Proposition 13 ([Medians of an equality body]). Let K⊂ⁿ , n≥1 , be an origin-symmetric convex body with P(K)=4ⁿ/n! , and put Q=K^. Every triple of points in (ⁿ,p_Q) has a metric median.

Write \|·\|=p_Q . Translation reduces the triple to 0,x,y . If two points coincide, that repeated point is a median, so assume that 0,x,y are distinct. Set p_1=\|x\|, p_2=\|y\|, p=\|x-y\|, s=p_1+p_2, a=p_1-p_2. All three distances are positive. The triangle and reverse triangle inequalities give s≥p≥|a| . Define A_+=(p+a)/(2), A_-=(p-a)/(2), c=(s-p)/(2). These numbers are nonnegative, with A_++A_-=p , c+A_+=p_1 , and c+A_-=p_2 . The compact set of aggregate directions. Let D_0= r_+-r_-: r_++r_-=x-y, \|r_+\|≤A_+, \|r_-\|≤A_- . Our goal is to prove x+y∈D_0+(s-p)Q . Such membership supplies r_+,r_- for which m=x-r_+=y+r_- has distances to 0,x,y bounded by c,A_+,A_- , respectively; the triangle inequalities then force the three median equalities. We shall prove this membership by estimating support functions in each direction. The set D_0 is compact and convex: it is the linear image of the intersection of A_+Q×A_-Q with the affine subspace r_++r_-=x-y . It is nonempty, since r_±=(A_±/p)(x-y) satisfy the constraints. This also covers A_+=0 or A_-=0 . Endpoint comparison. Fix u and let X_δ=(δu,1-δ) , 0<δ<1 . The lifted body B=K⊕_1[-1,1] contains X_δ in its interior because p_B(X_δ)=δp_K(u)+1-δ<1. Apply Proposition 12 ([A representation minimizing the lens cost]) with this X_δ and (q,a)=(x-y,p_1-p_2) . Its total weight is T=\|q\|,|a|=p . We obtain rows (b_i,σ_i)∈Q×-1,1 and weights α_i≥0 satisfying and . Their aggregates r_±=∑_ σ_i=±1 α_i b_i have r_++r_-=x-y . Since ∑_ σ_i=±1 α_i=(p±a)/2=A_± , they satisfy \|r_±\|≤A_± and hence r_+-r_-∈D_0 . No choice of these representations across different u or δ is required. For b∈Q and σ=±1 , evenness of λ gives λ(⟨(b,σ),X_δ⟩) =λ(1-δ(1-σ⟨b,u⟩)). The factors 1-σ⟨b,u⟩ belong to [1-p_K(u),1+p_K(u)] , a compact subinterval of (0,∞) . Lemma 6 ([Endpoint scaling]) therefore gives an error ε_u(δ)→0 such that | λ(⟨(b,σ),X_δ⟩) λ(1-δ) -1+σ⟨b,u⟩ |≤ε_u(δ) (b∈Q, σ=±1). In particular, the normalized cost of the optimal representation is at least p-⟨r_+-r_-,u⟩-pε_u(δ) ≥p-h_ D_0 (u)-pε_u(δ). Compare this cost, using, with the two-term representation (x-y,a)=p_1(x/p_1,1)+p_2(-y/p_2,-1). Both rows belong to Q×-1,1 . By , their normalized cost tends to s-⟨x+y,u⟩ . Letting δ0 yields ⟨x+y,u⟩≤h_ D_0 (u)+s-p (u). The error was uniform only over b,σ for the fixed u ; that is all this limit requires. Separation and the median. Continuity extends to u∈K . For any nonzero v∈^n , the point u=v/h_Q(v) lies on ∂K , by. Homogeneity of support functions gives ⟨x+y,v⟩≤h_ D_0 (v)+(s-p)h_Q(v). The same inequality holds at v=0 . The right side is the support function of the compact convex set D_0+(s-p)Q , with the convention 0Q= 0 . Convex separation therefore implies x+y∈D_0+(s-p)Q. Choose the corresponding pair r_+,r_- in , and put m=x-r_+=y+r_-. Then 2m=x+y-(r_+-r_-) , so gives \|m\|≤c, \|x-m\|≤A_+, \|y-m\|≤A_-. The triangle inequality forces all three bounds to be equalities: p_1, ≤\|m\|+\|x-m\|≤c+A_+=p_1,; p_2, ≤\|m\|+\|y-m\|≤c+A_-=p_2. Thus m lies between 0,x and between 0,y , and \|x-m\|+\|m-y\|=A_++A_-=p=\|x-y\|. This proves the median property. The argument permits c=0 or either A_±=0 , so saturated triangle inequalities require no limiting argument.