Non-uniform Random Sampling and Reconstruction in Signal Spaces with Finite Rate of Innovation

Abstract

We consider non-uniform random sampling in a signal space with finite rate of innovation $V^{2}(\varLambda,\varPhi) \subset{\mathrm {L}}^{2}(\mathbb {R}^{d})$ generated by a series of functions $\varPhi=(\phi_{\lambda})_{\lambda \in\varLambda}$. A subset $V_{R,\delta}^{2}(\varLambda,\varPhi)$ of $V^{2}(\varLambda,\varPhi)$ is consisting of functions concentrates at least $1-\delta$ of the whole energy in a cube with side lengths $R$. Under mild assumptions on the generators and the probability distribution, we show that for $R$ sufficiently large, taking $O(R^{d} \log(R^{d}))$ many samples with such the non-uniform distribution yields a sampling set for $V_{R,\delta}^{2}(\varLambda,\varPhi)$ with high probability. We impose compact support on the generators as an additional constraint for obtaining a reconstruction algorithm from non-uniform random sampling with high probability.

Appendix. Some Properties of the Signal Spaces \(V^{2}(\varLambda,\varPhi)\)

We assume that $\varPhi:=\{\phi_{\lambda}\}_{\lambda\in\varLambda}$ is a Riesz basis of $V^{2}(\varLambda,\varPhi)$, therefore there exists a dual Riesz basis $\widetilde{\varPhi}=\{\widetilde{\phi }_{\lambda}\}_{\lambda\in\varLambda}$. Thus we have that for all $f \in V^{2}(\varLambda,\varPhi)$, such that

$$ f = \sum_{\lambda\in\varLambda} \langle f, \widetilde{\phi}_{\lambda}\rangle\phi_{\lambda}= \sum _{\lambda\in\varLambda} \langle f,\phi_{\lambda}\rangle \widetilde{ \phi}_{\lambda}. $$

(21)

We can easily see that for all $1\leq q' \leq q \leq\infty$, $\infty \geq p' \geq p \geq1$ and $u'(x)\leq u(x), \forall x \in \mathbb {R}^{d}$, we have $\|\varPhi\|_{q',p',u'}\leq\|\varPhi\|_{q,p,u}$. Moreover, by [19, Theorem 4.1], we have that if ${\|\varPhi\| _{q,p,u}<\infty}$, then so is the same norm of the dual Riesz basis, or to say $\|\widetilde{\varPhi}\|_{q,p,u}<\infty$.

For convenience of the main proof, we will show some properties of the space $V^{2}(\varLambda,\varPhi)$ under the assumptions of the generators in Sect. 2.2.

Proposition 22

If $f \in{\mathrm{L}}^{2}(\mathbb {R}^{d})$, $\varPhi=\{\phi_{\lambda}\}$satisfying (A.0) and $\|\varPhi\|_{2,1,u_{0}}<\infty$, then we have:

$$ \sum_{\lambda\in\varLambda} \bigl\vert \langle f, \phi_{\lambda}\rangle \bigr\vert ^{2} \leq\|\varPhi \|_{2,1,u_{0}}^{2}\|f\|_{\mathrm{L}^{2}(\mathbb {R}^{d})}^{2} $$

(22)

and

$$ \sum_{\lambda\in\varLambda} \bigl\vert \langle f, \widetilde{\phi}_{\lambda}\rangle \bigr\vert ^{2} \leq \| \widetilde{\varPhi}\|_{2,1,u_{0}}^{2}\|f\|_{\mathrm{L}^{2}(\mathbb {R}^{d})}^{2} . $$

(23)

Proof

We firstly prove (22). In fact, we have

$$\begin{aligned} \sum_{\lambda\in\varLambda} \bigl\vert \langle f,\phi _{\lambda}\rangle \bigr\vert ^{2} = & \sum _{\lambda\in\varLambda} \biggl\vert \sum_{k\in \mathbb {Z}^{d}} \int_{k+[0,1]^{d}}f(x)\phi_{\lambda}(x)dx \biggr\vert ^{2} \\ \leq& \sum_{\lambda\in\varLambda} \biggl(\sum _{k\in \mathbb {Z}^{d}} \int_{k+[0,1]^{d}} \bigl\vert f(x) \bigr\vert \bigl\vert \phi _{\lambda}(x) \bigr\vert dx \biggr)^{2} \\ \leq& \sum_{\lambda\in\varLambda} \biggl(\sum _{k\in \mathbb {Z}^{d}} \bigl\Vert f(x) \bigr\Vert _{\mathrm {L}^{2}(k+[0,1]^{d})} \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})}dx \biggr)^{2} \\ = & \sum_{\lambda\in\varLambda} \biggl(\sum _{k\in \mathbb {Z}^{d}} \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})}^{1/2}\times \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})}^{1/2} \bigl\Vert f(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})} \biggr)^{2} \\ \leq& \sum_{\lambda\in\varLambda} \biggl(\sum _{k\in \mathbb {Z}^{d}} \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})} \biggr) \biggl(\sum_{k\in \mathbb {Z}^{d}} \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})} \bigl\Vert f(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})}^{2} \biggr) \\ \leq& \Vert \varPhi \Vert _{2,1,u_{0}}\sum_{\lambda\in\varLambda} \sum_{k\in \mathbb {Z}^{d}} \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})} \bigl\Vert f(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})}^{2} \\ = & \Vert \varPhi \Vert _{2,1,u_{0}} \sum_{k\in \mathbb {Z}^{d}} \bigl\Vert f(x) \bigr\Vert _{\mathrm {L}^{2}(k+[0,1]^{d})}^{2} \sum _{\lambda\in\varLambda} \bigl\Vert \phi_{\lambda}(x) \bigr\Vert _{\mathrm{L}^{2}(k+[0,1]^{d})} \\ \leq& \Vert \varPhi \Vert _{2,1,u_{0}}^{2} \sum _{k\in \mathbb {Z}^{d}} \bigl\Vert f(x) \bigr\Vert _{\mathrm {L}^{2}(k+[0,1]^{d})}^{2} = \Vert \varPhi \Vert _{2,1,u_{0}}^{2}\|f \|_{\mathrm{L}^{2}(\mathbb {R}^{d})}^{2} . \end{aligned}$$

A similar method is used in the proof of [18, Theorem 2.4(iii)]. Similar to the proof above, we can prove (23) as well. □

Proposition 23

For every $f \in V^{2}(\varLambda,\varPhi)$and every subset $\varGamma\subset \mathbb {R}^{d}$where the absolute covering index $N_{0}(\varGamma)$defined in (3) is finite, we have

$$ \sum_{\gamma\in\varGamma} \bigl\vert f(\gamma ) \bigr\vert ^{2}\leq D_{3} N_{0}( \varGamma)\|f\|_{2}^{2} . $$

(24)

Proof

First we consider $\varGamma$ with $N_{0}(\varGamma)=1$, that is, for any $k\in \mathbb {Z}^{d}$, there is at most one point in $\varGamma\cap (k+[0,1]^{d})$. Let $c(\lambda)=\langle f,\widetilde{\phi}_{\lambda}\rangle$, then $f=\sum_{\lambda\in\varLambda}c(\lambda)\phi _{\lambda}$. Similar to the proof of Proposition 22, we have

$$\begin{aligned} \sum_{\gamma\in\varGamma}\bigl|f(\gamma)\bigr|^{2} = & \sum _{\gamma\in\varGamma} \biggl\vert \sum _{\lambda\in\varLambda}c(\lambda)\phi_{\lambda}(\gamma) \biggr\vert ^{2} \\ \leq& \sum_{\gamma\in\varGamma} \biggl(\sum _{\lambda\in\varLambda} \bigl\vert c(\lambda) \bigr\vert \bigl\vert \phi_{\lambda}(\gamma) \bigr\vert \biggr)^{2} \\ \leq& \sum_{k\in \mathbb {Z}^{d}} \biggl(\sum _{\lambda\in\varLambda} \bigl\vert c(\lambda) \bigr\vert \Vert \phi _{\lambda} \Vert _{\mathrm{L}^{\infty}(k+[0,1]^{d})} \biggr)^{2} \\ = & \sum_{k\in \mathbb {Z}^{d}} \biggl(\sum _{\lambda\in\varLambda} \Vert \phi_{\lambda} \Vert _{\mathrm{L}^{\infty}(k+[0,1]^{d})}^{1/2}\times \Vert \phi_{\lambda} \Vert _{\mathrm{L}^{\infty}(k+[0,1]^{d})}^{1/2} \bigl\vert c(\lambda) \bigr\vert \biggr)^{2} \\ \leq& \sum_{k\in \mathbb {Z}^{d}} \biggl(\sum _{\lambda\in\varLambda} \Vert \phi_{\lambda} \Vert _{\mathrm{L}^{\infty}(k+[0,1]^{d})} \biggr) \biggl(\sum_{\lambda\in \varLambda} \Vert \phi_{\lambda} \Vert _{\mathrm{L}^{\infty}(k+[0,1]^{d})} \bigl\vert c(\lambda ) \bigr\vert ^{2} \biggr) \\ \leq& \Vert \varPhi \Vert _{\infty,1,u_{0}}\sum_{k\in \mathbb {Z}^{d}} \sum_{\lambda\in\varLambda} \Vert \phi _{\lambda} \Vert _{\mathrm{L}^{\infty}(k+[0,1]^{d})} \bigl\vert c(\lambda) \bigr\vert ^{2} \\ \leq& \Vert \varPhi \Vert _{\infty,1,u_{0}}^{2}\sum _{\lambda\in\varLambda} \bigl\vert c(\lambda) \bigr\vert ^{2} \\ \leq& \Vert \varPhi \Vert _{\infty,1,u_{0}}^{2}\|\widetilde{ \varPhi}\|_{2,1,u_{0}}^{2}\|f\|_{2}^{2} = D_{3}\|f\|_{2}^{2} , \end{aligned}$$

where the last inequality is derived from the conclusion of Proposition 22.

For a general subset $\varGamma$, we can split it into at most $N_{0}(\varGamma)$ non-intersect parts each of which has absolute covering index 1. Then sum the function value of each part respectively, sum all of them all together and we get the final conclusion. □

Proposition 24

$V^{2}(\varLambda,\varPhi)$is a reproducing kernel space, or to say there exists a family $(v_{x})_{x\in \mathbb {R}^{d}}\subset V^{2}(\varLambda,\varPhi)$such that $f(x)=\langle f,v_{x}\rangle ,\forall f\in V^{2}(\varLambda,\varPhi)$and $\|v_{x}\|_{2}^{2}\leq D_{2}$almost everywhere. Moreover, we have $\|f\|_{\infty}^{2}\leq D_{2}\|f\|_{2}^{2}$for all $f\in V^{2}(\varLambda,\varPhi)$.

Proof

We define that $v_{x}(y)=\sum_{\lambda\in\varLambda}\widetilde{\phi }_{\lambda}(y)\overline{\phi_{\lambda}(x)}$, and $w_{x}(y)=\sum_{\lambda \in\varLambda}\phi_{\lambda}(y)\overline{\widetilde{\phi}_{\lambda}(x)}$. For all $f\in V^{2}(\varLambda,\varPhi)$, we have that

$$\begin{aligned} f(x) = & \sum_{\lambda\in\varLambda}\langle f,\widetilde{\phi }_{\lambda}\rangle\phi_{\lambda}(x) \\ = & \sum_{\lambda\in\varLambda} \int_{\mathbb {R}^{d}}f(y)\overline{\widetilde{\phi}_{\lambda}(y)}dx\phi_{\lambda}(x) \\ = & \int_{\mathbb {R}^{d}}f(y)\overline{v_{x}(y)}dy = \langle f,v_{x}\rangle. \end{aligned}$$

Similarly, we have

$$\begin{aligned} f(x) = & \sum_{\lambda\in\varLambda}\langle f,\phi _{\lambda}\rangle\widetilde{\phi}_{\lambda}(x) \\ = & \int_{\mathbb {R}^{d}}f(y)\overline{w_{x}(y)}dy = \langle f,w_{x}\rangle. \end{aligned}$$

Therefore, $\langle f,w_{x}-v_{x}\rangle=0$ holds for all $f\in V^{2}(\varLambda,\varPhi)$. However, both $v_{x}$ and $w_{x}$ are in $V^{2}(\varLambda,\varPhi)$ by definition, then $w_{x}-v_{x}\in V^{2}(\varLambda,\varPhi)$. So $\langle w_{x}-v_{x},w_{x}-v_{x}\rangle=0$, or ${w_{x}-v_{x}=0}$. Therefore $w_{x}=v_{x}$, which implies $v_{x}(y)=\sum_{\lambda\in\varLambda }\phi_{\lambda}(y)\overline{\widetilde{\phi}_{\lambda}(x)}$ as well.

We now show that $v(x)_{x\in \mathbb {R}^{d}}$ is a reproducing kernel. For fixed $x\in \mathbb {R}^{d}$, we have

$$\begin{aligned} \|v_{x}\|_{2}^{2} = & \int_{\mathbb {R}^{d}}v_{x}(y)\overline{v_{x}(y)}dy \\ = & \int_{\mathbb {R}^{d}}\sum_{\lambda\in\varLambda}\sum _{\lambda'\in\varLambda}\widetilde{\phi}_{\lambda}(y) \overline{\phi_{\lambda}(x)}\;\overline{\phi_{\lambda'}(y)} \widetilde{\phi}_{\lambda'}(x)dy. \end{aligned}$$

By $\|\varPhi\|_{\infty,1,u_{0}}< \infty$, we can change the order of the sum and the integration. So we can obtain that

$$\begin{aligned} \|v_{x}\|_{2}^{2} = & \int_{\mathbb {R}^{d}}\sum_{\lambda\in\varLambda}\sum _{\lambda'\in\varLambda}\widetilde{\phi}_{\lambda}(y) \overline{\phi_{\lambda}(x)}\;\overline{\phi_{\lambda'}(y)} \widetilde{\phi}_{\lambda'}(x)dy \\ = & \sum_{\lambda\in\varLambda}\sum_{\lambda'\in\varLambda} \widetilde{\phi}_{\lambda'}(x)\overline{\phi_{\lambda}(x)} \int_{\mathbb {R}^{d}}\widetilde{\phi}_{\lambda}(y)\overline{ \phi_{\lambda'}(y)}dy \\ = & \sum_{\lambda\in\varLambda}\sum_{\lambda'\in\varLambda} \widetilde{\phi}_{\lambda'}(x)\overline{\phi_{\lambda}(x)} \langle \widetilde{\phi}_{\lambda},\phi_{\lambda'}\rangle \\ = & \sum_{\lambda\in\varLambda}\widetilde{\phi }_{\lambda}(x)\overline{\phi_{\lambda}(x)} \\ \ \leq& \biggl(\sum_{\lambda\in\varLambda}|\widetilde{\phi }_{\lambda}(x)|^{2} \biggr)^{\frac{1}{2}} \biggl(\sum _{\lambda\in\varLambda}|\phi_{\lambda}(x)|^{2} \biggr)^{\frac{1}{2}} \\ \leq& \|\varPhi\|_{\infty,2,u_{0}}\|\widetilde{\varPhi}\|_{\infty ,2,u_{0}} = D_{2}. \end{aligned}$$

Thus $\mathrm{ess}\,\mathrm{sup}_{x\in \mathbb {R}^{d}}\|v_{x}\|_{2}^{2}\leq D_{2}<\infty$. Therefore $(v_{x})_{x\in \mathbb {R}^{d}}$ is a reproducing kernel of $V^{2}(\varPhi,\varLambda)$. Moreover, we have

$$\begin{aligned} \|f\|_{\infty}^{2} = & \mathrm{ess}\,\mathrm{sup}_{x\in \mathbb {R}^{d}}\bigl|f(x)\bigr|^{2} \\ = & \mathrm{ess}\,\mathrm{sup}_{x\in \mathbb {R}^{d}}\bigl|\langle f,v_{x} \rangle \bigr|^{2} \\ \leq& \mathrm{ess}\,\mathrm{sup}_{x\in \mathbb {R}^{d}}\|f\|_{2}^{2} \|v_{x}\|_{2}^{2} \\ \leq& D_{2}\|f\|_{2}^{2}. \end{aligned}$$

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