Recovering finite parametric distributions and functions using the spherical mean transform

Abstract

The aim of the article is to recover a certain type of finite parametric distributions and functions using their spherical mean transform which is given on a certain family of spheres whose centers belong to a finite set $\Gamma$. For this, we show how the problem of reconstruction can be converted to a Prony’s type system of equations whose regularity is guaranteed by the assumption that the points in the set $\Gamma$ are in general position. By solving the corresponding Prony’s system we can extract the set of parameters which define the corresponding function or distribution.

Appendix

Lemma 5.1

Let g be a function, defined on ${\mathbb {R}}^{+}$, such that its radial extension belongs to the Schwartz space $\mathrm {S}\left( {\mathbb {R}}^{n}\right)$. Then g and its Hankel transform G are continuously differentiable and belong to $L^{1}\left( {\mathbb {R}}^{+}, r^{\frac{n - 1}{2}}\right)$.

Proof

Denote by $g_{0}$ and $G_{0}$ respectively the radial extensions of g and G to ${\mathbb {R}}^{n}$. Using a spherical coordinates system in ${\mathbb {R}}^{n}$, we have

$$\begin{aligned} \int _{|x| > 1}g_{0}(x)|x|^{-\frac{n - 1}{2}}dx = \int _{1}^{\infty }\int _{{\mathbb {S}}^{n - 1}}g_{0}(r\theta )r^{\frac{n - 1}{2}}d\theta dr = \frac{2\pi ^{\frac{n}{2}}}{\Gamma \left( \frac{n}{2}\right) }\int _{1}^{\infty }g(r)r^{\frac{n - 1}{2}}dr. \end{aligned}$$

Since $g_{0}$ belongs to $\mathrm {S}\left( {\mathbb {R}}^{n}\right)$ it follows that the integral in the left hand side converges and hence also the integral in the right hand side. Since g is also bounded on the interval [0, 1] (since $g_{0}$ is bounded in the unit disk), it follows that g is in $L^{1}\left( {\mathbb {R}}^{+}, r^{\frac{n - 1}{2}}dr\right)$. Denote by $\widehat{g}_{0}$ the Fourier transform of g, then using spherical coordinates again we have

$$\begin{aligned} \widehat{g}_{0}(\omega ) = \frac{1}{(2\pi )^{\frac{n}{2}}}\int _{{\mathbb {R}}^{n}}g_{0}(x)e^{-i\langle \omega , x\rangle }dx = \frac{1}{(2\pi )^{\frac{n}{2}}}\int _{0}^{\infty }\int _{{\mathbb {S}}^{n - 1}}g_{0}(r\theta )e^{-ir\langle \omega , \theta \rangle }d\theta r^{n - 1}dr. \end{aligned}$$

Denoting $\omega = \lambda \psi$, where $\lambda = |\omega |, \psi = \frac{\omega }{|\omega |}$ and using the identity

$$\begin{aligned} \int _{{\mathbb {S}}^{n - 1}}e^{-it\left\langle \psi , \theta \right\rangle } d\theta = (2\pi )^{\frac{n}{2}}j_{\frac{n}{2} - 1}(t) \end{aligned}$$

it follows that

$$\begin{aligned} \widehat{g}_{0}(\lambda \psi ) = \int _{0}^{\infty }g(r)j_{\frac{n}{2} - 1}(\lambda r)r^{n - 1}dr = G(\lambda ). \end{aligned}$$

Hence $\widehat{g}_{0}$ is the radial extension of G. Since the Fourier transform maps the Schwartz space $\mathrm {S}\left( {\mathbb {R}}^{n}\right)$ onto itself and since $g_{0}$ is in $\mathrm {S}\left( {\mathbb {R}}^{n}\right)$, it follows that the same is true for the radial extension of G. Hence, from the same arguments we used for g and its radial extension, it follows that G belongs to $L^{1}\left( {\mathbb {R}}^{+}, r^{\frac{n - 1}{2}}dr\right)$.

The assertion that g is continuously differentiable follows immediately from the fact that its radial extension if infinitely differentiable and the same is true for G. $\square$

Lemma 5.2

Let $n\ge 2$ and $\lambda _{1},\ldots ,\lambda _{n}$ be n real numbers satisfying $\lambda _{i}\ne \lambda _{j}, i\ne j$. Let $\sigma :\{1,2,\ldots ,n\}\rightarrow \{1,2,\ldots ,n\}$ be a permutation which is different from the identity. Define the following matrix

$$\begin{aligned} M = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \lambda _{1} - \lambda _{\sigma (1)} &{} \lambda _{2} - \lambda _{\sigma (2)} &{} \ldots &{} \lambda _{n} - \lambda _{\sigma (n)}\\ \lambda _{1}^{2} - \lambda _{\sigma (1)}^{2} &{} \lambda _{2}^{2} - \lambda _{\sigma (2)}^{2} &{} \ldots &{} \lambda _{n}^{2} - \lambda _{\sigma (n)}^{2}\\ \ldots \ldots \ldots \ldots \ldots \\ \lambda _{1}^{n} - \lambda _{\sigma (1)}^{n} &{} \lambda _{2}^{n} - \lambda _{\sigma (2)}^{n} &{} \ldots &{} \lambda _{n}^{n} - \lambda _{\sigma (n)}^{n} \end{array} \right) . \end{aligned}$$

(5.1)

Then, if $v = (v_{1},\ldots ,v_{n})\in {\mathbb {R}}^{n}$ satisfies $Mv^{T} = 0$ then there are two indices i and j, $i\ne j$ such that $v_{i} = v_{j}$.

Proof

The proof is by induction on $n\ge 2$. Since $\sigma$ is different from the identity then for $n = 2$ we have $\sigma (1) = 2$ and $\sigma (2) = 1$. In this case the vector $v = (1,1)$ is in the kernel of the matrix M as defined by (5.1) and indeed this vector has two equal components with different indices. There are no other independent vectors in the kernel of M since otherwise $M = 0$ which will imply that $\lambda _{1} = \lambda _{2}$.

Now we assume the induction hypothesis for every integer m satisfying $2\le m \le n - 1$ and our aim is to prove it for the integer n where we assume that $n\ge 3$.

First observe that we can assume that $\sigma$ does not have any fixed points. Indeed, suppose without loss of generality that $\sigma (n) = n$, then the matrix M has the following form

$$\begin{aligned} M = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \lambda _{1} - \lambda _{\sigma (1)} &{} \lambda _{2} - \lambda _{\sigma (2)} &{} \ldots &{} \lambda _{n - 1} - \lambda _{\sigma (n - 1)} &{} 0\\ \lambda _{1}^{2} - \lambda _{\sigma (1)}^{2} &{} \lambda _{2}^{2} - \lambda _{\sigma (2)}^{2} &{} \ldots &{} \lambda _{n - 1}^{2} - \lambda _{\sigma (n - 1)}^{2} &{} 0\\ \ldots \ldots \ldots \ldots \ldots \\ \lambda _{1}^{n - 1} - \lambda _{\sigma (1)}^{n - 1} &{} \lambda _{2}^{n - 1} - \lambda _{\sigma (2)}^{n - 1} &{} \ldots &{} \lambda _{n - 1}^{n - 1} - \lambda _{\sigma (n - 1)}^{n - 1} &{} 0\\ \lambda _{1}^{n} - \lambda _{\sigma (1)}^{n} &{} \lambda _{2}^{n} - \lambda _{\sigma (2)}^{n} &{} \ldots &{} \lambda _{n - 1}^{n} - \lambda _{\sigma (n - 1)}^{n} &{} 0 \end{array} \right) = \left( \begin{array}{c@{\quad }c} A &{} \overline{0}\\ * &{} 0 \end{array} \right) , \end{aligned}$$

where A is the $(n - 1)\times (n - 1)$ top left sub matrix of M. Hence, if $Mv^{T} = 0$ then v has the form $v = (v_{0}, *)$ where $v_{0}\in {\mathbb {R}}^{n - 1}$ is in the kernel of A. Now we can use our induction hypothesis on the matrix A. Indeed, if $\widetilde{\sigma }$ denotes the restriction of $\sigma$ to the set $\{1,2,\ldots ,n - 1\}$ then since $\sigma (n) = n$ it follows that $\widetilde{\sigma }(\{1,\ldots ,n - 1\}) = \{1,\ldots ,n - 1\}$ and since $\sigma$ is different from the identity then the same is true for $\widetilde{\sigma }$. Hence, using the induction hypothesis on the matrix A is follows that $v_{0}$ has at least two equal components with different indices and thus the same is true for v. $\square$

Now, with the assumption that $\sigma$ does not have any fixed points it follows that the set $\{1,2,\ldots ,n\}$ has a partition into mutually disjoint sets $A_{1},\ldots ,A_{k}$ which satisfy the following conditions:

• (i) $\left| A_{i}\right| \ge 2, 1\le i\le k$.

• (ii) $\sigma \left( A_{i}\right) = A_{i}, 1\le i\le k$.

• (iii) For every $1\le i\le k$ and every non-empty subset $B\subsetneqq A_{i}$ we have $\sigma (B)\ne B$.

We will distinguish between the cases when $k = 1$ and $k \ge 2$.

1.1 The case $k = 1$

If $k = 1$ we will show that the kernel of M is spanned by the vector $v = (1,1,\ldots ,1)$. Indeed, suppose that $u = (u_{1},\ldots ,u_{n})$ is in the kernel of M, we can assume without loss of generality that $u_{n} = 0$ (since otherwise we can replace u with $u - u_{n}v$). We will show that $u = \overline{0}$. Since u is in the kernel of M we have

$$\begin{aligned} Mu= & {} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \lambda _{1} - \lambda _{\sigma (1)} &{} \ldots &{} \lambda _{n - 2} - \lambda _{\sigma (n - 2)} &{} \lambda _{n - 1} - \lambda _{\sigma (n - 1)} &{} \lambda _{n} - \lambda _{\sigma (n)}\\ \lambda _{1}^{2} - \lambda _{\sigma (1)}^{2} &{} \ldots &{} \lambda _{n - 2}^{2} - \lambda _{\sigma (n - 2)}^{2} &{} \lambda _{n - 1}^{2} - \lambda _{\sigma (n - 1)}^{2} &{} \lambda _{n}^{2} - \lambda _{\sigma (n)}^{2}\\ \ldots \ldots \ldots \ldots \ldots \\ \lambda _{1}^{n} - \lambda _{\sigma (1)}^{n} &{} \ldots &{} \lambda _{n - 2}^{n} - \lambda _{\sigma (n - 2)}^{n} &{} \lambda _{n - 1}^{n} - \lambda _{\sigma (n - 1)}^{n} &{} \lambda _{n}^{n} - \lambda _{\sigma (n)}^{n} \end{array} \right) \left( \begin{array}{c} u_{1}\\ \ldots \\ u_{n - 2}\\ u_{n - 1}\\ 0 \end{array} \right) \nonumber \\= & {} \left( \rho _{1} - \rho _{\sigma (1)}\right) u_{1} + \cdots + \left( \rho _{n - 2} - \rho _{\sigma (n - 2)}\right) u_{n - 2} + \left( \rho _{n - 1} - \rho _{\sigma (n - 1)}\right) u_{n - 1} = \overline{0}\nonumber \\ \end{aligned}$$

(5.2)

where

$$\begin{aligned} \rho _{i} = \left( \lambda _{i},\lambda _{i}^{2}\ldots ,\lambda _{i}^{n}\right) ^{T},\quad 1 \le i \le n. \end{aligned}$$

(5.3)

For the case $k = 1$ condition (iii) implies that $\sigma \left( \{1,2,\ldots ,n - 1\}\right) \ne \{1,2,\ldots ,n - 1\}$. Hence, we can assume without loss of generality that $\sigma (k)\ne n - 1, 1 \le k \le n - 1$. Hence the vectors $\rho _{n - 1}$ and $\rho _{n}$ appear only once in the brackets of Eq. (5.2) and thus from Eq. (5.2) we have

$$\begin{aligned} a_{1}\rho _{1} + \cdots + a_{n - 2}\rho _{n - 2} + u_{n - 1}\rho _{n - 1} - u_{\sigma ^{- 1}(n)}\rho _{n} = \overline{0} \end{aligned}$$

(5.4)

where $a_{1},\ldots ,a_{n - 2}$ are constants which depend on $u_{1},\ldots ,u_{n - 1}$. Now observe that if non of the vectors $\rho _{1},\ldots ,\rho _{n}$ is equal to zero then by Eq. (5.2) it follows that they are linearly independent since they form the n columns of an $n\times n$ Vandermonde matrix and by our assumption $\lambda _{i}\ne \lambda _{j}$ if $i\ne j$. If one of these vectors is equal to zero then the other are different from zero and thus they are linearly independent since they form a sub matrix of the Vandermonde matrix. Hence since at least one of the vectors $\rho _{n - 1}$ or $\rho _{n}$ is different from zero it follows from Eq. (5.3) that either $u_{n - 1} = 0$ or $u_{\sigma ^{-1}(n)} = 0$.

Continuing in this way, assume that after $n - k - 1$ steps we can prove that $n - k - 1$ coefficients from the set $\left\{ u_{1},\ldots ,u_{n - 1}\right\}$ are equal to zero. Then from Eq. (5.2) we will have

$$\begin{aligned} \left( \rho _{i_{1}} - \rho _{\sigma (i_{1})}\right) u_{i_{1}} + \cdots + \left( \rho _{i_{k - 1}} - \rho _{\sigma (i_{k - 1})}\right) u_{i_{k - 1}} + \left( \rho _{i_{k}} - \rho _{\sigma (i_{k})}\right) u_{i_{k}} = \overline{0}. \end{aligned}$$

(5.5)

Since $\sigma \left( \left\{ i_{1},\ldots ,i_{k}\right\} \right) \ne \left\{ i_{1},\ldots ,i_{k}\right\}$ we can assume, without loss of generality, that $\sigma (i_{p})\ne i_{k}, 1\le p\le k$. Also, there exists an index l, different from any of the indices $i_{1},\ldots ,i_{k}$, such that $l\in \sigma \left( \left\{ i_{1},\ldots ,i_{k}\right\} \right)$. Hence the vectors $\rho _{i_{k}}$ and $\rho _{l}$ appear only once in the brackets of Eq. (5.4) and thus from Eq. (5.4) we have

$$\begin{aligned}&b_{1}\rho _{1} + \cdots + b_{i_{k} - 1}\rho _{i_{k} - 1} + u_{i_{k}}\rho _{i_{k}} + b_{i_{k} + 1}\rho _{i_{k + 1}}\\&\quad + \cdots + b_{l - 1}\rho _{l - 1} - u_{\sigma ^{-1}(l)}\rho _{l} + b_{l + 1}\rho _{l + 1} + \cdots + b_{n}\rho _{n} = \overline{0} \end{aligned}$$

where $b_{j}, 1\le j\le n, j\ne i_{k}, j\ne \sigma ^{-1}(l)$ are constants which depend on $u_{i_{1}},\ldots , u_{i_{k}}$. Since at least one of the vectors $\rho _{l}$ or $\rho _{i_{k}}$ is different from zero we can conclude, exactly as before, that either $u_{i_{k}} = 0$ or $u_{\sigma ^{-1}(l)} = 0$.

Hence after $n - 1$ steps we can conclude that $u_{1} = 0,\ldots ,u_{n - 1} = 0$.

1.2 The case $k \ge 2$

For the case $k\ge 2$ the set $\{1,2,\ldots ,n\}$ has a partition into disjoints subsets $A_{1},\ldots ,A_{k}$ which satisfy conditions (i), (ii) and (iii). Since at most one of the vectors $\rho _{1},\ldots ,\rho _{n}$ is equal to zero and $k\ge 2$ we can assume, without loss of generality, that all the vectors with indices in the set $A_{1}$ are different from zero and again without loss of generality we can assume that

$$\begin{aligned} A_{1} = \{1,\ldots ,m\},\quad 2\le m\le n - 2. \end{aligned}$$

Now assume again that $u = (u_{1},\ldots ,u_{n})$ is in the kernel of M, we will show that there are two different indices, i and j, such that $u_{i} = u_{j}$. From the definition of the matrix M it follows that

$$\begin{aligned}&\left( \rho _{1} - \rho _{\sigma (1)}\right) u_{1} + \cdots + \left( \rho _{m} - \rho _{\sigma (m)}\right) u_{m} + \left( \rho _{m + 1} - \rho _{\sigma (m + 1)}\right) u_{m + 1} \\&\quad + \cdots + \left( \rho _{n} - \rho _{\sigma (n)}\right) u_{n} =\overline{0}. \end{aligned}$$

The last equation can be rewritten as

$$\begin{aligned} (u_{1} - u_{\sigma ^{-1}(1)})\rho _{1} + \cdots + (u_{m} - u_{\sigma ^{-1}(m)})\rho _{m} + c_{m + 1}\rho _{m + 1} + \cdots + c_{n}\rho _{n} = \overline{0} \end{aligned}$$

(5.6)

where $c_{m + 1},\ldots ,c_{n}$ are constants which depend on $u_{m + 1},\ldots ,u_{n}$. Since, by our assumption, all the vectors $\rho _{1},\ldots ,\rho _{m}$ are different from zero, they form a sub matrix of the Vandermonde matrix which implies that they are linearly independent. Hence from Eq. (5.5) we have

$$\begin{aligned} u_{1} - u_{\sigma ^{-1}(1)} = 0,\ldots , u_{m} - u_{\sigma ^{-1}(m)} = 0. \end{aligned}$$

(5.7)

We will now show that the vector $\widetilde{u} = (u_{1},\ldots ,u_{m},0,\ldots ,0)$ is also in the kernel of M. Indeed, from the definition of M we have to prove that

$$\begin{aligned}&\left( \rho _{1} - \rho _{\sigma (1)}\right) u_{1} + \cdots + \left( \rho _{m} - \rho _{\sigma (m)}\right) u_{m}\\&\quad = \rho _{1}u_{1} + \cdots + \rho _{m}u_{m} - \underset{(*)}{\underbrace{\rho _{\sigma (1)}u_{1} - \cdots - \rho _{\sigma (m)}u_{m}}} = \overline{0}. \end{aligned}$$

Now by condition (ii) it follows that $\sigma \left( \{1,\ldots ,m\}\right) = \{1,\ldots ,m\}$ and thus we also have $\sigma ^{-1}\left( \{1,\ldots ,m\}\right) = \{1,\ldots ,m\}$. Thus the set $\{1,\ldots ,m\}$ is a partition of $\{\sigma ^{-1}(1),\ldots ,\sigma ^{-1}(m)\}$. Using this observation on $(*)$ we have

$$\begin{aligned}&\rho _{1}u_{1} + \cdots + \rho _{m}u_{m} - \rho _{\sigma (1)}u_{1} - \cdots - \rho _{\sigma (m)}u_{m}\\&\quad = \rho _{1}u_{1} + \cdots + \rho _{m}u_{m} - \rho _{\sigma (\sigma ^{-1}(1))}u_{\sigma ^{-1}(1)} - \cdots - \rho _{\sigma (\sigma ^{-1}(m))}u_{\sigma ^{-1}(m)}\\&\quad = \rho _{1}u_{1} + \cdots + \rho _{m}u_{m} - \rho _{1}u_{\sigma ^{-1}(1)} - \cdots - \rho _{m}u_{\sigma ^{-1}(m)}\\&\quad = \left( u_{1} - u_{\sigma ^{-1}(1)}\right) \rho _{1} + \cdots + \left( u_{m} - u_{\sigma ^{-1}(m)}\right) \rho _{m} \end{aligned}$$

and the last expression is equal to zero by (5.6).

Since ${\widetilde{u}}$ is in the kernel of M it follows that the vector $(u_{1},\ldots ,u_{m})$ is in the kernel of the top left $m\times m$ sub matrix of M. That is

$$\begin{aligned} \left( \begin{array}{c@{\quad }c@{\quad }c} \lambda _{1} - \lambda _{\sigma (1)} &{} \ldots &{} \lambda _{m} - \lambda _{\sigma (m)}\\ \ldots \ldots \ldots \ldots \ldots \\ \lambda _{1}^{m} - \lambda _{\sigma (1)}^{m} &{} \ldots &{} \lambda _{m}^{m} - \lambda _{\sigma (m)}^{m} \end{array} \right) \left( \begin{array}{c} u_{1}\\ \ldots \\ u_{m} \end{array} \right) = \left( \begin{array}{c} 0\\ \ldots \\ 0 \end{array} \right) . \end{aligned}$$

(5.8)

Now we can use the induction hypothesis on Eq. (5.7). Indeed, since $\sigma \left( \{1,\ldots ,m\}\right) = \{1,\ldots ,m\}$ and $\lambda _{i}\ne \lambda _{j}$ for $1\le i,j\le m, i\ne j$, all of the conditions for the sub matrix in the left hand side of Eq. (5.7) are satisfied. Hence we conclude that there are two different indices, i and j with $1\le i,j\le m$, such that $u_{i} = u_{j}$. This in particular implies that the vector u has two equal components with different indices. Thus the case $k\ge 2$ is also proved. $\square$