2.1 Convex bodies

Let K be a nonempty convex compact set. Its support function $h_K:\mathbb {R}^n \to \mathbb {R}$ is defined by

$$ \begin{align*}h_K(u)=\max\{x\cdot u:x\in K\}.\end{align*} $$

Sometimes, we also view it as a spherical function by restricting it to $\mathbb {S}^{n-1}$ . The support function is always sublinear. Conversely, a sublinear function uniquely determines a nonempty convex compact set.

The Hausdorff distance between two nonempty compact convex sets $K_1,K_2$ is defined by

$$ \begin{align*}d_H(K_1,K_2) = \|h_{K_1}-h_{K_2}\|_\infty.\end{align*} $$

Given a function $f\in C(\mathbb {S}^{n-1})$ , denote $W[f]$ to be the Wullf shape determined by f:

$$\begin{align*}W[f] = \Big\{x\in\mathbb{R}^n: x\cdot u \le f(u), \forall u\in \mathbb{S}^{n-1}\Big\}.\end{align*}$$

If $W[f]$ has interiors, and $f_i\in C(\mathbb {S}^{n-1}), i=1,\ldots ,$ is a sequence of continuous functions that converges to $ f$ uniformly, then $W[f_i]\to W[f]$ in the Hausdorff metric.

We say the surface area measure $S(K,\cdot )$ has a density, if it is absolutely continuous with respect to the spherical Lebesgue measure. In this case, we denote the Radon–Nikodym derivative by

$$\begin{align*}F_K(u) = \frac{dS(K,u)}{du},\end{align*}$$

and we call $F_K$ the curvature function.

Given a function $h\in C^2(\mathbb {S}^{n-1}),$ denote $D^2h$ by

$$\begin{align*}D^2h = \nabla^2 h + hI,\end{align*}$$

where $\nabla ^2 h$ is the Hessian of h with respect to the standard metric on $\mathbb {S}^{n-1}$ , and I denotes the identity map. If one extend the h to a positive-homogeneous function on $\mathbb {R}^n$ , then $D^2h(u)$ is exactly the restriction of its Hessian in $\mathbb {R}^n$ to the tangent space of $\mathbb {S}^{n-1}$ at u.

In fact, if $\partial K$ is $C^2$ and has positive Gauss curvature everywhere, then $S(K,\cdot )$ has a density,

$$ \begin{align*}F_K(u)=\frac{dS(K,u)}{du}= \det(D^2 h_K),\end{align*} $$

which is the reciprocal Gauss curvature at $\nu _K^{-1}(u)\in \partial K.$

2.2 Minkowski problem and the Monge–Ampère equation

Solution to the classical Minkowski problem: Suppose $\mu$ is a finite Borel measure on satisfying:

(1) $\mu $ is not concentrated in a half-sphere;

(2) the centroid of $\mu $ is the origin:

$$\begin{align*}\int_{\mathbb{S}^{n-1}} u d\mu(u) = 0.\end{align*}$$

Then, there is a unique (up to translations) convex body K in $\mathbb R^n$ so that $\mu (\cdot ) =S(K, \cdot )$ . Moreover, the convex body K is homothetic to the solution of the maximization problem

$$\begin{align*}\max V(Q), \quad \mathrm{subject~to~that}~ Q~\mathrm{is~a~convex~body ~and~}\int_{\mathbb{S}^{n-1}} h_Q d\mu =1.\end{align*}$$

If the measure $\mu $ has a density function $g :\mathbb S^{n-1}\rightarrow \mathbb R$ with respect to the Lebesgue measure of the unit sphere $\mathbb {S}^{n-1}$ , the Minkowski problem is equivalent to the study of solution to the following Monge–Ampère equation on $\mathbb S^{n-1}$ :

(2.1) $$ \begin{align} \det (D^2h) = \det(\nabla^2 h +h I)=g. \end{align} $$

Let $f\in L^1(\mathbb {S}^{n-1})$ be nonnegative, satisfying $\int _{\mathbb {S}^{n-1}} f>0$ and

$$\begin{align*}\int_{\mathbb{S}^{n-1}} uf(u)du =0.\end{align*}$$

Then, by the solution of the Minkowski problem, there is always a convex body K such that

$$\begin{align*}dS(K,u) = f(u)du.\end{align*}$$

2.3 Mixed discriminant and mixed volume

Let $A_1,\ldots ,A_m$ be self-adjoint linear transforms from a k-dimensional Hilbert V to itself. Then, the following is a k-homogeneous polynomial of $t_1,\ldots ,t_m \in \mathbb {R}$ :

$$\begin{align*}\det \Big(t_1A_1+\cdots t_mA_m\Big) = \sum_{i_1=1}^{m}\ldots\sum_{i_{k}=1}^{m} \mathrm{D} (A_{i_1},\ldots,A_{i_{k}})t_{i_1}\ldots t_{i_{k}}.\end{align*}$$

Its coefficients $\mathrm {D} (A_{i_1},\ldots ,A_{i_{k}})$ , which are symmetric functions of $A_{i_1},\ldots ,A_{i_{k}}$ , are called the mixed discriminants of the linear transforms.

If we choose an orthonormal basis in V, these $A_i$ can be represented by symmetric $k\times k$ matrices, and we still denote them by $A_i$ . Denote the matrices $A_{r}=\left (a_{i j}^{r}\right )_{i, j=1}^{k}, r=1, \ldots , k$ , and the mixed discriminant $\mathrm {D} (A_1,\ldots ,A_k)$ can be computed by

(2.2) $$ \begin{align} \mathrm{D} \left(A_1, \ldots, A_{k}\right)=\frac{1}{k!} \sum_{\sigma \in P_{k}}\left|\begin{array}{ccc} a_{11}^{\sigma(1)} & \cdots & a_{1 k}^{\sigma(k)} \\ \vdots & & \vdots \\ a_{k1}^{\sigma(1)} & \cdots & a_{k k}^{\sigma(k)} \end{array}\right|, \end{align} $$

where $P_k$ is the group of all permutations of the set $\{1,\ldots ,k\}.$ Clearly, $\mathrm {D} (A_1,\ldots ,A_k)$ is linear in each argument.

Let $K_1,\ldots ,K_m$ be convex bodies, and let $\lambda _1,\ldots ,\lambda _m\ge 0$ . Then, the volume of $\lambda _1K_1+\cdots \lambda _m K_m$ is an nth homogeneous polynomial of $\lambda _1,\ldots ,\lambda _m$ ,

$$ \begin{align*}V\left(\lambda_{1} K_{1}+\cdots+\lambda_{m} K_{m}\right)=\sum_{i_{1}, \ldots, i_{n}=1} \mathrm{V} \left(K_{i_{1}}, \ldots, K_{i_{n}}\right) \lambda_{i_{1}} \cdots \lambda_{i_{n}}. \end{align*} $$

The coefficients $\mathrm {V} \left (K_{i_{1}}, \ldots , K_{i_{n}}\right )$ , which will be called the mixed volumes, are nonnegative, symmetric in the indices, and dependent only on $K_{i_{1}}, \ldots , K_{i_{n}}$ .

If $\partial K$ is $C^2$ and has positive Gauss curvature everywhere, then

$$\begin{align*}V(K) = \frac1n \int_{\mathbb{S}^{n-1}} h_K(u) \det (D^2h_K)(u) du.\end{align*}$$

In addition, if $h_K=t_1h_1+\cdots t_mh_m$ , for some $h_i\in C^2$ and $t_i\in \mathbb {R}$ , we have

(2.3) $$ \begin{align}\qquad V(K) = \sum_{i_1=1}^{m}\ldots \sum_{i_n=1}^{m} \Big(\frac1n\int_{\mathbb{S}^{n-1}} h_{i_n}(u) \mathrm{D} (D^2h_{i_1},\ldots,D^2h_{i_{n-1}})(u)du\Big) t_{i_1}\ldots t_{i_{n}}.\end{align} $$

Here, $\mathrm {D} (D^2h_{i_1},\ldots ,D^2h_{i_{n-1}})(u)$ should be understood as the mixed discriminant of linear transforms restricted on $u^\perp $ .

Especially, notice that the spherical harmonics $(Y_{m,j})$ belong to $C^\infty (\mathbb {S}^{n-1})$ . When we are considering the numerical algorithm in Section 3, we represent the support function in finite-dimensional function space generated by the spherical harmonics in each step N, and (2.3) will be useful. Moreover, (2.3) also suggests the relation between mixed discriminant and mixed volume, which goes back to Hilbert–Minkowski–Aleksandrov (see [Reference Alexandrov1] for reference).

2.4 Sobolev space $H_k$

Let $k\ge 0$ be an integer. We denote $H_k(\mathbb S^{n-1})$ to be the Sobolev space, which is the completion of $C^\infty (\mathbb {S}^{n-1})$ with respect to the norm

$$\begin{align*}\|f\|_{H_k}=\sum_{m=0}^{k}\left(\int_{\mathbb{S}^{n-1}}\left|\nabla^{m} f\right|{}^{2} d u\right)^{1/2}, \quad f\in C^\infty(\mathbb{S}^{n-1}),\end{align*}$$

where $\nabla ^m f$ denotes the mth covariant derivative of f. For a function $g\in H_k(\mathbb {S}^{n-1})$ and $m\le k$ , $\nabla ^m g\in L^2(\mathbb {S}^{n-1})$ denotes its weak covariant derivative, which can be defined as the limit of $\nabla ^m f_k$ in $L^2$ , where $f_k\in C^\infty (\mathbb {S}^{n-1})$ ( $k\in \mathbb {N}$ ) is an arbitrary Cauchy sequence with respect to the norm $\|\cdot \|_{H_k}$ . See [Reference Hebey18] for reference.

Let $\Delta _S: C^\infty (\mathbb {S}^{n-1})\to C^\infty (\mathbb {S}^{n-1})$ be the Laplace operator on $\mathbb {S}^{n-1}$ , which is self-adjoint. Extend it to $\Delta _S: H_{k+2}(\mathbb {S}^{n-1})\to H_k(\mathbb {S}^{n-1}).$ Then, for $f,g\in H_{k+2}(\mathbb {S}^{n-1})$ , it satisfies

$$\begin{align*}(\Delta_S f, g) = (f,\Delta_S g).\end{align*}$$

Note that $(\cdot ,\cdot )$ always denotes the scalar product in $L^2(\mathbb {S}^{n-1})$ in this paper.

2.5 Spherical harmonics

The spherical harmonics are the eigenvectors (eigenfunctions) of spherical Laplace-Beltrami operator $\Delta _S:C^\infty (\mathbb {S}^{n-1})\to C^\infty (\mathbb {S}^{n-1})$ . It is well known that these eigenvectors can be chosen to be the restriction onto $\mathbb {S}^{n-1}$ of the harmonic polynomials (with respect to an orthonormal basis $e_1,\ldots ,e_n$ of $\mathbb {R}^n$ ) on $\mathbb {R}^n\backslash \{0\}.$ The eigenvalues of $\Delta _S$ are

$$\begin{align*}\lambda_m = -m(m+n-2),\quad m=0,1,\ldots.\end{align*}$$

With regard to $\lambda _m$ , the related eigenspace $E_m$ consists of all the harmonic polynomials on $\mathbb {R}^n\backslash \{0\}$ of degree m. The dimension of $E_m$ is

$$\begin{align*}N_m =\frac{(2m+n-2)\cdot(n+m-3)!}{(n-2)!\cdot m!}.\end{align*}$$

The compact operator theory in functional analysis tells that one can choose the eigenvectors to be a Hilbert basis of $L^2(\mathbb {S}^{n-1}).$ More precisely, we can choose $(Y_{m,j})$ such that the eigenspace of $\lambda _m$ is

$$\begin{align*}E_m = \mathrm{span} \Big\{Y_{m,1},\ldots,Y_{m,N_m} \Big\}\end{align*}$$

and

$$\begin{align*}\big(Y_{m,j},Y_{m,k}\big)=\int_{\mathbb{S}^{n-1}} Y_{m,j}(u)Y_{m,k}(u) du=\delta_{j}^{k}.\end{align*}$$

For distinct $m,m'$ , we automatically have $\big (Y_{m,j},Y_{m',j'}\big ) = 0.$ Especially, when $m=1$ , the eigenfunctions $\{Y_{1,j}\}_{j=1}^n$ are chosen to be

$$\begin{align*}Y_{1,j}(u) = \frac1{\sqrt{\omega_n}} e_j\cdot u,\quad \quad j=1,\ldots,n,\end{align*}$$

so that $(Y_{1,k},Y_{1,j})=\delta _j^k.$ From this, it is easy to obtain the following formula that describes the barycenter of an integrable function g:

(2.4) $$ \begin{align} v\cdot\Big( \frac1{\omega_n} \int_{\mathbb{S}^{n-1}} u g(u)du\Big) =& \sum_{j=1}^n \Big[\int_{\mathbb{S}^{n-1}}\Big(\frac{e_j}{\sqrt{\omega_n}}\cdot u\Big) g(u) du\Big] \Big(\frac{e_j}{\sqrt{\omega_n}}\cdot v\Big)\notag\\ =& \sum_{j=1}^n (g,Y_{1,j})Y_{1,j}(v). \end{align} $$

Since $(Y_{m,j})$ is a Hilbert basis, for each $g\in L^2(\mathbb S^{n-1})$ , we have

$$ \begin{align*}g=\sum_{m=0}^\infty\sum_{j=1}^{N_m}\left(g, Y_{m,j}\right) Y_{m,j}.\end{align*} $$

3.1 Optimization problem ( $\overline P_N$ )

Originally, our aim in Step N is to solve the following optimization problem ( $P_N$ ):

$$\begin{align*}\max ~ V(x_0,\ldots,x_N), \end{align*}$$

$$\begin{align*}\mathrm{subject~to} \quad(x_0,\ldots,x_N)\in \overline{\mathcal{C}}_N ,\end{align*}$$

where $\mathcal {C}_N$ is a convex set (will be proved later) defined by

$$\begin{align*}\overline{\mathcal{C}}_N = \left\{ (x_0,\ldots,x_N)\in \mathbb{R}^{N+1}: { D^2\left(\sum_{m=0}^N x_m Y_m \right)(u) \geq 0,~ \forall u\in\mathbb{S}^{n-1}}, \sum_{m=0}^N \mathrm{V}_m\cdot x_m =1 \right\}.\end{align*}$$

Solve the optimization problem ( $\overline P_N$ ) and normalize the solution.

However, the problem ( $P_N$ ) cannot be solved directly by computer, since the requirement $(x_0,\ldots ,x_N)\in \overline {\mathcal {C}}_N $ has actually infinitely many inequality constraints, although it can be shown to be a convex optimization problem. Therefore, we consider the following optimization problem ( $P_{N,k}$ ), for $N,k\in \mathbb {N}$ . We will show in Theorem 3.3 that the subproblem ( $P_{N,k}$ ) is also a convex optimization problem with only finitely many constraints when k is large, and it follows from Theorems 3.3 and 3.5 that its solution will converge to the solution of the Minkowski problem as $N,k\to \infty $ .

Now, we put our computational procedure Mink-Pro and the subproblem ( $P_{N,k}$ ) as follows.

For each $k\in \mathbb {N}$ , we assign a kth partition $T_k = \{O_1^k,\ldots ,O_{2^k}^k\}$ of $\mathbb {S}^{n-1}$ such that

(3.3) $$ \begin{align} O_i^k \subset \mathbb{S}^{n-1}, \quad \mathcal{H}^{n-1}(O_i^k) = \frac{n\omega_n}{2^k}, \quad \bigcup_{i=1}^{2^k} O_i^k = \mathbb{S}^{n-1}, \quad \mathrm{and} \quad T_{k+1} \subset T_k.\end{align} $$

Extract one point from each $O_i^k$ to obtain a set $U_k\subset \mathbb {S}^{n-1}$ such that $U_k$ contains $2^k$ elements and

(3.4) $$ \begin{align}U_{k+1}\subset U_k.\end{align} $$

Define

$$\begin{align*}\mathcal{C}_{N} \kern1pt{=}\kern1pt \Big\{ (x_0,\ldots,x_N)\in \mathbb{R}^{N+1} : D^2\left(\sum_{i=0}^N x_i Y_i\right) (u) \kern1pt{\geq}\kern1pt \frac IN , ~ \forall u\kern1pt{\in}\kern1pt \mathbb{S}^{n-1}, \mathrm{~and~}~ \sum_{m=0}^N \mathrm{V}_m\cdot x_m \kern1pt{=}\kern1.5pt 1 \!\Big\},\end{align*}$$

and

(3.5) $$ \begin{align}\mathcal{C}_{N,k} \kern1pt{=}\kern1pt \Big\{ (x_0,\ldots,x_N)\in \mathbb{R}^{N+1} : D^2\left(\sum_{i=0}^N x_i Y_i\right) (u) \kern1pt{\geq}\kern1pt \frac IN , ~ \forall u\kern1pt{\in}\kern1pt U_k, \mathrm{~and~}~ \sum_{m=0}^N \mathrm{V}_m\cdot x_m \kern1pt{=}\kern1.5pt 1 \!\Big\}. \end{align} $$

Clearly, $\mathcal {C}_{N,k+1}\subset \mathcal {C}_{N,k}$ , and

$$\begin{align*}\bigcap_{k=1}^\infty\mathcal{C}_{N,k} = \mathcal{C}_{N}.\end{align*}$$

Consider the modified optimization problem ( $ P_N^*$ ):

$$\begin{align*}\max ~ V(x_0,\ldots,x_N), \end{align*}$$

$$\begin{align*}\mathrm{subject~to} \quad(x_0,\ldots,x_N)\in \mathcal{C}_N.\end{align*}$$

Denote its solution by $(x_0^*,\ldots ,x_N^*)$ , and define

(3.6) $$ \begin{align}h_{N} = \left(\sum_{i=0}^N x_i^*Y_i\right) V(x_0^*,\ldots,x_N^*)^{- 1/({n-1})}.\end{align} $$

3.2 Minkowski Procedure

• • Input: Natural numbers $n,N,k$ , the general mixed volumes $\mathrm {V}_{i_1,\ldots ,i_n}$ and $ \mathrm {V}_i$ for the spherical harmonics $Y_0,\ldots ,Y_N$ defined by (3.1) and (3.2), the finite set $U_k$ satisfying (3.4) and (3.3).

• • Task: Compute the function $h_{N,k}$ .

• • Action:

  1. 1. Solve the following convex optimization problem ( $P_{N,k}$ ):

     $$\begin{align*}\sup ~ V (x_0,\ldots,x_N), \end{align*}$$
     $$\begin{align*}\mathrm{subject~to} \quad(x_0,\ldots,x_N)\in \mathcal{C}_{N,k}.\end{align*}$$

     Denote the solution of ( $P_{N,k}$ ) by $(x_0^k,\ldots ,x_N^k).$

  2. 2. Compute

     $$\begin{align*}h_{N,k} = \left(\sum_{i=0}^N x_i^kY_i\right) V(x_0^k,\ldots,x_N^k)^{- 1/({n-1})}.\end{align*}$$

Remark 3.1 1. We will show in Theorem 3.3 that ( $P_{N,k}$ ) and ( $P_{N}$ ) are convex optimization problems, for sufficiently large k. Thus, the maximizers $(x_0^k,\ldots ,x_0^k)$ and $(x_0^*,\ldots ,x_0^*)$ (theoretically) exist and are unique.

Especially, the approach to find the solution of ( $P_{N,k}$ ) is computationally tractable, since ( $P_{N,k}$ ) has only finitely many constraints. To write the detailed algorithm for the approach, see [Reference Boyd and Vandenberghe7, Reference Nocedal and Wright25] for references.

2. We will prove in Theorem 3.3(6) that the convex body, determined by $h_{N,k}$ ,

$$\begin{align*}K_{N,k}= \Big\{ z : z\cdot u \leq h_{N,k}(u), \quad \forall u\in\mathbb{S}^{n-1}\Big\}\end{align*}$$

converges to $K_N$ as $k\to \infty $ . In addition, we will prove in Theorem 3.5 that $K_N\rightarrow K$ , when $N\to \infty $ , where $K_N$ is the convex body determined by $h_N$ , and $S(K,\cdot ) = \mu $ .

3. Recalling equation (2.3), $V(x_0,\ldots ,x_N)$ is actually the volume of the convex body Q whose support function is

$$\begin{align*}h_Q=\sum_{i=0}^N x_iY_i,\end{align*}$$

for $(x_0,\ldots ,x_N)\in \mathcal {C}_{N}$ . The quantity $\sum _{i=0}^N \mathrm {V}_i\cdot x_i$ is $\int h_Q d\mu .$ This explains that Action 1 coincides with the optimization problem in solving the Minkowski problem. However, they are not completely the same, and we have to prove the convergence.

4. Although our procedure can reconstruct the body from a general measure, in practice, the smooth functions in the form

$$\begin{align*}F(u) = \sum_{i=0}^{k_0} z_i Y_i(u)\end{align*}$$

are more tractable. Since the $1$ -degree spherical harmonics $Y_{1,1},\ldots ,Y_{1,n}$ are removed from $Y_i$ ’s, such a function F always satisfies $\int _{\mathbb {S}^{n-1}} uF(u) du =0$ .

Lemma 3.1 Suppose K is a convex body such that

$$\begin{align*}\int_{\mathbb{S}^{n-1}} h_K(u)d\mu(u) = 1\end{align*}$$

and

(3.7) $$ \begin{align}\int_{\mathbb{S}^{n-1}} uh_K(u)du = 0.\end{align} $$

Then, there is a constant $c_1$ that only depends on $\mu $ , such that $\max \{|x|: x\in K\} \le c_1.$

Proof Since $\mu $ is not concentrated in any hemisphere, the function

$$\begin{align*}v \mapsto\int_{S^{n-1}} (u\cdot v)_+ d\mu(u) \end{align*}$$

is strictly positive on $S^{n-1}$ , and since the function is continuous on $S^{n-1}$ , which is compact, there exists $c_0>0$ such that

$$\begin{align*}\int_{S^{n-1}} (u\cdot v)_+ d\mu(u) \geq c_0> 0\qquad\text{for all}\ v\in S^{n-1}. \end{align*}$$

Denote $R_0 = \max \{|x| : x\in \bar K_N \}.$ Then, there exists $v_0\in \mathbb {S}^{n-1}$ such that $R_0\cdot v_0\in K.$ By (3.7), we must have $o\in K.$ Thus, we obtain

$$\begin{align*}R_0\cdot c_0\le\int_{\mathbb{S}^{n-1}} R_0( v_0 \cdot u)_+ d\mu(u) \leq \int_{\mathbb{S}^{n-1}}h_{K}(u)d\mu(u)= 1,\end{align*}$$

and hence $R_0\leq 1/c_0$ . Denote $c_1=1/c_0$ , and we complete the proof.

Proof (1) Since $Y_0$ is a constant, $(c,0,\ldots ,0)\in \mathcal {C}_N$ for some sufficiently large constant c. Since $\mathcal {C}_N \subset \overline {\mathcal {C}}_N$ and $\mathcal {C}_N\subset \mathcal {C}_{N,k}$ , they are all nonempty. Since $ D^2\Big (\sum _{i=0}^N x_iY_i\Big ) (u)$ is continuous in $(x_0,\ldots ,x_N)$ , $\overline {\mathcal {C}}_N, \mathcal {C}_N$ , and $\mathcal {C}_{N,k}$ are closed.

If, at some $u\in \mathbb {S}^{n-1}$ ,

$$\begin{align*}D^2\left(\sum_{i=0}^N x_iY_i\right) (u) \quad \mathrm{~and~} \quad D^2\left(\sum_{i=0}^N y_iY_i\right) (u)\end{align*}$$

are positive semidefinite, then

$$\begin{align*}D^2\left(\sum_{i=0}^N \big((1-\lambda)x_i+\lambda y_i\right)Y_i\Big) (u)\end{align*}$$

is also positive semidefinite at this u, $\forall \lambda \in [0,1]$ . It follows immediately that $\overline {\mathcal {C}}_N, \mathcal {C}_N$ , and $\mathcal {C}_{N,k}$ are convex.

(2) By Lemma 3.1, for any $(x_0,\ldots ,x_N)\in \mathcal {C}_N$ , we have

$$\begin{align*}|h_Q(u)| = \left|\sum_{m=0}^N x_mY_m(u)\right| \leq c_1, \quad \forall u\in\mathbb{S}^{n-1},\end{align*}$$

where $c_1$ is a constant that only depends on $\mu $ . Then,

$$ \begin{align*}|x_m|=|(h_Q, Y_m)| = \Big|\int_{\mathbb{S}^{n-1}} h_Q(u) Y_m(u) du\Big| \le c_1 \|Y_m\|_1.\end{align*} $$

Since $Y_m\in C^\infty (\mathbb {S}^{n-1})$ are fixed functions, we deduce that $\overline {\mathcal {C}}_N$ is bounded.

Proof (1) We will use the fact that a smooth support function is either the support function of a convex body or the support function of a fixed point. By our construction, $\sum _{i=0}^N x_i Y_i$ is always smooth, and $\{Y_i\}_{i=0}^N$ does not contain the spherical harmonics of degree $1$ . Thus, for $(x_0,\ldots ,x_N)\in \mathcal {C}_N$ , it cannot be the support function of a fixed point. Thus, V is always positive.

(2) Since $(x_0,\ldots ,x_N)\in \mathcal {C}_N$ ,

$$\begin{align*}\sum_{i=0}^N x_iY_i(u)\end{align*}$$

is a support function of some convex compact set, which we denote by $K(x_0,\ldots ,x_N)$ . Now,

$$ \begin{align*}V(x_0,\ldots,x_N) = V\big(K(x_0,\ldots,x_N)\big)\end{align*} $$

is its volume. Notice also that, for $(y_0,\ldots ,y_N)\in \mathcal {C}_N$ and $\lambda \in [0,1],$

$$\begin{align*}(1-\lambda)K(x_0,\ldots,x_N) + \lambda K(y_0,\ldots,y_N) = K\big((1-\lambda) x_0+ \lambda y_0,\ldots,(1-\lambda) x_N+ \lambda y_N\big).\end{align*}$$

Thus, statement (2) follows immediately from the Brunn–Minkowski inequality, statement (1), and the fact that

$$\begin{align*}\int_{\mathbb{S}^{n-1}} u h_{K(x_0,\ldots,x_N)}du = \int_{\mathbb{S}^{n-1}} uh_{K(y_0,\ldots,y_N)}(u)du =0.\end{align*}$$

(3) Since $\mathcal {C}_{N,k}$ is convex and closed for each $k\in \mathbb {N}$ ,

$$\begin{align*}\bigcap_{k=1}^\infty\mathcal{C}_{N,k} = \mathcal{C}_{N},\end{align*}$$

and $\mathcal {C}_N$ is compact, statement (3) follows immediately.

(4) Since $\mathcal {C}_{N,k} \to \mathcal {C}_N\subset \mathrm {int} \overline {\mathcal {C}}_N$ , and $ \overline {\mathcal {C}}_N$ is compact, V is still positive on $\mathcal {C}_{N,k}$ when k is sufficiently large. Moreover, $\mathcal {C}_{N,k}$ will also be compact. Recall that

$$\begin{align*}V(x_0,\ldots,x_N) = \sum_{i_1=0}^N\ldots\sum_{i_n=0}^N \mathrm{V}_{i_1,\ldots,i_n}\cdot x_{i_1}\ldots x_{i_n}\end{align*}$$

is a polynomial of degree n, where $\mathrm {V}_{i_1,\ldots ,i_n}$ are fixed constants. Thus, $V(\cdot ) C^\infty (\mathbb {R}^{N+1})$ . By the construction of $\mathcal {C}_{N,k}$ , when k is sufficiently large, there is a constant $\theta>0,$ such that the Hessian

$$ \begin{align*}D^2(V^{1/n})(x_0,\ldots,x_n)\ge \theta I_{N+1}\end{align*} $$

everywhere on $\mathcal {C}_{N,k}$ .

(5) As a consequence of (3), for sufficiently large k, $\mathcal {C}_{N,k}$ is also compact. Thus, any subsequence of $(x_0^k,\ldots ,x_N^k)$ has a subsubsequence, which is also denoted by $(x_0^k,\ldots ,x_N^k)$ , converging to a point $(\bar x_0,\ldots ,\bar x_N) \in \bigcap _k \mathcal {C}_{N,k} =\mathcal {C}_N$ . It suffices to show that $(\bar x_0,\ldots ,\bar x_N)=(x_0^*,\ldots ,x_N^*)$ .

Since $(x_0^*,\ldots ,x_N^*)\in \mathcal {C}_{N,k}$ , for each $k\in \mathbb {N}$ , and $(x_0^k,\ldots ,x_N^k)$ is the maximizer of ( $P_{N,k}$ ) for sufficiently large k, we have

(3.8) $$ \begin{align} V (x_0^k,\ldots,x_N^k) \geq V (x_0^*,\ldots,x_N^*). \end{align} $$

Taking limits in (3.8), we obtain

$$\begin{align*}V (\bar x_0,\ldots,\bar x_N) \geq V (x_0^*,\ldots,x_N^*).\end{align*}$$

Since $(\bar x_0,\ldots ,\bar x_N)\in \mathcal {C}_N$ , and $(x_0^*,\ldots ,x_N^*)$ is the unique maximizer of ( $P_N^*$ ), we must have $(\bar x_0,\ldots ,\bar x_N)=(x_0^*,\ldots ,x_N^*)$ . This completes the proof.

(6) By (5), we have

$$\begin{align*}\sum_{i=0}^N x_i^k Y_i \to \sum_{i=0}^N x_i^* Y_i\end{align*}$$

uniformly on $\mathbb {S}^{n-1}$ . Recall the fact collected in Section 2.1: $f_i\to f$ uniformly and that $W[f]$ has interior imply that $W[f_i]\to W[f]$ in the Hausdorff metric. Now, (6) follows immediately from this property, the continuity of volume, and the definitions of $h_N$ and $h_{N,k}$ .

Lemma 3.4 [Reference Seeley29, Theorems 4 and 5]

Suppose $F\in C^\infty (\mathbb {S}^{n-1})$ satisfying $\int F(u)udu =0.$ Denote

$$\begin{align*}F_N(u)= \sum_{m=0}^N (F,Y_{m})(u).\end{align*}$$

Then:

1. (1) $F_N$ converges to F uniformly on $\mathbb {S}^{n-1}$ ;

2. (2) $D^2F_N$ converges to $D^2F$ uniformly on $\mathbb {S}^{n-1}$ .

Remark 3.2 This Lemma follows immediately from the estimates in [Reference Seeley29, Theorems 4 and 5], which suggest the relationship between the max norm and the $L^2$ -norm of Fourier expansion into spherical harmonics. The assumption $\int F(u)udu =0$ is because our $\{Y_i\}_{i=0}^\infty $ does not contain the spherical harmonics $Y_{1,1},\ldots ,Y_{1,n}$ . Except for this, the only difference is that the $D^\alpha $ in [Reference Seeley29, Theorem 4(b)] is slightly different from ours.

Let $f\in C^\infty (\mathbb {S}^{n-1})$ . Denote $\bar D^2 f(x/|x|)$ to be the Euclidean Hessian, and $D^2 f$ to be our notation in Section 2.1. Then, restricting to $x^\perp $ ,

$$\begin{align*}\bar D^2 f(x/|x|) + f(x/|x|)I = D^2 f(x/|x|),\end{align*}$$

and $\bar D^2 f(x/|x|) =0$ on the line $\mathbb {R} x$ .

Theorem 3.5 Denote the solution of ( $P_N$ ) by $(x_0^*,\ldots ,x_N^*),$ and

$$\begin{align*}h_N = \left(\sum_{m=0}^N x_mY_m\right) V(x_0^*,x_1^*,\ldots,x_N^*)^{- 1/({n-1})}.\end{align*}$$

Let $K_N$ be the convex body determined by $h_N$ ,

$$\begin{align*}K_N = \{x: x\cdot u \le h_N(u),\ \forall u\in\mathbb{S}^{n-1}\}.\end{align*}$$

Then, $K_N$ converges to K in the Hausdorff metric, as $N\to \infty $ . Here, K is the unique (theoretical) solution of the Minkowski problem

$$\begin{align*}S(K,\cdot) = \mu(\cdot)\end{align*}$$

such that

$$\begin{align*}\int_{\mathbb{S}^{n-1}} uh_K(u)du=0.\end{align*}$$

Proof Denote

$$\begin{align*}h_{\bar K_N} = \sum_{m=0}^N x_m^*Y_m,\end{align*}$$

and denote $\bar K=\lambda K$ ( $\lambda>0$ ) to be the dilation of K such that $\int _{\mathbb {S}^{n-1}} h_{\bar K} d\mu = 1.$ To prove this theorem, it suffices to prove that $\bar K_N\to \bar K$ .

Step 1. Since $\{Y_m\}_{m=0}^\infty $ does not contain the $\{Y_{1,j}\}$ ’s, by equation (2.4), we always have

(3.9) $$ \begin{align}\int_{\mathbb{S}^{n-1}}uh_{\bar K_N}(u)du=0.\end{align} $$

Step 2. It follows immediately from Lemma 3.1 that the sequence $\{\bar K_N\}_{N=1}^\infty $ is bounded.

Step 3. By Step 2 and Blaschke’s selection theorem, any subsequence of $\{\bar K_N\}_{N=1}^\infty $ has a subsubsequence $\bar K_{N_i}$ converging to a convex body $K^*$ in the Hausdorff metric. In this step, we will show that $K^*$ satisfies

(3.10) $$ \begin{align}\qquad V(K^*) = \sup \Big\{V(Q): \int_{\mathbb{S}^{n-1}} h_Q(u)d\mu(u) =1,\ Q\ \mathrm{is\ a\ convex\ body}\Big\}, \end{align} $$

and

(3.11) $$ \begin{align}\int_{\mathbb{S}^{n-1}} h_{K^*}(u)d\mu(u)=1.\end{align} $$

The second equation follows immediately from the convergence. Let us consider (4.10).

• • Step 3.1. Suppose Q is a smooth convex body that has positive Gauss curvature, and $\int _{\mathbb {S}^{n-1}}h_Qd\mu =1$ . We will show that $V(K^*)\ge V(Q).$ By the assumption, $h_Q\in C^\infty (\mathbb {S}^{n-1})$ , and

  $$\begin{align*}\det (D^2h_Q)(u)> 0 , \quad \forall u\in\mathbb{S}^{n-1}.\end{align*}$$
  Denote
  $$\begin{align*}h_{Q_N} = \sum_{m=0}^N (h_Q,Y_{m})Y_m.\end{align*}$$
  Then, by Lemma 3.4(1), $h_{Q_N}$ converges to $h_Q$ uniformly. In addition, by Lemma 3.4(2), for sufficiently large N, we still have
  $$\begin{align*}\det (D^2h_{Q_N})(u)> 0, \quad \forall u\in\mathbb{S}^{n-1},\end{align*}$$
  and hence $h_{Q_N}$ is also support function. Therefore, if we denote
  $$\begin{align*}\bar Q_N =\frac1{\int_{\mathbb{S}^{n-1}}h_{Q_N}d\mu} Q_N, \end{align*}$$
  by the construction of $\bar K_N$ , we have $V(\bar K_N)\ge V(\bar Q_N)$ . Since $\bar K_N\to K^*$ , $\bar Q_N\to Q$ , we get
  $$\begin{align*}V(K^*)\ge V(Q).\end{align*}$$

• • Step 3.2. For any convex body L, we can choose a sequence of smooth convex bodies $Q_m$ that have positive curvatures, and such that $Q_m\to L$ . By the approximation argument, we complete the proof of (4.10).

By the solution of the Minkowski problem, the convex body satisfying (4.10) and (3.11) is unique up to translations. Recalling the fact that

$$\begin{align*}\int_{\mathbb{S}^{n-1}}uh_{\bar K}(u)du = \int_{\mathbb{S}^{n-1}}uh_{ K^*}(u)du =0,\end{align*}$$

we have $\bar K= K^*$ . Since we have proved that any subsequence of $\{\bar K_N\}$ has a convergent subsequence converging to $\bar K$ , we have actually completed the proof of this theorem.

4.1 Further applications of the Funk–Hecke theorem

The following is known as the Funk–Hecke theorem.

Lemma 4.1 [Reference Schneider28]

Let $f \in L_2[-1,1]$ be a function satisfying

$$ \begin{align*} \int^{1}_{-1}|f(t)|(1-t^2)^{\frac{n-3}{2}}<\infty. \end{align*} $$

If $Y_m \in \mathscr {F}^m$ is a spherical harmonic of degree m, then

$$\begin{align*} \int_{\mathbb{S}^{n-1}}f(u\cdot v) Y_m(u)du=\lambda_{m}(f)Y_m(v), \end{align*}$$

for $v\in \mathbb {S}^{n-1}$ , where

$$\begin{align*} \lambda_{m}(f)=(n-1)\omega_{n-1}[C_m^\nu(1)]^{-1}\int_{-1}^{1}f(t)C_m^\nu(t)(1-t^2)^{\frac{n-3}{2}}dt, \end{align*}$$

where

$$ \begin{align*}C_{m}^{\nu}(t)=a_{m}^{\nu}\left(1-t^{2}\right)^{-\nu+1 / 2}\left(\frac{d}{d t}\right)^{m}\left(1-t^{2}\right)^{m+\nu-1 / 2},\end{align*} $$

$$\begin{align*} a_{m}^{\nu}=C_{m}^{\nu}(1)\left(-\frac{1}{2}\right)^{m} \frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(m+\frac{n-1}{2}\right)},\quad \mathrm{and}\quad C_{m}^{\nu}(1)=\frac{\Gamma(m+n-2)}{\Gamma(n-2) \Gamma(m+1)}.\end{align*}$$

Lemma 4.2 [Reference Schneider28]

For the function $f(t)=|t|$ , the eigenvalue $\lambda _m(f)$ in the Funk–Hecke theorem is

$$ \begin{align*} \lambda_{m}(|t|)=\frac{(-1)^{\frac{m-2}{2}} \pi^{\frac{n-1}{2}}\Gamma{(m-1)}}{2^{m-2}\Gamma\left(\frac{m }{2}\right) \Gamma\left(\frac{m+n+1}{2}\right)}, \end{align*} $$

for even m, and $\lambda _{m}(|t|)=0$ for odd m. Furthermore, $|\lambda _m(|t|)^{-1}|=O\left (m^{(n+2)/2}\right )$ for even $m\rightarrow \infty .$

Since the lightness functions are defined in the form of partial cosine transform

$$\begin{align*}\mathcal{C}_- g(v) = \int_{\mathbb{S}^{n-1}}(v\cdot u)_-g(u)du,\end{align*}$$

we require the following Lemma to compute its eigenvalues.

Lemma 4.3 Let $f(t)=t_-$ , and denote $\beta _m $ to be the $\lambda _m(f)$ in the Funk–Hecke theorem. Then,

$$ \begin{align*} \beta_{m}=\frac{(-1)^{\frac{m-2}{2}} \pi^{\frac{n-1}{2}}\Gamma{(m-1)}}{2^{m-1}\Gamma\left(\frac{m }{2}\right) \Gamma\left(\frac{m+n+1}{2}\right)}, \end{align*} $$

for even m, $\beta _{m}=0$ if m is odd and $m\geq 3,$ and

$$ \begin{align*}\beta_1 =-\frac{\omega_n}2 \neq 0.\end{align*} $$

Applying Lemma 4.1, one can directly compute the coefficients $\beta _m$ , but it will be complicated when $m\ge 3$ is odd. However, we would like to provide the following simpler method, to understand the geometric fact that the partial cosine transform $\mathcal {C}_- g$ also defines a support function of a zonoid (which may not be symmetric with respect to the origin).

Proof Since $t_-= 1/2(|t| - t)$ , we have

(4.1) $$ \begin{align}\int_{\mathbb{S}^{n-1}} (v\cdot u)_- g(u) du =& \frac12\int_{\mathbb{S}^{n-1}} |v\cdot u| g(u) du - \frac12\int_{\mathbb{S}^{n-1}} (v\cdot u)g(u)du \notag\\ =& \frac12\int_{\mathbb{S}^{n-1}} |v\cdot u| g(u) du + v\cdot\Big( -\frac12\int_{\mathbb{S}^{n-1}} u g(u)du\Big). \end{align} $$

Denote $a_{m,j} = (g,Y_{m,j})$ . By the above equation and Lemmas 4.1 and 4.2, we have

(4.2) $$ \begin{align}\int_{\mathbb{S}^{n-1}} (v\cdot u)_- g(u) du = \frac12\sum_{m~\mathrm{ even}}\sum_{j=1}^{N_m}\lambda_m(|t|)a_{m,j}Y_{m,j}(v) - \frac12\int_{\mathbb{S}^{n-1}} (v\cdot u)g(u)du.\end{align} $$

Using Lemma 4.1 again, for the function $(v\cdot u)_-,$ we know that

(4.3) $$ \begin{align}\int_{\mathbb{S}^{n-1}} (v\cdot u)_- g(u) du = \sum_{m=0}^\infty\sum_{j=1}^{N_m}\beta_m a_{m,j}Y_{m,j}(v).\end{align} $$

By (2.4), we have

$$ \begin{align*} -\frac12\int_{\mathbb{S}^{n-1}} (v\cdot u) g(u)du =& -\frac{\omega_n}2\cdot\sum_{j=1}^n a_{1,j}Y_{1,j}(v).\end{align*} $$

By this and Lemma 4.2, comparing (4.2) with (4.3), we get the desired result.

Lemma 4.4 (Theorem 3.5.4 [Reference Schneider28])

If $\rho $ is an even signed measure on $\mathbb {S}^{n-1}$ with

$$ \begin{align*}\int_{\mathbb{S}^{n-1}}|(u\cdot v)| \mathrm{d} \rho(u)=0 \quad \forall~ v \in \mathbb{S}^{n-1}, \end{align*} $$

then $\rho =0$ .

If G is an even real function on $\mathbb {S}^{n-1}$ of differentiability class $C^k$ , where $k\geq n+2$ is even, then there exist an even continuous function g on $\mathbb {S}^{n-1}$ such that

(4.4) $$ \begin{align} G(v)=\int_{\mathbb{S}^{n-1}}|(u\cdot v)|g(u)du, \end{align} $$

for $v\in \mathbb {S}^{n-1}$ .

The above Lemma and its proof can be found in [Reference Schneider28]. Since we want to work with the nonsmooth convex bodies, we slightly improve this result to the functions in Sobolev spaces. Precisely, we consider the functions in ${H}_{k}(\mathbb {S}^{n-1})$ . Although the approach is standard, we did not find the result as follows. So, for completeness, we give a proof here.

Lemma 4.5 Let $k\ge n+1$ be an even integer. If $G \in H_k(\mathbb S^{n-1})$ is an even function, then there exists an even function $g\in L^2(\mathbb {S}^{n-1})$ such that

(4.5) $$ \begin{align} G(v)=\int_{\mathbb{S}^{n-1}}|(u\cdot v)| g(u)du, \end{align} $$

for all $v\in \mathbb {S}^{n-1}$ . Especially, g can be formulated by the convergent function series

(4.6) $$ \begin{align} \sum_{m~\mathrm{even}} \sum_{j=1}^{N_m} \lambda_m^{-1}(|t|)(G,Y_{m,j}) Y_{m,j}, \end{align} $$

where $\lambda _m^{-1}(|t|)$ is provided by the Funk–Hecke theorem.

Proof Denote $k=2l$ . Since $G\in H_k(\mathbb {S}^{n-1}),$ we have

$$ \begin{align*}G(v)= \sum_{m=0}^{\infty}\sum_{j=1}^{N_m}(G,Y_{m,j})Y_{m,j}(v).\end{align*} $$

We aim to prove that the series

(4.7) $$ \begin{align} \sum_{ ~m~\mathrm{even}} \lambda_m(|t|)^{-1}\left(\sum_{j=1}^{N_m}(G,Y_{m,j})Y_{m,j}\right) \end{align} $$

converges in $ L^2(\mathbb {S}^{n-1})$ .

Recall that

$$ \begin{align*}\Delta_{S} Y_{m,j} = -m(m+n-2)Y_{m,j}.\end{align*} $$

Since $\Delta _{S}$ is self-adjoint, we have

$$ \begin{align*} \left(G, Y_{m,j}\right)=\left[\frac{-1}{m(m+n-2)}\right]^l \left(\Delta_{S}^{l } G, Y_{m,j}\right). \end{align*} $$

By the Cauchy–Schwarz inequality, we have

$$\begin{align*}|(G, Y_{m,j})| \leq \left[\frac{1}{m(m+n-2)}\right]^l \cdot \|G\|_{H_k}.\end{align*}$$

As a result, the right-hand side of (4.7) satisfies

$$\begin{align*}\Big\|\lambda_m(|t|)^{-1}\left(\sum_{j=1}^{N_m}(G,Y_{m,j})Y_{m,j}\right) \Big\|_2^2\leq \|G\|_{H_k}^2 \left[\frac{1}{m(m+n-2)}\right]^{2l}\cdot \lambda_m(|t|)^{-2}\cdot {N_m}.\end{align*}$$

Since $G\in H_k$ , it suffices to prove that

(4.8) $$ \begin{align} {\left[m(m+n-2)\right]}^{-2l}\cdot \lambda_m(|t|)^{-2}\cdot {N_m} =\mathrm{O}\left(m^{\alpha}\right),\quad \mathrm{~for~some~} \alpha<-1. \end{align} $$

Here,

$$\begin{align*}\lambda_m(|t|)^{-1} = \mathrm{O}\left(m^{\frac{n+2}{2}}\right), \quad \mathrm{~and~}\quad N_m= \frac{(2m+n-2)\cdot(n+m-3)!}{(n-2)!\cdot m!} = \mathrm{O}\left(m^{\frac{n-2}{2}}\right).\end{align*}$$

If $l>n/2+1/4$ , the inequality (4.8) holds for

$$ \begin{align*}\alpha = {-4l+2n} <-1.\end{align*} $$

Therefore, when $k=2l\ge n+1$ , we have the desired result.

Lemmas 4.3 and 4.5 immediately lead to the following.

Corollary 4.6 Let $k\ge n+1$ be an even integer, and let $G\in H_k(\mathbb {S}^{n-1})$ . Suppose $g\in L^2(\mathbb {S}^{n-1})$ satisfies

$$\begin{align*}\int_{\mathbb{S}^{n-1}}(v\cdot u)_- g(u) du = G(v).\end{align*}$$

Then,

$$\begin{align*}g(u) + g(-u) = 2\sum_{m~\mathrm{even}} \sum_{j=1}^{N_m} \beta_m^{-1}(G,Y_{m,j}) Y_{m,j}\end{align*}$$

and

$$\begin{align*}\int_{\mathbb{S}^{n-1}}ug(u)du = -\frac2{\omega_n}\int_{\mathbb{S}^{n-1}}uG(u)du.\end{align*}$$

Moreover, $(G,Y_{m,j}) = 0$ , for each odd integer $m\geq 3$ .

Proof Denote $\tilde g(u) = g(u) + g(-u), a_{m,j} = (g,Y_{m,j}),$ and $b_{m,j} = (\tilde g,Y_{m,j}).$ When m is even, $Y_{m,j}$ is even, and thus $b_{m,j} = 2a_{m,j}$ . On one hand, by Lemmas 4.1 and 4.2, we have

$$\begin{align*}G(v)=\int_{\mathbb{S}^{n-1}}(v\cdot u)_- g(u) du = \sum_{m~\mathrm{even}}\sum_{j=1}^{N_m} \beta_m a_{m,j}Y_{m,j}(v) -\frac{\omega_n}2\sum_{j=1}^n a_{1,j}Y_{1,j}(v). \end{align*}$$

On the other hand,

$$\begin{align*}G(v)=\sum_{m=0}^\infty\sum_{j=1}^{N_m} (G,Y_{m,j})Y_{m,j}(v). \end{align*}$$

Comparing the coefficients, we have the following. For each odd integer $m\geq 3$ , we have $(G,Y_{m,j}) = 0$ . For $m=1$ , we have $(G,Y_{1,j}) =- ({\omega _n}/2) a_{1,j}.$ Applying (2.4) for g and G, respectively, we get

$$\begin{align*}-\frac{\omega_n}2 \int_{\mathbb{S}^{n-1}} u g(u)du = \int_{\mathbb{S}^{n-1}} u G(u)du. \end{align*}$$

For even m, we have

$$\begin{align*}(G,Y_{m,j}) = \frac{\beta_m }2b_{m,j}.\end{align*}$$

Since $\tilde g$ is an even function, we get

$$\begin{align*}\tilde g(u) = g(u) + g(-u) = 2\sum_{m~\mathrm{even}} \sum_{j=1}^{N_m} \beta_m^{-1}(G,Y_{m,j}) Y_{m,j}.\end{align*}$$

The convergence of the above function series follows from Lemma 4.5 and the fact that $2\beta _m = \lambda _m(|t|)$ .

4.2 The necessary and sufficient conditions of being a lightness function

Given a binary function $G: \mathbb {S}^{n-1}\times \mathbb {S}^{n-1} \rightarrow \mathbb {R}$ so that $G(\cdot ,w)$ is Sobolev for a fixed $w\in \mathbb {S}^{n-1}$ , we shall use the notation

$$\begin{align*}\left(G(\cdot,w),f\right) = \int_{\mathbb{S}^{n-1}} G(v,w)f(v) dv.\end{align*}$$

Proof Denote

$$\begin{align*}G_{K,i}(v) = Q_K(v,e_i)+Q_K(-v,e_i) + Q_K(v,-e_i)+Q_K(-v,-e_i),\end{align*}$$

which is an even function in $H_k(\mathbb {S}^{n-1})$ . On one hand,

$$\begin{align*}G_{K,i}(v) =\int_{\mathbb{S}^{n-1}}|u\cdot v| \left(f_{e_i}(u) + f_{-e_i}(u)\right) dS(K,u).\end{align*}$$

On the other hand, by Lemma 4.5, there is an even function $F_{K,i}\in L^2(\mathbb {S}^{n-1})$ such that

$$\begin{align*}G_{K,i}(v) = \int_{\mathbb{S}^{n-1}} |u\cdot v| F_{K,i}(u) du. \end{align*}$$

Since the cosine transform is injective (Lemma 4.4), we have

$$\begin{align*}\left(f_{e_i}(u) + f_{-e_i}(u)\right) dS(K,u) = F_{K,i}(u) du. \end{align*}$$

Note that, for each $u\in \mathbb {S}^{n-1},$ there exists an $i\in \{1,\ldots ,n\}$ such that $|u_i|\ge 1/\sqrt n.$ Since $|\tilde f_{e_i}|$ is positive and continuous in $\{u: |u_i|\geq 1/\sqrt n\}$ , it has a strict positive lower bound. Thus,

$$\begin{align*}dS(K,u) = \frac{F_{K,i}(u)}{f_{e_i}(u) + f_{-e_i}(u)} du\end{align*}$$

has a density in $L^2(\mathbb {S}^{n-1})$ , where the function $F_{K,i}/(f_{e_i}+ f_{-e_i})$ must be independent of i.

Proof By Lemma 4.7, $S(K,\cdot )$ has an $L^2(\mathbb {S}^{n-1})$ density $F_K:\mathbb {S}^{n-1}\rightarrow [0,\infty )$ . Substitute the function $f_w F_K$ into Corollary 4.6, and observe that

$$\begin{align*}Q_K(v,w) = \int_{\mathbb{S}^{n-1}} (v\cdot u)_- f_w(u) F_K(u) du.\end{align*}$$

Then, we immediately obtain (2) and (3), and

$$\begin{align*}\frac12\left(f_w(u) F_K(u) + f_w(-u) F_K(-u)\right) = \sum_{m \mathrm{~even}}\sum_{j=1}^{N_m} \beta_m^{-1}\big(Q_K(\cdot,w),Y_{m,j}\big) Y_{m,j}(u).\end{align*}$$

Since $f_w(u)$ is concentrated on $w^+, f_w(-u) F_K(-u)$ is just a symmetric copy of $f_w(u) F_K(u)$ in $w^-$ . Thus,

$$\begin{align*}\frac12 f_w(u) F_K(u) = 1_{w^+}(u)\cdot\sum_{m \mathrm{~even}}\sum_{j=1}^{N_m} \beta_m^{-1}\big(Q_K(\cdot,w),Y_{m,j}\big) Y_{m,j}(u).\end{align*}$$

This explains (1).

Theorem 4.9 Let $G: \mathbb {S}^{n-1}\times \mathbb {S}^{n-1} \rightarrow \mathbb {R}$ be a binary function, and let $k\ge n+1$ be an even integer. Suppose $G(\cdot ,w)\in H_k(\mathbb {S}^{n-1})$ , for any fixed $w\in \mathbb {S}^{n-1}$ . In addition, suppose G satisfies the following conditions:

• (1) The following function

  (4.9) $$ \begin{align} F(u) = \begin{cases} \sum\limits_{m~\mathrm{even}} \sum\limits_{k=1}^{N_m} \beta_m^{-1}\big(G(\cdot,w),Y_{m,k}\big) \frac{Y_{m,k}(u)}{f_w(u)} , \ \ & u\cdot w>0 \\ \phantom{\cdot} &\phantom{\cdot} \\ \sum\limits_{m~\mathrm{even}} \sum\limits_{k=1}^{N_m} \beta_m^{-1}\big(G(\cdot,-w),Y_{m,k}\big) \frac{Y_{m,k}(u)}{f_{-w}(u)} , \ \ & u\cdot w<0 \end{cases} \end{align} $$
  is nonnegative and independent of $w\in \mathbb {S}^{n-1}$ .

• (2) For each odd integer $m\geq 3$ , and for each fixed $w\in \mathbb {S}^{n-1}$ , the function $G(\cdot ,w): \mathbb {S}^{n-1}\rightarrow \mathbb {R}$ satisfies

  $$\begin{align*}\big(G(\cdot,w),Y_{m,j}\big)=0, \quad \forall k=1,\ldots,N_{m}. \end{align*}$$

• (3) The function F in (1) satisfies

  $$\begin{align*}\int_{\mathbb{S}^{n-1}} uF(u) du =0.\end{align*}$$

• (4) G and the function F in (1) satisfy

  $$\begin{align*}\frac 2{\omega_n}\int_{\mathbb{S}^{n-1}} vG(v,w)dv = \int_{\mathbb{S}^{n-1}} uf_w(u)F(u)du.\end{align*}$$

Then, there is a convex body K such that

$$\begin{align*}G = Q_K,\end{align*}$$

and K is unique up to translations.

Condition (1) means that for any $w_1,w_2\in \mathbb {S}^{n-1},$ the related two function series are equivalent almost everywhere.

Proof By Lemma 4.5, for each fixed $w\in \mathbb {S}^{n-1}$ , the function series

(4.10) $$ \begin{align}\sum_{m \mathrm{~even}} \sum_{j=1}^{N_m} \beta_m^{-1}\big(G(\cdot,w),Y_{m,j}\big) Y_{m,j}(u) \in L^2(\mathbb{S}^{n-1}).\end{align} $$

Let $e_1,\ldots ,e_n$ be an orthonormal basis of $\mathbb {R}^n$ , and denote

$$\begin{align*}\Omega_i = \Big\{ u\in\mathbb{S}^{n-1}: |u\cdot e_i| \geq 1/\sqrt n \Big\}. \end{align*}$$

By (4.9) and (4.10), we infer that $F \in L^2(\Omega _i)$ , $i=1,\ldots ,n.$ Since $\{\Omega _i\}_{i=1}^n$ covers $\mathbb {S}^{n-1}$ , we see that $F \in L^2(\mathbb {S}^{n-1}).$ Since $\mathbb {S}^{n-1}$ is compact, F is also in $L^1(\mathbb {S}^{n-1})$ . By conditions (1) and (3), F is independent of w and is of centroid $0$ . Then, there is a unique (up to translations) convex body K that solves the classical Minkowski problem

$$\begin{align*}dS(K,u) = F(u)du.\end{align*}$$

Notice that

$$\begin{align*}\frac12\left(f_w(u) F(u) + f_w(-u) F(-u)\right) = \sum_{m \mathrm{~even}}\sum_{j=1}^{N_m} \beta_m^{-1}\big(Q_K(\cdot,w),Y_{m,j}\big) Y_{m,j}(u).\end{align*}$$

For this K, by Corollary 4.6, Lemma 4.3, condition (4), equation (2.4), and finally by condition (2), we have

$$ \begin{align*} Q_K(v,w) &= \int_{\mathbb{S}^{n-1}} (v\cdot u)_- f_w(u) F(u) du\\ & = \frac12\int_{\mathbb{S}^{n-1}} (v\cdot u)_- \frac12\left(f_w(u) F(u) + f_w(-u) F(-u)\right) du \\&\quad- \frac12 \int_{\mathbb{S}^{n-1}} (u\cdot v) f_w(u) F(u) du\\ & = \int_{\mathbb{S}^{n-1}} (v\cdot u)_- \sum_{m~\mathrm{even}} \sum_{j=1}^{N_m} \beta_m^{-1} \big(G(\cdot,w),Y_{m,j}\big) Y_{m,j}(u) \\&\quad + \frac1{ {\omega_n}}\int_{\mathbb{S}^{n-1}}(u\cdot v) G(u,w)du\\ &= \sum_{m~\mathrm{even}} \sum_{j=1}^{N_m} \big(G(\cdot,w),Y_{m,j}\big) Y_{m,j}(v) + \frac1{ {\omega_n}}\int_{\mathbb{S}^{n-1}}(u\cdot v) G(u,w)du\\ &= \sum_{m~\mathrm{even}} \sum_{j=1}^{N_m} \big(G(\cdot,w),Y_{m,j}\big) Y_{m,j}(v) + \sum_{j=1}^{N_1} \big(G(\cdot,w),Y_{1,j}\big) Y_{1,j}(v)\\ &= \sum_{m=0}^\infty \sum_{j=1}^{N_m} \big(G(\cdot,w),Y_{m,j}\big) Y_{m,j}(v)\\ &= G(v,w). \end{align*} $$

That the convex body K is unique follows from Theorem 4.8 and the uniqueness of the Minkowski problem.