Petty’s projection-volume conjecture in dimensions at least four

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

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 \R^n. If a vector w=v/s(v) attains the maximum at such a point x, then

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

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 \Sph 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

are continuous on the relevant compact set. Comparing the maxima at t and 0, using v_t and v_0, therefore gives

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 \Sph. Dominated convergence now differentiates the volume formula and proves (4).□

Proof. Support functions add under Minkowski addition, so Lemma 2.1, 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 2.2 and homogeneity give

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

Thus the preceding volume inequality is an equality for every t>0. The equality case of Theorem 2.2, 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), \E 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 2.3 gives, for every symmetric convex body M,

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

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 \supp\eta_K failed to span \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

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

Induction on the degree now decomposes every homogeneous polynomial into terms \abs{x}^{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 \Sph. To see this without an approximation theorem, set

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:\abs{u-v}<\delta/2\}. Since (1+u\cdot v)/2=1-\abs{u-v}^2/4,

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 \gradS Y_{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

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 \norm{QH}_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\Sph. 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

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

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

For the gradient–gradient term, first integrate in u with w=\gradS f_2(v). Its equator term vanishes because v\cdot\gradS f_2(v)=0. What remains is

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\Sph 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

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

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

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, \lapS\abs{u\cdot v}=-p\abs{u\cdot v}; the jump of its normal derivative contributes 2d_0 times equator averaging. Thus, for any smooth Y,

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 3.3, the operator in (15) has multiplier one on constants and, on degree l, multiplier

Indeed, C_l-P_l=(\lambda_l/p)C_l. Put a=2/k+p/k^2>0. For even l\ge4,

Since also l-1<l+p+2, it follows that

Consequently \abs{\mu_l}\le\abs{\mu_4}=c_{n,k} on all even l\ge4. The constant term of the bilinear form is (\E f_1)(\E 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 3.1 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 \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

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,

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 2.1 supplies differentiability of g at almost every spherical direction. Equivalently, one may use its almost-everywhere differentiability in \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|_{\Sph} converge uniformly to g^k, and

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\Sph is \lapS g(u)+p g(u), and it is nonnegative by convexity. Multiplying by I(g)\ge0 and integrating by parts gives

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

The preceding argument and (12) therefore give

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 2.1. 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 4.1. In the following expectations, G,D,I denote their compositions with g. Set

The normalization gives \E e(g)=0, so m=\E M(g)\ge0, \E D=m-t, and \E 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 \E g^a\ge(\E g^{-n})^{-a/n}=1. Together with

this gives

For k=1 this simply says \ell=0.

Since Q annihilates constants, Q(g^k)=QG+QD. The Hilbert-space inequality

and \|QD\|_2^2\le\Var(D), followed by Lemma 4.2, give

The quadratic expression in parentheses is increasing for 0\le t\le m: its derivative is

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

Now (31) yields

as claimed. If g is not identically 1 on the sphere, continuity gives a set of positive spherical measure on which g\ne1. Lemma 4.3 then gives \E\mathcal L(g)<2\E W(g), so the final inequality is strict. The only constant norm on the sphere with \E g^{-n}=1 is g=1; in this case both sides of (23) vanish.□

Proof of Lemma 4.3. We first record the values of the spectral constant:

For n=4,5,6 the left side of (30) is, respectively,

These are smaller than 1,3/4,3/4, the respective right sides. For n\ge7 the left side is

Its difference from 3/4, with the latter first, is

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

Thus 2W(z)\ge(n+2k-1)(z-1)^2. The positive margins

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),

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

then substitution of (24)–(29) and collection of equal powers give

For clarity, the coefficients needed from P_n(1+x) are listed below; the constant and linear coefficients are zero.

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

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,

These lower bounds are positive. Indeed

Each follows from dt\le a+bt^2 when d^2\le4ab; the corresponding values of 4ab-d^2 are

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

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

where

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

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

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

Substituting H=18(n+4)(n+1)(n+3)/(n^2+5n+7) gives

The required signs hold throughout their stated ranges because, for t\ge0,

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

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

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

Substitution in (38) gives the componentwise lower bounds

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:

These finite substitutions can also be checked from the exact tuples (q_0,q_1,q_2,r_0,r_1):

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. The spanning hypothesis gives a norm on V:

Indeed, q_A(y)=0 implies that Ay is orthogonal to \supp\eta, and hence Ay=0; because y\in V, this forces y=0. For any normalized competitor L and y\in L, symmetry gives q_A(y)\le\tau_A(L)=1. Thus all normalized competitors have a common outer radius in V.

We also need a bound on their gauges along a minimizing sequence. For any nonzero seminorm g, let R=\max_{\Sph}g. By separation, g is the support function of the compact symmetric set \{z:z\cdot y\le g(y)\text{ for every }y\in\R^n\}. This set has maximal radius R: if Re is a point of maximal norm in it, then

Consequently,

Here \E|\log|u_1||^2<\infty: the first coordinate has a density proportional to (1-t^2)^{(n-3)/2} on (-1,1), and \int_0^{1/2}|\log t|^2\,dt<\infty. In particular every nonzero seminorm has a finite logarithmic objective.

Set b_A=\int|Ax|\,d\eta(x)>0. The ball of radius 1/b_A in V is a normalized competitor. Comparison with this ball bounds J above along a minimizing sequence, so the preceding inequalities bound R above. The bodies therefore contain a common inner ball in V, as well as lying in a common outer ball. Their gauges are uniformly bounded and equicontinuous on the unit sphere of V. A subsequence converges uniformly to the gauge of a body with the same two ball bounds. The support functions converge uniformly as well, so the normalization passes to the limit. The objective also passes to the limit: in the logarithmic case the extended gauges are uniformly comparable to |P_Vu|, whose logarithm is integrable by (45). This proves existence.

For uniqueness, let L_0,L_1 be two minimizers and L_t=(1-t)L_0+tL_1, where 0<t<1. Its normalization remains one. For P_Vu\ne0, write \rho_i(u)=1/g_{L_i}(P_Vu). Taking points on the ray through P_Vu gives

Both -\log\rho and 1/\rho are strictly convex decreasing functions of \rho>0. Thus the objective of L_t is at most the interpolated objectives, and the inequality is strict wherever \rho_0\ne\rho_1. Distinct bodies have different radial functions on an open set of directions in V, which gives a set of positive spherical measure after projection. They would therefore give a strictly smaller objective, a contradiction.□

Proof. Normalization turns the minimization problem into minimizing J(\tau_A(L)g_L) over all symmetric bodies. Put \tau=\tau_A(L) and I=\int\phi(Ax)\,d\eta(x). For small positive or negative t, define the body

The constraint is positive for sufficiently small |t|, so W_t is a symmetric convex body. Although that constraint need not be a support function, it bounds the actual support function above. Therefore

Both quantities are positive for small |t|. Since J increases under pointwise increase of a positive gauge, minimality gives

with equality at t=0.

Lemma 2.1 gives

almost everywhere, with uniform bounds that permit differentiation of the objective. The left side of (48) has a two-sided minimum at zero. Its derivative is consequently zero:

Equivalently,

Euler’s identity for the homogeneous gauge gives (47), and positive homogeneity of \phi then gives (46). The argument applies to both signs of t without imposing convexity on the varied constraint.□

Proof. Let A_j\to A in \mathcal A, and write V_j=\range(A_j), g_j=g_{A_j}, and b_j=\int|A_jx|\,d\eta(x). Then b_j\to b=\int|Ax|\,d\eta(x)>0. The normalized ball in V_j has extended gauge b_j|P_{V_j}u|, and hence objective at most \log b_j if k=1, and at most b_j if k=2. By (45), this bounds R_j=\max_{\Sph}g_j uniformly above. The minimizing body contains the ball of radius 1/R_j in V_j, so its normalization gives 1\ge b_j/R_j. Thus R_j\ge b_j, bounded away from zero.

The seminorms g_j have uniformly bounded Lipschitz constants. Every subsequence therefore has a further subsequence converging uniformly on \Sph to a nonzero seminorm g. Along such a subsequence,

For k=2 this is immediate. For k=1, choose unit vectors e_j with g_j(u)\ge R_j|u\cdot e_j|. The positive parts of \log g_j are uniformly bounded, and their negative parts have uniformly bounded squared integrals, by rotation invariance and \E|\log|u_1||^2<\infty. These logarithms are therefore uniformly integrable. They converge almost everywhere to \log g, since the kernel of a nonzero seminorm is a proper subspace. Truncating the logarithms below at -M and then sending M to infinity proves (50); the discarded integrals are uniformly O(M^{-1}) by the squared-integral bound.

We identify this subsequential limit. Set Z=\{g\le1\} and Z_j=\{g_j\le1\}, allowing infinite values for their support functions. If g(y)<1, then y\in Z_j for all sufficiently large j. Testing the support functions against y and then taking the supremum gives, for every x,

The supremum over g(y)<1 equals h_Z, by scaling towards zero. Since Z_j=L_{A_j}+V_j^\perp and A_jx\in V_j, its support function at A_jx equals that of L_{A_j}. All these support functions are nonnegative, so Fatou’s lemma yields

Let V=\range(A) and L_0=\overline{P_VZ}. This projection is bounded despite the possible unboundedness of Z. Indeed, for every z\in Z, symmetry and (52) give

The norm q_A bounds the radius in V. Moreover Z contains a Euclidean ball, so L_0 has interior in V. Its extended gauge \bar g=g_{L_0}\circ P_V satisfies \bar g\le g, and

For the opposite comparison, let N be any full-dimensional symmetric convex body in \R^n. The extended gauge of its projection to V_j is at most g_N, and this projected body’s normalization is \tau_{A_j}(N). Minimality and monotonicity give

Passing to the limit, using (50) and continuity of \tau_{A_j}(N), gives

Take N=N_R=L_A+RB_{V^\perp}, the orthogonal product of the minimizing body in V and a ball of radius R in V^\perp. Then \tau_A(N_R)=1 and

For k=1, the integrable function \log g_A is a lower bound for these logarithms, and their positive parts are uniformly bounded for R\ge1; hence the objectives converge. The case k=2 is immediate. Thus (54) implies J(g)\le J(g_A).

Finally, c_0\bar g is the extended gauge of the normalized body L_0/c_0. Since c_0\bar g\le g, we have

Strict increase of the objective integrands forces c_0\bar g=g almost everywhere, and then everywhere by continuity. Uniqueness in Lemma 5.1 gives c_0\bar g=g_A. Thus every subsequential limit is g_A, which proves the asserted continuity.□

Proof. Let the convex set be D, and work in its affine hull, of dimension d. The case d=0 is immediate. The nearest-point projection \pi onto D exists by compactness and is unique by strict convexity of squared distance along a segment between distinct minimizers. Minimizing along segments gives

Applying this twice yields

so \pi is continuous. If T:D\to D is continuous, choose a d-simplex \Delta containing D and set G=T\circ\pi:\Delta\to\Delta.

Brouwer’s theorem for a simplex gives v=G(v). Since G(\Delta)\subset D, this point lies in D, hence \pi(v)=v and T(v)=v.□

Proof. On \mathcal A define the symmetric traceless matrix

It is continuous by Lemma 5.3; its denominator is positive because g_A is a nonzero seminorm.

The direction -\mathcal F(A) points strictly into the positive cone along every missing direction of a singular matrix. To compute this, let A be singular and e a unit vector in \Ker A. For a standard Gaussian Y in \R^n, g_A(Y) depends only on the projection of Y to \range(A), and is independent of Y\cdot e. Separating the Gaussian radial and angular variables therefore gives

The last radial moment ratio follows by integration by parts in \int_0^\infty r^{n+k-1}e^{-r^2/2}\,dr. It follows, by polarization on \Ker A, that

The range–kernel block of \mathcal F(A) is zero: reflecting all kernel coordinates preserves g_A and reverses each mixed product. Consequently A-\epsilon\mathcal F(A) is positive definite for sufficiently small \epsilon>0. On the kernel this follows from (56); on the range it follows from the positive definiteness of the restriction of A. Its trace is one. The same perturbation is allowed at a positive definite A for sufficiently small \epsilon.

Use the Euclidean inner product \langle U,W\rangle=\tr(UW) on symmetric matrices, and consider the continuous map

Lemma 5.4 provides a fixed point A. The nearest-point condition there is

We may take B=A-\epsilon\mathcal F(A), as just shown. This gives -\epsilon\tr(\mathcal F(A)^2)\ge0, hence \mathcal F(A)=0. Formula (56) rules out singularity, so A is positive definite.

For every symmetric traceless matrix H, the vanishing field gives

These quadratics are exactly \mathcal H_2, since the Euclidean Laplacian of x^tHx is 2\tr H. Thus the required degree-two component vanishes.

Finally, for a>0 the correspondence L\mapsto L/a between normalized competitors for A and aA gives

Indeed, \tau_{aA}(L/a)=\tau_A(L), while multiplying a gauge by a adds \log a to the logarithmic objective and multiplies the linear objective by a. The minimizers therefore correspond. Taking a=(\det A)^{-1/n} makes the determinant one and preserves the vanishing degree-two component.□

Proof. Choose k as in (22) and write c=c_{n,k}. Apply Proposition 5.5 to \eta, obtaining a symmetric positive definite matrix A with determinant one. Replace \eta by its image under A and \zeta by its image under A^{-1}. Their pairing is unchanged, since (Ax)\cdot(A^{-1}y)=x\cdot y. Their support-functional inequalities also persist: for any linear map T,

and |T^tM|=|M| when |\det T|=1.

For the first transformed measure, the support functional at the identity is precisely the original \tau_A. Its unique minimizing body is therefore L_A, with the degree-two cancellation supplied by Proposition 5.5. For each transformed measure choose its minimizing body as in Proposition 5.2, and rescale that body to volume \kappa_n. Let the resulting gauges be g_1,g_2, and put

where \tau_i is the functional of the corresponding transformed measure. Thus

For the last inequality, if k=1 then \E g_i\ge1=t_i by the negative-moment normalization. If k=2, the same normalization gives \E g_i\ge1, and \E g_i^2\ge(\E g_i)^2\ge\E g_i=t_i. The first function f_1 has no degree-two harmonic component; scalar rescaling does not alter this property.

Apply (46) to |x\cdot y| successively in both variables. These are even continuous homogeneous tests, so the formula applies despite any asymmetry of the measures. It gives

The pointwise inequality in the last line uses the test \sgn(u\cdot v). Proposition 3.4 now gives

By Theorem 4.1 and (59),

Consequently the last lower bound is at least

Indeed a_1a_2-1\ge0 and

This proves (58).

Suppose equality holds. If a_i>1, the inequality s_i^2c\norm{Qf_i}_2^2\le a_i^2-1 is strict: this follows from Theorem 4.1 when g_i is nonconstant, and from its zero left side when g_i is constant. If both a_i exceed one, the product error in (61) is therefore strictly smaller than the displayed square root. If exactly one exceeds one, the error product vanishes and a_1a_2>1. Both alternatives contradict equality, so a_1=a_2=1. In particular, \E g_1^k\le\E g_1^{k-1}. For k=1, this forces \E g_1=1 and equality in the negative-moment power-mean bound; hence g_1=1. For k=2, the chain

again forces constancy and then g_1=1. Thus s_1=1, and the support-functional condition of the transformed first measure is sharp on B_2^n. Undoing the transformation makes the original condition sharp on A^tB_2^n, which is an origin-centered ellipsoid of volume \kappa_n.□

Proof of Theorem 1.1. Translate K so that the origin is in its interior, and first rescale it to satisfy |K|=\kappa_n. Proposition 2.5 supplies the admissible measure \eta_K. Set

The symmetric body D has volume \kappa_n, so it too has an admissible measure \eta_D. Its own support-functional integral is one: the volume derivative under D+tD=(1+t)D gives

Using (9) and Theorem 6.1,

Equation (11) yields r\ge\kappa_{n-1}, or |\Pi K|\ge\kappa_n\kappa_{n-1}^n under the current normalization. The projection body scales as \Pi(aK)=a^{n-1}\Pi K for a>0. Undoing the volume normalization proves (1).

If equality holds, the pairing of \eta_K and \eta_D equals b_n. Theorem 6.1 gives an ellipsoid on which the support-functional bound for \eta_K is sharp. The equality assertion of Proposition 2.5 then makes K a translate of a positive dilate of that ellipsoid.

For the converse, the quotient is invariant under translations and invertible linear maps. To verify the latter directly, the segment volume derivative gives, for every vector y and every invertible T,

Hence \Pi(TK)=|\det T|T^{-t}\Pi K and |\Pi(TK)|=|\det T|^{n-1}|\Pi K|, exactly matching the factor in |TK|^{n-1}. Finally, \Pi B_2^n=\kappa_{n-1}B_2^n gives equality for the ball, and therefore for every ellipsoid.□

Proof. Theorem 1.1 and the separately proved three-dimensional result [2, Theorem 1.1] establish Petty’s conjecture for every n\ge3. Lutwak’s conditional result [10], in the explicit formulation of Ortega-Moreno and Schuster [15, author version, Introduction, p. 2], 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

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

which is the projection-volume normalization above by Cauchy’s projection formula. His mixed-volume identity and the Aleksandrov–Fenchel inequality give

Here V denotes mixed volume, and Haddad’s \omega_m is our unit-ball volume \kappa_m; see [4, arXiv version, Introduction, p. 5]. Theorem 1.1 for n\ge4 and the separately proved three-dimensional case [2, Theorem 1.1] therefore give

Haddad’s Lemma 5.1 and the equivalence of Equations (18)–(21) [4, arXiv version, Section 5, pp. 16–17] 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 \norm{\mathbf1_K}_q=|K|^{(n-1)/n}; see [4, arXiv version, Equation (9) and Definition 2.2, pp. 8–9]. 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(\R^n) and in L^q. Define finite measures \nu_j on the sphere by

Reshetnyak’s continuity theorem [19, author version, Theorem 1.3, p. 3] 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 \norm{\mathbf1_E}_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, Definition 1(i), Equation (2), Section 5, and Theorem 12]. Minkowski’s mixed-volume inequality gives

The last step applies Theorem 1.1 to B^\circ for n\ge4, and the separate three-dimensional result [2, Theorem 1.1] 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.□