}^{n})}.\end{aligned}$$

• (E2) For $1< p < \infty$ the space $S_{p} :=L^{1}({\mathbb {R}}^{n}) \cap L^{p}({\mathbb {R}}^{n})$ is a star-symmetric Segal algebra with the norm

  $$\begin{aligned}\Vert f\Vert _{S_{p}} :=\Vert f\Vert _{L^{1}({\mathbb {R}}^{n})} + \Vert f\Vert _{L^{p}({\mathbb {R}}^{n})}.\end{aligned}$$

• (E3) Let $\mu$ be a positive and unbounded measure on ${\mathbb {R}}^{n}$, for example the Lebesgue measure $\textrm{d}x$. For $1 \le p < \infty$ we denote by

  $$\begin{aligned}S_{p}^{\mu } :=L^{1}({\mathbb {R}}^{n}) \cap {\mathcal {F}}^{-1}\left( L^{p}({\mathbb {R}}^{n}, \mu )\right) .\end{aligned}$$

  Then $S_{p}^{\mu }$ is a star-symmetric Segal algebra with the norm

  $$\begin{aligned}\Vert f\Vert _{S_{p}^{\mu }} :=\Vert f\Vert _{L^{1}({\mathbb {R}}^{n})} + \Vert {\mathcal {F}}(f)\Vert _{L^{p}({\mathbb {R}}^{n}, \mu )}.\end{aligned}$$

• (E4) Recall that Wiener’s algebra W on ${\mathbb {R}}^{n}$ is defined as the space of those continuous functions on ${\mathbb {R}}^{n}$ such that

  $$\begin{aligned} \Vert f\Vert _W' :=\sum _{m \in {\mathbb {Z}}^{n}} a_m(f) < \infty , \quad \text {where } a_m(f) = \sup _{x \in [0,1]^{n}} |f(m + x)|. \end{aligned}$$

  Clearly, $W \subset L^1({\mathbb {R}}^{n}) \cap C_0({\mathbb {R}}^{n})$. Letting the norm on W be defined by

  $$\begin{aligned} \Vert f\Vert _W :=\sup _{y \in {\mathbb {R}}^{n}} \Vert \alpha _y(f)\Vert _W', \end{aligned}$$

  turns W into a Segal algebra. It is immediately clear that W is also a module under pointwise multiplication by functions from $C_0({\mathbb {R}}^{n})$. Indeed, W is the smallest Segal with this property:

Theorem 2.4

([5]) Let $S \subset L^1({\mathbb {R}}^n)$ be a Segal algebra which is a $C_0({\mathbb {R}}^n)$ module with respect to pointwise multiplication. Then, S contains W and $\Vert f\Vert _W \le C\Vert f\Vert _S$ for all $f\in W$.

• (E5) The Feichtinger algebra ${\mathcal {S}}_{0}({\mathbb {R}}^n)$ in time-frequency analysis is typically defined through the short-time Fourier transform: One can define the Feichtinger algebra as the set of functions $f \in L^{1}({\mathbb {R}}^{n}) \cap L^{2}({\mathbb {R}}^{n})$ such that $V_{f}f \in L^{1}({\mathbb {R}}^{2n})$, where

  $$\begin{aligned} V_{g}f(x, \xi ) :=\int _{{\mathbb {R}}^{n}}f(t)\overline{g(t - x)}e^{- i \xi t} \, \textrm{d}t. \end{aligned}$$

  (4)

  Through this description, it is somewhat nontrivial to see that the Feichtinger algebra is a star-symmetric Segal algebra. However, one can also describe the Feichtinger algebra as the minimal Segal algebra that is (in a strong sense) closed under modulations

  $$\begin{aligned} M_{\xi }f(t) :=e^{i t \xi }f(t). \end{aligned}$$

  (5)

  Note that in [6], the term (strongly) character-invariant was used instead of modulation-invariant. We refer the reader to [16] for a careful examination of the Feichtinger algebra on locally compact abelian groups, together with the original paper [6].

Remark 3

The intersection of all Segal algebras is precisely the set of continuous functions in $L^{1}({\mathbb {R}}^{n})$ whose Fourier transform is compactly supported, we refer to [21, (vii) p. 26] for a proof. As such, not every Segal algebra contains the Schwartz functions ${\mathcal {S}}({\mathbb {R}}^{n})$ as a subspace.

Segal algebras on ${\mathbb {R}}^{n}$ are not unital algebras, since $(2\pi )^{n/2}\delta _0\not \in L^1({\mathbb {R}}^n)$. However, every Segal algebra contains an approximate unit that is normalized in $L^{1}({\mathbb {R}}^{n})$, see [21, Prop. 8.1]. As such, the Cohen-Hewitt factorization theorem implies that every element in a Segal algebra $g \in S$ can be factorized as $g = f *h$ for $f \in L^{1}({\mathbb {R}}^{n})$ and $h \in S$. We succinctly write

$$\begin{aligned} L^{1}({\mathbb {R}}^{n}) *S = S. \end{aligned}$$

(6)

It follows from [3, Prop. 3.1] and (6) applied to $S = L^{1}({\mathbb {R}}^{n})$ that no maximal ideal of $L^{1}({\mathbb {R}}^{n})$ can be dense. Hence Segal algebras are never maximal ideals. We note that $L^{1}({\mathbb {R}}^{n})$ has plenty of closed maximal ideals, e.g., for every $\xi \in {\mathbb {R}}^{n}$ the space

$$\begin{aligned}I_{\xi } :=\left\{ f \in L^{1}({\mathbb {R}}^{n}): {\mathcal {F}}f(\xi ) = 0\right\} ,\end{aligned}$$

is a maximal ideal. When it comes to ideals inside a Segal algebra $S \subset L^{1}({\mathbb {R}}^{n})$, then every closed ideal $I_{S}$ in S is on the form $I_{S} = I \cap S$, where I is a unique closed ideal of $L^{1}({\mathbb {R}}^{n})$ by [21, Thm. 9.1].

2.2 Quantum Harmonic Analysis

The operators in this article will always be on the Hilbert space ${\mathcal {H}}=L^2({\mathbb {R}}^n)$. Recall that the Weyl operators acting on ${\mathcal {H}}$ are given by

$$\begin{aligned} W_{(x,\,\xi )}\phi (y):=e^{i\xi y-ix\xi /2}\phi (y-x), \end{aligned}$$

where $x, \, \xi \in {\mathbb {R}}^n$. It is common to use the shorthand notation $W_z:=W_{(x,\,\xi )}$, where $z = (x,\, \xi )\in {\mathbb {R}}^{2n}$. The Weyl operators are unitary operators and satisfy $W_{z}^*= W_{-z}$. The name Weyl operators comes from the Weyl commutation relation, often called the CCR relation, given by

$$\begin{aligned} W_zW_{z'}=e^{-i\sigma (z,\,z')/2}W_{z+z'} =e^{-i\sigma (z,\,z')}W_{z'}W_{z}. \end{aligned}$$

Here, $\sigma$ denotes the symplectic form

$$\begin{aligned} \sigma (z, z')= \sigma ((x, \xi ), (x', \xi ')) :=\xi x' - x \xi '. \end{aligned}$$

We will now define the basic operations of quantum harmonic analysis, namely convolutions and Fourier transforms. Unless otherwise stated, the proof of the statements in Sect. 2.2.1 and Sect. 2.2.2 can be found in [25].

2.2.1 Convolutions

Given a bounded linear operator $A \in {\mathcal {L}}({\mathcal {H}})$, we define the translation of the operator A by $z \in {\mathbb {R}}^{2n}$ as

$$\begin{aligned}\alpha _z(A) :=W_z A W_{-z}.\end{aligned}$$

Using translations, one can define the function-operator convolution for $f \in L^1({\mathbb {R}}^{2n})$ and $A \in {\mathcal {L}}({\mathcal {H}})$ as the operator

$$\begin{aligned} f *A :=\frac{1}{(2\pi )^n}\int _{{\mathbb {R}}^{2n}} f(z) \alpha _z(A)\,\textrm{d}z=:A*f. \end{aligned}$$

(7)

The convolution (7) is interpreted as a weak integral and the resulting operator is bounded. We say that a linear space $X\subset {\mathcal {L}} ({\mathcal {H}})$ is translation-invariant if $\alpha _z (X)\subset X$ for all $z\in {\mathbb {R}}^{2n}$. A translation-invariant Banach space $(X,\Vert \cdot \Vert _X)$ of operators $X\subset {\mathcal {L}}({\mathcal {H}})$ is said to have a translation-invariant norm if the shifts act isometrically, i.e., $\Vert \alpha _z(A)\Vert _X = \Vert A\Vert _X$ and $z \mapsto \alpha _z(A)$ is continuous in $\Vert \cdot \Vert _X$ for all $A\in X$. Finally, a translation-invariant Banach space $(X, \Vert \cdot \Vert _X) \subset {\mathcal {L}}({\mathcal {H}})$ with translation-invariant norm is called strongly translation-invariant if ${\mathbb {R}}^{2n} \ni z \mapsto \alpha _z(A) \in X$ is continuous for each $A \in X$. Given a strongly translation-invariant Banach space $(X,\Vert \cdot \Vert _X)$ which continuously embeds into ${\mathcal {L}}({\mathcal {H}})$, then $f *A \in X$ for all $A\in X$ and $f \in L^1({\mathbb {R}}^{2n})$. Examples of such spaces are the p-Schatten class operators ${{\mathcal {T}}}^p = {{\mathcal {T}}}^p({\mathcal {H}})$ for $1 \le p \le \infty$. The special case $p=1$ is the trace class operators, while $p=2$ is the Hilbert-Schmidt operators. We use the convention that ${{\mathcal {T}}}^\infty$ is the compact operators with the operator norm.

We will use $P:{\mathcal {H}}\rightarrow {\mathcal {H}}$ to denote the parity operator defined by $P\phi (x) = \phi (-x)$. Notice that the parity operator satisfies

$$\begin{aligned}PW_z=W_{-z}P. \end{aligned}$$

For $A, B \in {{\mathcal {T}}}^1$ we define their operator-operator convolution as

$$\begin{aligned} A *B(z) :={{\,\textrm{tr}\,}}(A \alpha _z( P B P))=B *A(z), \quad z \in {\mathbb {R}}^{2n}. \end{aligned}$$

It is of central importance in quantum harmonic analysis that $A *B \in L^1({\mathbb {R}}^{2n})$. Furthermore, one has

$$\begin{aligned} \frac{1}{(2\pi )^n} \int _{{\mathbb {R}}^{2n}} A *B(z) \,\textrm{d}z = {{\,\textrm{tr}\,}}(A) {{\,\textrm{tr}\,}}(B). \end{aligned}$$

The convolutions satisfy the following associativity conditions

$$\begin{aligned}f*(A*B)=(f*A)*B, \qquad f*(g*A)=(f*g)*A\end{aligned}$$

for $f,g\in L^1({\mathbb {R}}^{2n})$ and $A,B\in {{\mathcal {T}}}^1$.

For $A \in {{\mathcal {T}}}^1$ we define the operator involution by

$$\begin{aligned} A^{*_{\text {QHA}}} :=PA^*P. \end{aligned}$$

The involution satisfies

$$\begin{aligned} (A*B)^*=(A^{*_{\text {QHA}}})*(B^{*_{\text {QHA}}}) \end{aligned}$$

(8)

and

$$\begin{aligned} (f*A)^{*_{\text {QHA}}}=(f^{*})*(A^{*_{\text {QHA}}}) \end{aligned}$$

(9)

for $f\in L^1({\mathbb {R}}^{2n})$ and $A,B\in {{\mathcal {T}}}^1$.

One can define a commutative product on $L^1 \oplus {{\mathcal {T}}}^1 :=L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$ by

$$\begin{aligned}(f,A)*(g,B)=(f*g+A*B, f*B+g*A),\end{aligned}$$

where $(f,A),\,(g,B)\in L^1 \oplus {{\mathcal {T}}}^1$. Defining the norm on $L^1 \oplus {{\mathcal {T}}}^1$ as

$$\begin{aligned} \Vert (f, A)\Vert = \Vert f\Vert _{L^1} + \Vert A\Vert _{{{\mathcal {T}}}^1} \end{aligned}$$

turns the space into a commutative Banach algebra. We can also make the space into a Banach $*$-algebra by defining the involution as

$$\begin{aligned} (f, A)^*= (f^*,A^{*_{\text {QHA}}}). \end{aligned}$$

Additionally, the involution is norm isometric:

$$\begin{aligned}\Vert (f,A)^*\Vert =\Vert (f,A)\Vert .\end{aligned}$$

We define the translation of $(f,A)\in L^1\oplus {{\mathcal {T}}}^1$ by an element $z\in {\mathbb {R}}^{2n}$ by

$$\begin{aligned}\alpha _z(f,A):=(\alpha _z f, \alpha _z A).\end{aligned}$$

Using the translations, we define a set $X\subset L^1\oplus {{\mathcal {T}}}^1$ to be translation-invariant if

$$\begin{aligned}\alpha _z(X)\subset X\text { for all } z\in {\mathbb {R}}^{2n}.\end{aligned}$$

2.2.2 Fourier Transforms

Quantum harmonic analysis comes with its own notion of Fourier transforms. We let ${\mathcal {F}}_{\sigma }(f)$ be the symplectic Fourier transform, i.e.,

$$\begin{aligned} {\mathcal {F}}_\sigma (f)(z') :=\frac{1}{(2\pi )^n}\int _{{\mathbb {R}}^{2n}}f(z) e^{-i\sigma (z', z)}\,\textrm{d}z, \quad z' \in {\mathbb {R}}^{2n},\ f \in L^{1}({\mathbb {R}}^{2n}).\qquad \end{aligned}$$

(10)

Further, we define the Fourier-Weyl transform by

$$\begin{aligned} {\mathcal {F}}_{W}(A)(z') :={{\,\textrm{tr}\,}}(AW_{z'}), \quad z' \in {\mathbb {R}}^{2n},\ A\in {{\mathcal {T}}}^1. \end{aligned}$$

We will also occasionally denote the symplectic Fourier transform of f by ${\widehat{f}} = {\mathcal {F}}_\sigma (f)$ and the Fourier-Weyl transform by ${\widehat{A}}={\mathcal {F}}_W(A)$. The Fourier-Weyl and symplectic Fourier transforms satisfy

$$\begin{aligned} {\mathcal {F}}_\sigma (A *B) = {\mathcal {F}}_W(A) \cdot {\mathcal {F}}_W(B) \end{aligned}$$

(11)

and

$$\begin{aligned} {\mathcal {F}}_W(f *A) = {\mathcal {F}}_\sigma (f) \cdot {\mathcal {F}}_W(A), \end{aligned}$$

(12)

for $A,B\in {{\mathcal {T}}}^1$ and $f\in L^1({\mathbb {R}}^{2n})$. Notice that

$$\begin{aligned} {\mathcal {F}}_W(A^{*_{\text {QHA}}}) = \overline{{\mathcal {F}}_W(A)}\quad \text { and }\quad {\mathcal {F}}_\sigma (f^{*}) = \overline{{\mathcal {F}}_\sigma (f)}. \end{aligned}$$

(13)

The map ${{\mathcal {T}}}^1 \ni A \mapsto {\mathcal {F}}_W(A)$ is injective and is in fact an isomorphism between ${{\mathcal {T}}}^2$ and $L^2({\mathbb {R}}^{2n})$. We define the Fourier transform on $L^1\oplus {{\mathcal {T}}}^1$ by

$$\begin{aligned} {\mathcal {F}}(f, A) = ({\mathcal {F}}_\sigma (f), {\mathcal {F}}_W(A)), \quad (f, A) \in L^1 \oplus {{\mathcal {T}}}^1. \end{aligned}$$

The inverse of the Fourier-Weyl transform denoted by ${\mathcal {F}}_W^{-1}$ is given by

$$\begin{aligned} {\mathcal {F}}_W^{-1}(f) = \int _{{\mathbb {R}}^{2n}} f(z) W_{-z}\,\textrm{d}z, \end{aligned}$$

understood as an integral in strong operator topology for $f \in L^1({\mathbb {R}}^{2n}).$ Using interpolation, one can define the inverse Fourier-Weyl transform on $L^p$ for $1\le p\le 2$. It should be noted that the inverse Fourier-Weyl transform can be generalized to tempered distributions, see [18]. In particular, this means that one can define the Fourier-Weyl transform of all compact operators.

The Fourier-Weyl transform satisfies a version of the Riemann-Lebesgue lemma and Hausdorff-Young inequality, see [25, Prop.  3.4] and [19, Prop. 6.5 and 6.6]. In particular, we have that

$$\begin{aligned} {\mathcal {F}}(f, A) \in C_0({\mathbb {R}}^{2n}) \oplus C_0({\mathbb {R}}^{2n}). \end{aligned}$$

(14)

Analogous results are true for ${\mathcal {F}}_W^{-1}$, as we show now.

Lemma 2.5

Let $1 \le p \le 2$ and $\frac{1}{p} + \frac{1}{q} = 1$. For $f \in L^p({\mathbb {R}}^{2n})$ one has ${\mathcal {F}}_W^{-1}(f) \in {{\mathcal {T}}}^q({\mathcal {H}})$ and ${\mathcal {F}}_W^{-1}$ satisfies the Hausdorff-Young inequality

$$\begin{aligned}\Vert {\mathcal {F}}_W^{-1}(f)\Vert _{{{\mathcal {T}}}^q} \le \Vert f\Vert _{L^p}.\end{aligned}$$

In particular, for $p=1$ we have the Riemann-Lebesgue lemma: ${\mathcal {F}}_W^{-1}(f) \in {{\mathcal {T}}}^\infty$, where ${{\mathcal {T}}}^\infty$ denotes the space of compact operators.

Proof

For the proof of the Riemann-Lebesgue lemma, note that ${\mathcal {F}}_W^{-1}(f) \in {{\mathcal {T}}}^{2}\subset {{\mathcal {T}}}^{\infty }$ for $f \in L^1({\mathbb {R}}^{2n}) \cap L^2({\mathbb {R}}^{2n})$. Hence, for arbitrary $f \in L^1({\mathbb {R}}^{2n})$ the operator ${\mathcal {F}}_W^{-1}(f)$ can be approximated by Hilbert-Schmidt operators in operator norm, hence it is compact.

Notice that for compact operators, the operator norm is given by $\Vert \cdot \Vert _{\textrm{Op}}=\Vert \cdot \Vert _{{{\mathcal {T}}}^\infty }$. Since $\Vert W_{-z}\Vert _{\textrm{Op}} = 1$ and $z \mapsto W_{-z}$ is continuous in the strong operator topology, clearly $\Vert {\mathcal {F}}_W^{-1}(f)\Vert _{\textrm{Op}} \le \Vert f\Vert _{L^1}$. Furthermore, we know that ${\mathcal {F}}_W^{-1}:L^2({\mathbb {R}}^{2n}) \rightarrow {{\mathcal {T}}}^2$ is an isometry. Hence, the result follows from applying complex interpolation. $\square$

One can also consider the convolution of a complex Radon measure with an operator. Denote by ${\text {Meas}}({\mathbb {R}}^{2n})$ the set of all finite complex Radon measures on ${\mathbb {R}}^{2n}$, i.e., the set of finite complex Borel measures that are both inner and outer regular. Then, for $\mu \in {\text {Meas}}({\mathbb {R}}^{2n})$ and $A \in {{\mathcal {T}}}^1$, we define

$$\begin{aligned} \mu *A :=\frac{1}{(2\pi )^n}\int _{{\mathbb {R}}^{2n}} \alpha _z(A) \,\textrm{d}\mu (z). \end{aligned}$$

The above expression is valid as a Bochner integral in ${{\mathcal {T}}}^1$. In particular, $(2\pi )^n\delta _0 *A = A$. One can verify that we still have the product formula

$$\begin{aligned} {\mathcal {F}}_W(\mu *A) = {\mathcal {F}}_\sigma (\mu ) {\mathcal {F}}_W(A), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {F}}_\sigma (\mu )(z') :=\frac{1}{(2\pi )^n}\int _{{\mathbb {R}}^{2n}} e^{-i\sigma (z', z)}\, \textrm{d}\mu (z). \end{aligned}$$

Then, if $A \in {{\mathcal {T}}}^1$ is such that ${\mathcal {F}}_W(A)(\xi ) \ne 0$ for every $\xi \in {\mathbb {R}}^{2n}$, we clearly obtain that the map

$$\begin{aligned} {\text {Meas}}({\mathbb {R}}^{2n}) \ni \mu \mapsto \mu *A \in {{\mathcal {T}}}^1 \end{aligned}$$

is injective.

2.2.3 Schwartz Operators and Weyl Quantization

The article [18] defines Schwartz and tempered operators, mirroring the Schwartz functions ${\mathcal {S}}({\mathbb {R}}^{2n})$ and tempered distributions ${\mathcal {S}}'({\mathbb {R}}^{2n})$ in harmonic analysis. We denote by ${\mathcal {S}}({\mathcal {H}}) \subset {{\mathcal {T}}}^1({\mathcal {H}})$ the space of Schwartz operators, which is a Frechét space that is continuously embedded into ${{\mathcal {T}}}^1({\mathcal {H}})$. Its topology can be described in terms of a countable family of seminorms, see [18, Sec. 3]. We will need the fact that the Fourier-Weyl transform is a topological isomorphism ${\mathcal {F}}_W:{\mathcal {S}}({\mathcal {H}}) \rightarrow {\mathcal {S}}({\mathbb {R}}^{2n})$. Hence, a trace class operator A is a Schwartz operator if and only if ${\mathcal {F}}_W(A)$ is a Schwartz function. Similarly, a Hilbert-Schmidt operator $A \in {{\mathcal {T}}}^2({\mathcal {H}})$ is a Schwartz operator if and only if its integral kernel is a Schwartz function.

Having ${\mathcal {S}}({\mathcal {H}})$ available, one can pass to the dual space ${\mathcal {S}}'({\mathcal {H}})$ of tempered operators. The space of bounded operators, ${\mathcal {L}}({\mathcal {H}})$, continuously embeds into ${\mathcal {S}}'({\mathcal {H}})$ via

$$\begin{aligned} \varphi _A(S) = {{\,\textrm{tr}\,}}(AS), \quad S \in {\mathcal {S}}({\mathcal {H}}) \end{aligned}$$

for $A \in {\mathcal {L}}({\mathcal {H}})$. The Fourier transform, being defined by duality, then extends to a topological isomorphism from ${\mathcal {S}}'({\mathcal {H}})$ to ${\mathcal {S}}'({\mathbb {R}}^{2n})$. In particular, we can talk about ${\mathcal {F}}_W(A)$ as a tempered distribution on ${\mathbb {R}}^{2n}$ for every $A \in {\mathcal {L}}({\mathcal {H}})$. The Fourier-Weyl transform satisfies (11) between ${\mathcal {S}}({\mathcal {H}})$ and ${\mathcal {S}}'({\mathcal {H}})$ and (12) between ${\mathcal {S}}({\mathbb {R}}^{2n})$ and ${\mathcal {S}}'({\mathcal {H}})$.

Define the Weyl quantization of an $L^1({\mathbb {R}}^{2n})$ function f to be the operator

$$\begin{aligned}A_f:=P{\mathcal {F}}_W^{-1}({\mathcal {F}}_\sigma (f))P={\mathcal {F}}_W^{-1}({\mathcal {F}}_\sigma ({\widetilde{f}})),\end{aligned}$$

where ${\widetilde{f}}(z)=f(-z)$. Notice that the Weyl quantization also makes sense for $f \in {\mathcal {S}}'({\mathbb {R}}^{2n})$. Thus, $f \mapsto A_f$ maps from ${\mathcal {S}}'({\mathbb {R}}^{2n})$ to ${\mathcal {S}}'({\mathcal {H}})$. Since the quantization map is also a topological isomorphism from ${\mathcal {S}}({\mathbb {R}}^{2n})$ to ${\mathcal {S}}({\mathcal {H}})$, we have

$$\begin{aligned} {\mathcal {S}}({\mathcal {H}}) \subset \{ A \in {{\mathcal {T}}}^1({\mathcal {H}}): A = A_f, ~f \in L^1({\mathbb {R}}^{2n})\}. \end{aligned}$$

Taking the Weyl quantization of the delta function

$$\begin{aligned}(2\pi )^n\delta _{0}\in {\mathcal {S}}'({\mathbb {R}}^{2n})\end{aligned}$$

gives $A_{(2\pi )^n\delta _{0}}=2^n P$, where P is the parity operator. Note that the quantization map satisfies $\alpha _z(A_f) = A_{\alpha _{-z}(f)}$. For $g \in L^p({\mathbb {R}}^{2n})$ where $1 \le p \le \infty$ and $f \in L^1({\mathbb {R}}^{2n})$ we have the convolution formulas

$$\begin{aligned} f *A_g = A_{{\widetilde{f}}*g}, \quad A_f *A_g = \widetilde{f *g}. \end{aligned}$$

(15)

This gives us the following quantization identity

$$\begin{aligned}2^nf*P=A_{{\widetilde{f}}},\qquad 2^nA_f*P={\widetilde{f}}\end{aligned}$$

for $f\in {\mathcal {S}}({\mathbb {R}}^{2n})$.

2.3 Modulation Invariance

In this section, we will develop a notion of modulation for operators. This allows us to define modulation-invariant spaces of operators. Motivated by the fact that the classical Fourier transform takes translations to modulations and vice versa, we have the following definition.

Definition 2.6

We say that a subspace ${\mathcal {A}} \subset {{\mathcal {T}}}^{1}$ is modulation-invariant if

$$\begin{aligned}{\mathcal {F}}_W({\mathcal {A}}) :=\left\{ {\mathcal {F}}_{W}(A): A \in {\mathcal {A}}\right\} \end{aligned}$$

is a translation-invariant subspace of $C_{0}({\mathbb {R}}^{2n})$.

Lemma 2.7

Let ${\mathcal {A}} \subset {{\mathcal {T}}}^1$ be a subspace. Then ${\mathcal {A}}$ is modulation-invariant if and only if $W_{z} A W_{z} \in {\mathcal {A}}$ for every $A \in {\mathcal {A}}$ and $z \in {\mathbb {R}}^{2n}$.

Proof

For $z' \in {\mathbb {R}}^{2n}$ the claim follows easily from the computation

$$\begin{aligned} \alpha _{z}\left( {\mathcal {F}}_{W}(A)(z')\right)&= {{\,\textrm{tr}\,}}(W_{z' - z}A) \\&= {{\,\textrm{tr}\,}}\left( W_{-\frac{z}{2}}AW_{-\frac{z}{2}} W_{z'}\right) \\&= {\mathcal {F}}_{W}\left( W_{-\frac{z}{2}}AW_{-\frac{z}{2}}\right) (z'). \end{aligned}$$

$\square$

Motivated by the proof of Lemma 2.7 we give the following definition for the modulation of an operator.

Definition 2.8

The modulation of an operator $A \in {{\mathcal {T}}}^{1}$ by $z \in {\mathbb {R}}^{2n}$ is defined by

$$\begin{aligned}\gamma _{z}(A):=W_{-z/2}AW_{-z/2}.\end{aligned}$$

For $f \in L^{1}({\mathbb {R}}^{2n})$ we define the modulation by

$$\begin{aligned} \gamma _{z}(f)(z') :=e^{-i\sigma (z', z)} f(z'). \end{aligned}$$

It is straightforward to verify the formulas

$$\begin{aligned} \gamma _{z}(f *g)&= (\gamma _{z}f) *(\gamma _{z} g), \\ \gamma _{z}(f *A)&= (\gamma _{z} f) *(\gamma _{z}A), \\ \gamma _{z}(A *B)&= (\gamma _{z} A) *(\gamma _{z}B), \end{aligned}$$

for $f, g \in L^{1}({\mathbb {R}}^{2n})$ and $A, B \in {{\mathcal {T}}}^1$. Notice that for $A \in {{\mathcal {T}}}^{1}$ and $f\in L^1({\mathbb {R}}^{2n})$ we can write

$$\begin{aligned}{\mathcal {F}}_{W}(A)(z) = {{\,\textrm{tr}\,}}(\gamma _{-z}A)\quad \text { and } \quad {\mathcal {F}}_{\sigma }(f)(z) = \int _{{\mathbb {R}}^{2n}}\gamma _{-z}(f)(z')\, \textrm{d} z'.\end{aligned}$$

Lemma 2.9

Let ${\mathcal {A}}$ be a translation-invariant subspace of ${{\mathcal {T}}}^1$. Then the following are equivalent:

• 1) ${\mathcal {A}}$ is modulation-invariant,

• 2) $W_{z} A \in {\mathcal {A}}$ for every $A\in {\mathcal {A}}$ and $z \in {\mathbb {R}}^{2n}$,

• 3) $AW_{z} \in {\mathcal {A}}$ for every $A \in {\mathcal {A}}$ and $z\in {\mathbb {R}}^{2n}$,

• 4) $W_{z'}AW_{z} \in {\mathcal {A}}$ for every $A \in {\mathcal {A}}$ and $z',z\in {\mathbb {R}}^{2n}$,

• 5) The space consisting of all integral kernels of $A\in {\mathcal {A}}$ is translation- and modulation-invariant.

Proof

The equivalence of 1)– 4) follows from translation-invariance together with the observation

$$\begin{aligned} \gamma _{z}\alpha _{z'}({\widehat{A}})&={\mathcal {F}}_W(\alpha _{z}\gamma _{z'}A)\\&={\mathcal {F}}_W(W_{z}W_{-z'/2}AW_{-z'/2}W_{z})\\&=e^{i\sigma (z',z)/2}{\mathcal {F}}_W(W_{z-z'/2}AW_{-z'/2-z}). \end{aligned}$$

For the last equivalence, recall that a trace class operator can be written as

$$\begin{aligned}A\phi (s)=\int _{{\mathbb {R}}^n}K_A(s,t)\phi (t)\,\textrm{d}t,\quad \phi \in {\mathcal {H}}.\end{aligned}$$

The integral kernel of $W_{z'}AW_z$ with $z = (x, \xi )$ and $z' = (x', \xi ')$ is given by

$$\begin{aligned} (s, t) \mapsto&e^{i\xi ' \cdot s - \frac{i}{2} \xi '\cdot x'}e^{i\xi \cdot t + \frac{i}{2} \xi \cdot x}K_A(s-x', t+x)\\&\quad =e^{ \frac{i}{2} \xi \cdot x - \frac{i}{2} \xi '\cdot x'}\gamma _{(\xi ,-\xi ')}\alpha _{(x',-x)}K_A(s, t). \end{aligned}$$

$\square$

Analogously to the case of the translation action $\alpha _x$, one also proves the following facts. We emphasize again that ${{\mathcal {T}}}^\infty ({\mathcal {H}})$ denotes the ideal of compact operators.

Lemma 2.10

The modulation $\gamma _x$ acts strongly continuous and isometrically on the Schatten classes ${{\mathcal {T}}}^p({\mathcal {H}})$ for every $1 \le p \le \infty$. It acts continuous in $\hbox {weak}^*$ topology on ${\mathcal {L}}({\mathcal {H}})$.

Proof

The proof is entirely analogous as for the action $\alpha _x$: Verify the statements first for rank one operators, using that $W_x$ is a strongly continuous projective unitary representation. Then, obtain the result on all of ${{\mathcal {T}}}^p({\mathcal {H}})$ by approximation through finite rank operators. $\square$

3 Structure Theory of $L^1 \oplus {{\mathcal {T}}}^1$

In Sect. 2.2.1 we defined a product structure for $L^1 \oplus {{\mathcal {T}}}^1$, making it into an involutive commutative Banach algebra. As such we can study the closed ideals of this algebra. We will start with the study of the regular maximal ideals, i.e., the Gelfand theory, and thereafter study the closed graded ideals of $L^1 \oplus {{\mathcal {T}}}^1$. For an introduction to Gelfand theory, see, e.g., [17].

3.1 Gelfand Theory for $L^1 \oplus {{\mathcal {T}}}^1$

In this section we will study the Gelfand theory of $L^1 \oplus {{\mathcal {T}}}^1$. We say that an ideal I is regular if the quotient $L^1 \oplus {{\mathcal {T}}}^1/I$ is unital, see [23, Sec. D4]. It is well known that the regular maximal ideals are given by the kernel of the multiplicative linear functionals.

Proposition 3.1

The set of nonzero multiplicative linear functionals on $L^1\oplus {{\mathcal {T}}}^1$ is parametrized by ${\mathbb {R}}^{2n} \times \mathbb Z_2$ as

$$\begin{aligned} \chi _{z, j}(f, A) = {\left\{ \begin{array}{ll} {\widehat{f}}(z) + {\widehat{A}}(z), &{}\quad j = 0,\\ {\widehat{f}}(z) - {\widehat{A}}(z), &{}\quad j = 1. \end{array}\right. } \end{aligned}$$

We write the set of all nonzero multiplicative linear functionals, the Gelfand spectrum, as $\chi _{{\mathbb {R}}^{2n} \times \mathbb Z_2}$. The Gelfand spectrum topology of $\chi _{{\mathbb {R}}^{2n} \times \mathbb Z_2}$ agrees with the standard product topology of the index set ${\mathbb {R}}^{2n}\times {\mathbb {Z}}_2$.

Proof

Let $\chi :L^1 \oplus {{\mathcal {T}}}^1 \rightarrow \mathbb C$ be a nonzero multiplicative linear functional. Note that by using linearity, $\chi$ is determined by its actions on functions and operators separately. By the Gelfand theory of $L^1({\mathbb {R}}^{2n})$, the multiplicative character $\chi$ acts on $L^1\oplus \{0\}$ by

$$\begin{aligned} \chi (f, 0) = {\widehat{f}}(z) \quad \text { for some }z \in {\mathbb {R}}^{2n}. \end{aligned}$$

For $A, B \in {{\mathcal {T}}}^1$ we have

$$\begin{aligned} \chi (0, A) \cdot \chi (0, B)&= \chi ((0, A) *(0, B)) = \chi ( A *B, 0) \\&= {\mathcal {F}}_\sigma (A *B)(z) = {\widehat{A}}(z) \cdot {\widehat{B}}(z). \end{aligned}$$

Letting $A = B$, we see that

$$\begin{aligned} \chi (0, A) = \pm {\widehat{A}}(z). \end{aligned}$$

Hence, we obtain that

$$\begin{aligned} \chi (f, A) = {\widehat{f}}(z) \pm {\widehat{A}}(z). \end{aligned}$$

As one easily verifies, every such $\chi$ is a multiplicative linear functional.

Let us now prove the equivalence of the topologies. A net $(z_\gamma ) \subset {\mathbb {R}}^{2n}$ converges to $z \in {\mathbb {R}}^{2n}$ if and only if $\chi _{z_\gamma , j} \rightarrow \chi _{z,j}$ in the Gelfand spectrum topology. Further, it can never happen that $\chi _{z_\gamma , j} \rightarrow \chi _{z, j+1}$. If so, we would have ${\widehat{A}}(z) = -{\widehat{A}}(z)$, implying that ${\widehat{A}}(z) = 0$ for every $A \in {{\mathcal {T}}}^1$. This can not happen, since there exist operators with nonvanishing Fourier-Weyl transform, e.g., $\phi \otimes \phi$ where $\phi (x)=e^{-x^2}$. $\square$

As a corollary, the maximal ideal space of $L^1\oplus {{\mathcal {T}}}^1$ consists of two disjoint copies of ${\mathbb {R}}^{2n}$.

Remark 4

Recall by (14) we have

$$\begin{aligned}{\mathcal {F}}(L^1 \oplus {{\mathcal {T}}}^1)\subset C_0({\mathbb {R}}^{2n}) \oplus C_0({\mathbb {R}}^{2n}).\end{aligned}$$

Using the convolution theorem, we can view the Fourier transform as a Banach algebra homomorphism

$$\begin{aligned} {\mathcal {F}}:L^1 \oplus {{\mathcal {T}}}^1 \rightarrow \ell ^1(\mathbb Z_2, C_0({\mathbb {R}}^{2n})). \end{aligned}$$

Denote by

$$\begin{aligned}\Gamma :L^1\oplus {{\mathcal {T}}}^1\rightarrow (\chi _{{\mathbb {R}}^{2n} \times \mathbb Z_2}\rightarrow {\mathbb {C}}),\end{aligned}$$

the Gelfand representation defined by

$$\begin{aligned}\Gamma (f,A)(\chi _{z,j}):=\chi _{z,j}(f,A)={\widehat{f}}(z)+(-1)^j{\widehat{A}}(z).\end{aligned}$$

We will refer to the map $\Gamma (f,A)$ is the Gelfand transform of (f, A).

Remark 5

The Gelfand representation can be viewed as a map $\Gamma :L^1\otimes {{\mathcal {T}}}^1\rightarrow C_0({\mathbb {R}}^{2n}\times {\mathbb {Z}}_2)$ given by

$$\begin{aligned}\Gamma (f,A)(z,j)={\widehat{f}}(z)+(-1)^j{\widehat{A}}(z).\end{aligned}$$

Since both Fourier transforms have dense range, it follows that the image of the Gelfand representation is dense in $C_0({\mathbb {R}}^{2n}\times {\mathbb {Z}}_2)$.

Recall that a ring is called Jacobson semisimple (or semiprimitive) if the Jacobson ideal is zero, i.e., the intersection of all the maximal ideals is zero. A property of the Gelfand transform is that $\ker (\Gamma )$ is precisely the Jacobson ideal. It follows from the Gelfand-Naimark theorem that any $C^{*}$-algebra is Jacobson semisimple. The following result shows that $L^1 \oplus {{\mathcal {T}}}^1$ shares this property.

Corollary 3.2

The algebra $L^1 \oplus {{\mathcal {T}}}^1$ is Jacobson semisimple.

Proof

Let (f, A) be such that $\Gamma (f, A) = 0$. Then for every $z \in {\mathbb {R}}^{2n}$ we obtain

$$\begin{aligned} {\widehat{f}}(z) + {\widehat{A}}(z) = 0 = {\widehat{f}}(z) - {\widehat{A}}(z). \end{aligned}$$

Hence ${\widehat{f}}={\widehat{A}} = 0$. Since both the symplectic Fourier transform and the Fourier-Weyl transform are injective, this yields that $(f, A) = 0$. Thus, the Jacobson radical is trivial. $\square$

We say that an involutive Banach algebra X is symmetric if its Gelfand representation satisfies

$$\begin{aligned}\Gamma (a^*)=\overline{\Gamma (a)} \quad \text { for all } a\in X.\end{aligned}$$

Using (13) we see that $L^1\oplus {{\mathcal {T}}}^1$ is symmetric since

$$\begin{aligned} \Gamma ((f, A)^*)(z, j) = \overline{{\mathcal {F}}_\sigma (f)(z)} + (-1)^j \overline{{\mathcal {F}}_W(A)(z)} = \overline{\Gamma (f, A)(z, j)}. \end{aligned}$$

Using that $L^1\oplus {{\mathcal {T}}}^1$ is symmetric and [9, Prop. 1.14c.], we get another proof of the denseness result in Remark 5.

3.2 Closed Ideals of $L^1 \oplus {{\mathcal {T}}}^1$

Having discussed Gelfand theory, we now turn to the task of understanding closed ideals of $L^1 \oplus {{\mathcal {T}}}^1$ in general. Recall that the closed ideals of $L^1({\mathbb {R}}^{2n})$ are precisely the closed translation-invariant subspaces of $L^1({\mathbb {R}}^{2n})$, see e.g., [9, Thm. 2.45]. Having a well-defined translation operator on $L^1\oplus {{\mathcal {T}}}^1$, one might hope that this carries over to closed ideals in this setting. However, the subspace $\{0\}\oplus {{\mathcal {T}}}^1$ is closed and translation-invariant, but not an ideal.

Definition 3.3

We say that a subspace $M\subset L^1\oplus {{\mathcal {T}}}^1$ is an $L^1$ module if for all $(f,A)\in M$ and $g\in L^1({\mathbb {R}}^{2n})$ we have

$$\begin{aligned}g*(f,A):=(g,0)*(f,A)=(g*f,g*A)\in M.\end{aligned}$$

Notice that all ideals of $L^1\oplus {{\mathcal {T}}}^1$ are $L^1$ modules.

Proposition 3.4

Let $M \subset L^1 \oplus {{\mathcal {T}}}^1$ be a closed subspace. Then, M an $L^1$ module if and only if it is translation-invariant.

Proof

If M is translation-invariant, then for every $(f, A) \in M$ and $g \in L^1$ the convolution satisfies

$$\begin{aligned} g *(f, A) = \int _{{\mathbb {R}}^{2n}} g(z) \alpha _z(f,A)\,\textrm{d}z\in M. \end{aligned}$$

On the other hand, if M is an $L^1$ module, then letting $\{g_t\}_{t>0}$ be a normalized approximate identity of $L^1({\mathbb {R}}^{2n})$ gives

$$\begin{aligned} \qquad \qquad \alpha _z(f, A) = \lim _{t \rightarrow 0} g_t *\alpha _z(f, A) = \lim _{t \rightarrow 0} \alpha _z(g_t) *(f, A) \in M.\qquad \qquad \end{aligned}$$

$\square$

As a consequence, for I being a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$, it is necessary to be translation-invariant. We continue our discussion with the following class of well-behaved ideals:

Definition 3.5

An ideal $I \subset L^1 \oplus {{\mathcal {T}}}^1$ is said to be graded if $I = I_{L^1} \oplus I_{{{\mathcal {T}}}^1}$, where $I_{L^1} = I \cap (L^1({\mathbb {R}}^{2n})\oplus \{0\})$ and $I_{{{\mathcal {T}}}^1} = I \cap (\{0\}\oplus {{\mathcal {T}}}^1)$.

We note that $I_{L^1}$ and $I_{{{\mathcal {T}}}^1}$ in the above definition are corresponding spaces in the sense of Werner [25, Sec. IV].

Note that a closed ideal I is a graded ideal if and only if $(f,0) \in I$ whenever $(f, A)\in I$. It follows that the regular maximal ideals

$$\begin{aligned} I_{(z, j)} :=\{ (f, A) \in L^1 \oplus {{\mathcal {T}}}^1: {\widehat{f}}(z) + (-1)^j {\widehat{A}}(z) = 0\} \end{aligned}$$

are never graded.

Lemma 3.6

A closed subspace $I \subset L^1 \oplus {{\mathcal {T}}}^1$ is a graded ideal if and only if $I = I_{L^1} \oplus \overline{({{\mathcal {T}}}^1 *I_{L^1})}$ for some closed ideal $I_{L^1}$ in $L^1({\mathbb {R}}^{2n})$.

Proof

Let us first quickly check that $I = I_{L^1} \oplus \overline{({{\mathcal {T}}}^1 *I_{L^1})}$ is indeed a closed ideal. If $(f, A *g) \in I$ and $(h, B) \in L^1 \oplus {{\mathcal {T}}}^1$, then

$$\begin{aligned} (h, B) *(f, A *g) = (h *f + B *A *g, A *g*h + f *B). \end{aligned}$$

Since $f, g \in I_{L^1}$ and $I_{L^1}$ is an ideal, we have $h *f \in I_{L^1}$ and $B *A *g \in I_{L^1}$, where we have used that $B *A \in L^1({\mathbb {R}}^{2n})$. Further, $A *g *h$ and $f *B$ are clearly contained in ${{\mathcal {T}}}^1 *I_{L^1}$, making I into an ideal. The fact that I is closed is inherited from $I_{L^1}$ being closed.

On the other hand, assume that $I= I_{L^1} \oplus I_{{{\mathcal {T}}}^1}$ is a graded closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$. Note that $I_{L^1}$ is a closed ideal of $L^1({\mathbb {R}}^{2n})$. Using the multiplication, we get $(\{0\}\oplus {{\mathcal {T}}}^1) *I_{L^1} \subset I_{{{\mathcal {T}}}^1}$ and $(\{0\}\oplus {{\mathcal {T}}}^1) *I_{{{\mathcal {T}}}^1} \subset I_{L^1}$. Further, as I is an $L^1$ module by Proposition 3.4, both $I_{L^1}$ and $I_{{{\mathcal {T}}}^1}$ are translation-invariant. Thus using [25, Thm. 4.1] gives that $I_{{{\mathcal {T}}}^1} = \overline{{{\mathcal {T}}}^1 *I_{L^1}}$, finishing the proof. $\square$

Remark 6

• (1) For $f \in L^1({\mathbb {R}}^{2n})$ we let Z(f) denote the closed set

  $$\begin{aligned} Z(f) :=\{ z \in {\mathbb {R}}^{2n}: {\widehat{f}}(z) = 0\}. \end{aligned}$$

  Then, for a closed ideal $I_{L^1}$ of $L^1({\mathbb {R}}^{2n})$, we set

  $$\begin{aligned}Z(I_{L^1}) :=\bigcap _{f \in I_{L^1}} Z(f),\end{aligned}$$

  which is closed. Malliavin’s theorem [23, Sec. 7.6] states that spectral synthesis fails for $L^1({\mathbb {R}}^{2n})$, that is: It is not true that every closed ideal $I_{L^1}$ is uniquely determined by $Z(I_{L^1})$.

  By analogy, for $(f, A) \in L^1 \oplus {{\mathcal {T}}}^1$ we let

  $$\begin{aligned} Z(f,A) :=\{ (z, j) \in {\mathbb {R}}^{2n} \times \mathbb Z_2: {\widehat{f}}(z) + (-1)^j {\widehat{A}}(z) = 0\}. \end{aligned}$$

  If I is a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$ we denote by

  $$\begin{aligned}Z(I) :=\bigcap _{(f,A)\in I} Z(f,A).\end{aligned}$$

  When I is a graded, by Lemma 3.6 we get that $Z(I) = Z(I_{L^1} ) \times \mathbb Z_2$. This means that for graded ideals the spectral synthesis again fails, i.e., I is not uniquely determined by this set Z(I).

• (2) Every proper closed graded ideal $I=I_{L^1} \oplus \overline{({{\mathcal {T}}}^1 *I_{L^1})}$ is contained in a regular maximal ideal. If $Z(I) \ne \emptyset$, we have that for $(z, j) \in Z(I)$ the inclusion $I \subset I_{(z, j)}$ holds. In the case $Z(I) = \emptyset$, it remains to show that $I=L^1 \oplus {{\mathcal {T}}}^1$. When $Z(I) = \emptyset$, we have $Z(I_{L^1})=\emptyset$. By the Tauberian theorem on $L^1({\mathbb {R}}^{2n})$, we have that $I_{L^1}=L^1({\mathbb {R}}^{2n})$. The statement follows from the fact that $\overline{{{\mathcal {T}}}^1*L^1({\mathbb {R}}^{2n})}={{\mathcal {T}}}^1$.

Every closed ideal contains a maximal graded ideal. To illustrate this, let us consider the regular maximal ideals.

Example 3.7

For $(z, j) \in {\mathbb {R}}^{2n} \times \mathbb Z_2$, we have

$$\begin{aligned} I_0&= I_{(z, j)} \cap ( L^1({\mathbb {R}}^{2n})\oplus \{0\} )= \{ f\in L^1({\mathbb {R}}^{2n}): {\widehat{f}}(z) = 0\}\\ I_1&= I_{(z, j)} \cap ( \{0\}\oplus {{\mathcal {T}}}^1 ) = \{ A \in {{\mathcal {T}}}^1: ~{\widehat{A}}(z) = 0\}. \end{aligned}$$

In this case, $I_0 \oplus I_1$ is a graded closed ideal given by

$$\begin{aligned} I_0 \oplus I_1 = I_{(z, 0)} \cap I_{(z, 1)}. \end{aligned}$$

The construction of graded subideals works more generally. Denote by J the map $J(f,A) :=(f, -A)$.

Lemma 3.8

Let I be a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$. Then

• 1) J(I) is a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$.

• 2) I is graded if and only if $J(I) = I$.

• 3) $I \cap J(I)$ is a closed graded ideal.

• 4) $Z(J(I)) = \{ (z, j+1) \in {\mathbb {R}}^{2n} \times \mathbb Z_2: ~(z, j) \in Z(I)\}$.

• 5) $Z(I \cap J(I)) = Z(I) \cup Z(J(I)) = \{ z \in {\mathbb {R}}^{2n}: ~(z, 0) \in Z(I) \text { or } (z, 1) \in Z(I)\} \times \mathbb Z_2$.

The proof of this lemma is straightforward and left to the reader. Note that the graded ideal $I \cap J(I)$ can be trivial, as the following example shows:

Example 3.9

We consider only $n = 1$, but the example can analogously be carried out for $n > 1$. We set

$$\begin{aligned} \Omega _j = \{ (x, y) \in {\mathbb {R}}^2: ~(-1)^jy \ge 0\}, \quad j\in {\mathbb {Z}}_2. \end{aligned}$$

Define the closed ideal

$$\begin{aligned} I :=\{ (f, A) \in L^1 \oplus {{\mathcal {T}}}^1: ~{\widehat{f}}(z) + (-1)^j {\widehat{A}}(z) = 0 \text { for } z \in \Omega _j,\, j\in {\mathbb {Z}}_2 \}. \end{aligned}$$

By the definition, we have $Z(I) = (\Omega _0 \times \{ 0\}) \cup (\Omega _1 \times \{ 1\})$. Thus, we clearly have

$$\begin{aligned} Z(I \cap J(I))=Z(\{(0,0)\}) = {\mathbb {R}}^2 \times \mathbb Z_2. \end{aligned}$$

Proposition 3.10

Let I be a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$. If $Z(I) = \emptyset$, then $I = L^1 \oplus {{\mathcal {T}}}^1$.

Proof

We clearly have $I \cap J(I) \subset I$. By Lemma 3.8 3), $I \cap J(I)$ is a closed graded ideal with $Z(I \cap J(I)) = \emptyset$. Thus, $I \cap J(I) = L^1 \oplus {{\mathcal {T}}}^1$ by Remark 6 (2). $\square$

Corollary 3.11

Every proper closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$ is contained in a regular maximal ideal.

Proof

By Proposition 3.10, if I is proper $Z(I) \ne \emptyset$. Hence, for $(z, j) \in Z(I)$, we have $I \subset I_{(z, j)}$. $\square$

Corollary 3.12

Every maximal closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$ is regular.

Proof

If I is a (proper) maximal closed ideal, then it is contained in a regular maximal ideal by the previous result. By maximality, both ideals have to agree. $\square$

Having looked at the maximal closed graded subideal of I, we now turn to the minimal closed graded ideal containing I.

Lemma 3.13

Let $I \subset L^1 \oplus {{\mathcal {T}}}^1$ be a closed ideal.

• 1. $\overline{I + J(I)}$ is a graded ideal.

• 2. $Z(\overline{I + J(I)} )= Z(I) \cap Z(J(I)) = \{ z \in {\mathbb {R}}^{2n}: ~(z, 0) \in Z(I) \text { and } (z, 1) \in Z(I)\} \times \mathbb Z_2$.

Again, we leave the proof of the lemma to the reader.

Example 3.14

In the case of the maximal ideal $I_{(z, j)}$, we have that

$$\begin{aligned}\overline{I_{(z, j)} + J(I_{(z, j)})}=L^1\oplus {{\mathcal {T}}}^1.\end{aligned}$$

This is a simple consequence of Proposition 3.10 together with $J(I_{(z, j)})=I_{(z, j+1)}$.

Let us now move on to prove some general properties of closed ideals. Let $V\subset L^1\oplus {{\mathcal {T}}}^1$, then

$$\begin{aligned}V':=\overline{(\{0\}\oplus {{\mathcal {T}}}^1 ) *V}.\end{aligned}$$

Proposition 3.15

Let $V \subset L^1 \oplus {{\mathcal {T}}}^1$ a closed, translation-invariant subspace.

• 1. $V'$ is a closed, translation-invariant subspace.

• 2. $Z(V') = Z(V)$.

• 3. $V = V''$.

• 4. $\overline{V + V'}$ is a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$.

• 5. $Z(\overline{V + V'}) = Z(V)$.

In particular, if $I \subset L^1 \oplus {{\mathcal {T}}}^1$ is a closed ideal then $I = \overline{I + I'}$.

Proof

• 1. By definition $V'$ is closed. Further, $(\{0\}\oplus {{\mathcal {T}}}^1 ) *V$ is translation-invariant, by the definition of convolution. Taking the closure preserves translation-invariance.

• 2. Let B be a regular operator, i.e., ${\mathcal {F}}_W(B)(z)$ is nonzero for all $z\in {\mathbb {R}}^{2n}$. Then by (11) and (12) we have that

  $$\begin{aligned} \widehat{A *B}(z) + (-1)^j \widehat{f *B}(z) =- {\widehat{B}}(z) ({\widehat{f}}(z) + (-1)^{j}{\widehat{A}}(z)), \quad (f,A)\in V. \end{aligned}$$

  Hence the result follows from the definition of $Z(V')$.

• 3. This follows from the fact that ${{\mathcal {T}}}^1 *{{\mathcal {T}}}^1 \subset L^1$ is dense, the assumptions on V and associativity of the convolutions.

• 4. Clearly, $\overline{V + V'}$ is an $L^1$ module. The definition of $V'$ makes $V'$ also invariant under convolution by ${{\mathcal {T}}}^1$.

• 5. The final property is an immediate consequence of 2.$\square$

The above result tells us how to construct ideals of $L^1 \oplus {{\mathcal {T}}}^1$: Pick any closed, translation-invariant subspace V of $L^1 \oplus {{\mathcal {T}}}^1$ and form $\overline{V + V'}$. Then, if $Z(V) \ne \emptyset$, we are guaranteed to obtain a proper closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$. Further, any closed ideal is of this form. Finally, if we let $V \subset L^1({\mathbb {R}}^{2n}) \oplus \{ 0\}$ be a closed translation-invariant subspace we obtain a graded ideal. Additionally, by Lemma 3.6 any closed graded ideal is of this form.

4 Quantum Segal Algebras

Moving on from the closed ideals and $L^1$ modules, we will in this section study dense $L^1$ modules endowed with a translation-invariant norm. In doing this, we will define a version of Segal algebras in $L^1\oplus {{\mathcal {T}}}^1$.

Definition 4.1

A quantum Segal algebra (QSA) is a pair $(QS, \Vert \cdot \Vert _{QS})$ where:

• (QS1) QS is a dense subspace of $L^1 \oplus {{\mathcal {T}}}^1$.

• (QS2) The space $(QS, \Vert \cdot \Vert _{QS})$ is a Banach algebra with multiplication inherited from $L^1 \oplus {{\mathcal {T}}}^1$.

• (QS3) The space QS is isometrically translation-invariant, i.e., $\alpha _z(f,A) \in QS$ for each $(f,A) \in QS$ and $z \in {\mathbb {R}}^{2n}$ with $\Vert \alpha _z(f,A)\Vert _{QS} = \Vert (f,A)\Vert _{QS}$.

• (QS4) The translations $\alpha _z$ act strongly continuous, i.e., for each $(f,A) \in QS$ the map ${\mathbb {R}}^{2n} \ni z \mapsto \alpha _z(f,A) \in QS$ is continuous.

If, additionally, the involution satisfies $QS^*=QS$ and further $\Vert (f,A)^*\Vert _{QS}=\Vert (f,A)\Vert _{QS}$ for all $(f,A)\in QS$, then we refer to $(QS, \Vert \cdot \Vert _{QS})$ as star-symmetric.

As a first consequence of the definition, we obtain that every quantum Segal algebra continuously embeds into the ambient space:

Proposition 4.2

Let $(QS, \Vert \cdot \Vert _{QS})$ be a quantum Segal algebra. Then, there is a constant $C > 0$ such that for each $(f, A) \in QS$ we have

$$\begin{aligned} \Vert (f, A)\Vert _{L^1 \oplus {{\mathcal {T}}}^1} \le C\Vert (f, A)\Vert _{QS}. \end{aligned}$$

The proposition is a direct application of the following well-known result on continuity of Banach algebra homomorphisms. Recall by Corollary 3.2 that $L^1 \oplus {{\mathcal {T}}}^1$ is Jacobson semisimple.

Lemma 4.3

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be commutative Banach algebras, where ${\mathcal {B}}$ is further assumed to be Jacobson semisimple. Then any algebra homomorphism $\varphi :{\mathcal {A}} \rightarrow {\mathcal {B}}$ is continuous.

Since the proof is very short, we present it for the reader’s convenience.

Proof

Let $(x_n)_{n \in \mathbb N}$ be a sequence in ${\mathcal {A}}$ converging to $0 \in {\mathcal {A}}$ and $y=\lim _{n\rightarrow \infty }\varphi (x_n)$. Denote by ${\mathcal {M}}({\mathcal {A}})$ and ${\mathcal {M}}({\mathcal {B}})$ the space of multiplicative linear functionals on ${\mathcal {A}}$ and ${\mathcal {B}}$, respectively. Then, for every $\chi \in {\mathcal {M}}({\mathcal {B}})$ we have $\chi \circ \varphi \in {\mathcal {M}}({\mathcal {A}})$, hence $\chi \circ \varphi$ is continuous. Thus,

$$\begin{aligned} \chi (y) = \chi (\lim _{n \rightarrow \infty } \varphi (x_n)) = \lim _{n \rightarrow \infty } \chi (\varphi (x_n)) = \chi (\varphi (0)) = 0. \end{aligned}$$

Hence, $\chi (y) = 0$ for every $\psi \in {\mathcal {M}}({\mathcal {B}})$. Since ${\mathcal {B}}$ is assumed to be Jacobson semisimple, it is $y = 0$. Thus, by the closed graph theorem, $\varphi$ is continuous. $\square$

Recall that every Segal algebra is an $L^1$ module by (2). Let $\{g_t\}_{t>0}$ be a normalized approximate identity for $L^1({\mathbb {R}}^{2n})$, then $(g_t,0)$ is an approximate identity for $L^1\oplus {{\mathcal {T}}}^1$. Hence, every quantum Segal algebra QS is an essential Banach module, and the Cohen-Hewitt factorization theorem implies that

$$\begin{aligned} QS = \{ g *(f, A): ~g \in L^1({\mathbb {R}}^{2n}), \, (f, A) \in QS\}. \end{aligned}$$

As in the case for Segal algebras, see, e.g., [4], the property of being an essential $L^1$ module uniquely determines quantum Segal algebras among dense subalgebras of $L^1 \oplus {{\mathcal {T}}}^1$ in the following sense.

Proposition 4.4

Let ${\mathcal {A}} \subset L^1 \oplus {{\mathcal {T}}}^1$ be a dense subalgebra, such that $({\mathcal {A}}, \Vert \cdot \Vert _{{\mathcal {A}}})$ is a Banach algebra. Then ${\mathcal {A}}$ is an essential $L^1$ Banach module if and only if it is a quantum Segal algebra.

Proof

Since ${\mathcal {A}}$ is essential, we have that every element of $(f, A)=h*(g,B)$, for $h \in L^1({\mathbb {R}}^{2n})$ and $(g, B) \in {\mathcal {A}}$. Hence ${\mathcal {A}}$ is translation-invariant since

$$\begin{aligned} \alpha _z(f, A) = \alpha _z(h *(g, B)) = \alpha _z(h) *(g, B) \in {\mathcal {A}} \text { for all } z\in {\mathbb {R}}^{2n}. \end{aligned}$$

Furthermore, for $z \rightarrow 0$ we have

$$\begin{aligned} \Vert \alpha _z(f, A) - (f, A)\Vert _{{\mathcal {A}}}&= \Vert \alpha _z(h) *(g, B) - h *(g,B)\Vert _{{\mathcal {A}}} \\&\le \Vert \alpha _z(h) - h\Vert _{L^1} \Vert (g, B)\Vert _{{\mathcal {A}}} \rightarrow 0. \end{aligned}$$

The only thing left is showing that the translations act isometric with respect to $\Vert \cdot \Vert _{{\mathcal {A}}}$. On the one hand, for $\{g_{t}\}_{t > 0} \subset L^1({\mathbb {R}}^{2n})$ a normalized approximate identity and $(f, A) \in {\mathcal {A}}$ we have

$$\begin{aligned} \Vert \alpha _z(f, A)\Vert _{{\mathcal {A}}}&= \lim _{t \rightarrow 0} \Vert \alpha _z(g_t) *(f, A)\Vert _{{\mathcal {A}}} \\&\le \limsup _{t \rightarrow 0} \Vert \alpha _z(g_t)\Vert _{L^1} \Vert (f, A)\Vert _{{\mathcal {A}}} \\&= \Vert (f, A)\Vert _{{\mathcal {A}}}. \end{aligned}$$

The other direction follows immediately by applying the same argument to $\alpha _{-z}(\alpha _{z}(f, A)) = (f, A)$. $\square$

We have seen that quantum Segal algebras are $L^1$ modules, however, they are not in general $L^1 \oplus {{\mathcal {T}}}^1$ ideals. Remark 9 below gives an example for this. In particular, this means that quantum Segal algebras are not abstract Segal algebras in the sense of Burnham [2]. Hence we should not hope for the full ideal theorem that Segal algebras satisfy. Nevertheless, there is a weak version:

Proposition 4.5

Let QS be a quantum Segal algebra.

• (1) For every closed ideal I of $L^1 \oplus {{\mathcal {T}}}^1$, the set $I \cap QS$ is a closed ideal of QS.

• (2) If I is a closed ideal of QS, then the $L^1 \oplus {{\mathcal {T}}}^1$-closure of I is a closed ideal of $L^1 \oplus {{\mathcal {T}}}^1$. Further, $I \subseteq {\overline{I}} \cap QS$.

When QS is an ideal of $L^1 \oplus {{\mathcal {T}}}^1$, then we have $I = {\overline{I}} \cap QS$ in (2).

Proof

The proof is nearly identical to the one given in [2, Thm. 1.1], and is hence omitted. $\square$

We have already seen in Lemma 3.6 that graded closed ideals of $L^1\oplus {{\mathcal {T}}}^1$ admit a particularly simple structure. This is also true for quantum Segal algebras. For completeness, we repeat the definition, which is the same as for closed ideals: A quantum Segal algebra QS is graded if $QS \cong (QS \cap (L^1 \oplus \{ 0\})) \oplus (QS \cap (\{ 0\} \oplus {{\mathcal {T}}}^1))$. It should be noted that not every quantum Segal algebra is graded, see Example 5.20. Note that for a graded quantum Segal algebra $(QS, \Vert \cdot \Vert _{QS})$, an easy application of the open map** theorem yields that its norm is equivalent to the sum of the subspace norms, i.e., for $(f, A) \in QS$ it holds true that:

$$\begin{aligned} c_1 (\Vert (f, 0)\Vert _{QS} + \Vert (0, A)\Vert _{QS})&\le \Vert (f, A)\Vert _{QS} \le c_2(\Vert (f, 0)\Vert _{QS} + \Vert (0, A)\Vert _{QS}). \end{aligned}$$

Even though we do not yet know if the complete ideal theorem for Segal algebras carries over, on the level of regular maximal ideals this is true for graded quantum Segal algebras.

Proposition 4.6

Let $(QS,\Vert \cdot \Vert _{QS})$ be a graded quantum Segal algebra.

• (1) Every $\chi \in \chi _{{\mathbb {R}}^{2n}\oplus {\mathbb {Z}}_2}$ is a multiplicative linear functional on QS.

• (2) For every $\chi \in {\mathcal {M}}(QS)$ there is some $(z, j) \in {\mathbb {R}}^{2n} \times \mathbb Z_2$ such that $\chi = \left. \chi _{z, j}\right| _{QS}$.

• (3) If $(z_1, j_1),\, (z_2, j_2) \in {\mathbb {R}}^{2n} \times \mathbb Z_2$ such that $\left. \chi _{z_1, j_1}\right| _{QS} = \left. \chi _{z_2, j_2}\right| _{QS}$ then $(z_1, j_1) = (z_2, j_2)$.

Hence, the Gelfand transform $\Gamma _{QS}$ of QS is simply the restriction $\Gamma _{L^1\oplus {{\mathcal {T}}}^1}|_{QS}$.

Proof

• (1) This is clear, as QS and $L^1 \oplus {{\mathcal {T}}}^1$ have the same algebraic operations.

• (2) Notice that $S=QS \cap L^1({\mathbb {R}}^{2n})\oplus \{0\}$ is a Segal algebra. Let $\chi \in {\mathcal {M}}(QS)$. Then, $\chi |_{S}$ is a nonzero multiplicative linear functional on S. As Segal algebras have the same Gelfand theory as $L^1({\mathbb {R}}^{2n})$, there is some $z \in {\mathbb {R}}^{2n}$ such that

  $$\begin{aligned} \chi |_{S}(f) = {\widehat{f}}(z), \quad f \in S. \end{aligned}$$

  From this, one concludes as in the proof of Proposition 3.1 that

  $$\begin{aligned} \chi (f, A) = {\widehat{f}}(z) +(-1)^j {\widehat{A}}(z)=\chi _{z,j}(f,A). \end{aligned}$$

• (3) The assumption yields

  $$\begin{aligned} {\widehat{f}}(z_1) = \chi _{z_1, j_1}(f, 0) = \chi _{z_2, j_2}(f,0) = {\widehat{f}}(z_2). \end{aligned}$$

  If $z_1 \ne z_2$, then there exists $f \in L^1({\mathbb {R}}^{2n})$ such that ${\widehat{f}}(z_1) \ne {\widehat{f}}(z_2)$ and a sequence $\{f_n\}_{n\in {\mathbb {N}}} \subset S$ with $f_n \rightarrow f$ in $L^1({\mathbb {R}}^{2n})$. Since point evaluations of the Fourier transform are continuous on $L^1({\mathbb {R}}^{2n})$, we thus would obtain

  $$\begin{aligned} {\widehat{f}}(z_1) =\lim _{n\rightarrow \infty } \widehat{f_n}(z_1) = \lim _{n\rightarrow \infty }\widehat{f_n}(z_2) ={\widehat{f}}(z_2), \end{aligned}$$

  which is a contradiction. Hence $z_1 = z_2$. Further, if $j_1 \ne j_2$, then ${\widehat{A}}(z_1) = 0$ for every $A \in QS \cap \{0\}\oplus {{\mathcal {T}}}^1$. Since there exists $B \in {{\mathcal {T}}}^1$ with ${\widehat{B}}(z_1) \ne 0$, we see that $j_1 = j_2$.

$\square$

Corollary 4.7

Let QS be a graded quantum Segal algebra. Then, QS is Jacobson semisimple.

5 Examples of Quantum Segal Algebras

While we know that $L^{1} \oplus {{\mathcal {T}}}^{1}$ is a quantum Segal algebra, we so far did not discuss any nontrivial examples. In this section we will look at several examples of quantum Segal algebras.

5.1 Induced Quantum Segal Algebras

We say that $A \in {{\mathcal {T}}}^{1}$ is a regular operator if it satisfies ${\mathcal {F}}_{W}(A)(z) \ne 0$ for all $z \in {\mathbb {R}}^{2n}$. Using this, we have the following construction.

Definition 5.1

Let $(S, \Vert \cdot \Vert _{S})$ denote a Segal algebra and fix a regular operator $A \in {{\mathcal {T}}}^{1}$. Define the subspace $S^{A} \subset L^{1} \oplus {{\mathcal {T}}}^{1}$ as

$$\begin{aligned}S^{A} :=\left\{ (f, g *A): f, g \in S \right\} ,\end{aligned}$$

and the norm on $S^{A}$ by

$$\begin{aligned} \Vert (f, g *A)\Vert _{S^{A}} :=\Vert f\Vert _{S} + \Vert A\Vert _{{{\mathcal {T}}}^{1}}\Vert g\Vert _{S}. \end{aligned}$$

(16)

We say that $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ is the induced quantum Segal algebra of $(S, \Vert \cdot \Vert _{S})$ and A.

The reason we need to require that $A \in {{\mathcal {T}}}^{1}$ is regular will be clear from the proof of the following result.

Theorem 5.2

The induced quantum Segal algebra $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ of $(S, \Vert \cdot \Vert _{S})$ and A is a quantum Segal algebra. Moreover, in the case that $(S, \Vert \cdot \Vert _{S})$ is star-symmetric and $A^{*_{\text {QHA}}} = A$, $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ is star-symmetric as well.

The final claim regarding star-symmetry follows immediately from (8) and (9). However, one can give the precise criterion for when $S^A$ is star-symmetric. We quickly discuss this before turning to the proof of Theorem 5.2:

Proposition 5.3

Let $(S^A, \Vert \cdot \Vert _{S^A})$ be an induced quantum Segal algebra. Then it is star-symmetric if and only if the Segal algebra S is star-symmetric and closed under convolution by the tempered distribution $\phi$ defined by

$$\begin{aligned} \varphi = {\mathcal {F}}_\sigma ( \overline{{\mathcal {F}}_W(A)}/{\mathcal {F}}_W(A)). \end{aligned}$$

Proof

We need to be able to solve the equation

$$\begin{aligned} (f_1, g_1 *A)^*= (f_2, g_2 *A) \end{aligned}$$

for $f_2, g_2 \in S$ whenever $f_1, g_1 \in S$ are given. Clearly, $f_2=f_1^*$ making S star-symmetric. Applying ${\mathcal {F}}_W$ to the second component yields

$$\begin{aligned} {\mathcal {F}}_\sigma (g_1^*) {\mathcal {F}}_W(A^{*_{\text {QHA}}}) = {\mathcal {F}}_\sigma (g_1^*) \overline{{\mathcal {F}}_W(A)} = {\mathcal {F}}_\sigma (g_2) {\mathcal {F}}_W(A). \end{aligned}$$

Since ${\mathcal {F}}_W(A)$ is nowhere zero, we can solve this for

$$\begin{aligned} {\mathcal {F}}_\sigma (g_2) = {\mathcal {F}}_\sigma (g_1^*) \overline{{\mathcal {F}}_W(A)}/{\mathcal {F}}_W(A), \end{aligned}$$

or equivalently

$$\begin{aligned} g_2 = g_1^**{\mathcal {F}}_\sigma ^{-1}(\overline{{\mathcal {F}}_W(A)}/{\mathcal {F}}_W(A)) \end{aligned}$$

in the sense of tempered distributions. $\square$

Remark 7

When $A = A^{*_{\text {QHA}}}$ we have $\overline{{\mathcal {F}}_W(A)}/{\mathcal {F}}_W(A) = 1$ implying that $\varphi = (2\pi )^n\delta _0$.

We break the verification of Theorem 5.2 into the following three lemmas:

Lemma 5.4

The space $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ is a Banach space.

Proof

The homogeneity and triangle inequality the norm is straightforward. For the positive definiteness, it is clear from (16) that $\Vert (f, g *A)\Vert _{S^{A}} = 0$ implies that $(f, g *A) = (0, 0)$. Conversely, assume that $(f, g *A) = (0,0)$. Then $f = 0$ and $g *A = 0$. By using (12) we have that

$$\begin{aligned}{\mathcal {F}}_{W}(g *A) = {\mathcal {F}}_{\sigma }(g)\cdot {\mathcal {F}}_{W}(A) = 0.\end{aligned}$$

Since A is regular, this forces ${\mathcal {F}}_{\sigma }(g)= 0$. Hence $g = 0$ and $\Vert \cdot \Vert _{S^{A}}$ is a norm on $S^{A}$. The completeness of $S^A$ follows easily from the completeness of S. $\square$

Lemma 5.5

The induced Segal algebra $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ satisfies (QS1) and (QS2).

Proof

To show (QS1), notice that $L^1({\mathbb {R}}^{2n}) *A$ is dense in ${{\mathcal {T}}}^1$ whenever A is regular by [25, Prop. 3.5]. By (S1) the induced Segal algebra $(S^{A}, \Vert \cdot \Vert _{S^{1}})$ is a dense subspace of $L^{1} \oplus {{\mathcal {T}}}^{1}$.

For (QS2), let us first show that $*$ is a well-defined operator on $S^{A}$. For two elements $(f_{1}, g_{1} *A)$ and $(f_{2}, g_{2} *A)$ we have

$$\begin{aligned}(f_{1}, g_{1} *A)&*(f_{2}, g_{2} *A) \\&= (f_{1} *f_{2} + (g_{1} *g_{2}) *(A *A), (f_{1} *g_{2} + f_{2} *g_{1}) *A).\end{aligned}$$

Since S is closed under convolution, we know that

$$\begin{aligned}f_{1} *f_{2},\, g_{1} *g_{2},\, f_{1} *g_{2} + f_{2} *g_{1}\in S.\end{aligned}$$

Moreover, we know that $A *A \in L^{1}({\mathbb {R}}^{2n})$. Since Segal algebras are ideals of $L^1({\mathbb {R}}^{2n})$ the product is well-defined. To see that $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ is a Banach algebra we compute

$$\begin{aligned}&\Vert (f_{1}, g_{1} *A) *(f_{2}, g_{2} *A)\Vert _{S^{A}} \\&\le \Vert f_{1} *f_{2}\Vert _{S} + \Vert g_{1} *g_{2}\Vert _{S} \Vert A *A\Vert _{L^{1}} + \Vert A\Vert _{{{\mathcal {T}}}^{1}}\Vert f_{1} *g_{2} + f_{2} *g_{1}\Vert _{S} \\&\le \Vert f_{1}\Vert _{S} \Vert f_{2}\Vert _{S} + \Vert g_{1}\Vert _{S} \Vert g_{2}\Vert _{S} \Vert A\Vert _{{{\mathcal {T}}}^{1}}^{2} + \Vert A\Vert _{{{\mathcal {T}}}^{1}}(\Vert f_{1}\Vert _{S} \Vert g_{2}\Vert _{S} + \Vert f_{2}\Vert _{S} \Vert g_{1}\Vert _{S}) \\&= \Vert (f_{1}, g_{1} *A)\Vert _{S^{A}} \cdot \Vert (f_{2}, g_{2} *A)\Vert _{S^{A}}. \end{aligned}$$

$\square$

Lemma 5.6

The induced Segal algebra $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ satisfies (QS3) and (QS4).

Proof

For any $z \in {\mathbb {R}}^{2n}$ we have that

$$\begin{aligned} \alpha _z(f, g *A) = (\alpha _z(f), \alpha _z(g) *A). \end{aligned}$$

Since the Segal algebra S is translation-invariant and translations are norm isometric, it follows that $(S^{A}, \Vert \cdot \Vert _{S^{A}})$ is translation-invariant and translations are norm isometric. Finally, the translations act strongly continuous since

$$\begin{aligned} \lim _{z\rightarrow 0}&\Vert \alpha _z(f, g *A) - (f, g *A)\Vert _{S^{A}}\\&= \lim _{z\rightarrow 0}\Vert \alpha _z(f) - f\Vert _{S} + \Vert A\Vert _{{{\mathcal {T}}}^1} \lim _{z\rightarrow 0}\Vert \alpha _z(g) - g\Vert _{S}\\&= 0. \end{aligned}$$

$\square$

Example 5.7

Let us assume that $A=\phi \otimes \psi$ is a rank one operator and $g\in S\subset L^1({\mathbb {R}}^{2n})$, where S is a Segal algebra. Define the localization operator by

$$\begin{aligned}{\mathcal {A}}^{\phi ,\psi }_g\eta :=\int _{{\mathbb {R}}^{2n}}g(z)\langle \eta ,W_z\psi \rangle W_z\phi \, \textrm{d}z.\end{aligned}$$

The localization operator can be rewritten as

$$\begin{aligned}g*(\phi \otimes \psi )={\mathcal {A}}^{\phi ,\psi }_g.\end{aligned}$$

Hence the induced quantum Segal algebra with respect to $\phi \otimes \psi$ and S is on the form

$$\begin{aligned}(f,{\mathcal {A}}^{\phi ,\psi }_g), \qquad f,g\in S.\end{aligned}$$

To get a quantum Segal algebra we need that ${\mathcal {F}}_W(\phi \otimes \psi )(z)\not =0$ for all $z\in {\mathbb {R}}^{2n}$. Computing the Fourier-Weyl transform of $\phi \otimes \psi$ we get

$$\begin{aligned}{\mathcal {F}}_W(\phi \otimes \psi )(z) = e^{i x\xi /2}V_{\psi }\phi (-z)\not = 0,\qquad z=(x,\xi )\in {\mathbb {R}}^{2n},\end{aligned}$$

where $V_{\psi }\phi$ is given in (4). Examples of functions satisfying $V_{\psi }\phi (z)\not = 0$ are given in [12].

Remark 8

Not every quantum Segal algebras is an induced Segal algebra. As an example, let ${\mathcal {S}}_0={\mathcal {S}}_0({\mathbb {R}}^{2n})$ denote the Feichtinger algebra defined in Example 2.3 (E5). Then for any regular operator A the set $L^1({\mathbb {R}}^{2n}) \oplus ({\mathcal {S}}_0 *A)$ is a quantum Segal algebra with the norm

$$\begin{aligned}\Vert (f,g*A)\Vert _{L^1({\mathbb {R}}^{2n}) \oplus ({\mathcal {S}}_0 *A)}=\Vert f\Vert _{L^1}+\Vert g\Vert _{{\mathcal {S}}_0}\Vert A\Vert _{{{\mathcal {T}}}^1}.\end{aligned}$$

Remark 9

Induced Segal algebras are in general not ideals of $L^1 \oplus {{\mathcal {T}}}^1$. As an example, let $n=1$ and consider the induced Segal algebra $L^1({\mathbb {R}}^2)^A$ with $A \in {{\mathcal {T}}}^1$ a regular operator. Then the set

$$\begin{aligned} X=\{ f *A: f \in L^1({\mathbb {R}}^2)\} = \{ B \in {{\mathcal {T}}}^{1}: (0, B) \in L^1({\mathbb {R}}^2)^A\} \end{aligned}$$

is dense in ${{\mathcal {T}}}^1$. Further, X is a proper subset of ${{\mathcal {T}}}^1$, since $A\not \in X$. This follows from $A = 2\pi \delta _0 *A$ together with the map $\mu \mapsto \mu *A$ being injective for all $\mu \in {\mathcal {S}}'({\mathbb {R}}^{2})$. Nevertheless, since ${{\mathcal {T}}}^1$ is an essential $L^1$ module, we have

$$\begin{aligned} f*B=A, \quad f\in L^1({\mathbb {R}}^2),\, B\in {{\mathcal {T}}}^1. \end{aligned}$$

We therefore conclude that $L^1({\mathbb {R}}^2)^A$ is not an $L^1 \oplus {{\mathcal {T}}}^1$ ideal, since $(f, 0)\in L^1({\mathbb {R}}^2)^A$ and

$$\begin{aligned}(0,B)*(f,0)=(0,A)\not \in L^1({\mathbb {R}}^2)^A.\end{aligned}$$

Remark 10

One might initially expect that if S is a modulation-invariant Segal algebra, then the induced quantum Segal algebra $S^{A}$ is modulation-invariant as well. However, this is not the case. As a simple counterexample, let $S = {\mathcal {S}}_{0}({\mathbb {R}}^{2})={\mathcal {S}}_{0}$ be the Feichtinger algebra and let A be the regular operator given by

$$\begin{aligned}{\mathcal {F}}_{W}(A)(z) = e^{-|z|^2}, \qquad z \in {\mathbb {R}}^{2}.\end{aligned}$$

Assume by contradiction that $S^{A}$ is modulation-invariant. Then for any $g_{1} \in {\mathcal {S}}_{0}$ and any $z' \in {\mathbb {R}}^{2}$ there should exist $g_{2} \in {\mathcal {S}}_{0}$ such that

$$\begin{aligned}\gamma _{z'}(g_{1} *A) = \gamma _{z'}(g_{1}) *\gamma _{z'}(A) = g_{2} *A.\end{aligned}$$

Taking the Fourier-Weyl transform of both sides gives

$$\begin{aligned}{\mathcal {F}}_{\sigma }(g_{1})(z - z')e^{-|z - z'|^2} = {\mathcal {F}}_{\sigma }(g_{2})(z)e^{-|z|^2}.\end{aligned}$$

Since ${\mathcal {F}}_{\sigma }({\mathcal {S}}_0)={\mathcal {S}}_0$, we have that $h_{1} = e^{-z'^2}{\mathcal {F}}_{\sigma }(g_{1}) \in {\mathcal {S}}_{0}$ and $h_{2} = {\mathcal {F}}_{\sigma }(g_{2}) \in {\mathcal {S}}_{0}$. By picking $z' = -\frac{1}{2}$ we obtain

$$\begin{aligned}h_{1}\left( z + \frac{1}{2}\right) e^{z} = h_{2}(z).\end{aligned}$$

Since $h_{1}$ is arbitrary, and $e^z{\mathcal {S}}_{0}\not = {\mathcal {S}}_{0}$ we have a contradiction. For example, for $h_{1}(z) = (1 + |z|^2)^{-1}$ we have no solution $h_2\in {\mathcal {S}}_0$.

5.2 Quantum Segal Algebras through Quantization

In this section, we will construct quantum Segal algebras using the Weyl quantization. Recall that ${\widetilde{f}}(z)=f(-z)$, which naturally extends to ${\mathcal {S}}({\mathbb {R}}^{2n})$. Given a set M of functions or distributions, we will use the notation

$$\begin{aligned}{\widetilde{M}} = \{{\widetilde{f}}: f\in M\}.\end{aligned}$$

Notice that if S is a Segal algebra, then so is also ${\widetilde{S}}$. Given a set of tempered operators ${\mathcal {A}} \subset {\mathcal {S}}'({\mathcal {H}})$, we denote the set of symbols by ${\textrm{Sym}}({\mathcal {A}}) \subset {\mathcal {S}}'({\mathbb {R}}^{2n})$, i.e.,

$$\begin{aligned}\widetilde{{\textrm{Sym}}({\mathcal {A}})} = {\mathcal {F}}_\sigma ({\mathcal {F}}_W({\mathcal {A}} )).\end{aligned}$$

Vice versa, given a set of tempered distributions $S \subset {\mathcal {S}}'({\mathbb {R}}^{2n})$, we denote by ${\textrm{Sym}}^{-1}(S)$ the set of quantized operators as a subset of ${\mathcal {S}}'({\mathcal {H}})$, i.e.,

$$\begin{aligned}{\textrm{Sym}}^{-1}(S) = P{\mathcal {F}}_W( {\mathcal {F}}_\sigma ({\widetilde{S}}))P.\end{aligned}$$

In the following result, we present a Segal algebra which plays an important role for the discussion of quantum Segal algebras. For this, we recall that a Segal algebra $S \subset L^1({\mathbb {R}}^n)$ is called strongly modulation-invariant (or strongly character-invariant, as it was called in [6]) if it $\gamma _x(f) \in S$ with $\Vert \gamma _x(f)\Vert _S = \Vert f\Vert _S$ for each $x \in {\mathbb {R}}^n$ and $x \mapsto \gamma _x(f) \in S$ is continuous for each $f \in S$.

Lemma 5.8

Denote

$$\begin{aligned} {{\mathcal {T}}} := \{ f \in L^1({\mathbb {R}}^{2n}): ~A_f \in {{\mathcal {T}}}^1\}. \end{aligned}$$

Endowed with the norm

$$\begin{aligned} \Vert f\Vert _{{{\mathcal {T}}}} := \Vert f\Vert _{L^1} + \Vert A_f\Vert _{{{\mathcal {T}}}^1}, \end{aligned}$$

the space ${{\mathcal {T}}}$ is a star-symmetric and strongly modulation-invariant Segal algebra.

Proof

Firstly, note that ${{\mathcal {T}}}$ contains ${\mathcal {S}}({\mathbb {R}}^{2n})$, so it is dense in $L^1({\mathbb {R}}^{2n})$. We have

$$\begin{aligned} \Vert \alpha _z(f)\Vert _{{{\mathcal {T}}}} = \Vert \alpha _z(f)\Vert _{L^1} + \Vert A_{\alpha _z(f)}\Vert _{{{\mathcal {T}}}^1} = \Vert \alpha _z(f)\Vert _{L^1} + \Vert \alpha _{-z}(A_f)\Vert _{{{\mathcal {T}}}^1} = \Vert f\Vert _{{{\mathcal {T}}}}. \end{aligned}$$

Similarly, the translations act strongly continuous on ${{\mathcal {T}}}$. Further, it is

$$\begin{aligned} \Vert f*g\Vert _{{{\mathcal {T}}}}&= \Vert f *g\Vert _{L^1} + \Vert A_{f *g}\Vert _{{{\mathcal {T}}}^1} = \Vert f*g\Vert _{L^1} + \Vert {\widetilde{f}} *A_g\Vert _{{{\mathcal {T}}}^1} \\&\le \Vert f\Vert _{L^1} \Vert g\Vert _{L^1} + \Vert f\Vert _{L^1} \Vert A_g\Vert _{{{\mathcal {T}}}^1} \le \Vert f\Vert _{{{\mathcal {T}}}} \Vert g\Vert _{{{\mathcal {T}}}} \end{aligned}$$

for $f, g \in {{\mathcal {T}}}$. Therefore, ${{\mathcal {T}}}$ is a Segal algebra. Regarding star-symmetry, note that

$$\begin{aligned} A_{f^*} = A_{\tilde{{\overline{f}}}} = P(A_{{\overline{f}}})P = PA_f^*P = A_f^{*_{\text {QHA}}}. \end{aligned}$$

Since $\Vert A_f\Vert _{{{\mathcal {T}}}^1} = \Vert A_f^{*_{\text {QHA}}}\Vert _{{{\mathcal {T}}}^1}$, and clearly $\Vert f\Vert _{L^1} = \Vert f^*\Vert _{L^1}$, we have $\Vert f\Vert _{{{\mathcal {T}}}} = \Vert f^*\Vert _{{{\mathcal {T}}}}$. The strong modulation invariance is now immediate from Lemma 2.10. $\square$

Proposition 5.9

Let $QS =S_{1} \oplus S_{2} \subset L^{1}({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^{1}$ be a graded quantum Segal algebra. Then both $S_{1}$ and ${\textrm{Sym}}(S_{2}) \cap L^{1}({\mathbb {R}}^{2n})$ are Segal algebras. Additionally,

$$\begin{aligned}\widetilde{{\textrm{Sym}}(S_{2})} *\widetilde{{\textrm{Sym}}(S_{2})} \subset S_{1}.\end{aligned}$$

In particular, when ${\textrm{Sym}}(S_{2}) \subset L^{1}({\mathbb {R}}^{2n})$ the set ${\textrm{Sym}}(S_{2})$ is a Segal algebra.

Proof

It is clear that $S_1$ is a Segal algebra. $S = {\textrm{Sym}}(S_{2}) \cap L^{1}({\mathbb {R}}^{2n})$ is a Banach space with the translation-invariant norm

$$\begin{aligned}\Vert f\Vert _{S}=\Vert (0, A_f)\Vert _{QS}+\Vert f\Vert _{L^1},\qquad f\in S.\end{aligned}$$

The set S is a Banach algebra, since by Proposition 3.4 and (15) we have

$$\begin{aligned}{\widetilde{S}}*{\widetilde{S}}\subset {\widetilde{S}}. \end{aligned}$$

It remains to show that S is dense in $L^1({\mathbb {R}}^{2n})$. For this, note that

$$\begin{aligned} {{\mathcal {T}}} *S_2 \subset L^1({\mathbb {R}}^{2n}) *S_2 = S_2. \end{aligned}$$

Further, we have

$$\begin{aligned} {{\mathcal {T}}}*S_2&={\textrm{Sym}}^{-1}({\textrm{Sym}}^{-1}({{\mathcal {T}}}) *{\widetilde{S}}_2)\subset {\textrm{Sym}}^{-1}( {{\mathcal {T}}}^1*{\widetilde{S}}_2)\\&\subset {\textrm{Sym}}^{-1}( {{\mathcal {T}}}^1*{{\mathcal {T}}}^1)\subset {\textrm{Sym}}^{-1}(L^1({\mathbb {R}}^{2n})), \end{aligned}$$

where $\tilde{S_2}={\textrm{Sym}}^{-1}(\widetilde{{\textrm{Sym}}(S_2)})$. Together, this shows ${{\mathcal {T}}} *S_2 \subset S$. Note that, since $S_2$ is dense in ${{\mathcal {T}}}^1$, either by Proposition 4.6 (3) or by Wiener’s approximation theorem for ${{\mathcal {T}}}^1$, cf. [25, Prop. 3.5], we have that

$$\begin{aligned} \{z\in {\mathbb {R}}^{2n}:{\mathcal {F}}_W(A)(z)=0 \text { for every } A \in S_2 \}=\emptyset .\end{aligned}$$

Now using (12) together with the fact that for every $z\in {\mathbb {R}}^{2n}$ there exist $f\in {{\mathcal {T}}}$ and $A\in S_2$ such that ${\mathcal {F}}_\sigma (f)\not =0$ and ${\mathcal {F}}_W(A)\not = 0$, we get that

$$\begin{aligned}\{z\in {\mathbb {R}}^{2n}:{\mathcal {F}}_W(A)(z)=0 \text { for every } A \in S \}=\emptyset .\end{aligned}$$

The set ${\mathcal {F}}_W(S)$ is an ideal of ${\mathcal {F}}_W({\textrm{Sym}}^{-1}({{\mathcal {T}}}))\subset C_0({\mathbb {R}}^{2n})$ with pointwise multiplication. Since ${{\mathcal {T}}}$ is a Segal algebra, the set ${\mathcal {F}}_\sigma ( {{\mathcal {T}}})$ is a standard algebra by [22, Re. 2.1.15]. Hence by [22, Prop. 2.1.14] we have that

$$\begin{aligned}C_c({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n}))\subset {\mathcal {F}}_W(S_2).\end{aligned}$$

Since ${\mathcal {F}}_\sigma (C_c({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n})))$ is dense in $L^1({\mathbb {R}}^{2n})$ the space S is a Segal algebra. The product structure on QS together with (15) gives

$$\begin{aligned}\widetilde{{\textrm{Sym}}(S_{2})}*\widetilde{{\textrm{Sym}}(S_{2})} \subset S_{1}. \end{aligned}$$

$\square$

The following result shows how to generate a wealth of examples of graded quantum Segal algebras.

Proposition 5.10

Let $S_{1}$ and $S_{2}$ be two Segal algebras with $\widetilde{S_{2}}\subset S_{1}$. Then

$$\begin{aligned}QS = S_{1} \oplus ({\textrm{Sym}}^{-1}(S_{2}) \cap {{\mathcal {T}}}^1)\end{aligned}$$

is a quantum Segal algebra with the norm

$$\begin{aligned}\Vert (f,A_g)\Vert _{QS}= \Vert f\Vert _{S_{1}}+C\Vert {\widetilde{g}}\Vert _{S_{2}} + C'\Vert A_g\Vert _{{{\mathcal {T}}}^1},\end{aligned}$$

for positive constants C, $C'$.

Proof

For property (QS1), we note that the set $A_0:= L^1({\mathbb {R}}^{2n}) \cap {\mathcal {F}}_\sigma (C_c({\mathbb {R}}^{2n}))$ is contained in ${{\mathcal {T}}}$, since ${{\mathcal {T}}}$ is a Segal algebra, cf. [22, Prop. 6.2.5]. Further, the set is dense in ${{\mathcal {T}}}$. Since ${\textrm{Sym}}^{-1}({{\mathcal {T}}})$ is dense in ${{\mathcal {T}}}^1$, we obtain that ${\textrm{Sym}}^{-1}(A_0)$ is dense in ${{\mathcal {T}}}^1$. But $A_0$ is also contained in $S_2$. Therefore, ${\textrm{Sym}}^{-1}(S_2) \cap {{\mathcal {T}}}^1$ is dense in ${{\mathcal {T}}}^1$.

The conditions (QS3) and (QS4) are straightforward to verify. For property (QS2), QS is easily seen to be a complete subspace of $L^1 \oplus {{\mathcal {T}}}^1$. The product is well-defined by (15). Hence, we only need to show that the Banach algebra property holds, that is for $(f_{1}, A_{g_{1}}), (f_{2}, A_{g_{2}}) \in QS$ we have

$$\begin{aligned}\Vert (f_1,A_{g_1}) *(f_2,A_{g_2})\Vert _{QS} \le \Vert (f_1,A_{g_1})\Vert _{QS} \cdot \Vert (f_2,A_{g_2})\Vert _{QS}.\end{aligned}$$

We know by (3) and Lemma 4.3 that there exist positive constants $c_1$ and $c_2$ such that

$$\begin{aligned} \Vert g\Vert _{S_{1}}&\le c_1 \cdot \Vert {\widetilde{g}}\Vert _{S_{2}} \\ \Vert f*{\widetilde{g}}\Vert _{S_{2}}&\le c_2 \cdot \Vert f\Vert _{S_{1}}\Vert {\widetilde{g}}\Vert _{S_{2}} \end{aligned}$$

for all $f\in S_{1}$ and $g\in S_{2}$. Denote by $C=\max \{c_1, c_2\}$. Further, by Lemma 4.3, there is $C' > 0$ with $\Vert f\Vert _{L^1} \le C' \Vert f\Vert _{S_1}$ for $f \in S_1$. We get

$$\begin{aligned} \Vert (f_1,A_{g_1}) *(f_2,A_{g_2})\Vert _{QS}&= \Vert f_1*f_2+\widetilde{g_1}*\widetilde{g_2}\Vert _{S_{1}}+\Vert f_1*\widetilde{g_2}+f_2*\widetilde{g_1}\Vert _{S_{2}} \\&\quad \quad + \Vert f_1 *A_{g_2} + f_2 *A_{g_1}\Vert _{{{\mathcal {T}}}^1}\\&\le \Vert f_1\Vert _{S_{1}}\Vert f_2\Vert _{S_{1}}+C^2\Vert \widetilde{g_1}\Vert _{S_{2}}\Vert \widetilde{g_2}\Vert _{S_{2}}\\&\quad \quad +C\Vert f_1\Vert _{S_{1}}\Vert \widetilde{g_2}\Vert _{S_{2}} + C\Vert f_2\Vert _{S_{1}}\Vert \widetilde{g_1}\Vert _{S_{2}} \\&\quad \quad + \Vert f_1\Vert _{L^1} \Vert A_{g_2}\Vert _{{{\mathcal {T}}}^1} + \Vert f_2\Vert _{L^1} \Vert A_{g_1}\Vert _{{{\mathcal {T}}}^1}\\&\le \Vert (f_1,A_{g_1})\Vert _{QS} \cdot \Vert (f_2,A_{g_2})\Vert _{QS}. \end{aligned}$$

$\square$

Remark 11

• 1. If, under the assumptions of the previous proposition, we further have ${\textrm{Sym}}^{-1}(S_2)\subset {{\mathcal {T}}}^1$, then one can instead let $C' = 0$, i.e., the norm becomes $\Vert (f, A_g)\Vert _{QS} = \Vert f\Vert _{S_1} + C\Vert {\widetilde{g}}\Vert _{S_2}$. This follows from essentially the same proof, simply omitting the additional terms.

• 2. By the previous result, we can consider

  $$\begin{aligned} {{\mathcal {T}}}^Q := {{\mathcal {T}}} \oplus {\textrm{Sym}}^{-1}({{\mathcal {T}}}), \end{aligned}$$

  which is a star-symmetric, strongly modulation-invariant quantum Segal algebra upon being endowed with the norm

  $$\begin{aligned} \Vert (f, A_g)\Vert _{{{\mathcal {T}}}^Q} = \Vert f\Vert _{{{\mathcal {T}}}} + \Vert g\Vert _{{{\mathcal {T}}}}. \end{aligned}$$

  While we won’t use this particular quantum Segal algebra in the following, it seems that it is a convenient framework to work within.

• 3. When we let $S_2 = \widetilde{S_1}$ in the above proposition, then simple computations show that the resulting quantum Segal algebra $QS = S_1 \oplus {\textrm{Sym}}^{-1}(\widetilde{S_1})$ is a module over ${{\mathcal {T}}}^Q$.

Corollary 5.11

Let G denote the set of all graded quantum Segal algebras. Then

$$\begin{aligned}{\mathcal {F}}\Big (\bigcap _{S_1\oplus S_2\in G}&S_1\oplus S_2\Big ) \\ {}&= (C_{c}({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n}))) \oplus (C_{c}({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n}))).\end{aligned}$$

Proof

We know that the intersection of all Segal algebras satisfies

$$\begin{aligned}{\mathcal {F}}_{\sigma }\Big (\bigcap _{S\in {\mathcal {S}}} S\Big ) = C_{c}({\mathbb {R}}^{2n}) \cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n})),\end{aligned}$$

where ${\mathcal {S}}$ denotes the set of all Segal algebras. By the proof of Proposition 5.9 we know that

$$\begin{aligned}{\mathcal {F}}\Big (\bigcap _{S_1\oplus S_2\in G}&S_1\oplus S_2\Big ) \\ {}&\supseteq (C_{c}({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n}))) \oplus (C_{c}({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n}))).\end{aligned}$$

The remaining inclusion follows from Proposition 5.10. Note that we can always enforce ${\textrm{Sym}}^{-1}( S_2) \subset {{\mathcal {T}}}^1$ by intersecting $S_2$ with the Feichtinger algebra ${\mathcal {S}}_0$ which satisfies ${\textrm{Sym}}^{-1}({\mathcal {S}}_0) \subset {{\mathcal {T}}}^1$, see [15, Thm. 3.5]. $\square$

5.3 The Quantum Feichtinger Algebra

In this section we will describe a quantum Segal algebra that we call the quantum Feichtinger algebra. It is a particular example of the quantum Segal algebras obtained by quantization of the Feichtinger algebra. We will show that this algebra is not on the form of an induced Segal algebra. Recall that ${\mathcal {S}}_{0} :={\mathcal {S}}_{0}({\mathbb {R}}^{2n})$ denotes the Feichtinger algebra defined in Example (E5).

Definition 5.12

We say that a bounded linear operator A on ${\mathcal {H}}$ is a Feichtinger operator if $A = A_{g}$ for $g \in {\mathcal {S}}_{0}({\mathbb {R}}^{2n})$. We denote the space of Feichtinger operators as ${\mathcal {S}}_0({\mathcal {H}})$.

Since ${\mathcal {S}}_0({\mathbb {R}}^{2n}) \subset L^2({\mathbb {R}}^{2n})$, there is no complication in talking about the integral kernel of a Feichtinger operator. As the discussions in [14, Sec. 7.4] shows, the integral kernel is contained in ${\mathcal {S}}_0({\mathbb {R}}^{2n})$ if and only if the Weyl symbol is in ${\mathcal {S}}_0({\mathbb {R}}^{2n})$, and, in this case, their norms in ${\mathcal {S}}_0({\mathbb {R}}^{2n})$ are equivalent. Therefore, $A_{g} \in {{\mathcal {T}}}^{1}$ whenever $g \in {\mathcal {S}}_{0}({\mathbb {R}}^{2n})$, which follows from [15, Thm. 3.5]. Clearly, the space ${\mathcal {S}}_0({\mathcal {H}})$ is a Banach space under the norm

$$\begin{aligned}\Vert A_{g}\Vert _{{\mathcal {S}}_0({\mathcal {H}})} :=\Vert g\Vert _{{\mathcal {S}}_{0}({\mathbb {R}}^{2n})}, \qquad A_{g} \in {\mathcal {S}}_0({\mathcal {H}}).\end{aligned}$$

We can create a norm on operators using the norm coming from the projective tensor project of ${\mathcal {S}}_0({\mathbb {R}}^n)\otimes {\mathcal {S}}_0({\mathbb {R}}^n)$. We let $\text {Fin}({\mathcal {S}}_{0})$ denote all finite-rank operators F satisfying $\Vert F\Vert _{\text {Fin}({\mathcal {S}}_{0})} < \infty$, where

$$\begin{aligned}\Vert F\Vert _{\text {Fin}({\mathcal {S}}_{0})} :=\inf \left\{ \sum _{i=1}^m|\alpha _{i}|\Vert \phi _{i}\Vert _{{\mathcal {S}}_{0}({\mathbb {R}}^n)}\Vert \psi _{i}\Vert _{{\mathcal {S}}_{0}({\mathbb {R}}^n)}: F = \sum _{i=1}^m\alpha _{i} \cdot \phi _{i} \otimes \psi _{i}\right\} .\end{aligned}$$

The Feichtinger operators inherits several equivalent characterizations from the Feichtinger algebra.

Theorem 5.13

Let $A \in {{\mathcal {T}}}^{1}$. Then the following characterizations are equivalent:

• 1) The operator A is a Feichtinger operator.

• 2) The integral kernel $K_{A}$ of A is in the Feichtinger algebra ${\mathcal {S}}_0({\mathbb {R}}^{2n})$.

• 3) The operator A is in the completion of $\text {Fin}({\mathcal {S}}_{0})$ with the norm $\Vert \cdot \Vert _{\text {Fin}({\mathcal {S}}_{0})}$, i.e., in the projective tensor product ${\mathcal {S}}_0({\mathbb {R}}^n) {\hat{\otimes }}_\pi {\mathcal {S}}_0({\mathbb {R}}^n)$.

• 4) The operator A satisfies ${\mathcal {F}}_{W}(A) \in {\mathcal {S}}_{0}({\mathbb {R}}^{2n})$.

• 5) The modulation of A satisfies $\int _{{\mathbb {R}}^{2n}} \Vert (\gamma _z A) *A\Vert _{L^1} \, \textrm{d}z < \infty$.

• 6) As a function of $(z, z') \in {\mathbb {R}}^{2n}\times {\mathbb {R}}^{2n}$, it is ${{\,\textrm{tr}\,}}(A (\alpha _z \gamma _{z'}A)^*) \in L^1({\mathbb {R}}^{2n} \times {\mathbb {R}}^{2n})$.

Introduce the notation

$$\begin{aligned} \Vert A\Vert _{B, \gamma }:=\int _{{\mathbb {R}}^{2n}} \Vert (\gamma _z A) *B\Vert _{L^1} \, \textrm{d}z, \ \Vert A\Vert _{B, \alpha \gamma } :=\Vert {{\,\textrm{tr}\,}}(A (\gamma _{z'}\alpha _z B)^*)\Vert _{L^1({\mathbb {R}}^{2n} \times {\mathbb {R}}^{2n})}, \end{aligned}$$

where $B \in {\mathcal {S}}_0({\mathcal {H}})\setminus \{0\}$ is a fixed operator. Then ${\mathcal {S}}_0({\mathcal {H}})$ can be given the following equivalent norms

$$\begin{aligned} \Vert A\Vert _{{\mathcal {S}}_0({\mathcal {H}})}&\cong \Vert {\mathcal {F}}_W(A)\Vert _{{\mathcal {S}}_0({\mathbb {R}}^{2n})} \cong \Vert K_A\Vert _{{\mathcal {S}}_0({\mathbb {R}}^{2n})} \\&\cong \Vert A\Vert _{{\mathcal {S}}_0({\mathbb {R}}^n) {\hat{\otimes }}_\pi {\mathcal {S}}_0({\mathbb {R}}^n)} \cong \Vert A\Vert _{B, \gamma } \cong \Vert A\Vert _{B, \alpha \gamma }. \end{aligned}$$

Remark 12

Note that $V_B(A)(z, z') :={{\,\textrm{tr}\,}}(A\alpha _z \gamma _{z'}(B))$ serves as an operator analogue to the short-time Fourier transform. To the authors’ knowledge, the operator STFT is not present in the literature. It could serve as an interesting object for further studies.

Proof

The equivalence between 1) and 2) has already been discussed above. The equivalence between 2) and 3) is a reformulation of the tensor factorization property of the Feichtinger algebra, see e.g., [16, Thm. 7.4]. Since for $A=A_g$ we have ${\mathcal {F}}_{\sigma }({\widetilde{g}}) = {\mathcal {F}}_{W}(A_g)$ and ${\mathcal {F}}_{\sigma }$ is a bijective isometry on ${\mathcal {S}}_{0}$ the equivalence between 1) and 4) follows. For the equivalence between 1) and  we have

$$\begin{aligned}(\gamma _z A_{g})*A_{g} = A_{\gamma _{-z}g}*A_{g} = \widetilde{(\gamma _{-z}g) *g}.\end{aligned}$$

Hence  holds if and only if

$$\begin{aligned}\int _{{\mathbb {R}}^{2n}} \Vert \gamma _z g *g\Vert _{L^1} \, \textrm{d}z < \infty .\end{aligned}$$

This is a reformulation of g being in ${\mathcal {S}}_{0}$, see [16, Thm. 4.7]. Finally, for seeing that 4) and  are equivalent, write $f_1 = {\mathcal {F}}_W(A)$, and $f_2 = {\mathcal {F}}_W(B)$. Then

$$\begin{aligned} {{\,\textrm{tr}\,}}(A (\alpha _z \gamma _{z'}B)^*)&= \langle A, \alpha _z \gamma _{z'}B\rangle _{{{\mathcal {T}}}^2} = \langle {\mathcal {F}}_W( A), {\mathcal {F}}_W(\alpha _z \gamma _{z'}B)\rangle _{L^2} \\&= \langle f, \gamma _{z}\alpha _{z'} g\rangle _{L^2} = \int _{{\mathbb {R}}^{2n}} f(v) e^{-i\sigma (z, v)}{\overline{g}}(v-z')\,\textrm{d}v. \end{aligned}$$

It is not hard to see that

$$\begin{aligned} \int _{{\mathbb {R}}^{2n}} \int _{{\mathbb {R}}^{2n}}&\left| \int _{{\mathbb {R}}^{2n}} f(v) e^{-i\sigma (z, v)}{\overline{f}}(v-z')\,\textrm{d}v \right| \,\textrm{d}z\,\textrm{d}w \\&= \int _{{\mathbb {R}}^{2n}} \int _{{\mathbb {R}}^{2n}} | V_f(f)(z',z) | \,\textrm{d}z\,\textrm{d}w, \end{aligned}$$

where $V_f(f)$ denotes the STFT.

The equivalences of the norms are straightforward and from the analogous results for the Feichtinger algebra. $\square$

We can now combine the spaces ${\mathcal {S}}_{0}({\mathbb {R}}^{2n})$ and ${\mathcal {S}}_0({\mathcal {H}})$ to form the following algebra.

Definition 5.14

The quantum Feichtinger algebra ${\mathcal {S}}_{0}^{Q} \subset L^{1} \oplus {{\mathcal {T}}}^{1}$ is given by

$$\begin{aligned} {\mathcal {S}}_{0}^{Q} :={\mathcal {S}}_{0}({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_0({\mathcal {H}}) = \left\{ (f, A_g) \in L^{1} \oplus {{\mathcal {T}}}^{1}: f,g \in {\mathcal {S}}_{0}\right\} \end{aligned}$$

with the norm

$$\begin{aligned}\Vert (f,A_g)\Vert _{{\mathcal {S}}_0^Q}=\Vert f\Vert _{{\mathcal {S}}_0}+\Vert g\Vert _{{\mathcal {S}}_0}.\end{aligned}$$

We say that a quantum Segal algebra $(QS, \Vert \cdot \Vert _{QS})$ is strongly modulation-invariant if $\Vert \gamma _z(f, A)\Vert _{QS} = \Vert (f, A)\Vert _{QS}$ and $z \mapsto \gamma _z$ are continuous in the strong sense on $(QS, \Vert \cdot \Vert _{QS})$, i.e., ${\mathbb {R}}^{2n} \ni z \mapsto \alpha _z((f, A)) \in QS$ is a continuous map for each $(f, A) \in QS$.

Proposition 5.15

The quantum Feichtinger algebra ${\mathcal {S}}_{0}^{Q}$ is a star-symmetric and strongly modulation-invariant quantum Segal algebra.

Proof

The fact that ${\mathcal {S}}^Q_0$ is a quantum Segal algebra follows from Proposition 5.10. The star-symmetry follows from the property

$$\begin{aligned}(f, A_{g})^{*} = (f^{*}, PA_{g}^{*}P) = (f^{*}, A_{g^*}).\end{aligned}$$

Finally, the continuity of the modulation invariance is immediate since for $z \in {\mathbb {R}}^{2n}$ and $f, g \in {\mathcal {S}}_{0}$ we have

$$\begin{aligned}\qquad \qquad \gamma _{z}(f, A_{g}) = (\gamma _{z}(f), \gamma _{z}(A_{g})) = (\gamma _{z}(f), A_{\gamma _{-z}g}).\qquad \qquad \qquad \qquad \end{aligned}$$

$\square$

The Feichtinger algebra is yet another example of a quantum Segal algebra which is not of the induced type:

Proposition 5.16

The quantum Feichtinger algebra is not an induced quantum Segal algebra.

Proof

If ${\mathcal {S}}_0^Q$ was an induced quantum Segal algebra there would exist a regular trace class operator A such that every Feichtinger operator B can be written as

$$\begin{aligned}B=f*A,\end{aligned}$$

for some $f\in {\mathcal {S}}_0$. By Theorem 5.13 4) together with the fact that ${\mathcal {F}}_\sigma ({\mathcal {S}}_0)={\mathcal {S}}_0$ this is equivalent to that every function $g\in {\mathcal {S}}_0$ can be written uniquely as

$$\begin{aligned}g=f\cdot {\mathcal {F}}_W(A),\end{aligned}$$

for some $f\in {\mathcal {S}}_0$. Or equivalently, the map $\phi (f)= f/{\mathcal {F}}_W(A)$ is a bijection from ${\mathcal {S}}_0$ to itself. We will show that the surjectivity of $\phi$ contradicts the fact that ${\mathcal {F}}_W(A)\in L^2({\mathbb {R}}^{2n})$.

Denote by

$$\begin{aligned}h(z)=\prod _{j=1}^n\frac{1}{1+|x_j|^2}\frac{1}{1+|\xi _j|^2}\in {\mathcal {S}}_0, \qquad z=(x,\xi )\in {\mathbb {R}}^{2n}.\end{aligned}$$

Then consider $\phi ^4(h)=h/({\mathcal {F}}_W(A))^4\in {\mathcal {S}}_0$. Since ${\mathcal {S}}_0\subset C_0({\mathbb {R}}^{2n})$ there exists a $C>0$ such that for all $z\in {\mathbb {R}}^{2n}$ we have

$$\begin{aligned}|h(z)|/|{\mathcal {F}}_W(A)(z)|^{4}\le C,\end{aligned}$$

or equivalently

$$\begin{aligned}\sqrt{|h(z)|}\le C^{1/4}|{\mathcal {F}}_W(A)(z)|^{2}.\end{aligned}$$

However, $\sqrt{|h(z)|}\not \in L^1({\mathbb {R}}^{2n})$ implying that ${\mathcal {F}}_W(A)\not \in L^2({\mathbb {R}}^{2n})$. $\square$

Remark 13

The Feichtinger algebra ${\mathcal {S}}_0({\mathbb {R}}^{n})$ is well-known for being the smallest strongly modulation-invariant Segal algebra. The same is indeed true for graded quantum Segal algebras: Denote by $(X,\Vert \cdot \Vert _X)$ a Banach space where $X \subset L^1 \oplus {{\mathcal {T}}}^1$ is a graded subspace, which is both strongly translation-invariant and strongly modulation-invariant. Further, assume that there is at least one element $(f, A) \in X$ with both $f \ne 0$ and $A \ne 0$ and $(f, A) \in {\mathcal {S}}_0^Q$. By Corollary 5.11 such elements always exists for X a graded quantum Segal algebra. Then, we endow ${\mathcal {F}}(X) \subset C_0({\mathbb {R}}^{2n}) \oplus C_0({\mathbb {R}}^{2n})$ with the norm $\Vert ({\mathcal {F}}_\sigma (f), {\mathcal {F}}_W(A))\Vert = \Vert (f, A)\Vert _X$. This turns both components of ${\mathcal {F}}(X)$ into a strongly translation-invariant and strongly modulation-invariant space that contains at least one nontrivial element of ${\mathcal {S}}_0({\mathbb {R}}^{2n})$. By [16, Thm. 7.3] we thus have ${\mathcal {S}}_0({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_0({\mathbb {R}}^{2n}) \subset {\mathcal {F}}(X)$.

One of the useful applications of the Feichtinger algebra is as a space of test functions. Its dual space ${\mathcal {S}}_0'({\mathbb {R}}^{2n})$ is sometimes referred to as the space of mild distributions. It contains, in addition to all $L^p$ spaces, some distributions such as the point measures $\delta _x$ and Dirac combs $\sum _{k \in \mathbb Z^{d}} \delta _k$. Having defined the Feichtinger operators, we can also consider the Banach space dual of ${\mathcal {S}}_0({\mathcal {H}})$, which we will denote by ${\mathcal {S}}_0'({\mathcal {H}})$. Since the Schwartz operators ${\mathcal {S}}({\mathcal {H}})$ embeds continuously and densely into ${\mathcal {S}}_0({\mathcal {H}})$, the dual ${\mathcal {S}}_0'({\mathcal {H}})$ is continuously embedded into $S'({\mathcal {H}})$. As is the case for the Feichtinger operators, the space ${\mathcal {S}}_0'({\mathcal {H}})$ can be equipped with many equivalent norms. As an example, define the norm

$$\begin{aligned} \Vert A_g\Vert _{{\mathcal {S}}_0'({\mathcal {H}})} :=\Vert g \Vert _{{\mathcal {S}}_0'({\mathbb {R}}^{2n})}. \end{aligned}$$

Here, $\Vert \cdot \Vert _{{\mathcal {S}}_0'({\mathbb {R}}^{2n})}$ is any of the equivalent norms of ${\mathcal {S}}_0'({\mathbb {R}}^{2n})$. We will denote by $\langle \cdot , \cdot \rangle _{{\mathcal {S}}_0',{\mathcal {S}}_0}$ the sesquilinear pairing of an element in ${\mathcal {S}}_0'({\mathcal {H}})$ and ${\mathcal {S}}_0({\mathcal {H}})$. The following statement is, in principle, the dual statement of Theorem 5.13.

Theorem 5.17

Let $A \in {\mathcal {S}}'({\mathcal {H}})$ and let $B \in {\mathcal {S}}_0({\mathcal {H}}){\setminus }\{0\}$ be any fixed Feichtinger operator. Denote

$$\begin{aligned} \Vert A\Vert _{\infty , B, \gamma }:=\sup _{z \in {\mathbb {R}}^{2n}} \Vert (\gamma _z A) *B\Vert _{L^1}, \ \Vert A\Vert _{\infty , B, \gamma \alpha } :=\sup _{z,z' \in {\mathbb {R}}^{2n}} |\langle A, \gamma _{z'}\alpha _z(B)\rangle _{{\mathcal {S}}_0',{\mathcal {S}}_0}|. \end{aligned}$$

Then, the following characterizations are equivalent:

• 1) The operator A can be written as $A = A_{g}$ for some $g \in {\mathcal {S}}_{0}'({\mathbb {R}}^{2n})$.

• 2) There exists an element $K_{A}\in {\mathcal {S}}_{0}'({\mathbb {R}}^{2n})$ such that

  $$\begin{aligned} \langle Af_2, f_1\rangle _{{\mathcal {S}}_0'({\mathbb {R}}^n),{\mathcal {S}}_0({\mathbb {R}}^n)}= \langle K_A, f_1\otimes f_2\rangle _{{\mathcal {S}}_0'({\mathbb {R}}^{2n}),{\mathcal {S}}_0({\mathbb {R}}^{2n})} \end{aligned}$$

  for all $f_1, f_2\in {\mathcal {S}}_0({\mathbb {R}}^n)$.

• 3) The Fourier-Weyl transform of A satisfies ${\mathcal {F}}_{W}(A) \in {\mathcal {S}}_{0}'({\mathbb {R}}^{2n})$.

• 4) A satisfies $\Vert A\Vert _{\infty , B, \gamma } < \infty$.

• 5) A satisfies $\Vert A\Vert _{\infty , B, \gamma \alpha } < \infty$.

Moreover, the following are equivalent norms on ${\mathcal {S}}_0'({\mathcal {H}})$:

$$\begin{aligned}\Vert A\Vert _{{\mathcal {S}}_0'({\mathcal {H}})} \cong \Vert {\mathcal {F}}_W(A)\Vert _{{\mathcal {S}}_0'({\mathbb {R}}^{2n})} \cong \Vert K_A\Vert _{{\mathcal {S}}_0'({\mathbb {R}}^{2n})} \cong \Vert A\Vert _{\infty , B, \gamma } \cong \Vert A\Vert _{\infty , B, \gamma \alpha }. \end{aligned}$$

We omit the proof, as it is very similar to that of Theorem 5.13, replacing characterizations of ${\mathcal {S}}_0({\mathbb {R}}^{2n})$ with those of ${\mathcal {S}}_0'({\mathbb {R}}^{2n})$, cf. [16]. Note that ${\mathcal {L}}({\mathcal {H}})$ continuously embeds into ${\mathcal {S}}_0'({\mathcal {H}})$. Further, by the kernel theorem for ${\mathcal {S}}_0({\mathbb {R}}^n)$, cf. [16, Thm. 9.3] and characterization 2), ${\mathcal {S}}_0'({\mathcal {H}})$ agrees exactly with those elements of ${\mathcal {S}}'({\mathcal {H}})$ which extend to bounded linear operators from ${\mathcal {S}}_0({\mathbb {R}}^n)$ to ${\mathcal {S}}_0'({\mathbb {R}}^n)$. Hence, among all the operators given by kernels/symbols in ${\mathcal {S}}'({\mathbb {R}}^{2n})$, characterization 4) and  5) from the proposition yield characterizations through quantum harmonic analysis methods of those which are well-behaved in the Feichtinger setting.

Remark 14

We now briefly discuss how key properties of the mild distributions carry over to ${\mathcal {S}}_0'({\mathcal {H}})$ and ${\mathcal {S}}_0'({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_0'({\mathcal {H}})$. First note that ${\mathcal {S}}_0'({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_0'({\mathcal {H}})$ is not an algebra, but it is a module over ${\mathcal {S}}_0({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_0({\mathcal {H}})$.

• 1) The space ${\mathcal {S}}_{0}^{Q}$ can through the Weyl quantization be identified with ${\mathcal {S}}_{0}({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_{0}({\mathbb {R}}^{2n})$ as a Banach space. As such, the Banach space dual ${\mathcal {S}}_{0}^{Q}{}':=({\mathcal {S}}_{0}^{Q})'$ of ${\mathcal {S}}_{0}^{Q}$ can be identified with ${\mathcal {S}}_{0}'({\mathbb {R}}^{2n}) \oplus {\mathcal {S}}_{0}'({\mathbb {R}}^{2n})$. Hence bounded linear operators

  $$\begin{aligned}A:{\mathcal {S}}_{0}^{Q} \rightarrow {\mathcal {S}}_{0}^{Q}{}'\end{aligned}$$

  can be decomposed into four operators $A_{ij}:{\mathcal {S}}_{0}({\mathbb {R}}^{2n}) \rightarrow {\mathcal {S}}_{0}'({\mathbb {R}}^{2n})$ for $i,j =1,2$. Each of these operators can be represented with an integral kernel $K_{ij} \in {\mathcal {S}}_{0}'({\mathbb {R}}^{4n})$ by [16, Thm. 9.3]. Thus for $f_{1},f_{2},g_{1},g_{2} \in {\mathcal {S}}_{0}({\mathbb {R}}^{2n})$ we can write A as

  $$\begin{aligned} \langle A(f_{1}, A_{g_{1}}), (f_{2}, A_{g_{2}}) \rangle _{{\mathcal {S}}_{0}^{Q}{}', {\mathcal {S}}_{0}^{Q}} = \langle K_{11}, f_{2} \otimes f_{1} \rangle _{{\mathcal {S}}_0',{\mathcal {S}}_0} + \langle K_{21}, f_{2} \otimes g_{1} \rangle _{{\mathcal {S}}_0',{\mathcal {S}}_0}\\ + \langle K_{12}, g_{2} \otimes f_{1} \rangle _{{\mathcal {S}}_0',{\mathcal {S}}_0} + \langle K_{22}, g_{2} \otimes g_{1} \rangle _{{\mathcal {S}}_0',{\mathcal {S}}_0}. \end{aligned}$$

  From here, it is not difficult to formulate a kernel theorem for bounded linear maps from ${\mathcal {S}}_0^Q$ to ${\mathcal {S}}_0^{Q}{}'$. We leave the details to the interested reader.

• 2) Recall that elements from ${\mathcal {S}}_0({\mathbb {R}}^n)$ satisfy Poisson’s summation formula, see e.g., [8, Thm. 4.2]. Indeed, there is a more general form with summation over different lattices, but for simplicity we stick to the following basic formulation: It is

  $$\begin{aligned} \sum _{k \in \mathbb Z^n} f(k) =(2\pi )^{n/2} \sum _{k \in 2\pi {\mathbb {Z}}^n} {\mathcal {F}}(f)(k). \end{aligned}$$

  In even dimensions, the right-hand side can of course be replaced by the symplectic Fourier transform. For $A = A_g \in {\mathcal {S}}_0({\mathcal {H}})$, we therefore of course have

  $$\begin{aligned} \sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_W(A_g)(k) = \sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_\sigma (g)(k) = \frac{1}{(2\pi )^n}\sum _{k \in \mathbb Z^{2n}} g(k), \end{aligned}$$

  which can be interpreted as a Poisson summation formula for quantum harmonic analysis. By Theorem 5.13 2) we know that every $A \in {\mathcal {S}}_0({\mathcal {H}})$ is of the form

  $$\begin{aligned}Af(s) = \int _{{\mathbb {R}}^n} f(t) K_A(s,t)\,\textrm{d}y\end{aligned}$$

  for $K_A \in {\mathcal {S}}_0$. For $(x,\xi ) \in {\mathbb {R}}^{2n}$, it is not hard to verify that the integral kernel of $AW_{(x, \xi )}$ is now

  $$\begin{aligned} K_{AW_{(x, \xi )}}(s, t) = e^{i\xi \cdot t+i\xi \cdot x/2} K_A(s,t+x). \end{aligned}$$

  By [7, Cor. 3.15] we have

  $$\begin{aligned} {\mathcal {F}}_W(A)(x, \xi ) = {{\,\textrm{tr}\,}}(AW_{(x, \xi )}) = \int _{{\mathbb {R}}^n} e^{i\xi \cdot s+i\xi \cdot x/2}K_A(s,s+x)\,\textrm{d}s. \end{aligned}$$

  On the other hand, if $A=A_g$ then g can be computed by

  $$\begin{aligned} g(x,\xi ) = \int _{{\mathbb {R}}^n} K_A\left( x+\frac{y}{2}, x-\frac{y}{2}\right) e^{- i \xi \cdot y}\,\textrm{d}y \in {\mathcal {S}}_0({\mathbb {R}}^{2n}). \end{aligned}$$

  Hence, the above Poisson summation formula shows

  $$\begin{aligned} \sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_W(A_g)(k)&= \sum _{(k_1, k_2) \in \mathbb Z^{2n}} \int _{{\mathbb {R}}^n} e^{2\pi i k_2 \cdot s+2\pi ^2 ik_1k_2}K_A(s,s+2\pi k_1)\,\textrm{d}s \\&=\frac{1}{(2\pi )^n} \sum _{k \in \mathbb Z^{2n}} g(k) \\&= \frac{1}{(2\pi )^n}\sum _{(k_1, k_2) \in \mathbb Z^{2n}} \int _{{\mathbb {R}}^n} K_A(k_1+\frac{y}{2}, k_1-\frac{y}{2}) e^{- i k_2 \cdot y}\,\textrm{d}y. \end{aligned}$$

  A similar formula can be obtained by expressing the integral kernel in terms of the symbol.

• 3) Another point of view on the Poisson summation formula is the following: It is

  $$\begin{aligned} \frac{1}{(2\pi )^n} \sum _{k \in \mathbb Z^{2n}} \langle \delta _k, f\rangle = \frac{1}{(2\pi )^n} \sum _{k \in \mathbb Z^{2n}} f(k) = \sum _{k \in 2\pi \mathbb Z^{2n}} {\widehat{f}}(k) = \sum _{k \in 2\pi \mathbb Z^{2n}} \langle \delta _k, {\widehat{f}}\rangle , \end{aligned}$$

  where $\sum _{k \in \mathbb Z^{2n}} \delta _k, \ \sum _{k \in 2\pi \mathbb Z^{2n}} \widehat{\delta _k}$ are in $S_0'({\mathbb {R}}^{2n})$. Hence

  $$\begin{aligned} \frac{1}{(2\pi )^n} \sum _{k \in \mathbb Z^{2n}} \alpha _k(\delta _0)&= \sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_\sigma (\alpha _k(\delta _0)) = \sum _{k \in 2\pi \mathbb Z^{2n}} \gamma _k {\mathcal {F}}_\sigma (\delta _0) \\&= \sum _{k \in 2\pi \mathbb Z^{2n}} \frac{1}{(2\pi )^n}\gamma _k(1). \end{aligned}$$

  Applying the inverse Fourier-Weyl transform, using that ${\mathcal {F}}_W^{-1}(\delta _0) = {\textrm{Id}}$ and ${\mathcal {F}}_W^{-1}(1) = 2^nP$, we get

  $$\begin{aligned} \sum _{k \in \mathbb Z^{2n}}{\mathcal {F}}_W^{-1} (\alpha _k(\delta _0))&= \sum _{k \in \mathbb Z^{2n}} \gamma _k {\mathcal {F}}_W^{-1}(\delta _0) = \sum _{k \in \mathbb Z^{2n}} \gamma _k({\textrm{Id}}) = \sum _{k \in \mathbb Z^{2n}} W_{k},\\ \sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_W^{-1}(\gamma _k(1))&= \sum _{k \in 2\pi \mathbb Z^{2n}} \alpha _k {\mathcal {F}}_W^{-1}(1) = 2^n\sum _{k \in 2\pi \mathbb Z^{2n}} \alpha _k(P) \\&= 2^n\sum _{k \in 2\pi \mathbb Z^{2n}} W_{2k}P. \end{aligned}$$

  Hence, we obtain in ${\mathcal {S}}_0'({\mathcal {H}})$:

  $$\begin{aligned} 2^n \sum _{k \in 2\pi \mathbb Z^{2n}} W_{2k}P = \sum _{k \in \mathbb Z^{2n}} W_{k}. \end{aligned}$$

  Applying this to some $A \in {\mathcal {S}}_0({\mathcal {H}})$ gives the somewhat ominous result:

  $$\begin{aligned} \sum _{k \in \mathbb Z^{2n}} {\mathcal {F}}_W(A)(k) = 2^n\sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_W(AP)(2k). \end{aligned}$$

  This equality can conceptually be explained a little better by doing the following: Recall the parity operator is both symmetric and unitary, hence P has spectrum contained in $\{ -1, 1\}$. The spectral decomposition of ${\mathcal {H}}$ is given by ${\mathcal {H}} = L_{\textrm{even}}^2 \oplus L_{\textrm{odd}}^2$, the decomposition into even and odd functions. Now, $P|_{L_{\textrm{even}}^2} = {\textrm{Id}}$ and $P|_{L_{\textrm{odd}}^2} = -{\textrm{Id}}$. We define a unitary square root V of P by letting $V|_{L_{\textrm{even}}^2} = {\textrm{Id}}$ and $V|_{L_{\textrm{odd}}^2} = i{\textrm{Id}}$. Then $VW_{(x,\xi )} = W_{(\xi , -x)}V$ and $W_z V = VW_{(-\xi , x)}$. Thus, we get

  $$\begin{aligned} \sum _{k \in 2\pi \mathbb Z^{2n}} {\mathcal {F}}_W(AP)(2k)&= \sum _{k_1,k_2 \in 2\pi \mathbb Z^{n}} {\mathcal {F}}_W(V A V)(2k_1,-2k_1) \\&= \sum _{k \in 4\pi \mathbb Z^{2n}} {\mathcal {F}}_W(V A V)(k). \end{aligned}$$

  Since passing from A to PAP plays the role of a reflection (analogously to passing from f(z) to $f(-z)$, $z \in \mathbb C$), passing from A to VAV can be thought of as rotating the argument by $\frac{\pi }{2}$ (analogously to passing from f(z) to f(iz)). Recalling now that taking the Fourier transform can be thought of as rotating the phase space by $\frac{\pi }{2}$, it is somewhat more reasonable that the Poisson summation formula should relate the A and VAV.

5.4 Miscellaneous examples

We have thus far seen examples of quantum Segal algebras that are

not ideals of $L^1 \oplus {{\mathcal {T}}}^1$, and will now describe a class of examples that are ideals. They are natural analogues of the Segal algebras from Example 2.3 (E3).

Example 5.18

Let $\mu$ be any Radon measure on ${\mathbb {R}}^{2n}$. Define the set $S_p^\mu$ by

$$\begin{aligned} S_p^\mu = \{ (f, A) \in L^1 \oplus {{\mathcal {T}}}^1: ~\Vert {\widehat{f}}\Vert _{L^p(\mu )} + \Vert {\widehat{A}}\Vert _{L^p({\mathbb {R}}^{2n},\mu )} < \infty \}, \end{aligned}$$

with the norm

$$\begin{aligned} \Vert (f, A)\Vert _{S_p^\mu } = \Vert f\Vert _{L^1} + \Vert {\widehat{f}}\Vert _{L^p({\mathbb {R}}^{2n},\mu )} + \Vert A\Vert _{{{\mathcal {T}}}^1} + \Vert {\widehat{A}}\Vert _{L^p({\mathbb {R}}^{2n},\mu )}. \end{aligned}$$

Since $S_p^\mu$ contains any $(f, A) \in L^1 \oplus {{\mathcal {T}}}^1$ such that ${\widehat{f}}$ and ${\widehat{A}}$ both have compact support, it is a dense subspace of $L^1 \oplus {{\mathcal {T}}}^1$. Further, it is not hard to verify that $S_p^\mu$ is both a graded quantum Segal algebra and an ideal of $L^1 \oplus {{\mathcal {T}}}^1$. In the case that $\mu$ is a finite measure we have that $S_p^\mu = L^1 \oplus {{\mathcal {T}}}^1$, but for appropriately infinite measures $S_p^\mu$ is a strict subset.

Note that, as in [21, (vii), p. 26], these $S_p^\mu$ also have the following theoretical value: Given $A \in {{\mathcal {T}}}^1$ such that ${\widehat{A}}$ is not compactly supported, then one can explicitly write down a $\mu$ such that the graded quantum Segal algebra $S_1^\mu$ does not contain A.

We note that quantum Segal algebras, which are ideals of $L^1 \oplus {{\mathcal {T}}}^1$, are abstract Segal algebras in the sense of Burnham [2, Def. 1.1]. In particular, such quantum Segal algebras have the same closed ideals as $L^1 \oplus {{\mathcal {T}}}^1$. We refrain from giving the precise result here and only refer to [2, Thm. 1.1]. Instead, we want to give another result on such quantum Segal algebras, which is a close relative to our Corollary 5.11.

Proposition 5.19

Let G denote the set of all graded quantum Segal algebras which are ideals of $L^1({\mathbb {R}}^{2n}) \oplus {{\mathcal {T}}}^1({\mathcal {H}})$. Then

$$\begin{aligned}{\mathcal {F}}\Big (\bigcap _{S_1\oplus S_2\in G}&S_1\oplus S_2\Big ) \\ {}&= (C_{c}({\mathbb {R}}^{2n})\cap {\mathcal {F}}_\sigma (L^1({\mathbb {R}}^{2n}))) \oplus (C_{c}({\mathbb {R}}^{2n})\cap {\mathcal {F}}_W({{\mathcal {T}}}^1({\mathcal {H}}))).\end{aligned}$$

Proof

As described above, the algebras $S_1^\mu$ can be utilized to show the inclusion “$\subseteq$”. For the reverse inclusion, note that for any $f \in L^1({\mathbb {R}}^{2n})$ with ${\widehat{f}} \in C_c({\mathbb {R}}^{2n})$ we have that (f, 0) is contained in the intersection by Corollary 5.11. Let $A \in {{\mathcal {T}}}^1$ such that ${\widehat{A}}$ is compactly supported. Then, there exists $g \in {\mathcal {S}}({\mathbb {R}}^{2n})$ such that ${\widehat{g}}$ is compactly supported and ${\widehat{g}} \equiv 1$ on the support of ${\widehat{A}}$. In particular, (g, 0) is contained in every graded quantum Segal algebra. Thus, the convolution $(g, 0) *(0, A)$ is contained in every graded quantum Segal algebra which is an ideal in $L^1 \oplus {{\mathcal {T}}}^1$. By comparing Fourier transforms, we find that $(g, 0) *(0, A) = (0, A)$. Therefore, the element (0, A) is contained in every graded quantum Segal algebra which is an ideal of $L^1 \oplus {{\mathcal {T}}}^1$. $\square$

We still owe the reader an example justifying the special treatment of graded quantum Segal algebras, i.e., showing that not every quantum Segal algebra is graded. Here it is:

Example 5.20

Assume that $S_1$ and $S_2$ are two Segal algebras contained in the Feichtinger algebra ${\mathcal {S}}_0({\mathbb {R}}^{2n})$ such that there exist $f_1\in S_1$ and $g_1\in S_2$ with $f_1\not \in S_2$ and $g_1\not \in S_1$. Let

$$\begin{aligned}S_T=\{(f,A_{{\tilde{g}}}): f, g \in \mathcal S_0(\mathbb R^{2n}) \text { s.th. } f+g \in S_1, ~f-g \in S_2\}\end{aligned}$$

with the norm

$$\begin{aligned}\Vert (f,A_{{\tilde{g}}})\Vert _{S_T}:=\Vert f+g\Vert _{S_1}+\Vert f-g\Vert _{S_2}.\end{aligned}$$

Then

$$\begin{aligned} \Vert (f,A_{{\tilde{g}}})*(h,A_{{\tilde{j}}})\Vert _{S_T}&= \Vert (f+g)*(h+j)\Vert _{S_1}+\Vert (f-g)*(h-j)\Vert _{S_2}\\&\le \Vert (f+g)\Vert _{S_1}\Vert (h+j)\Vert _{S_1}+\Vert (f-g)\Vert _{S_2}\Vert (h-j)\Vert _{S_2}\\&\le \Vert (f,A_{{\tilde{g}}})\Vert _{S_T}\Vert (h,A_{{\tilde{j}}})\Vert _{S_T}, \end{aligned}$$

showing that $S_T$ is a normed algebra. Further, it is not hard to see that it is complete and satisfy (QS3) and (QS4). The property (QS1) follows from the fact that $C_0({\mathbb {R}}^{2n})\oplus C_0({\mathbb {R}}^{2n})\subset {\mathcal {F}}(S_T)$. If we set $f+g=f_1$ and $f-g=g_1$ we have an example $(f, A_g) \in S_T$ where (f, 0) cannot be in $S_T$.

We give such an example of two Segal algebras $S_1$, $S_2$ in the case $n=1$, and leave it to the reader to carry out the necessary adaptations for obtaining such subalgebras of ${\mathcal {S}}_0({\mathbb {R}}^{2n})$. Let $S_1={\mathcal {S}}_0({\mathbb {R}}^2) \cap {\mathcal {F}}^{-1}(L^2({\mathbb {R}}^2, (1+x^2)^{10}))$ and $S_2={\mathcal {S}}_0({\mathbb {R}}^2) \cap {\mathcal {F}}^{-1}(L^2({\mathbb {R}}^2, e^{2x}))$. By Example 2.3, these are intersections of Segal algebras, hence themselves Segal algebras. Consider the function

$$\begin{aligned}f_a(x,\xi )=\frac{1}{x^2+a^2}\frac{1}{\xi ^2+a^2}\end{aligned}$$

with Fourier transform

$$\begin{aligned}{\mathcal {F}}(f_a)(x,\xi )=\frac{\pi }{2a^2}e^{-a(|x|+|\xi |)}.\end{aligned}$$

Using that $f_a\in {\mathcal {S}}_0({\mathbb {R}}^{2})$ for all a, we observe that $f_1\in S_1$ but $f\not \in S_2$. On the other hand, the function

$$\begin{aligned}g(x,\xi )=\frac{1-e^{-x^2}}{1+x^2}\frac{1-e^{-\xi ^2}}{1+\xi ^2}H(-x)H(-\xi ),\end{aligned}$$

with H the Heaviside function, is smooth and in $L^1({\mathbb {R}}^2)$ with the first two derivatives in $L^1({\mathbb {R}}^2)$. Hence, its Fourier transform is in $L^1({\mathbb {R}}^2)$ as well. We know that ${\mathcal {S}}_0({\mathbb {R}}^2)*L^1({\mathbb {R}}^2) \subset {\mathcal {S}}_0({\mathbb {R}}^2)$, hence

$$\begin{aligned}g_1={\mathcal {F}}^{-1}(f_1)*{\mathcal {F}}^{-1}\left( g\right) \in {\mathcal {S}}_0({\mathbb {R}}^2).\end{aligned}$$

Then $g_1\in S_2$ while $g_1\not \in S_1$.