PROJECTIONS OF CONVEX BODIES AND THE FOURIER TRANSFORM

A. KOLDOBSKY, D. RYABOGIN, AND A. ZVAVITCH

Abstract. The Fourier analytic approach to sections of convex bodies has recently been developed and has led to several results, including a complete analytic solution to the Busemann-Petty problem, characterizations of intersection bodies, extremal sections of l_{p}-balls. In this article, we extend this approach to projections of convex bodies and show that the projection counterparts of the results mentioned above can be proved using similar methods. In particular, we present a Fourier analytic proof of the recent result of Barthe and Naor on extremal projections of l_{p}-balls, and give a Fourier analytic solution to Shephard’s problem, originally solved by Petty and Schneider and asking whether symmetric convex bodies with smaller hyperplane projections necessarily have smaller volume. The proofs are based on a formula expressing the volume of hyperplane projections in terms of the Fourier transform of the curvature function.

1. Introduction

The study of geometric properties of bodies based on information about sections and projections of these bodies has important applications to many areas of mathematics and science. A new approach to sections of convex bodies, based on methods of Fourier analysis, has recently been developed and has led to several results including an analytic solution to the Busemann-Petty problem (see the survey [K4] for a brief description of this approach). The goal of this article is to extend this method to projections of convex bodies and show that several results on projections can be proved in the same spirit as their section counterparts.

The Fourier transform approach to sections of convex bodies is based on certain formulas connecting the volume of sections with the Fourier transform of powers of the Minkowski functional. For instance, it was proved in [K5] that, for any symmetric convex body K in ℝ^{n} and every ξ ∈ S^{n−1},

Vol_{n−1}(K ∩ ξ^{⊥}) = \frac{1}{π(n − 1)}(∥ · ∥_{K}^{−n+1})^{∧}(ξ), (1)

where ∥x∥_{K} = min{a ≥ 0 : x ∈ aK} is the Minkowski functional of K, ξ^{⊥} is the central hyperplane orthogonal to ξ, and the Fourier transform is considered in the sense of distributions.

In Section 2 we prove the projection analog of this formula (see Theorem 2 below): if the surface area measure of K is absolutely continuous, then

Vol_{n−1} (K|θ^{⊥}) = −\frac{1}{π} f̂_{K}(θ) ∀θ ∈ S^{n−1}, (2)

where K|θ^{⊥} is the orthogonal projection of K onto the hyperplane θ^{⊥}, f_{K} is the curvature function of the body K extended to a homogeneous of degree −n − 1 function on ℝ^{n}, and the Fourier transform is in the sense of distributions.

In Section 3 we apply formula (2) to give a Fourier analytic proof of the recent result of Barthe and Naor [BN] that, for p ≥ 2, the minimal and maximal hyperplane projections of the unit balls B_{p}^{n} of the spaces l_{p}^{n} are the ones corresponding to the vectors θ_{1} = (1, 0, ..., 0) and θ_{n} = (1/\sqrt{n}, ..., 1/\sqrt{n}), respectively. The proof in [BN] is based on a certain formula for the volume of projections (see formula (8)) that is obtained using probabilistic arguments. We show that this formula is a particular case of formula (2) and can be derived by a direct computation of the Fourier transform of the curvature function of l_{p}- balls. This makes the proof of the result of Barthe and Naor similar to the proof of the corresponding result for hyperplane sections of the balls B_{p}^{n} with 0 < p ≤ 2 in [K5].

Another application is to the Shephard problem. The problem (see [Sh]) reads as follows. Let K, L be convex symmetric bodies in ℝ^{n} and suppose that, for every θ ∈ S^{n−1},

Vol_{n−1} (K|θ^{⊥}) ≤ Vol_{n−1} (L|θ^{⊥}) . (3)

Does it follow that

Vol_{n}(K) ≤ Vol_{n}(L)? (4)

The problem was solved independently by Petty [P] and Schneider [Sc1], who showed that the answer is affirmative if n ≤ 2 and negative if n ≥ 3. Further results were obtained by Ball [B], who proved that it is necessary to multiply Vol_{n}(L) by \sqrt{n} to make the answer affirmative in all dimensions, and by Goodey and Zhang [GZ], who solved the generalized Shephard’s problem with lower dimensional projections. The section counterpart of Shephard’s problem is the Busemann-Petty problem asking whether symmetric convex bodies with larger central hyperplane sections necessarily have greater volume. The solution to the Busemann-Petty problem has recently been completed (see [GKS] and [Zh] for historical details) and the answer is affirmative if n ≤ 4 and negative if n ≥ 5.

In Section 5, we present a new solution of the Shephard problem, which is completely Fourier analytic, and, along with formula (2), is based on a certain spherical version of Parseval’s formula. This solution is in the spirit of the solution to the Busemann-Petty problem from [K3].

In Section 4, we give a Fourier analytic characterization of projection bodies, which, in particular, leads to the following connection between projections and sections: if n is an even integer and L is a symmetric convex body in ℝ^{n} with infinitely smooth (on S^{n−1}) support function, then L is a projection body if and only if, for every ξ ∈ S^{n−1}, we have (−1)^{n/2}A_{L^{∗},ξ}^{(n)}(0) ≥ 0, where A_{L^{∗},ξ} is the parallel section function of the polar body L^{∗} in the direction of ξ. If n is odd the condition is slightly more complicated, but is also expressed in terms of sections. These characterization of projection bodies are similar to that of intersection bodies in [K6]. In particular, Theorem 1 from [K6], in conjunction with formula (15), implies that an infinitely smooth symmetric convex body L in ℝ^{n} (with even n) is an intersection body if and only if, for every ξ ∈ S^{n−1}, (−1)^{(n−2)/2}A_{L,ξ}^{(n−2)}(0) ≥ 0. These conditions explain why the transition in Shephard’s problem occurs between the dimensions 2 and 3, while in the Busemann-Petty problem it happens between 4 and 5: convexity allows to control the derivatives only up to the second order.

2. Volume of projections via Fourier transform.

Our main tool is the Fourier transform of distributions. We denote by S the space of rapidly decreasing infinitely differentiable functions (test functions) on ℝ^{n} with values in ℂ. By S′ we denote the space of distributions over S. Every locally integrable real valued function f on ℝ^{n} with power growth at infinity represents a distribution acting by integration: for every ϕ ∈ S, ⟨f, ϕ⟩ = ∫_{ℝ^{n}} f(x)ϕ(x)dx. The Fourier transform of a distribution f is defined by ⟨f̂, ϕ̂⟩ = (2π)^{n}⟨f, ϕ⟩, for every test function ϕ.

Let μ be a finite Borel measure on the unit sphere S^{n−1}. A distribution μ_{e} is called the extended measure of μ if, for every even test function ϕ ∈ S(ℝ^{n}),

⟨μ_{e}, ϕ⟩ = \frac{1}{2} ∫_{S^{n−1}} ⟨r^{−2}, ϕ(rξ)⟩dμ(ξ). (5)

In most cases we are only interested in test functions supported outside of the origin, for which ⟨r^{−2}, ϕ(rξ)⟩ = ∫_{ℝ} r^{−2} ϕ(rξ) dr, see [GS] page 52. Observe that μ_{e} is a homogeneous distribution of degree −n − 1.

If μ is absolutely continuous with the density g ∈ L_{1}(S^{n−1}), we define the extension g(x), x ∈ ℝ^{n} \ {0} as a homogeneous function of degree −n − 1: g(x) = |x|^{−n−1} g(x/|x|) and identify μ̂_{e} with ĝ.

Throughout the paper, we write that two homogeneous distributions are equal on the sphere meaning that their homogeneous extensions are equal as distributions on ℝ^{n}. For instance, the distributions in both sides of (1) are homogeneous of degree -1, while in (2) the degree of homogeneity is 1. Recall that the Fourier transform of an even homogeneous distribution of degree p is an even homogeneous distribution of degree −n − p.

The following fact is well-known, see for example [Se], formula 4.7. We include here a proof similar to that from [K2], Lemma 2.

Theorem 1. For every θ ∈ S^{n−1},

μ̂_{e}(θ) = −\frac{π}{2} ∫_{S^{n−1}} |(θ, y)|dμ(y).

Proof : Let ϕ ∈ S(ℝ^{n}) be an even test function so that 0 ∉ supp(ϕ̂). Then, by the definition of μ_{e},

⟨μ̂_{e}, ϕ⟩ = ⟨μ_{e}, ϕ̂⟩ = \frac{1}{2} ∫_{S^{n−1}} dμ_{e}(θ) ∫_{−∞}^{∞} r^{−2}ϕ̂(rθ)dr.

By Lemma 2.1 from [K2]

∫_{ℝ^{n}} |(θ, x)|ψ̂(x)dx = (2π)^{n−1} \frac{4\sqrt{π}}{Γ(−1/2)} ∫_{ℝ} r^{−2}ψ(rθ)dr

for any even ψ ∈ S(ℝ^{n}) with 0 ∉ supp(ψ).

This gives (with ψ̂ = ϕ)

⟨μ̂_{e}, ϕ⟩ = −\frac{π}{2} ∫_{S^{n−1}} dμ(θ) ∫_{R^{n}} |(θ, ξ)|ϕ(ξ)dξ

= −\frac{π}{2} ∫_{R^{n}} ϕ(ξ)dξ ∫_{S^{n−1}} |(θ, ξ)|dμ(θ) = ⟨−\frac{π}{2} ∫_{S^{n−1}} |(θ, ξ)|dμ(θ), ϕ⟩.

Since ϕ is an arbitrary even test function with 0 ∉ supp(ϕ̂), the distributions μ̂_{e} and −π/2 ∫_{S^{n−1}} |(θ, ξ)|dμ(θ) can differ by a polynomial only. But both distributions are even and homogeneous of degree one, so the polynomial must be equal to zero, and μ̂_{e} coincides with the continuous function −π/2 ∫_{S^{n−1}} |(θ, ξ)|dμ(θ).

□

To apply Theorem 1 to the study of volumes of projections of convex bodies, we use the well-known Cauchy formula ([G], page 361):

Vol_{n−1} (L|θ^{⊥}) = \frac{1}{2} ∫_{S^{n−1}} |θ · v|dS_{n−1}(L, v), θ ∈ S^{n−1}. (6)

Here S_{n−1}(L, ·) is the surface area measure of L ([G], page 351). A convex body L is said to have a curvature function

f_{L}(·) : S^{n−1} → ℝ,

if its surface area measure S_{n−1}(L, ·) is absolutely continuous with respect to Lebesgue measure σ_{n−1} on S^{n−1}, and

\frac{dS_{n−1}(L, ·)}{dσ_{n−1}} = f_{L}(·) ∈ L_{1}(S^{n−1}).

The next statement follows from the Cauchy formula (6) and Theorem 1. We denote by S_{e}(L) the extended measure of the surface area measure S_{n−1}(L, ·), see (5). Theorem 2. Let L be a convex origin symmetric body in ℝ^{n}. Then

{ \atop S_{e}(L)}(θ) = −πVol_{n−1} (L|θ^{⊥}) , ∀θ ∈ S^{n−1}.

In particular, if the body L has a curvature function then

f̂_{L}(θ) = −πVol_{n−1} (L|θ^{⊥}) , ∀θ ∈ S^{n−1}.

3. Projections of the unit ball of ℓ_{p}^{n}

In order to show similarities between the section and projection cases, we first recall how was formula (1) applied in [K5] to the problem of finding the extremal sections of the unit balls B_{p}^{n} of the spaces l_{p}^{n}, 0 < p ≤ 2. The result of [K5] is as follows: if 0 < p ≤ 2 then, for every θ ∈ S^{n−1},

Vol_{n−1} (B_{p}^{n} ∩ θ_{n}^{⊥}) ≤ Vol_{n−1} (B_{p}^{n} ∩ θ^{⊥}) ≤ Vol_{n−1} (B_{p}^{n} ∩ θ_{1}^{⊥}) ,

where θ_{1} = (1, 0, …, 0) and θ_{n} = (1/\sqrt{n}, …, 1/\sqrt{n}).

In the case of l_{p}-balls the Fourier transform in the right-hand side of (1) can be expressed in terms of the function γ_{p} which is the Fourier transform of the function z → exp(−|z|^{p}), z ∈ ℝ. Formula (1) turns into

(7) Vol_{n−1}(B_{p} ∩ ξ^{⊥}) = \frac{p}{π(n − 1)Γ((n − 1)/p)} ∫_{0}^{∞} ∏_{k=1}^{n} γ_{p}(tξ_{k}) dt,

which is true for every p > 0 and ξ ∈ S^{n−1}. Note that the latter formula was first proved by Meyer and Pajor [MP] for 1 ≤ p ≤ 2 using different methods, and that it was used in [MP] to find the extremal sections of B_{1}^{n}.

The second step in [K5] is to prove that, for 0 < p ≤ 2, the function γ_{p}(\sqrt{·}) is log-convex on [0, ∞). The proof is based on the fact that the function exp(−| · |^{α}) is completely monotonic on [0, ∞) for every α ∈ (0, 1].

The log-convexity implies that for any choice of numbers 0 < ξ_{1} < η_{1} < η_{2} < ξ_{2} with ξ_{1}^{2} + ξ_{2}^{2} = η_{1}^{2} + η_{2}^{2} = 1 and any t > 0, one has

γ_{p}(tξ_{1})γ_{p}(tξ_{2}) ≥ γ_{p}(tη_{1})γ_{p}(tη_{2}).

It is now clear that the integrand in the right-hand side of (7) is minimal if all the coordinates of ξ are equal, and maximal when ξ has only one non-zero coordinate, which finishes the proof.

The result of Barthe and Naor [BN] on the extremal hyperplane projections of the balls B_{p}^{n} with p ≥ 2 is as follows:

Theorem 3. [BN] Let p ≥ 2 and θ ∈ S^{n−1}, then

Vol_{n−1} (B_{p}^{n}|θ_{1}^{⊥}) ≤ Vol_{n−1} (B_{p}^{n}|θ^{⊥}) ≤ Vol_{n−1} (B_{p}^{n}|θ_{n}^{⊥}) ,

where θ_{1} = (1, 0, …, 0) and θ_{n} = (1/\sqrt{n}, …, 1/\sqrt{n}).

The proof in [BN] is based on a formula similar to (7): for every ξ ∈ S^{n−1},

(8)Vol_{n−1} (B_{p}^{n}|ξ^{⊥}) = \frac{2^{n}Γ^{n}(1/p)}{πp^{n−1}Γ(n − \frac{n−1}{p^{∗}})} ∫_{0}^{∞} \frac{1 − ∏_{k=1}^{n} \frac{p^{∗}}{2Γ(1/p)}β_{p^{∗}}(tξ_{k})}{t^{2}}dt,

where 1/p + 1/p^{∗} = 1 and β_{p^{∗}}(u) is the Fourier transform of the function |x|^{p^{∗}−2}e^{−|x|^{p^{∗}}} on ℝ. The proof of this formula in [BN] uses probabilistic arguments. The rest of the proof of Theorem 3 is similar to that for sections. For p ≥ 2, the function β_{p^{∗}}(\sqrt{·}) is log-convex on [0, ∞), which is proved in a way similar to the case of the function γ_{p} in [K5]. Theorem 3 immediately follows from this and (8).

In this section we show that formula (8) is a particular case of formula (2) and holds for every p > 1. In particular, this allows us to unify methods in the section and projection cases. We do that by computing the Fourier transform of the curvature function of l_{p}-balls using techniques from differential geometry (see [Th]) and a trick for calculating the Fourier transform from [K5].

Lemma 1. Let B_{p}^{n} = {x ∈ ℝ^{n} : ∑_{i=1}^{n} |x_{i}|^{p} ≤ 1}, 1 < p < ∞. Then

f_{B_{p}^{n}}(θ) = (p^{∗} − 1)^{n−1} (∏_{i=1}^{n} |θ_{i}|^{p^{∗}−2}) ∥θ∥_{p^{∗}}^{(n−1)−np^{∗}}, θ ∈ S^{n−1},

where 1/p^{∗} + 1/p = 1. Proof : Recall that f_{B_{p}^{n}} is the reciprocal Gauss curvature, viewed as a function of the unit normal vector (see [Sc2], page 419), so to find f_{B_{p}^{n}}(θ) we first compute the Gaussian curvature K_{p}(x) of B_{p}^{n}, and then we invert the Gaussian map.

Let F (x_{1}, …, x_{n}) = ∑_{i=1}^{n} |x_{i}|^{p}, then the boundary ∂B_{p}^{n} = {x ∈ ℝ^{n} :

F (x_{1}, …, x_{n}) = 1}. Let

Z(x) = −\frac{1}{p}∇F (x) = − (|x_{1}|^{p−1} sign x_{1}, …, |x_{n}|^{p−1} sign x_{n}) .

We use the following formula for the Gaussian curvature (see, for example, [Th], Theorem 5, page 89):

(9) K(x) = (−1)^{n−1}det ( ∇_{v_{1}}Z ; ⋮ ; ∇_{v_{n−1}}Z ; Z(x) ) /|Z(x)|^{n−1}det ( v_{1} ; ⋮ ; v_{n−1} ; Z(x) ) .

Here vectors v_{1}, …, v_{n−1} can be chosen as any basis in the tangent space at the corresponding point, and

∇_{v_{i}}Z = (∇_{v_{i}}Z_{1}, …, ∇_{v_{i}}Z_{n}) = (v_{i} · ∇Z_{1}, …, v_{i} · ∇Z_{n}).

We choose v_{1}, …, v_{n−1} in the following way: for every k, 1 ≤ k ≤ n − 1,

v_{k} = (|x_{k+1}|^{p−1} sign x_{k+1}, \underbrace{0, …, 0}_{k−1 times}, −|x_{1}|^{p−1} sign x_{1}, 0, …, 0,).

Then

∇_{v_{k}}Z(x) = (p − 1)|x_{1} x_{k+1}|^{p−2}(−x_{k+1}, 0, …, 0, \underbrace{x_{1}}_{(k+1)s place} , 0, …, 0,).

Substituting this in (9) (computation of the determinants is standard), we get

K_{p}(x) = (p − 1)^{n−1}|Z(x)|^{−n−1} ∏_{i=1}^{n} |x_{i}|^{p−2} (10)

for almost all x ∈ ∂B_{p}^{n}.

It remains to invert the Gaussian map θ = Z(x)/|Z(x)|. We see that θ and Z(x) are collinear vectors in ℝ^{n}, i.e. θ = λZ(x) for some λ > 0. Next

x_{i} = λ^{1/(p−1)}|θ_{i}|^{1/(p−1)} sign θ_{i} ,

and using ∥x∥_{p} = 1 we get

λ = \frac{1}{(∑_{i=1}^{n} |θ_{i}|^{p/(p−1)})^{(p−1)/p}} .

Finally

x_{i} = − (|θ_{i}|^{1/(p−1)} sign θ_{i}) / (∑_{i=1}^{n} |θ_{i}|^{p/(p−1)})^{1/p} .

To finish the proof, we substitute the last expression into (10) and use p − 1 = (p^{∗} − 1)^{−1}.

□

In the next statement we present a new proof of formula (8), originally proved in [BN]. Theorem 4. Formula (8) holds for every p > 1. Proof : According to Theorem 2, we have to compute the Fourier transform of f_{B_{p}^{n}}(x) = |x|^{−n−1} f_{B_{p}^{n}}(x/|x|), x ∈ ℝ^{n} \ {0}. From the definition of the Gamma function

∥x∥_{p^{∗}}^{−q} = \frac{p^{∗}}{Γ(q/p^{∗})} ∫_{0}^{∞}y^{q−1}e^{−y^{p^{∗}}(|x_{1}|^{p^{∗}}+···+|x_{n}|^{p^{∗}})}dy (11)

for any q > 0 and x ∈ ℝ^{n} \ {0}.

Let ϕ ∈ S(ℝ^{n}) be even and such that 0 ∉ supp(ϕ̂). Using (11), we get

⟨f̂_{B_{p}^{n}}, ϕ⟩ = ⟨f_{B_{p}^{n}}, ϕ̂⟩

= \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(q/p^{∗})} ∫_{0}^{∞}y^{q−1}dy ∫_{ℝ^{n}} ∏_{k=1}^{n} |x_{k}|^{p^{∗}−2}e^{−y^{p^{∗}}|x_{k}|^{p^{∗}}} ϕ̂(x)dx

= \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(q/p^{∗})} ∫_{0}^{∞}y^{q−1}dy ∫_{ℝ^{n}} (∏_{k=1}^{n} |x_{k}|^{p^{∗}−2}e^{−y^{p^{∗}}|x_{k}|^{p^{∗}}})^{∧} (ξ)ϕ(ξ)dξ

= \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(q/p^{∗})} ∫_{0}^{∞}y^{q−1}dy ∫_{ℝ^{n}} ∏_{k=1}^{n} (|t|^{p^{∗}−2}e^{−y^{p^{∗}}|t|^{p^{∗}}})^{∧} (ξ_{k})ϕ(ξ)dξ.

This gives

⟨f̂_{B_{p}^{n}}, ϕ⟩ = \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(q/p^{∗})} ∫_{0}^{∞}y^{q−1+(1−p^{∗})n}dy ∫_{ℝ^{n}} ∏_{k=1}^{n} β_{p^{∗}}(\frac{ξ_{k}}{y}) ϕ(ξ)dξ.

Now replacing q by np^{∗} −(n−1), and using the fact that 0 ∉ supp(ϕ̂) implies ∫_{ℝ^{n}} ϕ(ξ)dξ = 0, we obtain

⟨f̂_{B_{p}^{n}}, ϕ⟩ = \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(n − (n − 1)/p^{∗})} ∫_{0}^{∞}dy ∫_{ℝ^{n}} (∏_{k=1}^{n} β_{p^{∗}}(\frac{ξ_{k}}{y}) − β_{p^{∗}}^{n}(0)) ϕ(ξ)dξ

= \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(n − (n − 1)/p^{∗})} ∫_{ℝ^{n}} ϕ(ξ)dξ ∫_{0}^{∞} ∏_{k=1}^{n} β_{p^{∗}}(\frac{ξ_{k}}{y}) − β_{p^{∗}}^{n}(0)dy,

where

β_{p^{∗}}(0) = ∫_{−∞}^{∞} |u|^{p^{∗}−2}e^{−|u|^{p^{∗}}}du = \frac{2}{p^{∗}} ∫_{0}^{∞}v^{−1/p^{∗}}e^{−v}dv = \frac{2}{p^{∗}} Γ(1 − 1/p^{∗}).

Making a substitution ξ_{k}/y = t, we finally get

⟨f̂_{B_{p}^{n}}, ϕ⟩ = \frac{(p^{∗} − 1)^{n−1}p^{∗}}{Γ(n − (n − 1)/p^{∗})} ∫_{ℝ^{n}} ϕ(ξ)dξ ∫_{0}^{∞} \frac{∏_{k=1}^{n} β_{p^{∗}}(tξ_{k}) − β_{p^{∗}}^{n}(0)}{t^{2}}dt

= \frac{2^{n}Γ^{n}(1/p)}{p^{n−1}Γ(n − (n − 1)/p^{∗})} ∫_{ℝ^{n}} ϕ(ξ)dξ ∫_{0}^{∞} \frac{∏_{k=1}^{n} \frac{p^{∗}}{2Γ(1/p)}β_{p^{∗}}(tξ_{k}) − 1}{t^{2}}dt.

Since ϕ is an arbitrary even test function with 0 ∉ supp(ϕ̂), the distributions in the left and right-hand sides of the latter equality are equal (they are both even homogeneous of degree 1). Now the result follows from Theorem 2.

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4. Projection bodies

Let K be an origin symmetric convex body in ℝ^{n}. The projection body ΠK of K is defined as an origin symmetric convex body in ℝ^{n} whose support function in every direction is equal to the volume of the hyperplane projection of K in this direction: for every θ ∈ S^{n−1},

h_{ΠK}(θ) = Vol_{n−1}(K|θ^{⊥}) = \frac{1}{2} ∫_{S^{n−1}} |θ · v|dS_{n−1}(K, v). (12)

We extend h_{ΠK} to a homogeneous function of degree 1 on ℝ^{n}. It immediately follows from Theorem 1 that the Fourier transform of h_{ΠK} is equal (up to a constant) to the extended surface area measure of K :

ĥ_{ΠK} = −(2π)^{n−1}S_{e}(K). (13)

On the other hand, if L is an origin symmetric convex body so that ĥ_{L} = −(2π)^{n−1}μ_{e} is an extended measure of some even finite Borel measure μ on S^{n−1} then, again by Theorem 1, one has

h_{L}(θ) = \frac{1}{2} ∫_{S^{n−1}} |θ · v|dμ(v).

Since L is an n-dimensional body, the measure μ cannot be supported in any great subsphere of S^{n−1}, and, by Minkowski’s existence theorem ([G], page 356), the measure μ is the surface area measure of some origin symmetric convex body K. By Cauchy’s formula, we have L = ΠK. We have proved the following Fourier analytic characterization of projection bodies (this fact can also be shown as a combination of several well-known facts about zonoids and subspaces of L_{1}):

Proposition 1. An origin symmetric convex body L in ℝ^{n} is the projection body of some origin symmetric convex body if and only if there exists a measure μ on S^{n−1} so that

ĥ_{L} = −(2π)^{n−1}μ_{e}. (14)

The measure μ serves as the surface area measure of an origin symmetric convex body K so that L = ΠK.

The condition that L is a projection body is equivalent to L being an origin symmetric zonoid (see [G], p. 134), which, in turn, is equivalent to the polar body L^{∗} being the unit ball of a subspace of L_{1}. It is well-known that every origin symmetric convex body in ℝ^{2} is a projection body (or it is the unit ball of a subspace of L_{1}, see [He], [Fe], [Li]). As proved in [K1], for L = B_{p}^{n}, 0 < p < 2, n ≥ 3 the function h_{B_{p}^{n}} = ∥ · ∥_{p^{∗}} has a sign-changing Fourier transform, which implies that the spaces l_{q}^{n}, q > 2, n ≥ 3 do not embed isometrically in L_{1} (this fact was first proved by other methods by Dor [D]).

To get a characterization of projection bodies in terms of sections of the polar body, we use a formula from [GKS], p. 693. Let D be an infinitely smooth origin symmetric convex body in ℝ^{n}, k ∈ ℕ ∪ {0}, k ≠ n − 1, ξ ∈ S^{n−1}. We denote by

A_{D,ξ}(t) = Vol_{n−1}(D ∩ {ξ^{⊥} + tξ}), t ∈ ℝ

the parallel section function of D in the direction of ξ. Then: (i) If k is even

(∥ · ∥_{D}^{−n+k+1})^{∧}(ξ) = (−1)^{k/2}π(n − k − 1)A_{D,ξ}^{(k)}(0); (15)

(ii) If k is odd

(∥ · ∥_{D}^{−n+k+1})^{∧}(ξ) = (16)

c_{n,k} ∫_{0}^{∞} \frac{A_{D,ξ}(z) − A_{D,ξ}(0) − ... − A_{D,ξ}^{(k−1)}(0)\frac{z^{k−1}}{(k−1)!}}{z^{k+1}} dz,

where c_{n,k} = (−1)^{(k+1)/2}2(n − k − 1)k!, A_{D,ξ}^{(k)} stands for the derivative of the order k of the function A_{D,ξ}, and, as before, the equality of distributions on the sphere is understood as the equality of their homogeneous extensions.

Suppose now that an origin symmetric convex body L in ℝ^{n} has the property that h_{L} is an infinitely differentiable function on the sphere S^{n−1}. This means that the polar body L^{∗} is infinitely smooth (recall that h_{L} = ∥ · ∥_{L^{∗}}). Putting k = n, D = L^{∗} in the formulas above, we get that if n is even then, for every ξ ∈ S^{n−1},

ĥ_{L}(ξ) = (−1)^{1+n/2}πA_{L^{∗},ξ}^{(n)}(0) (17)

and if n is odd then, for every ξ ∈ S^{n−1},

ĥ_{L}(ξ) = (18)

(−1)^{(n−1)/2}2n! ∫_{0}^{∞} \frac{A_{L^{∗},ξ}(z) − A_{L^{∗},ξ}(0) − ... − A_{L^{∗},ξ}^{(n−1)}(0)\frac{z^{n−1}}{(n−1)!}}{z^{n+1}} dz.

We can now characterize projection bodies in terms of sections of the polar body. Proposition 2. Let L be an origin symmetric convex body in ℝ^{n} so that h_{L} is infinitely differentiable on S^{n−1}. The body L is a projection body of some convex body K if and only if for every ξ ∈ S^{n−1},

(i) if n is even

(−1)^{n/2}A_{L^{∗},ξ}^{(n)}(0) ≥ 0;

(ii) if n is odd

(−1)^{(n+1)/2} ∫_{0}^{∞} \frac{A_{L^{∗},ξ}(z) − A_{L^{∗},ξ}(0) − ... − A_{L^{∗},ξ}^{(n−1)}(0)\frac{z^{n−1}}{(n−1)!}}{z^{n+1}} dz ≥ 0.

Using formulas (17), (18), one can express the volume of a convex body in terms of the (n − 1)-dimensional volumes of its hyperplane projections. If L^{∗} is an infinitely smooth body and n is even then

Vol_{n}(L) = (−1)^{n/2} \frac{π^{2}}{(2π)^{n}n} ∫_{S^{n−1}} A_{L^{∗},ξ}^{(n)}(0)Vol_{n}(L|ξ^{⊥})dξ. (19)

This formula is an analog of the formula expressing the volume of a body in terms of volumes of central hyperplane sections (see the remark before Theorem 4.5 in [K4]).

To prove (19) we recall that (see [G], page 354, [Sc2], page 275)

Vol_{n}(L) = \frac{1}{n} ∫_{S^{n−1}} h_{L}(θ)f_{L}(θ)dθ. (20)

Using Parseval’s formula on the sphere (see Appendix, Lemma 4, at the end of the paper) we have

Vol_{n}(L) = \frac{1}{(2π)^{n}n} ∫_{S^{n−1}} ĥ_{L}(ξ)f̂_{L}(ξ)dξ,

which, together with Theorem 2 and (17), gives (19).

5. A Fourier analytic solution to the Shephard problem

Let h_{L}(x) = max{x · y : y ∈ L} be the support function of a convex body L. By an approximation argument (see [Sc2], pages 158-160), we may assume in the formulation of Shephard’s problem that the bodies K and L are such that h_{K}, h_{L} are infinitely smooth functions on ℝ^{n} \{0}. Using, for example, formulas (15), (16), we get in this case that the Fourier transforms ĥ_{K}, ĥ_{L} are the extensions of infinitely differentiable functions on the sphere to homogeneous distributions on ℝ^{n} of degree −n − 1. Moreover, by the same approximation argument, we may assume that our bodies have absolutely continuous surface area measures. Therefore, in the rest of this paper, K an L are convex symmetric bodies with infinitely smooth support functions and absolutely continuous surface area measures. Theorem 5.

(i) If a body L is such that ĥ_{L}(θ) ≤ 0 for all θ ∈ S^{n−1} then the Shephard problem has an affirmative answer for this L and any K. (ii) If f_{K}(θ) > 0 for every θ ∈ S^{n−1} and ĥ_{K} is positive on an open subset of S^{n−1} then there exists a body L giving together with K a counterexample in Shephard’s problem.

Proof : This theorem will follow from the next two lemmas and the fact that the condition

Vol_{n−1} (K|θ^{⊥}) ≤ Vol_{n−1} (L|θ^{⊥})

is equivalent (see Theorem 2) to

f̂_{K}(θ) ≥ f̂_{L}(θ), θ ∈ S^{n−1}.

Lemma 2. If ĥ_{L}(θ) ≤ 0, and f̂_{K}(θ) ≥ f̂_{L}(θ), ∀θ ∈ S^{n−1}, then

Vol_{n}(K) ≤ Vol_{n}(L).

Proof : From f̂_{K}(θ) ≥ f̂_{L}(θ) and ĥ_{L}(θ) ≤ 0 we get

∫_{S^{n−1}} ĥ_{L}(θ)f̂_{K}(θ)dθ ≤ ∫_{S^{n−1}} ĥ_{L}(θ)f̂_{L}(θ)dθ.

Using Parseval’s formula on the sphere (see Appendix) and (20), we get

V_{1}(K, L) ≤ Vol_{n}(L).

where V_{1}(K, L) is the mixed volume (see [Sc2], page 275). Now we apply the Minkowski inequality ([Sc2], page 317):

V_{1}(K, L) ≥ Vol_{n}(L)^{\frac{1}{n}} Vol_{n}(K)^{\frac{n−1}{n}}

to get

Vol_{n}(L) ≥ Vol_{n}(L)^{\frac{1}{n}} Vol_{n}(K)^{\frac{n−1}{n}} .
□

Lemma 3. Let K be such that f_{K}(θ) > 0 ∀θ ∈ S^{n−1}. If ĥ_{K} is positive on an open subset of S^{n−1}, then there exists a convex symmetric body L in ℝ^{n}, such that

f̂_{K} ≥ f̂_{L},

but

Vol_{n}(K) > Vol_{n}(L).

Proof : Let Ω = {θ ∈ S^{n−1} : ĥ_{K}(θ) > 0} and let v ∈ C^{∞}(S^{n−1}) be a non-positive even function supported on Ω, v is not identically zero. We extend v to a homogeneous function rv(θ) of degree 1 on ℝ^{n}. Then the Fourier transform of rv(θ) is a homogeneous function of degree −n − 1: { \atop rv(θ)} = r^{−n−1}g(θ), where g is an infinitely smooth function on S^{n−1} (see Lemma 5 from [K3]).

Since g is bounded on S^{n−1}, one can choose a small ε > 0 so that, for every θ ∈ S^{n−1} and r > 0,

f_{L}(rθ) = f_{K}(rθ) + εr^{−n−1}g(θ) > 0.

By Minkowski’s existence theorem (see [G], page 356, [Sc2], page 389), f_{L}(θ) defines a convex symmetric body L ∈ ℝ^{n}. By the definition of the function v,

f̂_{L}(rθ) = f̂_{K}(rθ) + εrv(θ) ≤ f̂_{K}(rθ).

Next, since v is supported and is non-positive in the set where ĥ_{K} > 0,

∫_{S^{n−1}} ĥ_{K}(θ)f̂_{L}(θ)dθ = ∫_{S^{n−1}} ĥ_{K}(θ)f̂_{K}(θ)dθ + ∫_{S^{n−1}} ĥ_{K}(θ)εv(θ)dθ <

< ∫_{S^{n−1}} ĥ_{K}(θ)f̂_{K}(θ)dθ = (2π)^{n} ∫_{S^{n−1}} h_{K}(θ)f_{K}(θ)dθ = n(2π)^{n}Vol_{n}(K).

But again by Parseval’s formula,

∫_{S^{n−1}} ĥ_{K}(θ)f̂_{L}(θ)dθ = (2π)^{n} ∫_{S^{n−1}} h_{K}(θ)f_{L}(θ)dθ = n(2π)^{n}V_{1}(L, K),

so

V_{1}(L, K) < Vol_{n}(K).

As in the previous lemma, this implies

Vol_{n}(K) > Vol_{n}(L).
□

Theorem 6. The Shephard problem has an affirmative answer in ℝ^{2} and negative answer in ℝ^{n}, n ≥ 3. Proof : Let n = 2. In view of Theorem 5 , it is enough to show that for any convex symmetric body L in ℝ^{2} such that h_{L} is infinitely smooth, we have ĥ_{L} ≤ 0. This follows from the fact that every two-dimensional normed space embeds in L_{1}, as mentioned in the previous section. We, however, prefer to give a direct Fourier analytic argument, similar to that in the solution of the Busemann-Petty problem: by formula (17) with n = k = 2, for every ξ ∈ S^{n−1},

ĥ_{L}(ξ) = πAL^{∗},ξ^{′′}(0) ≤ 0,

where the last inequality holds because the central section has maximal volume among all hyperplane sections perpendicular to a fixed direction, so the function A_{L^{∗},ξ} has maximum at zero.

Let n ≥ 3. To apply Theorem 5, we need to give an example of a symmetric convex body K so that h_{K} is infinitely differentiable, positive on an open subset of S^{n−1} and f_{K} > 0 on S^{n−1}. Let K′ be the unit ball of ℓ_{4/3}^{n}. Then the function h_{K′}(x) = ∥x∥_{ℓ_{4}^{n}} is infinitely smooth and ĥ_{K′} is positive on some open subset of S^{n−1} (see [K1], §4).

Define the body K : h_{K} = h_{K′} + εh_{B_{2}^{n}}, where ε > 0 is such that ĥ_{K} is still positive on an open subset of S^{n−1}. Then f_{K}(θ) > 0, ∀θ ∈ S^{n−1}, and the result follows from Theorem 5.

□

Remarks. (i) The result of Theorem 5 can be translated into the language of geometry using the Fourier transform characterization of projection bodies, see Proposition 1. What we get are the results of Petty and Schneider connecting Shephard’s problem with projection bodies.

(ii) To explain why the transition in Shephard’s problem occurs between the dimensions 2 and 3, let us look again at formulas (17) and (18). The sign of ĥ_{L} basically depends on the sign of the derivative A_{L^{∗},ξ}^{(n)}(0). If n = 2 we can control this sign, thanks to the convexity of L^{∗}. However, when n ≥ 3 convexity does not control the derivative of order n, which allows to construct counterexamples. Another way to do that is to follow (with very minor changes; just replace the exponent 1/4 in the definition of the function f_{ϵ} by 1/2) the counterexample to the Busemann-Petty problem from [GKS], Theorem 4.

Appendix: A version of Parseval’s formula on the sphere

Lemma 4.

∫_{S^{n−1}} ĥ_{L}(θ)f̂_{K}(θ)dθ = (2π)^{n} ∫_{S^{n−1}} h_{L}(θ)f_{K}(θ)dθ.

Proof : We follow [K3], Lemmas 2, 3.

Proposition 3. Let E(t) = t^{4}e^{−t^{4}}, μ_{L}(ξ) = { \atop (E(h_{L})}(ξ). Then for almost all (with respect to Lebesgue measure) θ ∈ S^{n−1},

∫_{0}^{∞}t^{n}|μ_{L}(tθ)|dt < ∞.

Proof : Since μ_{L} is a bounded function (h_{L} is homogeneous of degree 1, so E(h_{L}) ∈ L_{1}(ℝ^{n})),

∫_{0}^{1} t^{n}|μ_{L}(tθ)|dt < ∞ ∀θ ∈ S^{n−1}.

It remains to show that for almost all θ

∫_{1}^{∞}t^{n}|μ_{L}(tθ)|dt < ∞. (21)

But for an even n one may see that

△^{\frac{n+2}{2}} [h_{L}^{4}e^{−h_{L}^{4}}] ∈ L_{1}(ℝ^{n}). (22)

In fact, after differentiation, the function in front of the exponent is the sum of homogeneous functions of degrees greater than −n, and each of them is continuous on the unit sphere.

Now, (22) implies that ξ ↦ |ξ|^{n+2}μ_{L}(ξ), ξ ∈ ℝ^{n} is a bounded function, since it is the Fourier transform of an L_{1}−function, hence

∫_{|ξ|>1} |ξ||μ_{L}(ξ)|dξ < ∞.

Passing to the polar coordinates we get (21).

Finally, one can put \frac{n+1}{2} in place of \frac{n+2}{2} to prove result for odd n.

□

Proposition 4. Let μ_{L} be as above and ϕ ∈ S with 0 ∉ supp(ϕ). Then

(23) ∫_{ℝ^{n}} μ_{L}(x)dx ∫_{0}^{∞}ϕ(rx)\frac{dr}{r^{2}} = \frac{Γ(3/4)}{4} ∫_{S^{n−1}} ĥ_{L}(θ)dθ ∫_{0}^{∞}ϕ(rθ)\frac{dr}{r^{2}}.

Proof : Observe that { \atop E(h_{L})}(rx)(ξ) = r^{−n}μ_{L}(ξ/r) for every r > 0. We have

∫_{ℝ^{n}} μ_{L}(x)dx ∫_{0}^{∞}r^{−2}ϕ(rx)dr = ∫_{0}^{∞}r^{−2}dr ∫_{ℝ^{n}} ϕ(rx)μ_{L}(x)dx

= ∫_{0}^{∞}r^{−2}dr ∫_{ℝ^{n}} ϕ(ξ)μ_{L}(ξ/r)r^{−n}dξ = ∫_{0}^{∞}r^{−2}dr ∫_{ℝ^{n}} ϕ̂(y)E(r h_{L}(y))dy

= ∫_{ℝ^{n}} ϕ̂(y)dy ∫_{0}^{∞}r^{−2}E(r h_{L}(y))dr.

By making a substitution r h_{L}(y) = t in the last integral and using the fact that

∫_{0}^{∞}t^{2}e^{−t^{4}}dt = \frac{Γ(3/4)}{4},

we get the desired result

∫_{ℝ^{n}} μ_{L}(x)dx ∫_{0}^{∞}r^{−2}ϕ(rx)dr = \frac{Γ(3/4)}{4} ∫_{ℝ^{n}} h_{L}(y)ϕ̂(y)dy

= \frac{Γ(3/4)}{4}⟨h_{L}, ϕ̂⟩ = \frac{Γ(3/4)}{4}⟨ĥ_{L}, ϕ⟩

= \frac{Γ(3/4)}{4} ∫_{ℝ^{n}} ĥ_{L}(ξ)ϕ(ξ)dξ = ∫_{S^{n−1}} ĥ_{L}(θ)dθ ∫_{0}^{∞}r^{−2}ϕ(rθ)dr.

□

Proposition 5. Let μ_{L} be as above. Then

∫_{0}^{∞}r^{n}μ_{L}(rθ)dr = \frac{Γ(3/4)}{4}ĥ_{L}(θ), ∀θ ∈ S^{n−1}. (24)

Proof : Let ϕ ∈ S be such that 0 ∉ supp(ϕ). Passing to the polar coordinates, using Proposition 3 and (23), we get

∫_{ℝ^{n}} μ_{L}(x)dx ∫_{0}^{∞}ϕ(rx)\frac{dr}{r^{2}} = ∫_{S^{n−1}} (∫_{0}^{∞}t^{n}μ_{L}(tθ)dt) (∫_{0}^{∞}ϕ(rθ)\frac{dr}{r^{2}} ) dθ

= \frac{Γ(3/4)}{4} ∫_{S^{n−1}} ĥ_{L}(θ) (∫_{0}^{∞}ϕ(rθ)\frac{dr}{r^{2}} ) dθ.

Now we put ϕ(rθ) = u(r)v(θ), where v is any infinitely smooth function on S^{n−1} and u is a non-negative test function on ℝ, such that 0 ∉ supp u. This gives:

∫_{S^{n−1}} ( ∫_{0}^{∞}t^{n}μ_{L}(tθ)dt)v(θ)dθ = \frac{Γ(3/4)}{4} ∫_{S^{n−1}} ĥ_{L}(θ)v(θ)dθ.

for every v ∈ C^{∞}(S^{n−1}). □

Proposition 6. Under the same notation,

∫_{ℝ^{n}} f_{K}(x)E(h_{L})(x)dx = \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} f̂_{K}(ξ)μ_{L}(ξ) dξ.

Proof : First, note that both integrals in the latter formula converge absolutely, because h_{L} is a homogeneous function of degree 1 and by Proposition 3. Since E(h_{L})(0) = 0, we have

∫_{ℝ^{n}} f_{K}(x)E(h_{L})(x)dx = ∫_{S^{n−1}} f_{K}(θ) dθ ∫_{0}^{∞} \frac{E(h_{L})(rθ) − E(h_{L})(0)}{r^{2}}dr.

Let γ_{ε} be the standard Gaussian density with variance ε. Then the convolution E(h_{L}) ∗ γ_{ε} is an even test function. Hence, the integral

∫_{0}^{∞} \frac{(E(h_{L}) ∗ γ_{ε})(rθ) − (E(h_{L}) ∗ γ_{ε})(0)}{r^{2}} dr

is well defined for any ε > 0, and is equal to 1/2 ⟨r^{−2}, (E(h_{L})∗γ_{ε})(rθ)⟩ (see [GS], page 52, formula (7)). Splitting it into two integrals over [0, 1] and [1, ∞) and using the fact that the derivatives up to order 4 of the function F (r) ≡ (E(h_{L}) ∗ γ_{ε})(rθ) are uniformly (with respect to ϵ) bounded, we get

lim_{ϵ→0} ∫_{0}^{∞} \frac{(E(h_{L}) ∗ γ_{ε})(rθ) − (E(h_{L}) ∗ γ_{ε})(0)}{r^{2}} dr = ∫_{0}^{∞} \frac{E(h_{L})(rθ)}{r^{2}} dr

This, together with the definition of extended measure (5) and the dominated convergence theorem gives:

∫_{ℝ^{n}} f_{K}(x)E(h_{L})(x) dx = lim_{ε→0} \frac{1}{2} ∫_{S^{n−1}} ⟨r^{−2}, (E(h_{L}) ∗ γ_{ε})(rθ)⟩ f_{K}(θ)dθ

= lim_{ε→0} ⟨f_{K}, E(h_{L}) ∗ γ_{ε}⟩ = lim_{ε→0} \frac{1}{(2π)^{n}} ⟨f̂_{K}, (E(h_{L}) ∗ γ_{ε})^{∧}⟩

= lim_{ε→0} \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} f̂_{K}(ξ) { \atop E(h_{L})}(ξ) γ̂_{ε}(ξ) dξ = \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} f̂_{K}(ξ)μ_{L}(ξ) dξ.

□

Proposition 7. Under the same notation,

\frac{Γ(3/4)}{4} ∫_{S^{n−1}} f_{K}(θ)h_{L}(θ)dθ = \frac{1}{(2π)^{n}} ∫_{S^{n−1}} f̂_{K}(θ) ∫_{0}^{∞}r^{n}μ_{L}(rθ)drdθ.

Proof : Note that

\frac{Γ(3/4)}{4}h_{L}(θ) = ∫_{0}^{∞}r^{−2}E(h_{L})(rθ)dr. (25)

Passing to polar coordinates and using (25), Proposition 6, and the fact that f_{K} is a homogeneous function of degree −n − 1 on ℝ^{n}, we get

\frac{Γ(3/4)}{4} ∫_{S^{n−1}} f_{K}(θ)h_{L}(θ)dθ = ∫_{ℝ^{n}} f_{K}(x)E(h_{L})(x)dx

= \frac{1}{(2π)^{n}} ∫_{ℝ^{n}} f̂_{K}(ξ)μ_{L}(ξ)dξ = \frac{1}{(2π)^{n}} ∫_{S^{n−1}} { \atop f_{K}(θ)}dθ ∫_{0}^{∞}r^{n}μ_{L}(rθ)dr.

□

We finish the proof of Lemma 4 by comparing Propositions 5 and 7.

□

Acknowledgments: We thank Mark Rudelson for useful discussions. The first named author was supported in part by the NSF grants DMS-9996431 and DMS-0136022. The second named author was partially supported by a grant from the University of Missouri.

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Alexander Koldobsky, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA

  E-mail address: koldobsk@math.missouri.edu

Dmitry Ryabogin, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA

  E-mail address: ryabs@math.missouri.edu

Artem Zvavitch, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA

  E-mail address: zvavitch@math.missouri.edu