Proof. Choose \(0<m\le s(v)\le M\). For small \(|t|\) the constraints \(s+t\phi\) lie between \(m/2\) and some fixed finite constant \(M'\). Consequently \[(m/2)B_2^n\subset W_t\subset M'B_2^n.\] The defining inequalities show that \(x\in aW_t\) precisely when \(x\cdot v\le a(s(v)+t\phi(v))\) for every \(v\). Taking the least such \(a\) gives the maximum formula in (2). The radial function of \(W_t\) is \(1/g_t(u)\), so polar integration, with spherical area \(n\kappa_n\), gives the volume formula. If \(s=h_L\), separation identifies the defining intersection with \(L\).

For completeness, the maximum formula also gives the differentiability needed here without any regularity assumption on the boundary of \(W_0\). The function \(g_0\) is convex and Lipschitz: it is the maximum of linear functions whose coefficient vectors \(v/s(v)\) form a compact set. On every coordinate line, a finite convex function has a derivative except on a set of one-dimensional measure zero. Slicing therefore shows that all coordinate partial derivatives of \(g_0\) exist at almost every point of \(\mathbb R^n\). If a vector \(w=v/s(v)\) attains the maximum at such a point \(x\), then \[g_0(x+he_i)\ge g_0(x)+h w_i\qquad(h\in\mathbb R).\] The two signs of \(h\) imply \(w_i=\partial_i g_0(x)\) for every \(i\). Thus the maximizing vector is unique.

Uniqueness implies differentiability. Indeed, if \(w_z\) maximizes at \(x+z\) and \(w\) maximizes at \(x\), compactness and continuity imply \(w_z\to w\) as \(z\to0\). The maximum formula gives \[0\le g_0(x+z)-g_0(x)-w\cdot z \le (w_z-w)\cdot z=o(|z|).\] The nondifferentiability set away from the origin is invariant under positive dilations, by homogeneity. Since it has Lebesgue measure zero, polar integration shows that its intersection with \(S^{n-1}\) has \(\sigma\)-measure zero.

Fix a differentiability direction \(u\), and let \(v_0\) maximize at \(t=0\). Uniqueness of \(v_0/s(v_0)\) implies uniqueness of \(v_0\) itself, since \(s\) is positive. Every maximizing direction \(v_t\) for \(g_t(u)\) tends to \(v_0\) as \(t\to0\). The derivatives with respect to \(t\) of the functions \[(u,v,t)\longmapsto\frac{u\cdot v}{s(v)+t\phi(v)}\] are continuous on the relevant compact set. Comparing the maxima at \(t\) and \(0\), using \(v_t\) and \(v_0\), therefore gives \[\left.\frac{d}{dt}\right|_{0}g_t(u) =-\frac{(u\cdot v_0)\phi(v_0)}{s(v_0)^2} =-g_0(u)\phi\!\left(\frac{v_0}{s(v_0)}\right).\] This is (3). Notice that the argument uses neither convexity nor evenness of the function \(s+t\phi\).

The same uniform bound on the derivatives of the displayed quotients bounds \((g_t-g_0)/t\) uniformly in \(u\). The ball inclusions bound \(g_t\) above and away from zero on \(S^{n-1}\). Dominated convergence now differentiates the volume formula and proves (4). ◻

Proof. Support functions add under Minkowski addition, so Lemma 2, with \(s=h_K\) and \(\phi=h_M\), gives the derivative identity. This applies equally to a compact convex set \(M\) of lower dimension. For full-dimensional \(M\), Theorem 3 and homogeneity give \[F(t):=|K+tM|^{1/n}\ge |K|^{1/n}+t|M|^{1/n} \qquad(t\ge0).\] In addition, \(F\) is concave, by applying that theorem to \(K+sM\) and \(K+tM\). Its right derivative at zero therefore satisfies \(F'(0+)\ge |M|^{1/n}\), which is the asserted lower bound.

If equality holds in this derivative bound, concavity gives \[F(t)\le F(0)+tF'(0+)=|K|^{1/n}+t|M|^{1/n}.\] Thus the preceding volume inequality is an equality for every \(t>0\). The equality case of Theorem 3, applied for example to \(K+M\), forces homothety. Conversely, if \(M=a+cK\) with \(c>0\), then \(K+tM=ta+(1+tc)K\), which verifies equality in the derivative bound. ◻

Proof. By (2), \(\mathbb E_{\sigma}g_K^{-n}=1\), so (8) defines a probability measure. The active vectors \(v/h_K(v)\) lie in a bounded set because \(h_K\) is positive on the sphere. Hence \(\eta_K\) has bounded support. Applying Corollary 4 gives, for every symmetric convex body \(M\), \[\tau_{\eta_K}(M) =\frac1{n\kappa_n} \left.\frac{d}{dt}\right|_{0+}|K+tM| \ge\left(\frac{|M|}{\kappa_n}\right)^{1/n},\] with precisely the stated homothety equality condition. The same derivative identity with \(M=K\), using \(|K+tK|=(1+t)^n\kappa_n\), proves (10); this identity does not require \(K\) to be symmetric.

Now let \(y\ne0\) and take \(M=[-y,y]\). Every nonempty fiber of \(K\) parallel to \(y\) is a compact interval. Adding \(t[-y,y]\) increases its length by \(2t|y|\), without changing its projection onto \(y^\perp\). Fubini’s theorem therefore gives \[|K+t[-y,y]|=|K|+ 2t|y|\operatorname{vol}_{n-1} (\operatorname{proj}_{y^\perp}K) =|K|+2t h_{\Pi K}(y).\] Since the support function of the segment is \(x\mapsto|x\cdot y|\), the derivative identity yields (9). The formula is also true at \(y=0\). The projection of a full-dimensional convex body has positive \((n-1)\)-volume in every direction. If \(\mathop{\mathrm{supp}}\eta_K\) failed to span \(\mathbb R^n\), a nonzero vector perpendicular to its span would make the integral in (9) vanish. This contradiction proves the spanning property and hence admissibility.

Finally, for \(K=B_2^n\) one has \(g_K(u)=1\) and \(\nabla g_K(u)=u\) on the sphere, so \(\eta_K=\sigma\). Its projection volume in every unit direction is \(\kappa_{n-1}\). Substituting this into (9) proves (11). ◻

Proof. The polar-coordinate formula for the Euclidean Laplacian gives \[\Delta_{\mathbb R^n}\big(r^lY(u)\big) =r^{l-2}\big(l(l+n-2)Y(u)+\Delta_SY(u)\big).\] This proves the eigenvalue formula. Integration by parts on the sphere then proves orthogonality of spaces with distinct degrees.

We give the completeness argument explicitly. If \(Y_l\) is homogeneous and harmonic, direct differentiation yields \[\Delta_{\mathbb R^n}\big(\left|x\right|^{2j}Y_l(x)\big) =2j(2l+2j+n-2)\left|x\right|^{2j-2}Y_l(x).\] Induction on the degree now decomposes every homogeneous polynomial into terms \(\left|x\right|^{2j}Y_l(x)\). Indeed, decompose its Laplacian by the induction hypothesis and use the displayed identity to find a sum of such terms with the same Laplacian. The difference is harmonic. Restricting to the unit sphere shows that every polynomial restriction belongs to the span of the \(\mathcal H_l\).

Polynomial restrictions uniformly approximate continuous functions on \(S^{n-1}\). To see this without an approximation theorem, set \[K_j(u,v)= \frac{((1+u\cdot v)/2)^j} {\displaystyle\int_{S^{n-1}}((1+u\cdot w)/2)^j\,d\sigma(w)}.\] The denominator is positive and independent of \(u\), so \(T_jH(u)=\int K_j(u,v)H(v)\,d\sigma(v)\) is a polynomial in \(u\). For \(0<\delta<2\), let \(m_\delta>0\) be the measure of the cap \(\{v:\left|u-v\right|<\delta/2\}\). Since \((1+u\cdot v)/2=1-\left|u-v\right|^2/4\), \[\int_{\left|u-v\right|\ge\delta}K_j(u,v)\,d\sigma(v) \le \frac1{m_\delta} \left(\frac{1-\delta^2/4}{1-\delta^2/16}\right)^j.\] Uniform continuity of \(H\) therefore gives \(T_jH\to H\) uniformly. Continuous functions are dense in \(L^2(\sigma)\): approximate the sets in a simple function from inside by compact sets and from outside by open sets, then interpolate their indicators continuously using distance to these sets. The squared error is bounded by the measure of the intervening sets. This proves completeness.

Choose real orthonormal bases \(Y_{l,a}\) of the spaces \(\mathcal H_l\), with \(Y_{0,1}=1\). Integration by parts shows that the tangent fields \(\nabla_SY_{l,a}/\sqrt{\lambda_l}\), for \(l\ge1\), are orthonormal. Bessel’s inequality, which follows by expanding the squared distance from any finite orthogonal sum, consequently gives \[\mathbb E_{\sigma}\left|\nabla_SH\right|^2 \ge \sum_{l\ge1,a} \left|\mathbb E_{\sigma}\,\nabla_SH\cdot \frac{\nabla_SY_{l,a}}{\sqrt{\lambda_l}}\right|^2 =\sum_{l\ge1,a}\lambda_l\left|\mathbb E_{\sigma}(HY_{l,a})\right|^2.\] No differentiation of an infinite series is needed. If \(H\) is even, its odd-degree coefficients vanish. Completeness identifies its variance with the sum of the squared nonconstant coefficients and \(\left\lVert QH\right\rVert_2^2\) with the corresponding sum over even \(l\ge4\). Using \(\lambda_l\ge\lambda_2\) for even \(l\ge2\) and \(\lambda_l\ge\lambda_4\) for \(l\ge4\) proves (12). ◻

Proof. Let \(d_0\) be the density at zero of \(u\cdot v\) for uniform \(u\) and fixed \(v\in S^{n-1}\). Equivalently, \(d_0\) is the ratio of the surface areas of \(S^{n-2}\) and \(S^{n-1}\). For a fixed ambient vector \(w\), its tangential projection has divergence \(-p\,u\cdot w\). Integrating separately on the two hemispheres gives \[\begin{align*} &\int_{S^{n-1}}\mathop{\mathrm{sgn}}(u\cdot v)\,\nabla_Sf(u)\cdot w\,d\sigma(u) \\ &\quad=p\int_{S^{n-1}}\mathop{\mathrm{sgn}}(u\cdot v)f(u)\,u\cdot w\,d\sigma(u) -2d_0\int_{v^\perp\cap S^{n-1}}f(u)\,v\cdot w \,d\sigma_{v^\perp}(u). \tag{16}\end{align*}\] The outward conormals on the equator are \(-v\) and \(v\), respectively; the sign change between the hemispheres makes their contributions add with the displayed negative sign. Taking \(f=1\) and \(w=v\) yields \[2d_0=p b_n.\]

The radial–radial term in (14), before division by \(b_n\), is \(b_n\langle f_1,Cf_2\rangle\). Apply (16) with \(w=v\) and then integrate against \(f_2(v)\). Each of the two mixed radial–gradient terms, before its factor \(k^{-1}\), equals \[p b_n\langle f_1,(C-P)f_2\rangle.\] For the gradient–gradient term, first integrate in \(u\) with \(w=\nabla_Sf_2(v)\). Its equator term vanishes because \(v\cdot\nabla_Sf_2(v)=0\). What remains is \[p\iint f_1(u)\mathop{\mathrm{sgn}}(u\cdot v)\, u\cdot\nabla_Sf_2(v)\,d\sigma(u)d\sigma(v).\] A second application of (16), now in \(v\) with \(w=u\), makes this \(p^2b_n\langle f_1,(C-P)f_2\rangle\). Adding the four terms proves (15). ◻

Proof. Fix \(u\in S^{n-1}\) and a homogeneous harmonic polynomial \(Y_{2j}\). Let \(Z\) be a standard Gaussian in the \(p\)-dimensional space \(u^\perp\). Its direction is uniform on the equator and independent of its radius, so polar integration gives \[\mathbb E Y_{2j}(Z) =p(p+2)\cdots(p+2j-2)\,(PY_{2j})(u).\] The radial moment here follows by successive integration by parts in \(\int_0^\infty r^{p+2j-1}e^{-r^2/2}\,dr\). The one-dimensional Gaussian moment recurrence \(\mathbb E Z_1^{2a}=(2a-1)\mathbb E Z_1^{2a-2}\), together with independence of coordinates, also gives \[\mathbb E Y_{2j}(Z) =\frac{(\Delta_{u^\perp})^jY_{2j}(0)}{2^j j!}.\] This identity follows term by term on monomials; monomials with an odd exponent have both sides zero, and the other coefficients are exactly the multinomial coefficients in the power of the Laplacian. Harmonicity and commutation of constant-coefficient derivatives imply \[(\Delta_{u^\perp})^jY_{2j} =(-1)^j\partial_u^{2j}Y_{2j}, \qquad \partial_u^{2j}Y_{2j}(0)=(2j)!Y_{2j}(u).\] Since \((2j)!/(2^j j!)=1\cdot3\cdots(2j-1)\), the formula for \(P_{2j}\) follows.

To obtain \(C\), integrate by parts on the two hemispheres in the variable \(v\). Off the equator, \(\Delta_S\left|u\cdot v\right|=-p\left|u\cdot v\right|\); the jump of its normal derivative contributes \(2d_0\) times equator averaging. Thus, for any smooth \(Y\), \[\int_{S^{n-1}}\left|u\cdot v\right|\Delta_SY(v)\,d\sigma(v) =-p\int_{S^{n-1}}\left|u\cdot v\right|Y(v)\,d\sigma(v) +2d_0(PY)(u).\] For \(Y\in\mathcal H_l\), division by \(b_n\) gives \(-\lambda_l CY=p(PY-CY)\). The denominator \(p-\lambda_l=-(l-1)(l+p)\) is nonzero for every even \(l\), proving the second formula in (17). Its first three values give those in (18); division of successive values, using \(P_{l+2}/P_l=-(l+1)/(l+p)\), gives the recurrence. ◻

Proof. By Lemma 9, the operator in (15) has multiplier one on constants and, on degree \(l\), multiplier \[\mu_l=C_l\left[1+\left(\frac2k+\frac p{k^2}\right)\lambda_l\right].\] Indeed, \(C_l-P_l=(\lambda_l/p)C_l\). Put \(a=2/k+p/k^2>0\). For even \(l\ge4\), \[(l+p+2)\lambda_l-(l-1)\lambda_{l+2} =(p-1)l^2+(p^2-1)l+2(p+1)>0.\] Since also \(l-1<l+p+2\), it follows that \[\frac{\left|\mu_{l+2}\right|}{\left|\mu_l\right|} =\frac{l-1}{l+p+2}\, \frac{1+a\lambda_{l+2}}{1+a\lambda_l}\le1.\] Consequently \(\left|\mu_l\right|\le\left|\mu_4\right|=c_{n,k}\) on all even \(l\ge4\). The constant term of the bilinear form is \((\mathbb E_{\sigma}f_1)(\mathbb E_{\sigma}f_2)\), and its degree-two term vanishes by hypothesis. Cauchy–Schwarz on the remaining harmonic coefficients proves (20). This expansion is justified by the completeness in Lemma 7 and the boundedness of \(C\) and \(P\).

We finish by proving the passage to norms, retaining the details needed when their unit balls have corners. Let \(g\) be a norm and let \(g_\epsilon=g*\rho_\epsilon\) on \(\mathbb R^n\), where \(\rho_\epsilon\) is a smooth, even, nonnegative function of integral one, supported in the ball of radius \(\epsilon\). The functions \(g_\epsilon\) are smooth, even and convex. If \(L\) is the Euclidean Lipschitz constant of \(g\), then \[\left|g_\epsilon-g\right|\le L\epsilon, \qquad \left|\nabla g_\epsilon\right|\le L.\] At a differentiability point \(x\) of \(g\), these gradients converge to \(\nabla g(x)\). Indeed, every convergent subsequence of the bounded gradients has a limit \(a\), and convexity gives, on passing to the limit, \[g(y)\ge g(x)+a\cdot(y-x)\qquad(y\in\mathbb R^n).\] Taking \(y=x\pm th\) and then \(t\downarrow0\) forces \(a\cdot h=\nabla g(x)\cdot h\) for every \(h\), so the only possible limit is \(\nabla g(x)\).

Lemma 2 supplies differentiability of \(g\) at almost every spherical direction. Equivalently, one may use its almost-everywhere differentiability in \(\mathbb R^n\) and homogeneity: the exceptional set is a union of rays, so polar integration makes its intersection with the sphere null. On the sphere \(g\) is bounded away from zero. Thus the restrictions \(f_\epsilon=g_\epsilon^k|_{S^{n-1}}\) converge uniformly to \(g^k\), and \[\nabla_Sf_\epsilon(u) =k g_\epsilon(u)^{k-1} (I_n-uu^t)\nabla g_\epsilon(u) \longrightarrow k g(u)^{k-1}(I_n-uu^t)\nabla g(u)\] almost everywhere and in \(L^2(\sigma)\), by bounded convergence. Euler’s identity \(u\cdot\nabla g(u)=g(u)\), obtained by differentiating \(g(tu)=tg(u)\), identifies the limiting vector field with (21).

The approximants \(g_\epsilon\) need not be homogeneous; only their restrictions and tangential gradients are used here. The fields \(X_{f_\epsilon}\) converge in \(L^2\), so the bounded sign kernel in (14) permits passage to the limit. The right side of (15) also converges, because \(C\) and \(P\) are bounded on \(L^2\). This proves the operator identity for norm powers. Finally apply its harmonic multiplier estimate directly to the limiting functions. The approximants themselves need not have vanishing degree-two part: that hypothesis is used only for the limiting \(f_1\). This proves (20) in the stated nonsmooth setting. ◻

Proof. Suppose first that \(g\) is smooth away from the origin. The trace of its Euclidean Hessian on the tangent space at \(u\in S^{n-1}\) is \(\Delta_Sg(u)+p g(u)\), and it is nonnegative by convexity. Multiplying by \(I(g)\ge0\) and integrating by parts gives \[\begin{align*} \mathbb E_{\sigma}|\nabla_SG(g)|^2 &=\beta^2\mathbb E_{\sigma}[I'(g)|\nabla_Sg|^2] =-\beta^2\mathbb E_{\sigma}[I(g)\Delta_Sg] \le p\beta^2\mathbb E_{\sigma}[gI(g)]. \end{align*}\] The spectral gap estimate (12) proves (28) in this case.

Here is a passage to arbitrary norms that does not assume the approximants are homogeneous. Convolve the ambient norm with smooth, even, nonnegative mollifiers of mass one and shrinking compact support, and write the resulting smooth convex functions as \(g_\varepsilon\). On the unit sphere their tangent Hessian trace is \[\Delta_Sg_\varepsilon+p\partial_r g_\varepsilon\ge0.\] The preceding argument and (12) therefore give \[2n\mathop{\mathrm{Var}}(G(g_\varepsilon))+d_*\|QG(g_\varepsilon)\|_2^2 \le p\beta^2\mathbb E_{\sigma}[I(g_\varepsilon)\partial_r g_\varepsilon].\] The functions \(g_\varepsilon\) converge uniformly to \(g\) on compact sets, are uniformly Lipschitz there, and their gradients converge to \(\nabla g\) at every differentiability point of \(g\). To see the last assertion, every limit of their bounded gradients satisfies the subgradient inequality for \(g\) by convexity and uniform convergence; differentiability makes that subgradient unique. The exceptional directions have spherical measure zero, as in Lemma 2. Homogeneity of \(g\) then gives \(\partial_r g_\varepsilon(u)\to\nabla g(u)\cdot u=g(u)\) almost everywhere. On the sphere the approximants stay in a fixed positive compact interval, while their radial derivatives are bounded. Uniform convergence on the left and dominated convergence on the right prove (28). ◻

Proof of Theorem 11. In the following expectations, \(G,D,I\) denote their compositions with \(g\). Set \[m=\mathbb E_{\sigma}g^k-1,\qquad \ell=\mathbb E_{\sigma}g^{k-1}-1,\qquad t=\mathbb E_{\sigma}G.\] The normalization gives \(\mathbb E_{\sigma}e(g)=0\), so \(m=\mathbb E_{\sigma}M(g)\ge0\), \(\mathbb E_{\sigma}D=m-t\), and \(\mathbb E_{\sigma}W(g)=m-\ell\). Since \(0\le G\le\beta M\), we have \(0\le t\le\beta m\). For every \(a>0\), Jensen’s inequality applied to the convex function \(y\mapsto y^{-a/n}\) yields \(\mathbb E_{\sigma}g^a\ge(\mathbb E_{\sigma}g^{-n})^{-a/n}=1\). Together with \[z^{k-1}\le\frac{(k-1)z^k+1}{k},\] this gives \[ 0\le\ell\le\frac{k-1}{k}\,m. \tag{32}\] For \(k=1\) this simply says \(\ell=0\).

Since \(Q\) annihilates constants, \(Q(g^k)=QG+QD\). The Hilbert-space inequality \[\|a+b\|^2\le\frac{\|a\|^2}{w} +\frac{\|b\|^2}{1-w} \qquad(0<w<1)\] and \(\|QD\|_2^2\le\mathop{\mathrm{Var}}(D)\), followed by Lemma 12, give \[\begin{align*} c\|Q(g^k)\|_2^2 &\le\frac{c}{wd_*} \left(p\beta^2\mathbb E_{\sigma}[gI(g)]-2n\mathop{\mathrm{Var}}(G)\right) +\frac{c}{1-w}\mathop{\mathrm{Var}}(D)\\ &=\mathbb E_{\sigma}\mathcal L(g)+c\left( \frac{2n}{wd_*}t^2-\frac{(m-t)^2}{1-w}\right). \tag{33}\end{align*}\] The quadratic expression in parentheses is increasing for \(0\le t\le m\): its derivative is \[\frac{4n}{wd_*}t+\frac{2(m-t)}{1-w}\ge0.\] It is therefore no larger than its value at \(t=\beta m\). Using (30) and (32), we conclude that the final term of (33) is at most \[\frac{2k-1}{k^2}m^2\le m^2-\ell^2.\] Now (31) yields \[\begin{align*} c\|Q(g^k)\|_2^2 &\le 2\mathbb E_{\sigma}W(g)+m^2-\ell^2\\ &=2(m-\ell)+m^2-\ell^2 =(\mathbb E_{\sigma}g^k)^2-(\mathbb E_{\sigma}g^{k-1})^2, \end{align*}\] as claimed. If \(g\) is not identically \(1\) on the sphere, continuity gives a set of positive spherical measure on which \(g\ne1\). Lemma 13 then gives \(\mathbb E_{\sigma}\mathcal L(g)<2\mathbb E_{\sigma}W(g)\), so the final inequality is strict. The only constant norm on the sphere with \(\mathbb E_{\sigma}g^{-n}=1\) is \(g=1\); in this case both sides of (23) vanish. ◻

Proof of Lemma 13. We first record the values of the spectral constant: \[ c=\frac{121}{35}\quad(n=4),\qquad c=\frac{n^2+5n+7}{(n+1)(n+3)}\quad(n\ge5). \tag{34}\] For \(n=4,5,6\) the left side of (30) is, respectively, \[\frac{11}{12},\qquad \frac{589}{864},\qquad \frac{1387}{1890}.\] These are smaller than \(1,3/4,3/4\), the respective right sides. For \(n\ge7\) the left side is \[\frac{2(3n-4)(n^2+5n+7)}{9(n+1)(n+3)(n+4)}.\] Its difference from \(3/4\), with the latter first, is \[\frac{3n^3+128n^2+505n+548}{36(n+1)(n+3)(n+4)}>0.\] This proves (30) for every dimension.

The interval \(0<z\le1\). Here \(G=I=0\) and \(|D(z)|=|z^k-1|\le k|z-1|\). Also \(W(1)=W'(1)=0\) and \[W''(z)=\begin{cases} (n+1)z^{-n-2},&k=1,\\ 2+(n+1)z^{-n-2},&k=2. \end{cases}\] Thus \(2W(z)\ge(n+2k-1)(z-1)^2\). The positive margins \[ n+2k-1-\frac{ck^2}{1-w} =\begin{cases} 3/5,&n=4,\\ 7/8,&n=5,\\ 43/21,&n=6,\\ n+3-8c,&n\ge7 \end{cases} \tag{35}\] give the desired strict inequality for \(z<1\). In the last case the margin is \(9/10\) at \(n=7\) and increases thereafter: regarding \(n\) as a real variable in (34), \[c'(n)=-\frac{n^2+8n+13}{(n+1)^2(n+3)^2}<0.\] At \(z=1\) both sides of (31) vanish.

The interval \(z>1\) in dimensions \(4,5,6\). Set \(x=z-1>0\). Multiplication by a positive factor clears the negative powers of \(z\): if \[P_n(z)=s_nz^{2n}\bigl(2W(z)-\mathcal L(z)\bigr), \qquad (s_4,s_5,s_6)=(2880,712800,221130),\] then substitution of (24)–(29) and collection of equal powers give \[\begin{align*} P_4(z)={}&462z^{10}-983z^9+93z^8 +3399z^5-3378z^4+407,\\ P_5(z)={}&151525z^{14}-273570z^{12}-680800z^{11} +938157z^{10}\\ &\quad+225720z^7-462264z^5+101232,\\ P_6(z)={}&48399z^{16}-75894z^{14}-218004z^{13} +285961z^{12}\\ &\quad+56940z^8-127478z^6+30076. \tag{36}\end{align*}\] For clarity, the coefficients needed from \(P_n(1+x)\) are listed below; the constant and linear coefficients are zero. \[\begin{array}{c|rrr} [x^j]&n=4&n=5&n=6\\ \hline j=2&1728&623700&452790\\ j=3&-1446&-1505900&679380\\ j=4&-6711&-5802775&-132249\\ j=5&1173&12849430&7800780\\ j=6&17052&86314305&51269452\\ j=7&20796&191506920&146058120 \end{array}\] All remaining coefficients are nonnegative, as can be seen without listing them. Let \(a_nz^{d_n}\) be the leading term in (36). At \(j=8\), its contribution \(a_n\binom{d_n}{8}\) minus the contributions of all negative terms is \[11943,\qquad 207280425,\qquad 114414300 \quad\text{for } n=4,5,6,\] respectively. For every \(i<d_n\), the ratio \(\binom{i}{j}/\binom{d_n}{j}\) is nonincreasing in \(j\); while it is nonzero, its ratio at successive indices is \((i-j)/(d_n-j)\le1\). Once \(j>i\) it is zero. Thus the same leading term covers all negative contributions for every \(8\le j\le d_n\).

It follows that \(P_n(1+x)/x^2\) is bounded below by, respectively, \[\begin{align*} U_4(x)&=1728-1446x-6711x^2+17052x^4,\\ U_5(x)&=623700-1505900x-5802775x^2+86314305x^4,\\ U_6(x)&=452790-132249x^2+51269452x^4. \end{align*}\] These lower bounds are positive. Indeed \[\begin{align*} 1446x&\le400+1400x^2,& 8111x^2&\le1000+17052x^4,\\ 1505900x&\le200000+2900000x^2,& 8702775x^2&\le230000+86314305x^4. \end{align*}\] Each follows from \(dt\le a+bt^2\) when \(d^2\le4ab\); the corresponding values of \(4ab-d^2\) are \[149084,\quad 2419679,\quad 52265190000,\quad 3670867899375.\] The first row gives \(U_4(x)\ge328\) and the second gives \(U_5(x)\ge193700\). Finally \(U_6\), regarded as a quadratic in \(x^2\), has positive leading coefficient and negative discriminant, since \[4\cdot452790\cdot51269452-132249^2=92839690886319>0.\] Therefore \(P_n(z)>0\) for \(z>1\) in all three dimensions.

The interval \(z>1\) in all dimensions \(n\ge7\). Put \(H=18(n+4)/c\). Direct collection of powers in (29) now gives \[ \frac{9(n+4)}{c}z^{2n}(2W-\mathcal L) =z^{2n}(A_4z^4+A_2z^2+A_1z+A_0) +z^n(B_2z^2+B_0)+T_0, \tag{37}\] where \[\begin{align*} A_4&=\frac{2n-8}{3},& A_2&=H-12n-32-\frac{64}{n},\\ A_1&=-H+\frac{16(n-1)}3+32-\frac{16(n-1)}{2n+1},&&\\ A_0&=-\frac Hn+6n+40+\frac{64}{n}-\frac{128}{n^2},&&\\ B_2&=16+\frac{64}{n},& B_0&=\frac Hn-48-\frac{64}{n}+\frac{256}{n^2},\\ T_0&=\frac{16(n-1)}{2n+1}-\frac{128}{n^2}.&& \tag{38}\end{align*}\] We prove that the polynomial on the right of (37), expressed in \(x=z-1\), has nonnegative coefficients and a positive quadratic coefficient.

Write its two parentheses as \(\sum_{i=0}^4q_ix^i\) and \(\sum_{i=0}^2r_ix^i\). Thus \[\begin{align*} q_0&=A_4+A_2+A_1+A_0,&q_1&=4A_4+2A_2+A_1,\\ q_2&=6A_4+A_2,&q_3&=4A_4,\qquad q_4=A_4,\\ r_0&=B_2+B_0,&r_1&=2B_2,\qquad r_2=B_2. \end{align*}\] In particular \(q_3,q_4,r_1,r_2>0\). The constant and linear coefficients of the full polynomial vanish. Its quadratic coefficient is \[ \frac{9(n+4)}c\,(n+3-8c)>0. \tag{39}\] One can obtain these three coefficients either from (38) or directly from the order of vanishing at \(z=1\): \(G=O(x^2)\), \(I=O(x^3)\), and \(D=2x+O(x^2)\), so that \(2W-\mathcal L=(n+3-8c)x^2+O(x^3)\). Positivity is (35).

The remaining signs admit explicit polynomial certificates. Set \[\begin{align*} V_0(n)&=34n^5-95n^4-1666n^3-5701n^2-8052n-2688,\\ V_1(n)&=2n^5+9n^4-162n^3-845n^2-1220n-448,\\ V_2(n)&=5n^3+8n^2-41n-56,\\ R(n)&=7n^4+8n^3-187n^2-748n-896. \end{align*}\] Substituting \(H=18(n+4)(n+1)(n+3)/(n^2+5n+7)\) gives \[\begin{align*} q_0+\frac{r_0}{8} &=\frac{V_0(n)}{4n^2(2n+1)(n^2+5n+7)},\\ q_1&=\frac{2V_1(n)}{n(2n+1)(n^2+5n+7)},\\ q_2&=\frac{2(n+4)V_2(n)}{n(n^2+5n+7)},& r_0&=-\frac{2R(n)}{n^2(n^2+5n+7)}. \tag{40}\end{align*}\] The required signs hold throughout their stated ranges because, for \(t\ge0\), \[\begin{align*} V_0(10+t)={}&34t^5+1605t^4+28534t^3 +227319t^2\\ &\quad+698128t+130692,\\ V_1(10+t)={}&2t^5+109t^4+2198t^3+19695t^2 +69280t+30852,\\ V_2(7+t)={}&5t^3+113t^2+806t+1764,\\ R(7+t)={}&7t^4+204t^3+2039t^2+7414t+4256. \tag{41}\end{align*}\] Consequently \(q_2>0\) and \(r_0<0\) for every \(n\ge7\), while \(q_0+r_0/8>0\) and \(q_1>0\) for every \(n\ge10\).

Use the convention that \(\binom mj=0\) when \(j<0\) or \(j>m\). For \(3\le j\le2n\), comparison of the factors in the binomial products gives \[ \binom nj\le 2^{-j}\binom{2n}{j}. \tag{42}\] Indeed \((n-a)/(2n-a)\le1/2\) for \(0\le a<n\); if \(j>n\) the left side is zero. For \(n\ge10\), the sum of the \(q_0\) and \(r_0\) contributions to the coefficient of \(x^j\) is therefore at least \[\left(q_0+\frac{r_0}{2^j}\right)\binom{2n}{j} \ge\left(q_0+\frac{r_0}{8}\right)\binom{2n}{j}>0.\] All other contributions are nonnegative. For \(j>2n\), the \(q_0\) and \(r_0\) contributions vanish and every remaining term is nonnegative. Together with (39), this proves coefficient positivity for every \(n\ge10\).

It remains to treat \(n=7,8,9\), where \(q_1\) can be negative. For \(3\le j\le2n\), divide the coefficient of \(x^j\) by \(\binom{2n}{j}\), discard the nonnegative \(q_3,q_4,r_2\) contributions, and use (42) for the negative \(r_0\) term. The resulting lower bound is \[ \begin{split} q_0+\frac{r_0}{2^j} +\frac{j}{2n-j+1} \left(q_1+q_2\frac{j-1}{2n-j+2}\right) +r_1\frac{\binom n{j-1}}{\binom{2n}{j}}. \end{split} \tag{43}\] Substitution in (38) gives the componentwise lower bounds \[\begin{array}{c|rrrrr} n&q_0&q_1&q_2&r_0&r_1\\ \hline 7&-2&-11&60&-2&50\\ 8&-1&-6&72&-6&46\\ 9&-1&-6&72&-6&46 \end{array}\] All weights in (43) are nonnegative, so these bounds may be substituted directly. They give the following positive numbers at \(j=3,4\); at \(j=5\) we have already omitted the final \(r_1\) term: \[\begin{array}{c|ccc} n&j=3&j=4&j=5\ \text{without the final term}\\ \hline 7&5/26&1233/1144&589/176\\ 8&37/28&10707/3640&1153/208\\ 9&101/136&3851/2040&395/112 \end{array}\] These finite substitutions can also be checked from the exact tuples \((q_0,q_1,q_2,r_0,r_1)\): \[\begin{align*} n=7:&\quad \left(-\frac{5984}{3185},-\frac{4832}{455}, \frac{792}{13},-\frac{1216}{637},\frac{352}{7}\right),\\ n=8:&\quad \left(-\frac{421}{629},-\frac{3736}{629}, \frac{2688}{37},-\frac{145}{37},48\right),\\ n=9:&\quad \left(\frac{2416}{10773},-\frac{2192}{1197}, \frac{100568}{1197},-\frac{57968}{10773},\frac{416}{9}\right). \end{align*}\] For \(5\le j\le2n\), the parenthesis \(q_1+q_2(j-1)/(2n-j+2)\) is positive and increasing in \(j\). In fact the displayed componentwise bounds give at \(j=5\) the positive lower bounds \(119/11\), \(210/13\), and \(66/5\), respectively. The prefactor \(j/(2n-j+1)\) is positive and increasing too, while \(r_0/2^j\) increases because \(r_0<0\). The expression (43) with its last term omitted is therefore increasing throughout this range. Positivity at \(j=5\) proves positivity at every subsequent index through \(2n\).

For \(j>2n\) all \(r_i\) contributions vanish, since \(n+2<2n+1\). The only possibly negative term is the \(q_1\) contribution at \(j=2n+1\). At this index the \(q_1\) and \(q_2\) contributions sum to \(q_1+2nq_2\), which the same componentwise bounds make at least \(829,1146,1290\) for \(n=7,8,9\), respectively. For \(j\ge2n+2\) only the nonnegative \(q_2,q_3,q_4\) terms remain. This handles every coefficient, including the endpoints.

We have proved that the polynomial in (37) has nonnegative coefficients and a strictly positive coefficient of \(x^2\) for every \(n\ge7\). It is consequently positive for every \(x>0\). Its multiplier \(9(n+4)z^{2n}/c\) is positive, so \(2W(z)-\mathcal L(z)>0\) for \(z>1\). This finishes the proof of (31) and its strictness in every dimension. ◻

Proof. Theorem 1 and the separately proved three-dimensional result [2] establish Petty’s conjecture for every \(n\ge3\). Lutwak’s conditional result [10], in the explicit formulation of Ortega-Moreno and Schuster [15], states that Petty’s conjecture implies the Euclidean-ball lower bound for these ratios for every \(1\le i\le n-1\). Applying that implication gives the inequality. For its explicit value, the Steiner normalization gives \[V_j(B_2^m)=\binom{m}{j}\frac{\kappa_m}{\kappa_{m-j}}, \qquad \Pi_iB_2^n=V_i(B_2^{n-1})B_2^n.\] The \((i+1)\)-homogeneity of \(V_{i+1}\) now gives the displayed ball ratio. Only the inequality conclusion of the conditional result is used. ◻

Proof. Write \(c_n=\kappa_{n-1}^{\,n}\kappa_n^{\,2-n}\). Haddad’s geometric functional \(\widetilde I_1(K_1,\ldots,K_n)\) integrates the absolute determinant against the surface-area measures of the convex bodies \(K_i\). He uses \[h_{\Pi K}(u)=\frac12\int_{S^{n-1}}|\left\langle u,v\right\rangle|\,dS_K(v),\] which is the projection-volume normalization above by Cauchy’s projection formula. His mixed-volume identity and the Aleksandrov–Fenchel inequality give \[\frac1{n!}\widetilde I_1(K_1,\ldots,K_n) =V(\Pi K_1,\ldots,\Pi K_n) \ge\prod_{i=1}^n|\Pi K_i|^{1/n}.\] Here \(V\) denotes mixed volume, and Haddad’s \(\omega_m\) is our unit-ball volume \(\kappa_m\); see [4]. Theorem 1 for \(n\ge4\) and the separately proved three-dimensional case [2] therefore give \[\widetilde I_1(K_1,\ldots,K_n) \ge n!c_n\prod_{i=1}^n|K_i|^{(n-1)/n}.\]

Haddad’s Lemma 5.1 and the equivalence of Equations (18)–(21) [4] transfer this \(p=1\) geometric inequality to the functional inequality for \(\mathcal D_n\), with critical exponent \(p^*=n/(n-1)=q\). At this endpoint his normalized weak profile for \(K\) is \(\mathbf1_K\), with the same surface-area measure as \(K\) and \(\left\lVert\mathbf 1_K\right\rVert_q=|K|^{(n-1)/n}\); see [4]. Consequently the functional and geometric sharp constants agree, as also follows from the constant relations after his Equation (21). This proves Equation (63); taking all \(f_i=f\) gives Equation (64), the \(p=1\) case of his Equation (22).

For sharpness, take nonnegative compactly supported smooth mollifications \(f_j\) of the indicator of an ellipsoid \(E\). They converge to \(\mathbf1_E\) strictly in \(BV(\mathbb R^n)\) and in \(L^q\). Define finite measures \(\nu_j\) on the sphere by \[\int\psi\,d\nu_j =\int_{\{\nabla f_j\ne0\}} \psi\!\left(-\frac{\nabla f_j}{|\nabla f_j|}\right) |\nabla f_j|\,dx, \qquad \psi\in C(S^{n-1}).\] Reshetnyak’s continuity theorem [19] gives \(\nu_j\rightharpoonup S_E\); the minus sign selects the outward normal in \(D\mathbf1_E\). The product measures also converge weakly on the compact product sphere. Since the absolute determinant is continuous and homogeneous in each variable, this gives \(\mathcal D_n(f_j,\ldots,f_j)\to \widetilde I_1(E,\ldots,E)=n!|\Pi E|\). Since \(\left\lVert\mathbf 1_E\right\rVert_q^n=|E|^{n-1}\), ellipsoid equality in Petty’s inequality gives the stated constants in the limit. This is the weak \(p=1\) interpretation of sharpness. ◻

Proof. The fixed-norm isoperimetrix formula and the equivalence with symmetric unpolarized Petty are classical; see Martini–Mustafaev [12]. Minkowski’s mixed-volume inequality gives \[\frac{\mathcal A_B(\partial C)^n}{\mathcal V_B(C)^{n-1}} \ge n^n\frac{\kappa_n^{n-1}}{\kappa_{n-1}^{n}}R_n(B^\circ) \ge n^n\kappa_n.\] The last step applies Theorem 1 to \(B^\circ\) for \(n\ge4\), and the separate three-dimensional result [2] for \(n=3\). Equality in the first step means that \(C\) is homothetic to \(I_B\), while equality in the last makes \(B^\circ\), hence \(B\), an ellipsoid. Then \(I_B\) is homothetic to \(B\), giving exactly the stated equality cases. ◻