On the Weyl transform for rotationally invariant symbols

Abstract

Several formulas for the eigenvalues $\lambda _j$ of the Weyl transforms $W_\sigma$ of symbols $\sigma$ given by radially symmetric distributions are derived. These yield criteria for the boundedness and the compactness, respectively, of the pseudo-differential operators $W_\sigma .$ We investigate some examples by analyzing the asymptotic behavior of $\lambda _j$ for $j\rightarrow \infty$.

1 Introduction and notation

Our starting point for this study were the Theorems 24.5 and 26.1 in the monograph [20]. Therein, criteria for the boundedness and for the compactness, respectively, of Weyl transforms for rotation invariant symbols in the space of “temperate functions” $L^1_{-\infty }({\mathbb {R}}^2)=\cup _{j\in {\mathbb {N}}_0} (1+x^2+y^2)^j\cdot L^1({\mathbb {R}}^2_{xy})$ are formulated. However, M. Wong’s interesting example of the Dirac measure on the circle $\rho {\mathbb {S}}^1,$$\rho >0,$ i.e., $\sigma =\delta (\rho -\sqrt{x^2+y^2}),$ already shows that there are symbols $\sigma$ not belonging to $L^1_{-\infty }({\mathbb {R}}^2),$ but nevertheless leading to compact Weyl transforms, see [20, Thm. 28.1, p. 131].

In this paper, we slightly generalize [20, Thm. 24.5, p. 115] to the space ${\mathcal {S}}'_r({\mathbb {R}}^2)$ of radially symmetric temperate distributions (see Proposition 3), and similarly [20, Thm. 26.1, p. 123] to the space $\mathcal {D}'_{L^1,-1}({\mathbb {R}}^2)$ of weighted integrable distributions (see the Corollary to Proposition 4). In Proposition 5 and its corollary, we formulate the straight-forward generalization to n dimensions. In the preparative Proposition 1, we repeat some classical results and in Proposition 2, we give a constructive proof of B. Simon’s theorem of the existence of $L^q\!$-functions, $q>2,$ without bounded Weyl transforms, cf. [20, Sect. 13, p. 67].

In Sect. 5, we present three more advanced examples: We characterize the boundedness and the compactness of the Weyl transforms for the symbols $\sigma _z=(x^2+y^2)^z,$$\sigma _z=(x^2+y^2-\rho ^2)_+^z,$ and $\sigma _{p,z}=\chi (x^2+y^2)(x^2+y^2)^z\exp ({\mathrm{i}}(x^2+y^2)^p),$ respectively, for $z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}}),$$\rho >0,$$p>1$ and $\chi \in {\mathcal {C}}^\infty ({\mathbb {R}})$ with $\chi (t)=1$ for $t\ge 2,$$\chi (t)=0$ for $t\le 1,$ in dependence on the parameters z and p. (Note that the second example contains the Dirac measure on the circle as a limit case since $\lim _{z\rightarrow -1}t_+^z/\Gamma (1+z)=\delta (t)$ in ${\mathcal {S}}'({\mathbb {R}}_t^1).)$ In these examples, the determination of the asymptotic behavior of the eigenvalues of the Weyl transforms plays an essential rôle.

Let us introduce some notation. Besides the familiar Banach spaces $L^p({\mathbb {R}}^n),$$1\le p\le \infty ,$ we use the Banach spaces of integrable measures $\mathcal {M}^1({\mathbb {R}}^n),$ see [6, p. 345], and of bounded and of compact linear operators on $L^2({\mathbb {R}}^n),$ respectively, which we denote by ${\mathcal {L}}(L^2({\mathbb {R}}^n))$ and by ${\text {Com}}(L^2({\mathbb {R}}^n)),$ respectively, compare [11].

We employ the standard notation for the distribution spaces $\mathcal {D}',\,{\mathcal {S}}',\,{\mathcal {E}}',$ the dual spaces of the spaces $\mathcal {D},\,{\mathcal {S}},\,{\mathcal {E}}$ of “test functions”, of “rapidly decreasing functions” and of $C^\infty$ functions, respectively, see [4, 6, 15]. ${\mathcal {S}}'_r({\mathbb {R}}^n)$ denotes the closed subspace of ${\mathcal {S}}'({\mathbb {R}}^n)$ consisting of radially symmetric temperate distributions. In order to display the active variable in a distribution, say $x\in {\mathbb {R}}^n,$ we use notation as T(x) or $T\in \mathcal {D}'({\mathbb {R}}^n_x).$ Furthermore, we use at one instance the space $s'({\mathbb {N}}_0)$ of slowly increasing sequences, which is the dual of the space $s({\mathbb {N}}_0)$ of fast decreasing sequences.

The spaces $\mathcal {D}_{L^p},\mathcal {D}'_{L^p},\ 1\le p\le \infty ,$ were introduced in [15, Ch. VI, Sect. 8, p. 199], and we need, just for $p=1$ and $p=\infty ,$ so-called “weighted” $\mathcal {D}_{L^p}\!$ and $\mathcal {D}'_{L^p}\!$-spaces, i.e.,

$$\begin{aligned}&\mathcal {D}_{L^p,\mu }:=\mathcal {D}_{L^p}\cdot (1+|x|^2)^{-\mu /2},\ \mathcal {D}'_{L^p,\mu }:=\mathcal {D}'_{L^p}\cdot (1+|x|^2)^{-\mu /2},\nonumber \\&\quad 1\le p\le \infty ,\ \mu \in {\mathbb {R}}, \end{aligned}$$

see [7, 8]. (The first appearance of a weighted ${\mathcal {D}}'_{L^p}\!$-space seems to be in [13, Sect. 3, p. 7].) In Sect. 6, we also employ the spaces

$$\begin{aligned} \mathcal {D}_{L^\infty ,1,\dots ,1}:=\mathcal {D}_{L^\infty }({\mathbb {R}}^n_t)\cdot \prod _{j=1}^n\left( 1+t_j^2\right) ^{-1/2},\ \mathcal {D}'_{L^1,-1,\dots ,-1}:=\mathcal {D}'_{L^1}({\mathbb {R}}^n_t)\cdot \prod _{j=1}^n\sqrt{1+t_j^2}. \end{aligned}$$

For the evaluation of a distribution $T\in E'$ on a test function $\phi \in E$ (where E is a normal space of distributions, see [6, Ch. IV, Sect. 2, Def. 3, p. 319]), we use angle brackets, i.e., $\langle \phi ,T\rangle$ or, more precisely ${}_E\langle \phi , T\rangle _{E'}.$ More generally, if $\phi \in E\,{\hat{\otimes }}\, F$ and $T\in E'$ for distribution spaces E, F,  then ${}_{E\,{\hat{\otimes }}\, F}\langle \phi , T\rangle _{E'}$ symbolizes the vector-valued scalar product $(E\,{\hat{\otimes }}\, F)\times E'\rightarrow F,$ see [14] for more information on vector-valued distributions. (In all tensor products of this study, both factors are complete and at least one of the factors is nuclear and hence $E\,{\hat{\otimes }}_\pi \, F=E\,{\hat{\otimes }}_\epsilon \, F$ and we simply write $E\,{\hat{\otimes }}\, F.)$

The Heaviside function is denoted by Y,  see [15, p. 36], and we write $\delta _\tau (t)\in \mathcal {D}'({\mathbb {R}}^1_t),\ \tau \in {\mathbb {R}},$ for the delta distribution with support in $\tau ,$ which is the derivative of $Y(t-\tau ),$ i.e., $\langle \phi ,\delta _\tau \rangle =\phi (\tau )$ for $\phi \in \mathcal {D}({\mathbb {R}}^1).$ In contrast, $\delta \in \mathcal {D}'({\mathbb {R}}^n)$ without any subscript stands for the delta distribution at the origin. For $z\in {\mathbb {C}}$ with ${\text {Re}}\,z>-1,$ we write $t_+^z$ for the locally integrable function $Y(t)t^z$ and we obtain, by analytic continuation, the analytic distribution-valued function

$$\begin{aligned} t_+^z:{\mathbb {C}}{\setminus }(-{\mathbb {N}})\longrightarrow {\mathcal {S}}'\left( {\mathbb {R}}^1_t\right) :z\longmapsto t_+^z, \end{aligned}$$

see [9, Ex. 1.4.8, p. 49]. The distribution $t_-^z$ is defined by $t_-^z=t_+^z(-t).$

The pull-back $h^*T=T\circ h\in \mathcal {D}'(\Omega )$ of a distribution T in one variable t with respect to a submersive ${\mathcal {C}}^\infty$ function $h:\Omega \rightarrow {\mathbb {R}},\ \Omega \subset {\mathbb {R}}^n{\text { open}},$ is defined as in [4, Equ. (7.2.4/5), p. 82] or in [9, Def. 1.2.12, p. 19], i.e.,

$$\begin{aligned} \langle \phi ,h^*T\rangle =\Big \langle \frac{\text {d}}{\text {d}t}\biggl (\int _\Omega Y(t-h(x)) \phi (x)\,{\text {d}}x\biggr ), T\Big \rangle ,\qquad \phi \in \mathcal {D}(\Omega ). \end{aligned}$$

We use the Fourier transform ${\mathcal {F}}$ in the form

$$\begin{aligned} ({\mathcal {F}}\phi )(\xi ):= \int _{{\mathbb {R}}^n}{\mathrm{e}}^{-{\mathrm{i}}\xi x}\phi (x)\,{\text {d}}x,\qquad \phi \in {\mathcal {S}}({\mathbb {R}}^n), \end{aligned}$$

this being extended to ${\mathcal {S}}'$ by continuity. For the partial Fourier transforms of a distribution $T\in {\mathcal {S}}'({\mathbb {R}}^{m+n}_{xy})$ with respect to x or y,  respectively, we use the notation ${\mathcal {F}}_xT$ and ${\mathcal {F}}_yT,$ respectively.

The Hermite polynomials $H_j$ and Hermite functions $h_j,h_\alpha$ are defined as usually:

$$\begin{aligned} H_j(x)=(-1)^j{\mathrm{e}}^{x^2}\cdot \left( {\mathrm{e}}^{-x^2}\right) ^{(j)},\quad h_j(x)={\mathrm{e}}^{-x^2/2}H_j(x),\quad j\in {\mathbb {N}}_0,\ x\in {\mathbb {R}}, \end{aligned}$$

and

$$\begin{aligned} h_\alpha (x)=\prod _{j=1}^n h_{\alpha _j}(x_j),\quad \alpha \in {\mathbb {N}}_0^n,\ x\in {\mathbb {R}}^n, \end{aligned}$$

see [5, Sect. 8.95], [20, Sects. 18, 19]. Similarly, the Laguerre polynomials $L^\alpha _j(x)$ of order $\alpha$ are defined by

$$\begin{aligned} L^\alpha _j(x):=j!^{-1} x^{-\alpha }{\mathrm{e}}^x\cdot \left( {\mathrm{e}}^{-x}x^{\alpha +j}\right) ^{(j)},\quad x>0,\ \alpha >-1,\ j\in {\mathbb {N}}_0, \end{aligned}$$

see [5, Sect. 8.97], [20, Sect. 20, p. 95]. We write $L_j(x):=L^0_j(x)$ for the Laguerre polynomials of order 0.

2 Review of definitions and classical results

The Weyl transformation W can be described as a partial Fourier transform composed with a linear transformation in ${\mathbb {R}}^{2n}_{xy},$ to wit

$$\begin{aligned} W:{\mathcal {S}}'({\mathbb {R}}^{2n})\longrightarrow {\mathcal {S}}'({\mathbb {R}}^{2n}):\sigma (x,y)\longmapsto W_\sigma (x,y). \end{aligned}$$

where

$$\begin{aligned} W_\sigma (x,y)=(2\pi )^{-n}({\mathcal {F}}_y \sigma )\Bigl (\frac{x+y}{2},y-x\Bigr ), \end{aligned}$$

see [3, p. 79], [20, Equ.(6.7), p. 30]. Hence if $\sigma \in {\mathcal {S}}'({\mathbb {R}}^{2n}),$ then its Weyl transform $W_\sigma$ is the pseudo-differential operator represented by the continuous linear mapping $W_\sigma :{\mathcal {S}}({\mathbb {R}}^n)\rightarrow {\mathcal {S}}'({\mathbb {R}}^n)$ with the kernel $W_\sigma (x,y)\in {\mathcal {S}}'({\mathbb {R}}^{2n}),$ i.e.,

$$\begin{aligned} (W_\sigma \phi )(x)=\langle \phi (y),W_\sigma (x,y)\rangle \in {\mathcal {S}}'({\mathbb {R}}^n_x),\quad \phi \in {\mathcal {S}}({\mathbb {R}}^n). \end{aligned}$$

For $\sigma \in {\mathcal {S}}({\mathbb {R}}^{2n})$ and $\phi \in {\mathcal {S}}({\mathbb {R}}^n),$ this implies

$$\begin{aligned} \begin{aligned} (W_\sigma \phi )(x)&=(2\pi )^{-n}\int \phi (y)({\mathcal {F}}_y \sigma )\Bigl (\frac{x+y}{2},y-x\Bigr ){\text {d}}y\\&=(2\pi )^{-n}\iint {\mathrm{e}}^{{\mathrm{i}}(x-y)\eta }\phi (y)\sigma (\tfrac{x+y}{2},\eta ) \,{\text {d}}y\text{ d }\eta , \end{aligned} \end{aligned}$$

compare [20, Equ. (4.2), p. 19].

In this study, we are mainly concerned with the question for which $\sigma$ the Weyl transform $W_\sigma$ can be extended to a bounded linear mapping $W_\sigma :L^2({\mathbb {R}}^n)\rightarrow L^2({\mathbb {R}}^n),$ i.e., $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n)).$ In other words, we investigate for which $\sigma$ the norm $\Vert W_\sigma \phi \Vert _2$ can be bounded by a constant multiple of $\Vert \phi \Vert _2$ for $\phi \in {\mathcal {S}}({\mathbb {R}}^n).$ If this condition is satisfied, we look for criteria guaranteeing that $W_\sigma$ is a compact operator in $L^2({\mathbb {R}}^n),$ i.e., that $W_\sigma \in {\text {Com}}(L^2({\mathbb {R}}^n)).$

Usually, W is expressed by the “Wigner transform,” which is a continuous sesquilinear mapping

$$\begin{aligned} w:{\mathcal {S}}({\mathbb {R}}^n)\times {\mathcal {S}}({\mathbb {R}}^n)\longrightarrow {\mathcal {S}}({\mathbb {R}}^{2n}). \end{aligned}$$

The mappings w and W are connected by

$$\begin{aligned} \langle {\bar{\psi }},W_\sigma \phi \rangle =\langle w( \phi ,\psi ),\sigma \rangle ,\quad \phi ,\psi \in {\mathcal {S}}({\mathbb {R}}^n). \end{aligned}$$

This implies

$$\begin{aligned} \begin{aligned} \langle {\bar{\psi }},W_\sigma \phi \rangle&=(2\pi )^{-n} \big \langle \overline{\psi (x)}\phi (y), ({\mathcal {F}}_y \sigma )\bigl (\tfrac{x+y}{2},y-x\bigr )\big \rangle \\&=(2\pi )^{-n} \big \langle \overline{\psi \bigl (x-\tfrac{y}{2}\bigr )}\phi \bigl (x+\tfrac{y}{2}\bigr ),{\mathcal {F}}_y \sigma \big \rangle \\&=(2\pi )^{-n} \Big \langle {\mathcal {F}}_y\Bigl ( \overline{\psi \bigl (x-\tfrac{y}{2}\bigr )}\phi \bigl (x+\tfrac{y}{2}\bigr )\Bigr ), \sigma \Big \rangle =\langle w( \phi ,\psi ),\sigma \rangle \end{aligned} \end{aligned}$$

and hence

$$\begin{aligned} \bigl (w(\phi ,\psi )\bigr )(x,y)=(2\pi )^{-n}\int _{{\mathbb {R}}^n}{\mathrm{e}}^{-{\mathrm{i}}y\eta }\phi (x+\tfrac{\eta }{2})\overline{\psi (x-\tfrac{\eta }{2})}\,{\text {d}}\eta . \end{aligned}$$

Note that we use for w a normalization constant different from that in [20, Equ. (3.12), p. 15], and that we also changed, for clarity, the notation for the Wigner transform from $W(\phi ,\psi )$ to $w(\phi ,\psi ).$

Next note that the properties of w mirror those of W. In fact, the bilinear mapping $(\phi ,\psi )\mapsto w(\phi ,{\bar{\psi }})$ extends from the tensor product ${\mathcal {S}}({\mathbb {R}}^n)\otimes {\mathcal {S}}({\mathbb {R}}^n)$ to its completion ${\mathcal {S}}({\mathbb {R}}^{2n}).$ This gives rise to the linear mapping

$$\begin{aligned} {\tilde{w}}:{\mathcal {S}}({\mathbb {R}}^{2n})\longrightarrow {\mathcal {S}}({\mathbb {R}}^{2n}):\Phi (x,y) \longmapsto (2\pi )^{-n}{\mathcal {F}}_y\Bigl (\Phi \bigl (x+\tfrac{y}{2},x-\tfrac{y}{2}\bigr )\Bigr ) \end{aligned}$$

which trivially extends to continuous linear mappings from $L^2({\mathbb {R}}^{2n})$ into itself, and from ${\mathcal {S}}'({\mathbb {R}}^{2n})$ into itself, respectively. We shall write again $w(\phi ,\psi )$ for ${\tilde{w}}(\phi \otimes {\bar{\psi }})$ if $\phi ,\psi$ both belong to $L^2({\mathbb {R}}^n)$ or to ${\mathcal {S}}'({\mathbb {R}}^n),$ respectively.

Furthermore, from

$$\begin{aligned} \begin{aligned} \Vert w(\phi ,\psi )\Vert _\infty&\le (2\pi )^{-n}\sup _{x\in {\mathbb {R}}^n}\Vert \phi (x+\tfrac{y}{2})\overline{\psi (x-\tfrac{y}{2})}\Vert _{L^1({\mathbb {R}}^n_y)} \\&\le (2\pi )^{-n}\Vert \phi \Vert _{2}\cdot \Vert \psi \Vert _{2},\quad \phi ,\psi \in L^2({\mathbb {R}}^n), \end{aligned} \end{aligned}$$

(2.1)

and the density of ${\mathcal {S}}({\mathbb {R}}^n)$ in $L^2({\mathbb {R}}^n),$ we conclude that

$$\begin{aligned} w: L^2({\mathbb {R}}^n)\times L^2({\mathbb {R}}^n)\longrightarrow C_0({\mathbb {R}}^{2n}) \end{aligned}$$

is well-defined and continuous with respect to the supremum norm on $C_0({\mathbb {R}}^{2n}).$ This in turn leads to the following proposition.

Proposition 1

• (a) If $\sigma \in L^2({\mathbb {R}}^{2n}),$ then $W_\sigma \in {\mathrm{Com}}(L^2({\mathbb {R}}^n));$

• (b) if $\sigma \in {\mathcal {M}}^1({\mathbb {R}}^{2n}),$ then $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n));$ if $\sigma \in L^1({\mathbb {R}}^{2n}),$ then $W_\sigma \in {\mathrm{Com}}(L^2({\mathbb {R}}^n));$

• (c) if $\sigma \in \bigcup _{p\in [1,2]} L^p({\mathbb {R}}^{2n}),$ then $W_\sigma \in {\mathrm{Com}}(L^2({\mathbb {R}}^n));$

• (d) if $\sigma \in \bigcup _{p\in [1,2]} {\mathcal {F}}\bigl (L^p({\mathbb {R}}^{2n})\bigr ),$ then $W_\sigma \in {\mathrm{Com}}(L^2({\mathbb {R}}^n));$

• (e) if $\sigma \in {\mathcal {F}}\bigl ({\mathcal {M}}^1({\mathbb {R}}^{2n})\bigr ),$ then $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n)).$

Proof

• (a) Since $W_\sigma (x,y)\in L^2({\mathbb {R}}^{2n})$ for $\sigma \in L^2({\mathbb {R}}^{2n}),$ it operates as a Hilbert–Schmidt operator on $L^2({\mathbb {R}}^n).$ Of course, the inequality

  $$\begin{aligned} |\langle W_\sigma \phi ,\psi \rangle |=|\langle W_\sigma ,\psi \otimes \phi \rangle | \le \Vert W_\sigma \Vert _{2}\cdot \Vert \phi \Vert _{2}\cdot \Vert \psi \Vert _{2}, \quad \phi ,\psi \in {\mathcal {S}}({\mathbb {R}}^n), \end{aligned}$$

  implies that $W_\sigma$ can be extended to a continuous linear mapping in $L^2({\mathbb {R}}^n),$ i.e., $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n)).$ Moreover, Hilbert–Schmidt operators are compact since the functions in $L^2({\mathbb {R}}^n)\otimes L^2({\mathbb {R}}^n)$ (which lead to finite rank operators) are dense in $L^2({\mathbb {R}}^{2n}).$

• (b) If $\sigma \in {\mathcal {M}}^1({\mathbb {R}}^{2n})$ and $\phi ,\psi \in {\mathcal {S}}({\mathbb {R}}^n),$ then the inequality in (2.1) implies

  $$\begin{aligned} |\langle W_\sigma \phi ,\psi \rangle |= & {} |\langle w(\phi ,{\bar{\psi }}),\sigma \rangle |\\\le & {} \Vert \sigma \Vert _{{\mathcal {M}}^1}\cdot \Vert w(\phi , {\bar{\psi }})\Vert _\infty \le (2\pi )^{-n} \Vert \sigma \Vert _{{\mathcal {M}}^1}\cdot \Vert \phi \Vert _2\cdot \Vert \psi \Vert _2 \end{aligned}$$

  and hence $\Vert W_\sigma \phi \Vert _2\le (2\pi )^{-n}\Vert \sigma \Vert _{{\mathcal {M}}^1}\cdot \Vert \phi \Vert _2.$ Therefore $W_\sigma$ belongs to ${\mathcal {L}}(L^2({\mathbb {R}}^n))$ and $\Vert W_\sigma \Vert _{{\mathcal {L}}(L^2({\mathbb {R}}^n))}\le (2\pi )^{-n}\Vert \sigma \Vert _{{\mathcal {M}}^1}.$ From the density of ${\mathcal {S}}({\mathbb {R}}^{2n})$ in $L^1({\mathbb {R}}^{2n})$ and since the linear mapping

  $$\begin{aligned} L^1({\mathbb {R}}^{2n})\hookrightarrow {\mathcal {M}}^1({\mathbb {R}}^{2n})\longrightarrow {\mathcal {L}}(L^2({\mathbb {R}}^n)) \end{aligned}$$

  is continuous, we also conclude that $W_\sigma$ is a compact operator for $\sigma \in L^1({\mathbb {R}}^{2n}).$ Note, however, that $L^1({\mathbb {R}}^{2n})$ is closed and thus not dense in ${\mathcal {M}}^1({\mathbb {R}}^{2n}).$ Therefore $W_\sigma$ is not compact for general $\sigma \in {\mathcal {M}}^1({\mathbb {R}}^{2n}).$ E.g., if $\sigma =\delta ,$ then $(W_\sigma \phi )(x)=\pi ^{-n}\phi (-x)$ and this operator is clearly not compact.

• (c) This follows from (a) and (b) on account of the inclusion

  $$\begin{aligned} \bigcup _{p\in [1,2]} L^p({\mathbb {R}}^{2n})\subset L^1({\mathbb {R}}^{2n}) +L^2({\mathbb {R}}^{2n}). \end{aligned}$$

• (d) and (e) If $\sigma \in {\mathcal {S}}'({\mathbb {R}}^{2n})$ and $\sigma (x,y)=\tau (-\frac{y}{2},\frac{x}{2}),$ then

  $$\begin{aligned} W_{{\mathcal {F}}\sigma }(x,y)=({\mathcal {F}}_x\sigma )(\tfrac{x+y}{2},x-y)=2^n({\mathcal {F}}_y\tau )(\tfrac{y-x}{2},x+y). \end{aligned}$$

  Hence $W_{{\mathcal {F}}\sigma }\phi =(4\pi )^n(W_\tau \phi )(-x)$ for $\phi \in {\mathcal {S}}({\mathbb {R}}^n)$ and (d), (e) follow from (b) and (c). The proof is complete. $\square$

Remark

The statements (a), (c) and (d) of Proposition 1 correspond to the Theorems 11.1 and 11.3 in [20, pp. 55, 57]. In [20, Thm. 28.1, p. 131], $W_\sigma$ is considered for $\sigma =\delta (\rho -\sqrt{x^2+y^2})\in {\mathcal {M}}^1({\mathbb {R}}^2),$$\rho >0.$ That $W_\sigma$ is a bounded operator on $L^2({\mathbb {R}})$ follows from Proposition 1 (b). The compactness of $W_\sigma$ will be shown in Sect. 3.

By the Hausdorff–Young inequality, ${\mathcal {F}}\bigl (L^p({\mathbb {R}}^{2n})\bigr )\subset L^q({\mathbb {R}}^{2n})$ if $1\le p\le 2$ and $\frac{1}{p}+\frac{1}{q}=1.$ The statement in Prop. 1 (d) asserts that $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n))$ holds for $\sigma$ belonging to the subspace ${\mathcal {F}}\bigl (L^p({\mathbb {R}}^{2n})\bigr )$ of $L^q({\mathbb {R}}^{2n}),$$2< q\le \infty .$ That $W_\sigma$ is not bounded on $L^2({\mathbb {R}}^n)$ for general $\sigma$ in $L^q({\mathbb {R}}^{2n}),$$2< q\le \infty ,$ is easily seen if $q=\infty .$ In fact the bounded function $\sigma ={\mathrm{e}}^{2{\mathrm{i}}xy}$ yields $W_\sigma =2^{-n}\delta \otimes 1$ which clearly does not belong to ${\mathcal {L}}(L^2({\mathbb {R}}^n)),$ compare [3, p. 92, (iii)], [20, Thm. 12.4, p. 61].

The case $2<q<\infty$ is trickier; it was settled in [16]. The detailed elaboration in [20, Sect. 13, p. 63] consists in an indirect proof relying on the uniform boundedness principle. We shall give a constructive proof in Proposition 2 below by means of an explicit example of $\sigma \in \bigcap _{q\in [2,\infty ]} L^q({\mathbb {R}}^2)$ such that $W_\sigma \not \in {\mathcal {L}}(L^2({\mathbb {R}})).$ (The n-dimensional case follows therefrom by tensoring.)

Proposition 2

The function

$$\begin{aligned} \sigma (x,y)=\mathrm{e}^{2{\mathrm{i}}xy}\cdot \frac{\Gamma (1-{\mathrm{i}}y)\,{\mathrm{e}}^{\pi y/2}-1}{y\,\sqrt{1+|x|}} \end{aligned}$$

fulfills $\sigma \in C_0({\mathbb {R}}^2)\cap \bigcap _{q>2} L^q({\mathbb {R}}^2)$ and $W_\sigma {\mathrm{e}}^{-x^2}\not \in L^2({\mathbb {R}}).$ In particular, $W_\sigma \not \in {\mathcal {L}}(L^2({\mathbb {R}})).$

Proof

Evidently, the function $(1+|x|)^{-1/2}$ belongs to $C_0({\mathbb {R}})\cap \bigcap _{q>2} L^q({\mathbb {R}}).$ The same holds true for

$$\begin{aligned} f(x)=\frac{\Gamma (1-{\mathrm{i}} x){\mathrm{e}}^{\pi x/2}-1}{x} \end{aligned}$$

according to Stirling’s formula, see [9, Ex. 1.6.8, (c), p. 91]. There it is also shown that $({\mathcal {F}} f)(x)=2\pi {\mathrm{i}}(Y(x)-{\mathrm{e}}^{{\mathrm{i}}\exp (-x)}).$ This implies

$$\begin{aligned} (2\pi )^{-1}{\mathcal {F}}_y({\mathrm{e}}^{2{\mathrm{i}}xy} f(y))={\mathrm{i}}(Y(y-2x)-{\mathrm{e}}^{{\mathrm{i}}\exp (2x-y)}) \end{aligned}$$

and

$$\begin{aligned} W_\sigma (x,y)=\frac{\sqrt{2}\,{\mathrm{i}}}{\sqrt{2+|x+y|}}\cdot (Y(-x)-{\mathrm{e}}^{\mathrm{i}}\exp (2x)). \end{aligned}$$

For $\phi ={\mathrm{e}}^{-x^2}\in {\mathcal {S}}({\mathbb {R}}),$ we conclude that

$$\begin{aligned} (W_\sigma \phi )(x)=\sqrt{2}\,{\mathrm{i}}(Y(-x)-{\mathrm{e}}^{\mathrm{i}}\exp (2x))\cdot \Bigl ({\mathrm{e}}^{-x^2}*\frac{1}{\sqrt{2+|x|}}\Bigr ) \end{aligned}$$

and hence, for $x>1,$

$$\begin{aligned} |(W_\sigma \phi )(x)|=\sqrt{2}\,{\mathrm{e}}^{-x^2}*\frac{1}{\sqrt{2+|x|}}\ge \frac{\sqrt{\log 2}}{\sqrt{2}\sqrt{2+x}}. \end{aligned}$$

This shows that $W_\sigma \phi$ is not square integrable and thus $W_\sigma \not \in {\mathcal {L}}(L^2({\mathbb {R}})).$ The proof is complete. $\square$

Remark

The known asymptotic of the $p\!$-norm (for $1\le p<2)$ of the Laguerre functions $L_j(2t){\mathrm{e}}^{-t},$ see [17, Lemma 1.5.4, p. 28], implies, due to Proposition 3 below and via the uniform boundedness principle, that $\sigma \in L^q({\mathbb {R}}^2)$ (for $2<q\le \infty )$ could even be chosen radially symmetric and such that $W_\sigma$ is unbounded on $L^2({\mathbb {R}}).$ We are not aware, however, of an explicit construction for such a function $\sigma .$

3 Rotationally invariant symbols

We assume first that $n=1$ and we postpone the straight-forward generalization to higher dimensions n to Sect. 6. We hence consider radially symmetric symbols $\sigma \in {\mathcal {S}}'({\mathbb {R}}^2)$ and we aim at establishing criteria for the boundedness or compactness, respectively, of the operators $W_\sigma$ on $L^2({\mathbb {R}}).$

For motivation, let us first describe the two lines of reasoning in [20]: In [20, Sect. 24], $\sigma (x,y)$ is a temperate radial function and it is shown that the Hermite functions $h_j\in {\mathcal {S}}({\mathbb {R}})$ (for the notation see Sect. 1) are eigenfunctions of $W_\sigma ,$ i.e., $W_\sigma h_j=\lambda _jh_j,$$\lambda _j\in {\mathbb {C}},$$j\in {\mathbb {N}}_0.$ Therefore the mapping properties of $W_\sigma$ are characterized by the eigenvalues $\lambda _j$ and it is shown that

$$\begin{aligned} \lambda _j=2(-1)^j\int _0^\infty \sigma (\rho ,0) L_j\left( 2\rho ^2\right) \,{\mathrm{e}}^{-\rho ^2}\rho \,\text{ d }\rho \end{aligned}$$

(3.1)

where $L_j$ denotes Laguerre polynomials of order 0, see [20, Thm. 24.5, p. 115, and Equ. (24.13), p. 116].

On the other hand, if $\sigma$ has the representation $\sigma =({\mathcal {F}}g)(x^2+y^2)$ with $g\in L^1({\mathbb {R}})\cup L^2({\mathbb {R}}),$ then $W_\sigma$ is compact according to [20, Thm. 26.1, p. 123]. We shall take up this second line of reasoning in Sect. 4 below, and we shall generalize the last-mentioned theorem in [20, p. 123] in order to give a characterization of boundedness and compactness, respectively, of $W_\sigma .$ Thereupon, we shall analyze some illustrative examples in Sect. 5.

In this section, we slightly generalize Theorem 24.5 in [20, p. 115] (and in particular formula (3.1) for the eigenvalues $\lambda _j)$ from temperate radial functions to temperate radial distributions.

Since the mapping

$$\begin{aligned} h:{\mathbb {R}}^2{\setminus }\{0\}\longrightarrow (0,\infty ):(x,y)\longmapsto x^2+y^2 \end{aligned}$$

(3.2)

is submersive, the pullback $h^*S,$$S\in \mathcal {D}'((0,\infty )),$ is well-defined (see Sect. 1), and we obtain in fact an isomorphism

$$\begin{aligned} \mathcal {D}'((0,\infty ))\overset{\sim }{\longrightarrow }\mathcal {D}'_r({\mathbb {R}}^2{\setminus }\{0\}):S\longmapsto S\circ h:=h^*S \end{aligned}$$

(3.3)

if $\mathcal {D}'_r({\mathbb {R}}^2{\setminus }\{0\})$ denotes the radially symmetric distributions on ${\mathbb {R}}^2{\setminus }\{0\}.$

Analogously to [12], see also [18, Sect. 3.10, p. 249], the map in (3.3) yields the following isomorphism:

$$\begin{aligned} \Phi :{\mathcal {S}}'([0,\infty ))\overset{\sim }{\longrightarrow }{\mathcal {S}}'_r({\mathbb {R}}^2):S\longmapsto \left( \phi \mapsto \tfrac{1}{2}\langle \int _{{\mathbb {S}}^1}\phi (\sqrt{t}\,\omega )\,\text{ d }\omega , S_t\rangle \right) . \end{aligned}$$

(3.4)

Here ${\mathcal {S}}'\bigl ([0,\infty )\bigr )$ is the dual of ${\mathcal {S}}\bigl ([0,\infty )\bigr )=\{\phi |_{[0,\infty )}:\phi \in {\mathcal {S}}({\mathbb {R}})\}$ equipped with the seminorms

$$\begin{aligned} \Vert \phi \Vert _{k,l}=\Vert (1+x)^k\phi ^{(l)}(x)\Vert _{L^\infty ([0,\infty ))},\quad k,l\in {\mathbb {N}}_0, \end{aligned}$$

and the proof is based on the fact that $\int _{{\mathbb {S}}^1}\phi (\sqrt{t}\,\omega )\,\text{ d }\omega \in {\mathcal {S}}([0,\infty ))$ for $\phi \in {\mathcal {S}}({\mathbb {R}}^2).$

Furthermore, we have an isomorphism

$$\begin{aligned} {\mathcal {S}}'\bigl ([0,\infty )\bigr )\overset{\sim }{\longrightarrow }\{S\in {\mathcal {S}}'({\mathbb {R}});\, \text {supp}\,S\subset [0,\infty )\}:S\longmapsto (\phi \mapsto \langle \phi |_{[0,\infty )},S\rangle ). \end{aligned}$$

E.g., the isomorphism $\Phi$ in (3.4) yields

$$\begin{aligned} \Phi (\delta )=\pi \delta \in {\mathcal {S}}'_r({\mathbb {R}}^2),\quad \Phi (\delta ')=-\tfrac{\pi }{4}\Delta \delta \in {\mathcal {S}}'_r({\mathbb {R}}^2). \end{aligned}$$

In this setting, the following holds.

Proposition 3

Let $h_j$ and $L_j,$$j\in {\mathbb {N}}_0,$ denote the Hermite functions and Laguerre polynomials, respectively, see Sect. 1, and let $\Phi$ be as in (3.4).

If $S\in {\mathcal {S}}'([0,\infty ))$ and $\sigma =\Phi (S)\in {\mathcal {S}}'_r({\mathbb {R}}^2),$ then $W_\sigma h_j=\lambda _j h_j$ where

$$\begin{aligned} \lambda _j=(-1)^j\cdot {}_{{\mathcal {S}}([0,\infty ))}\left\langle L_j(2t)\,\mathrm{e}^{-t}, S\right\rangle _{{\mathcal {S}}'([0,\infty ))}, \quad j\in {\mathbb {N}}_0. \end{aligned}$$

(3.5)

In particular, $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}))$ if and only if the sequence $(\lambda _j)_{j\in {\mathbb {N}}_0}$ is bounded, and $W_\sigma \in \mathrm{Com}(L^2({\mathbb {R}}))$ if and only if $\lim _{j\rightarrow 0}\lambda _j=0.$

Proof

For a temperate function S,  we have $\sigma =\Phi (S)=S(x^2+y^2)$ and formula (3.1) from [20, Sect. 24] implies $W_\sigma h_j=\lambda _j h_j$ with

$$\begin{aligned} \begin{aligned} \lambda _j&=2(-1)^j\int _0^\infty S(\rho ^2) L_j(2\rho ^2)\,{\mathrm{e}}^{-\rho ^2}\rho \,\text{ d }\rho \\&=(-1)^j\int _0^\infty S(t) L_j(2t)\,{\mathrm{e}}^{-t}\,\text {d}t=(-1)^j\langle L_j(2t)\mathrm{e}^{-t}, S\rangle ,\quad j\in {\mathbb {N}}_0. \end{aligned} \end{aligned}$$

Therefore, formula (3.5) holds for general $S\in {\mathcal {S}}'([0,\infty ))$ by the density of temperate functions in ${\mathcal {S}}'([0,\infty )).$

For the remaining assertions in Proposition 3, we follow the reasoning in [20, Sect. 24]. We note that $e_j:=\Vert h_j\Vert _2^{-1}h_j,$$j\in {\mathbb {N}}_0,$ yields an orthonormal basis in $L^2({\mathbb {R}})$ such that $\sum _{j=0}^\infty \langle \phi , \overline{e_j}\rangle e_j$ converges not just in $L^2({\mathbb {R}})$ but also in ${\mathcal {S}}({\mathbb {R}})$ to $\phi$ for $\phi \in {\mathcal {S}}({\mathbb {R}}),$ compare [15, Ch. VII, Sect. 7, Ex. 7, pp. 260, 262] or [20, Thm. 24.2, p. 114]. Clearly, if the sequence $(\lambda _j)_{j\in {\mathbb {N}}_0}$ is not bounded, then $W_\sigma$ cannot be extended to a bounded operator on $L^2({\mathbb {R}}).$ On the other hand, if the sequence $(\lambda _j)_{j\in {\mathbb {N}}_0}$ is bounded, then the continuity of $W_\sigma$ on ${\mathcal {S}}({\mathbb {R}})$ implies

$$\begin{aligned} W_\sigma \phi =\sum _{j=0}^\infty \lambda _j \langle \phi , \overline{e_j}\rangle e_j,\quad \phi \in {\mathcal {S}}({\mathbb {R}}), \end{aligned}$$

and thus $\Vert W_\sigma \phi \Vert _2\le \Vert \phi \Vert _2\cdot \sup _{j\in {\mathbb {N}}_0}|\lambda _j|,$$W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}))$ and $\Vert W_\sigma \Vert _{{\mathcal {L}}(L^2({\mathbb {R}}))}=\sup _{j\in {\mathbb {N}}_0}|\lambda _j|.$ Finally, it is well-known that a bounded diagonal operator is compact if and only if the series of its eigenvalues converges to 0. This completes the proof. $\square$

Examples

For the measure $\sigma =\delta (\rho -\sqrt{x^2+y^2})=\delta (\rho ^2-x^2-y^2)\cdot 2\rho ,$$\rho >0,$ we have to set $S=2\rho \,\delta _{\rho ^2}$ and we obtain $\lambda _j=2(-1)^j\rho L_j(2\rho ^2){\mathrm{e}}^{-\rho ^2}.$ Due to the asymptotic behavior of the Laguerre polynomials, see [5, Equ. 8.978.3],

$$\begin{aligned} \lambda _j=\frac{2(-1)^j\sqrt{\rho }}{\sqrt{\pi }\root 4 \of {2j}}\cdot \cos \left( 2\rho \sqrt{2j}-\tfrac{\pi }{4}\right) + O(j^{-3/4}),\quad j\rightarrow \infty , \end{aligned}$$

and hence $W_\sigma \in \text {Com}(L^2({\mathbb {R}}))$ by Proposition 3, see also [20, Thm. 28.1, p. 131].

Instead, if $\sigma =\delta ,$ then $S=\frac{1}{\pi }\,\delta$ and $\lambda _j=\frac{1}{\pi }(-1)^j$ in accordance with $(W_\sigma \phi )(x)=\frac{1}{\pi }\,\phi (-x).$ Hence $W_\delta \in {\mathcal {L}}(L^2({\mathbb {R}})){\setminus }\text {Com}(L^2({\mathbb {R}})).$ Similarly, for $\sigma =\Delta \delta ,$ we have $S=-\frac{4}{\pi }\,\delta '$ and $\lambda _j=(-1)^{j+1}\frac{4}{\pi }(2j+1)$ and $W_{\Delta \delta }\not \in {\mathcal {L}}(L^2({\mathbb {R}})).$

Also note that the orthogonality relation

$$\begin{aligned} \int _0^\infty L_j(x)L_k(x)\,{\mathrm{e}}^{-x}\,{\text {d}}x=\left\{ \begin{aligned}&0:&j\ne k,\\&1:&j=k,\end{aligned}\right. \end{aligned}$$

see [5, Equ. 7.414.3], [20, Prop. 20.1, p. 96], implies, as a consequence of formula (3.5), that, if $(\lambda _j)\in s'({\mathbb {N}}_0),$ then the symbol

$$\begin{aligned} \sigma (x,y)=2{\mathrm{e}}^{-x^2-y^2}\sum _{j=0}^\infty (-1)^j\lambda _j \, L_j(2x^2+2y^2) \end{aligned}$$

(3.6)

(where the sum converges in ${\mathcal {S}}'({\mathbb {R}}^2))$ has the Weyl transform with eigenvalues $\lambda _j.$ In particular, if $(\lambda _j)_{j\in {\mathbb {N}}_0}$ is a bounded sequence, then $W_\sigma (x,y)=\sum _{j=0}^\infty \lambda _j e_j(x)e_j(y)\in {\mathcal {L}}(L^2({\mathbb {R}})).$ As in the proof above, we have set here $e_j=\Vert h_j\Vert _2^{-1} h_j,$$j\in {\mathbb {N}}_0.$

The series expansions in [5, Equs. 8.975, 8.976] and in [1, Equ. 5.11.1] therefore provide us with several examples, e.g.,

$$\begin{aligned} \sigma (x,y)= & {} 2 J_0\bigl (2^{3/2}\sqrt{-z(x^2+y^2)}\bigr )\, {\mathrm{e}}^{-z-x^2-y^2}\in {\mathcal {S}}({\mathbb {R}}^2) \Longrightarrow \lambda _j=\frac{z^j}{j!},\quad z\in {\mathbb {C}}; \\ \sigma (x,y)= & {} -2{\mathrm{e}}^{x^2+y^2} \,\text {Ei}(-2x^2-2y^2)\in L^2({\mathbb {R}}^2)\Longrightarrow \lambda _j=\dfrac{(-1)^j}{j+1}; \\ \sigma (x,y)= & {} K_0(x^2+y^2)\in L^2({\mathbb {R}}^2)\Longrightarrow \lambda _j=\dfrac{(-1)^j}{2j+1}. \end{aligned}$$

Note that the Weyl transforms $W_\sigma$ in the last three examples are even Hilbert–Schmidt operators.

4 Symbols represented by radial Fourier transformation

Since the mapping $h(x,y)=x^2+y^2$ (compare (3.2)) is not submersive at 0,  some care has to be exercised when composing a distribution $S\in \mathcal {D}'({\mathbb {R}})$ with h. In Sect. 3, we took recourse to the space ${\mathcal {S}}'([0,\infty )),$ here we shall consider $S={\mathcal {F}}T$ for $T\in \mathcal {D}'_{L^1,-1}({\mathbb {R}}).$ Then we can define $\sigma =h^*S=({\mathcal {F}}T)(x^2+y^2)\in {\mathcal {S}}'({\mathbb {R}}^2_{xy})$ by the formula

$$\begin{aligned} \langle \phi ,\sigma \rangle ={}_{\mathcal {D}'_{L^1,-1}({\mathbb {R}})\otimes {\mathcal {S}}({\mathbb {R}}^2)}\langle T\otimes \phi , {\mathrm{e}}^{-\text{ i }t (x^2+y^2)}\rangle _{\mathcal {D}_{L^\infty ,1}({\mathbb {R}}){\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}}^2)},\quad \phi \in {\mathcal {S}}({\mathbb {R}}^2),\nonumber \\ \end{aligned}$$

(4.1)

since ${\mathrm{e}}^{-\text{ i }t (x^2+y^2)}\in \mathcal {D}_{L^\infty ,1}({\mathbb {R}}^1_t)\,{\hat{\otimes }}\, {\mathcal {S}}'({\mathbb {R}}^2_{xy})$ by [10, Lemma 2.2, p. 266]. As a short-hand for (4.1), we write $\sigma =({\mathcal {F}}T)(x^2+y^2)= {}_{\mathcal {D}'_{L^1,-1}}\!\langle T(t),\mathrm{e}^{-{\mathrm{i}}t (x^2+y^2)}\rangle _{\mathcal {D}_{L^\infty ,1}{\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}}^2)},$ see the notation in Sect. 1. Of course, in ${\mathbb {R}}^2{\setminus }\{0\},$ the distribution $\sigma$ coincides with the pullback $h^*S$ defined in (3.3).

Proposition 4

Let $T\in \mathcal {D}'_{L^1,-1}({\mathbb {R}})$ and set

$$\begin{aligned} \sigma =({\mathcal {F}}T)(x^2+y^2) ={}_{\mathcal {D}'_{L^1,-1}}\!\langle T(t),\mathrm{e}^{-\mathrm{i}t (x^2+y^2)}\rangle _{\mathcal {D}_{L^\infty ,1}{\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}}^2)}\in {\mathcal {S}}'({\mathbb {R}}^2). \end{aligned}$$

Let $h_j$ denote the Hermite functions as in Sect. 1. Then $W_\sigma h_j=\lambda _j h_j$ where

$$\begin{aligned} \lambda _j={}_{\mathcal {D}_{L^\infty ,1}}\!\big \langle \frac{(1-\mathrm{i} t)^j}{(1+\mathrm{i} t)^{j+1}}, T\rangle _{\mathcal {D}'_{L^1,-1}},\quad j\in {\mathbb {N}}_0. \end{aligned}$$

(4.2)

Furthermore,

$$\begin{aligned} W_\sigma (x,y)={}_{\mathcal {D}'_{L^1,-1}}\!\big \langle T(t), \frac{1}{\sqrt{4\pi \mathrm{i}t}}\,\mathrm{e}^{-\mathrm{i}t (x+y)^2/4+\mathrm{i}(x-y)^2/(4t))} \big \rangle _{\mathcal {D}_{L^\infty ,1}{\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}}^2)}\in {\mathcal {S}}'({\mathbb {R}}^2). \end{aligned}$$

In particular, $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}))$ if and only if the sequence $(\lambda _j)_{j\in {\mathbb {N}}_0}$ is bounded, and $W_\sigma \in \mathrm{Com}(L^2({\mathbb {R}}))$ if and only if $\lim _{j\rightarrow 0}\lambda _j=0.$

Proof

Under somewhat stronger assumptions on $\sigma ,$ formula (4.2) for the eigenvalues $\lambda _j$ was already deduced in [20, Sects. 25, 26] from (3.5) by Fourier transformation. We prefer to give a direct, independent proof of (4.2), which in turn would then also yield (3.5) by the residue theorem.

If $t\ne 0$ and $\sigma _t(x,y)={\mathrm{e}}^{-{\mathrm{i}} t(x^2+y^2)},$ then

$$\begin{aligned} W_{\sigma _t}(x,y)=\frac{1}{\sqrt{4\pi {\mathrm{i}} t}}\,\exp \bigl (-{\mathrm{i}}\alpha (t)(x^2+y^2)-2{\mathrm{i}}\beta (t) xy\bigr ) \end{aligned}$$

(4.3)

for $\alpha (t)=\frac{1}{4}(t-t^{-1}),$$\beta (t)=\frac{1}{4}(t+ t^{-1}).$ Since the Weyl transformation

$$\begin{aligned} W:{\mathcal {S}}'({\mathbb {R}}^2)\longrightarrow {\mathcal {S}}'({\mathbb {R}}^2):\tau \longmapsto W_\tau \end{aligned}$$

is an isomorphism and since ${\mathrm{e}}^{-\text{ i }t (x^2+y^2)}\in \mathcal {D}_{L^\infty ,1}({\mathbb {R}})\,{\hat{\otimes }}\, {\mathcal {S}}'({\mathbb {R}}^2)$ by [10, Lemma 2.2, p. 266], we conclude that the kernel $K(t,x,y)=W({\mathrm{e}}^{-\text{ i }t (x^2+y^2)})$ also belongs to $\mathcal {D}_{L^\infty ,1}({\mathbb {R}})\,{\hat{\otimes }}\, {\mathcal {S}}'({\mathbb {R}}^2).$ For $t\ne 0,$ it is given by the ${\mathcal {C}}^\infty$ function in (4.3). (Of course, K also depends $C^\infty$ on $t\in {\mathbb {R}}$ with values in ${\mathcal {S}}'({\mathbb {R}}^2)$ and $K(0,x,y)=W_1(x,y)=\delta (y-x).)$ This implies that

$$\begin{aligned} W_\sigma ={}_{\mathcal {D}'_{L^1,-1}}\!\langle T(t),K(t,x,y)\rangle _{\mathcal {D}_{L^\infty ,1}{\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}}^2)}\in {\mathcal {S}}'({\mathbb {R}}^2). \end{aligned}$$

When applying $W_\sigma$ to a test function $\phi \in {\mathcal {S}}({\mathbb {R}})$ we therefore conclude by means of Fubini’s Theorem, see [14, Cor., p. 136], that

$$\begin{aligned} W_\sigma \phi ={}_{\mathcal {D}'_{L^1,-1}}\!\langle T(t),W_{\sigma _t}\phi \rangle _{\mathcal {D}_{L^\infty ,1}{\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}})}\in {\mathcal {S}}'({\mathbb {R}}). \end{aligned}$$

We shall show below that

$$\begin{aligned} W_{\sigma _t}h_j=\frac{(1-\mathrm{i} t)^j}{(1+\mathrm{i} t)^{j+1}}\cdot h_j, \quad t\in {\mathbb {R}}, \end{aligned}$$

(4.4)

and this will imply then formula (4.2). The remaining statements in Proposition 4 follow therefrom as in Proposition 3.

For $t=0,$ formula (4.4) is immediate, and for $t\ne 0,$ we infer from (4.3) the following:

$$\begin{aligned} \begin{aligned} (W_{\sigma _t}h_j)(x)&=\frac{1}{\sqrt{4\pi {\mathrm{i}} t}}\int _{{\mathbb {R}}} h_j(y){\mathrm{e}}^{-{\mathrm{i}}\alpha (t)(x^2+y^2)-2{\mathrm{i}}\beta (t) xy}\,{\text {d}}y \\&=\frac{{\mathrm{e}}^{-{\mathrm{i}}\alpha (t)x^2}}{\sqrt{4\pi {\mathrm{i}} t}}\int _{{\mathbb {R}}} H_j(y) \exp \bigl (-(\tfrac{1}{2}+{\mathrm{i}}\alpha (t))y^2-2{\mathrm{i}}\beta (t) xy\bigr )\,{\text {d}}y \\&=\frac{1}{\sqrt{4\pi {\mathrm{i}} t}}\int _{{\mathbb {R}}} H_j(y) \exp \Bigl (-(\tfrac{1}{2}+{\mathrm{i}}\alpha (t))\Bigl [y+\frac{{\mathrm{i}}\beta (t) x}{\frac{1}{2}+{\mathrm{i}}\alpha (t)}\Bigr ]^2\Bigr )\,\text{ d }y \\&\quad \times \exp \Bigl (x^2\Bigl [-{\mathrm{i}}\alpha (t)-\frac{\beta (t)^2}{\frac{1}{2}+{\mathrm{i}}\alpha (t)}\Bigr ]\Bigr ). \end{aligned} \end{aligned}$$

Due to $\alpha (t)^2-\beta (t)^2=-\frac{1}{4},$ we have $-{\mathrm{i}}\alpha (t)-\frac{\beta (t)^2}{\frac{1}{2}+{\mathrm{i}}\alpha (t)}=-\frac{1}{2}.$ Furthermore, [5, Equ. 7.374.10] yields

$$\begin{aligned} \begin{aligned} (W_{\sigma _t}h_j)(x)&=\frac{{\mathrm{e}}^{-x^2/2}}{\sqrt{4\pi \text{ i }t}}\cdot \sqrt{\pi }\cdot \frac{({\mathrm{i}}\alpha (t)-\frac{1}{2})^{j/2}}{({\mathrm{i}}\alpha (t)+\frac{1}{2})^{(j+1)/2}}\cdot H_j\biggl (-\frac{{\mathrm{i}}\beta (t)x}{\frac{1}{2}+{\mathrm{i}}\alpha (t)}\sqrt{\frac{{\mathrm{i}}\alpha (t)+\frac{1}{2}}{{\mathrm{i}}\alpha (t)-\frac{1}{2}}}\biggr ) \\&=\frac{(1-\mathrm{i} t)^j}{(1+\mathrm{i} t)^{j+1}}\cdot h_j(x) \end{aligned} \end{aligned}$$

due to ${\mathrm{i}}\alpha (t)+\tfrac{1}{2}=\frac{(1+\text{ i }t)^2}{4\text{ i }t}$ and ${\mathrm{i}}\alpha (t)-\tfrac{1}{2}=\frac{(1-\text{ i }t)^2}{4\text{ i }t}.$ This concludes the proof.$\square$

Corollary

For $T\in \mathcal {D}'_{L^1,-1}({\mathbb {R}})$ let $\sigma =({\mathcal {F}}T)(x^2+y^2)\in {\mathcal {S}}'({\mathbb {R}}^2).$

• (a) If $T\in \sqrt{1+t^2}\cdot {\mathcal {M}}^1({\mathbb {R}}),$ then $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}));$

• (b) if $T\in \sqrt{1+t^2}\cdot L^1({\mathbb {R}}),$ then $W_\sigma \in \mathrm{Com}(L^2({\mathbb {R}}));$

• (c) if $T\in \bigcup _{p\in [1,\infty )} L^p({\mathbb {R}}),$ then $W_\sigma \in \mathrm{Com}(L^2({\mathbb {R}})).$

Proof

• (a) In fact, if $T=\sqrt{1+t^2}\cdot \mu$ with $\mu \in {\mathcal {M}}^1({\mathbb {R}}),$ then the eigenvalues $\lambda _j$ of $W_\sigma$ are bounded by $\Vert \mu \Vert _{{\mathcal {M}}^1}$ due to formula (4.2):

  $$\begin{aligned} |\lambda _j|=\Big |\big \langle \frac{(1-\text{ i }t)^j}{(1+\text{ i }t)^{j+1}}\cdot \sqrt{1+t^2},\mu \big \rangle \Big |\le \Vert \mu \Vert _{{\mathcal {M}}^1} \end{aligned}$$

  since the function $(1-\text{ i }t)^j\cdot (1+\text{ i }t)^{-j-1}\cdot \sqrt{1+t^2}$ has absolute value 1 for each real t. Hence $\Vert W_\sigma \Vert _{{\mathcal {L}}(L^2({\mathbb {R}}))}\le \Vert \mu \Vert _{{\mathcal {M}}^1}.$

• (b) Since ${\mathcal {S}}({\mathbb {R}})$ is dense in $\sqrt{1+t^2}\cdot L^1({\mathbb {R}})$ and since $W_\sigma$ is a compact operator for $T\in {\mathcal {S}}({\mathbb {R}}),$ we conclude that $W_\sigma$ is compact for each T in $\sqrt{1+t^2}\cdot L^1({\mathbb {R}}).$

• (c) If $p\in [1,\infty )$ and $q\in (1,\infty ]$ such that $\frac{1}{p}+\frac{1}{q}=1,$ then $(1+t^2)^{-1/2}\in L^q({\mathbb {R}})$ and hence $(1+t^2)^{-1/2}\cdot L^p({\mathbb {R}})\subset L^1({\mathbb {R}}).$ This implies

  $$\begin{aligned} L^p({\mathbb {R}})=\sqrt{1+t^2}\cdot (1+t^2)^{-1/2}\cdot L^p({\mathbb {R}})\subset \sqrt{1+t^2}\cdot L^1({\mathbb {R}}) \end{aligned}$$

  and therefore $W_\sigma \in \text {Com}(L^2({\mathbb {R}}))$ for $T\in L^p({\mathbb {R}})$ due to (b). The proof is complete.

$\square$

Remark

The last corollary generalizes Theorem 26.1 in [20, p. 123]. In fact, there the compactness of $W_\sigma$ is proven for $T\in \cup _{p\in [1,2]} L^p({\mathbb {R}})$ and for $T\in \cup _{p\in [1,2]} {\mathcal {F}}(L^p({\mathbb {R}})).$ Note that the first assertion is contained in (c) of the corollary. The second one follows directly from Proposition 1 (c) since $T={\mathcal {F}}f$ with$f\in L^p({\mathbb {R}}),$$1\le p\le 2,$ implies $\sigma (x,y)=2\pi \cdot f(-x^2-y^2)$ and hence $\sigma \in L^p({\mathbb {R}}^2).$

5 Three examples

In the following three examples, the radially symmetric distributions $\sigma _z(x,y)\in {\mathcal {S}}'({\mathbb {R}}^2_{xy})$ will depend meromorphically on the complex parameter z,  see [8] for the theoretical background. Since the mapping W is linear and continuous, we infer that the eigenvalues $\lambda _j(z)$ are also meromorphic functions of z and can be determined by analytic continuation.

Example 1

Let us start with $\sigma _z(x,y)=(x^2+y^2)^z.$ This yields an analytic function

$$\begin{aligned} {\mathbb {C}}{\setminus }(- {\mathbb {N}})\longrightarrow {\mathcal {S}}'({\mathbb {R}}^2_{xy}):z\longmapsto \sigma _z \end{aligned}$$

with simple poles at $z\in -{\mathbb {N}}$ and residues

$$\begin{aligned} \hbox {Res}_{z=-1-j}=\frac{\pi \,\Delta ^j\delta }{2^{2j}\,j!^2},\quad j\in {\mathbb {N}}_0, \end{aligned}$$

see [8, Ex. 2.3.1, p. 41].

For $z\in {\mathbb {C}}{\setminus }{\mathbb {Z}},$ we have

$$\begin{aligned} {\mathcal {F}}\Bigl (\frac{t_+^{-z-1}}{\Gamma (-z)}\Bigr )=({\mathrm{i}} t)^z={\mathrm{e}}^{{\mathrm{i}} z\pi /2} t_+^z+{\mathrm{e}}^{-{\mathrm{i}} z\pi /2} t_-^z, \end{aligned}$$

see [9, Equ. (1.6.2), p. 87]. If $\text {Re}\,z>-1,$ then

$$\begin{aligned} T_z:={\mathrm{e}}^{-\text{ i }z\pi /2}\,\frac{t_+^{-z-1}}{\Gamma (-z)}\in \mathcal {D}'_{L^1,-1}({\mathbb {R}}^1_t) \end{aligned}$$

and hence $\sigma _z(x,y)=({\mathcal {F}} T_z)(x^2+y^2)$ holds in the sense of formula (4.1). Therefore, formula (4.2) in Proposition 4 furnishes, if $z\in {\mathbb {C}}{\setminus } {\mathbb {N}}_0$ with $\text {Re}\,z>-1,$ the representation

$$\begin{aligned} \lambda _j(z)={\mathrm{e}}^{-\text{ i }z\pi /2}\,{}_{\mathcal {D}_{L^\infty ,1}}\!\big \langle \frac{(1-\mathrm{i} t)^j}{(1+\mathrm{i} t)^{j+1}}, \frac{t_+^{-z-1}}{\Gamma (-z)}\rangle _{\mathcal {D}'_{L^1,-1}},\quad j\in {\mathbb {N}}_0, \end{aligned}$$

for the eigenvalues of $W_{\sigma _z}.$

For $-1<\text {Re}\,z<0,$ this is an absolutely convergent integral we can easily evaluate:

$$\begin{aligned} \begin{aligned} \lambda _j(z)&=\frac{{\mathrm{e}}^{-\text{ i }z\pi /2}}{\Gamma (-z)}\cdot \int _0^\infty (1-\text{ i }t)^j(1+\text{ i }t)^{-j-1} t^{-z-1}\,\text{ d }t \\&=\frac{(-1)^j{\mathrm{e}}^{-{\mathrm{i}}(z+1)\pi /2}}{\Gamma (-z)}\cdot \int _0^\infty t^{-z-1}(t-{\mathrm{i}})^{-j-1}(t+{\mathrm{i}})^j\,\text{ d }t \\&={\mathrm{e}}^{-\text{ i }z\pi }\Gamma (1+z)\, {}_2F_1(j+1,-z;1;2)=\Gamma (1+z)\, {}_2F_1(-j,-z;1;2) \end{aligned} \end{aligned}$$

(5.1)

by [5, Equs. 3.197.1 and 9.131.1]. The result in formula (5.1) holds for each $z\in {\mathbb {C}}{\setminus }(- {\mathbb {N}})$ by analytic continuation and it shows, in particular, that $\lambda _j(z)/\Gamma (1+z)$ is a polynomial of degree j in z.

For $-1<\text {Re}\,z<0,$ the function $T_z$ is contained in $\sqrt{1+t^2}\cdot L^1({\mathbb {R}})$ and hence $W_{\sigma _z}\in \text {Com}( L^2({\mathbb {R}}))$ by the Corollary to Proposition 4. E.g., if $z=-\frac{1}{2},$ then [2, Equ. 7.3.8.2] yields $\lambda _j(-\frac{1}{2})=0$ if j is odd and $\lambda _j(-\frac{1}{2})=\sqrt{\pi }\, 2^{-j}j!(\frac{j}{2})!^{-2}$ if j is even. By means of Stirling’s formula, we infer the asymptotic expansion $\lambda _j(-\frac{1}{2})\sim \sqrt{2/j}$ for j even and $j\rightarrow \infty .$ Also note that [5, Equ. 9.131.1] furnishes the equation

$$\begin{aligned} \Gamma (z)\lambda _j(-z)=(-1)^j\Gamma (1-z)\lambda _j(z-1),\quad j\in {\mathbb {N}}_0,\ z\in {\mathbb {C}}{\setminus }{\mathbb {Z}}, \end{aligned}$$

which also implies that $\lambda _j(-\frac{1}{2})$ must vanish if j is an odd number.

For $z=0,$ formula (5.1) yields $\lambda _j(0)=1$ in accordance with $\sigma _0=1$ and $W_1=\delta (x-y);$ similarly, for $z\rightarrow -1,$ we have

$$\begin{aligned} \lim _{z\rightarrow -1}\frac{\sigma _z}{\Gamma (1+z)}=\hbox {Res}_{z=-1}\sigma _z=\pi \delta \Longrightarrow \lim _{z\rightarrow -1}\frac{W_{\sigma _z}}{\Gamma (1+z)}=W_{\pi \delta }=\delta (x+y) \end{aligned}$$

in accordance with

$$\begin{aligned} \hbox {Res}_{z=-1}\lambda _j(z)=\lim _{z\rightarrow -1}\frac{\lambda _j(z)}{\Gamma (1+z)}={}_2F_1(-j,1;1;2)=(-1)^j,\quad j\in {\mathbb {N}}_0. \end{aligned}$$

Let us finally investigate the asymptotic behavior of $\lambda _j(z)$ for general z. Upon inserting the standard integral representation for the hypergeometric function, see [5, Equ. 9.111], we obtain, by analytic continuation,

$$\begin{aligned} \lambda _j(z)={}_{{\mathcal {E}}'({\mathbb {R}}^1_t)}\langle \tfrac{t_+^{-z-1}}{\Gamma (-z)}\cdot (1-t)_+^z,(1-2t)^j\rangle _{{\mathcal {C}}^\infty ({\mathbb {R}}^1_t)}, \quad j\in {\mathbb {N}}_0,\ z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}}).\nonumber \\ \end{aligned}$$

(5.2)

Note that $t_+^{-z-1}/\Gamma (-z)$ is an entire distribution-valued function of z and that $(1-t)_+^z$ is holomorphic for $z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}}).$ We then approximate $(1-2t)^j$ by $(1-t)^{2j}$ near zero and by $(-1)^jt^{2j}$ near one, respectively, and employ Stirling’s formula to conclude from formula (5.2) that

$$\begin{aligned} \lambda _j(z)\sim \left\{ \begin{aligned}&(2j)^z&:\text {Re}\,z>-\tfrac{1}{2},\\&(2j)^z+(-1)^j\tfrac{\Gamma (1+z)}{\Gamma (-z)}\,(2j)^{-z-1}&:\text {Re}\,z=-\tfrac{1}{2},\\&(-1)^j\tfrac{\Gamma (1+z)}{\Gamma (-z)}\,(2j)^{-z-1}&:\text {Re}\,z<-\tfrac{1}{2}, \end{aligned}\right. \end{aligned}$$

for $j\rightarrow \infty$ and fixed $z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}}).$ Here the precise meaning of $a_j\sim b_j$ is $a_j-b_j=o(j^{-1/2+|\text {Re}\,z+1/2|})$ for $j\rightarrow \infty .$ (Let us mention that this asymptotic expansion could also be inferred from [19].) In particular, $W_{\sigma _z}\in \text {Com}(L^2({\mathbb {R}}))$ if and only if $-1<\text {Re}\,z<0$ and $W_{\sigma _z}\in {\mathcal {L}}(L^2({\mathbb {R}}))$ if and only if $-1\le \text {Re}\,z\le 0,$$z\ne -1.$

Example 2

Let us fix $\rho >0.$ Since $(t-\rho ^2)_+^z\in {\mathcal {S}}'({\mathbb {R}})$ depends analytically on $z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}}),$ see [9, Ex. 1.4.8, p. 49], and since the mapping $h(x,y)=x^2+y^2$ is submersive at the circle $h=\rho ^2,$ we obtain an analytic function

$$\begin{aligned} \sigma :{\mathbb {C}}{\setminus }(-{\mathbb {N}})\longrightarrow {\mathcal {S}}'({\mathbb {R}}^2):z\longmapsto \sigma _z(x,y)=h^*((t-\rho ^2)_+^z)=(x^2+y^2-\rho ^2)_+^z. \end{aligned}$$

At $z=-k,$$k\in {\mathbb {N}},$ the function $\sigma$ has simple poles with the residues

$$\begin{aligned} \hbox {Res}_{z=-k} \sigma _z=\frac{(-1)^{k-1}}{(k-1)!}\,\delta ^{(k-1)}(x^2+y^2-\rho ^2), \end{aligned}$$

see [9, Equ. (1.4.1), p. 49].

In order to study the asymptotic behavior of the eigenvalues $\lambda _j(z),$$j\in {\mathbb {N}}_0,$ of $W_{\sigma _z},$ we first assume that $\text {Re}\, z>-1.$ Then $(t-\rho ^2)^z={\mathcal {F}}T$ for

$$\begin{aligned} T=\frac{\Gamma (1+z)}{2\pi }\, (-{\mathrm{i}} t)^{-1-z}\,{\mathrm{e}}^{{\mathrm{i}}\rho ^2t}\in \mathcal {D}'_{L^1,-1}({\mathbb {R}}^1_t), \end{aligned}$$

see [9, Ex. 1.6.7, p. 86], and hence formula (4.2) in Proposition 4 yields

$$\begin{aligned} \lambda _j(z)=\frac{\Gamma (1+z)}{2\pi }\,\left\langle \frac{(1-\text{ i }t)^j}{(1+\text{ i }t)^{j+1}}\, {\mathrm{e}}^{{\mathrm{i}}\rho ^2t},(-\text{ i }t)^{-1-z}\right\rangle ,\quad j\in {\mathbb {N}}_0. \end{aligned}$$

(5.3)

For $-1<\text {Re}\,z<0,$ formula (5.3) represents $\lambda _j(z)$ by an absolutely convergent integral. If we use

$$\begin{aligned} \Bigl (\frac{1-\text{ i }t}{1+\text{ i }t}\Bigr )^j=(-1)^j{\mathrm{e}}^{2\text{ i }j\,\text {arccot}\, t} \end{aligned}$$

and substitute $t=\cot x,$ then we obtain

$$\begin{aligned} \lambda _j(z)&=\frac{(-1)^j\Gamma (1+z)}{2\pi }\int _{-\pi /2}^{\pi /2}{\mathrm{e}}^{2\text{ i }jx+{\mathrm{i}}\rho ^2\cot x} \bigl ((-{\mathrm{i}}\cot x)^{-1-z}+(-{\mathrm{i}}\cot x)^{-z}\bigr )\,{\text {d}}x \nonumber \\&=(-1)^j\Gamma (1+z){\mathcal {F}}_x^{-1}\bigl [Y(\tfrac{\pi }{2}-|x|){\mathrm{e}}^{{\mathrm{i}}\rho ^2\cot x} \bigl ((-{\mathrm{i}}\cot x)^{-1-z}\nonumber \\&\quad +(-{\mathrm{i}}\cot x)^{-z}\bigr )\bigr ](2j) \end{aligned}$$

(5.4)

Formula (5.4) remains valid for ${\mathbb {C}}{\setminus }(-{\mathbb {N}})$ if the distribution S in square brackets is defined by analytic continuation with respect to z.

Therefore the asymptotic behavior of $\lambda _j(z)$ is determined by the behavior of the distribution S at its singular support, i.e., at the three points $\pm \frac{\pi }{2}$ and 0. The points $\pm \frac{\pi }{2}$ yield, similarly as in Example 1, the asymptotic behavior $(2j)^z.$ Let us consider now the origin. For $j\rightarrow \infty ,$ it contributes the same asymptotic behavior as

$$\begin{aligned} \begin{aligned} (-1)^j\Gamma (1+z)&{\mathcal {F}}_x^{-1}\bigl [{\mathrm{e}}^{{\mathrm{i}}\rho ^2/x}(\text{ i }x)^z\bigr ](2j) =\frac{(-1)^j}{2\pi }\,\Gamma (1+z)\int _{-\infty }^\infty {\mathrm{e}}^{{\mathrm{i}}(2jx+\rho ^2/x)}(\text{ i }x)^z\,{\text {d}}x \\&=(-1)^j\Gamma (1+z)\Bigl (\frac{\rho }{\sqrt{2j}}\Bigr )^{1+z} J_{-1-z}(2\rho \sqrt{2j}) \\&\sim \frac{(-1)^j}{\sqrt{\pi }}\,\Gamma (1+z)\rho ^{1/2+z}(2j)^{-z/2-3/4}\cos \bigl (2\rho \sqrt{2j}+\tfrac{\pi z}{2}+\tfrac{\pi }{4}\bigr ) \end{aligned} \end{aligned}$$

according to [5, Equs. 8.421.8 and 8.451.1]. (Though the integral in the first formula line is convergent only for $-2<\text {Re}\,z<0,$ the result remains valid for $z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}})$ by analytic continuation.) Altogether we obtain for $\lambda _j(z)$ the following asymptotic behavior if $j\rightarrow \infty$ and $z\in {\mathbb {C}}{\setminus }(-{\mathbb {N}})$ is fixed:

$$\begin{aligned} \lambda _j(z)\sim \left\{ \begin{aligned}&(2j)^z&:\text {Re}\,z>-\tfrac{1}{2},\\&(2j)^z+(-1)^j\tfrac{\Gamma (1+z)}{\sqrt{\pi }}\,\rho ^{1/2+z}(2j)^{-z/2-3/4} \cos (2\rho \sqrt{2j}+\tfrac{\pi z}{2}+\tfrac{\pi }{4})&:\text {Re}\,z=-\tfrac{1}{2},\\&(-1)^j\tfrac{\Gamma (1+z)}{\sqrt{\pi }}\,\rho ^{1/2+z}(2j)^{-z/2-3/4}\cos (2\rho \sqrt{2j}+\tfrac{\pi z}{2}+\tfrac{\pi }{4})&:\text {Re}\,z<-\tfrac{1}{2}. \end{aligned}\right. \end{aligned}$$

In particular, this implies that $W_{\sigma _z}\in \text {Com}(L^2({\mathbb {R}}))$ if and only if $-\frac{3}{2}<\text {Re}\,z<0$ and $W_{\sigma _z}\in {\mathcal {L}}(L^2({\mathbb {R}}))$ if and only if $-\frac{3}{2}\le \text {Re}\,z\le 0,$$z\ne -1.$ As a control, we can consider the limit

$$\begin{aligned} \lim _{z\rightarrow -1}\frac{\sigma _z}{\Gamma (1+z)}=\hbox {Res}_{z=-1}\sigma _z=\delta _{\rho ^2}(x^2+y^2) \end{aligned}$$

and obtain the correct asymptotic behavior, compare the example in Sect. 3.

Example 3

Let us finally consider the symbols

$$\begin{aligned} \sigma _{p,z}=\chi (x^2+y^2)(x^2+y^2)^z\exp \bigl ({\mathrm{i}}(x^2+y^2)^p\bigr ),\quad z\in {\mathbb {C}}, \ p>1, \end{aligned}$$

if $\chi \in {\mathcal {C}}^\infty ({\mathbb {R}})$ with $\chi (t)=1$ for $t\ge 2,$$\chi (t)=0$ for $t\le 1.$ (Since $(x^2+y^2)^z$ has already been investigated in Example 1, we are interested here only in the contributions stemming from the behavior of $\sigma _{p,z}$ near infinity and we therefore employ the excision function $\chi .)$

Since $\sigma _{p,z}(x,y)=({\mathcal {F}}T)(x^2+y^2)$ with

$$\begin{aligned} T={\mathcal {F}}^{-1}\bigl (\chi (t) t^z{\mathrm{e}}^{{\mathrm{i}} t^p}\bigr )\in {\mathcal {F}}^{-1}({\mathcal {O}}_M)={\mathcal {O}}_C'\subset {\mathcal {D}}'_{L^1,-1}({\mathbb {R}}^1_t), \end{aligned}$$

we can apply formula (4.2) in Proposition 4. The distribution T is again a fast oscillating ${\mathcal {C}}^\infty$ function whose behavior at infinity is readily obtained by the method of stationary phase:

If $t<0$ and $h(s)=s^p+st,$$\frac{1}{p}+\frac{1}{q}=1$ and $h'(s_0)=0,$ then

$$\begin{aligned} \begin{aligned} T(t)&=\frac{1}{2\pi }\int _0^\infty \chi (s) s^z {\mathrm{e}}^{\text{ i }h(s)}\,\text{ d }s \sim \frac{1}{2\pi }\,{\mathrm{e}}^{{\mathrm{i}} h(s_0)}\sqrt{\frac{2\pi {\mathrm{i}}}{h''(s_0)}} \\&=\sqrt{\frac{{\mathrm{i}}}{2\pi p(p-1)}}\, \Bigl (\frac{|t|}{p}\Bigr )^{(2-p+2z)/(2(p-1))} \exp \Bigl (-{\mathrm{i}}(p-1)\Bigl (\frac{|t|}{p}\Bigr )^q\Bigr ) \end{aligned} \end{aligned}$$

for $t\rightarrow -\infty .$ For $t\rightarrow \infty ,$ the function T(t) is fast decreasing.

With the same substitution as in formula (5.4), we therefore obtain for the eigenvalues $\lambda _j(p,z)$ of $W_{\sigma _{p,z}}$ the following:

$$\begin{aligned} \begin{aligned} \lambda _j(p,z)&=(-1)^j\int _{-\pi /2}^{\pi /2} {\mathrm{e}}^{2{\mathrm{i}} jx}\, T(\cot x)\cdot (1-{\mathrm{i}}\cot x)\,{\text {d}}x \\&\sim \frac{(-1)^j {\mathrm{i}}^{3/2} p^{-(1+2z)/(2(p-1))}}{\sqrt{2\pi (p-1)}}\int _0^\infty \bigl (1-\chi (x)\bigr ) {\mathrm{e}}^{-2{\mathrm{i}} jx} x^{-q/2-z/(p-1)}{\mathrm{e}}^{-{\mathrm{i}}(p-1)(px)^{-q}}\,{\text {d}}x \end{aligned} \end{aligned}$$

(For the convergence of the last integral, we have to stipulate that ${\text {Re}}\,z<\frac{3}{2}p-1.$ The final result in formula (5.5) below can, as before, be extended to $z\in {\mathbb {C}}$ by analytic continuation. This is done by representing the integral as the Fourier transform of a distribution S and by employing the equation $({\mathcal {F}} S')(2j)=2{\mathrm{i}} j\,{\mathcal {F}}S.)$

Upon using again the method of stationary phase with $h(x)=2jx+(p-1)(px)^{-q}$ and $px_0=(2j/p^2)^{(1-p)/(2p-1)},$ this finally implies

$$\begin{aligned} \begin{aligned} \lambda _j(p,z)&\sim \frac{(-1)^j {\mathrm{i}}^{3/2}\,p^{-(1+2z)/(2(p-1))}}{\sqrt{2\pi (p-1)}}\, x_0^{-q/2-z/(p-1)}{\mathrm{e}}^{-\text{ i }h(x_0)} \sqrt{\frac{2\pi }{{\mathrm{i}} h''(x_0)}} \\&=\frac{(-1)^j {\mathrm{i}}}{\sqrt{2p-1 }}\, p^{-(1+2z)/(2p-1)} (2j)^{(z+1-p)/(2p-1)} \exp \Bigl (-{\mathrm{i}}(2p-1)\Bigl (\frac{2j}{p^2}\Bigr )^{p/(2p-1)}\Bigr ). \end{aligned}\nonumber \\ \end{aligned}$$

(5.5)

From formula (5.5), we can conclude that $W_{\sigma _{p,z}}$ is bounded if and only if $\text {Re}\,z\le p-1$ and that $W_{\sigma _{p,z}}$ is compact if and only if $\text {Re}\, z<p-1.$

6 Generalization to n dimensions

Let us consider here the variables in the ordering $x_1,y_1,x_2,y_2,\dots ,x_n,y_n.$ We denote by ${\mathcal {S}}'_{pr}({\mathbb {R}}^{2n})$ the set of all partially radially symmetric $\sigma \in {\mathcal {S}}'({\mathbb {R}}^{2n}),$ i.e., those temperate distributions $\sigma$ that are rotationally symmetric with respect to all $(x_j,y_j)-\!$planes, i.e., such that

$$\begin{aligned} \sigma \Bigl (A_1\left( {\begin{array}{c}x_1\\ y_1\end{array}}\right) ,\dots ,A_n\left( {\begin{array}{c}x_n\\ y_n\end{array}}\right) \Bigr )=\sigma \end{aligned}$$

for all two-dimensional rotations $A_1,\dots ,A_n.$ As in Sect. 3, we obtain an isomorphism

$$\begin{aligned} \Phi _{pr}:{\mathcal {S}}'([0,\infty )^n)\longrightarrow & {} {\mathcal {S}}'_{pr}({\mathbb {R}}^{2n})\nonumber \\ S\longmapsto & {} \bigl (\phi \mapsto 2^{-n}\big \langle \int _{({\mathbb {S}}^1)^n}\phi (\sqrt{ t_1}\omega _1,\dots ,\sqrt{t_n}\omega _n)\,\text{ d }\omega , S_t\big \rangle \bigr )\qquad \end{aligned}$$

(6.1)

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

Alternatively, we can consider $\sigma =({\mathcal {F}}T)(x_1^2+y_1^2,\dots ,x_n^2+y_n^2)$ for

$$\begin{aligned} T\in \mathcal {D}'_{L^1,-1,\dots ,-1}= \mathcal {D}'_{L^1}({\mathbb {R}}^n)\cdot \prod _{j=1}^n \sqrt{1+t_j^2}. \end{aligned}$$

As in formula (4.1), $\sigma$ is defined by

$$\begin{aligned} \langle \phi ,\sigma \rangle ={}_{\mathcal {D}'_{L^1,-1,\dots ,-1}\otimes {\mathcal {S}}({\mathbb {R}}^{2n})}\langle T\otimes \phi , {\mathrm{e}}^{-\text{ i }t_1(x_1^2+y_1^2)-\dots -\text{ i }t_n(x_n^2+y_n^2)}\rangle _{\mathcal {D}_{L^\infty ,1,\dots ,1} {\hat{\otimes }} {\mathcal {S}}'({\mathbb {R}}^{2n})},\nonumber \\ \end{aligned}$$

(6.2)

for $\phi \in {\mathcal {S}}({\mathbb {R}}^{2n}).$ Then Propositions 3 and  4 generalize to the following.

Proposition 5

Let $h_\alpha ,$$\alpha \in {\mathbb {N}}_0^n,$ and $L_j,$$j\in {\mathbb {N}}_0,$ denote the Hermite functions and Laguerre polynomials, respectively, see Sect. 1, and let $\Phi _{pr}$ be as in (6.1).

If $S\in {\mathcal {S}}'([0,\infty )^n)$ and $\sigma =\Phi _{pr}(S)\in {\mathcal {S}}'_{pr}({\mathbb {R}}^{2n}),$ then $W_\sigma h_\alpha =\lambda _\alpha h_\alpha$ where

$$\begin{aligned} \lambda _\alpha =(-1)^{|\alpha |}\cdot {}_{{\mathcal {S}}([0,\infty )^n)}\big \langle \prod _{j=1}^n L_{\alpha _j}(2t_j)\,\mathrm{e}^{-t_j}, S\big \rangle _{{\mathcal {S}}'([0,\infty )^n)}, \quad \alpha \in {\mathbb {N}}_0^n. \end{aligned}$$

(6.3)

If instead $\sigma =({\mathcal {F}}T)(x_1^2+y_1^2,\dots ,x_n^2+y_n^2)\in {\mathcal {S}}'_{pr}({\mathbb {R}}^{2n})$ for $T\in \mathcal {D}'_{L^1,-1,\dots ,-1}({\mathbb {R}}^n)$ according to (6.2), then $W_\sigma h_\alpha =\lambda _\alpha h_\alpha$ where

$$\begin{aligned} \lambda _{\alpha }={}_{\mathcal {D}_{L^\infty ,1,\dots ,1}}\!\big \langle \prod _{j=1}^n\frac{(1-\mathrm{i} t_j)^{\alpha _j}}{(1+\mathrm{i} t_j)^{\alpha _j+1}}, T\big \rangle _{\mathcal {D}'_{L^1,-1,\dots ,-1}},\quad \alpha \in {\mathbb {N}}_0^n. \end{aligned}$$

(6.4)

In particular, $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n))$ if and only if the sequence $(\lambda _\alpha )_{\alpha \in {\mathbb {N}}_0^n}$ is bounded, and $W_\sigma \in \mathrm{Com}(L^2({\mathbb {R}}^n))$ if and only if $\lim _{|\alpha |\rightarrow \infty }\lambda _\alpha =0.$

Proof

If $\sigma _t(x,y)=\prod _{j=1}^n {\mathrm{e}}^{-{\mathrm{i}} t_j(x_j^2+y_j^2)}$ for $t\in {\mathbb {R}}^n,$ then Proposition 4 yields

$$\begin{aligned} W_{\sigma _t}h_\alpha =\prod _{j=1}^n(W_{\sigma _{t_j}} h_{\alpha _j})(x_j)=\prod _{j=1}^n \frac{(1-\mathrm{i} t_j)^{\alpha _j}}{(1+\mathrm{i} t_j)^{\alpha _j+1}}\, h_{\alpha _j}(x_j) \end{aligned}$$

and this implies formula (6.4) as in the proof of Proposition 4.

Formula (6.3) can be deduced from formula (6.4); in fact, if $\sigma =\otimes _{j=1}^n \delta (\rho _j^2-x_j^2-y_j^2)$ for $\rho _1>0,\dots ,\rho _n>0,$ then $\sigma =({\mathcal {F}}T)(x_1^2+y_1^2,\dots ,x_n^2+y_n^2)$ for $T=(2\pi )^{-n} \prod _{j=1}^n {\mathrm{e}}^{{\mathrm{i}} t_j\rho _j^2}.$ Though $T\not \in \mathcal {D}'_{L^1,-1,\dots ,-1}({\mathbb {R}}^n),$ approximation yields that formula (6.4) can be applied in the form

$$\begin{aligned} \lambda _{\alpha }=\prod _{j=1}^n\frac{1}{2\pi }\int _{-\infty }^\infty {\mathrm{e}}^{{\mathrm{i}} t_j\rho _j^2} \frac{(1-\mathrm{i} t_j)^{\alpha _j}}{(1+\mathrm{i} t_j)^{\alpha _j+1}}\,\text{ d }t_j \end{aligned}$$

where the integrals are conditionally convergent. Eventually, formula (6.3) follows upon employing the residue theorem, see the details in the proof of the corollary below.

Alternatively, formula (6.3) follows directly from Proposition 3 for tensor products $S\in {\mathcal {S}}'([0,\infty ))\otimes \cdots \otimes {\mathcal {S}}'([0,\infty ))$ and then must hold generally by density. This completes the proof. $\square$

Let us finally specialize to symbols $\sigma$ in the space

$$\begin{aligned} {\mathcal {S}}'_r({\mathbb {R}}^{2n})=\{\sigma \in {\mathcal {S}}'({\mathbb {R}}^{2n});\, \sigma \text { is radially symmetric with respect to all variables}\}. \end{aligned}$$

According to [12] or [18, Sect. 3.10, p. 249], the isomorphism

$$\begin{aligned} \mathcal {D}'((0,\infty ))\overset{\sim }{\rightarrow }\mathcal {D}'_r({\mathbb {R}}^{2n}{\setminus }\{0\}):S\longmapsto (t^{1-n}S)\circ (|x|^2+|y|^2) \end{aligned}$$

can be extended to ${\mathbb {R}}^{2n}$ in the following way:

$$\begin{aligned} \Phi _r:{\mathcal {S}}'([0,\infty ))\overset{\sim }{\rightarrow }{\mathcal {S}}'_r({\mathbb {R}}^{2n}):S\longmapsto \bigl (\phi \mapsto \tfrac{1}{2}\left\langle \int _{{\mathbb {S}}^{2n-1}}\phi (\sqrt{t}\,\omega )\,\text{ d }\omega , S_t\right\rangle \bigr )\nonumber \\ \end{aligned}$$

(6.5)

Furthermore, if $T\in \mathcal {D}'_{L^1,-n}({\mathbb {R}}),$ then $\sigma =({\mathcal {F}}T)(|x|^2+|y|^2)\in {\mathcal {S}}'_r({\mathbb {R}}^{2n})$ is well-defined, see [10, Lemma 2.2, p. 266].

Corollary

Let $h_\alpha ,$$\alpha \in {\mathbb {N}}_0^n,$ and $L_j^{n-1},$$j\in {\mathbb {N}}_0,$ denote the Hermite functions and Laguerre polynomials, respectively, see Sect. 1, and let $\Phi _r$ be as in (6.5).

If $S\in {\mathcal {S}}'([0,\infty ))$ and $\sigma =\Phi _r(S)\in {\mathcal {S}}'_{r}({\mathbb {R}}^{2n}),$ then $W_\sigma h_\alpha =\lambda _{|\alpha |} h_\alpha$ where

$$\begin{aligned} \lambda _j=\frac{(-1)^j\,j!}{(j+n-1)!}\cdot {}_{{\mathcal {S}}([0,\infty ))}\langle L_j^{n-1}(2t)\,\mathrm{e}^{-t}, S\rangle _{{\mathcal {S}}'([0,\infty ))}, \quad j\in {\mathbb {N}}_0. \end{aligned}$$

(6.6)

If instead $\sigma =({\mathcal {F}}T)(|x|^2+|y|^2)\in {\mathcal {S}}'_{r}({\mathbb {R}}^{2n})$ for $T\in \mathcal {D}'_{L^1,-n}({\mathbb {R}}),$ then $W_\sigma h_\alpha =\lambda _{|\alpha |} h_\alpha$ where

$$\begin{aligned} \lambda _j={}_{\mathcal {D}_{L^\infty ,n}}\!\big \langle \frac{(1-\mathrm{i} t)^j}{(1+\mathrm{i} t)^{j+n}}, T\rangle _{\mathcal {D}'_{L^1,-n}},\quad j\in {\mathbb {N}}_0. \end{aligned}$$

(6.7)

Proof

We first prove formula (6.7). If $T\in \mathcal {D}'_{L^1,-n}({\mathbb {R}}),$ then

$$\begin{aligned} S=\delta (t_1-t_2)\cdots \delta (t_{n-1}-t_n)T(t_n)\in \mathcal {D}'_{L^1,-1,\dots ,-1}({\mathbb {R}}^n) \end{aligned}$$

is defined as the composition of $\delta _{{\mathbb {R}}^{n-1}}\otimes T$ with a non-degenerate linear mapping and

$$\begin{aligned} ({\mathcal {F}}S)(x_1^2+y_1^2,\dots ,x_n^2+y_n^2)=({\mathcal {F}} T)(|x|^2+|y|^2). \end{aligned}$$

Therefore formula (6.4) implies formula (6.7):

$$\begin{aligned} \lambda _{\alpha }=\big \langle \prod _{j=1}^n\frac{(1-\mathrm{i} t_j)^{\alpha _j}}{(1+\mathrm{i} t_j)^{\alpha _j+1}},S\big \rangle = \big \langle \frac{(1-\mathrm{i} t)^{|\alpha |}}{(1+\mathrm{i} t)^{|\alpha |+n}},T\big \rangle . \end{aligned}$$

In the same way, formula (6.6) could be inferred from formula (6.3) by induction employing [5, Equ. 7.413]. It appears to be more illuminating to deduce (6.6) from (6.7) by the residue theorem. We first observe that linear combinations of $\delta _{\rho ^2},$$\rho >0,$ are dense in ${\mathcal {S}}'([0,\infty )).$ Hence it suffices to prove (6.6) for $S=\delta _{\rho ^2},$$\rho >0.$ In this case,

$$\begin{aligned} \sigma =\Phi _r(\delta _{\rho ^2})=\rho ^{2(1-n)}\delta _{\rho ^2}(|x|^2+|y|^2)=({\mathcal {F}}T)(|x|^2+|y|^2) \end{aligned}$$

where $T=(2\pi )^{-1}\rho ^{2(1-n)}\,{\mathrm{e}}^{{\mathrm{i}}\rho ^2 t}\in \mathcal {D}'_{L^1,-n}({\mathbb {R}}).$ Hence, if $\alpha \in {\mathbb {N}}_0^n$ and $|\alpha |=j,$ then formula (6.7) furnishes the following:

$$\begin{aligned} \begin{aligned} \lambda _{\alpha }&=\frac{\rho ^{2(1-n)}}{2\pi }\int _{-\infty }^\infty \frac{(1-\text{ i }t)^j}{(1+{\mathrm{i}} t)^{j+n}}\,{\mathrm{e}}^{{\mathrm{i}}\rho ^2 t}\,\text{ d }t ={\mathrm{i}} \rho ^{2(1-n)}\hbox {Res}_{t={\mathrm{i}}}\Bigl [ {\mathrm{e}}^{{\mathrm{i}}\rho ^2 t} \frac{(1-\text{ i }t)^j}{(1+{\mathrm{i}} t)^{j+n}}\Bigr ]\\&={\mathrm{i}} \rho ^{2(1-n)}{\mathrm{e}}^{-\rho ^2}\hbox {Res}_{s=0}\Bigl [ {\mathrm{e}}^{{\mathrm{i}}\rho ^2 s} \frac{(-{\mathrm{i}})^j(s+2{\mathrm{i}})^j}{({\mathrm{i}} s)^{j+n}}\Bigr ]\\&=(-1)^j{\mathrm{e}}^{-\rho ^2}\hbox {Res}_{s=0}\Bigl [ \sum _{k=0}^j\sum _{l=0}^\infty \left( {\begin{array}{c}j\\ k\end{array}}\right) (2{\mathrm{i}})^k \frac{({\mathrm{i}}\rho ^2)^{l+1-n}}{l!}\, s^{l-k-n}\Bigr ]\\&=(-1)^j{\mathrm{e}}^{-\rho ^2} \sum _{k=0}^j\frac{(-2\rho ^2)^k}{(k+n-1)!}\,\left( {\begin{array}{c}j\\ k\end{array}}\right) \\&=\frac{(-1)^j\, j!}{(j+n-1)!}\,{\mathrm{e}}^{-\rho ^2}\, L_j^{n-1}(2\rho ^2) =\frac{(-1)^j\, j!}{(j+n-1)!}\,\left\langle L_j^{n-1}(2t){\mathrm{e}}^{-t},S \right\rangle , \end{aligned} \end{aligned}$$

compare [20, Equ. (20.2), p. 95] or [5, Equ. 8.970.1] for the last line. This completes the proof.$\square$

Remark

For $\sigma =({\mathcal {F}}T)(|x|^2+|y|^2)\in {\mathcal {S}}'({\mathbb {R}}^{2n}),$ we can conclude from the corollary to Proposition 5, in the same way as in the one-dimensional case, that $W_\sigma \in {\mathcal {L}}(L^2({\mathbb {R}}^n))$ if $T\in (1+t^2)^{n/2}\cdot {\mathcal {M}}^1({\mathbb {R}})$ and that $W_\sigma \in \text {Com}(L^2({\mathbb {R}}^n))$ if $T\in (1+t^2)^{n/2}\cdot L^1({\mathbb {R}}).$

Examples

If $n\ge 2$ and

$$\begin{aligned} \sigma (x,y)=\delta (\rho -\sqrt{|x|^2+|y|^2})=2\rho \,\delta (\rho ^2-|x|^2-|y|^2),\quad \rho >0, \end{aligned}$$

then $\sigma =({\mathcal {F}}T)(|x|^2+|y|^2)$ with $T(t)=\tfrac{\rho }{\pi }\,{\mathrm{e}}^{{\mathrm{i}}\rho ^2t}\in (1+t^2)^{n/2} L^1({\mathbb {R}})$ and hence $W_\sigma \in \text {Com}(L^2({\mathbb {R}}^n))$ by the above remark. More precisely, we have $\sigma =\Phi _r(2\rho ^{2n-1}\delta _{\rho ^2})$ and therefore $W_\sigma h_{\alpha }=\lambda _{|\alpha |}h_\alpha$ where

$$\begin{aligned} \lambda _j=\frac{2(-1)^j \,j!\rho ^{2n-1}}{(j+n-1)!}\,L_j^{n-1}(2\rho ^2){\mathrm{e}}^{-\rho ^2}\sim \frac{2(-1)^j \rho ^{n-1/2}}{\sqrt{\pi }\,(2j)^{(2n-1)/4}}\,\cos \bigl (2\rho \sqrt{2j}-\tfrac{(2n-3)\pi }{4}\bigr ) \end{aligned}$$

for $j\rightarrow \infty ,$ see [5, Equ. 8.978.3]. The kernel of the pseudo-differential operator $W_\sigma$ is given by

$$\begin{aligned} W_\sigma (x,y)= & {} \frac{\rho Y(2\rho -|x+y|)}{(2\pi )^{n/2}}\\&\cdot \frac{(\rho ^2-|x+y|^2/4)^{n/4-1/2}}{|x-y|^{n/2-1}}\, J_{n/2-1}\biggl (|x-y|\sqrt{\rho ^2-\frac{|x+y|^2}{4}}\biggr ). \end{aligned}$$

Let us finally mention that the calculation of the “oscillator semigroup”, cf. [3, Ch. 5, p. 223], becomes deceptively easy upon employing formula (6.7). In fact, for

$$\begin{aligned} \sigma _\zeta (x,y)=\exp \bigl (-\zeta (|x|^2+|y|^2)\bigr ),\quad (x,y)\in {\mathbb {R}}^{2n},\ \zeta \in {\mathbb {C}},\ \text {Re}\,\zeta \ge 0, \end{aligned}$$

we have $\sigma _\zeta =({\mathcal {F}}T_\zeta )(|x|^2+|y|^2)$ with

$$\begin{aligned} T_\zeta =\left\{ \begin{aligned}&\tfrac{\zeta }{\pi (\zeta ^2+t^2)}&:\zeta \ne 0,\\&\delta&:\zeta =0\end{aligned}\right\} \in {\mathcal {D}}'_{L^1,-n}({\mathbb {R}}) \end{aligned}$$

and hence $W_{\sigma _\zeta } h_\alpha =\lambda _{|\alpha |}(\zeta )h_\alpha ,$$\alpha \in {\mathbb {N}}_0^n,$ where (for $\text {Re}\,\zeta \ge 0,\ \zeta \ne 0,\ j\in {\mathbb {N}}_0)$

$$\begin{aligned} \begin{aligned} \lambda _j(\zeta )&=\int _{-\infty }^\infty \frac{(1-{\mathrm{i}} t)^j}{(1+{\mathrm{i}} t)^{j+n}}\cdot \frac{\zeta }{\pi (\zeta ^2+t^2)}\,\text{ d }t \\&=-2{\mathrm{i}}\zeta \hbox {Res}_{t=-{\mathrm{i}}\zeta }\biggl ( \frac{(1-{\mathrm{i}} t)^j}{(1+{\mathrm{i}} t)^{j+n}(\zeta ^2+t^2)}\biggr )= \frac{(1-\zeta )^j}{(1+\zeta )^{j+n}} \end{aligned} \end{aligned}$$

in accordance with [3, Equ. (5.30), p. 238].