The minimizers of the $p$-frame potential

TL;DR

This paper determines the unique set of vectors that minimizes the $p$-frame potential for $N=d+1$ and $p$ in (0,2), confirming a previous conjecture and solving the minimization problem in this case.

Contribution

It establishes the unique minimizer of the $p$-frame potential for $N=d+1$ and $p$ in (0,2), confirming a conjecture and completing the minimization analysis for this scenario.

Findings

Identifies the unique minimizer of the $p$-frame potential for $N=d+1$ and $p$ in (0,2).

Confirms a conjecture by Chen et al. regarding the minimization problem.

Provides a complete solution to the minimization problem in the specified case.

Abstract

For any positive real number $p$, the $p$-frame potential of $N$ unit vectors $X:=\{\mathbf x_1,\ldots,\mathbf x_N\}\subset \mathbb R^d$ is defined as ${\rm FP}_{p,N,d}(X)=\sum_{i\neq j}|\langle \mathbf x_i,\mathbf x_j\rangle |^p$. In this paper, we focus on the special case $N=d+1$ and establish the unique minimizer of ${\rm FP}_{p,d+1,d}$ for $p\in (0,2)$. Our results completely solve the minimization problem of $p$-frame potential when $N=d+1$, which confirms a conjecture posed by Chen, Gonzales, Goodman, Kang and Okoudjou.

Equations

$${\mathrm{FP}}_{p,N,d}(X):=\sum _{i=1}^N\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩{|}^p,$$

$$\underset{X\in S(N,d)}{\mathrm{argmin}}{\mathrm{FP}}_{p,N,d}(X),$$

$${\mathrm{FP}}_{2k,N,d}(X)\ge N^2\frac{1\cdot 3\cdot 5\dots (2k-1)}{d(d+2)\dots (d+2k-2)}-N$$

$${\mathrm{FP}}_{p,N,d}(X)\ge N(N-1){\left(\frac{N-d}{d(N-1)}\right)}^{\frac{p}{2}},$$

$${\mathrm{FP}}_{p,N,d}(X)\ge 2(N-d)\frac{1}{p^{\frac{p}{2}}(2-p{)}^{\frac{2-p}{2}}},$$

$$|\langle{{\mathbf{x}}_{i},{\mathbf{x}}_{j}}\rangle\rvert:=\left\{\begin{array}[]{cl}\frac{1}{k}&i,j\in\{1,\ldots,k+1\},i\neq j\\ 1&i=j\\ 0&else\end{array}\right..$$

$$M_{\alpha ,d+1}(z_1,\dots ,z_{d+1}):=\sum _{i=1}^{d+1}\sum _{j\ne i}{z}_i^{\alpha }{z}_j^{\alpha }$$

$$\underset{(z_1,\dots ,z_{d+1})}{\mathrm{argmax}}{M}_{\alpha ,d+1}(z_1,\dots ,z_{d+1}),s.t.z_1+\cdots +z_{d+1}=1,z_1\ge 0,\dots ,z_{d+1}\ge 0,$$

$$a_{k}:=\left\{\begin{array}[]{cl}\infty&k=0\\ \frac{1}{2}\cdot\frac{\ln(k+2)-\ln(k)}{\ln(k+2)-\ln(k+1)}&k\in\{1,2,\ldots,d-1\}\\ 1&k=d\end{array}\right..$$

$$G=(⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩)\in {\mathbb{R}}^{(d+1)\times (d+1)}.$$

$$0=(G\mathbf{y}{\mathbf{y}}^T{)}_{i,i}=\sum _{j=1}^{d+1}⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩\cdot y_i{y}_j=y_i^2+\sum _{j\ne i}⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩\cdot y_i{y}_j,$$

$$y_i^2=|\sum _{j\ne i}⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩\cdot y_i{y}_j|\le \sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|\cdot |y_i||y_j|.$$

$$1=\sum _{i=1}^{d+1}{y}_i^2\le \sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|\cdot |y_i||y_j|.$$

$$|y_i||y_j|\le \frac{y_i^2+y_j^2}{2}\le \frac{1}{2},\text{ for all }i\ne j$$

$$1\le \sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|\cdot |y_i||y_j|\le \frac{1}{2}\cdot \sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|,$$

$$\sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|\ge 2.$$

$$|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩{|}^p\ge |⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|,\forall i\ne j$$

$$\sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩{|}^p\ge \sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|\ge 2.$$

$$1\le \sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩|\cdot |y_i||y_j|\le (\sum _{i=1}^{d+1}\sum _{j\ne i}|⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩{|}^p{)}^{\frac{1}{p}}(\sum _{i=1}^{d+1}\sum _{j\ne i}|y_i{|}^q|y_j{|}^q{)}^{\frac{1}{q}}$$

$$c\cdot |⟨{\mathbf{x}}_i,{\mathbf{x}}_j⟩{|}^{p-1}=|y_i||y_j|,\text{ for all }i\ne j.$$

$${\mathrm{FP}}_{p,d+1,d}(X)\ge \frac{1}{(\sum _{i=1}^{d+1}\sum _{j\ne i}|y_i{|}^q|y_j{|}^q{)}^{\frac{p}{q}}}.$$

$${\mathrm{FP}}_{p,d+1,d}(X)\ge \frac{1}{(M_{\alpha ,d+1}(z_1,\dots ,z_{d+1}){)}^{\frac{p}{q}}},$$

$${\mathrm{FP}}_{p,d+1,d}(X)\ge \frac{1}{(k(k+1{)}^{1-2\alpha }{)}^{\frac{p}{q}}}=(k+1)k^{1-p}$$

$$|\langle{{\mathbf{x}}_{i},{\mathbf{x}}_{j}}\rangle\rvert=\left\{\begin{array}[]{cl}\frac{1}{k}&i,j\in\{1,\ldots,k+1\}\\ 0&else\end{array}\right.,$$

$${\mathrm{FP}}_{p_k,d+1,d}(X)\ge \frac{1}{(k(k+1{)}^{1-2a_k}{)}^{\frac{p_k}{2\cdot a_k}}}=(k+1)k^{1-p_k}$$

$$f_{m_1,\alpha ,d+1}(t):=M_{\alpha ,d+1}\left(\underset{m_1}{\underbrace{t,\dots ,t}},\underset{d+1-m_1}{\underbrace{s,\dots ,s}}\right)=(m_1\cdot t^{\alpha }+(d+1-m_1)\cdot s^{\alpha }{)}^2-(m_1\cdot t^{2\alpha }+(d+1-m_1)\cdot s^{2\alpha }),$$

$$r_i(z_1,\dots ,z_{d+1}):=-z_i,i=1,2,\dots ,d+1.$$

$$\nabla M_{\alpha ,d+1}(w_1,\dots ,w_{d+1})$$

$$r_0(w_1,\dots ,w_{d+1})$$

$$r_i(w_1,\dots ,w_{d+1})$$

$${\mu }_i{r}_i(w_1,\dots ,w_{d+1})$$

$${\mu }_i$$

Full text

The project was supported by NSFC grant (91630203, 11688101), Beijing Natural Science Foundation (Z180002).

1. Introduction

1.1. The $p$-frame potential

Minimal potential energy problems have been actively discussed over the last decades in connection with applications in physics, signal analysis and numerical integration. Generally, one aims to find distributions of $N$ points on the unit sphere which minimize the potential energy over all sized $N$ configurations [18, 5, 2].

One of the most interesting potential energies is the $p$-frame potential. Assume that $X:=\{{\mathbf{x}}_{i}\}_{i=1}^{N}$ where ${\mathbf{x}}_{i}\in{\mathbb{R}}^{d}$ with $\|{\mathbf{x}}_{i}\|_{2}=1,i=1,\ldots,N$. For $p>0$, the value

[TABLE]

is called the $p$*-frame potential * of $X$ (see [7, 4]). The minimization problem of the $p$-frame potential is to solve

[TABLE]

where $S(N,d)$ denotes all sets of $N$ unit-norm vectors in ${\mathbb{R}}^{d}$. This problem actually has a long history and has attracted much attention. For $N\leq d$, the set of $N$ orthogonal vectors in ${\mathbb{R}}^{d}$ is always the minimizer of (2) for any positive $p$ and hence we only consider the case where $N\geq d+1$. We also note that the value of ${\rm FP}_{p,N,d}(X)$ does not change if we replace ${\mathbf{x}}_{i}$ by $c_{i}U{\mathbf{x}}_{i}$ for each $i\in\{1,2,\cdots,N\}$, where $U$ is an orthogonal matrix and $c_{i}\in\{1,-1\}$. Thus, for convenience we say the minimizer of (2) is unique if the solution to (2) is unique up to a common orthogonal transformation and a real unimodular constant for each vector.

1.2. Related work

There are many results which presented a lower bound of ${\rm FP}_{p,N,d}(X)$. When $p=2k\in{2\mathbb{Z}}_{+}$, the following bound

[TABLE]

was presented by Sidelnikov in [17]. The equality in (3) holds precisely when $X\cup-X$ is a spherical $2k$-design, see [11, 17]. The set $X=\{{\mathbf{x}}_{i}\}_{i=1}^{N}$ attaining the bound in (3) can also be identified as a projective $k$-design since we consider ${\mathbf{x}}_{i}$ and $-{\mathbf{x}}_{i}$ as the same point (see [12] for the definition of projective designs). For the special case $k=1$, i.e. $p=2$, the minimizers of the $2$-frame potential are projective $1$-designs [11, 17]. Noting the fact that finite unit-norm tight frames (FUNTFs) are in one-to-one correspondence with projective 1-designs [6], we know that the equality in (3) holds for $k=1$ and all $N\geq d$ as long as $X$ is a FUNTF (see also [1]). However, when $k>1$, the bound in (3) is not tight for small $N$ since the existence of spherical designs requires $N$ to be large enough [6].

For any $p>2$, Ehler and Okoudjou provided another bound in [7]:

[TABLE]

where equality holds if and only if $X$ is an equiangular tight frame (ETF) in ${\mathbb{R}}^{d}$ [8, 13]. We take $N=d+1$ as an example. Since there always exist $d+1$ unit vectors in ${\mathbb{R}}^{d}$ forming an ETF [8, 15], namely the regular simplex of size $d+1$, the set of these vectors uniquely minimizes the $p$-frame potential for $p>2$.

When $p\in(0,2)$, not much is known about minimizers of this value except in a few special cases. In [7], Ehler and Okoudjou solved the simplest case where $d=2$ and $N=3$ and also proved that the minimizer of the $p$-frame potential is exactly $n$ copies of an orthonormal basis if $N=nd$ where $n$ is a positive integer. In [9], Glazyrin provided a lower bound for any $1\leq p\leq 2$:

[TABLE]

but the condition under which equality holds is strict. In [4], Chen, Gonzales, Goodman, Kang and Okoudjou considered this special case where $N=d+1$. Particularly, [4] showed through numerical experiments that the sets $L_{k}^{d}$, which they call lifted ETFs, seem to be minimizers of the $p$-frame potential for certain $k$ depending on $p$. Here, $L_{k}^{d}=\{{\mathbf{x}}_{1},\ldots,{\mathbf{x}}_{d+1}\}\subset{\mathbb{R}}^{d}$ is defined as a set of $d+1$ unit vectors in ${\mathbb{R}}^{d}$ satisfying

[TABLE]

Note that $\{{\mathbf{x}}_{i}\}_{i=1}^{k+1}\subset L_{k}^{d}$ actually forms an ETF in some subspace $W\subset{\mathbb{R}}^{d}$ with dimension $k$ and the rest of $d-k$ vectors form an orthonormal basis in the orthogonal complement of $W$.

More precisely, the following conjecture is proposed in [4]:

Conjecture 1.1**.**

Suppose $d\geq 2$. Set $p_{0}:=0$, $p_{d}:=2$ and $p_{k}:=\frac{\ln(k+2)-\ln(k)}{\ln(k+1)-\ln(k)}$ for each $k\in\{1,2,\ldots,d-1\}$. Then, when $p\in(p_{k-1},p_{k}]$, $k=1,2,\ldots,d$, the set $L_{k}^{d}$ minimizes the $p$-frame potential when $N=d+1$.

The cases $d=2$ and $p=2$ for Conjecture 1.1 are already solved in [7] and [1], respectively. The first new result for Conjecture 1.1 was obtained by Glazyrin in [10] who showed that an orthonormal basis in ${\mathbb{R}}^{d}$ plus a repeated vector minimizes ${\rm FP}_{p,d+1,d}(X)$ for any $p\in(0,2(\frac{\ln{3}}{\ln{2}}-1)]$. Combining Glazyrin’s result with the previous ones, the minimizer of ${\rm FP}_{p,d+1,d}(X)$ is only known for $p\in(0,2(\frac{\ln{3}}{\ln{2}}-1)]\cup[2,\infty)$. Recently, Park extented Glazyrin’s result to the case $N=d+m$ where $1\leq m<d$, and showed that an orthonormal basis plus $m$ repeated vectors is the minimizer for any $p\in[1,2\frac{\ln{(2m+1)}-\ln{(2m)}}{\ln{(m+1)}-\ln{(m)}}]$ (see [16]). But Conjecture 1.1 remains open when $d>2$.

1.3. Our contributions

The aim of this paper is to confirm Conjecture 1.1 and to show that the minimizers are unique provided $p\neq p_{k}$. Our main result is the following theorem which completely solves the minimal $p$-frame potential problem for the case where $N=d+1$.

Theorem 1.2**.**

Let $d\geq 2$ be an integer. Set $p_{0}:=0$, $p_{d}:=2$ and $p_{k}:=\frac{\ln(k+2)-\ln(k)}{\ln(k+1)-\ln(k)}$ for each $k\in\{1,2,\ldots,d-1\}$. Assume that $p\in(0,2)$ is a real number. Let $X=\{{\mathbf{x}}_{1},\ldots,{\mathbf{x}}_{N}\}$ be a set of $N$ unit vectors in ${\mathbb{R}}^{d}$, where $N=d+1$.

1. (i)

For $p\in(p_{k-1},p_{k}),k=1,2,\ldots,d$, and for any $X\in S(d+1,d)$ we have ${\rm FP}_{p,d+1,d}(X)\geq(k+1)k^{1-p}$ and equality holds if and only if $X=L_{k}^{d}$. 2. (ii)

For $p=p_{k},\,k=1,\ldots,d-1$, and for any $X\in S(d+1,d)$ we have ${\rm FP}_{p,d+1,d}(X)\geq(k+1)k^{1-p_{k}}$ and equality holds if and only if $X=L_{k}^{d}$ or $X=L_{k+1}^{d}$.

Based on the previous results and Theorem 1.2 in this paper, in Table 1.3, we list optima for the minimal $p$-frame potential problem when $N=d+1$. Note that $2(\frac{\ln{3}}{\ln{2}}-1)\approx 1.16993$ and $\frac{\ln{3}}{\ln{2}}\approx 1.58496$. Hence, $(0,2(\frac{\ln{3}}{\ln{2}}-1)]$ is a subinterval in $(0,\frac{\ln{3}}{\ln{2}})$. In Table 1.3, we also use the fact that $L_{1}^{d}$ is essentially an orthonormal basis plus a repeated vector and $L_{d}^{d}$ forms an ETF in ${\mathbb{R}}^{d}$.

1.4. Organization

The paper is organized as follows. In Section 2, we prove Theorem 1.2 based on Lemma 2.1. The proof of Lemma 2.1 is presented in Section 3.

2. Proof of Theorem 1.2

In this section we present a proof of Theorem 1.2. The following lemma plays a key role in our proof of Theorem 1.2. We postpone its proof to Section 3. To this end, we set

[TABLE]

where $\alpha>1$. We consider

[TABLE]

where $\alpha>1$. Noting that $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ is a symmetric function on $d+1$ variables $z_{1},\ldots,z_{d+1}$, we identify solutions to (7) only up to permutations of $z_{i}$.

Lemma 2.1**.**

Suppose that $d\geq 1$ is an integer. Set

[TABLE]

1. (i)

If $\alpha\in(a_{k},a_{k-1})$ then the unique solution to (7) is $\left(\underbrace{\frac{1}{k+1},\ldots,\frac{1}{k+1}}_{k+1},\underbrace{0,\ldots,0}_{d-k}\right)$ where $k=1,2,\ldots,d$. 2. (ii)

Assume that $\alpha=a_{k}$ where $k=1,\ldots,d-1$. Then (7) has exactly two solutions: $\left(\underbrace{\frac{1}{k+1},\ldots,\frac{1}{k+1}}_{k+1},\underbrace{0,\ldots,0}_{d-k}\right)$ and $\left(\underbrace{\frac{1}{k+2},\ldots,\frac{1}{k+2}}_{k+2},\underbrace{0,\ldots,0}_{d-k-1}\right)$.

We next state a proof of Theorem 1.2. Our method of estimating the $p$-frame potential in this proof is motivated by the work of Bukh and Cox [3]. For a finite set $X=\{{\mathbf{x}}_{i}\}_{i=1}^{N}\in S(N,d)$, Bukh and Cox derived a new lower bound on $\mu(X):=\max\limits_{j\neq l}\lvert\langle{\mathbf{x}}_{j},{\mathbf{x}}_{l}\rangle\rvert$ with the help of an orthonormal basis in the null space of the Gram matrix of $X$ (see also [14]). We borrow their idea of considering the null space of the Gram matrix of $X$. Noting the corresponding null space in our case is a one-dimensional subspace in ${\mathbb{R}}^{N}$, we pick a unit vector ${\mathbf{y}}$ in this subspace, showing that we can use the coordinates of ${\mathbf{y}}$ to provide an estimation on the value of ${\rm FP}_{p,d+1,d}(X)$.

Proof of Theorem 1.2.

(i) Note that ${\rm FP}_{p,d+1,d}(L_{k}^{d})=(k+1)k^{1-p}$. To this end, it is enough to show that ${\rm FP}_{p,d+1,d}(X)\geq(k+1)k^{1-p}$ when $p\in(p_{k-1},p_{k})$ and $L_{k}^{d}$ is the unique minimizer for each $k\in\{1,2,\ldots,d\}$.

Recall that $X=\{{\mathbf{x}}_{i}\}_{i=1}^{d+1}\subset{\mathbb{R}}^{d}$ is a set of $d+1$ unit-norm vectors. Set

[TABLE]

Note that ${\rm rank}(G)\leq d$. Thus, there exists a unit vector ${\mathbf{y}}=(y_{1},\ldots,y_{d+1})^{T}\in{\mathbb{R}}^{d+1}$ such that $G{\mathbf{y}}=0$. We compute the value of $(i,i)$-entry of the matrix $G{\mathbf{y}}{\mathbf{y}}^{T}$ and obtain

[TABLE]

which implies

[TABLE]

Summing up the above inequality from $1$ to $d+1$, we obtain

[TABLE]

We next present a proof for (i) with dividing the proof into two cases:

Case 1: $p\in(0,1]$. Note that $(0,1]\subset(p_{0},p_{1})$. It is enough to prove that the unique solution to $\mathop{\rm argmin}\limits_{X\in S(d+1,d)}{\rm FP}_{p,d+1,d}(X)$ is $X=L_{1}^{d}$ for any $p\in(0,1]$. We first consider the case where $p=1$. Since

[TABLE]

we obtain

[TABLE]

which implies

[TABLE]

The equality in (9) holds if and only if there exist $i_{1},i_{2}\in\{1,2,\ldots,d+1\}$ with $i_{1}\neq i_{2}$ such that $|\langle{\mathbf{x}}_{i_{1}},{\mathbf{x}}_{i_{2}}\rangle|=1$ and the remaining terms in the sum $\sum\limits_{i=1}^{d+1}\sum\limits_{j\neq i}|\langle{\mathbf{x}}_{i},{\mathbf{x}}_{j}\rangle|$ are all zero. We arrive at the conclusion.

We next turn to the case $p\in(0,1)$. Noting $|\langle{{\mathbf{x}}_{i},{\mathbf{x}}_{j}}\rangle\rvert\leq 1$, we have

[TABLE]

for any $p\in(0,1)$. Thus,

[TABLE]

The equality holds if and only if $|\langle{\mathbf{x}}_{i},{\mathbf{x}}_{j}\rangle|=0$ or 1 for any distinct $i,j$. Thus, the minimizer of $1$-frame potential is also the unique minimizer of $p$-frame potential for any $p\in(0,1)$.

Case 2: $1<p<2$. For $1<p<2$, we use H$\ddot{\text{o}}$lder’s inequality to obtain

[TABLE]

where $q>2$ satisfies $\frac{1}{p}+\frac{1}{q}=1$. The second equality in (11) holds if and only if there exists a constant $c\in{\mathbb{R}}$ such that

[TABLE]

Equation (11) implies

[TABLE]

Let $\alpha=\frac{q}{2}$ and $z_{i}=|y_{i}|^{2}$ for $i=1,2,\ldots,d+1$. Then we can rewrite the inequality in (13) as

[TABLE]

where $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})=\sum\limits_{i=1}^{d+1}\sum\limits_{j\neq i}z_{i}^{\alpha}z_{j}^{\alpha}$, $z_{1}+\cdots+z_{d+1}=1,z_{i}\geq 0,i=1,\ldots,d+1$.

Note that $\alpha=\frac{q}{2}=\frac{1}{2}+\frac{1}{2}\frac{1}{p-1}$. If $p\in(p_{k-1},p_{k})\cap(1,2)$ where $k\in\{1,\ldots,d\}$, then $\alpha\in(a_{k},a_{k-1})$. Here, $a_{k}$ is defined in (8). According to Lemma 2.1, $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ attains its maximum, $k(k+1)^{1-2\alpha}$, only when $z_{i}=\frac{1}{k+1}$ for $i=1,\ldots,k+1$ and $z_{i}=0$ for $i\geq k+2$. Thus, we obtain

[TABLE]

when $p\in(p_{k-1},p_{k})\cap(1,2)$, $k=1,\ldots,d$. Combining with equation (12), the equality in (15) holds if and only if for $i\neq j$

[TABLE]

which implies that $X=L_{k}^{d}$. Combining the analysis in Case 1, we arrive at the conclusion (i) for Theorem 1.2.

(ii) Note that ${\rm FP}_{p,d+1,d}(L_{k}^{d})={\rm FP}_{p,d+1,d}(L_{k+1}^{d})=(k+1)k^{1-p}$ when $p=p_{k}$, $k=1,2,\ldots,d-1$. To this end, it is enough to prove that ${\rm FP}_{p_{k},d+1,d}(X)\,\,\geq\,\,(k+1)k^{1-p_{k}}$ and that the minimizers are $L_{k}^{d}$ and $L_{k+1}^{d}$. Since $p_{k}\in(1,2)$ for each $k\in\{1,2,\ldots,d-1\}$, we follow our analysis for the proof of (i).

If $p=p_{k}$ where $k\in\{1,\ldots,d-1\}$, then $\alpha$ in (14) is equal to $a_{k}$. According to Lemma 2.1, $M_{a_{k},d+1}(z_{1},\ldots,z_{d+1})$ attains its maximum, which is $k(k+1)^{1-2a_{k}}$, at exactly two points: $\left(\underbrace{\frac{1}{k+1},\ldots,\frac{1}{k+1}}_{k+1},\underbrace{0,\ldots,0}_{d-k}\right)$ and $\left(\underbrace{\frac{1}{k+2},\ldots,\frac{1}{k+2}}_{k+2},\underbrace{0,\ldots,0}_{d-k-1}\right)$. Thus, we obtain

[TABLE]

for $k=1,2,\ldots,d-1$. According to (12), the equality in (17) holds if and only if $X=L_{k}^{d}$ or $L_{k+1}^{d}$, which implies the conclusion (ii).

∎

Remark 2.2**.**

For convenience, we state Theorem 1.2 and its proof for the real case. In fact, it is easy to extend the result in Theorem 1.2 to complex case. Moreover, the method which is employed to prove Theorem 1.2 can be used to estimate the matrix potential, i.e. $\sum\limits_{i\neq j}|A_{i,j}|^{p}$, where $A_{i,j}$ is the $(i,j)$-entry of any matrix $A\in{\mathbb{C}}^{(d+1)\times(d+1)}$ whose rank is $d$ and diagonal elements are equal to 1 (see [10]).

3. Proof of Lemma 2.1

In this section we present the proof of Lemma 2.1. We begin with introducing the following lemma, which portrays the main feature of local extrema for (7).

For convenience, we set

[TABLE]

where $s:=\frac{1-m_{1}t}{d+1-m_{1}},m_{1}\in[1,\frac{d+1}{2}]\cap{\mathbb{Z}}$.

Lemma 3.1**.**

Assume that $(w_{1},\ldots,w_{d+1})$ is a local maxima of $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ subject to the constraints in (7) and $w_{i}>0$ for each $i\in\{1,2,\dots,d+1\}$. Then

1. (i)

The maxima $(w_{1},\ldots,w_{d+1})$ is in the form $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d+1-m_{1}}\right)$ up to a permutation where $m_{1}\in[1,\frac{d+1}{2}]\cap{\mathbb{Z}},t_{0}\in(0,\frac{1}{m_{1}})$ and $s_{0}=\frac{1-m_{1}t_{0}}{d+1-m_{1}}$. 2. (ii)

The value $t_{0}$ is a local maxima of $f_{m_{1},\alpha,d+1}(t)$.

Proof.

(i) We claim that $w_{1},\ldots,w_{d+1}$ can take at most two different values. Note that $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ is a symmetric function on $z_{1},\ldots,z_{d+1}$. Hence, up to a permutation, we can write $(w_{1},\ldots,w_{d+1})$ as $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d+1-m_{1}}\right)$ for some $t_{0}\in(0,\frac{1}{m_{1}})$ and $s_{0}=\frac{1-m_{1}t_{0}}{d+1-m_{1}}$. It remains to prove the claim. Set $r_{0}(z_{1},\ldots,z_{d+1}):=z_{1}+\cdots+z_{d+1}-1$ and

[TABLE]

Since $(w_{1},\ldots,w_{d+1})$ is a local extreme point, according to KKT conditions, there exist constants $\lambda$ and $\mu_{i},i=1,2,\ldots,d+1$, which are called KKT multipliers, such that the followings hold:

[TABLE]

Combining $w_{i}>0$ and (19d), we can obtain $\mu_{i}=0,i=1,2,\ldots,d+1$. Substituting $\mu_{i}=0$ into (19a), we obtain

[TABLE]

which implies that $\lambda>0$ and

[TABLE]

Hence, we obtain

[TABLE]

where $f(x):=x^{\alpha}+\frac{\lambda}{2\alpha}\cdot\frac{1}{x^{\alpha-1}}$. Set $w_{0}:=(\frac{\alpha-1}{2\alpha^{2}}\cdot\lambda)^{\frac{1}{2\alpha-1}}$. Noting that $f^{\prime}(x)=\alpha x^{\alpha-1}-\frac{\lambda(\alpha-1)}{2\alpha}x^{-\alpha}$, we obtain that $f^{\prime}(x)<0,x\in(0,w_{0})$, $f^{\prime}(w_{0})=0$ and $f^{\prime}(x)>0,x\in(w_{0},\infty)$, which implies that, for any $c\in{\mathbb{R}}$, the cardinality of the set $\{x:f(x)=c,x>0\}$ is less than or equal to $2$ . Hence, equation (21) implies that $w_{1},\ldots,w_{d+1}$ can take at most two different values.

(ii) Combining

[TABLE]

with $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d+1-m_{1}}\right)$ being a local maxima of $M_{\alpha,d+1}\left(\underbrace{t,\ldots,t}_{m_{1}},\underbrace{s,\ldots,s}_{d+1-m_{1}}\right)$, we obtain the conclusion immediately. ∎

Lemma 3.2**.**

Let $m_{1}\in[1,\frac{d+1}{2}]\cap{\mathbb{Z}}$ and $m_{2}=d+1-m_{1}$ where $d\geq 2$ is an integer. Set

[TABLE]

where $\alpha>1$. We use $h^{\prime}(x)$ to denote the derivative with respect to $x$ of $h(x)$. Then

1. (i)

The function $h^{\prime}(x)$ has at most two zeros on $(0,\infty)$, and hence $h(x)$ has at most two extreme points on $(0,\infty)$; 2. (ii)

If $\alpha<1+\frac{1}{d-1}$, then there exist $\hat{x}_{1}\in(0,1)$, $\hat{x}_{2}\in(1,\infty)$ such that $h^{\prime}(x)>0$ for $x\in(0,\hat{x}_{1})\cup(\hat{x}_{2},\infty)$ and $h^{\prime}(x)<0$ for $x\in(\hat{x}_{1},\hat{x}_{2})$; 3. (iii)

If $\alpha\geq 1+\frac{1}{d-1}$, then $h(x)$ is positive and monotonically increasing on $(1,\infty)$; 4. (iv)

If $\alpha=1+\frac{1}{d-1}$ and $m_{1}=\frac{d+1}{2}$, then $h(x)$ is monotonically increasing on $(0,\infty)$; 5. (v)

If $\alpha=1+\frac{1}{d-1}$ and $m_{1}<\frac{d+1}{2}$, then there exists $\hat{x}_{3}\in(0,1)$ such that $h^{\prime}(x)>0$ for $x\in(0,\hat{x}_{3})\cup(1,\infty)$ and $h^{\prime}(x)<0$ for $x\in(\hat{x}_{3},1)$.

Proof.

We split the proof, proving each claim separately.

(i) By computation, we have

[TABLE]

where $h_{1}(x)=(4\alpha-2)\cdot(m_{2}-1)x^{2\alpha}-2\alpha\cdot m_{2}\cdot x^{2}+(2\alpha-2)\cdot m_{1}$.

Set

[TABLE]

Noting that $h_{1}^{\prime}(x)<0,x\in(0,x_{0})$, $h_{1}^{\prime}(x)>0,x\in(x_{0},\infty)$ and $h_{1}^{\prime}(x_{0})=0$, $h_{1}(x)=0$ has at most two distinct solutions on $(0,\infty)$. According to (22), $h^{\prime}(x)=0$ also has at most two distinct solutions on $(0,\infty)$, which implies the conclusion.

(ii) When $\alpha<1+\frac{1}{d-1}$, we obtain $h_{1}(1)=2\alpha(d-1)-2d<0$. Then we have

[TABLE]

Observing that $m_{2}>1$ and $\alpha>1$, we obtain

[TABLE]

[TABLE]

Thus, combining (24), (25) and (26), we obtain that $h_{1}(x)=0$ has exactly two solutions $\hat{x}_{1}$, $\hat{x}_{2}$, where $\hat{x}_{1}\in(0,1)$, $\hat{x}_{2}\in(1,\infty)$. By the monotonicity of $h_{1}(x)$, we also know that $h_{1}(x)<0$, $x\in(\hat{x}_{1},\hat{x}_{2})$ and $h_{1}(x)>0$, $x\in(0,\hat{x}_{1})\cup(\hat{x}_{2},\infty)$. According to (22), we obtain that $h^{\prime}(x)<0$, $x\in(\hat{x}_{1},\hat{x}_{2})$ and $h^{\prime}(x)>0$, $x\in(0,\hat{x}_{1})\cup(\hat{x}_{2},\infty)$.

(iii) Note that

[TABLE]

where we use $m_{2}=d+1-m_{1}\geq\frac{d+1}{2}$ and $\alpha\geq 1+\frac{1}{d-1}$. So the function $h_{1}(x)$ is monotonically increasing when $x>1$. Noting that $h_{1}(1)=2\alpha(d-1)-2d\geq 0$, we have $h_{1}(x)>0$ on $(1,\infty)$, which implies that $h(x)$ is monotonically increasing on $(1,\infty)$. Since $h(1)=0$, we conclude that $h(x)>0$ when $x>1$.

(iv) When $\alpha=1+\frac{1}{d-1}$ and $m_{1}=\frac{d+1}{2}$, we have $h_{1}(1)=0$ and $x_{0}=1$ from (23), which implies that $h_{1}(x_{0})=0$. Since $x_{0}$ is the minimum point of $h_{1}(x)$, we obtain $h_{1}(x)\geq 0$ on $(0,\infty)$. Finally, from (22) we see that $h^{\prime}(x)\geq 0$ on $(0,\infty)$, which implies the conclusion.

(v) Noting that $x_{0}\neq 1$ provided $\alpha=1+\frac{1}{d-1}$ and $m_{1}<\frac{d+1}{2}$, we have

[TABLE]

Since $\alpha=1+\frac{1}{d-1}$, from (iii) we have that $h(x)$ is monotonically increasing on $(1,\infty)$. Noting that (25) and (26) also hold for $\alpha=1+\frac{1}{d-1}$, we conclude that $h_{1}(x)=0$ has exactly two solutions $\hat{x}_{3}$ and 1, where $\hat{x}_{3}\in(0,1)$. From (22), we obtain that $h^{\prime}(x)<0$, for $x\in(\hat{x}_{3},1)$ and $h^{\prime}(x)>0$, for $x\in(0,\hat{x}_{3})\cup(1,\infty)$.

∎

We next study the local maxima of $f_{m_{1},\alpha,d+1}(t)$ for each $m_{1}\in[1,\frac{d+1}{2}]\cap{\mathbb{Z}}$ and $\alpha\in(1,1+\frac{1}{d-1}]$. The following lemma shows that if $1<\alpha\leq 1+\frac{1}{d-1}$, then $f_{m_{1},\alpha,d+1}(t)$ attains a local maximum at $t_{0}$ only if $t_{0}\in\{0,\frac{1}{d+1},\frac{1}{m_{1}}\}$.

Lemma 3.3**.**

Assume $d\geq 2$ is an integer and $m_{1}\in[1,\frac{d+1}{2}]\cap{\mathbb{Z}}$.

1. (i)

Assume that $1<\alpha<1+\frac{1}{d-1}$, $t_{0}\in[0,\frac{1}{m_{1}}]$ and $f_{m_{1},\alpha,d+1}(t)$ has a local maximum at $t_{0}$. Then $t_{0}\in\left\{0,\frac{1}{d+1},\frac{1}{m_{1}}\right\}.$ 2. (ii)

Assume that $\alpha=1+\frac{1}{d-1}$, $t_{0}\in[0,\frac{1}{m_{1}}]$ and $f_{m_{1},\alpha,d+1}(t)$ has a local maximum at $t_{0}$. Then $t_{0}\in\left\{0,\frac{1}{m_{1}}\right\}.$

Proof.

For convenience, let $m_{2}:=d+1-m_{1}>1$. Recall that

[TABLE]

where $s=\frac{1-m_{1}\cdot t}{m_{2}}$. Noting that $t,s\geq 0$ and $m_{1}\cdot t+m_{2}\cdot s=1$, we can set $t=\frac{\cos^{2}\theta}{m_{1}}$, $s=\frac{\sin^{2}\theta}{m_{2}}$, where $\theta\in[0,\frac{\pi}{2}]$. We use the substitution $t=\frac{\cos^{2}\theta}{m_{1}},s=\frac{\sin^{2}\theta}{m_{2}}$ to transform the function from $f_{m_{1},\alpha,d+1}(t)$ to

[TABLE]

To this end, it is enough to study the local maxima of $g$ on $[0,\frac{\pi}{2}]$. A simple calculation shows that

[TABLE]

We can rewrite $g^{\prime}(\theta)$ as

[TABLE]

where $v:=\sqrt{\frac{s}{t}}=\sqrt{\frac{m_{1}}{m_{2}}}\cdot\frac{\sin\theta}{\cos\theta}$ and $h(v):=(m_{2}-1)v^{4\alpha-2}-m_{2}\cdot v^{2\alpha}+m_{1}\cdot v^{2\alpha-2}-(m_{1}-1)$. Particularly, when $\theta=\theta_{*}:=\arctan(\sqrt{\frac{m_{2}}{m_{1}}})$, we have $v=\sqrt{\frac{m_{1}}{m_{2}}}\cdot\frac{\sin\theta_{*}}{\cos\theta_{*}}=1$.

Noting that $\alpha>1$, $m_{1}\geq 1$ and $m_{2}>1$, we obtain

[TABLE]

Since $4\alpha\cdot\frac{m_{1}}{m_{1}^{2\alpha}}\cdot(\cos\theta)^{4\alpha-1}\sin\theta$ is positive for any $\theta\in(0,\frac{\pi}{2})$, to study the monotonicity of $g(\theta)$, it is enough to consider the sign of $h(v)$ with $v>0$.

(i) First we consider the case $1<\alpha<1+\frac{1}{d-1}$.

Lemma 3.2 shows that there exists $\hat{v}_{1}\in(0,1)$ and $\hat{v}_{2}\in(1,\infty)$ such that $h^{\prime}(v)>0$ for $v\in(0,\hat{v}_{1})\cup(\hat{v}_{2},\infty)$ and $h^{\prime}(v)<0$ for $v\in(\hat{v}_{1},\hat{v}_{2})$. Noting that $h(1)=0$ and $\hat{v}_{1}<1<\hat{v}_{2}$, we obtain that $h(\hat{v}_{1})>0$ and $h(\hat{v}_{2})<0$. Combining Lemma 3.2 and the results above, we obtain that $h(v)=0$ has exactly one solution on $[0,\hat{v}_{1})$ , say $v_{1}$. Similarly, $h(v)=0$ also has exactly one solution on $(\hat{v}_{2},\infty)$, say $v_{2}$. Let $\theta_{1}:=\arctan(v_{1}\sqrt{\frac{m_{2}}{m_{1}}})$ and $\theta_{2}:=\arctan(v_{2}\sqrt{\frac{m_{2}}{m_{1}}})$.

If $m_{1}=1$, then we have $h(0)=0$ and hence $v_{1}=0$. From the monotonicity of $h(v)$, we obtain that $h(v)<0$, $v\in(1,v_{2})$, $h(v)>0$, $v\in(0,1)\cup({v}_{2},\infty)$ and $h(v)=0$, $v\in\{0,1,v_{2}\}$. Then from (31) it is easy to check that $g^{\prime}(\theta)<0$, $\theta\in(\theta_{*},\theta_{2})$, $g^{\prime}(\theta)>0$, $\theta\in(0,\theta_{*})\cup(\theta_{2},\frac{\pi}{2})$ and $g^{\prime}(\theta)=0$, $\theta\in\{0,\theta_{*},\theta_{2},\frac{\pi}{2}\}$, which implies $g(\theta)$ has only two local maxima: $\theta_{*}$ and $\frac{\pi}{2}$.

If $m_{1}>1$, then $h(0)<0$, which means $v_{1}\in(0,\hat{v}_{1})$. Thus, by the monotonicity of $h(v)$ we conclude that $h(v)<0$, $v\in(0,v_{1})\cup(1,v_{2})$, $h(v)>0$, $v\in({v}_{1},1)\cup({v}_{2},\infty)$ and $h(v)=0$, $v\in\{v_{1},1,v_{2}\}$. We can use (31) to transform these results to $g^{\prime}(\theta)$. Hence, we obtain that $g^{\prime}(\theta)<0$, $\theta\in(0,\theta_{1})\cup(\theta_{*},\theta_{2})$, $g^{\prime}(\theta)>0$, $\theta\in(\theta_{1},\theta_{*})\cup(\theta_{2},\frac{\pi}{2})$ and $g^{\prime}(\theta)=0$, $\theta\in\{0,\theta_{1},\theta_{*},\theta_{2},\frac{\pi}{2}\}$, which implies $g(\theta)$ has only three local maxima: 0, $\theta_{*}$ and $\frac{\pi}{2}$.

(ii) We next consider the case where $\alpha=1+\frac{1}{d-1}$. We divided the proof into two cases.

Case 1: $m_{1}=\frac{d+1}{2}$ . Lemma 3.2 implies that $h(v)$ is monotonically increasing on $(0,\infty)$. Noting that $h(0)=-(m_{1}-1)<0$ and $h(1)=0$, we have $h(v)<0$, $v\in(0,1)$ and $h(v)>0$, $v\in(1,\infty)$. We use (31) to transform the result to $g^{\prime}(\theta)$ and obtain that $g^{\prime}(\theta)<0$, $\theta\in(0,\theta_{*})$, $g^{\prime}(\theta)>0$, $\theta\in(\theta_{*},\frac{\pi}{2})$ and $g^{\prime}(\theta)=0$, $\theta\in\{0,\theta_{*},\frac{\pi}{2}\}$, which implies $g(\theta)$ has only two local maxima: 0 and $\frac{\pi}{2}$.

Case 2: $m_{1}<\frac{d+1}{2}$ . According to Lemma 3.2, there exists $\hat{v}_{3}\in(0,1)$ such that $h^{\prime}(v)>0$ for $v\in(0,\hat{v}_{3})\cup(1,\infty)$ and $h^{\prime}(v)<0$ for $v\in(\hat{v}_{3},1)$.

If $m_{1}=1$, then $h(0)=h(1)=0$. According to the sign of $h^{\prime}(v)$, we obtain that $h(v)\geq 0$, $v\in[0,\infty)$. Equation (31) implies that $g^{\prime}(\theta)$ is always non-negative on $[0,\frac{\pi}{2}]$, which means $\frac{\pi}{2}$ is the only local maxima of $g(\theta)$.

If $1<m_{1}<\frac{d+1}{2}$, then $h(0)<0$. So there exists $v_{3}\in(0,\hat{v}_{3})$ such that $h(v)<0$, $v\in(0,v_{3})$ and $h(v)\geq 0$, $v\in[v_{3},\infty)$. Set $\theta_{3}:=\arctan(v_{3}\sqrt{\frac{m_{2}}{m_{1}}})$. According to (31), we have $g^{\prime}(\theta)<0$, $\theta\in(0,\theta_{3})$, $g^{\prime}(\theta)>0$, $\theta\in(\theta_{3},\frac{\pi}{2})$ and $g^{\prime}(\theta)=0$, $\theta\in\{0,\theta_{3},\frac{\pi}{2}\}$, which implies $g(\theta)$ has only two local maxima: 0 and $\frac{\pi}{2}$.

∎

Remark 3.4**.**

When $1<\alpha\leq 1+\frac{1}{d-1}$, combining Lemma 3.1 and Lemma 3.3, we obtain that $(\frac{1}{d+1},\ldots,\frac{1}{d+1})$ is the only local maxima of $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ with the constraints $z_{1}+\cdots+z_{d+1}=1$ and $z_{i}>0$, $i=1,2,\ldots,d+1$.

We deal with the case $\alpha>1+\frac{1}{d-1}$ in the next lemma.

Lemma 3.5**.**

Assume that $\alpha>1+\frac{1}{d-1}$ and $d\geq 2$. Assume that $(w_{1},w_{2},\ldots,w_{d+1})$ is a local maxima of $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ with the constraints in (7). Then there exists $k_{0}\in\{1,\ldots,d+1\}$ such that $w_{k_{0}}=0$.

Proof.

We proceed by contradiction, supposing that $w_{i}>0$ for $i\in\{1,\ldots,d+1\}$. According to Lemma 3.1, $(w_{1},\ldots,w_{d+1})$ is in the form $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d+1-m_{1}}\right)$ up to a permutation where $m_{1}\in[1,\frac{d+1}{2}]\cap{\mathbb{Z}}$, $t_{0}\in(0,\frac{1}{m_{1}})$ and $s_{0}=\frac{1-m_{1}t_{0}}{d+1-m_{1}}$. Lemma 3.1 also implies that $t_{0}$ is a local maxima of $f_{m_{1},\alpha,d+1}(t)$. So, it is enough to show the following claim:

Claim 1: When $\alpha>1+\frac{1}{d-1}$, if $t_{0}\in(0,\frac{1}{m_{1}})$ is a local maxima of $f_{m_{1},\alpha,d+1}(t)$, then $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d+1-m_{1}}\right)$ is not a local maxima of $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ with the constraints in (7).

Claim 1 contradicts $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d+1-m_{1}}\right)$ being a local maxima of $M_{\alpha,d+1}(z_{1},\ldots,z_{d+1})$ with the constraints in (7). Hence, there exists $k_{0}\in\{1,\ldots,d+1\}$ such that $w_{k_{0}}=0$.

It remains to prove Claim 1. For convenience, set $m_{2}:=d+1-m_{1}$. Since $m_{1}\leq\frac{d+1}{2}$ and $d\geq 2$, we have $m_{2}\geq 2$. Set

[TABLE]

where $l=m_{2}-1$ and $\varepsilon\in(-\frac{s_{0}}{l},s_{0})$. We aim to show that $\varepsilon=0$ is not a local maxima of $F(\varepsilon)$. In fact, we can prove this by showing that $\varepsilon=0$ is a local minima of $F(\varepsilon)$.

A simple calculation shows that

[TABLE]

Noting $l=m_{2}-1$, we can check that

[TABLE]

We claim $F^{\prime\prime}(0)>0$ and hence $\varepsilon=0$ is a local minima of $F(\varepsilon)$.

It remains finally to prove $F^{\prime\prime}(0)>0$. Note that

[TABLE]

Since $t_{0}\notin\{0,\frac{1}{m_{1}}\}$ is a local maxima of $f_{m_{1},\alpha,d+1}(t)$, from equation (31) we know that $\sqrt{\frac{s_{0}}{t_{0}}}$ is a root of $h(v)=0$, where $h(v)=(m_{2}-1)v^{4\alpha-2}-m_{2}\cdot v^{2\alpha}+m_{1}\cdot v^{2\alpha-2}-(m_{1}-1)$. According to Lemma 3.2, $h(v)>0$ for $v>1$ provided $\alpha\geq 1+\frac{1}{d-1}$, which implies that $\sqrt{\frac{s_{0}}{t_{0}}}\leq 1$ and hence $s_{0}\leq t_{0}$. Combining $s_{0}>0$ and $l^{2}+m_{2}-1\geq 2$, we have

[TABLE]

Noting that $\alpha>1+\frac{1}{d-1}$, we obtain

[TABLE]

∎

We next present the proof of Lemma 2.1.

Proof of Lemma 2.1.

We prove Lemma 2.1 by induction on $d$. First, we consider the case $d=1$. For $d=1$, we have only two non-negative variables $z_{1},z_{2}$ which satisfy $z_{1}+z_{2}=1$. For any $\alpha>1$ we have

[TABLE]

where equality holds if and only if $z_{1}=z_{2}=\frac{1}{2}$. Hence, the solution to (7) is $(\frac{1}{2},\frac{1}{2})$ which implies Lemma 2.1 holds for $d=1$. We assume that Lemma 2.1 holds for $d=d_{0}-1$ and hence we know the solution to (7) for $d=d_{0}-1$. So, we consider the case where $d=d_{0}$.

Assume that $(w_{1},\ldots,w_{d_{0}+1})$ is a solution to (7) with $d=d_{0}$. Recall that $a_{0}=\infty$, $a_{d_{0}}=1$, $a_{k}=\frac{1}{2}\cdot\frac{\ln(k+2)-\ln(k)}{\ln(k+2)-\ln(k+1)}$, $k=1,2,\ldots,d_{0}-1$. For convenience, we set ${\mathbf{e}}_{k+1}:=(\frac{1}{k+1},\ldots,\frac{1}{k+1})\in{\mathbb{R}}^{k+1}$ and ${\mathbf{0}}_{d_{0}-k}:=(0,\ldots,0)\in{\mathbb{R}}^{d_{0}-k}$. We also set

[TABLE]

We first show

[TABLE]

dividing the proof into two cases.

Case 1: $\alpha\in(1+\frac{1}{d_{0}-1},\infty)$ .

According to Lemma 3.5, at least one of the entries in $(w_{1},\ldots,w_{d_{0}+1})$ is 0. Without loss of generality, we assume $w_{d_{0}+1}=0$. Since $M_{\alpha,d_{0}+1}(w_{1},\ldots,w_{d_{0}},0)=M_{\alpha,d_{0}}(w_{1},\ldots,w_{d_{0}})$, $(w_{1},\ldots,w_{d_{0}})$ is the solution to (7) with $d=d_{0}-1$. Hence, by induction we conclude that (35) holds.

Case 2: $\alpha\in(1,1+\frac{1}{d_{0}-1}]$.

If one of entries in $(w_{1},\ldots,w_{d_{0}+1})$ is 0, we can show that (35) holds using a similar argument as above. So, we consider the case where $w_{i}>0$ for each $i\in\{1,\ldots,d_{0}+1\}$. Lemma 3.1 shows that $(w_{1},\ldots,w_{d_{0}+1})$ is in the form $\left(\underbrace{t_{0},\ldots,t_{0}}_{m_{1}},\underbrace{s_{0},\ldots,s_{0}}_{d_{0}+1-m_{1}}\right)$ up to a permutation where $m_{1}\in[1,\frac{d_{0}+1}{2}]\cap{\mathbb{Z}}$, $t_{0}\in(0,\frac{1}{m_{1}})$ and $s_{0}=\frac{1-m_{1}t_{0}}{d_{0}+1-m_{1}}$ . Lemma 3.1 also implies that $t_{0}$ is a local maxima of the function $f_{m_{1},\alpha,d_{0}+1}(t)$, where $f_{m_{1},\alpha,d_{0}+1}(t)$ is defined in (18). According to Lemma 3.3, we obtain $t_{0}=\frac{1}{d_{0}+1}$. Hence $(w_{1},\ldots,w_{d_{0}+1})=(\frac{1}{d_{0}+1},\ldots,\frac{1}{d_{0}+1})$ , which implies (35).

It is now enough to compare the values among $M_{\alpha,d_{0}+1}\left({\mathbf{e}}_{k+1},{\mathbf{0}}_{d_{0}-k}\right),k=1,\ldots,d_{0}.$ Setting $H(x):=x^{1-2\alpha}(x-1)$, we obtain $M_{\alpha,d_{0}+1}\left({\mathbf{e}}_{k+1},{\mathbf{0}}_{d_{0}-k}\right)=H(k+1)$ for each $k\in\{1,2,\ldots,d_{0}\}$. A simple calculation shows that $H(x)$ is monotonically increasing on $(0,1+\frac{1}{2\alpha-2})$ and monotonically decreasing on $(1+\frac{1}{2\alpha-2},\infty)$. Hence, the sequence $H(k+1),k=1,\ldots,d_{0}$, is unimodal.

(i) Firstly, we consider the case where $\alpha\in(a_{k},a_{k-1})$, $k=1,2,\ldots,d_{0}-1$. Noting that $H(k)<H(k+1)$ and $H(k+1)>H(k+2)$, we obtain

[TABLE]

where equality holds if and only if $x=k$. For the case $\alpha\in(a_{d_{0}},a_{d_{0}-1})$, noting that $H(d_{0})<H(d_{0}+1)$, we obtain

[TABLE]

To summarize, $\left({\mathbf{e}}_{k+1},{\mathbf{0}}_{d_{0}-k}\right)$ is the unique solution to (7) with $d=d_{0}$ when $\alpha\in(a_{k},a_{k-1})$, $k=1,2,\ldots,d_{0}$.

(ii) It remains finally to check the case where $\alpha=a_{k}$, $k=1,2,\ldots,d_{0}-1$. Noting $H(k+1)=H(k+2)$, $H(k)<H(k+1)$ and $H(k+2)>H(k+3)$ provided $\alpha=a_{k},\ k=1,2,\ldots,d_{0}-2$, we obtain that (7) has two solutions which are $\left({\mathbf{e}}_{k+2},{\mathbf{0}}_{d_{0}-k-1}\right)$ and $\left({\mathbf{e}}_{k+1},{\mathbf{0}}_{d_{0}-k}\right)$ with $d=d_{0}$ . When $\alpha=a_{d_{0}-1}$, noting $H(d_{0})=H(d_{0}+1)$ and $H(d_{0}-1)<H(d_{0})$, we obtain that (7) also has two solutions which are ${\mathbf{e}}_{d_{0}+1}$ and $\left({\mathbf{e}}_{d_{0}},{\mathbf{0}}_{1}\right)$ with $d=d_{0}$ . Hence, the conclusion also holds for $d=d_{0}$ which completes the proof. ∎

Acknowledgement. We would like to thank the anonymous reviewers for their comments which help improve this paper substantially.