Proof. First take \(u\) to be a unit vector. Orthogonal projection from the hyperplane of \(F\) to \(u^\perp\) has absolute Jacobian \(|\langle\nu_F,u\rangle|\). Indeed, for an orthonormal tangent basis \(v_1,\ldots,v_{d-1}\), that Jacobian is \(|\det(v_1,\ldots,v_{d-1},u)|\), which equals the stated inner product in absolute value.

Apart from a set of \((d-1)\)-dimensional measure zero, a point in the projection of \(P\) lies on the projections of exactly two facet interiors: the entry and exit points of its line parallel to \(u\). To see that the exceptions are null, discard the boundary of the projection and the projections of all faces of dimension at most \(d-2\). Also discard projections of facets whose hyperplanes are parallel to \(u\); these projections have dimension at most \(d-2\). Every remaining line meets \(P\) in a nondegenerate interval with its two endpoints in facet interiors. Integrating this multiplicity gives (3) for unit \(u\), and homogeneity gives the general case.

The support function of \([-v/2,v/2]\) is \(\tfrac12|\langle v,u\rangle|\). Adding these support functions proves (4). ◻

Proof. The face exposed by a linear functional \((u,v)\) on \(A\times B\) is the product of the faces exposed by \(u\) and \(v\). Dimensions add under Cartesian products. A nonzero linear functional exposes a proper face of a full-dimensional body, so if both \(u\) and \(v\) are nonzero, the product face has codimension at least two. A facet therefore has one facet factor and one whole factor. Thus the facets are \(F\times B\) and \(A\times G\), with area-normal vectors \[(|B|s_F\nu_F,0) \quad\hbox{and}\quad (0,|A|s_G\nu_G),\] respectively. The area factors are products because the two coordinate spaces are orthogonal. Lemma 2 now gives \[h_{\Pi(A\times B)}(u,v) =|B|h_{\Pi A}(u)+|A|h_{\Pi B}(v).\] This is the support function of the right side of (5), proving that identity. Hence \[\begin{split} R_{r+s}(A\times B) &=\frac{|B|^r|A|^s|\Pi A||\Pi B|} {(|A||B|)^{r+s-1}}\\ &=\frac{|\Pi A|}{|A|^{r-1}} \frac{|\Pi B|}{|B|^{s-1}}, \end{split}\] as required. ◻

Proof. First suppose that \(P\) is a polytope. The transformed outward unit normal of \(F\) is \(L^{-t}\nu_F/\|L^{-t}\nu_F\|\). Its area is \[s_{LF}=|\det L|\,\|L^{-t}\nu_F\|\,s_F.\] For the area formula, take a prism of height one over \(F\), extending in direction \(\nu_F\). Its image has volume \(|\det L|s_F\) and height \(1/\|L^{-t}\nu_F\|\) over \(LF\); dividing volume by height gives the formula. Thus each area-normal vector transforms by \(|\det L|L^{-t}\), including when \(\det L<0\). Apply (4). The absolute determinant of this map is \(|\det L|^{d-1}\), exactly the scaling of \(|LP|^{d-1}\). Translation invariance follows directly from projection volumes.

For completeness, we justify the passage to an arbitrary convex body and the existence of its projection body. Translate an interior point of \(P\) to the origin, choose \(R>0\) with \(P\subset RB_2^d\), and take increasing inscribed polytopes \(P_m\) containing a fixed ball \(aB_2^d\), \(a>0\), such that \[P_m\subset P\subset P_m+\varepsilon_mB_2^d \subset(1+\varepsilon_m/a)P_m, \qquad \varepsilon_m\longrightarrow0.\] Take, for example, convex hulls of cumulative finite nets in \(P\) and a fixed inscribed polytope containing \(aB_2^d\). Put \(\lambda_m=1+\varepsilon_m/a\), and for unit \(u\) write \(F(u)=\mathop{\mathrm{vol}}_{d-1}(\mathop{\mathrm{proj}}_{u^\perp}P)\) and \(f_m(u)=h_{\Pi P_m}(u)\). Projection and volume monotonicity give \[0\le F(u)-f_m(u) \le (\lambda_m^{d-1}-1)\kappa_{d-1}R^{d-1}.\] Thus \(f_m\to F\) uniformly. The increasing bodies \(\Pi P_m\) lie in a fixed ball and contain \(\kappa_{d-1}a^{d-1}B_2^d\), so their union closure is a convex body with support function \(F\), namely \(\Pi P\). The support inequalities also give \[\Pi P_m\subset\Pi P\subset\lambda_m^{d-1}\Pi P_m, \qquad |\Pi P_m|\le|\Pi P|\le\lambda_m^{d(d-1)}|\Pi P_m|.\] Likewise \(|P_m|\le|P|\le\lambda_m^d|P_m|\). Applying \(L\) to the original sandwich gives the same convergence for \(LP_m\). Passing to the limit in the polytope identity proves covariance and, by these volume inequalities, the asserted invariance of \(R_d\). ◻

Proof. The simplex volume follows by iterated base and height. Put \(w=e_1+\cdots+e_d\). The coordinate facets have area-normal vectors \(-e_i/(d-1)!\). The remaining facet has unit normal \(w/\sqrt d\). Dropping its last coordinate projects it onto the standard \((d-1)\)-simplex with Jacobian \(1/\sqrt d\); its area-normal vector is therefore \(w/(d-1)!\). In the centered segments of Lemma 2, changing the sign of a generator changes no segment. Translating each segment to start at the origin therefore shows that \(\Pi T_d\) is a translate of \[\frac{[0,1]^d+[0,w]}{(d-1)!}.\]

To compute the numerator’s volume, let \(Q=[0,1]^d\). Every nonempty fiber of \(Q\) parallel to \(w\) is an interval, and adding \([0,w]\) increases its length by \(\|w\|\). The base projection is unchanged. Fubini therefore gives \[|Q+[0,w]| =|Q|+\|w\|\mathop{\mathrm{vol}}_{d-1}(\mathop{\mathrm{proj}}_{w^\perp}Q) =1+h_{\Pi Q}(w).\] The cube has two facets of area one in each coordinate direction, so (3) gives \(h_{\Pi Q}(w)=d\). Thus \(|Q+[0,w]|=d+1\); Figure 1 illustrates the two-dimensional case. Dilation by \(1/(d-1)!\) and division by \(|T_d|^{d-1}\) now yield \[R_d(T_d)=\frac{(d+1)(d!)^{d-1}}{((d-1)!)^d} =\frac{(d+1)d^d}{d!}.\] Every full-dimensional simplex is an invertible affine image of \(T_d\), so Lemma 4 proves the final assertion. ◻

Proof of Theorem 1. The product \(K=T_{10}\times T_{10}\) is compact and convex with nonempty interior in \(\mathbb R^{20}\). Propositions 3 and 5 give \[R_{20}(K)=R_{10}(T_{10})^2 =\left(\frac{11\cdot10^{10}}{10!}\right)^2.\] Consequently, \[\frac{R_{20}(K)}{c_{20}} =\frac{121\cdot10^{20}\cdot20!} {(10!)^2\cdot21\cdot20^{20}} =\frac{121\binom{20}{10}}{21\cdot2^{20}} =\frac{22\,355\,476}{22\,020\,096}>1.\] Here \(\binom{20}{10}=184\,756\), and the exact integer certificate is \[121\cdot184\,756-21\cdot1\,048\,576=335\,380>0.\] This is an exact strict violation of (2). ◻

Proof. We use repeated ten-dimensional factors to obtain exponential growth, and one factor of dimension between ten and nineteen to reach every sufficiently large dimension. For \(r\in\{0,\ldots,9\}\), put \(d_r=10+r\). Every integer \(n\ge20\) has a unique representation \(n=10k+d_r\) with \(k\ge1\). In the orthogonal coordinate product space, take \[K_n=T_{10}^{\,k}\times T_{d_r}.\] Here \(T_{10}^{\,k}\) denotes the \(k\)-fold Cartesian product. All factors and their successive products are full-dimensional convex polytopes, with each factor dimension at least ten. Thus Propositions 3 and 5 apply iteratively and give \[R_n(K_n)=c_{10}^k c_{d_r}.\]

To compare this value with \(c_n\), we bound the simplex constants by \(n+1\) times a fixed exponential, whose rate is smaller than that supplied by the ten-dimensional factors. Put \(\beta=11/4\). The exponential series gives \[e=\frac52+\sum_{j=3}^{\infty}\frac1{j!} <\frac52+\frac16\sum_{j=0}^{\infty}3^{-j} =\beta,\] since the successive ratios in the tail are strictly less than \(1/3\). The positive term \(n^n/n!\) in the series for \(e^n\) is strictly smaller than the full sum, so \[c_n<(n+1)e^n<(n+1)\beta^n\qquad(n\ge2).\] For the fixed ten-dimensional block, exact arithmetic yields \[c_{10}=\frac{17\,187\,500}{567},\qquad A:=\frac{c_{10}}{\beta^{10}} =\frac{1\,638\,400\,000\,000}{1\,336\,956\,340\,797}>1.\] Choose a fixed \(\lambda\) with \(1<\lambda<A^{1/10}\). Combining these estimates with the formula for \(R_n(K_n)\) gives \[\frac{R_n(K_n)}{R_n(T_n)\lambda^n} >c_{d_r}\beta^{-d_r}\lambda^{-d_r} \frac{(A/\lambda^{10})^k}{10k+d_r+1}.\] For each fixed \(r\), the right side tends to infinity as \(k\to\infty\), because \(A/\lambda^{10}>1\). A threshold can therefore be chosen in each of the ten residue classes; their maximum, enlarged to be at least twenty, gives one \(N\) for all \(n\ge N\). ◻