The logarithmic Brunn–Minkowski conjecture

Proof of Theorem 1.1 from Proposition 2.1. Suppose first that n\ge2. Fix nonzero spanning vectors p_1,\ldots,p_N and positive endpoint widths h_i^0,h_i^1. For t\in[0,1], define

The spanning condition makes each P(t) bounded; positivity of the widths gives nonempty interior. Use these widths in (2.1) and denote the resulting integral by Z_{q,\eps}(t). For fixed q,\eps, its first two derivatives may be taken under the integral: on a compact interval of t, the differentiated integrands are bounded by a polynomial times e^{-\eps|x|^2/2}. With b_i=-q\log(h_i^1/h_i^0) one has

and hence

by Proposition 2.1. Therefore

Keep \eps>0 fixed and let q tend to infinity through even integers. For each fixed t, the integrands converge almost everywhere to e^{-\eps|x|^2/2}\mathbf1_{P(t)}. The exceptional set lies in the finitely many hyperplanes \langle p_i,x\rangle=\pm h_i(t) and has measure zero. The common integrable bound e^{-\eps|x|^2/2} gives

Apply this at t=0,\lambda,1 in (2.3), then let \eps\downarrow0. Monotone convergence gives

Figure 1 illustrates how additional slab directions refine an outer approximation. To pass to all directions, let K,L be as in Theorem 1.1. Choose nested finite sets D_m\subset S^{n-1} that contain the coordinate vectors and whose union is dense in the sphere. Define

The bodies P_m(t) decrease with m and are all contained in the bounded body P_1(t). By continuity of the support functions and their evenness,

Indeed any point satisfying the dense set of inequalities satisfies all of them by continuity in the direction. Conversely, the Wulff inequalities imply the absolute-value constraints by evenness. At t=0,1 these intersections are K,L, respectively, by the supporting-half-space characterization of a convex body. Continuity of measure from above, applied at 0,\lambda,1 in (2.4), proves (1.1).

Finally, in dimension one write K=[-a,a] and L=[-b,b] with a,b>0. Their Wulff interpolation is the interval with radius a^{1-\lambda}b^\lambda, whose length is (2a)^{1-\lambda}(2b)^\lambda. Thus (1.1) is equality. The endpoints \lambda=0,1 are equality in every dimension.□

Proof. Evenness and uniform convexity give V(x)\ge V(0)+\eps|x|^2/2, so 0<Z<\infty. For R\ge2, write B_R=\{x:|x|<R\} and set

We first use two compact-support results. The moment theorem of Berman–Berndtsson [2, Theorem 1.1], applied to the body \overline B_R and the positive smooth density e^{-V}/Z_R, gives a smooth convex potential \varphi_R, unique up to translation, for which \nabla\varphi_R:\R^n\to B_R is a diffeomorphism and

The centering hypothesis holds by evenness. Integrating the equation through this diffeomorphism shows that d\nu_R=e^{-\varphi_R}\,dz is a probability and (\nabla\varphi_R)_\#\nu_R=\mu_R. Klartag’s estimate [19, Proposition 3.1 and Remark 3.5] gives

Indeed, the ball has smooth positively curved boundary, and \rho_R=V+\log Z_R is convex and smooth, with itself and all its derivatives bounded on B_R, and D^2\rho_R\ge\eps I. The constant C is independent of R.

Translate the unique preimage of 0 to 0. It is the unique minimum of \varphi_R. Reflection gives a second potential for the same target with its minimum at 0; uniqueness up to translation therefore makes \varphi_R even. For each fixed R, take the preimages of the slopes \pm(R/2)e_i. Their supporting planes give

with a finite, possibly R-dependent constant b_R. Thus \nu_R has an exponential tail. In particular, since |\nabla\varphi_R|\le R, integration of \operatorname{div}(z e^{-\varphi_R}) with expanding compact cutoffs is justified and yields

We now obtain bounds independent of R. Put m_R=\varphi_R(0) and g_R=\varphi_R-m_R\ge0. The Hessian bound implies g_R(y)\le C|y|^2/2 and |\nabla\varphi_R(y)|\le C|y|. For arbitrary y,z, set x=\nabla\varphi_R(z). Supporting planes at z and -z give the single pointwise inequality

Integrate with respect to \nu_R, discard the nonnegative \E_{\nu_R}g_R, and use (3.3):

On B_1 the density of \mu_R is at least d_0=Z^{-1}\exp(-\max_{\overline B_1}V)>0. Rotational symmetry of the ball therefore bounds the integral below by c|y|, where c=d_0\int_{B_1}|x_1|\,dx>0. Hence

This bounds m_R on both sides and gives one integrable exponential majorant for all the densities e^{-\varphi_R}.

It remains to obtain a smooth limit and to verify that its gradient reaches every point. On each fixed ball, the preceding estimates bound \varphi_R and \nabla\varphi_R uniformly. Since Z_2\le Z_R\le Z, equation (3.2) bounds \det D^2\varphi_R below by a positive constant there. Together with D^2\varphi_R\le CI, this gives a uniform local bound \delta I\le D^2\varphi_R\le CI: if the determinant is at least d>0, every eigenvalue is at least d/C^{n-1}. The right-hand sides

of the logarithmic equations are uniformly Lipschitz on that ball, because \nabla f_R=D^2\varphi_R\nabla V(\nabla\varphi_R) -\nabla\varphi_R is uniformly bounded.

To apply an interior uniformly elliptic estimate, choose a smooth concave scalar function s equal to \log on a slightly larger interval than [\delta,C], with s' bounded above and bounded below by positive constants on \R. Such an extension is obtained by smoothly continuing the decreasing derivative 1/t to positive constant tails. For symmetric matrices Q, the spectral function \mathcal F(Q)=\tr s(Q) is smooth, concave and uniformly elliptic: its derivative has eigenvalues s'(\lambda_i(Q)), and its second variation has coefficients s''(\lambda_i) and the nonpositive divided differences of s'. It agrees with \log\det on the Hessians under consideration. The nonhomogeneous Evans–Krylov estimate [33, Corollary 2.3] applied to \mathcal F(D^2\varphi_R)=f_R gives uniform interior C^{2,\alpha} bounds for some \alpha>0.

Each \varphi_R is already smooth. Its derivative w_{R,k}=\partial_k\varphi_R satisfies

where Q:N=\tr(Q^TN). The principal coefficients and the right-hand side are uniformly C^\alpha on smaller balls. Interior Schauder estimates [16, Theorem 2.20 and Corollary 2.21] give uniform C^{3,\alpha} bounds for \varphi_R, and iteration gives bounds for every derivative on compact sets. A diagonal subsequence, still indexed by R, converges locally smoothly to an even \varphi with positive definite Hessian, the same upper cap, and the logarithmic equation in (3.1). The bound \varphi_R(z)\ge c|z|-b, with b independent of R, also passes to \varphi.

The common exponential majorant from (3.4) proves \int e^{-\varphi}=1 by dominated convergence. For each bounded continuous h, it also gives

Thus (\nabla\varphi)_\#\nu=\mu. Positive definite Hessian makes the gradient locally invertible and injective, since for z\ne w,

Its image is dense: an open ball missed by the image would have zero pushforward mass, contrary to the everywhere positive density of \mu. Fix x_0\in\R^n. Density permits finitely many image points y_j=\nabla\varphi(z_j) whose convex hull contains a ball B_r(x_0) with r>0; one may take small perturbations of the points x_0\pm e_i. Their supporting planes imply

This function attains a minimum, at which \nabla\varphi=x_0. The gradient is therefore onto, and its local smooth inverses combine to a global smooth inverse.□

Proof. Write \Gamma(f,h)=\nabla f\cdot\tau\nabla h and \Gamma(f)=\Gamma(f,f). The integration formula also holds when one factor is smooth and the other has compact support, by inserting a cutoff around that support. In particular \E Ak=0 for compactly supported smooth k. Since A preserves parity, the displayed range consists of centered even functions.

Let g be a centered even L^2(\mu) function orthogonal to this range. It also annihilates A(A-1)f for odd f, because that image is odd. Formal symmetry therefore gives

Indeed, if \mathcal P^t denotes the transpose with respect to Lebesgue measure and d\mu=\rho\,dx, weighted formal symmetry gives \mathcal P^t(\rho g)=\rho\mathcal P g in distributions. Orthogonality makes the left side zero, and smooth positivity of \rho gives the displayed equation. The principal symbol of \mathcal P is (\xi\cdot\tau(x)\xi)^2, which is positive for every \xi\ne0. Interior elliptic regularity consequently gives g\in C^\infty [14, Theorem 14.2]. Only positivity on compact sets is needed here.

Splitting the equation in L^2. Choose 0\le\eta_r\le1 smooth, equal to one on B_r, zero outside B_{2r}, with |\nabla\eta_r|\le C/r and |D^2\eta_r|\le C/r^2, where r\ge1. The upper bound on \tau gives constants B,K, independent of r, such that

For the second estimate, use |x|\le2r on the support of the cutoff derivatives. Those derivatives, including A\eta_r, vanish outside B_{2r}\setminus B_r.

Put v=Ag, initially just a smooth function, and write

Here G is finite by hypothesis, while X_r,Y_r are finite by compact support. Testing \mathcal P g=0 against \eta_r^4g gives \E[v(A-1)(\eta_r^4g)]=0. The product rule A(ab)=aAb+bAa-2\Gamma(a,b) yields the exact expansion

Thus Cauchy–Schwarz and (5.1) imply

The first inequality is immediate if X_r=0; otherwise divide the resulting estimate for X_r^2 by X_r. For the second, integrate by parts in \E[\eta_r^2gAg]:

Combining (5.2) gives Y_r^2\le aG^2+10BGY_r, so both Y_r and X_r are bounded uniformly in r. Exhaustion by balls proves v=Ag\in L^2(\mu). We may now set

The harmonic part is constant. Testing Aw=0 with \eta_r^2w and using (5.1) gives

On any fixed ball, let r\to\infty. Then \Gamma(w)=0 there; strict positivity of \tau shows that w is constant.

The eigenvalue-one part is linear. The analogous test of Av=v gives, with Z_r^2=\E[\eta_r^2\Gamma(v)],

Hence \E\Gamma(v)<\infty. For f_r=\eta_rv\in C_c^\infty, the product rule now proves directly that

Indeed,

The middle term has norm at most K\|v\mathbf{1}_{B_{2r}\setminus B_r}\|_2\to0. For the last term, use

Apply (4.7) to these compactly supported functions:

On each fixed ball, f_r=v for all sufficiently large r. Positivity of the density and of \tau therefore forces D^2v=0 on that ball. Thus v is affine, and Av=v removes its constant term.

We have proved that every L^2 distributional solution of \mathcal P g=0 is affine. Our g=w+v is even, so it is constant, and its mean is zero. Hence g=0. Taking the orthogonal complement in the centered even subspace proves the lemma.□

Proof. All test fields used in the integrations by parts below have compact support. Weighted Bochner calculations underlie many variance estimates; see Bakry–Émery [1, Proposition 3] and the transport-Hessian formulas in [20, Section 4.3]. Here the required identity is for a vector field, so we derive it explicitly and retain the matrix trace \tr((D_xW)^2). First, for any such vector field W and u=\delta_\mu W,

Pairing with W_i and using adjunction gives the weighted identity

Thus it remains to prove (6.1). Its proof compares two expansions, integrates their derivative terms by parts, and writes the resulting difference as a sum of squares.

Recall h=D_x^2f and P=\tau h from (4.4), and let (C_i)_{kl}=C_{ikl}, where C=D_z^3\varphi. The moment equation (3.1) reads

Its first derivative gives

Since \partial_jM=-MC_jM, differentiating once more gives

To combine the fourth-derivative term and the term involving \nabla_xV into a weighted divergence, note that

Full symmetry of D_z^4\varphi therefore yields

For the other expansion, the commutator [\partial_l,\partial_k^*]=\tau_{kl} gives

Both D_zW and B are symmetric, so R is symmetric. Since D_xW=(B+R)M and MBM=h, expansion followed by one integration by parts gives

The last term is nonnegative: \tr(MRMR)=\|M^{1/2}RM^{1/2}\|_{\mathrm{HS}}^2. On the right side of (6.1), the coefficient of \tau H\tau in the trace contraction is

Substituting (6.4) and integrating its divergence term by parts gives

The common first term cancels. We now evaluate the three remaining contracted terms in coordinates where \tau=I at the point under consideration.

This normalization uses a constant dual change of coordinates. For an invertible constant matrix T, set

The corresponding normalized potentials and partition function are

They preserve both probability measures by change of variables, x'=\nabla_{z'}\varphi', and the moment equation. The chain rule gives

Each z-derivative index, including all three indices of C, transforms covariantly with T; the same law holds for \partial_k^*. Also D_{x'}W'=T^T(D_xW)T^{-T}. It follows that every contracted scalar in (6.5) and (6.6) is invariant. For example, in (\partial_kh_{al})(\partial_lP_{ak}) each index pairs one covariant and one contravariant factor. At a fixed point we may therefore take T=\tau^{-1/2}, using the value of \tau at that point. All derivatives are computed under this constant transformation. No moving frame is differentiated; the integrations by parts have already been completed.

In these normalized coordinates, put

The tensor S is fully symmetric, and J is symmetric in its last two indices. Write J^{(12)}_{ijk}=J_{jik}, and use the Euclidean tensor norm and inner product in this frame. The three contractions are

For (6.7), use \partial_kh_{al}=S_{alk} and \partial_lP_{ak}=C_{ail}h_{ik}+S_{akl}. For (6.8), differentiating the two factors h in Q=h\tau h gives the two copies of \langle S,J\rangle. Differentiating its middle factor gives

Finally, (6.9) follows by expanding Q=h^2 and the trace, using the full symmetry of C.

Thus, apart from the nonnegative R term, the integrand of (6.5) minus (6.6) is

Full symmetrization of J is

Its norm satisfies

Consequently the preceding integrand equals

This pointwise calculation proves (6.1); combining it with (6.3) proves (6.2).□

Proof of Proposition 2.1. The potential is smooth and even, and D^2V\ge\eps I. Lemma 3.1 therefore supplies the moment coordinates, with 0<\tau\le (4/\eps)I. Put

Each H_i is positive semidefinite, and a_i>0 since p_i\ne0 and \mu has positive density everywhere. Homogeneity gives

The adjunction identity of Section 4 also gives

To justify it for these noncompact functions, apply \E x_j k=\E\tau_{jk}\partial_{x_k}k to k=\eta_R\partial_{x_j}\psi_i and sum in j, where \eta_R is a smooth cutoff equal to one on B_R and zero outside B_{2R}. The extra term is bounded in absolute value by C R^{-1}\E[|\nabla\psi_i|\mathbf1_{\{|x|\ge R\}}], which tends to zero. The other terms converge because \tau is bounded and all polynomial moments are finite.

Take an even f\in C_c^\infty(\R^n) and use

Applying (4.5) to F=\psi_i gives, for each i,

Two Cauchy–Schwarz estimates will give complementary bounds for these same numbers d_i.

First apply the Hilbert–Schmidt Cauchy–Schwarz inequality, including integration, to the matrix fields H_i^{1/2}\tau^{1/2} and H_i^{1/2}B\tau^{-1/2}. Their pairing is \tr(H_iB), so

This does not require B to be positive semidefinite. Second, (7.1)–(7.3) give

Cauchy–Schwarz for the positive semidefinite form (X,Y)\mapsto\E X^TH_iY therefore yields

Since H=D^2V=\eps I+\sum_iH_i, Lemma 6.1 and the weighted Bochner identity (6.3) imply

The discarded terms are nonnegative, since M is positive definite and B is symmetric. This is the required estimate on the dense class of double divergences.

By Lemma 5.1, such u are dense in the centered even subspace of L^2(\mu). Each \psi_i\in L^2(\mu), so all pairings u\mapsto\E u\psi_i are continuous. Thus (7.6) holds for every centered even u\in L^2(\mu). For F=\sum_i b_i\psi_i, choose u=F-\E F. Then

If \|u\|_2=0, the conclusion is immediate; otherwise division and squaring prove (2.2). This completes the analytic assertion and, by Section 2, Theorem 1.1.□

Proof of Corollary 1.2. At \lambda=0,1, the assertion follows from \mathcal W[h_K]=K and \mathcal W[h_L]=L. Suppose 0<\lambda<1. Set

so that |K_0|=|L_0|=1. Define

Then 1-\theta=(1-\lambda)a^p/c^p. For every u\in S^{n-1}, the support identities h_K=a h_{K_0} and h_L=b h_{L_0}, followed by the weighted arithmetic–geometric mean inequality, give

All support values here are positive. The Wulff definition is monotone in its support bound and satisfies \mathcal W[cf]=c\mathcal W[f] for c>0. Hence

By Theorem 1.1, the unscaled body on the left has volume at least 1, because K_0,L_0 are origin-symmetric and have unit volume. The body on the right therefore has volume at least c^n. Raising this bound to the positive power p/n gives (1.2), since c^p=(1-\lambda)|K|^{p/n}+\lambda|L|^{p/n}.□

Proof. Fix 0<\lambda<1 and write D=\mathcal W[h_K^{1-\lambda}h_L^\lambda]. The assertion is immediate when \mu is the zero measure. Otherwise, choose a closed ball C centered at the origin large enough that K\cup L\cup D\subset C and 0<\mu(C)<\infty. This is possible because the three bodies are compact and \mu is a nonzero Radon measure. The probability measure

is even and log-concave: for compact A,B, convexity of C gives

and the defining inequality for \mu applies. Since K,L,D\subset C, the common factor \mu(C) cancels from (8.1). It therefore suffices to prove that inequality for \nu.

Let E be the affine hull of \operatorname{supp}\nu. Evenness gives E=-E, so E is a linear subspace. If \dim E=0, then \nu=\delta_0; all three bodies contain 0, and equality holds. Suppose d=\dim E\ge1 and identify E isometrically with \R^d. By inner regularity, the compact-set definition of log-concavity for \nu is equivalent to the Borel-set formulation using inner measure. Borell’s characterization [4, Theorem 1.1, with s=0], applied to \nu on E, represents it as an affine image of a log-concave density in a dimension k\le d. Since its support spans E, the affine map has rank d. Thus k=d, the map is invertible, and \nu has a log-concave density f with respect to Lebesgue measure on E. Evenness of \nu gives f(x)=f(-x) almost everywhere; replacing f(x) by \sqrt{f(x)f(-x)} gives an even log-concave representative of the same density.

Set K_E=K\cap E and L_E=L\cap E. These are origin-symmetric convex bodies in E, since K and L contain neighborhoods of 0. Define their relative support functions for every v\in E by

The relative Wulff body in E is

Using all v\in E rather than only unit vectors gives the same body by homogeneity. Let P_E be the orthogonal projection onto E. For x\in D_E and u\in S^{n-1},

Indeed, h^E_{K_E}(P_Eu)=\max_{y\in K\cap E}\langle y,u\rangle\le h_K(u), and likewise for L; if P_Eu=0, the middle product is 0. Thus

Theorem 1.1 in dimension d, followed by Saroglou’s transfer theorem [29, Theorem 3.1] in that same dimension, gives

Since \nu is supported on E, inclusion (8.3) yields

as required. At \lambda=0,1, the Wulff bodies are K,L, respectively, and the endpoint convention gives equality.

Finally, for s,t\in\R, the support identity h_{e^sK}=e^s h_K gives

Applying (8.1) to e^sK and e^tK proves

This is exactly the log-concavity in (8.2), and is the scalar-dilation argument of Saroglou [29, Corollary 3.2].□