1 Introduction
Let
$K$
be a compact subset of
$\mathbb{R}^{n}$
which contains the origin
$o$
as an interior point and is star-shaped with respect to
$o$
. Such a set
$K$
is called a star body if the radial function
$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D703})=\max \{r\geqslant 0:r\unicode[STIX]{x1D703}\in K\},\quad \unicode[STIX]{x1D703}\in S^{n-1},\end{eqnarray}$$
that determines the shape of
$K$
is continuous (see Gardner [Reference Gardner2]). We denote by
$G_{n,k}$
,
$2\leqslant k\leqslant n-1$
, the Grassmann manifold of all
$k$
-dimensional linear subspaces
$\unicode[STIX]{x1D709}$
of
$\mathbb{R}^{n}$
. Passing to spherical coordinates, one can evaluate the
$k$
-dimensional volume of
$K\cap \unicode[STIX]{x1D709}$
as
$$\begin{eqnarray}\text{vol}_{k}(K\cap \unicode[STIX]{x1D709})=\frac{1}{k}\int _{S^{n-1}\cap \unicode[STIX]{x1D709}}\unicode[STIX]{x1D70C}_{K}^{k}(\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703},\end{eqnarray}$$
where
$d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703}$
stands for the corresponding
$(k-1)$
-dimensional surface element. The right-hand side of (1.2) is a constant multiple of the Funk-type transform
$$\begin{eqnarray}(F_{k}f)(\unicode[STIX]{x1D709})=\int _{S^{n-1}\cap \unicode[STIX]{x1D709}}f(\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703},\quad \unicode[STIX]{x1D709}\in G_{n,k},\end{eqnarray}$$
that can be explicitly inverted in different ways provided that
$f$
is an even integrable function on
$S^{n-1}$
(see, e.g., [Reference Helgason7, Reference Rubin8, Reference Rubin11]). This fact makes it possible to determine the shape of
$K$
from the knowledge of the volumes of
$K\cap \unicode[STIX]{x1D709}$
for all
$\unicode[STIX]{x1D709}\in G_{n,k}$
provided that
$K$
is origin-symmetric. The star bodies which are not origin-symmetric cannot be uniquely reconstructed from the integrals (1.2) (see, e.g., Gardner [Reference Gardner2, §7.2]). This statement agrees with a known fact that the kernel of
$F_{k}$
on
$L^{1}(S^{n-1})$
consists of odd functions.
In the case
$k=n-1$
, Groemer [Reference Groemer6], who followed some ideas from Backus [Reference Backus1], suggested replacing the hyperplane central sections by the “half-sections”
$K\cap H(u,v)$
, where
$$\begin{eqnarray}H(u,v)=\{x\in \mathbb{R}^{n}:x\cdot u=0,x\cdot v\geqslant 0\},\quad u\bot v,\end{eqnarray}$$
is a half-plane determined by mutually orthogonal unit vectors
$u$
and
$v$
, so that the origin
$o$
lies on the boundary of
$H(u,v)$
. Thus (1.2) is substituted by the lower-dimensional hemispherical integral
$$\begin{eqnarray}\text{vol}_{n-1}(K\cap H(u,v))=\frac{1}{n-1}\int _{S^{n-1}\cap H(u,v)}\unicode[STIX]{x1D70C}_{K}^{n-1}(\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703}.\end{eqnarray}$$
One of the main results of [Reference Groemer6] states that if the star bodies
$K$
and
$L$
satisfy
$$\begin{eqnarray}\text{vol}_{n-1}(K\cap H(u,v))=\text{vol}_{n-1}(L\cap H(u,v))\end{eqnarray}$$
for all
$u,v\in S^{n-1}$
,
$u\bot v$
, then
$K=L$
. The case of half-sections of lower dimensions is a direct consequence of Groemer’s result, as was pointed out by Goodey and Weil [Reference Goodey and Weil5].
To the best of our knowledge, the following important questions remain open.
Question 1. How can we explicitly reconstruct the shape of
$K$
(or the radial function
$\unicode[STIX]{x1D70C}_{K}$
) from volumes (1.5) or, more generally, from the corresponding
$k$
-dimensional volumes?
Question 2. How can we eliminate overdeterminedness of the inversion problem in Question 1?
Regarding Question 2, we observe that (1.4) parameterizes the corresponding half-sections by the elements of the Stiefel manifold
$V_{n,2}=\{(u,v):u,v\in S^{n-1},u\bot v\}$
, so that
$\dim V_{n,2}=2n-3>n-1=\dim S^{n-1}$
if
$n>2$
. In other words, the dimension of the target space is greater than the dimension of the source space. The latter means that the inversion problem in Question 1 can be overdetermined. In the case
$k<n-1$
, the difference between the dimensions of the target space and source space is bigger because the corresponding lower-dimensional Funk transform (1.3) is overdetermined itself. This situation is pursuant to Gel’fand’s celebrated question [Reference Gel’fand3] about how to reduce overdeterminedness of transformations in integral geometry. In our case, it means the following problem.
Problem.
Find an
$(n-1)$
-dimensional submanifold
$\mathfrak{M}$
of the manifold of all
$k$
-dimensional central half-sections of
$K$
so that
$K$
could be recovered from the volumes of the half-sections belonging to
$\mathfrak{M}$
only.
In the present article, we solve this problem and give an answer to Questions 1 and 2. The basic idea is to consider separately the half-sections lying in the open half-spaces
$$\begin{eqnarray}H_{\pm }=\{x=(x_{1},\ldots ,x_{n})\in \mathbb{R}^{n}:\pm x_{n}>0\}.\end{eqnarray}$$
The case of all
$2\leqslant k\leqslant n-1$
is considered in §2. Here our inversion formulas remain overdetermined if
$k<n-1$
. This overdeterminedness is eliminated in §3.
2 Inversion formulas
We consider the following hemispherical modifications of the Funk-type transform (1.3) when a function
$f\in L^{1}(S^{n-1})$
is integrated over
$(k-1)$
-dimensional hemispheres
$S_{\pm }^{n-1}\cap \unicode[STIX]{x1D709}$
,
$\unicode[STIX]{x1D709}\in G_{n,k}$
, lying in the
$(n-1)$
-dimensional hemispheres
$$\begin{eqnarray}S_{\pm }^{n-1}=\{\unicode[STIX]{x1D703}=(\unicode[STIX]{x1D703}_{1},\ldots ,\unicode[STIX]{x1D703}_{n})\in S^{n-1}:\pm \unicode[STIX]{x1D703}_{n}>0\},\end{eqnarray}$$
respectively. Specifically, for any subspace
$\unicode[STIX]{x1D709}\in G_{n,k}$
not orthogonal to
$e_{n}=(0,\ldots ,0,1)$
, we set
$$\begin{eqnarray}\displaystyle (F_{k}^{+}f)(\unicode[STIX]{x1D709}) & = & \displaystyle \int _{S_{+}^{n-1}\cap \unicode[STIX]{x1D709}}f(\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703}\qquad \biggl(\!=\int _{S_{-}^{n-1}\cap \unicode[STIX]{x1D709}}f(-\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703}\biggr),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle (F_{k}^{-}f)(\unicode[STIX]{x1D709}) & = & \displaystyle \int _{S_{-}^{n-1}\cap \unicode[STIX]{x1D709}}f(\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703}\qquad \biggl(\!=\int _{S_{+}^{n-1}\cap \unicode[STIX]{x1D709}}f(-\unicode[STIX]{x1D703})\,d_{\unicode[STIX]{x1D709}}\unicode[STIX]{x1D703}\biggr).\end{eqnarray}$$
Clearly,
$$\begin{eqnarray}\displaystyle (F_{k}^{+}f)(\unicode[STIX]{x1D709}) & = & \displaystyle {\textstyle \frac{1}{2}}(F_{k}f_{1})(\unicode[STIX]{x1D709}),\qquad f_{1}(\unicode[STIX]{x1D703})=\left\{\begin{array}{@{}ll@{}}f(\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{+}^{n-1},\\ f(-\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{-}^{n-1},\end{array}\right.\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle (F_{k}^{-}f)(\unicode[STIX]{x1D709}) & = & \displaystyle {\textstyle \frac{1}{2}}(F_{k}f_{2})(\unicode[STIX]{x1D709}),\qquad f_{2}(\unicode[STIX]{x1D703})=\left\{\begin{array}{@{}ll@{}}f(-\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{+}^{n-1},\\ f(\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{-}^{n-1},\end{array}\right.\end{eqnarray}$$
where
$f_{i}$
(
$i=1,2$
) are integrable even functions on
$S^{n-1}$
that can be reconstructed from
$\unicode[STIX]{x1D711}^{\pm }=F_{k}^{\pm }f$
by the formulas
$f_{1}=2F_{k}^{-1}\unicode[STIX]{x1D711}^{+}$
and
$f_{2}=2F_{k}^{-1}\unicode[STIX]{x1D711}^{-}$
. Combining these formulas, we obtain
$$\begin{eqnarray}f(\unicode[STIX]{x1D703})=\left\{\begin{array}{@{}ll@{}}2(F_{k}^{-1}\unicode[STIX]{x1D711}^{+})(\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{+}^{n-1},\\ 2(F_{k}^{-1}\unicode[STIX]{x1D711}^{-})(\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{-}^{n-1}.\end{array}\right.\end{eqnarray}$$
Thus we have proved the following theorem.
Theorem 2.1. For
$2\leqslant k\leqslant n-1$
, a function
$f\in L^{1}(S^{n-1})$
can be recovered from the integrals
$\unicode[STIX]{x1D711}^{\pm }=F_{k}^{\pm }f$
by the formula (2.5).
Some comments are in order.
(i) Let
$\unicode[STIX]{x1D709}\in G_{n,k}$
be a subspace which is not orthogonal to
$e_{n}=(0,\ldots ,0,1)$
. We define
$\unicode[STIX]{x1D709}_{\pm }=\unicode[STIX]{x1D709}\cap \{x\in \mathbb{R}^{n}:\pm \,x_{n}>0\}$
. Then Theorem 2.1 implies the following corollary.
Corollary 2.2. The radial function
$\unicode[STIX]{x1D70C}_{K}$
of the star body
$K$
can be recovered from the volumes
$v^{\pm }(\unicode[STIX]{x1D709})=\text{vol}_{k}(K\cap \unicode[STIX]{x1D709}_{\pm })$
by the formula
$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{K}^{k}(\unicode[STIX]{x1D703})=\left\{\begin{array}{@{}ll@{}}2k(F_{k}^{-1}v^{+})(\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{+}^{n-1},\\ 2k(F_{k}^{-1}v^{-})(\unicode[STIX]{x1D703}) & \text{if}~\unicode[STIX]{x1D703}\in S_{-}^{n-1}.\end{array}\right.\end{eqnarray}$$
For
$\unicode[STIX]{x1D703}_{n}=0$
, it can be determined from (2.6) by continuity.
(ii) In the case
$k=n-1$
, we can set
$\unicode[STIX]{x1D709}=u^{\bot }$
,
$u\in S^{n-1}$
and write (1.3) as the usual Funk transform
$$\begin{eqnarray}(Ff)(u)=\int _{\{\unicode[STIX]{x1D703}\in S^{n-1}:u\cdot \unicode[STIX]{x1D703}=0\}}f(\unicode[STIX]{x1D703})\,d_{u}\unicode[STIX]{x1D703}.\end{eqnarray}$$
Similar notation
$(F^{\pm }f)(u)$
can be used for the hemispherical transforms (2.1) and (2.2).
(iii) The above reasoning shows that to reconstruct
$f$
on
$S_{+}^{n-1}$
(or
$S_{-}^{n-1}$
) the knowledge of
$\unicode[STIX]{x1D711}^{+}=F_{k}^{+}f$
(or
$\unicode[STIX]{x1D711}^{-}=F_{k}^{-}f$
, respectively) is sufficient.
(iv) A theory of the Funk transform provides a variety of inversion formulas for
$F_{k}$
(see, e.g., [Reference Gelfand, Gindikin and Graev4, Reference Helgason7–Reference Rubin9, Reference Rubin11]). The functions
$f_{1}$
and
$f_{2}$
in (2.3) and (2.4) can have discontinuity on the equator
$\unicode[STIX]{x1D703}_{n}=0$
. It means that we cannot apply inversion formulas for
$F_{k}$
(at least, not straightforwardly) in which the smoothness is crucial. However, if
$f\in L^{p}(S^{n-1})$
,
$1\leqslant p<\infty$
, then
$f_{1}$
and
$f_{2}$
also belong to
$L^{p}(S^{n-1})$
and can be reconstructed, for example, by the method of mean value operators as follows.
For
$r=\cos \unicode[STIX]{x1D713}$
,
$\unicode[STIX]{x1D713}\in [0,\unicode[STIX]{x1D70B}/2]$
and
$\unicode[STIX]{x1D703}\in S^{n-1}$
, consider the shifted dual transform
$$\begin{eqnarray}(F_{k,\unicode[STIX]{x1D703}}^{\ast }\unicode[STIX]{x1D711})(r)=\int _{d(\unicode[STIX]{x1D703},\unicode[STIX]{x1D709})=\unicode[STIX]{x1D713}}\unicode[STIX]{x1D711}(\unicode[STIX]{x1D709})\,d\unicode[STIX]{x1D707}(\unicode[STIX]{x1D709}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D711}$
is a function on
$G_{n,k}$
,
$d(\cdot ,\cdot )$
denotes the geodesic distance and
$d\unicode[STIX]{x1D707}(\unicode[STIX]{x1D709})$
stands for the corresponding canonical measure (see [Reference Rubin9, §5] for details).
Theorem 2.3 [Reference Rubin9, Theorem 5.3].
An even function
$f\in L^{p}(S^{n-1})$
,
$1\leqslant p$
${<}\infty$
, can be recovered from
$\unicode[STIX]{x1D711}=F_{k}f$
by the formula
$$\begin{eqnarray}f(\unicode[STIX]{x1D703})=\lim _{s\rightarrow 1}\biggl(\frac{1}{2s}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}s}\biggr)^{k}\biggl[\frac{\unicode[STIX]{x1D70B}^{-k/2}}{\unicode[STIX]{x1D6E4}(k/2)}\int _{0}^{s}(s^{2}-r^{2})^{k/2-1}(F_{k,\unicode[STIX]{x1D703}}^{\ast }\unicode[STIX]{x1D711})(r)r^{k}\,dr\biggr].\end{eqnarray}$$
In particular, for
$k$
even,
$$\begin{eqnarray}f(\unicode[STIX]{x1D703})=\lim _{s\rightarrow 1}\frac{1}{2\unicode[STIX]{x1D70B}^{k/2}}\biggl(\frac{1}{2s}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}s}\biggr)^{k/2}[s^{k-1}(F_{k,\unicode[STIX]{x1D703}}^{\ast }\unicode[STIX]{x1D711})(s)].\end{eqnarray}$$
Alternatively,
$$\begin{eqnarray}f(\unicode[STIX]{x1D703})=\lim _{s\rightarrow 1}\biggl(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}s}\biggr)^{k}\biggl[\frac{2^{-k}\unicode[STIX]{x1D70B}^{-k/2}}{\unicode[STIX]{x1D6E4}(k/2)}\int _{0}^{s}(s^{2}-r^{2})^{k/2-1}(F_{k,\unicode[STIX]{x1D703}}^{\ast }\unicode[STIX]{x1D711})(r)\,dr\biggr].\end{eqnarray}$$
The limit in these formulas is understood in the
$L^{p}$
-norm.
3 The case
$k<n-1$
If
$k<n-1$
, the inversion problem for
$F_{k}^{\pm }$
and
$F_{k}$
is overdetermined because the dimension of the target space is greater than the dimension of the source space: that is,
$$\begin{eqnarray}\dim G_{n,k}=k(n-k)>n-1=\dim S^{n-1}.\end{eqnarray}$$
Below we eliminate this overdeterminedness by choosing an
$(n-1)$
-dimensional submanifold
$\mathfrak{M}$
of
$G_{n,k}$
which is sufficient to reconstruct
$f$
from
$(F_{k}^{\pm }f)(\unicode[STIX]{x1D709})$
with
$\unicode[STIX]{x1D709}\in \mathfrak{M}$
. We proceed as in [Reference Rubin10].
Suppose that
$\{e_{1},\ldots ,e_{n}\}$
is a standard orthonormal basis in
$\mathbb{R}^{n}$
and denote
$$\begin{eqnarray}\displaystyle & \displaystyle \mathbb{R}^{n-k}=\mathbb{R}e_{1}\oplus \cdots \oplus \mathbb{R}e_{n-k},\qquad \mathbb{R}^{k}=\mathbb{R}e_{n-k+1}\oplus \cdots \oplus \mathbb{R}e_{n}, & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle & \displaystyle \mathbb{R}^{k+1}=\mathbb{R}e_{n-k}\oplus \mathbb{R}^{k},\qquad S^{k}=S^{n-1}\cap \mathbb{R}^{k+1},\qquad S_{\pm }^{k}=S_{\pm }^{n-1}\cap \mathbb{R}^{k+1}. & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
Given a point
$v\in S^{n-k-1}=S^{n-1}\cap \mathbb{R}^{n-k}$
, we fix an orthogonal transformation
$\unicode[STIX]{x1D6FE}_{v}$
in
$\mathbb{R}^{n-k}$
so that
$\unicode[STIX]{x1D6FE}_{v}e_{n-k}=v$
. Let
$$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6FE}}_{v}=\left[\begin{array}{@{}ll@{}}\unicode[STIX]{x1D6FE}_{v} & 0\\ 0 & I_{k}\end{array}\right],\end{eqnarray}$$
where
$I_{k}$
is the identity
$k\times k$
matrix. We denote by
$G_{k}(\tilde{\unicode[STIX]{x1D6FE}}_{v}\mathbb{R}^{k+1})$
the Grassmannian of all
$k$
-dimensional linear subspaces of
$\tilde{\unicode[STIX]{x1D6FE}}_{v}\mathbb{R}^{k+1}$
and set
$$\begin{eqnarray}\mathfrak{M}=\mathop{\bigcup }_{v\in S^{n-k-1}}G_{k}(\tilde{\unicode[STIX]{x1D6FE}}_{v}\mathbb{R}^{k+1}).\end{eqnarray}$$
The restrictions
$\tilde{F}_{k}^{\pm }f$
of
$F_{k}^{\pm }f$
onto
$\mathfrak{M}$
can be identified with functions on the set
$\tilde{S}_{n,k}=\{(v,w):v\in S^{n-k-1},w\in S^{k}\}$
. These functions are represented by the hemispherical integrals
$$\begin{eqnarray}(\tilde{F}_{k}^{\pm }f)(v,w)=\int _{\{\unicode[STIX]{x1D702}\in S_{\pm }^{k}:\unicode[STIX]{x1D702}\cdot w=0\}}f_{v}(\unicode[STIX]{x1D702})\,d_{w}\unicode[STIX]{x1D702},\quad f_{v}(\unicode[STIX]{x1D702})=f(\tilde{\unicode[STIX]{x1D6FE}}_{v}\unicode[STIX]{x1D702}),\end{eqnarray}$$
over
$(k-1)$
-dimensional hemispheres on
$S^{k}$
. Thus
$\tilde{F}_{k}^{\pm }$
can be inverted as in the previous section.
Let us explain the details. We equip
$\tilde{S}_{n,k}$
with the product measure
$dv\,dw$
, where
$dv$
and
$dw$
stand for the corresponding surface elements on
$S^{n-k-1}$
and
$S^{k}$
. Using [Reference Rubin10, Theorem 3.2] and taking into account that the restrictions of
$f_{v}$
onto
$S_{\pm }^{k}$
and their even extensions (cf. (2.3) and (2.4)) belong to the same
$L^{p}$
spaces, we obtain the following existence result.
Theorem 3.1. Let
$1\leqslant k<n-1$
,
$f\in L^{p}(S^{n})$
and
$n-k<p\leqslant \infty$
. Then
$(\tilde{F}_{k}^{\pm }f)(v,w)$
are finite for almost all
$(v,w)\in \tilde{S}_{n,k}$
. If
$p\leqslant n-k$
, then there are functions
$\tilde{f}_{\pm }\in L^{p}(S^{n-1})$
for which
$(\tilde{F}_{k}^{\pm }f_{\pm })(v,w)=\infty$
.
Due to (2.5), to reconstruct
$f$
from
$\unicode[STIX]{x1D711}_{v}^{\pm }=\tilde{{\mathcal{F}}}_{k}^{\pm }f$
, it suffices to invert the usual codimension one Funk transform
$\tilde{F}$
on
$S^{k}$
. This gives
$$\begin{eqnarray}f_{v}(\unicode[STIX]{x1D702})=\left\{\begin{array}{@{}ll@{}}2(\tilde{F}^{-1}\unicode[STIX]{x1D711}_{v}^{+})(\unicode[STIX]{x1D702}) & \text{if}~\unicode[STIX]{x1D702}\in S_{+}^{k},\\ 2(\tilde{F}^{-1}\unicode[STIX]{x1D711}_{v}^{-})(\unicode[STIX]{x1D702}) & \text{if}~\unicode[STIX]{x1D702}\in S_{-}^{k}.\end{array}\right.\end{eqnarray}$$
If
$f$
is a continuous function, its value at a point
$\unicode[STIX]{x1D703}\in S^{n-1}$
can be found as follows. We interpret
$\unicode[STIX]{x1D703}$
as a column vector
$\unicode[STIX]{x1D703}=(\unicode[STIX]{x1D703}_{1},\ldots ,\unicode[STIX]{x1D703}_{n+1})^{T}$
and set
$$\begin{eqnarray}\unicode[STIX]{x1D703}^{\prime }=(\unicode[STIX]{x1D703}_{1},\ldots ,\unicode[STIX]{x1D703}_{n-k})^{T}\in \mathbb{R}^{n-k},\qquad \unicode[STIX]{x1D703}^{\prime \prime }=(\unicode[STIX]{x1D703}_{n-k+1},\ldots ,\unicode[STIX]{x1D703}_{n})^{T}\in \mathbb{R}^{k}.\end{eqnarray}$$
Suppose that
$\unicode[STIX]{x1D703}^{\prime }\neq 0$
and set
$$\begin{eqnarray}v=\unicode[STIX]{x1D703}^{\prime }/|\unicode[STIX]{x1D703}^{\prime }|\in S^{n-k-1},\qquad \unicode[STIX]{x1D702}=(0,\ldots ,0,|\unicode[STIX]{x1D703}^{\prime }|,\unicode[STIX]{x1D703}^{\prime \prime })^{T}\in S^{k}.\end{eqnarray}$$
Then
$\tilde{\unicode[STIX]{x1D6FE}}_{v}\unicode[STIX]{x1D702}=\unicode[STIX]{x1D703}$
and we get
$f(\unicode[STIX]{x1D703})=({\mathcal{F}}^{-1}\unicode[STIX]{x1D711}_{v})(\unicode[STIX]{x1D702})$
for
$v$
and
$\unicode[STIX]{x1D702}$
, as in (3.7). If
$\unicode[STIX]{x1D703}^{\prime }=0$
, then
$f(\unicode[STIX]{x1D703})$
can be reconstructed by continuity from its values at the neighboring points.
If
$f$
is an arbitrary function in
$L^{p}(S^{n})$
,
$n-k<p\leqslant \infty$
, then
$f$
can be explicitly reconstructed at almost all points on almost all spheres
$S_{v}^{k}=\tilde{\unicode[STIX]{x1D6FE}}_{v}S^{k}$
by making use of known inversion formulas for the Funk transform on this class of functions (see, e.g., [Reference Rubin10, p. 297]).
The above reasoning is obviously applicable to reconstruction of the radial function
$\unicode[STIX]{x1D70C}_{K}$
of the star body
$K$
from the volumes
$v^{\pm }(\unicode[STIX]{x1D709})=\text{vol}_{k}(K\cap \unicode[STIX]{x1D709}_{\pm })$
with
$\unicode[STIX]{x1D709}\in \mathfrak{M}$
, as in Corollary 2.2.
Acknowledgements
The author is grateful to professors Richard Gardner, Paul Goodey and Wolfgang Weil for useful discussions.
