On a Conjecture of Bahri-Xu
aa r X i v : . [ m a t h . C A ] J a n ON A CONJECTURE OF BAHRI-XU
HONG CHEN, JIANQUAN GE, KAI JIA, AND ZHIQIN LU
Abstract.
In order to study the Yamabe changing-sign problem, Bahri and Xuproposed a conjecture which is a universal inequality for p points in R m . They haveverified the conjecture for p ≤
3. In this paper, we first simplify this conjectureby giving two sufficient and necessary conditions inductively. Then we prove theconjecture for the basic case m = 1 with arbitrary p . In addition, for the cases when p = 4 , m ≥
2, we manage to reduce them to the basic case m = 1 and thusprove them as well. Introduction
In the study of Yamabe problem on S , we consider a semi-linear equation(1.1) ∆ R u + u = 0 , u > . It is well known that δ ( a, λ ) = c √ λ (1 + λ | x − a | ) / for any real number λ and vector a ∈ R (with appropriate constant c ) is a solution tothe above equation. Moreover, for λ i large enough, the combinations P pi =1 δ ( a i , λ i ) are almost solutions to the equation.From PDE point of view, it would be interesting to study Equation (1.1) withoutthe positivity assumption on u as well. In the book [1], Bahri and Xu introducedthe problem of studying the Morse Lemma at infinity. Morse Lemma is interestingin this situation because even for positive solutions, Equation (1.1) is a variationalproblem with defects. In order to study these defects, in [1], the following question wasintroduced.We consider the functional J ( u ) defined by J ( u ) = (cid:18)Z R u d x (cid:19) − Mathematics Subject Classification.
Primary: 58C40; Secondary: 58E35.
Key words and phrases.
Yamabe problem; Bahri-Xu Conjecture.The second author is partially supported by Beijing Natural Science Foundation (Z190003). Thelast author is partially supported by National Sciences Foundation of USA (DMS-19-08513). on the space Σ = { w | Z R ( |∇ w | + w ) dx < ∞ , Z R |∇ w | dx = 1 } . Let ¯ w , · · · , ¯ w p be p (possibly sign changing) solutions of (1.1). Let a , · · · , a p ∈ R be p points. Let λ , · · · , λ p > J that corresponds to the following linear combination p X i =1 α i p λ i ¯ w i ( λ i ( x − a i ))of solutions. More precisely, we wish to establish an asymptotic expansion of thefollowing J p X i =1 α i p λ i ¯ w i ( λ i ( x − a i )) + v , where α i are constants and v is a function satisfying Condition (Vo) in [1, Page 3].In order to establish such an asymptotic expansion, Bahri and Xu made two ad-ditional assumptions (conjectures), one of which is purely linear algebraic, and can bestated as follows Conjecture 1.1.
Let x , . . . , x p ( p ≥
2) be distinct vectors in R m . Then there is apositive constant c = c ( p, m ) depending only on p and m , such that | AU | + sup ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) U T (cid:18) ∂A∂x i (cid:19) U (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c ( p, m ) X i,jj = i u j | x i − x j | (1.2)for any U = (cid:0) u , . . . , u p (cid:1) T ∈ R p , where A = ( a ij ) ≤ i,j ≤ p is a p × p matrix with entries a ij = , i = j | x i − x j | , i = j that is, A = | x − x | · · · | x − x p | | x − x | · · · | x − x p | ... ... . . . ... | x p − x | | x p − x | · · · , and ∂ A∂x k = ∂ a ij ∂x k ! ≤ i,j ≤ p is a vector-valued p × p matrix. N A CONJECTURE OF BAHRI-XU 3
Equivalently, (1.2) can be rewritten as X ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c ( p, m ) X i,jj = i u j | x i − x j | . (1.3)Let I , I denote the two hands of (1.3): I = X ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ,I = X i,jj = i u j | x i − x j | . Remark 1.
The conjecture was stated in [1, page 4, Conjecture 2], where m = 3.We extend the conjecture from 3 dimensional to arbitrary dimensional. We found thateven in the case m = 1, that is, when all x i are real numbers instead of vectors, theconjecture is still interesting and open. Remark 2.
Due to the symmetries of (1.3), we only need to consider the conjecturefor m ≤ p −
1. In fact, if m ≥ p , we may assume that x , . . . , x p lie in R p − ֒ → R m since any p points in R m must lie in some affine subspace of dimension p − R m .The main purpose of this paper is to study Conjecture 1.1. If p = 2, then | AU | = u + u | x − x | = I , and thus the conjecture is valid for c ( p, m ) = 1.In [2], Xu proved the case p = m = 3. By the above remark, Xu’s result actuallyimplies the cases of p = 3 with arbitrary m .From now on, we are only concerned with p ≥ m = 1 with arbitrary p ; and p = 4 , m .In Section 2 we give the following equivalent characterization for Conjecture 1.1. Theorem 1.1.
Let p ≥ . Conjecture 1.1 holds for any ≤ p ≤ p , if and only if forany ≤ p ≤ p and distinct x , . . . , x p ∈ R m , the equations X j = i u j | x i − x j | = 0 , ≤ i ≤ p, (1.4) and, u i X j = i x i − x j | x i − x j | u j = 0 , ≤ i ≤ p, (1.5) H. CHEN, J.Q. GE, K. JIA, AND Z.Q. LU about u , . . . , u p have NO non-zero solution. In Section 3 we firstly prove the following simplified equivalent conditions, usingwhich we then prove the special cases mentioned before.
Theorem 1.2.
Let p ≥ . Conjecture 1.1 holds for ≤ p ≤ p , if and only if for ≤ p ≤ p , given any distinct y , . . . , y p ∈ S m , the equations X j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p (1.6) for v , . . . , v p have NO non-zero solution. Proof of Theorem 1.1
In this section, we prove Theorem 1.1 by induction on p . Assuming that Conjec-ture 1.1 holds for p ≤ p − p ≥ p = p .The difficulty lies in the analysis when there are singularities, namely, one or morepairs of vectors with distance | x i − x j | going to zero or infinity. This would be alsoan obstacle if one tried to prove the equivalence by contradiction. Hence from theinductive viewpoint, we treat with these two cases of singularities in Lemmas 2.2 and2.1 separately.First, observe that the inequality is homogeneous with respect to x , . . . , x p . Thus,without loss of generality, we may assume that | x − x | = min i = j | x i − x j | = 1, and1 = | x − x | ≤ | x − x | ≤ · · · ≤ | x − x p | . The following Lemma 2.1 deals with unbounded cases.
Lemma 2.1.
Assume that Conjecture 1.1 holds for p ≤ p − . When p = p ≥ ,let ≤ s ≤ p − . For any β , . . . , β s − ≥ , there exists M s > , such that for any x , . . . , x s satisfying | x − x i +1 | ≤ β i , ≤ i ≤ s − , ( ⋆ ) Conjecture 1.1 holds if | x − x s +1 | > M s .Proof. Fix any β , . . . , β s − ≥
1, then | x − x i +1 | ≤ β i , for i = 1 , , . . . , s −
1. Let α = β s − , then 1 ≤ | x − x s | ≤ α . In this proof, we consider only the case | x − x s +1 | ≥ α . N A CONJECTURE OF BAHRI-XU 5
Firstly, we give some important inequalities that would be useful in the followingproof. For j ≤ s, i ≥ s + 1, | x i − x j | ≥ | x − x i | − | x − x j |≥ α − | x − x j |≥ | x − x j | − | x − x j | = | x − x j | ;(2.1) | x i − x s +1 | ≤ | x i − x | + | x − x s +1 |≤ | x i − x |≤ | x i − x j | + 2 | x j − x |≤ | x i − x j | ;(2.2) | x j − x s +1 | ≤ | x i − x j | . (2.3)Now, we deal with I . Since ( a + 2 b ) = a +4 ab +4 b = 2[( a + b ) − ( a / − b )] ≥ X ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j ≤ s +1 u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s +1 u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.4)Since sup i ≤ s +1 | a i + b i | ≥ | a k + b k | ≥ | a k | − | b k | ≥ | a k | − sup i ≤ s +1 | b i | , ∀ k ≤ s + 1 , we have sup i ≤ s +1 | a i + b i | ≥ sup i ≤ s +1 | a i | − sup i ≤ s +1 | b i | . Thus, 2 sup ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j ≤ s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j>s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.5) H. CHEN, J.Q. GE, K. JIA, AND Z.Q. LU
By (2.4) and (2.5), we have I ≥ X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j ≤ s +1 u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j ≤ s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s +1 u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j>s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since s + 1 ≤ p −
1, by the inductive assumption, there exists constant c = c ( p, m ) >
0, such that I ≥ c X i,j ≤ s +1 ,j = i u j | x i − x j | − X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s +1 u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j>s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.6)Consider the middle term in (2.6), by the special case of the power means inequal-ity, we have 1 p − s − X j>s +1 u j | x i − x j | ≤ vuut p − s − X j>s +1 u j | x i − x j | , thus, X i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s +1 u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( p − s − X j>s +1 ,i ≤ s +1 u j | x i − x j | . (2.7)Consider the last term in (2.6). When i = s + 1, then2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s +1 u s +1 x s +1 − x j | x s +1 − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X j>s +1 | u s +1 u j || x s +1 − x j | ≤ X j>s +1 u s +1 | x s +1 − x j | + X j>s +1 u j | x s +1 − x j | . (2.8)When i ≤ s , since | x i − x j | ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X j>s +1 | u i u j || x i − x j | ≤ X j>s +1 | u i || u j || x i − x j | . (2.9)By the inequality of arithmetic and geometric means,1 a | u i | + a | u j | | x i − x j | ≥ | u i || u j || x i − x j | , N A CONJECTURE OF BAHRI-XU 7 taking a = ( p − s −
1) 2 α c >
0, we get X j>s +1 | u i || u j || x i − x j | ≤ X j>s +1 a u i + X j>s +1 a u j | x i − x j | ≤ c α u i + a X j>s +1 ,i ≤ s u j | x i − x j | . (2.10)Now take i : 1 ≤ i ≤ s , such that u i = max ≤ i ≤ s u i . If i = 1, we take the term i = 2 , j = 1 in the sum c P i,j ≤ s +1 ,j = i u j | x i − x j | , then12 c X i,j ≤ s +1 ,j = i u j | x i − x j | ≥ c u | x − x | ≥ c α u i . If 2 ≤ i ≤ s , we take the term i = 1 , j = i , then12 c X i,j ≤ s +1 ,j = i u j | x i − x j | ≥ c u i | x i − x | ≥ c α u i . In either case, we have c α u i ≤ c X i,j ≤ s +1 ,j = i u j | x i − x j | (2.11)By (2.8),(2.9),(2.10) and (2.11), the last term in (2.6) has the following upper bound:2 sup ≤ i ≤ s +1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j>s +1 u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c X i,j ≤ s +1 ,j = i u j | x i − x j | + c X j>s +1 ,i ≤ s +1 u j | x i − x j | + X j>s +1 u s +1 | x s +1 − x j | , (2.12)where c = max { , a } > I ≥ c X i,j ≤ s +1 ,j = i u j | x i − x j | − ( p − s − c ) X j>s +1 ,i ≤ s +1 u j | x i − x j | + X j>s +1 | u s +1 | | x s +1 − x j | ≥ c X i,j ≤ s +1 ,j = i u j | x i − x j | − c X j>s +1 ,i ≤ s +1 u j | x i − x j | + X j>s +1 u s +1 | x s +1 − x j | , (2.13)where c = max { p − s − c , } = p − s − c ≥ H. CHEN, J.Q. GE, K. JIA, AND Z.Q. LU
Now, consider the minus term in (2.13). Rewrite and then use inequalities (2.1),(2.2) and (2.3): X j>s +1 ,i ≤ s +1 u j | x i − x j | + X j>s +1 u s +1 | x s +1 − x j | = X j>s +1 ,i ≤ s u j | x i − x j | + X j>s +1 ,i = s +1 u j | x i − x j | + X i>s +1 ,j = s +1 u j | x i − x j | ≤ X j>s +1 ,i ≤ s u j | x s +1 − x j | + X j>s +1 ,i = s +1 u j | x i − x j | + X i>s +1 ,j = s +1 u j | x i − x j | = (16 s + 1) X j>s +1 ,i = s +1 u j | x i − x j | + X i>s +1 ,j = s +1 u j | x i − x j | ≤ (16 s + 1) X i,j ≥ s +1 ,i = j u j | x i − x j | . Set c = (16 s + 1) c >
0, then (2.13) becomes I ≥ c X i,j ≤ s +1 ,j = i u j | x i − x j | − c X i,j ≥ s +1 ,j = i u j | x i − x j | . (2.14)By similar argument as in (2.4) and (2.5), we have I ≥ X i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j = i,j>s u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j>s,j = i u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − X i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . By the inductive assumption, there exists c > I ≥ c X i,j>s,j = i u j | x i − x j | − X i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (2.15)We claim that given any ε >
0, there exists b ε ≥
1, such that X i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ εI (2.16)when | x i − x j | > b ε for i > s, j ≤ s . N A CONJECTURE OF BAHRI-XU 9
In fact, by homogeneity we can assume | u i | ≤ i and thus (2.16) can beestimated as follows: the left hand is X i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i>s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j ≤ s u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X i>s (cid:16) s X j ≤ s u j | x i − x j | + 2 s X j ≤ s u j | x i − x j | (cid:17) ≤ s ( p − s + 1) b ε X j ≤ s u j , while the right hand is I ≥ X j ≤ s (cid:16) X i = j | x i − x j | (cid:17) u j ≥ α X j ≤ s u j . Therefore, setting b ε = r s ( p − s + 1)4 α ε , we get the required inequality (2.16).By (2.15) and (2.16), I ≥ c X i,j>s,j = i u j | x i − x j | − εI . (2.17)Now, let δ = c c , then δ · (2.14) + (2.17) :(1 + δ ) I ≥ c δ X i,j ≤ s +1 ,j = i u j | x i − x j | + c X i,j>s,j = i u j | x i − x j | − εI ≥ c X i,j ≤ s +1 ,j = i u j | x i − x j | + X i,j>s,j = i u j | x i − x j | − εI . (2.18)where c = min n c δ , c o > . By (2.2) and (2.3), X j>s +1 ,i ≤ s u j | x i − x j | ≤ s X j>s +1 u j | x j − x s +1 | ≤ s X i,j>s,i = j u j | x i − x j | , X j ≤ s,i>s +1 u j | x i − x j | ≤ X j ≤ s,i>s +1 u j | x j − x s +1 | = 25( p − s − X j ≤ s u j | x j − x s +1 | ≤ p − s − X i,j ≤ s +1 ,i = j u j | x i − x j | . Thus, I ≤ (16 s + 1) X i,j>s u j | x i − x j | + (cid:0) p − s −
1) + 1 (cid:1) X i,j ≤ s +1 ,i = j u j | x i − x j | ≤ c X i,j>s u j | x i − x j | + X i,j ≤ s +1 ,i = j u j | x i − x j | , (2.19)where c = max (cid:8) s + 1 , p − s −
1) + 1 (cid:9) > ε = c c , choose M s = M s ( ε ) ≥ α . By (2.18) and (2.19), we have I ≥ c δ ) c I = cI , where c = c δ ) c > p and m . (cid:3) Now, let s = 2 and take β = 1. Then by Lemma 2.1, there exists M > ⋆ ) holds (automatically) and | x − x | > M . Then itsuffices to consider the case | x − x | ≤ M .Similarly, let s = 3 and take β = M . Then by Lemma 2.1, there exists M > ⋆ ) holds and | x − x | > M . Then it suffices toconsider the case | x − x | ≤ M .Continuing the process repeatedly, we reach s = p − β , . . . , β p − . Again, by Lemma 2.1, there exists M p − > ⋆ ) holds and | x − x p − | > M p − . It suffices to consider the case | x − x p − | ≤ M p − .Therefore, from now on, we may assume that1 ≤ | x i − x j | ≤ β, ≤ i, j ≤ p − . N A CONJECTURE OF BAHRI-XU 11
Given x , . . . , x p − , let c ( p, m, x , . . . , x p − ) denote the infimum of X i,j ≤ p − i = j u j | x i − x j | + ˜ u p − × X i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p − ,j = i, u j | x i − x j | + ˜ u p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p − ,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , where u + · · · + u p − + ˜ u p = 1. Lemma 2.2.
Assume Conjecture 1.1 holds for p ≤ p − . If c ( p, m, x , . . . , x p − ) > holds for any x , . . . , x p − , then Conjecture 1.1 is valid for p = p .Proof. Since c ( p, m, x , . . . , x p − ) is continuous relative to x , . . . , x p − , there existspositive number c ( p, m ) depending only on p, m such that c ( p, m, x , . . . , x p − ) ≥ c ( p, m ) > . Thus, X i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p − ,j = i u j | x i − x j | + ˜ u p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p − ,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c ( p, m ) X i,j ≤ p − ,j = i u j | x i − x j | + ˜ u p , (2.20)for u + · · · + u p − + ˜ u p = 1. Let u p = ˜ u p | x − x p | .Similar to the argument as in the proof of Lemma 2.1, we have X i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p,j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ X i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p − ,j = i u j | x i − x j | + ˜ u p + u p | x i − x p | − ˜ u p !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ X i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p − ,j = i u j | x i − x j | + ˜ u p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − X j ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − x p || x j − x p | − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ u p , (2.21) andsup i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ sup i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p − ,j = i x i − x j | x i − x j | u j + u i x i − x p | x i − x p | u p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ sup i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p − ,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − sup i ≤ p − | x − x p || x i − x p | | ˜ u p u i |≥ sup i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p − ,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − X j ≤ p − | x − x p || x i − x p | | ˜ u p u i | . (2.22)Thus, by (2.20), (2.21) and (2.22), X i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p,j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ X i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p − ,j = i u j | x i − x j | + ˜ u p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p − ,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − X j ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − x p || x j − x p | − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ u p − X j ≤ p − | x − x p || x j − x p | | ˜ u p u j |≥ c ( p, m ) X i,j ≤ p − ,j = i u j | x i − x j | + ˜ u p − X j ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − x p || x j − x p | + 1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ u p + 2 X j ≤ p − | x − x p || x j − x p | | ˜ u p u j | . (2.23)Let β ′ > β + 1 > β β ′ − β + β (1 − ββ ′ )( β ′ − β ) < c ( p, m )4 p , and 1 − ββ ′ > √ ≈ . . Note that, the minus term in (2.23) can be written as2 X j ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − x p || x j − x p | − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ u p + 2 X j ≤ p − | x − x p || x j − x p | | ˜ u p u j | = 2 X ≤ j ≤ p − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − x p || x j − x p | − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ u p + | x − x p || x j − x p | | ˜ u p u j | + 2 | ˜ u p u || x − x p | . (2.24) N A CONJECTURE OF BAHRI-XU 13
Considering the case | x − x p | > β ′ , we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | x − x p || x j − x p | − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˜ u p + | x − x p || x j − x p | | ˜ u p u j |≤ (cid:0) | x − x p | − | x j − x p | (cid:1) | x j − x p | ˜ u p + | x − x p || x j − x p | ( ˜ u p + u j ) ≤ | x − x j | | x j − x p | ˜ u p + | x − x p || x j − x p | ˜ u p + | x − x j | | x − x p || x j − x p | u j | x − x j | ≤ β ( β ′ − β ) ˜ u p + | x − x j | | x − x p || x p − x j | ˜ u p + u j | x − x j | ! ≤ β ( β ′ − β ) ˜ u p + | x − x j | (cid:16) − | x − x j || x − x p | (cid:17) (cid:0) | x − x p | − | x − x j | (cid:1) ˜ u p + u j | x − x j | ! ≤ β ( β ′ − β ) ˜ u p + β (cid:16) − ββ ′ (cid:17) ( β ′ − β ) ˜ u p + u j | x − x j | ! ≤ β ( β ′ − β ) + β (cid:16) − ββ ′ (cid:17) ( β ′ − β ) ˜ u p + u j | x − x j | ! , and | ˜ u p u || x − x p | ≤ β ′ (cid:16) ˜ u p + u (cid:17) = 1 β ′ ˜ u p + u | x − x | ! . Thus,(2.24) ≤ β ( β ′ − β ) + β (cid:16) − ββ ′ (cid:17) ( β ′ − β ) X ≤ j ≤ p − ˜ u p + u j | x − x j | ! + 2 β ′ ˜ u p + u | x − x | ! ≤ β β ′ − β + β (cid:16) − ββ ′ (cid:17) ( β ′ − β ) ( p −
2) ˜ u p + X ≤ j ≤ p − u j | x − x j | + 2 β β ′ − β ˜ u p + u | x − x | ! ≤ β β ′ − β + β (cid:16) − ββ ′ (cid:17) ( β ′ − β ) × ( p −
1) ˜ u p + X ≤ j ≤ p − u j | x − x j | + u | x − x | ≤ c ( p, m )2 p ( p −
1) ˜ u p + X i,j ≤ p − ,i = j u j | x i − x j | ≤ c ( p, m )2 ˜ u p + X i,j ≤ p − ,i = j u j | x i − x j | . (2.25)By (2.23) and (2.25), X i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p,j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c ( p, m )2 X i,j ≤ p − ,j = i u j | x i − x j | + ˜ u p . (2.26)Note that for 1 ≤ j ≤ p − u j | x p − x j | ≤ u j (cid:0) | x p − x | − | x j − x | (cid:1) ≤ u j ( β ′ − β ) ≤ u j β ≤ u j | x − x j | . That is, X j ≤ p − u j | x p − x j | ≤ X j ≤ p − u j | x − x j | . (2.27)For 1 ≤ i ≤ p − | x − x p | | x i − x p | ≤ | x − x p | (cid:0) | x − x p | − | x − x i | (cid:1) = 1 (cid:16) − | x − x i || x − x p | (cid:17) ≤ (cid:16) − ββ ′ (cid:17) < . Thus, 1 | x i − x p | ≤ | x − x p | , ≤ i ≤ p − . Sum over i from 1 to p − X i ≤ p − | x i − x p | ≤ p − | x − x p | Thus, X i ≤ p − u p | x i − x p | ≤ p u p | x − x p | = 2 p ˜ u p . (2.28) N A CONJECTURE OF BAHRI-XU 15
Thus, by (2.27) and (2.28) X i,j ≤ p,i = j u j | x i − x j | = X i,j ≤ p − ,i = j u j | x i − x j | + X j ≤ p − ,i = p u j | x p − x j | + X i ≤ p − ,j = p u p | x i − x p | ≤ X i,j ≤ p − ,i = j u j | x i − x j | + 19 X j ≤ p − u j | x − x j | + 2 p ˜ u p ≤ p X i,j ≤ p − ,i = j u j | x i − x j | + ˜ u p . (2.29)Therefore, by (2.26) and (2.29), we have X i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p,j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c ( p, m )2 X i,j ≤ p − ,i = j u j | x i − x j | + ˜ u p ≥ c ( p, m )4 p X i,j ≤ p,i = j u j | x i − x j | . Thus, Conjecture 1.1 holds for | x − x p | > β ′ .Now, consider the case | x − x p | ≤ β ′ . we have1 ≤ | x i − x j | ≤ β ′ , ≤ i, j ≤ p. Let c ( p, m, x , . . . , x p ) be the infimum of X i,ji = j u j | x i − x j | − × X i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ≤ p,j = i, u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) u i X j ≤ p,j = i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) where u + . . . u p = 1. By the continuity of c ( p, m, x , . . . , x p ) relative to x , . . . , x p , itsuffices to prove that c ( p, m, x , . . . , x p ) > x , . . . , x p .Argue by contradiction. Assume that c ( p, m, x , . . . , x p ) = 0 for some given x , . . . , x p , then by the continuity relative to u , . . . , u p , there exist real numbers u , . . . , u p satisfying u + · · · + u p = 1 such that X j = i u j | x i − x j | = 0 , ≤ i ≤ p (2.30) u i X j = i x i − x j | x i − x j | u j = 0 , ≤ i ≤ p (2.31) Since the inequality is invariant under translation of x , . . . , x p , we may assume that x p = 0. Let t i = x i | x i | , v i = u i | x i | , ≤ i ≤ p − ,t p = ∞ , v p = u p . Note that | t i − t j | = | x i − x j || x i || x j | , ≤ i, j ≤ p − , i = j. When 1 ≤ i ≤ p −
1, we can rewrite (2.30) as0 = | x i | · X j ≤ p,j = i u j | x i − x j | = X j ≤ p − ,j = i v j | t i − t j | + v p , ≤ i ≤ p − , (2.32)that is, X j ≤ p − ,j = i | t i | − h t i , t j i + | t j | | t i − t j | v j + v p = 0 , ≤ i ≤ p − . (2.33)Similarly, 1 ≤ i ≤ p −
1, we can rewrite (2.31) as0 = u i X j = i x i − x j | x i − x j | u j = u i X j = i,j ≤ p − x i − x j | x i − x j | u j + u i x i | x i | u p = v i | t i | X j = i,j ≤ p − | t i | | t j | | t i − t j | t i | t i | − t j | t j | ! v j | t j | + v i t i v p , ≤ i ≤ p − , that is, v i X j = i,j ≤ p − | t j | t i − | t i | t j | t i − t j | v j + v i t i v p = 0 , ≤ i ≤ p − . (2.34)Taking inner product with t i , we have v i X j = i,j ≤ p − | t j | − h t i , t j i| t i − t j | v j + v i v p = 0 , ≤ i ≤ p − . (2.35)Taking the calculation v i × (2.33) − × (2.35), we get: v i X j ≤ p − ,j = i | t i | − | t j | | t i − t j | v j = v i v p , ≤ i ≤ p − N A CONJECTURE OF BAHRI-XU 17
Taking the calculation 1 | t i | ( t i × (2.36) + (2.34)), we get: v i X j ≤ p − ,j = i t i − t j | t i − t j | v j = 0 , ≤ i ≤ p − . (2.37)By (2.32) and (2.37), and by the assumption that c ( p, m, x , . . . , x p − ) >
0, wehave v i = 0 , ∀ i = 1 , . . . , p . Thus u i = 0 , ∀ i = 1 , . . . , p , which is contrary to theassumption that u + · · · + u p = 1. (cid:3) Before we apply Lemma 2.2, we need to define a useful transform.
Definition 1.
The Kelvin transform about some point N ∈ R m is defined as follows: K N : x N + x − N k x − N k , where x ∈ R m . Proposition 2.3.
Equations (1.4) and (1.5) about u , . . . , u p X j = i u j | x i − x j | = 0 , ≤ i ≤ p (1.4) u i X j = i x i − x j | x i − x j | u j = 0 , ≤ i ≤ p (1.5) are invariant under Kelvin transform.Proof. Fix N = x i , ∀ ≤ i ≤ p . Let y i = K N ( x i ) = N + x i − N | x i − N | , y ′ i = y i − N = x i − N | x i − N | , ≤ i ≤ p. Note that | y ′ i | = | x i − N | and that | y ′ i − y ′ j | = h y ′ i − y ′ j , y ′ i − y ′ j i = 1 | x i − N | | x j − N | (cid:16) | x i − N | − h x i − N, x j − N i + | x j − N | (cid:17) = | x i − x j || x i − N || x j − N | ! , thus, | x i − x j | = | y ′ i − y ′ j || y ′ i || y ′ j | . (1.4) and (1.5) will be changed to: | y ′ i | X j = i | y ′ i − y ′ j | | y ′ j | u j = 0 , ≤ i ≤ p (2.38) | y ′ i | u i X j = i y ′ i | y ′ j | − y ′ j | y ′ i | | y ′ i − y ′ j | | y ′ j | u j = 0 , ≤ i ≤ p (2.39)Taking the calculation u i y ′ i × (2.38) + (2.39) − y ′ i | y ′ i | × h (2.39) , y ′ i i , we get: | y ′ i | u i X j = i y ′ i − y ′ j | y ′ i − y ′ j | | y ′ j | u j = 0(2.40)Let v i = | y ′ i | u i for i = 1 , . . . , p , then (2.38) and (2.40) give X j = i | y i − y j | v j = 0 , ≤ i ≤ p (2.41) v i X j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p. (2.42)Therefore, the equations (1.4) and (1.5) are invariant under Kelvin transform. (cid:3) Sometimes we need to send some point to infinity, e.g., N = x p . Again, let y i = K x p ( x i ) and y ′ i = y i − x p , namely, y i = x p + x i − x p | x i − x p | , y ′ i = x i − x p | x i − x p | , ≤ i ≤ p − , and y p = ∞ . Then (1.4) will be changed to: | y ′ i | X ≤ j ≤ p − ,j = i | y ′ i − y ′ j | | y ′ j | u j + u p = 0 , ≤ i ≤ p − p − X j =1 | y ′ j | u j = 0 , i = p. (2.44)And (1.5) will be changed to: | y ′ i | u i X ≤ j ≤ p − ,j = i y ′ i | y ′ j | − y ′ j | y ′ i | | y ′ i − y ′ j | | y ′ j | u j + y ′ i u p = 0 , ≤ i ≤ p − u p p − X j =1 y ′ j | y ′ j | u j = 0 , i = p. (2.46) N A CONJECTURE OF BAHRI-XU 19
By a similar argument as above, taking the calculation u i y ′ i × (2.43) + (2.45) − y ′ i | y ′ i | ×h (2.45) , y ′ i i gives | y ′ i | u i X ≤ j ≤ p − ,j = i y ′ i − y ′ j | y ′ i − y ′ j | | y ′ j | u j = 0 , ≤ i ≤ p − . (2.47)Let v i = | y ′ i | u i for 1 ≤ i ≤ p −
1, and v p = u p , then (2.43) and (2.47) give: X ≤ j ≤ p − ,j = i | y i − y j | v j + v p = 0 , ≤ i ≤ p − v i X ≤ j ≤ p − ,j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p − . Since Kelvin transform is invertible, we have the following corollary.
Corollary 2.4.
Given any x , . . . , x p , the equations about u , . . . , u p X j = i u j | x i − x j | = 0 , u i X j = i x i − x j | x i − x j | u j = 0 , ≤ i ≤ p have only zero solution if and only if given any y , . . . , y p − , the equations about v , . . . , v p X ≤ j ≤ p − ,j = i | y i − y j | v j + v p = 0 , v i X ≤ j ≤ p − ,j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p − . have only zero solution. (cid:3) Theorem 2.5.
Let p ≥ . If Conjecture 1.1 is valid for p ≤ p − , then when p = p ,the conjecture is valid if and only if for any x , . . . , x p , equations (1.4) and (1.5) X j = i u j | x i − x j | = 0 , ≤ i ≤ p (1.4) u i X j = i x i − x j | x i − x j | u j = 0 , ≤ i ≤ p (1.5) about u , . . . , u p have NO non-zero solution.Proof. The only if direction is obvious due to the positivity of I . Otherwise, assumefor some x , . . . , x p , the equations (1.4) and (1.5) about u , . . . , u p have a non-zerosolution ( u , . . . , u p ). Substitute x i and u i , 1 ≤ i ≤ p in I , I : I = X ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u j | x i − x j | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 sup ≤ i ≤ p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j = i u i x i − x j | x i − x j | u j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 0 ,I = X i,jj = i u j | x i − x j | > , which is a contradiction.Conversely, assume that Conjecture 1.1 is NOT valid, by Lemma 2.2 there exist y , . . . , y p − , ∈ R m , such that c ( p, m, y , . . . , y p − , ) = 0 . By the definition of c ( p, m, y , . . . , y p − , ) and the property of continuous function ona compact region, there exist v , . . . , v p ∈ R satisfying P pi =1 v i = 1, such that X ≤ j ≤ p − ,j = i | y i − y j | v j + v p = 0 , v i X ≤ j ≤ p − ,j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p − . By Corollary 2.4, there exist x , . . . , x p ∈ R m , such that the equations (1.4) and (1.5)have non-zero solutions. (cid:3) Now, we may prove Theorem 1.1.
Proof of Theorem 1.1.
We argue by induction on p .When p ≤
3, Conjecture 1.1 holds due to Xu [2], so we may apply Theorem 2.5 for p = 4. Then we see that Conjecture 1.1 holds for p = 4 if and only if (1.4) and (1.5)have no non-zero solution for p = 4. This proves Theorem 1.1 for p = 4.Now assume the equivalence of Theorem 1.1 holds for 4 ≤ p ≤ p −
1. We wantto show that Conjecture 1.1 holds for 4 ≤ p ≤ p is equivalent to that (1.4) and (1.5)have no non-zero solution for 4 ≤ p ≤ p .The necessity is obvious due to the positivity of I . It is left to consider thesufficiency. By the given condition for 4 ≤ p ≤ p − ≤ p ≤ p −
1. Applying Theorem 2.5, by the given conditionfor p = p , we see that Conjecture 1.1 holds for p = p , hence for p ≤ p . (cid:3) Note that equations (1.4) and (1.5) can be written in matrix form, i.e., AU = 0 , U T ∂ A∂x i U = 0 , i = 1 , . . . , p. (2.48)Let 1 ≤ α ≤ p be a natural number, and denote by U α the column vector obtainedby crossing out the α -th row of U , and A α the ( p − × ( p −
1) matrix obtained bycrossing out the α -th row and α -th column of A .To prove Conjecture 1.1 inductively by contradiction, one needs only to verifywhether there is a non-zero solution U = ( u , . . . , u p ) T of equation (2.48) with u i = 0for all i = 1 , · · · , p . Since if u α = 0 for some 1 ≤ α ≤ p , then (2.48) yields A α U α = 0 , U T α ∂ A α ∂x i U α = 0 , ≤ i = α ≤ p, N A CONJECTURE OF BAHRI-XU 21 which has the same form as (2.48). By the inductive assumption that the conjectureholds for p ≤ p − U α = 0, hence U = 0. The claim followsfrom the contradiction.In summary, we have the following corollary. Corollary 2.6.
If the conjecture holds for p ≤ p − and does not hold for p = p ,then any component of a non-zero solution to (2.48) is non-zero. Further equivalent conditions and partial results
In this section, we shall continue to use Kelvin transform to simplify the conjectureeven further and prove it for the basic case m = 1 with arbitrary p , and p = 4 , m . Again, assuming that the conjecture holds for p ≤ p − p ≥ p = p .Note that we can always embed the ambient Euclidean space R m , where the p points x , . . . , x p lie in, into the first m -components of a larger space R m +1 . Then weapply the Kelvin transform about the point N = (0 , . . . , , ∈ R m +1 \ R m . Notethat, this map is, in fact, a stereographic projection. The image of R m ֒ → R m +1 is an m -sphere centered at N = (0 , . . . , , ) with radius in R m +1 .Let y = K N ( x ) denote the image of x . By Proposition 2.3, equations (1.4) and(1.5) become equations (2.41) and (2.42): X j = i | y i − y j | v j = 0 , ≤ i ≤ p, (2.41) v i X j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p. (2.42)Since these equations are invariant under translation and scaling of y , . . . , y n , we mayassume that y , . . . , y p lie on the unit m -sphere S m .Now, we prove the following theorem. Theorem 3.1.
Let p ≥ . If the conjecture holds for p ≤ p − , then when p = p ,the conjecture is valid if and only if given any distinct y , . . . , y p ∈ S m , the equations X j = i y i − y j | y i − y j | v j = 0 , ≤ i ≤ p (1.6) for v , . . . , v p have only zero solution.Proof. Assume that the conjecture is false when p = p , then by Theorem 2.5, equations(1.4) and (1.5) have a non-zero solution, say U = ( u , . . . , u p ) T . By Corollary 2.6, eachcomponent of U is non-zero. By Proposition 2.3, equations (2.41) and (2.42) have anon-zero solution V = ( v , . . . , v p ) T , and each component v i is also non-zero. Thus we can derive equation (1.6) from (2.42) by dividing v i on both sides of the i -th equationfor 1 ≤ i ≤ p . Therefore, equation (1.6) have a non-zero solution.Conversely, assume that equation (1.6) have a non-zero solution, say V = ( v , . . . , v p ) T ,we are going to show that the conjecture is false when p = p by showing the existenceof non-zero solutions for equations (2.41) and (2.42). In fact, V clearly satisfies equation(2.42). For (2.41), note that | y i − y j | = h y i − y j , y i − y j i = 2 − h y i , y j i . Taking inner product of (1.6) with 2 y i , we get0 = X j = i − h y i , y j i| y i − y j | v j = X j = i | y i − y j | | y i − y j | v j = X j = i | y i − y j | v j , ≤ i ≤ p Thus, V satisfies (2.41) as well. In conclusion, V is also non-zero a solution for equations(2.41) and (2.42). (cid:3) Proof of Theorem 1.2.
The proof is similar to that of Theorem 1.1 and thus left to thereader. (cid:3)
Case 1. m = 1 . In this subsection, we prove Conjecture 1.1 when m = 1 usingTheorem 3.1. Theorem 3.2.
Conjecture 1.1 is valid for m = 1 and arbitrary p .Proof. It suffices to prove the case p = p ≥
4, under the assumption that the cases p ≤ p − y k = e i α k ∈ S ֒ → C for 1 ≤ k ≤ p with0 ≤ α < α < · · · < α p < π. We shall prove, the equations X j = k y k − y j | y k − y j | v j = 0 , ≤ k ≤ p (3.1)for v , . . . , v p have only zero solution. We prove this by contradiction.Assume that V = ( v , . . . , v p ) T is a non-zero solution for equation (3.1), then eachcomponent of V is non-zero by a similar argument as in Corollary 2.6. Notice the N A CONJECTURE OF BAHRI-XU 23 following identities: y k − y j = e i α k − e i α j = (cos α k − cos α j ) + i(sin α k − sin α j )= − α k + α j α k − α j α k + α j α k − α j
2= 2 sin α k − α j − sin α k + α j α k + α j α k − α j i αk + αj + π , | y k − y j | = | e i α k − e i α j | = 2 sgn( k − j ) sin α k − α j > . Substituting the identities above in (3.1), we obtain: X j = k sgn( k − j ) e i αj (sin α k − α j ) v j = 0 , ≤ k ≤ p. For each 1 ≤ s ≤ p , multiplying e i − αs on both sides of all equations but the s -thone, we havesgn( k − s ) 1(sin α k − α s ) v s + X j = k,s sgn( k − j ) e i αj − αs (sin α k − α j ) v j = 0 , ≤ k = s ≤ p. Taking the imaginary part, we get: X j = k,s sgn( k − j ) sin α j − α s (sin α k − α j ) v j = 0 , ≤ k = s ≤ p. (3.2)Note that (3.1) is also equivalent to X j = k sgn( k − j ) e i α k − e i α j (sin α k − α j ) v j = 0 , ≤ k ≤ p. Multiplying v k sin α k − α s e − i αk on the equation above, and setting w k = e − i αk v k , ≤ k ≤ p, we get:sgn( k − s ) e i α k − e i α s (sin α k − α s ) v s w k + X j = k,s sgn( k − j ) sin α k − α s (sin α k − α j ) i αj v j v k = 0 , ≤ k = s ≤ p. Taking sum over k , we get0 = X k = s sgn( k − s ) e i α k − e i α s (sin α k − α s ) v s w k + X k = s X j = k,s sgn( k − j ) sin α k − α s (sin α k − α j ) i αj v j v k Exchange the order ============= v s X k = s sgn( k − s ) e i α k − e i α s (sin α k − α s ) w k + X j = s X k = j,s sgn( k − j ) sin α k − α s (sin α k − α j ) i αj v j v k = v s X k = s sgn( k − s ) e i α k − e i α s (sin α k − α s ) w k + X j = s i αj v j X k = j,s sgn( k − j ) sin α k − α s (sin α k − α j ) v k (3.2) ==== v s X k = s sgn( k − s ) e i α k − e i α s (sin α k − α s ) w k . That is, v s X k = s sgn( k − s ) e i α k − e i α s (sin α k − α s ) w k = 0 , ≤ s ≤ p, or equivalently, X k = s sgn( k − s ) y k − y s | y k − y s | w k = 0 , ≤ s ≤ p. Taking inner product with 2 y s gives(3.3) X k = s sgn( k − s ) w k = 0 , ≤ s ≤ p. In other words, setting C = (sgn( k − s )) k,s = . . . − . . . − − . . . − − − . . . , W = ( w , w , . . . , w p ) T ∈ C p , we have CW = 0 by (3.3).When p is even, one easily sees that det C = 1 by expanding the determinant, thus W = 0, and hence V = 0, a contradiction.When p is odd, det C = 0. Evidently, we can find a non-zero ( p − × ( p − C = p −
1. We have CW = C e i α v e i α v ...e i αp v p = 0 , N A CONJECTURE OF BAHRI-XU 25 in this case. Consider the real and imaginary parts of W : C cos α v cos α v ...cos α p v p = 0 , C sin α v sin α v ...sin α p v p = 0 . Since V = 0, cos α v cos α v ...cos α p v p , and sin α v sin α v ...sin α p v p are two solutions for CZ = 0, where Z = ( z , . . . , z p ) is a real variable. But since (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) cos α v cos α v sin α v sin α v (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = sin α − α v v = 0 , the two solutions are linearly independent, which is contradictory to the previous con-clusion that rank C = p − V =( v , . . . , v p ) T is a non-zero solution for equation (3.1) is false, proving the conjecture. (cid:3) Case 2. p = 4 , . In this subsection, we deal with the cases p = 4 ,
5. We manageto reduce them down to m = 1. Theorem 3.3.
Conjecture 1.1 is valid for p = 4 , and arbitrary m .Proof. Assume that U = ( u , . . . , u p ) T is a non-zero solution for equations (1.4) and(1.5) for some x , . . . , x p ∈ R m . It suffices to transform x , . . . , x p to lie on the sameline, yielding a contradiction by Theorems 1.1 and 3.2.3.2.1. p = 4 . We may assume that m = 3 by Remark 2, and that each component of U is non-zero by Corollary 2.6. Thus, we have X j = i u j | x i − x j | = 0 , ≤ i ≤ p (3.4) X j = i x i − x j | x i − x j | u j = 0 , ≤ i ≤ p. (3.5)Choose the x -axis of R as the line passing x and x . If x and x are on x -axis, thenwe have completed the proof by contradiction. If x , x are not on the x -axis, then from (3.5), we claim that x , . . . , x mustbe on the same 2-dimensional plane. Otherwise, if x is outside the plane defined by x , x , x , then u must be zero by equation (3.5) when i = 1, a contradiction.Since equations (3.4) and (3.5) are invariant under the Kelvin transform by Propo-sition 2.3, we can make x , x , x co-line by transforming the circle through them intoa line. If x is not on the line of x , x , x , then u = 0 in the above equation for i = 4,a contradiction again. Finally we still come to the co-line case and thus complete theproof for p = 4.3.2.2. p = 5 . We may assume that m ≤
4. Since the case p = 4 has been verified, wemay assume that U ∈ R p ( u i = 0) is a solution to equations (3.4) and (3.5) again.Denote E := Span { x − x i | i = 2 , , , } .If dim E = 4, i.e., x − x / ∈ E := Span { x − x i | i = 2 , , } , then dist( x − x , E ) :=inf y ∈ E | y − ( x − x ) | can be achieved at some point y ∈ E . So y − ( x − x ) ⊥ E .Let i = 1 in (3.5), and make inner product between it and y − ( x − x ), we have u = 0, a contradiction. So x − x ∈ E , and thus E = E , dim E ≤ E = 3, then we can assume the 5 points are in the same 3-dimensionalsubspace and we can pick up 3 points such that the other 2 locate at the same side ofthe plane determined by the 3 points. Suppose x − x and x − x are on the sameside of the plane E := Span { x − x , x − x } . We can write x − x = c ( x − x ) + a ( x − x ) + b ( x − x )for constants a, b and c >
0. And we can find y ∈ E such that α := y − ( x − x ) ⊥ E .Let i = 1 , , α , then h α, x − x i| x − x | u + h α, x − x i| x − x | u = 0 , h α, x − x i| x − x | u + h α, x − x i| x − x | u = 0 , h α, x − x i| x − x | u + h α, x − x i| x − x | u = 0 . Note that h α, x − x i = c h α, x − x i <
0, thus we have | x − x || x − x | = | x − x || x − x | = | x − x || x − x | := K, u = − c K u . Let i = 4 , u | x − x | + u | x − x | + u | x − x | − K u c | x − x | = 0 ,u K | x − x | + u K | x − x | + u K | x − x | + u | x − x | = 0 . N A CONJECTURE OF BAHRI-XU 27
Then we have c = − K <
0, a contradiction. This shows that x − x ∈ E , i.e., E = E = E . So dim E ≤ U = 0 by Theorem 3.2, hence a contradiction to Theorem 1.1.This completes the proof for p = 5. (cid:3) Acknowledgements .
The authors would like to thank the referees for their valuablecomments.
References [1] A. Bahri and Y. Xu, Recent progress in conformal geometry, ICP Advanced Texts in Mathematics,vol.1, Imperial College Press, London, 2007.[2] Y.Xu, Note on an inequality, Ann. Inst. H. Poincare Anal. Non Lineaire (2006), no.5 629–639. School of Mathematical Sciences, Beijing Normal University, Beijing 100875, P. R.China
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