# Independent verification by finite approximation Date: 2026-09-22 (America/Los_Angeles). ## Claim and verdict Let \(K\subset\mathbb R^n\) be compact, convex, and have nonempty interior, with \(n\ge1\). Write \[ g_K=\frac1{|K|}\int_K x\,dx,\qquad C_2(K)=\frac1{|K|}\int_K |x|^2\,dx. \] Assume \(r>0\) and \(|v|\ge r\) for every extreme point \(v\) of \(K\). The candidate's exact claim is valid: \[ \Delta_r(K):=C_2(K)-\frac{r^2+(n+1)|g_K|^2}{n+2}\ge0, \] and equality holds exactly for full-dimensional simplices with all vertices on \(rS^{n-1}\). The verification below uses only finite polytope decompositions and continuity of volume integrals. In particular, it does not assume the candidate's countable decomposition lemma. The central independent check is a strictly positive lower bound that survives approximation whenever there is an extreme point outside an initial extreme simplex. Mathematical verification completion estimate: **100% for the stated theorem**. This is not a priority or literature verdict. ## 1. Finite identity For a full-dimensional simplex \(A=\operatorname{conv}(a_0,\ldots,a_n)\), the elementary barycentric integral gives \[ C_2(A)=\frac{q_A+(n+1)|c_A|^2}{n+2},\qquad q_A=\frac1{n+1}\sum_{i=0}^n|a_i|^2,\quad c_A=\frac1{n+1}\sum_{i=0}^n a_i. \] Indeed, uniform barycentric coordinates satisfy \(E\lambda_i^2=2/((n+1)(n+2))\) and \(E\lambda_i\lambda_j=1/((n+1)(n+2))\) for \(i\ne j\). Suppose a polytope \(P\) is partitioned into finitely many full-dimensional simplices \(A\), modulo their boundaries, and all their vertices have norm at least \(r\). Set \(w_A=|A|/|P|\). The simplex identity and the finite variance identity yield \[ \Delta_r(P)=\frac1{n+2}\sum_A w_A(q_A-r^2) +\frac{n+1}{n+2}\sum_A w_A|c_A-g_P|^2. \tag{1} \] All summands are nonnegative. This proves the finite inequality and, more importantly, allows fixed summands to be retained through limits. ## 2. Approximation retaining selected simplices Choose affinely independent extreme points \(v_0,\ldots,v_n\), and put \(S=\operatorname{conv}(v_0,\ldots,v_n)\). Such points exist because the extreme points affinely span \(K\). Take a finite or countable dense set of distinct extreme points that contains these points, and enumerate the remaining points as \(u_1,u_2,\ldots\). Its nested finite convex hulls, starting with \(S\), converge to \(K\) in Hausdorff distance. This follows from the finite-dimensional extreme-point representation of compact convex sets and compactness. Each new point lies outside the preceding finite hull: an extreme point cannot be a convex combination of other points of \(K\). On inserting a point \(u\) outside a full-dimensional polytope \(P\), its new region is a finite union of pyramids \(\operatorname{conv}(u,F)\), where \(F\) ranges over the strictly visible facets of \(P\). Triangulating each facet with its existing vertices partitions each pyramid into full-dimensional simplices. Their interiors are disjoint, and their interiors lie outside \(P\). Consequently, every simplex already present can be retained unchanged at every later finite stage. For clarity, no compatibility of triangulations along lower-dimensional faces is needed here: boundaries have zero \(n\)-dimensional volume, and all applications are identities for volume integrals. If \(K\ne S\), some extreme point \(u\) lies outside \(S\); otherwise the closed convex hull of all extreme points would be contained in \(S\). Insert this \(u\) first. At least one facet \(F_i\) of \(S\) is strictly visible from \(u\). If \(v_i\) is the opposite vertex, then \[ T=\operatorname{conv}\bigl(u,\{v_j:j\ne i\}\bigr) \] is a full-dimensional simplex in the first shell. It has positive volume and interior disjoint from \(S\). The two centroids satisfy \[ c_T-c_S=\frac{u-v_i}{n+1}\ne0. \tag{2} \] Both \(S\) and \(T\) are retained in every subsequent finite decomposition. Every vertex introduced is an extreme point of \(K\), so the lower norm bound needed for (1) holds throughout. ## 3. Strictness survives the limit Write \(a=|S|>0\), \(b=|T|>0\), and \(V_m=|P_m|\). Keeping the two centroid terms from (1), and completing the square, gives \[ \begin{aligned} \Delta_r(P_m) &\ge\frac{n+1}{n+2}\left( \frac a{V_m}|c_S-g_{P_m}|^2+ \frac b{V_m}|c_T-g_{P_m}|^2\right)\\ &\ge\frac{n+1}{n+2} \frac{ab}{V_m(a+b)}|c_S-c_T|^2\\ &\ge\frac{n+1}{n+2} \frac{ab}{|K|(a+b)}|c_S-c_T|^2>0. \end{aligned} \tag{3} \] The second inequality is exactly the elementary identity \[ a|p-z|^2+b|q-z|^2 =\frac{ab}{a+b}|p-q|^2 +(a+b)\left|\frac{ap+bq}{a+b}-z\right|^2. \] Hausdorff convergence of the nested full-dimensional convex polytopes to \(K\) implies convergence of their volumes and of their integrals of \(x\) and \(|x|^2\). One direct justification is that a convex union with closure \(K\) contains \(\operatorname{int}K\), the boundary of a convex body has zero volume, and all functions are bounded on the common compact set \(K\). Dominated convergence then applies. Thus \(\Delta_r(P_m)\to\Delta_r(K)\). The strict lower bound in (3) is independent of \(m\). Therefore, if \(K\ne S\), \[ \boxed{\quad \Delta_r(K)\ge \frac{ab\,|u-v_i|^2} {(n+1)(n+2)|K|(a+b)}>0. \quad} \tag{4} \] In particular, equality forces \(K=S\). Applying the simplex identity to \(S\) then says equality is equivalent to \(\frac1{n+1}\sum_i |v_i|^2=r^2\). Since every term is at least \(r^2\), this is equivalent to \(|v_i|=r\) for every \(i\). Conversely, that vertex condition gives equality immediately. The inequality itself follows by taking limits in (1), without needing to assume \(K\ne S\). ## Boundary and hidden-assumption audit - **Dimension one.** Every convex body is an interval \([a,b]\), \(a