A Geometric Characterization of a Class of Poisson Type Distributions

Abstract

Tempered distributions on $\mathbb {R}^{n}$ which have Dirac comb structure are characterized in terms of sparseness properties of its support and spectrum.

Appendix

Proof

of Proposition 4. For any $\psi \in \mathrm {S}\left( E^{*}\right) ,$

$$\begin{aligned} F\left( u*_{e}v^{*}\right) \left( \psi \right)&=\left[ u*_{e}v^{*}\right] \left( \hat{\psi }\right) =\left[ u\times v^{*}\right] \left( e\left( x-y\right) \hat{\psi }\left( \frac{x+y}{2}\right) \right) \\&=F\left( u\times v^{*}\right) \left( F^{-1}\left( e\left( x-y\right) \hat{\psi }\left( \frac{x+y}{2}\right) \right) \right) ,\\&=F\left( u\times v^{*}\right) \left( \tilde{e}\left( \frac{\rho }{2}\right) \psi \left( \omega \right) \right) , \end{aligned}$$

since the coordinates $\rho /2,\omega$ are dual to $x-y,\left( x+y\right) /2$ where $\rho =\xi -\eta ,$ $\omega =\xi +\eta \$ We have $F\left( u\times v^{*}\right) =F\left( u\right) \times F\left( v^{*}\right) =\hat{u} \times \overline{\hat{v}}$ and finally

$$\begin{aligned} F\left( u*_{e}v^{*}\right) \left( \psi \right) =\left[ \hat{u} \times \overline{\hat{v}}\right] \left( \tilde{e}\left( \frac{\rho }{2}\right) \psi \left( \omega \right) \right) =\left[ \hat{u}*_{\tilde{e}}\overline{\hat{v}}\right] \left( \psi \right) \end{aligned}$$

which proves (7).

Suppose now that $\hat{u}=\sum \hat{u}_{\sigma }\left( D\right) \delta _{\sigma },\ \hat{v}=\sum \hat{v}_{\sigma }\left( D\right) \delta _{\sigma }$ where the sums are taken over $\Sigma$ and calculate

$$\begin{aligned} F\left( u*_{e}v^{*}\right) \left( \psi \right)&=\sum _{\sigma ,\tau \in \Sigma }\left[ \hat{u}_{\sigma }\left( D_{\xi }\right) \delta _{\sigma }\times \overline{\hat{v}}_{\tau }\left( -D_{\eta }\right) \delta _{\tau }\right] \left( \tilde{e}\left( \frac{\rho }{2}\right) \psi \left( \omega \right) \right) \nonumber \\&=\sum _{\sigma ,\tau }\left[ \delta _{\sigma }\times \delta _{\tau }\right] \hat{u}_{\sigma }\left( -D_{\xi }\right) \overline{\hat{v}}_{\tau }\left( D_{\eta }\right) \left( \tilde{e}\left( \frac{\rho }{2}\right) \psi \left( \omega \right) \right) , \end{aligned}$$

(15)

where

$$\begin{aligned} \ \delta _{\sigma }\left( \xi \right) \times \delta _{\tau }\left( \eta \right) =\delta _{\sigma -\tau }\left( \frac{\rho }{2}\right) \delta _{\sigma +\tau }\left( \omega \right) ,\ D_{\xi }=D_{\rho }+D_{\omega },\ D_{\eta }=D_{\omega }-D_{\rho }. \end{aligned}$$

The right hand side of (15) is equal to

$$\begin{aligned} \sum _{\sigma ,\tau }\left[ \delta _{\sigma -\tau }\left( \frac{\rho }{2}\right) \times \delta _{\sigma +\tau }\left( \omega \right) \right] \hat{u}_{\sigma }\left( -D_{\omega }-D_{\rho }\right) \overline{\hat{v}}_{\tau }\left( D_{\omega }-D_{\rho }\right) \left( \tilde{e}\left( \frac{\rho }{2}\right) \psi \left( \omega \right) \right) . \end{aligned}$$

The set $\Sigma -\Sigma$ is discrete and for any $\rho \in \Sigma -\Sigma ,$ the sum over the set $\left\{ \left( \sigma ,\tau \right) ;\ \sigma -\tau =\rho \right\}$ converges to a distribution supported at $\rho$, since by (4) $\left\| \hat{u}_{\sigma }\overline{\hat{v}}_{\tau }\right\| =O\left( \left| \sigma \right| ^{q}\right)$ as $\left| \sigma \right| \rightarrow \infty$ for some q whereas all the derivatives of $\psi$ decrease fast as $\left| \omega \right| =\left| \sigma +\tau \right| \rightarrow \infty .\$This implies the second statement. $\square$

Proof

of Theorem 3. We have

$$\begin{aligned} \left[ u*_{e_{i}}v\right] \left( \varphi \right)&=\left[ \sum _{p\in \Lambda }u_{p}\left( D_{x}\right) \delta _{p}\left( x\right) \times \sum _{q\in \Lambda }v_{q}\left( -D_{y}\right) \delta _{-q}\left( y\right) \right] e_{i}\left( x-y\right) \varphi \left( \frac{x+y}{2}\right) \\&=\sum _{p\in \Lambda }\sum _{q\in \Lambda }u_{p}\left( D_{x}\right) v_{q}\left( -D_{y}\right) \delta _{p-q}\left( e_{i}\left( x-y\right) \varphi \left( \frac{x+y}{2}\right) \right) \left. {}\right| _{x=p,y=q}, \end{aligned}$$

where the terms with $p-q\ne 0$ vanish since the support of e is too small. The right hand side equals

$$\begin{aligned} \sum _{p\in \Lambda }u_{p}\left( -D_{x}\right) v_{p}\left( D_{y}\right) \left( e_{i}\left( x-y\right) \varphi \left( \frac{x+y}{2}\right) \right) \left. {}\right| _{x=y=p}. \end{aligned}$$

By changing the variables $s=x+y,\ r=x-y$ we write

$$\begin{aligned} w_{p}\left( D_{s},D_{r}\right)&\doteqdot u_{p}\left( -D_{x}\right) v_{p}\left( D_{y}\right) =u_{p}\left( -D_{s}-D_{r}\right) v_{p}\left( D_{s}-D_{r}\right) ,\\ z_{p}\left( D_{s},D_{r}\right)&\doteqdot u_{p}\left( D_{x}\right) v_{p}\left( -D_{y}\right) =u_{p}\left( D_{s}+D_{r}\right) v_{p}\left( -D_{s}+D_{r}\right) . \end{aligned}$$

By the Leibniz formula

$$\begin{aligned} u_{p}\left( -D_{x}\right) v_{p}\left( D_{y}\right) \left( e_{i}\left( r\right) \varphi \left( \frac{s}{2}\right) \right) _{r=0}=w_{p}\left( D_{s} ,D_{r}\right) \left( e_{i}\left( r\right) \varphi \left( \frac{s}{2}\right) \right) _{r=0}\\ =\sum _{j}\frac{1}{j!}D^{j}e_{i}\left( 0\right) w_{p}^{\left( j\right) }\left( D_{s},0\right) \varphi \left( \frac{s}{2}\right) =\frac{1}{i!}w_{p}^{\left( i\right) }\left( D_{s},0\right) \varphi \left( \frac{s}{2}\right) , \end{aligned}$$

where

$$\begin{aligned} w^{\left( i\right) }\left( \sigma ,\rho \right) \doteqdot \frac{\partial ^{i}w\left( \sigma ,\rho \right) }{\partial ^{i}\rho } \end{aligned}$$

for $i\in \mathbb {Z}_{+}^{n}.$ Note that $w^{\left( i\right) }\left( \sigma ,0\right) =\left( -1\right) ^{\left| i\right| }z^{\left( i\right) }\left( 0,-\sigma \right)$ which implies

$$\begin{aligned} \mathrm {A}_{t}\left( u,v\right)&\doteqdot \sum _{i\in \mathbb {Z}_{+}^{n} }\left( D-t\right) ^{i}\left( u*_{e_{i}}v\right) \\&=\sum _{p\in \Lambda }\sum _{i}u_{p}\left( -D_{x}\right) v_{p}\left( D_{y}\right) \left( \frac{\left( x-y\right) ^{i}}{i!}\left( -D-t\right) ^{i}\varphi \left( \frac{x+y}{2}\right) \right) \left. {}\right| _{x=y=p}\\&=\sum _{p}\sum _{i}w_{p}^{\left( i\right) }\left( D_{s},0\right) \left( -D_{s}-t\right) ^{i}\varphi \left( \frac{s}{2}\right) \left. {}\right| _{s=p}\\&=\sum _{p}\sum _{i}z_{p}^{\left( i\right) }\left( 0,-D_{s}\right) \left( D_{s}+t\right) ^{i}\varphi \left( \frac{s}{2}\right) \left. {}\right| _{s=p} =\sum _{p}z_{p}\left( 0,t\right) \varphi \left( p\right) , \end{aligned}$$

since by the Taylor formula

$$\begin{aligned} \sum _{i}z_{p}^{\left( i\right) }\left( 0,-D_{s}\right) \left( D_{s}+t\right) ^{i}=z_{p}\left( 0,t\right) =u_{p}\left( t\right) v_{p}\left( t\right) . \end{aligned}$$

This yields (6). $\square$