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1 Introduction

The following inequality has recently been proved in [12]. Let K be an origin symmetric convex body in \(\mathbb{R}^{n},\) and let μ be a measure on K with even continuous non-negative density f so that μ(B) = ∫ B f for every Borel subset of K. Then

$$\displaystyle{ \mu (K)\ \leq \ \sqrt{n} \frac{n} {n - 1}c_{n}\max _{\xi \in S^{n-1}}\mu (K \cap \xi ^{\perp })\ \vert K\vert ^{1/n}\, }$$
(1)

where \(c_{n} = \left \vert B_{2}^{n}\right \vert ^{\frac{n-1} {n} }/\left \vert B_{2}^{n-1}\right \vert <1,\) \(B_{2}^{n}\) is the unit Euclidean ball in \(\mathbb{R}^{n},\) and | K | stands for volume of proper dimension. Note that c n  < 1 for every n. 

In the case of volume, when f = 1 everywhere on K, inequality (1) was proved in [17, p. 96]. Another argument follows from [6, Theorem 8.2.13]; in [6] this argument is attributed to Rolf Schneider. Also, in the case of volume the constant \(\sqrt{n}\) can be improved to Cn 1∕4, where C is an absolute constant, as shown by Klartag [9] who improved an earlier estimate of Bourgain [4]. These results are much more involved. The question of whether n 1∕4 can also be removed in the case of volume is the matter of the hyperplane conjecture [1–3, 17]; see the book [5] for the current state of the problem.

In this note we prove analogs of inequality (1) for sections of lower dimensions and in the complex case; see Theorems 2 and 4, respectively. As in [12], the proofs are based on certain stability results for generalizations of intersection bodies.

2 Lower Dimensional Sections

We need several definitions and facts. A closed bounded set K in \(\mathbb{R}^{n}\) is called a star body if every straight line passing through the origin crosses the boundary of K at exactly two points different from the origin, the origin is an interior point of K, and the Minkowski functional of K defined by

$$\displaystyle{\|x\|_{K} =\min \{ a \geq 0:\ x \in aK\}}$$

is a continuous function on \(\mathbb{R}^{n}.\)

The radial function of a star body K is defined by

$$\displaystyle{\rho _{K}(x) =\| x\|_{K}^{-1},\qquad x \in \mathbb{R}^{n}.}$$

If x ∈ S n−1 then ρ K (x) is the radius of K in the direction of x. 

If μ is a measure on K with even continuous density f, then

$$\displaystyle{ \mu (K) =\int _{K}f(x)\ \mathit{dx} =\int \limits _{S^{n-1}}\left (\int \limits _{0}^{\|\theta \|_{K}^{-1} }r^{n-1}f(r\theta )\ \mathit{dr}\right )d\theta. }$$
(2)

Putting f = 1, one gets

$$\displaystyle{ \vert K\vert = \frac{1} {n}\int _{S^{n-1}}\rho _{K}^{n}(\theta )d\theta = \frac{1} {n}\int _{S^{n-1}}\|\theta \|_{K}^{-n}d\theta. }$$
(3)

For 1 ≤ k ≤ n − 1, denote by Gr n−k the Grassmanian of (n − k)-dimensional subspaces of \(\mathbb{R}^{n}.\) The (n − k)-dimensional spherical Radon transform \(R_{n-k}: C(S^{n-1})\mapsto C(\mathit{Gr}_{n-k})\) is a linear operator defined by

$$\displaystyle{R_{n-k}g(H) =\int _{S^{n-1}\cap H}g(x)\ \mathit{dx},\quad \forall H \in \mathit{Gr}_{n-k}}$$

for every function g ∈ C(S n−1). 

The polar formulas (2) and (3), applied to sections of K, express volume in terms of the spherical Radon transform:

$$\displaystyle\begin{array}{rcl} \mu (K \cap H)\quad & =& \int _{K\cap H}f =\int _{S^{n-1}\cap H}\left (\int _{0}^{\|\theta \|_{K}^{-1} }r^{n-k-1}f(r\theta )\ \mathit{dr}\right )d\theta \\ & =& R_{n-k}\left (\int _{0}^{\|\cdot \|_{K}^{-1} }r^{n-k-1}f(r\ \cdot )\ \mathit{dr}\right )(H). {}\end{array}$$
(4)

and

$$\displaystyle{ \vert K \cap H\vert = \frac{1} {n - k}\int _{S^{n-1}\cap \xi ^{\perp }}\|\theta \|_{K}^{-n+k}d\theta = \frac{1} {n - k}R_{n-k}(\|\cdot \|_{K}^{-n+k})(H). }$$
(5)

The class of intersection bodies was introduced by Lutwak [15] and played a crucial role in the solution of the Busemann-Petty problem; see [6, 10] for definition and properties. A more general class of bodies was introduced by Zhang [18] in connection with the lower dimensional Busemann-Petty problem. Denote

$$\displaystyle{R_{n-k}\left (C(S^{n-1})\right ) = X \subset C(\mathit{Gr}_{ n-k}).}$$

Let M +(X) be the space of linear positive continuous functionals on X, i.e. for every ν ∈ M +(X) and non-negative function f ∈ X, we have ν(f) ≥ 0.

An origin symmetric star body K in \(\mathbb{R}^{n}\) is called a generalized k-intersection body if there exists a functional ν ∈ M +(X) so that for every g ∈ C(S n−1),

$$\displaystyle{ \int _{S^{n-1}}\|x\|_{K}^{-k}g(x)\ \mathit{dx} =\nu (R_{ n-k}g). }$$
(6)

When k = 1 we get the class of intersection bodies. It was proved by Grinberg and Zhang [7, Lemma 6.1] that every intersection body in \(\mathbb{R}^{n}\) is a generalized k-intersection body for every k < n. More generally, as proved later by Milman [16], if m divides k, then every generalized m-intersection body is a generalized k-intersection body.

We need the following stability result for generalized k-intersection bodies.

Theorem 1.

Suppose that 1 ≤ k ≤ n − 1, K is a generalized k-intersection body in \(\mathbb{R}^{n},\) f is an even continuous function on K, f ≥ 1 everywhere on K, and \(\varepsilon> 0.\) If

$$\displaystyle{ \int _{K\cap H}f\ \leq \ \vert K \cap H\vert +\varepsilon,\qquad \forall H \in \mathit{Gr}_{n-k}, }$$
(7)

then

$$\displaystyle{ \int _{K}f\ \leq \ \vert K\vert + \frac{n} {n - k}\ c_{n,k}\ \vert K\vert ^{k/n}\varepsilon, }$$
(8)

where \(c_{n,k} = \vert B_{2}^{n}\vert ^{\frac{n-k} {n} }/\vert B_{2}^{n-k}\vert <1.\)

Proof.

Use polar formulas (4) and (5) to write the condition (7) in terms of the (n − k)-dimensional spherical Radon transform: for all H ∈ Gr n−k

$$\displaystyle{R_{n-k}\left (\int _{0}^{\|\cdot \|_{K}^{-1} }r^{n-k-1}f(r\ \cdot )\ \mathit{dr}\right )(H) \leq \frac{1} {n - k}R_{n-k}\left (\|\cdot \|_{K}^{-n+k}\right )(H) +\varepsilon.}$$

Let ν be the functional corresponding to K by (6), apply ν to both sides of the latter inequality (the direction of the inequality is preserved because ν is a positive functional) and use (6). We get

$$\displaystyle\begin{array}{rcl} & & \int _{S^{n-1}}\|\theta \|_{K}^{-k}\left (\int _{ 0}^{\|\theta \|_{K}^{-1} }r^{n-k-1}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & \qquad \leq \frac{1} {n - k}\int _{S^{n-1}}\|\theta \|_{K}^{-n}\ d\theta +\varepsilon \nu (1). {}\end{array}$$
(9)

Split the integral in the left-hand side into two integrals and then use f ≥ 1 as follows:

$$\displaystyle\begin{array}{rcl} & & \int _{S^{n-1}}\left (\int _{0}^{\|\theta \|_{K}^{-1} }r^{n-1}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & +\int _{S^{n-1}}\left (\int _{0}^{\|\theta \|_{K}^{-1} }(\|\theta \|_{K}^{-k} - r^{k})r^{n-k-1}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & \geq \int _{K}f +\int _{S^{n-1}}\left (\int _{0}^{\|\theta \|_{K}^{-1} }(\|\theta \|_{K}^{-k} - r^{k})r^{n-k-1}\ \mathit{dr}\right )d\theta \\ & & =\int _{K}f + \frac{k} {n - k}\vert K\vert. {}\end{array}$$
(10)

Now estimate ν(1) by first writing \(1 = R_{n-k}1/\vert S^{n-k-1}\vert\) and then using definition (6), Hölder’s inequality and \(\vert S^{n-1}\vert = n\vert B_{2}^{n}\vert\):

$$\displaystyle\begin{array}{rcl} & & \nu (1) = \frac{1} {\left \vert S^{n-k-1}\right \vert }\nu (R_{n-k}1) = \frac{1} {\left \vert S^{n-k-1}\right \vert }\int _{S^{n-1}}\|\theta \|_{K}^{-k}\ d\theta \\ & & \qquad \quad \leq \frac{1} {\left \vert S^{n-k-1}\right \vert }\left \vert S^{n-1}\right \vert ^{\frac{n-k} {n} }\left (\int _{S^{n-1}}\|\theta \|_{K}^{-n}\ d\theta \right )^{\frac{k} {n} } \\ & & = \frac{1} {\left \vert S^{n-k-1}\right \vert }\left \vert S^{n-1}\right \vert ^{\frac{n-k} {n} }n^{k/n}\vert K\vert ^{k/n} = \frac{n} {n - k}c_{n,k}\vert K\vert ^{k/n}.{}\end{array}$$
(11)

Combining (9)–(11) we get

$$\displaystyle{\int _{K}f + \frac{k} {n - k}\vert K\vert \leq \frac{n} {n - k}\vert K\vert + \frac{n} {n - k}c_{n,k}\vert K\vert ^{k/n}\varepsilon.}$$

 □ 

It was proved in [13] (generalizing the result for k = 1 from [11]) that if L is a generalized k-intersection body and μ is a measure with even continuous density, then

$$\displaystyle{\mu (L)\, \leq \, \frac{n} {n - k}c_{n,k}\max _{H\in \mathit{Gr}_{n-k}}\mu (L \cap H)\ \vert L\vert ^{k/n}.}$$

We show now that it is possible to extend this inequality to arbitrary origin symmetric convex bodies in \(\mathbb{R}^{n}\) at the expense of an extra constant n k∕2. 

Theorem 2.

Suppose that L is an origin symmetric convex body in \(\mathbb{R}^{n},\) and μ is a measure with even continuous non-negative density g on L. Then

$$\displaystyle{ \mu (L)\ \leq \ n^{k/2} \frac{n} {n - k}c_{n,k}\max _{H\in \mathit{Gr}_{n-k}}\mu (L \cap H)\ \vert L\vert ^{k/n}. }$$
(12)

Proof.

By John’s theorem [8], there exists an origin symmetric ellipsoid K such that

$$\displaystyle{ \frac{1} {\sqrt{n}}K \subset L \subset K.}$$

The ellipsoid K is an intersection body [6, Corollary 8.1.7], and every intersection body is a generalized k-intersection body for every k [7, Lemma 6.1]. Let \(f =\chi _{K} + g\chi _{L},\) where \(\chi _{K},\ \chi _{L}\) are the indicator functions of K and L, then f ≥ 1 everywhere on K. Put

$$\displaystyle{\varepsilon =\max _{H\in \mathit{Gr}_{n-k}}\left (\int _{K\cap H}f -\vert K \cap H\vert \right ) =\max _{H\in \mathit{Gr}_{n-k}}\int _{L\cap H}g.}$$

Now we can apply Theorem 1 to \(f,K,\varepsilon\) (the function f is not necessarily continuous on K, but the result holds by a simple approximation argument). We get

$$\displaystyle\begin{array}{rcl} \mu (L)& =& \int _{L}g =\int _{K}f -\ \vert K\vert {}\\ &\leq & \frac{n} {n - k}c_{n,k}\vert K\vert ^{k/n}\max _{ H\in \mathit{Gr}_{n-k}}\int _{L\cap H}g {}\\ & \leq & n^{k/2}\ \frac{n} {n - k}c_{n,k}\vert L\vert ^{k/n}\max _{ H\in \mathit{Gr}_{n-k}}\mu (L \cap H), {}\\ \end{array}$$

because \(K \subset \sqrt{n}L,\) so \(\vert K\vert \leq n^{n/2}\vert L\vert.\) □ 

3 The Complex Case

Origin symmetric convex bodies in \(\mathbb{C}^{n}\) are the unit balls of norms on \(\mathbb{C}^{n}.\) We denote by \(\|\cdot \|_{K}\) the norm corresponding to the body K:

$$\displaystyle{K =\{ z \in \mathbb{C}^{n}:\ \| z\|_{ K} \leq 1\}.}$$

In order to define volume, we identify \(\mathbb{C}^{n}\) with \(\mathbb{R}^{2n}\) using the standard mapping

$$\displaystyle{\xi = (\xi _{1},\ldots,\xi _{n}) = (\xi _{11} + i\xi _{12},\ldots,\xi _{n1} + i\xi _{n2})\mapsto (\xi _{11},\xi _{12},\ldots,\xi _{n1},\xi _{n2}).}$$

Since norms on \(\mathbb{C}^{n}\) satisfy the equality

$$\displaystyle{\|\lambda z\| = \vert \lambda \vert \|z\|,\quad \forall z \in \mathbb{C}^{n},\ \forall \lambda \in \mathbb{C},}$$

origin symmetric complex convex bodies correspond to those origin symmetric convex bodies K in \(\mathbb{R}^{2n}\) that are invariant with respect to any coordinate-wise two-dimensional rotation, namely for each θ ∈ [0, 2π] and each \(\xi = (\xi _{11},\xi _{12},\ldots,\xi _{n1},\xi _{n2}) \in \mathbb{R}^{2n}\)

$$\displaystyle{ \|\xi \|_{K} =\| R_{\theta }(\xi _{11},\xi _{12}),\ldots,R_{\theta }(\xi _{n1},\xi _{n2})\|_{K}, }$$
(13)

where R θ stands for the counterclockwise rotation of \(\mathbb{R}^{2}\) by the angle θ with respect to the origin. We shall say that K is a complex convex body in \(\mathbb{R}^{2n}\) if K is a convex body and satisfies Eq. (13). Similarly, complex star bodies are R θ -invariant star bodies in \(\mathbb{R}^{2n}.\)

For \(\xi \in \mathbb{C}^{n},\vert \xi \vert = 1,\) denote by

$$\displaystyle{H_{\xi } =\{ z \in \mathbb{C}^{n}:\ (z,\xi ) =\sum _{ k=1}^{n}z_{ k}\overline{\xi _{k}} = 0\}}$$

the complex hyperplane through the origin, perpendicular to ξ. Under the standard mapping from \(\mathbb{C}^{n}\) to \(\mathbb{R}^{2n}\) the hyperplane H ξ turns into a (2n − 2)-dimensional subspace of \(\mathbb{R}^{2n}.\)

Denote by \(C_{c}(S^{2n-1})\) the space of R θ -invariant continuous functions, i.e. continuous real-valued functions f on the unit sphere S 2n−1 in \(\mathbb{R}^{2n}\) satisfying f(ξ) = f(R θ (ξ)) for all ξ ∈ S 2n−1 and all θ ∈ [0, 2π]. The complex spherical Radon transform is an operator \(\mathcal{R}_{c}: C_{c}(S^{2n-1}) \rightarrow C_{c}(S^{2n-1})\) defined by

$$\displaystyle{\mathcal{R}_{c}f(\xi ) =\int _{S^{2n-1}\cap H_{\xi }}f(x)\mathit{dx}.}$$

We say that a finite Borel measure μ on S 2n−1 is R θ -invariant if for any continuous function f on S 2n−1 and any θ ∈ [0, 2π],

$$\displaystyle{\int _{S^{2n-1}}f(x)d\mu (x) =\int _{S^{2n-1}}f(R_{\theta }x)d\mu (x).}$$

The complex spherical Radon transform of an R θ -invariant measure μ is defined as a functional \(\mathcal{R}_{c}\mu\) on the space \(C_{c}(S^{2n-1})\) acting by

$$\displaystyle{\left (\mathcal{R}_{c}\mu,f\right ) =\int _{S^{2n-1}}\mathcal{R}_{c}f(x)d\mu (x).}$$

Complex intersection bodies were introduced and studied in [14]. An origin symmetric complex star body K in \(\mathbb{R}^{2n}\) is called a complex intersection body if there exists a finite Borel R θ -invariant measure μ on S 2n−1 so that \(\|\cdot \|_{K}^{-2}\) and \(\mathcal{R}_{c}\mu\) are equal as functionals on \(C_{c}(S^{2n-1}),\) i.e. for any \(f \in C_{c}(S^{2n-1})\)

$$\displaystyle{ \int _{S^{2n-1}}\|x\|_{K}^{-2}f(x)\ \mathit{dx} =\int _{ S^{2n-1}}\mathcal{R}_{c}f(\theta )d\mu (\theta ). }$$
(14)

Theorem 3.

Suppose that K is a complex intersection body in \(\mathbb{R}^{2n},\) f is an even continuous R θ -invariant function on K, f ≥ 1 everywhere on K, and \(\varepsilon> 0.\) If

$$\displaystyle{ \int _{K\cap H_{\xi }}f\ \leq \ \vert K \cap H_{\xi }\vert +\varepsilon,\qquad \forall \xi \in S^{2n-1}, }$$
(15)

then

$$\displaystyle{ \int _{K}f\ \leq \ \vert K\vert + \frac{n} {n - 1}\ d_{n}\ \vert K\vert ^{1/n}\varepsilon, }$$
(16)

where \(d_{n} = \vert B_{2}^{2n}\vert ^{\frac{n-1} {n} }/\vert B_{2}^{2n-2}\vert <1.\)

Proof.

Use the polar formulas (4) and (5) to write the condition (15) in terms of the complex spherical Radon transform: for all ξ ∈ S 2n−1

$$\displaystyle{\mathcal{R}_{c}\left (\int _{0}^{\|\cdot \|_{K}^{-1} }r^{2n-3}f(r\ \cdot )\ \mathit{dr}\right )(\xi ) \leq \frac{1} {2n - 2}\mathcal{R}_{c}\left (\|\cdot \|_{K}^{-2n+2}\right )(\xi ) +\varepsilon.}$$

Let μ be the measure on S 2n−1 corresponding to K by (14). Integrate the latter inequality over S 2n−1 with the measure μ and use (14):

$$\displaystyle\begin{array}{rcl} & & \int _{S^{2n-1}}\|\theta \|_{K}^{-2}\left (\int _{ 0}^{\|\theta \|_{K}^{-1} }r^{2n-3}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & \quad \leq \frac{1} {2n - 2}\int _{S^{2n-1}}\|\theta \|_{K}^{-2n}\ d\theta +\varepsilon \int _{ S^{2n-1}}d\mu (\xi ) \\ & & \quad = \frac{n} {n - 1}\vert K\vert +\varepsilon \int _{S^{2n-1}}d\mu (\xi ). {}\end{array}$$
(17)

Recall (2), (3) and the assumption that f ≥ 1. We estimate the integral in the left-hand side of (17) as follows:

$$\displaystyle\begin{array}{rcl} & & \int _{S^{2n-1}}\|\theta \|_{K}^{-2}\left (\int _{ 0}^{\|\theta \|_{K}^{-1} }r^{2n-3}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & \quad =\int _{S^{2n-1}}\left (\int _{0}^{\|\theta \|_{K}^{-1} }r^{2n-1}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & \quad \quad +\int _{S^{2n-1}}\left (\int _{0}^{\|\theta \|_{K}^{-1} }(\|\theta \|_{K}^{-2} - r^{2})r^{2n-3}f(r\theta )\ \mathit{dr}\right )d\theta \\ & & \quad \geq \int _{K}f +\int _{S^{2n-1}}\left (\int _{0}^{\|\theta \|_{K}^{-1} }(\|\theta \|_{K}^{-2} - r^{2})r^{2n-3}\ \mathit{dr}\right )d\theta \\ & & \quad =\int _{K}f + \frac{1} {2(n - 1)n}\int _{S^{2n-1}}\|\theta \|_{K}^{-2n}\ d\theta =\int _{ K}f + \frac{1} {n - 1}\vert K\vert.{}\end{array}$$
(18)

Let us estimate the second term in the right-hand side of (17) by adding the complex spherical Radon transform of the unit constant function under the integral (\(\mathcal{R}_{c}1(\xi ) = \left \vert S^{2n-3}\right \vert\) for every ξ ∈ S 2n−1), using again (14) and then applying Hölder’s inequality:

$$\displaystyle\begin{array}{rcl} \varepsilon \int _{S^{2n-1}}d\mu (\xi )& =& \frac{\varepsilon } {\left \vert S^{2n-3}\right \vert }\int _{S^{2n-1}}\mathcal{R}_{c}1(\xi )\ d\mu (\xi ) \\ & =& \frac{\varepsilon } {\left \vert S^{2n-3}\right \vert }\int _{S^{2n-1}}\|\theta \|_{K}^{-2}\ d\theta \\ & \leq & \frac{\varepsilon } {\left \vert S^{2n-3}\right \vert }\left \vert S^{2n-1}\right \vert ^{\frac{n-1} {n} }\left (\int _{S^{2n-1}}\|\theta \|_{K}^{-2n}\ d\theta \right )^{ \frac{1} {n} } \\ & =& \frac{\varepsilon } {\left \vert S^{2n-3}\right \vert }\left \vert S^{2n-1}\right \vert ^{\frac{n-1} {n} }(2n)^{1/n}\vert K\vert ^{1/n} = \frac{n} {n - 1}d_{n}\vert K\vert ^{1/n}\varepsilon.{}\end{array}$$
(19)

In the last step we used \(\vert S^{2n-1}\vert = 2n\vert B_{2}^{2n}\vert.\) Combining (17)–(19) we get

$$\displaystyle{\int _{K}f + \frac{1} {n - 1}\vert K\vert \leq \frac{n} {n - 1}\vert K\vert + \frac{n} {n - 1}d_{n}\vert K\vert ^{1/n}\varepsilon.}$$

 □ 

It was proved in [14] that if K is a complex intersection body in \(\mathbb{R}^{2n}\) and γ is an arbitrary measure on \(\mathbb{R}^{2n}\) with even continuous density, then

$$\displaystyle{\gamma (K) \leq \frac{n} {n - 1}d_{n}\max _{\xi \in S^{2n-1}}\gamma (K \cap H_{\xi })\ \vert K\vert ^{\frac{1} {n} }.}$$

In Theorem 4 below, we remove the condition that K is a complex intersection body at the expense of an extra constant. We use a result from [14, Theorem 4.1] that a complex star body is a complex intersection body if and only if \(\|\cdot \|_{K}^{-2}\) is a positive definite distribution, i.e. its Fourier transform in the sense of distributions assumes non-negative values on non-negative test functions. We refer the reader to [10, 14] for details.

Theorem 4.

Suppose that L is an origin symmetric complex convex body in \(\mathbb{R}^{2n}\) and γ is an arbitrary measure on \(\mathbb{R}^{2n}\) with even continuous density g, then

$$\displaystyle{\gamma (L) \leq 2n \frac{n} {n - 1}d_{n}\max _{\xi \in S^{2n-1}}\gamma (L \cap H_{\xi })\ \vert L\vert ^{\frac{1} {n} }.}$$

Proof.

By John’s theorem [8], there exists an origin symmetric ellipsoid K such that

$$\displaystyle{ \frac{1} {\sqrt{2n}}K \subset L \subset K.}$$

Construct a new body K c by

$$\displaystyle{\|x\|_{K_{c}}^{-2} = \frac{1} {2\pi }\int _{0}^{2\pi }\|R_{\theta }x\|_{ K}^{-2}d\theta.}$$

Clearly, K c is R θ -invariant, so it is a complex star body. For every θ ∈ [0, 2π] the distribution \(\|R_{\theta }x\|_{K}^{-2}\) is positive definite, because this is a linear transformation of the Euclidean norm. So \(\|x\|_{K_{c}}^{-2}\) is also a positive definite distribution, and, by Koldobsky et al. [14, Theorem 4.1], K c is a complex intersection body. Since \(\frac{1} {\sqrt{2n}}K \subset L \subset K\) and L is R θ -invariant as a complex convex body, we have

$$\displaystyle{ \frac{1} {\sqrt{2n}}R_{\theta }K \subset L \subset R_{\theta }K,\quad \forall \theta \in [0,2\pi ],}$$

so

$$\displaystyle{ \frac{1} {\sqrt{2n}}K_{c} \subset L \subset K_{c}.}$$

Let \(f =\chi _{K_{c}} + g\chi _{L},\) where \(\chi _{K_{c}},\ \chi _{L}\) are the indicator functions of K c and L. Clearly, f is R θ -invariant and f ≥ 1 everywhere on K. Put

$$\displaystyle{\varepsilon =\max _{\xi \in S^{2n-1}}\left (\int _{K_{c}\cap H_{\xi }}f -\vert K_{c} \cap H_{\xi }\vert \right ) =\max _{\xi \in S^{2n-1}}\int _{L\cap H_{\xi }}g}$$

and apply Theorem 3 to \(f,K_{c},\varepsilon\) (the function f is not necessarily continuous on K c , but the result holds by a simple approximation argument). We get

$$\displaystyle\begin{array}{rcl} \mu (L)& =& \int _{L}g =\int _{K_{c}}f -\ \vert K_{c}\vert {}\\ &\leq & \frac{n} {n - 1}d_{n}\vert K_{c}\vert ^{1/n}\max _{ \xi \in S^{2n-1}}\int _{L\cap H_{\xi }}g {}\\ & \leq & 2n\ \frac{n} {n - 1}d_{n}\vert L\vert ^{1/n}\max _{ \xi \in S^{2n-1}}\mu (L \cap H_{\xi }), {}\\ \end{array}$$

because \(\vert K_{c}\vert ^{1/n} \leq 2n\ \vert L\vert ^{1/n}.\) □ 

Theorem 4 shows that if bodies have additional symmetries then maximum in the slicing inequality can be taken over a rather small set of subspaces.