Each result cited is universally quantified over the data in its own statement.
Preliminaries. Let ΠN be the nonempty finite index set of the definition of the logarithmic energy, let M be its number of elements, and fix a bijection σ:[M]→ΠN, which exists as recorded in Sum over a Finite Index Set; by that definition ∑p∈ΠNg(p)=∑m=1Mg(σ(m)) for every map g:ΠN→R. Each p∈ΠN is written p=(p1,p2), so that p1<p2, and for x∈WN the number xp1−xp2 is positive, as recorded in that definition. We use the antisymmetry
ajk(x)=−akj(x)(x∈WN, k,j∈[N]).(A)
Indeed, if k=j both sides are 0; if k=j, then xj−xk=−(xk−xj) and (−(xk−xj))(−(xk−xj)−1)=1 by claim 2 of Zero Products and Elementary Identities in a Field, so (xj−xk)−1=−(xk−xj)−1.
Claim 1. Let x∈WN and put tm=xσ(m)1−xσ(m)2 for m∈[M], a positive number. Let μ1=t1 and, for m<M, let μm+1 be an element with μm+1≤μm, μm+1≤tm+1 and μm+1∈{μm,tm+1}, given by claim 9 of Elementary Order Arithmetic in an Ordered Field. By induction on m, μm is one of t1,…,tm, hence positive, and μm≤tm′ for every m′≤m, by transitivity of ≤. Since σ is surjective, μ=μM satisfies 0<μ and μ≤xi−xj for every (i,j)∈ΠN. Let δ=μ⋅2−1; by claim 8 of Elementary Order Arithmetic in an Ordered Field, 0<δ and δ+δ=μ.
Let y lie in the open ball BdE(x,δ), so ∥x−y∥=dE(x,y)<δ by claim 2 of Elementary Properties of the Euclidean Norm on Rn. The kth coordinate of x−y is xk−yk (Difference, Dot Product, and Orthogonality in Rn), so claim 4 of Elementary Properties of the Euclidean Norm on Rn and mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field) give ∣xk−yk∣<δ, that is −δ<xk−yk<δ by claim 9 of Properties of the Absolute Value in an Ordered Field, for every k∈[N]. Let (i,j)∈ΠN. Adding suitable terms (claim 1 of Elementary Order Arithmetic in an Ordered Field) gives xi−δ<yi and yj<xj+δ, hence −(xj+δ)<−yj by claim 4 there, and claim 3 there yields
(xi−xj)−μ=(xi−δ)−(xj+δ)<yi−yj.
As μ≤xi−xj, the left side is nonnegative by claim 3 of Elementary Arithmetic in an Ordered Field, so 0<yi−yj by claim 2 of Elementary Order Arithmetic in an Ordered Field, and yj<yi by claim 1 there. Thus y∈WN by the definition of WN, so BdE(x,δ)⊆WN. Hence WN is open in (RN,dE), which is openness in the sense of the setting (by Euclidean Openness Agrees with Metric Openness on Rn).
Let q∈RN have kth coordinate N−k+1. If 1≤i<j≤N, then i<j as real numbers, so −j<−i and N+1−j<N+1−i by claims 4 and 1 of Elementary Order Arithmetic in an Ordered Field; that is qj<qi, and q∈WN.
Claim 2. Let J=(0,∞), an interval as recorded in The Natural Logarithm. Every t∈J is an interior point of J, since t⋅2−1 and t+1 lie in J and t⋅2−1<t<t+1, by claims 8, 6, 1 and 3 of Elementary Order Arithmetic in an Ordered Field. For p∈ΠN and k∈[N] put ck(p)=1 if k=p1, ck(p)=−1 if k=p2, and ck(p)=0 otherwise (well defined as p1=p2), and define functions on WN, which is open by Claim 1, by
Lp(x)=xp1−xp2,ρp(x)=Lp(x)−1,φp(x)=logLp(x).
(a) Partial derivatives along slices. Let f:J→R be differentiable at every point of J, with derivative f′. We show that for x∈WN and k∈[N] the partial derivative of f∘Lp with respect to the kth variable exists at x and
∂k(f∘Lp)(x)=ck(p)f′(Lp(x)).(2.1)
By claim 1 of Slice Function and the Partial Derivative there are r0>0 and the interval I={s:xk−r0<s<xk+r0}, with xk an interior point, such that x[s]∈WN for s∈I, where x[s] is x with kth coordinate replaced by s. Put γ(s)=Lp(x[s]) for s∈I; then γ(s)∈J, and checking the cases k=p1, k=p2 and k∈/{p1,p2} gives γ(s)=Lp(x)+ck(p)(s−xk). For h=0 with xk+h∈I the difference quotient (γ(xk+h)−γ(xk))/h equals ck(p), so γ is differentiable at xk with γ′(xk)=ck(p) by Derivative at an Interior Point. Since γ(xk)=Lp(x) is an interior point of J, Chain Rule for One-Dimensional Derivatives shows that f∘γ is differentiable at xk with derivative f′(Lp(x))ck(p). As f∘γ is the slice function of f∘Lp at x in the kth variable, claim 2 of Slice Function and the Partial Derivative gives (2.1).
Applying (2.1) to f=log, whose derivative is log′(t)=t−1 by The Natural Logarithm, gives ∂kφp(x)=ck(p)ρp(x). Applying it to f=r, r(t)=t−1, which is differentiable at every t∈J with r′(t)=−(t−1)2 by claim 2 of Reciprocal Rule for One-Dimensional Derivatives (as 0∈/J), and noting ρp=r∘Lp, gives ∂lρp(x)=−cl(p)ρp(x)2 for l∈[N]. Since ∂kφp=ck(p)ρp as functions on WN, claim 1 of Constants, Coordinate Functions, Sums and Products of Ck Functions on a Euclidean Open Set gives
∂l∂kφp(x)=−ck(p)cl(p)ρp(x)2(x∈WN, k,l∈[N]).(2.2)
(b) Continuity. Lp=πp1+(−1)πp2 is smooth on WN by claims 2 and 3 of Constants, Coordinate Functions, Sums and Products of Ck Functions on a Euclidean Open Set, hence continuous on WN relative to WN as a map into (R,dR) by claim 3 of Euclidean Space is Open in Itself, and Ck Maps are Continuous. The function log is continuous on J relative to J by Differentiability at an Interior Point Implies Continuity There, applied at every point of J, so φp=log∘Lp is continuous on WN by claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map. The function ρp is continuous on WN by claim 2 of Continuity of the Reciprocal of a Nonvanishing Real-Valued Function on a Metric Space, hence so are ck(p)ρp and −ck(p)cl(p)ρpρp by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. By claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions, all these functions are continuous in the Euclidean sense at every point of WN.
(c) Regularity of φp. By (a), (b) and clause 1 of C^k Maps on a Euclidean Open Set, φp is of class C1 on WN (it is continuous and ∂kφp=ck(p)ρp exists and is continuous), and so is each ∂kφp=ck(p)ρp (it is continuous, and its partial derivatives (2.2) exist and are continuous). By clauses 2 and 3 there, φp is of class C2 on WN.
(d) Regularity of H. For m∈[M] let Ψm(x)=∑m′=1mφσ(m′)(x) on WN. By the recursion of claim 1 of Properties of Finite Sums (Ψ1=φσ(1) and Ψm+1=Ψm+φσ(m+1)) and claims 1 and 3 of Constants, Coordinate Functions, Sums and Products of Ck Functions on a Euclidean Open Set, applied to Ψm and φσ(m+1) and then to ∂kΨm and ∂kφσ(m+1), induction on m shows that Ψm is of class C2 on WN with ∂kΨm(x)=∑m′=1m∂kφσ(m′)(x) and ∂k∂kΨm(x)=∑m′=1m∂k∂kφσ(m′)(x). By the definition of H and the choice of σ, H=(−β)ΨM, so claims 1 and 3 of Constants, Coordinate Functions, Sums and Products of Ck Functions on a Euclidean Open Set show that H is of class C2 on WN, in the sense of the setting (see clause 2 of the calculus setting), and, reading the sums along σ back as sums over ΠN and using (a) and (2.2) with l=k,
∂kH(x)=−βp∈ΠN∑ck(p)ρp(x),∂k∂kH(x)=βp∈ΠN∑(ck(p)ρp(x))2,(2.3)
where the second form uses claim 4 of Properties of a Sum over a Finite Index Set with λ=−1 and claim 2 of Zero Products and Elementary Identities in a Field.
(e) Reindexing. Fix x∈WN, k∈[N] and F:R→R with F(0)=0. We show
p∈ΠN∑F(ck(p)ρp(x))=j=1∑NF(akj(x)).(2.4)
Let E={p∈ΠN:p1=k or p2=k} and G=[N]∖{k}; G is nonempty since 1=2 both lie in [N] as N≥2. Define θ:G→E by θ(j)=(k,j) if k<j and θ(j)=(j,k) if j<k. The coordinates of θ(j) are k and j, so θ is injective; and if p∈E with p1=k then p=θ(p2), while if p2=k then p=θ(p1), so θ is a bijection and E is nonempty. For j∈G we have ck(θ(j))ρθ(j)(x)=akj(x): if k<j this is 1⋅(xk−xj)−1, and if j<k it is −(xj−xk)−1=akj(x) by (A). For p∈ΠN∖E, ck(p)=0, so the term is F(0)=0 (claim 1 of Zero Products and Elementary Identities in a Field); likewise F(akk(x))=F(0)=0. Hence, by claim 4 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, claim 2 of Properties of a Sum over a Finite Index Set along θ, claim 4 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set again, and claim 1 of Properties of a Sum over a Finite Index Set,
p∈ΠN∑F(ck(p)ρp(x))=p∈E∑F(ck(p)ρp(x))=j∈G∑F(akj(x))=j∈[N]∑F(akj(x))=j=1∑NF(akj(x)).
Taking F(t)=t and F(t)=t2 in (2.4), (2.3) becomes ∂kH(x)=−β∑j=1Nakj(x) and ∂k∂kH(x)=β∑j=1Nakj(x)2. By Trace of a Real Square Matrix and Hessian Matrix of a C^2 Function, and claim 3 of Properties of Finite Sums,
tr(D2H(x))=k=1∑N∂k∂kH(x)=k=1∑Nβj=1∑Nakj(x)2=βS(x).
Claim 3. Fix x∈WN and z∈RN, write akj=akj(x), and let Q=∑k=1N∑j=1Nakjzk. By Interchange of a Finite Double Sum and renaming the bound indices, then (A) and claim 3 of Properties of Finite Sums with λ=−1,
Q=j=1∑Nk=1∑Nakjzk=k=1∑Nj=1∑Najkzj=−k=1∑Nj=1∑Nakjzj.
Adding the two expressions for Q and using claims 2 and 3 of Properties of Finite Sums, Q+Q=∑k=1N∑j=1Nakj(zk−zj); since Q+Q=2Q and 2−1 exists (claim 8 of Elementary Order Arithmetic in an Ordered Field), Q=2−1∑k∑jakj(zk−zj). By Gradient of a Real-Valued Function on a Euclidean Open Set, Difference, Dot Product, and Orthogonality in Rn, Claim 2 and claim 3 of Properties of Finite Sums,
DH(x)⋅z=k=1∑N∂kH(x)zk=k=1∑N(−βj=1∑Nakj)zk=−βQ=−2βk=1∑Nj=1∑Nakj(zk−zj).
Claim 4. Let x,y∈WN and z=x−y, so zk=xk−yk. For k,j∈[N] put dkj=xk−xj, ekj=yk−yj and wkj=zk−zj=dkj−ekj. By claim 3 of Bilinearity and Symmetry of the Dot Product on Rn, Claim 3 at x and at y, and claims 2 and 3 of Properties of Finite Sums,
(DH(x)−DH(y))⋅z=DH(x)⋅z−DH(y)⋅z=2βk=1∑Nj=1∑N(akj(y)−akj(x))wkj.
Each summand is nonnegative. If k=j it is 0⋅wkk=0 (claim 1 of Zero Products and Elementary Identities in a Field). Let k=j and write d=dkj, e=ekj. If k<j, then (k,j)∈ΠN and d,e are positive, so 0<de by claim 5 of Elementary Order Arithmetic in an Ordered Field; if j<k, then −d,−e are positive and de=(−d)(−e) is positive by claim 2 of Zero Products and Elementary Identities in a Field and the same claim 5. In both cases (de)−1 is positive (claim 7 there), and field arithmetic gives e−1−d−1=(d−e)(de)−1, so
(akj(y)−akj(x))wkj=(e−1−d−1)(d−e)=(d−e)2(de)−1.
Since 0≤(d−e)2 by claim 2 of Nonnegativity of Squares in an Ordered Field, claim 5 of Elementary Arithmetic in an Ordered Field (with x=0) and claim 1 of Zero Products and Elementary Identities in a Field give 0≤(d−e)2(de)−1. By claim 5 of Properties of Finite Sums, applied to the inner and then the outer sum, the double sum is nonnegative; as 2β is positive (claim 8 of Elementary Order Arithmetic in an Ordered Field), claim 5 of Elementary Arithmetic in an Ordered Field gives 0≤(DH(x)−DH(y))⋅(x−y).
Claim 5. Fix x∈WN, write akj=akj(x) and Bk=∑j=1Nakj. By claim 1 of Elementary Properties of the Euclidean Norm on Rn, Gradient of a Real-Valued Function on a Euclidean Open Set, Claim 2, claim 2 of Zero Products and Elementary Identities in a Field and claim 3 of Properties of Finite Sums,
∥DH(x)∥2=k=1∑N(∂kH(x))2=k=1∑N(−βBk)2=β2k=1∑NBk2.
By claim 3 of Properties of Finite Sums, used twice, Bk2=∑j=1NakjBk=∑j=1N∑l=1Nakjakl. For k,j,l∈[N] let bkjl=akjakl if j=l and bkjl=0 if j=l. For fixed k,j, the numbers akjakl−bkjl (l∈[N]) vanish for l=j and equal akj2 for l=j, so by claims 7 and 2 of Properties of Finite Sums, ∑lakjakl=akj2+∑lbkjl. Summing over j and k with claim 2 there,
k=1∑NBk2=S(x)+T,T=k=1∑Nj=1∑Nl=1∑Nbkjl.
We show bkjl+bjkl+blkj=0 for all k,j,l∈[N]. Each of the three terms has the form buvw with (u,v,w) a rearrangement of (k,j,l). If two of k,j,l coincide, then in each term either v=w, so buvw=0 by definition, or v=w and u∈{v,w}, so buvw=auvauw has a factor auu=0 and vanishes by claim 1 of Zero Products and Elementary Identities in a Field. If k,j,l are distinct, put A=xk, B=xj, C=xl, which are pairwise distinct; using (A) and field arithmetic over the common nonzero denominator D=(A−B)(A−C)(B−C) (nonzero by claim 3 of Zero Products and Elementary Identities in a Field),
bkjl+bjkl+blkj=(A−B)(A−C)1+(B−A)(B−C)1+(C−A)(C−B)1=D(B−C)−(A−C)+(A−B)=0.
By Interchange of a Finite Double Sum, applied to the outer two sums, and renaming the bound indices, ∑k∑j∑lbjkl=∑j∑k∑lbjkl=T. Applying the same lemma first to the inner two sums and then to the outer two, and renaming, ∑k∑j∑lblkj=∑k∑l∑jblkj=∑l∑k∑jblkj=T. Summing the three-term identity over k,j,l with claim 2 of Properties of Finite Sums (a sum of zeros being 0 by claim 3 there with λ=0 and claim 1 of Zero Products and Elementary Identities in a Field) gives T+T+T=0, that is (1+1+1)T=0. Since 0<1+1+1 by claims 6, 8 and 3 of Elementary Order Arithmetic in an Ordered Field, claim 3 of Zero Products and Elementary Identities in a Field gives T=0. Hence ∥DH(x)∥2=β2S(x), and by Claim 2, βtr(D2H(x))=β⋅βS(x)=β2S(x)=∥DH(x)∥2.
Claim 6. Let x∈WN. By Claim 3 with z=x,
DH(x)⋅x=−2βk=1∑Nj=1∑Nakj(x)(xk−xj).
For k=j, akj(x)(xk−xj)=(xk−xj)−1(xk−xj)=1=ϵkj; for k=j, akk(x)(xk−xk)=0=ϵkk by claim 1 of Zero Products and Elementary Identities in a Field. So the double sum is dN, termwise, and DH(x)⋅x=−2βdN.