# Reverse Alexandrov–Fenchel inequalities for zonoids

NN  Károly J. Böröczky111Alfréd Rényi Institute of Mathematics, Reáltanoda u.
13-15, H-1053 Budapest, Hungary.
Central European University, Nador utca 9, H-1051 Budapest, Hungary.
E-mail<EMAIL_ADDRESS>Daniel Hug222Karlsruhe Institute of
Technology (KIT), D-76128 Karlsruhe, Germany. E-mail<EMAIL_ADDRESS>
###### Abstract

The Alexandrov–Fenchel inequality bounds from below the square of the mixed
volume $V(K_{1},K_{2},K_{3},\ldots,K_{n})$ of convex bodies
$K_{1},\ldots,K_{n}$ in $\mathbb{R}^{n}$ by the product of the mixed volumes
$V(K_{1},K_{1},K_{3},\ldots,K_{n})$ and $V(K_{2},K_{2},K_{3},\ldots,K_{n})$.
As a consequence, for integers $\alpha_{1},\ldots,\alpha_{m}\in\mathbb{N}$
with $\alpha_{1}+\cdots+\alpha_{m}=n$ the product
$V_{n}(K_{1})^{\frac{\alpha_{1}}{n}}\cdots
V_{n}(K_{m})^{\frac{\alpha_{m}}{n}}$ of suitable powers of the volumes
$V_{n}(K_{i})$ of the convex bodies $K_{i}$, $i=1,\ldots,m$, is a lower bound
for the mixed volume $V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])$, where
$\alpha_{i}$ is the multiplicity with which $K_{i}$ appears in the mixed
volume. It has been conjectured by Ulrich Betke and Wolfgang Weil that there
is a reverse inequality, that is, a sharp upper bound for the mixed volume
$V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])$ in terms of the product of the
intrinsic volumes $V_{\alpha_{i}}(K_{i})$, for $i=1,\ldots,m$. The case where
$m=2$, $\alpha_{1}=1$, $\alpha_{2}=n-1$ has recently been settled by the
present authors (2020). The case where $m=3$, $\alpha_{1}=\alpha_{2}=1$,
$\alpha_{3}=n-2$ has been treated by Artstein-Avidan, Florentin, Ostrover
(2014) under the assumption that $K_{2}$ is a zonoid and $K_{3}$ is the
Euclidean unit ball. The case where $\alpha_{2}=\cdots=\alpha_{m}=1$, $K_{1}$
is the unit ball and $K_{2},\ldots,K_{m}$ are zonoids has been considered by
Hug, Schneider (2011). Here we substantially generalize these previous
contributions, in cases where most of the bodies are zonoids, and thus we
provide further evidence supporting the conjectured reverse Alexandrov–Fenchel
inequality. The equality cases in all considered inequalities are
characterized. More generally, stronger stability results are established as
well.

Keywords. Geometric inequality, Brunn–Minkowski theory, Alexandrov–Fenchel
inequality, mixed volume, intrinsic volume, zonoid, stability result

MSC. Primary 52A39, 52A40, 52A10; Secondary 52A25, 52A38.

##  1 Introduction

Mixed volumes in Euclidean space $\mathbb{R}^{n}$ are a cornerstone of the
Brunn–Minkowski theory of convex bodies. They arise in a natural way as the
coefficients of the polynomial expansion of the volume of a Minkowski
combination of convex bodies, thus providing a far reaching generalization of
the basic Steiner formula. Let $K_{1},\ldots,K_{n}$ be (non-empty) compact
convex sets (convex bodies) in $\mathbb{R}^{n}$ and write $\mathcal{K}^{n}$
for the space of convex bodies in $\mathbb{R}^{n}$. By polarization, the mixed
volume $V(K_{1},\ldots,K_{n})$ of $K_{1},\ldots,K_{n}\in\mathcal{K}^{n}$ can
be defined explicitly by

$V(K_{1},\ldots,K_{n})=\frac{1}{n!}\sum_{k=1}^{n}(-1)^{n+k}\sum_{1\leq
i_{1}<\cdots<i_{k}\leq n}V_{n}(K_{i_{1}}+\cdots+K_{i_{k}}),$

where $V_{n}$ denotes the volume functional. In the following, we write

$V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])=V(K_{1},\ldots,K_{1},\ldots,K_{m},\ldots,K_{m})$

for the mixed volume of $\alpha_{i}$ copies of $K_{i}$, $i=1,\ldots,m$, where
the integers $\alpha_{1},\ldots,\alpha_{m}\in\mathbb{N}_{0}$ satisfy
$\alpha_{1}+\cdots+\alpha_{m}=n$ (if $\alpha_{i}=0$, then $K_{i}$ is omitted).
We refer to [49, Chapter 5] (see also [29, Chapter 3.3]) for an introduction
to mixed volumes, a detailed study of their properties and further references.

As functionals of convex bodies, mixed volumes provide far reaching conceptual
generalizations of basic functionals such as surface area and mean width or
other intrinsic volumes. The explicit connection to the $i$th intrinsic volume
$V_{i}$ in $\mathbb{R}^{n}$ is given by the relation

$V_{i}(K)=\frac{\binom{n}{i}}{\kappa_{n-i}}V(K[i],B^{n}[n-i]),\quad
i\in\\{0,\ldots,n\\},$

where $K\subset\mathbb{R}^{n}$ is a convex body, $B^{n}$ denotes the Euclidean
unit ball of $\mathbb{R}^{n}$, and
$\kappa_{j}=\pi^{\frac{j}{2}}/\Gamma(\frac{j}{2}+1)$ is the $j$-dimensional
volume of the $j$-dimensional unit ball in $\mathbb{R}^{j}$. The intrinsic
volumes can also be introduced independently of their connection to mixed
volumes. For another approach, let $K$ be a convex body in $\mathbb{R}^{n}$.
We write $K|A$ for the orthogonal projection of $K$ onto an affine subspace
$A$ of $\mathbb{R}^{n}$. If $\operatorname{G}(n,i)$ is the linear Grassmannian
of $i$-dimensional linear subspaces of $\mathbb{R}^{n}$ and $\nu_{i}$ denotes
the Haar probability measure on $\operatorname{G}(n,i)$, then

$V_{i}(K)=\binom{n}{i}^{-1}\frac{\kappa_{i}\kappa_{n-i}}{\kappa_{n}}\int_{\operatorname{G}(n,i)}V_{i}(C|L)\,\nu_{i}(dL),\quad
i\in\\{0,\ldots,n\\}.$

In addition, we have $V_{0}(K)=1$, $V_{n}(K)=V(K[n])=V(K,\ldots,K)=:V(K)$ is
the volume (the $n$-dimensional Lebesgue measure) of $K$, and $V_{i}(K)$ is
the $i$-dimensional Lebesgue measure $\lambda_{i}(K)$ of $K$ if $\dim(K)\leq
i$.

Mixed volumes exhibit deep and surprising connections and have applications to
seemingly unrelated subjects such as combinatorics, algebraic geometry or
probability. One of the fundamental results for mixed volumes is the
Alexandrov–Fenchel inequality which states that

$V(K_{1},K_{2},K_{3},\ldots,K_{n})^{2}\geq V(K_{1}[2],K_{3},\ldots,K_{n})\cdot
V(K_{2}[2],K_{3},\ldots,K_{n}).$

Various approaches to this inequality have been developed (see [49, 29, 17,
52, 55] and the literature cited there). However, despite considerable effort
(see Schneider [45, 49]) and substantial recent progress (see Shenfeld, van
Handel [52, 53, 54]), the confirmation of the conjectured classification of
all equality cases by Schneider [44, 47, 48] remains a challenging open
problem. It follows from the Alexandrov–Fenchel inequality that if
$K_{1},\ldots,K_{m}$ are compact convex sets in $\mathbb{R}^{n}$ and
$\alpha_{1},\ldots,\alpha_{m}\in\mathbb{N}$ satisfy
$\alpha_{1}+\cdots+\alpha_{m}=n$, then

$V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])\geq
V_{n}(K_{1})^{\frac{\alpha_{1}}{n}}\cdots
V_{n}(K_{m})^{\frac{\alpha_{m}}{n}}.$

As a reverse counterpart of the Alexandrov–Fenchel inequality, we state the
following conjecture.

######  Conjecture 1.1.

If $K_{1},\ldots,K_{m}$, $m\leq n$, are compact convex sets in
$\mathbb{R}^{n}$, and $\alpha_{1},\ldots,\alpha_{m}\in\mathbb{N}$ satisfy
$\alpha_{1}+\cdots+\alpha_{m}=n$, then

$\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])\leq
V_{\alpha_{1}}(K_{1})\cdots V_{\alpha_{m}}(K_{m}).$

If $\dim(K_{i})\geq\alpha_{i}$ for $i=1,\ldots,m$, then equality holds if and
only if $\dim(K_{i})=\alpha_{i}$ for $i=1,\ldots,m$ and the affine hulls of
$K_{1},\ldots,K_{m}$ are pairwise orthogonal.

We note that $V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])>0$ if and only if
there exist $\alpha_{i}$-dimensional convex sets $C_{i}\subset K_{i}$ for
$i=1,\ldots,m$ such that $C_{1}+\cdots+C_{m}$ is $n$-dimensional.

Conjecture 1.1 was suggested by Betke and Weil [10, Section 6] in a less
explicit form and verified in the plane (but see also [1, 6] for related
recent work). For $n\geq 3$, the conjecture is proved by Böröczky and Hug [12]
in the particular case where $m=2$, $\alpha_{1}=1$ and $\alpha_{2}=n-1$, which
provides a reverse form of the Minkowski inequality.

######  Theorem 1.2 (Reverse Minkowski Inequality, [12]).

If $K_{1},K_{2}$ are compact convex sets in $\mathbb{R}^{n}$, then

$nV(K_{1}[1],K_{2}[n-1])\leq V_{1}(K_{1})\cdot V_{n-1}(K_{2}).$

If $\dim(K_{1})\geq 1$ and $\dim(K_{2})\geq n-1$, then equality holds if and
only if $\dim(K_{1})=1$, $\dim(K_{2})=n-1$ and the affine hulls of
$K_{1},K_{2}$ are pairwise orthogonal.

Let $\langle\cdot\,,\cdot\rangle$ denote the Euclidean scalar product. For a
convex body $K$ in $\mathbb{R}^{n}$, the support function
$h(K,\cdot):\mathbb{R}^{n}\to\mathbb{R}$ of $K$ is defined by
$h(K,u)=\max\\{\langle x,u\rangle:x\in K\\}$ for $u\in\mathbb{R}^{n}$. It is a
convex and (positively) homogeneous function on $\mathbb{R}^{n}$, where
(positive) homogeneity means that $h(K,\lambda u)=\lambda\,h(K,u)$ for
$\lambda\geq 0$ and $u\in\mathbb{R}^{n}$. It is known that any convex and
homogeneous function on $\mathbb{R}^{n}$ is the support function of a uniquely
determined non-empty compact convex set. We say that a compact convex set $Z$
in $\mathbb{R}^{n}$ is a zonoid if there exists a non-trivial finite even
generating Borel measure $\varrho(Z,\cdot)$ on $\mathbb{S}^{n-1}$ such that

$h(Z,u)=\int_{\mathbb{S}^{n-1}}|\langle u,v\rangle|\,\varrho(Z,dv),\qquad
u\in\mathbb{R}^{n}.$

In fact, we do not have to assume that zonoids are centred at the origin $o$
of $\mathbb{R}^{n}$, but could also consider arbitrary translates in the
following, since all relevant functionals are translation invariant.

In this paper, we verify Conjecture 1.1 if all the compact convex sets but
possibly one (say $K_{1}$ with multiplicity $\alpha_{1}$) are zonoids or if
$K_{1},K_{2}$ are arbitrary compact convex sets, $K_{1}$ (say) has
multiplicity $\alpha_{1}=1$, $K_{2}$ has arbitrary multiplicity and the
remaining bodies are zonoids. The following partial confirmation of Conjecture
1.1 admits the additional inclusion of several copies of the unit ball
$B^{n}$. Thus we obtain generalizations of partial results available in the
literature (see the subsequent comments).

######  Theorem 1.3.

Let integers $0\leq\gamma\leq\beta\leq n$, $0\leq m\leq n$ and
$\alpha_{1},\ldots,\alpha_{m}\geq 1$ with
$\alpha_{1}+\cdots+\alpha_{m}=n-\beta$ be given. If $Z_{1},\ldots,Z_{m}$ are
zonoids and $K$ is a compact convex set in $\mathbb{R}^{n}$, then

$\displaystyle\binom{n}{\gamma,\beta-\gamma,\alpha_{1},\ldots,\alpha_{m}}V(K[\gamma],B^{n}[\beta-\gamma],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle\qquad\qquad\leq\kappa_{\beta-\gamma}V_{\gamma}(K)V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m}).$ (1)

If $\dim(Z_{i})\geq\alpha_{i}$ for $i=1,\ldots,m$ and $\dim(K)\geq{\gamma}$,
then equality holds if and only if the affine hulls of $K,Z_{1},\ldots,Z_{m}$
are pairwise orthogonal.

If $\gamma=\beta$, $\dim(Z_{i})\geq\alpha_{i}$ for $i=1,\ldots,m$ and
$\dim(K)\geq{\gamma}$, then equality holds if and only if the affine hulls of
$K,Z_{1},\ldots,Z_{m}$ are pairwise orthogonal. Then we also have
$\dim(K)=\gamma$ and $\dim(Z_{i})=\alpha_{i}$ for $i=1,\ldots,m$.

######  Remarks 1.4.

  1. (a)

In the special case where $\gamma=\beta$ we thus confirm the conjecture in the
case where one of the bodies is arbitrary and the others are zonoids.

  2. (b)

Since $B^{n}$ is a zonoid, Theorem 1.3 also yields that

$\displaystyle\binom{n}{\gamma,\beta-\gamma,\alpha_{1},\ldots,\alpha_{m}}V(K[\gamma],B^{n}[\beta-\gamma],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle\qquad\qquad\leq
V_{\gamma}(K)V_{\beta-\gamma}(B^{n})V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m}),$

However, since $\kappa_{\beta-\gamma}\leq V_{\beta-\gamma}(B^{n})$ with strict
inequality unless $\beta-\gamma=n$ (see Lemma 3.1 below), this is weaker than
(1).

  3. (c)

In the special case $\gamma=1$, $\beta=n-1$ and $m=\alpha_{1}=1$, and writing
$Z_{1}=Z$, Theorem 1.3 yields

$V(K,Z,B^{n}[n-2])\leq\frac{\kappa_{n-2}}{n(n-1)}V_{1}(K)V_{1}(Z).$

If $K,Z$ are at least one-dimensional, then equality holds if and only if $K$
and $Z$ lie in orthogonal affine subspaces. Since
$\kappa_{n-1}V_{1}(K)=nV(K,B^{n}[n-1])$, this recovers Theorem 1.2 in [1].

  4. (d)

The special case where $\gamma=0$, $\beta=n-m$ and
$\alpha_{1}=\cdots=\alpha_{m}=1$, which involves only zonoids with
multiplicity one (except for the unit ball $B^{n}$), has been established in
[30, Theorem 2]. However, the main contribution in [30] was the proof of the
inequality

$V(Z_{1},\ldots,Z_{m},B^{n}[n-m])\geq 2^{m}\kappa_{n-m}$ (2)

for zonoids $Z_{1},\ldots,Z_{m}$ in $\mathbb{R}^{n}$ with isotropic generating
measures (which amounts to zonoids in special positions). Here equality holds
for $m\geq 2$ if and only if $Z_{1}=\cdots=Z_{m}$ is a cube of side length $2$
(for $m=1$ the inequality is an equality). From a very special case of (2),
inequalities for intrinsic volumes which are reverse to classical ones, have
been derived in [30] (see [15, 41] for related recent investigations). Such
reverse inequalities are based on an intermediate maximization or minimization
of the relevant functionals over affine equivalence classes of convex bodies,
as suggested for instance already in classical work by Behrend [9] (see also
[30, Introduction] and [35]).

  5. (e)

Reverse inequalities obtained by an affine optimization process as indicated
in (d) or involving only affine invariant functionals (see [35, 40] or [49,
Chapter 10] and the literature cited there) had a substantial impact on the
Brunn–Minkowski theory. The major breakthrough came with Ball’s reverse
isoperimetric inequality [2]. Since then, this line of research has been
explored repeatedly (see, e.g., [3, 4, 7, 8, 33, 36, 37, 38, 43] and the
literature cited there). It should be pointed out, however, that Conjecture
1.1 is not concerned with an affine invariant inequality although mixed
volumes are invariant with respect to the simultaneous application of a volume
preserving affine map to all convex bodies involved.

  6. (f)

Motivated by the interpretation of the classical Bézout inequality in
algebraic geometry in terms of mixed volumes, Saroglou, Soprunov, Zvavitch
[42] showed that

$V(L_{1},\ldots,L_{n})V(K)\leq n\,V(L_{1},K[n-1])V(L_{2},\ldots,L_{n},K)$ (3)

for all convex bodies $K,L_{1},\ldots,L_{n}\subset\mathbb{R}^{n}$, by
combining an inequality due to Discant (which provides a lower bound for the
relative inradius of one convex body with respect to another convex body) with
the monotonicity of the mixed volumes. More general inequalities in this
spirit have been found by Xiao [56]. Although (some of) these inequalities
turn out to be sharp, they do not imply the reverse Minkowski inequality by
special choices of $L_{i},K$. For instance, the natural choice $K=B^{n}$,
$L_{1}=:K_{1}$, $L_{2}=\cdots=L_{n}=:K_{2}$ yields

$V(K_{1},K_{2}[n-1])\leq\frac{2\kappa_{n-1}}{\kappa_{n}}\cdot\frac{1}{n}\,V_{1}(K_{1})\,V_{n-1}(K_{2})$

with

$\frac{2\kappa_{n-1}}{\kappa_{n}}\sim\sqrt{\frac{2}{\pi}}\cdot\sqrt{n}.$

In addition to Theorem 1.3, which provides support to the conjectured reverse
Alexandrov–Fenchel inequality, we also verify the following generalization of
the reverse Minkowski inequality from [12], stated as Theorem 1.2.

######  Theorem 1.5.

Let $K_{1},\ldots,K_{m}$ be compact convex sets in $\mathbb{R}^{n}$, $2\leq
m\leq n$, and let the integers $\alpha_{1},\ldots,\alpha_{m}\geq 1$ satisfy
$\alpha_{1}+\cdots+\alpha_{m}=n$. If $\alpha_{1}=1$ and $K_{3},\ldots,K_{m}$
are zonoids, then

$\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])\leq
V_{\alpha_{1}}(K_{1})\cdots V_{\alpha_{m}}(K_{m}).$ (4)

If $\dim(K_{i})\geq\alpha_{i}$ for $i=1,\ldots,m$, then equality holds if and
only if $\dim(K_{i})=\alpha_{i}$ for $i=1,\ldots,m$ and the affine hulls of
$K_{1},\ldots,K_{m}$ are pairwise orthogonal.

Stability results provide strengthened versions of geometric inequalities. The
first stability forms of the Brunn–Minkowski inequality were due to Minkowski
himself (see Groemer [24]). If the distance of the convex bodies is measured
in terms of the so-called Hausdorff distance, then Diskant [18] and Groemer
[23] provided close to optimal stability versions (see Groemer [24]). However,
the natural distance is in terms of the volume of the symmetric difference,
and the essentially optimal result is due to Figalli, Maggi, Pratelli [21,
22]. It has been slightly improved subsequently by Segal [51], and even
further by Kolesnikov, Milman [31]. The paper Eldan, Klartag [19] discusses
“isomorphic” stability versions of the Brunn–Minkowski inequality.

We note that the stability of the Brunn–Minkowski inequality has been
considered even if the sets may not be convex. The case $n=1$ is clarified by
Frieman (see Christ [16]), and the case $n=2$ by van Hintum, Spink, Tiba [28].
If $n\geq 3$, then partial results with essentially optimal error term are due
to Barchiesi, Julin [5] (when one of the sets is convex) and to Hintum, Spink,
Tiba [27] (when the two sets coincide), and a stability formula (with not
optimal error term) for any pair of bounded measurable sets is provided by
Figalli, Jerison [20].

Schneider [46] and Martinez-Maure [39] provide stability versions of the
Alexandrov–Fenchel inequality if the bodies involved have $C^{2}_{+}$
boundaries. For some additional recent related stability results, see [34, 11,
13, 14].

In the following, we establish stability results for Theorems 1.3 and 1.5. As
a preparation, we need the following notation. For integers
$\alpha_{1},\ldots,\alpha_{m}\in\mathbb{N}$ with
$\alpha_{1}+\cdots+\alpha_{m}=n$ and for linear subspaces
$L_{1},\ldots,L_{m}\subset\mathbb{R}^{n}$ with $\text{dim}(L_{i})=\alpha_{i}$,
the bracket $[L_{1},\ldots,L_{m}]$ is defined as

$[L_{1},\ldots,L_{m}]=|\det(u_{1},\ldots,u_{\alpha_{1}},\ldots,u_{n-\alpha_{m}+1},\ldots,u_{n})|,$

where $u_{1},\ldots,u_{\alpha_{1}}$ is an orthonormal basis of $L_{1}$, …,
$u_{n-\alpha_{m}+1},\ldots,u_{n}$ is an orthonormal basis of $L_{m}$. We
observe that $0\leq[L_{1},\ldots,L_{m}]\leq 1$, with $[L_{1},\ldots,L_{m}]>0$
if and only if $L_{1},\ldots,L_{m}$ span $\mathbb{R}^{n}$, and
$[L_{1},\ldots,L_{m}]=1$ if and only if $L_{1},\ldots,L_{m}$ are pairwise
orthogonal.

For $i=1,\ldots,n$ we denote by $r_{i}(K)$ the largest radius of
$i$-dimensional ball contained in $K$. In particular, $2r_{1}(K)$ is the
diameter of $K$.

We recall that Böröczky and Hug [12] proved a stability version of the reverse
Minkowski inequality (see Theorem 1.2). We verify various stability versions
of Theorems 1.3 and 1.5 in Sections 4 and 5. In order to avoid technical
details at this point, here we only quote the stability version of Theorem 1.3
in the case where all compact convex sets are zonoids.

######  Theorem 1.6.

Let $\varepsilon\in[0,1]$ and $2\leq m\leq n$. Let $Z_{1},\ldots,Z_{m}$ be
zonoids in $\mathbb{R}^{n}$, and let $\alpha_{1},\ldots,\alpha_{m}\geq 1$ be
integers with $\alpha_{1}+\cdots+\alpha_{m}=n$. If

$0<V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m})\leq(1+\varepsilon)\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(Z_{1}[\alpha_{1}],\ldots,Z_{\alpha_{m}}[\alpha_{m}]),$

then there exist $L_{i}\in\operatorname{G}(n,\alpha_{i})$ for $i=1,\ldots,m$
such that

$[L_{1},\ldots,L_{m}]\geq 1-n^{10}2^{\frac{n}{2}}\sqrt{\varepsilon}$

and

$Z_{i}\subset
Z_{i}|L_{i}+n^{\frac{9}{2}}2^{\frac{n}{2}}r_{\alpha_{i}}(Z_{i})\sqrt{\varepsilon}\,B^{n}\quad\text{for
}i=1,\ldots,m.$

##  2 Mixed volumes involving zonoids

In the following, we use results from [49, Section 5.3.3]. Let $Z$ be a zonoid
in $\mathbb{R}^{n}$ with centre at $o$ and (non-negative) even generating
measure $\rho(Z,\cdot)$ on $\mathbb{S}^{n-1}$. This means that

$h(Z,u)=\int_{\mathbb{S}^{n-1}}|\langle u,v\rangle|\,\varrho(Z,dv),\quad
u\in\mathbb{R}^{n}.$

The generating measure of $Z$ is uniquely determined by this relation. It
follows that

$V_{n}(Z)=\frac{2^{n}}{n!}\int_{\mathbb{S}^{n-1}}\cdots\int_{\mathbb{S}^{n-1}}|\det(v_{1},\ldots,v_{n})|\,\varrho(Z,dv_{1})\cdots\varrho(Z,dv_{n}).$
(5)

Given the zonoids $Z_{1},\ldots,Z_{n}$ in $\mathbb{R}^{n}$, an application of
(5) to the Minkowski combination $Z=\sum_{i=1}^{n}\lambda_{i}Z_{i}$, for
$\lambda_{1},\ldots,\lambda_{n}\geq 0$, leads to

$V(Z_{1},\ldots,Z_{n})=\frac{2^{n}}{n!}\int_{\mathbb{S}^{n-1}}\cdots\int_{\mathbb{S}^{n-1}}|\det(v_{1},\ldots,v_{n})|\,\varrho(Z_{1},dv_{1})\cdots\varrho(Z_{n},dv_{n}).$
(6)

Clearly, (5) is recovered as a special case of (6).

For $u_{1},\ldots,u_{k}\in\mathbb{S}^{n-1}$, we write
$D_{k}(u_{1},\ldots,u_{k})$ to denote the $k$-volume of the parallelepiped
$[o,u_{1}]+\cdots+[o,u_{k}]$, which can be computed as the absolute value of
the determinant of the vectors $u_{1},\ldots,u_{k}$ in the $k$-dimensional
linear subspace they span (or as zero, if they are linearly dependent).
Alternatively, $D_{k}(u_{1},\ldots,u_{k})$ is also equal to the norm of the
alternating product of the vectors $u_{1},\ldots,u_{k}$.

More generally, let $U_{1},\ldots,U_{m}$ be linear subspaces of
$\mathbb{R}^{n}$ with $\dim(U_{i})=\alpha_{i}\geq 1$, where
$\alpha_{1}+\cdots+\alpha_{m}=n-\beta\leq n$ and $\beta\in\\{0,\ldots,n-1\\}$.
Then the bracket $[U_{1},\ldots,U_{m}]_{n-\beta}$ is defined as

$[U_{1},\ldots,U_{m}]_{n-\beta}=|\det(u_{1},\ldots,u_{\alpha_{1}},\ldots,u_{n-\beta-\alpha_{m}+1},\ldots,u_{n-\beta})|,$

where $u_{1},\ldots,u_{\alpha_{1}}$ is an orthonormal basis of $U_{1}$, …,
$u_{n-\beta-\alpha_{m}+1},\ldots,u_{n-\beta}$ is an orthonormal basis of
$U_{m}$, and the determinant is calculated in the subspace
$U_{1}+\cdots+U_{m}$ with the induced Euclidean structure, if the linear
subspaces form a direct sum of dimension $n-\beta$, and as zero otherwise.
This is equivalent to the definition in [50, Section 4.1]. Moreover, by basic
(multilinear) linear algebra we have

$D_{n-\beta}(u_{1},\ldots,u_{n-\beta})=[U_{1},\ldots,U_{m}]_{n-\beta}D_{\alpha_{1}}(u_{1},\ldots,u_{\alpha_{1}})\cdots
D_{\alpha_{m}}(u_{n-\beta-\alpha_{m}+1},\ldots,u_{n-\beta}).$ (7)

Next we recall from [49, Section 5.3.3] a formula for the mixed volume of
convex bodies and zonoids (if at least one zonoid is involved). As usual, we
write $v^{(\alpha)}(\cdot,\ldots,\cdot)$ to denote the mixed volume of compact
convex sets contained in some $\alpha$-dimensional linear (or affine)
subspace. For a $k$-dimensional affine subspace $A$ of $\mathbb{R}^{n}$ with
$A=L+t$ for some $t\in\mathbb{R}^{n}$ and $L\in\operatorname{G}(n,k)$, we
write $A^{\bot}$ to denote the linear subspace
$L^{\perp}\in\operatorname{G}(n,n-k)$, the orthogonal complement of $L$. For
$u\in\mathbb{S}^{n-1}$ we set $u^{\bot}=({\rm lin}\,u)^{\bot}$, where ${\rm
lin}\,u$ is the linear subspace spanned by $u$.

Let $1\leq j\leq n$, $Z_{1},\ldots,Z_{j}\subset\mathbb{R}^{n}$ be zonoids and
let $K_{1},\ldots,K_{n-j}$ be general compact convex sets. Then (see [49,
Theorem 5.3.2])

$\displaystyle V(K_{1},\ldots,K_{n-j},Z_{1},\ldots,Z_{j})$
$\displaystyle=\frac{2^{j}(n-j)!}{n!}\int_{\mathbb{S}^{n-1}}\ldots\int_{\mathbb{S}^{n-1}}D_{j}(u_{1},\ldots,u_{j})$
$\displaystyle\qquad\times
v^{(n-j)}\left(K_{1}|\operatorname{lin}\\{u_{1},\ldots,u_{j}\\}^{\perp},\ldots,K_{n-j}|\operatorname{lin}\\{u_{1},\ldots,u_{j}\\}^{\perp}\right)$
(8)
$\displaystyle\qquad\times\varrho(Z_{1},du_{1})\ldots\varrho(Z_{j},du_{j}).$

The $j$th projection generating measure $\varrho_{(j)}(Z,\cdot)$ of a zonoid
$Z$ in $\mathbb{R}^{n}$ with generating measure $\varrho(Z,\cdot)$ is a non-
negative measure on the linear Grassmannian $\operatorname{G}(n,j)$ of
$j$-dimensional linear subspaces of $\mathbb{R}^{n}$ and defined by

$\varrho_{(j)}(Z,\cdot)=\frac{2^{j}}{j!\kappa_{j}}\int_{(\mathbb{S}^{n-1})^{j}}\mathbf{1}\\{\text{lin}\\{u_{1},\ldots,u_{j}\\}\in\cdot\\}D_{j}(u_{1},\ldots,u_{j})\,\rho(Z,\cdot)^{\otimes
j}(d(u_{1},\ldots,u_{j})).$ (9)

Now we combine (7), (2) and (9) to obtain a representation of mixed volumes of
general convex bodies and of zonoids counted with multiplicities. For this,
let $K_{1},\ldots,K_{\beta}\subset\mathbb{R}^{n}$ be compact convex sets, and
let $Z_{1},\ldots,Z_{m}$ be zonoids with multiplicities
$\alpha_{1},\ldots,\alpha_{m}\geq 1$, where
$\alpha_{1}+\cdots+\alpha_{m}=n-\beta$ and $\beta\in\\{0,\ldots,n-1\\}$. Then
we obtain

$\displaystyle
V(K_{1},\ldots,K_{\beta},Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle=\frac{2^{n-\beta}\beta!}{n!}\int_{\mathbb{S}^{n-1}}\ldots\int_{\mathbb{S}^{n-1}}D_{n-\beta}(u_{1},\ldots,u_{n-\beta})$
$\displaystyle\qquad\times
v^{(\beta)}\left(K_{1}|\operatorname{lin}\\{u_{1},\ldots,u_{n-\beta}\\}^{\perp},\ldots,K_{\beta}|\operatorname{lin}\\{u_{1},\ldots,u_{n-\beta}\\}^{\perp}\right)$
$\displaystyle\qquad\times\varrho(Z_{1},du_{1})\ldots\varrho(Z_{1},du_{\alpha_{1}})\ldots\varrho(Z_{m},du_{n-\beta-\alpha_{m}+1})\ldots\varrho(Z_{m},du_{n-\beta})$
$\displaystyle=\frac{2^{n-\beta}\beta!}{n!}\frac{\alpha_{1}!\kappa_{\alpha_{1}}}{2^{\alpha_{1}}}\cdots\frac{\alpha_{m}!\kappa_{\alpha_{m}}}{2^{\alpha_{m}}}\int_{\operatorname{G}(n,\alpha_{1})}\ldots\int_{\operatorname{G}(n,\alpha_{m})}[U_{1},\ldots,U_{m}]_{n-\beta}$
$\displaystyle\qquad\times
v^{(\beta)}\left(K_{1}|(U_{1}+\cdots+U_{m})^{\perp},\ldots,K_{\beta}|(U_{1}+\cdots+U_{m})^{\perp}\right)$
$\displaystyle\qquad\times\varrho_{(\alpha_{1})}(Z_{1},dU_{1})\ldots\varrho_{(\alpha_{m})}(Z_{m},dU_{m}),$

hence

$\displaystyle
V(K_{1},\ldots,K_{\beta},Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle=\binom{n}{\beta,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\alpha_{1}}\cdots\kappa_{\alpha_{m}}\int_{\operatorname{G}(n,\alpha_{1})}\ldots\int_{\operatorname{G}(n,\alpha_{m})}[U_{1},\ldots,U_{m}]_{n-\beta}$
$\displaystyle\qquad\times
v^{(\beta)}\left(K_{1}|(U_{1}+\cdots+U_{m})^{\perp},\ldots,K_{\beta}|(U_{1}+\cdots+U_{m})^{\perp}\right)$
(10)
$\displaystyle\qquad\times\varrho_{(\alpha_{1})}(Z_{1},dU_{1})\ldots\varrho_{(\alpha_{m})}(Z_{m},dU_{m}).$

In the special case where $K_{1}=\cdots=K_{\beta}=B^{n}$, we obtain

$\displaystyle V(Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}],B^{n}[\beta])$
$\displaystyle=\binom{n}{\beta,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\beta}\kappa_{\alpha_{1}}\cdots\kappa_{\alpha_{m}}\int_{\operatorname{G}(n,\alpha_{1})}\ldots\int_{\operatorname{G}(n,\alpha_{m})}[U_{1},\ldots,U_{m}]_{n-\beta}$
$\displaystyle\qquad\times\varrho_{(\alpha_{1})}(Z_{1},dU_{1})\ldots\varrho_{(\alpha_{m})}(Z_{m},dU_{m}).$
(11)

Since $\kappa_{n-i}V_{i}(K)=\binom{n}{i}V(K[i],B^{n}[n-i])$, the very special
case $m=1$ of (11) shows that

$V_{\alpha_{j}}(Z_{j})=\kappa_{\alpha_{j}}\rho_{\alpha_{j}}(Z_{j},\operatorname{G}(n,\alpha_{j})).$
(12)

Using (11),

$[U_{1},\ldots,U_{m}]_{n-\beta}\leq 1$ (13)

and (12), we get

$V(Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}],B^{n}[\beta])\leq\binom{n}{\beta,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\beta}V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m}).$ (14)

The inequality (14) is sharp. Suppose that $\text{dim}(Z_{j})\geq\alpha_{j}$
for $j=1,\ldots,k$. Then equality holds if and only if the zonoids
$Z_{1},\ldots,Z_{k}$ lie in orthogonal linear subspaces of dimensions
$\alpha_{1},\ldots,\alpha_{k}$. In fact, this inequality and the description
of the equality case is a special case of Theorem 1.3.

##  3 Proofs of Theorem 1.3 and Theorem 1.5

We start with an important observation which allows us to compare the
intrinsic volume $V_{\alpha}(K)$ of a convex body $K$ to the intrinsic volume
$V_{\alpha}(K|A)$ of a projection of $K$ to a subspace $A$ of dimension
$\beta\geq\alpha$ (see Lemma 3.1) and to characterize the equality case. In
the next section, when the stability of Theorem 1.3 and Theorem 1.5 is
discussed, we establish a stronger version in the special case where
$\alpha=\beta$.

For a compact convex set $K$, $u\in\mathbb{S}^{n-1}$ and
$H=u^{\bot}\in\operatorname{G}(n,n-1)$, the Steiner symmetral $S_{H}K$ of $K$
is defined as

$S_{H}K=\left\\{x+\mbox{$\frac{t-s}{2}$}\,u:\,x\in K|H\mbox{ and
}x+tu,\;x+su\in K\right\\}.$

It is known (see Schneider [49]) that $S_{H}K$ is a compact convex set
symmetric through $H$ and $S_{H}K|H=K|H$. Moreover, it is known that Steiner
symmetrization does not increase the intrinsic volumes, that is,

$V_{\alpha}(S_{H}K)\leq V_{\alpha}(K)\quad\text{for }\alpha=1,\ldots,n.$ (15)

This can be found in [32, Satz 18.5] or in Hadwiger’s monograph [26]. In
addition,

$V_{n}(S_{H}K)=V_{n}(K).$ (16)

For a subset $X\subset\mathbb{R}^{n}$, we write ${\rm lin}\,X$ and ${\rm
conv}\,X$ to denote the linear hull and the convex hull of $X$, respectively,
and for a compact convex set $K$, we write ${\rm aff}\,K$ and ${\rm
relint}\,K$ to denote the affine hull and the relative interior with respect
to the affine full of $K$, respectively.

Let $\operatorname{A}(n,\beta)$ denote the affine Grassmannian of
$\beta$-dimensional affine subspaces ($\beta$-flats, for short) of
$\mathbb{R}^{n}$.

######  Lemma 3.1.

Let $1\leq\alpha\leq\beta\leq n-1$. Let $K\in\mathcal{K}^{n}$ and
$A\in\operatorname{A}(n,\beta)$. Then the following is true.

  1. (a)

$V_{\alpha}(K|A)\leq V_{\alpha}(K)$.

  2. (b)

If $\dim(K)\geq\alpha$, then $V_{\alpha}(K|A)=V_{\alpha}(K)$ if and only if
$K\subset A+z$ for some $z\in\mathbb{R}^{n}$ (even $K=K|A$ provided $A\cap
K\neq\emptyset$).

###### Proof.

For the proof we can assume that $K$ has its centroid at the origin $o$ and
$A\in\operatorname{G}(n,\beta)$.

In a first step, we consider the case where $\beta=n-1$.

(a) Recall that $V_{\alpha}(S_{A}K)\leq V_{\alpha}(K)$ (see [32, Satz 18.5]).
Since $K|A\subset S_{A}K$, we get

$V_{\alpha}(K|A)\leq V_{\alpha}(S_{A}K)\leq V_{\alpha}(K).$ (17)

(b) For the non-trivial direction, we assume that
$V_{\alpha}(K|A)=V_{\alpha}(K)$. Then (17) implies that

$V_{\alpha}(K|A)=V_{\alpha}(S_{A}K)=V_{\alpha}(K).$ (18)

Aiming at a contradiction, we assume that $K\not\subset A$. Suppose that
$\dim(K)=\alpha$. Then $V_{\alpha}(K|A)=V_{\alpha}(K)>0$ by (18), hence
$\dim(K|A)=\alpha$. Since $K\not\subset A$, it follows that
$V_{\alpha}(K)=\lambda_{\alpha}(K)>\lambda_{\alpha}(K|A)=V_{\alpha}(K|A)$,
which contradicts (18). This shows that $\dim(K)\geq\alpha+1$. Since $0\leq
n-\dim(K+A^{\perp})<n-\alpha<n$ and $V_{\alpha}(S_{A}K)=V_{\alpha}(K)$, [32,
Satz 18.5, (444b)] now yields that $K$ is symmetric with respect to $A$ (here
we use that $K$ has its centroid at the origin), in particular we have
$K|A=K\cap A\subsetneqq K$ since $K\not\subset A$. Since $\dim(K)>\alpha$,
there is an open set of affine flats $E\in\operatorname{A}(n,n-\alpha)$ for
which $K\cap E\neq\emptyset$ and $(K|A)\cap E=K\cap A\cap E=\emptyset$. Let
$\mu_{n-\alpha}$ denote the suitably normalized motion invariant Haar measure
on $\operatorname{A}(n,n-\alpha)$ and let $\chi$ denote the Euler
characteristic. Then the Crofton formula (see [29, Chapter 5] , [49, Section
4.4]) implies that

$\displaystyle V_{\alpha}(K)$
$\displaystyle=c_{n,\alpha}\int_{\operatorname{A}(n,n-\alpha)}\chi(K\cap
E)\,\mu_{n-\alpha}(dE)$
$\displaystyle>c_{n,\alpha}\int_{\operatorname{A}(n,n-\alpha)}\chi((K|A)\cap
E)\,\mu_{n-\alpha}(dE)$ $\displaystyle=V_{\alpha}(K|A),$

where $c_{n,\alpha}$ is a constant depending only on $n,\alpha$. This is the
required contradiction.

Now we turn to the general case.

(a) We choose $A_{i}\in\operatorname{G}(n,i)$ for $i=\beta,\ldots,n-1$ such
that $A=A_{\beta}\subsetneqq A_{\beta+1}\subsetneqq\cdots\subsetneqq A_{n-1}$.
By the first step, we have $V_{\alpha}(K|A_{n-1})\leq V_{\alpha}(K)$. Since
$K|A_{n-2}=(K|A_{n-1})|A_{n-2}$, we can apply the result from the first step
in $A_{n-1}$ to $K|A_{n-1}$ with respect to $A_{n-2}$ to get

$V_{\alpha}(K|A_{n-2})=V_{\alpha}((K|A_{n-1})|A_{n-2})\leq
V_{\alpha}(K|A_{n-1})\leq V_{\alpha}(K).$

Observe that by the intrinsic nature of $V_{\alpha}$ the respective ambient
space need not be indicated. Using $K|A_{n-3}=(K|A_{n-2})|A_{n-3}$, we obtain
in the same way that

$V_{\alpha}(K|A_{n-3})=V_{\alpha}((K|A_{n-2})|A_{n-3})\leq
V_{\alpha}(K|A_{n-2})\leq V_{\alpha}(K|A_{n-1})\leq V_{\alpha}(K).$

By induction, we thus get

$V_{\alpha}(K|A)=V_{\alpha}(K|A_{\beta})\leq
V_{\alpha}(K|A_{\beta+1})\leq\ldots\leq V_{\alpha}(K|A_{n-1})\leq
V_{\alpha}(K).$ (19)

(b) Now we assume that $V_{\alpha}(K|A)=V_{\alpha}(K)$. From (19), we deduce
that

$V_{\alpha}(K|A)=V_{\alpha}(K|A_{\beta})=V_{\alpha}(K|A_{\beta+1})=\ldots=V_{\alpha}(K|A_{n-1})=V_{\alpha}(K).$

By the result of the first step and $V_{\alpha}(K|A_{n-1})\leq V_{\alpha}(K)$
we conclude that $K\subset A_{n-1}$, hence $K|A_{n-1}=K$. Since
$K|A_{n-2}=(K|A_{n-1})|A_{n-2}$ and
$V_{\alpha}(K|A_{n-2})=V_{\alpha}(K|A_{n-1})$, it follows from the first step,
applied in $A_{n-1}$, that $K=K|A_{n-1}\subset A_{n-2}$. Continuing
inductively in this way it follows that $K\subset A_{\beta}=A$, as asserted. ∎

For $u_{1},\ldots,u_{k}\in\mathbb{R}^{n}$, we write ${\rm
lin}\\{u_{1},\ldots,u_{k}\\}^{\bot}$ to denote the orthogonal complement of
the linear hull of $u_{1},\ldots,u_{k}$.

For the proof of Theorems 1.3 and 1.5 we start with some preparation.

Note that Theorem 1.3 holds trivially with equality if $\beta=n$. Since
Theorem 1.5 boils down to Theorem 1.2 if no zonoids are involved, it is
sufficient in the following to consider the following range for the involved
parameters.

Let $0\leq\gamma_{1},\gamma_{2}$, $\gamma:=\gamma_{1}+\gamma_{2}$,
$\alpha_{1},\ldots,\alpha_{m}\geq 1$, $\gamma\leq\beta\leq n-1$ such that
$\alpha_{1}+\cdots+\alpha_{m}=n-\beta$, $m\geq 1$. Let
$K_{1},K_{2}\subset\mathbb{R}^{n}$ be arbitrary compact convex sets and let
$Z_{1},\ldots,Z_{m}\subset\mathbb{R}^{n}$ be zonoids.

A special case of (2) gives

$\displaystyle
V(K_{1}[\gamma_{1}],K_{2}[\gamma_{2}],B^{n}[\beta-\gamma],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle=\binom{n}{\beta,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\alpha_{1}}\cdots\kappa_{\alpha_{m}}\int_{\operatorname{G}(n,\alpha_{1})}\ldots\int_{\operatorname{G}(n,\alpha_{m})}[U_{1},\ldots,U_{m}]_{n-\beta}$
$\displaystyle\quad\times
v^{(\beta)}\left(K_{1}|(U_{1}+\cdots+U_{m})^{\bot}[\gamma_{1}],K_{2}|(U_{1}+\cdots+U_{m})^{\bot}[\gamma_{2}],B^{n}|(U_{1}+\cdots+U_{m})^{\bot}[\beta-\gamma]\right)$
$\displaystyle\quad\times\varrho_{(\alpha_{1})}(Z_{1},dU_{1})\ldots\varrho_{(\alpha_{m})}(Z_{m},dU_{m}).$
(20)

###### Proof of Theorem 1.3.

Here we have $\gamma_{2}=0$, and hence $\gamma_{1}=\gamma\leq\beta$, and we
set $K_{1}=:K$. If $\dim(U_{1}+\cdots+U_{m})=n-\beta$, then

$\displaystyle
v^{(\beta)}\left(K_{1}|(U_{1}+\cdots+U_{m})^{\bot}[\gamma],B^{n}|(U_{1}+\cdots+U_{m})^{\bot}[\beta-\gamma]\right)$
$\displaystyle=V_{\gamma}(K|(U_{1}+\cdots+U_{m})^{\perp})\frac{\kappa_{\beta-\gamma}}{\binom{\beta}{\gamma}}$
$\displaystyle\leq
V_{\gamma}(K)\frac{\kappa_{\beta-\gamma}}{\binom{\beta}{\gamma}}$ (21)

by Lemma 3.1. Moreover, if $\dim(K)\geq\gamma$, then equality holds if and
only if $K$ is contained in a translate of $(U_{1}+\cdots+U_{m})^{\perp}$.

Now we insert this estimate into (3). This leads to

$\displaystyle
V(K[\gamma],B^{n}[\beta-\gamma],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle\leq\binom{n}{\gamma,\beta-\gamma,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\beta-\gamma}\kappa_{\alpha_{1}}\cdots\kappa_{\alpha_{m}}V_{\gamma}(K)$
(22)
$\displaystyle\qquad\times\int_{\operatorname{G}(n,\alpha_{1})}\ldots\int_{\operatorname{G}(n,\alpha_{m})}[U_{1},\ldots,U_{m}]_{n-\beta}\,\varrho_{(\alpha_{1})}(Z_{1},dU_{1})\ldots\varrho_{(\alpha_{m})}(Z_{m},dU_{m})$
$\displaystyle=\binom{\beta}{\gamma}^{-1}\frac{\kappa_{\beta-\gamma}}{\kappa_{\beta}}V_{\gamma}(K)V(Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}],B^{n}[\beta]).$

If we use the upper bound $[U_{1},\ldots,U_{m}]_{n-\beta}\leq 1$ and (12), we
derive from (3) that

$\displaystyle
V(K[\gamma],B^{n}[\beta-\gamma],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle\leq\binom{n}{\gamma,\beta-\gamma,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\beta-\gamma}V_{\gamma}(K)V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m}).$ (23)

This proves (1).

Now we assume that equality holds. Let $u_{0}\in\operatorname{lin}(K)$ (the
linear subspace parallel to the affine hull of $K$) and
$u_{i}\in\operatorname{lin}(Z_{i})$ for $i=1,\ldots,m$. Since
$\dim(Z_{i})\geq\alpha_{i}$,
$\operatorname{lin}(Z_{i})=\operatorname{lin}\operatorname{supp}(\varrho(Z_{i},\cdot))$
and

$\operatorname{lin}\\{u:u\in
U\in\operatorname{supp}(\varrho_{(\alpha_{i})}(Z_{i},\cdot))\\}=\operatorname{lin}(Z_{i}),$

there are $U_{i}\in\operatorname{supp}(\varrho_{(\alpha_{i})}(Z_{i},\cdot))$
with $u_{i}\in U_{i}$, for $i=1,\ldots,m$. Since in the transition from (3) to
(3) equality holds, we must have $[U_{1},\ldots,U_{m}]_{n-\beta}=1$ if
$U_{i}\in\operatorname{supp}(\varrho_{(\alpha_{i})}(Z_{i},\cdot))$ for
$i=1,\ldots,m$ (note that
$(U_{1},\ldots,U_{m})\mapsto[U_{1},\ldots,U_{m}]_{n-\beta}$ is continuous),
and hence $U_{1},\ldots,U_{m}$ are pairwise orthogonal. Moreover, equality in
(3) implies that we must also have equality in (3), hence it follows that
$u_{0}\in(U_{1}+\cdots+U_{m})^{\perp}$. Thus we have shown that
$u_{0},u_{1},\ldots,u_{m}$ are pairwise orthogonal, and hence
$K,\operatorname{lin}(Z_{1}),\ldots,\operatorname{lin}(Z_{m})$ lie in pairwise
orthogonal affine subspaces.

The reverse implication is clear from the preceding argument, since equality
holds in each step. ∎

###### Proof of Theorem 1.5.

Now we have $\gamma_{1}=1$, $\gamma_{2}=\gamma-1\geq 1$, $\beta=\gamma$ and
$\alpha_{1}+\cdots+\alpha_{m}=n-\gamma=n-\beta$. We apply Theorem 1.2 to the
mixed volume in the integrand of (3) for $\beta=\gamma$ in
$(U_{1}+\cdots+U_{m})^{\perp}$ as the ambient space. Here we can assume that
$\dim(U_{1}+\cdots+U_{m})=n-\beta$, since otherwise the integrand is zero.
This gives

$\displaystyle
v^{(\beta)}\left(K_{1}|(U_{1}+\cdots+U_{m})^{\bot},K_{2}|(U_{1}+\cdots+U_{m})^{\bot}[\gamma-1]\right)$
$\displaystyle\leq\frac{1}{\beta}V_{1}\left(K_{1}|(U_{1}+\cdots+U_{m})^{\perp}\right)V_{\beta-1}\left(K_{2}|(U_{1}+\cdots+U_{m})^{\perp}\right)$
(24) $\displaystyle\leq\frac{1}{\beta}V_{1}(K_{1})V_{\beta-1}(K_{2}),$ (25)

where Lemma 3.1 was applied twice in the last step. Since $\dim(K_{1})\geq 1$
and $\dim(K_{2})\geq\beta-1$, equality holds in (25) if and only if $K_{1}$
and $K_{2}$ are contained in translates of $(U_{1}+\ldots+U_{m})^{\perp}$.
Then we deduce from (3) that

$\displaystyle
V(K_{1},K_{2}[\gamma-1],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle\leq\binom{n}{\beta,\alpha_{1},\ldots,\alpha_{m}}^{-1}\kappa_{\alpha_{1}}\cdots\kappa_{\alpha_{m}}\frac{1}{\beta}V_{1}(K_{1})V_{\beta-1}(K_{2})$
$\displaystyle\qquad\times\int_{\operatorname{G}(n,\alpha_{1})}\ldots\int_{\operatorname{G}(n,\alpha_{m})}[U_{1},\ldots,U_{m}]_{n-\beta}\,\varrho_{(\alpha_{1})}(Z_{1},dU_{1})\ldots\varrho_{(\alpha_{m})}(Z_{m},dU_{m})$
$\displaystyle\leq\binom{n}{1,\beta-1,\alpha_{1},\ldots,\alpha_{m}}^{-1}V_{1}(K_{1})V_{\beta-1}(K_{2})V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m}),$ (26)

where $[U_{1},\ldots,U_{m}]_{n-\beta}\leq 1$ and (12) were used in the last
step. This proves (4).

Next we assume that

$\displaystyle
V(K_{1},K_{2}[\gamma-1],Z_{1}[\alpha_{1}],\ldots,Z_{m}[\alpha_{m}])$
$\displaystyle=\binom{n}{1,\beta-1,\alpha_{1},\ldots,\alpha_{m}}^{-1}V_{1}(K_{1})V_{\beta-1}(K_{2})V_{\alpha_{1}}(Z_{1})\cdots
V_{\alpha_{m}}(Z_{m}).$

Then equality holds in (26), (25) and (24). Let
$v_{1}\in\operatorname{lin}(K_{1})$, $v_{2}\in\operatorname{lin}(K_{2})$ and
$u_{i}\in\operatorname{lin}(Z_{i})$ for $i=1,\ldots,m$. As in the proof of
Theorem 1.3 we can choose
$U_{i}\in\operatorname{supp}(\varrho_{(\alpha_{i})}(Z_{i},\cdot))$ with
$u_{i}\in U_{i}$ for $i=1,\ldots,m$. Then from equality in (26) it follows
that $[U_{1},\ldots,U_{m}]_{n-\beta}=1$, therefore $U_{1},\ldots,U_{m}$ are
pairwise orthogonal and $\dim(U_{1}+\cdots+U_{m})^{\perp}=\beta$. From (25) we
conclude that $v_{1},v_{2}\in\operatorname{lin}(U_{1}+\cdots+U_{m})^{\perp}$,
and from (24) we then see that $v_{1},v_{2}$ are orthogonal. This shows that
$v_{1},v_{2},u_{1},\ldots,u_{m}$ are pairwise orthogonal. This finally implies
that the affine hulls of $K_{1},K_{2},Z_{1},\ldots,Z_{m}$ are pairwise
orthogonal.

The reverse statement is then also clear from the preceding argument, since
equality holds in each step. ∎

##  4 The stability version when all bodies are zonoids

The main goal of this section is a to prove Theorem 1.6. Although this theorem
is stated for zonoids, most of the preparatory statements are valid for
arbitrary convex bodies.

For $m=1,\ldots,n$, the $m$-dimensional size of a compact convex set $K$ can
be effectively measured by the largest radius $r_{m}(K)$ of $m$-dimensional
balls contained in $K$.

######  Lemma 4.1.

Let $1\leq m\leq d\leq n$, and let $E$ be a $d$-dimensional ellipsoid in
$\mathbb{R}^{n}$ with half-axes $a_{1}\geq\ldots\geq a_{d}>0$.

  1. (i)

$r_{m}(E)=a_{m}$.

  2. (ii)

There exists some $A\in\operatorname{A}(n,m-1)$ such that $E\subset
A+r_{m}(E)B^{n}$ and even $r_{m-1}(E|A)=r_{m-1}(E)$ provided $m\geq 2$. If
$o\in E$, then $A$ can be chosen as a linear subspace.

  3. (iii)

For any $L\in\operatorname{G}(n,m)$, we have $r_{m}(E|L)\leq r_{m}(E)$.

###### Proof.

We may assume that the origin is the center of $E$ and $d=n$. Let
$e_{1},\ldots,e_{n}$ be an orthonormal basis of $\mathbb{R}^{n}$ such that

$E=\left\\{\sum_{i=1}^{n}t_{i}e_{i}:\,\sum_{i=1}^{n}\frac{t_{i}^{2}}{a_{i}^{2}}\leq
1\right\\}.$ (27)

For (i), there is an $m$-ball $B$ of radius $r_{m}(E)$ contained in $E$ and
centered at the origin $o$ because $E$ is $o$-symmetric. It follows that there
exists a $y\in B\cap\operatorname{lin}\\{e_{m},\ldots,e_{n}\\}$ with
$\|y\|=r_{m}(E)$. Since $\|y\|\leq a_{m}$ by (27), we conclude (i).

For (ii), we can take $A=\\{o\\}$ if $m=1$, and
$A=\operatorname{lin}\\{e_{1},\ldots,e_{m-1}\\}$ if $m\geq 2$ by (i) and (27).

For (iii), let $L\in\operatorname{G}(n,m)$. Since $E$ is a linear image of
$B^{n}$, there exists an $L_{0}\in\operatorname{G}(n,m)$ such that $E|L=(E\cap
L_{0})|L$; therefore, $r_{m}(E|L)\leq r_{m}(E\cap L_{0})\leq r_{m}(E)$. ∎

According to John’s theorem (see Schneider [49, Theorem 10.12.2]), for any
$d$-dimensional compact convex set $K$ in $\mathbb{R}^{n}$, $1\leq d\leq n$,
there exists a $z\in K$ and a $d$-dimensional ellipsoid $E$ centred at $o$
such that

$z+E\subset K\subset z+d\cdot E.$ (28)

######  Lemma 4.2.

Let $K$ be a compact convex set in $\mathbb{R}^{n}$, and let
$m\in\\{1,\ldots,n-1\\}$.

  1. (i)

If $A\in\operatorname{A}(n,m)$ and $K\subset A+\varrho B^{n}$ for some
$\varrho\geq 0$, then $r_{m+1}(K)\leq\varrho$.

  2. (ii)

There exists some $A_{0}\in\operatorname{A}(n,m)$ such that $K\subset
A_{0}+nr_{m+1}(K)B^{n}$ and $r_{m}(K|A_{0})\geq\frac{1}{n}\,r_{m}(K)$. where
$o\in A_{0}$ If $K$ is $o$-symmetric, then $A_{0}$ can be chosen as a linear
subspace.

  3. (iii)

If $L\in\operatorname{G}(n,m)$, then $r_{m}(K|L)\leq nr_{m}(K)$.

###### Proof.

For (i), let $B$ be an $(m+1)$-dimensional ball of radius $r_{m+1}(K)$ and
center $z$ contained in $K$. It follows that $(z+A^{\bot})\cap B$ contains a
segment $s$ of length $2r_{m+1}(K)$. Since $s$ is a subset of the
$(n-m)$-dimensional ball $(z+A^{\bot})\cap(A+\varrho B^{n})$ of radius
$\varrho$, we have $r_{m+1}(K)\leq\varrho$.

For (ii), let $z\in K$ and let $E$ be an ellipsoid centred at $o$ with ${\rm
dim}\,E={\rm dim}\,K$ and satisfying (28). It follows from Lemma 4.1 (ii) that
there exists $L_{0}\in\operatorname{G}(n,m)$ such that $E\subset
L_{0}+r_{m+1}(E)B^{n}$ and $r_{m}(E|L_{0})=r_{m}(E)$. Since $r_{m+1}(E)\leq
r_{m+1}(K)$ and $E\subset K-z\subset nE$, we conclude that with
$A_{0}:=z+L_{0}$ we have $K\subset A_{0}+nr_{m+1}(K)B^{n}$ and
$r_{m}(K|A_{0})\geq r_{m}(E|L_{0})=r_{m}(E)\geq\frac{1}{n}\,r_{m}(K)$.

For the proof of (iii), we choose $z,E$ as for (ii). Then it follows from
Lemma 4.1 (iii) that

$r_{m}(K|L)\leq r_{m}((z+nE)|L)\leq r_{m}(z+nE)=n\,r_{m}(z+E)\leq nr_{m}(K),$

which yields the assertion. ∎

The main tool used to verify stability versions of Theorems 1.3 and 1.5 is
Proposition 4.3.

######  Proposition 4.3.

Let $n\geq 2$ and $1\leq\beta\leq n-1$. If $K$ is a compact convex set in
$\mathbb{R}^{n}$ with $\dim(K)\geq\beta$, then

$V_{\beta}(K)\geq\left(1+\frac{1}{2^{n+2}n^{5}}\cdot\frac{r_{\beta+1}(K)^{2}}{r_{\beta}(K)^{2}}\right)\max\\{V_{\beta}(K|L):L\in\operatorname{G}(n,\beta)\\}.$

######  Remark 4.4.

$(\beta+1)$-dimensional cones show that the order of the estimate is optimal.

###### Proof.

We may assume that $r_{\beta+1}(K)>0$ (compare Lemma 3.1) and that the origin
$o$ is the center of mass of $K$.

Let $L_{\beta}\in\operatorname{G}(n,\beta)$ be such that
$V_{\beta}(K|L_{\beta})=\max\\{V_{\beta}(K|L):L\in\operatorname{G}(n,\beta)\\}$,
and let $B$ be a $(\beta+1)$-dimensional ball of radius $r_{\beta+1}(K)$
contained in $K$. Writing $z$ to denote the center of $B$, it follows that
there exists a segment $s\subset B\cap(z+L_{\beta}^{\bot})$ of length
$2r_{\beta+1}(K)$, where $s=[z-r_{\beta+1}(K)v,z+r_{\beta+1}(K)v]$ for a
suitable $v\in L_{\beta}^{\bot}\cap\mathbb{S}^{n-1}$. On the other hand,
according to Lemma 4.2, there exists a $(\beta-1)$-dimensional affine subspace
$A_{\beta-1}\subset L_{\beta}$ (note that this is obviously true if $\beta=1$)
such that

$K|L_{\beta}\subset A_{\beta-1}+nr_{\beta}(K|L_{\beta})B^{n}\subset
A_{\beta-1}+n^{2}r_{\beta}(K)B^{n}.$ (29)

Let $L_{\beta-1}\subset L_{\beta}$ be the $(\beta-1)$-dimensional linear
subspace parallel to $A_{\beta-1}$.

We consider the $(\beta+1)$-dimensional linear subspace $L_{\beta+1}={\rm
lin}\\{L_{\beta},v\\}$ and an orthonormal basis $u_{1},\ldots,u_{n}$ of
$\mathbb{R}^{n}$, where $u_{1},\ldots,u_{\beta}\in L_{\beta}$,
$L_{\beta-1}=u_{\beta}^{\bot}\cap L_{\beta}$ and $u_{\beta+1}=v$. For
$i=0,\ldots,n$, we define $K_{i}$ by reverse induction; namely, $K_{n}=K$ and
$K_{i-1}=S_{u_{i}^{\bot}}K_{i}$ for $i=1,\ldots,n$. We deduce from (15) and
reverse induction on $i$ that

$V_{\beta}(K_{i})\leq V_{\beta}(K)\mbox{ \ for $i=0,\ldots,n$.}$ (30)

We observe that

$K|L_{\beta+1}=K_{\beta+1}|L_{\beta+1}\subset K_{\beta+1}$

and

$[(z|L_{\beta+1})-r_{\beta+1}(K)v,(z|L_{\beta+1})+r_{\beta+1}(K)v]\subset
K_{\beta+1}\cap L_{\beta+1},$

which in turn yields

$K|L_{\beta}=K_{\beta}|L_{\beta}\subset K_{\beta}\mbox{ \ and \
}[(z|L_{\beta})-r_{\beta+1}(K)v,(z|L_{\beta})+r_{\beta+1}(K)v]\subset
K_{\beta}\cap L_{\beta+1}.$

By construction, $K_{0}$ is unconditional with respect to the basis
$u_{1},\ldots,u_{n}$. Moreover, (29) and an application of (16) in $L_{\beta}$
show that

$\displaystyle V_{\beta}(K_{0}\cap L_{\beta})$ $\displaystyle=$ $\displaystyle
V_{\beta}(K|L_{\beta}),$ (31) $\displaystyle K_{0}\cap L_{\beta}$
$\displaystyle\subset$ $\displaystyle L_{\beta-1}+n^{2}r_{\beta}(K)B^{n},$
(32) $\displaystyle[-r_{\beta+1}(K)v,r_{\beta+1}(K)v]$ $\displaystyle\subset$
$\displaystyle K_{0}.$ (33)

We now transform $K_{0}$ into a convex body whose sections parallel to
$L_{\beta-1}$ are $(\beta-1)$-dimensional balls, by possibly further
symmetrizing $K_{0}$. If $\beta=1$, then let $K_{0}^{(\beta-1)}=K_{0}$. If
$\beta\geq 2$, then by applying repeated Steiner symmetrizations with respect
to suitable subspaces $u^{\bot}$ with $u\in L_{\beta-1}$ and taking limits, we
obtain a compact convex set $K_{0}^{(\beta-1)}$ (see [25] or [32, §19]) such
that

$\displaystyle K_{0}^{(\beta-1)}|L_{\beta-1}^{\bot}$ $\displaystyle=$
$\displaystyle K_{0}|L_{\beta-1}^{\bot},$ $\displaystyle
K_{0}^{(\beta-1)}\cap(x+L_{\beta-1})$ $\displaystyle{\rm is}$
$\displaystyle\mbox{a $(\beta-1)$-ball centered at $x$ if $x\in(\text{\rm
relint }K_{0})|L_{\beta-1}^{\bot}$},$ $\displaystyle
V_{\beta-1}(K_{0}^{(\beta-1)}\cap(x+L_{\beta-1}))$ $\displaystyle=$
$\displaystyle V_{\beta-1}(K_{0}\cap(x+L_{\beta-1}))\mbox{ if $x\in(\text{\rm
relint }K_{0})|L_{\beta-1}^{\bot}$}.$

From the properties (15) and (16) of the Steiner symmetrization, and from
(30), (31), (32) and (33) we deduce that

$\displaystyle V_{\beta}(K_{0}^{(\beta-1)})$ $\displaystyle\leq$
$\displaystyle V_{\beta}(K),$ (34) $\displaystyle
V_{\beta}(K_{0}^{(\beta-1)}\cap L_{\beta})$ $\displaystyle=$ $\displaystyle
V_{\beta}(K|L_{\beta}),$ (35) $\displaystyle K_{0}^{(\beta-1)}\cap L_{\beta}$
$\displaystyle\subset$ $\displaystyle L_{\beta-1}+n^{2}r_{\beta}(K)B^{n},$
(36) $\displaystyle[-r_{\beta+1}(K)v,r_{\beta+1}(K)v]$ $\displaystyle\subset$
$\displaystyle K_{0}^{(\beta-1)}.$ (37)

It follows from (34), (35) and (37) that the unconditional
$(\beta+1)$-dimensional compact convex set

$\widetilde{K}=[K_{0}^{(\beta-1)}\cap
L_{\beta},-r_{\beta+1}(K)v,r_{\beta+1}(K)v]$ (38)

satisfies

$V_{\beta}(\widetilde{K})\leq V_{\beta}(K)\mbox{ \ and \
}V_{\beta}(\widetilde{K}\cap L_{\beta})=V_{\beta}(K|L_{\beta}).$ (39)

We define the set

$\widetilde{K}_{0}=\widetilde{K}\cap L_{\beta}=K_{0}^{(\beta-1)}\cap
L_{\beta},$ (40)

which has axial rotational symmetry around $\mathbb{R}u_{\beta}$ and according
to (36) satisfies

$\widetilde{K}_{0}\subset L_{\beta-1}+n^{2}r_{\beta}(K)B^{n}.$ (41)

According to (39), Proposition 4.3 will follow once we have shown that

$V_{\beta}(\widetilde{K})\geq\left(1+\frac{1}{2^{n+2}n^{5}}\cdot\frac{r_{\beta+1}(K)^{2}}{r_{\beta}(K)^{2}}\right)V_{\beta}(\widetilde{K}_{0}).$
(42)

For the proof, we will use (38), (40) and (41). We distinguish the cases
$\beta=1$ and $\beta\geq 2$.

Case 1: $\beta=1$. In this case, $\widetilde{K}$ is a rhombus, one of the
diagonals is $\widetilde{K}_{0}=\widetilde{K}\cap L_{1}=K_{0}\cap L_{1}$,
which has length

$V_{1}(\widetilde{K}_{0})=V_{1}(K_{0}\cap L_{1})=V_{1}(K|L_{1})\leq
2r_{1}(K|L_{1})\leq 2nr_{1}(K),$

and the other one is of length $2r_{2}(K)$. Hence it follows that

$\displaystyle V_{1}(\widetilde{K})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\cdot
4\cdot\sqrt{\left(\frac{V_{1}(\widetilde{K}_{0})}{2}\right)^{2}+r_{2}(K)^{2}}=V_{1}(\widetilde{K}_{0})\sqrt{1+\frac{4r_{2}(K)^{2}}{V_{1}(\widetilde{K}_{0})^{2}}}$
$\displaystyle\geq$ $\displaystyle
V_{1}(\widetilde{K}_{0})\sqrt{1+\frac{r_{2}(K)^{2}}{n^{2}r_{1}(K)^{2}}}\geq
V_{1}(\widetilde{K}_{0})\left(1+\frac{r_{2}(K)^{2}}{4n^{2}r_{1}(K)^{2}}\right).$

Case 2: $\beta\geq 2$. We write $\partial\widetilde{K}$ to denote the relative
boundary of $\widetilde{K}$ in $L_{\beta+1}$, $\partial\widetilde{K}_{0}$ to
denote the relative boundary of $\widetilde{K}_{0}$ in $L_{\beta}$ and define
$f:\,\widetilde{K}_{0}\to\mathbb{R}_{\geq 0}$ to be the concave function
satisfying $x+f(x)v\in\partial\widetilde{K}$ for $x\in\widetilde{K}_{0}$. In
particular, we have

$V_{\beta}(\widetilde{K})=\int_{\widetilde{K}_{0}}\sqrt{1+\|\nabla
f(x)\|^{2}}\,dx,$ (43)

where the integration is with respect to the $\beta$-dimensional Lebesgue
measure on $L_{\beta}$, and the derivative $\nabla f$ of $f$ is well-defined
for almost all $x\in\widetilde{K}_{0}$.

Let $x\in\text{\rm relint }\widetilde{K}_{0}$ be such that $\nabla f(x)$
exists. Hence $x\neq o$ and there exist $z\in\partial\widetilde{K}_{0}$ and
$t\in(0,1)$ such that $x=tz$. Since $f$ is concave and $\widetilde{K}$ is a
double cone, we deduce that

$\displaystyle f(x)+\langle\nabla f(x),y-x\rangle$ $\displaystyle\geq$
$\displaystyle f(y)\mbox{ \ for $y\in\widetilde{K}_{0}$},$ $\displaystyle
f(x)+\langle\nabla f(x),-x\rangle$ $\displaystyle=$ $\displaystyle f(o),$
$\displaystyle f(x)+\langle\nabla f(x),z-x\rangle$ $\displaystyle=$
$\displaystyle f(z)=0.$

Using also (38), we deduce that if $x=tz$ for $z\in\partial\widetilde{K}_{0}$
and $t\in(0,1)$, and $\nabla f(x)$ exists, then

$\displaystyle f(o)$ $\displaystyle=$ $\displaystyle r_{\beta+1}(K),$ (44)
$\displaystyle f(o)+\langle\nabla f(x),y\rangle\geq f(y)$ $\displaystyle\geq$
$\displaystyle 0\mbox{ \ for $y\in\widetilde{K}_{0}$},$ (45) $\displaystyle
f(o)+\langle\nabla f(x),z\rangle=f(z)$ $\displaystyle=$ $\displaystyle 0.$
(46)

Next we define

$\displaystyle\Xi$ $\displaystyle=$
$\displaystyle\\{z\in\partial\widetilde{K}_{0}:\,z|L_{\beta-1}\in\mbox{$\frac{1}{2}$}\,\widetilde{K}_{0}\\}$
$\displaystyle\widetilde{\Xi}$ $\displaystyle=$
$\displaystyle\\{tz:\,z\in\Xi\mbox{ and }t\in[0,1]\\}.$

We claim that if $x\in\widetilde{\Xi}$ and $\nabla f(x)$ exists, then

$\sqrt{1+\|\nabla f(x)\|^{2}}\geq
1+\frac{r_{\beta+1}(K)^{2}}{16n^{4}r_{\beta}(K)^{2}}.$ (47)

For the proof, we can write $x$ as $x=tz$ for $z\in\Xi$ and $t\in(0,1]$ and we
have $z_{0}=z|L_{\beta-1}\in\frac{1}{2}\,\widetilde{K}_{0}$. There exists an
$s\in\mathbb{R}$ such that $z-z_{0}=su_{\beta}$, and hence
$y=2z_{0}\in\widetilde{K}_{0}$ satisfies

$2su_{\beta}=2z-y.$ (48)

We deduce from (45), (46) and (48) that

$f(o)+\langle\nabla f(x),2su_{\beta}\rangle=2\left[f(o)+\langle\nabla
f(x),z\rangle\right]-\left[f(o)+\langle\nabla f(x),y\rangle\right]\leq 0.$
(49)

Here $0<|s|\leq n^{2}r_{\beta}(K)$ by (41) and (49). Therefore (44) and (49)
imply that

$\|\nabla
f(x)\|\geq\frac{f(o)}{2|s|}\geq\frac{r_{\beta+1}(K)}{2n^{2}r_{\beta}(K)}.$

Using $\sqrt{1+t}\geq 1+\frac{1}{4}\,t$ for $t\in[0,1]$, we thus obtain (47).
Let $\lambda_{\beta}$ be the Lebesgue-measure in $L_{\beta}$. Combining (43)
and (47), we get

$\displaystyle V_{\beta}(\widetilde{K})$
$\displaystyle=\int_{\widetilde{K}_{0}}\sqrt{1+\|\nabla f(x)\|^{2}}\,dx$
$\displaystyle=\int_{\widetilde{K}_{0}\setminus\widetilde{\Xi}}\sqrt{1+\|\nabla
f(x)\|^{2}}\,dx+\int_{\widetilde{\Xi}}\sqrt{1+\|\nabla f(x)\|^{2}}\,dx$
$\displaystyle\geq\lambda_{\beta}(\widetilde{K}_{0}\setminus\widetilde{\Xi})+\left(1+\frac{1}{4}\left(\frac{r_{\beta+1}(K)}{2n^{2}r_{\beta}(K)}\right)^{2}\right)\lambda_{\beta}(\widetilde{\Xi})$
$\displaystyle=V_{\beta}(\widetilde{K}_{0})+\frac{r_{\beta+1}(K)^{2}}{16n^{4}r_{\beta}(K)^{2}}\cdot\lambda_{\beta}(\widetilde{\Xi}).$
(50)

What is left to find is a lower bound for the ratio
$\lambda_{\beta}(\widetilde{\Xi})/V_{\beta}(\widetilde{K}_{0})$. As
$\widetilde{K}_{0}$ is unconditional within $L_{\beta}$, we have

${\rm
conv}\,\widetilde{\Xi}=\\{x\in\widetilde{K}_{0}:\,x|L_{\beta-1}\in\mbox{$\frac{1}{2}$}\,\widetilde{K}_{0}\\},$

which is again an unconditional set. We observe that the linear transformation
$\Phi:\,L_{\beta}\to L_{\beta}$ with $\Phi u_{\beta}=u_{\beta}$ and $\Phi
u_{i}=\frac{1}{2}\,u_{i}$ for $i=1,\ldots,\beta-1$ satisfies
$\Phi\widetilde{K}_{0}\subset{\rm conv}\,\widetilde{\Xi}$, therefore

$V_{\beta}({\rm conv}\,\widetilde{\Xi})\geq
2^{-(\beta-1)}V_{\beta}(\widetilde{K}_{0}).$ (51)

As $\widetilde{K}_{0}$ has axial rotational symmetry around
$\mathbb{R}u_{\beta}$, the same holds for $\widetilde{\Xi}$. In particular,
there exist $\varrho>0$ and $s_{0}>0$ such that

$\displaystyle L_{\beta-1}\cap{\rm conv}\,\widetilde{\Xi}$ $\displaystyle=$
$\displaystyle L_{\beta-1}\cap\varrho B^{n},$ $\displaystyle Z$
$\displaystyle\subset$ $\displaystyle{\rm conv}\,\widetilde{\Xi}\mbox{ \ for
the cylinder $(L_{\beta-1}\cap\varrho
B^{n})+[-s_{0}u_{\beta},s_{0}u_{\beta}]$,}$ $\displaystyle({\rm
conv}\,\widetilde{\Xi})\setminus Z$ $\displaystyle\subset$
$\displaystyle\widetilde{\Xi}.$

Since $\widetilde{\Xi}\cap Z$ is the union of two cones with height $s_{0}$,
using (51) we deduce that

$\frac{\lambda_{\beta}(\widetilde{\Xi})}{V_{\beta}(\widetilde{K}_{0})}\geq\frac{\lambda_{\beta}(\widetilde{\Xi})}{2^{\beta-1}V_{\beta}({\rm
conv}\,\widetilde{\Xi})}=\frac{1}{2^{\beta-1}}\frac{\lambda_{\beta}(\widetilde{\Xi}\setminus
Z)+\lambda_{\beta}(\widetilde{\Xi}\cap Z)}{V_{\beta}(\widetilde{\Xi}\setminus
Z)+V_{\beta}(Z)}\geq\frac{\lambda_{\beta}(\widetilde{\Xi}\cap
Z)}{2^{\beta-1}V_{\beta}(Z)}=\frac{1}{\beta\,2^{\beta-1}}.$

It follows from (4) and $\beta\leq n-1$ that

$V_{\beta}(\widetilde{K})\geq\left(1+\frac{1}{\beta\,2^{\beta-1}16n^{4}}\cdot\frac{r_{\beta+1}(K)^{2}}{r_{\beta}(K)^{2}}\right)V_{\beta}(\widetilde{K}_{0})\geq\left(1+\frac{1}{2^{n+2}n^{5}}\cdot\frac{r_{\beta+1}(K)^{2}}{r_{\beta}(K)^{2}}\right)V_{\beta}(\widetilde{K}_{0}).$

We conclude (42), and in turn Proposition 4.3. ∎

######  Proposition 4.5.

Let $K_{1}$ be a compact convex set in $\mathbb{R}^{n}$, let
$Z_{2},\ldots,Z_{m}$, $2\leq m\leq n$, be zonoids in $\mathbb{R}^{n}$, and let
$\alpha_{1},\ldots,\alpha_{m}\in\mathbb{N}$ be such that
$\alpha_{1}+\cdots+\alpha_{m}=n$. Let $\varepsilon\in[0,1]$. Suppose that

$\displaystyle 0$
$\displaystyle<V_{\alpha_{1}}(K_{1})\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i})\leq(1+\varepsilon)\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],Z_{2}[\alpha_{2}],\ldots,Z_{m}[\alpha_{m}]).$
(52)

Then there exists an affine subspace $A_{1}\in\operatorname{A}(n,\alpha_{1})$
such that

$K_{1}\subset A_{1}+c\sqrt{\varepsilon}\,r_{\alpha_{1}}(K_{1})B^{n}\quad\mbox{
and }\quad r_{\alpha_{1}}(K_{1}|A_{1})\geq\frac{r_{\alpha_{1}}(K_{1})}{n},$

where $c=2^{\frac{n+2}{2}}n^{\frac{7}{2}}$, and $A_{1}$ can be chosen as a
linear subspace if $K_{1}$ is a zonoid.

###### Proof.

Let $L_{1}\in\operatorname{G}(n,\alpha_{1})$ be such that

$V_{\alpha_{1}}(K_{1}|L_{1})=\max\left\\{V_{\alpha_{1}}(K_{1}|L):L\in\operatorname{G}(n,\alpha_{1})\right\\}.$

A special case of (3), (13) and (12) imply that

$\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],Z_{2}[\alpha_{2}],\ldots,Z_{m}[\alpha_{m}])\leq
V_{\alpha_{1}}(K_{1}|L_{1})\cdot V_{\alpha_{2}}(Z_{2})\cdots
V_{\alpha_{m}}(Z_{\alpha_{m}}).$

It follows from the assumption (52) in Proposition 4.5 that
$V_{\alpha_{1}}(K_{1})\leq(1+\varepsilon)V_{\alpha_{1}}(K_{1}|L_{1})$, and
hence Proposition 4.3 yields

$r_{\alpha_{1}+1}(K_{1})\leq
2^{\frac{n+2}{2}}n^{\frac{5}{2}}r_{\alpha_{1}}(K_{1})\sqrt{\varepsilon}.$

From Lemma 4.2 we deduce the existence of an affine subspace
$A_{1}\in\operatorname{A}(n,\alpha_{1})$ such that

$K_{1}\subset
A_{1}+2^{\frac{n+2}{2}}n^{\frac{7}{2}}r_{\alpha_{1}}(K_{1})\sqrt{\varepsilon}\,B^{n}\quad\mbox{
and }\quad r_{\alpha_{1}}(K_{1}|A_{1})\geq\frac{r_{\alpha_{1}}(K_{1})}{n},$

where we can choose $A_{1}\in\operatorname{G}(n,\alpha_{1})$ if $K_{1}$ is a
zonoid. ∎

######  Lemma 4.6.

Let $2\leq m\leq n$, let $K_{1},\ldots,K_{m}$ be compact, convex sets in
$\mathbb{R}^{n}$, and let $\alpha_{1},\ldots,\alpha_{m}\geq 1$ be integers
satisfying $\alpha_{1}+\cdots+\alpha_{m}=n$. Let $\varepsilon\in[0,1]$. Assume
that for $i=1,\ldots,m$ there exist $q_{i}\in K_{i}$ and linear subspace
$L_{i}\in\operatorname{G}(n,\alpha_{i})$ such that $K_{i}|L_{i}$ is a zonoid
for $i=2,\ldots,m$,

$K_{i}-q_{i}\subset(K_{i}|L_{i})+\varepsilon
r_{\alpha_{i}}(K_{i})\,B^{n}\quad\mbox{ and }\quad
r_{\alpha_{i}}(K_{i}|L_{i})\geq\frac{r_{\alpha_{i}}(K_{i})}{n}>0,$ (53)

and $K_{1},\ldots,K_{m}$ satisfy

$\displaystyle\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])$
$\displaystyle\qquad\qquad\leq\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\leq(1+\varepsilon)\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}]).$
(54)

Then

$[L_{1},\ldots,L_{m}]\geq 1-n^{5}\varepsilon.$

###### Proof.

If $\varepsilon\geq n^{-4}$, then $1-n^{5}\varepsilon\leq-1$ and hence there
is nothing to show. In the following, we may assume that $\varepsilon<n^{-4}$.

By assumption, the sets $C_{i}=K_{i}|L_{i}$ are zonoids for $i=2,\ldots,m$.
Using first (4.6) and then (53) we get

$\displaystyle\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})$
$\displaystyle\leq(1+\varepsilon)\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])$
$\displaystyle\leq\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\ldots,K_{m}[\alpha_{m}])+\varepsilon\cdot\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})$
$\displaystyle\leq\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V((C_{1}+\varepsilon\,r_{\alpha_{1}}(K_{1})\,B^{n})[\alpha_{1}],\ldots,(C_{m}+\varepsilon\,r_{\alpha_{m}}(K_{m})\,B^{n})[\alpha_{m}])$
$\displaystyle\quad+\varepsilon\cdot\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})$
$\displaystyle=\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(C_{1}[\alpha_{1}],\ldots,C_{m}[\alpha_{m}])+\Omega+\varepsilon\cdot\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})$
$\displaystyle=\left(\prod_{i=1}^{m}V_{\alpha_{i}}(C_{i})\right)[L_{1},\ldots,L_{m}]+\Omega+\varepsilon\cdot\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i}),$
(55)

where with

$\Gamma=\\{(\gamma_{1},\ldots,\gamma_{m}):\,\gamma_{i}\in\\{0,\ldots,\alpha_{i}\\}\mbox{
for }i=1,\ldots,m\mbox{ and }\sum_{i=1}^{m}\gamma_{i}\geq 1\\}$

and by the linearity of mixed volumes we have

$\displaystyle\Omega$
$\displaystyle=\binom{n}{\alpha_{1},\ldots,\alpha_{m}}\big{[}V((C_{1}+\varepsilon\,r_{\alpha_{1}}(K_{1})\,B^{n})[\alpha_{1}],\ldots,(C_{m}+\varepsilon\,r_{\alpha_{m}}(K_{m})\,B^{n})[\alpha_{m}])$
$\displaystyle\qquad\qquad\qquad\qquad-V(C_{1}[\alpha_{1}],\ldots,C_{m}[\alpha_{m}])\big{]}$
$\displaystyle=\binom{n}{\alpha_{1},\ldots,\alpha_{m}}\sum_{(\gamma_{1},\ldots,\gamma_{m})\in\Gamma}\varepsilon^{\sum_{i=1}^{m}\gamma_{i}}\left(\prod_{i=1}^{m}r_{\alpha_{i}}(K_{i})^{\gamma_{i}}\right)\prod_{i=1}^{m}\binom{\alpha_{i}}{\gamma_{i}}$
$\displaystyle\qquad\times
V\left(C_{1}[\alpha_{1}-\gamma_{1}],\ldots,C_{m}[\alpha_{m}-\gamma_{m}],B^{n}\left[\gamma_{1}+\cdots+\gamma_{m}\right]\right).$
(56)

For each $(\gamma_{1},\ldots,\gamma_{m})\in\Gamma$, it follows from Theorem
1.3 (note that $K$ in the statement of Theorem 1.3 need not be a zonoid, hence
here we do not have to assume that $C_{1}$ is a zonoid) that

$\displaystyle
V\left(C_{1}[\alpha_{1}-\gamma_{1}],\ldots,C_{m}[\alpha_{m}-\gamma_{m}],B^{n}\left[\gamma_{1}+\cdots+\gamma_{m}\right]\right)$
$\displaystyle\qquad\leq\binom{n}{\alpha_{1}-\gamma_{1},\ldots,\alpha_{m}-\gamma_{m},\gamma_{1}+\cdots+\gamma_{m}}^{-1}\kappa_{\gamma_{1}+\cdots+\gamma_{m}}\left(\prod_{i=1}^{m}V_{\alpha_{i}-\gamma_{i}}(C_{i})\right).$
(57)

Using the notation $B_{i}^{\alpha_{i}}=B^{n}\cap L_{i}$ for $i=1,\ldots,m$, it
follows from (53) that
$p_{i}+\frac{r_{\alpha_{i}}(K_{i})}{n}\,B_{i}^{\alpha_{i}}\subset C_{i}$ for
some $p_{i}\in C_{i}$. Therefore, using Lemma 3.1 at the end, we have

$\displaystyle
r_{\alpha_{i}}(K_{i})^{\gamma_{i}}V_{\alpha_{i}-\gamma_{i}}(C_{i})$
$\displaystyle=\frac{\binom{\alpha_{i}}{\gamma_{i}}}{\kappa_{\gamma_{i}}}\cdot
r_{\alpha_{i}}(K_{i})^{\gamma_{i}}v^{(\alpha_{i})}(C_{i}[\alpha_{i}-\gamma_{i}],B_{i}^{\alpha_{i}}[\gamma_{i}])$
$\displaystyle=\frac{n^{\gamma_{i}}\binom{\alpha_{i}}{\gamma_{i}}}{\kappa_{\gamma_{i}}}\cdot
v^{(\alpha_{i})}\left(C_{i}[\alpha_{i}-\gamma_{i}],\frac{r_{\alpha_{i}}(K_{i})}{n}\,B_{i}^{\alpha_{i}}[\gamma_{i}]\right))$
$\displaystyle\leq\frac{n^{\gamma_{i}}\binom{\alpha_{i}}{\gamma_{i}}}{\kappa_{\gamma_{i}}}\cdot
v^{(\alpha_{i})}\left(C_{i}[\alpha_{i}-\gamma_{i}],C_{i}[\gamma_{i}]\right)$
$\displaystyle=\frac{n^{\gamma_{i}}\binom{\alpha_{i}}{\gamma_{i}}}{\kappa_{\gamma_{i}}}\cdot
V_{\alpha_{i}}(C_{i})\leq\frac{n^{\gamma_{i}}\binom{\alpha_{i}}{\gamma_{i}}}{\kappa_{\gamma_{i}}}\cdot
V_{\alpha_{i}}(K_{i}).$ (58)

In the following, we use the rough upper bounds

$\binom{\sum_{i=1}^{m}\gamma_{i}}{\gamma_{1},\ldots,\gamma_{m}}\leq
n^{\sum_{i=1}^{m}\gamma_{i}},\qquad\frac{\kappa_{\sum_{i=1}^{m}\gamma_{i}}}{\prod_{i=1}^{m}\kappa_{\gamma_{i}}}\leq
1$

and $t\leq e^{t}-1\leq 2t$, where the right inequality holds for $t\in[0,1)$.
Combining these facts with (56), (4), (58), we thus obtain for $n\geq 3$ that

$\displaystyle\Omega$
$\displaystyle\leq\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\sum_{(\gamma_{1},\ldots,\gamma_{m})\in\Gamma}\frac{\binom{n}{\alpha_{1},\ldots,\alpha_{m}}}{\binom{n}{\alpha_{1}-\gamma_{1},\ldots,\alpha_{m}-\gamma_{m},\gamma_{1}+\cdots+\gamma_{m}}}\frac{\kappa_{\sum_{i=1}^{m}\gamma_{i}}}{\prod_{i=1}^{m}\kappa_{\gamma_{i}}}n^{\sum_{i=1}^{m}\gamma_{i}}\prod_{i=1}^{m}\binom{\alpha_{i}}{\gamma_{i}}^{2}\varepsilon^{\sum_{i=1}^{m}\gamma_{i}}$
$\displaystyle\leq\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\sum_{(\gamma_{1},\ldots,\gamma_{m})\in\Gamma}\binom{\gamma_{1}+\cdots+\gamma_{m}}{\gamma_{1},\ldots,\gamma_{m}}n^{\sum_{i=1}^{m}\gamma_{i}}\varepsilon^{\sum_{i=1}^{m}\gamma_{i}}\prod_{i=1}^{m}\binom{\alpha_{i}}{\gamma_{i}}$
$\displaystyle\leq\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\sum_{(\gamma_{1},\ldots,\gamma_{m})\in\Gamma}\left(n^{2}\varepsilon\right)^{\sum_{i=1}^{m}\gamma_{i}}\prod_{i=1}^{m}\binom{\alpha_{i}}{\gamma_{i}}$
$\displaystyle=\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\left[\prod_{i=1}^{m}\left(1+n^{2}\varepsilon\right)^{\alpha_{i}}-1\right]$
$\displaystyle\leq\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\left[\exp\left(n^{2}\varepsilon(\alpha_{1}+\cdots+\alpha_{m})\right)-1\right]=\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\left(\exp\left(n^{3}\varepsilon\right)-1\right)$
$\displaystyle\leq\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot
2n^{3}\varepsilon\leq(n^{5}-1)\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\cdot\varepsilon,$
(59)

where $n\geq 2$ was used in the last step.

Substituting (4) into (55), for $\varepsilon<n^{-4}$ we get

$\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\leq\left(\prod_{i=1}^{m}V_{\alpha_{i}}(C_{i})\right)[L_{1},\ldots,L_{m}]+n^{5}\left(\prod_{i=1}^{m}V_{\alpha_{i}}(K_{i})\right)\cdot\varepsilon.$

Since $V_{\alpha_{i}}(C_{i})\leq V_{\alpha_{i}}(K_{i})$, according to Lemma
3.1, and $\dim(K_{i})\geq\alpha_{i}$, $i=1,\ldots,m$, we deduce that

$1-n^{5}\varepsilon\leq[L_{1},\ldots,L_{m}],$

which completes the argument. ∎

###### Proof of Theorem 1.6.

We deduce from the assumptions and by repeated application of Proposition 4.5
that for $i=1,\ldots,m$, there exist linear subspaces
$L_{i}\in\operatorname{G}(n,\alpha_{i})$ such that

$Z_{i}\subset(Z_{i}|L_{i})+cr_{\alpha_{i}}(Z_{i})\sqrt{\varepsilon}\,B^{n}\mbox{
and }r_{\alpha_{i}}(Z_{i}|L_{i})\geq\frac{r_{\alpha_{i}}(Z_{i})}{n}>0\mbox{ \
for $c=2^{\frac{n}{2}}n^{\frac{9}{2}}$}.$ (60)

It follows from (60) and Lemma 4.6 that

$[L_{1},\ldots,L_{m}]\geq
1-n^{5}2^{\frac{n}{2}}n^{\frac{9}{2}}\sqrt{\varepsilon}\geq
1-n^{10}2^{\frac{n}{2}}\sqrt{\varepsilon},$

if $2^{\frac{n}{2}}n^{\frac{9}{2}}\sqrt{\varepsilon}\leq 1$. But if this is
not satisfied, then the requested lower bound is trivially satisfied. Thus we
conclude Theorem 1.6. $\Box$ ∎

##  5 The stability version of Theorem 1.3

This section is devoted to the proof of the following theorem.

######  Theorem 5.1.

Let $\varepsilon\in[0,n^{-16}2^{-2n}]$. Let $K_{1}\subset\mathbb{R}^{n}$ be a
compact convex set and let $Z_{2},\ldots,Z_{m}\subset\mathbb{R}^{n}$ be
zonoids for $2\leq m\leq n$. If $\alpha_{1},\ldots,\alpha_{m}\geq 1$ are
integers with $\alpha_{1}+\cdots+\alpha_{m}=n$ and

$0<V_{\alpha_{1}}(K_{1})\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i})\leq(1+\varepsilon)\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],Z_{2}[\alpha_{2}],\ldots,Z_{{m}}[\alpha_{m}]),$

then there exist linear subspace $L_{i}\in\operatorname{G}(n,\alpha_{i})$ for
$i=1,\ldots,m$ and $q\in K_{1}$ such that

$[L_{1},\ldots,L_{m}]\geq 1-n^{13}2^{2n+4}\varepsilon^{\frac{1}{8}}$

and

$\displaystyle K_{1}-q$ $\displaystyle\subset$ $\displaystyle
L_{1}+n^{\frac{7}{2}}2^{\frac{n+2}{2}}\,r_{\alpha_{1}}(K_{1})\,\varepsilon^{\frac{1}{2}}\,B^{n},$
$\displaystyle Z_{i}$ $\displaystyle\subset$ $\displaystyle
L_{i}+n^{8}2^{2n+4}\,r_{\alpha_{i}}(Z_{i})\,\varepsilon^{\frac{1}{8}}\,B^{n},\quad
i=2,\ldots,m.$

We prepare the proof of Theorem 5.1 with two observations, Lemma 5.2 and Lemma
5.3, where the first one is based on basic properties of the $\Gamma$ function
$\Gamma$.

######  Lemma 5.2.

If $n\geq 2$ and $j=1,\ldots,n$, then

$\frac{V_{j}(B^{n})}{\kappa_{j}}\leq 2^{\frac{n}{2}}.$

###### Proof.

We repeatedly use the duplication formula for the Gamma function

$\frac{\Gamma(2x)}{\Gamma(x)}=\frac{2^{2x-1}}{\sqrt{\pi}}\cdot\Gamma\left(x+\frac{1}{2}\right)\quad\mbox{
for $x>0$}.$

By the logarithmic convexity of the Gamma function, which yields

$\Gamma\left(\frac{x+y}{2}\right)\leq\sqrt{\Gamma(x)\Gamma(y)}\quad\text{ for
$x,y>0$,}$

the functional equation $\Gamma(x+1)=x\Gamma(x)$,
$\binom{n}{j}=\frac{\Gamma(n+1)}{\Gamma(n-j+1)\Gamma(j+1)}$ and
$\kappa_{d}=\frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+1)}$, we get

$\displaystyle\frac{V_{j}(B^{n})}{\kappa_{j}}$
$\displaystyle=\frac{\binom{n}{j}\kappa_{n}}{\kappa_{j}\kappa_{n-j}}=\frac{\Gamma(n+1)}{\Gamma(\frac{n+2}{2})}\cdot\frac{\Gamma(\frac{j+2}{2})}{\Gamma(j+1)}\cdot\frac{\Gamma(\frac{n-j+2}{2})}{\Gamma(n-j+1)}$
$\displaystyle=\sqrt{\pi}\cdot\frac{\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(\frac{n-j}{2}+\frac{1}{2})\Gamma(\frac{j}{2}+\frac{1}{2})}\leq\sqrt{\pi}\cdot\frac{\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(\frac{n}{4}+\frac{1}{2})^{2}}$
$\displaystyle=\frac{\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(\frac{n}{2}+1)}\cdot\frac{\Gamma(\frac{n}{4}+1)}{\Gamma(\frac{n}{4}+\frac{1}{2})}\cdot\frac{\sqrt{\pi}\cdot\Gamma(\frac{n}{2}+1)}{\Gamma(\frac{n}{4}+\frac{1}{2})\Gamma(\frac{n}{4}+1)}=2^{\frac{n}{2}}\cdot\frac{\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(\frac{n}{2}+1)}\cdot\frac{\Gamma(\frac{n}{4}+1)}{\Gamma(\frac{n}{4}+\frac{1}{2})}$
$\displaystyle\leq
2^{\frac{n}{2}}\cdot\frac{\sqrt{\Gamma(\frac{n}{2})\Gamma(\frac{n}{2}+1)}}{\Gamma(\frac{n}{2}+1)}\cdot\frac{\sqrt{\Gamma(\frac{n}{4}+\frac{1}{2})\Gamma(\frac{n}{4}+\frac{3}{2})}}{\Gamma(\frac{n}{4}+\frac{1}{2})}=2^{\frac{n-1}{2}}\sqrt{\frac{n+2}{n}}\leq
2^{\frac{n}{2}},$

which proves Lemma 5.2. ∎

######  Lemma 5.3.

If $1\leq\alpha<d\leq n$, $0<\eta<\frac{1}{\alpha n}$ and $M$ is a compact
convex set in $\mathbb{R}^{d}$ such that there exists an affine subspace
$A\in\operatorname{A}(n,\alpha)$ such that

$M\subset A+\eta r_{\alpha}(M)\,B^{d}\quad\mbox{ and }\quad
r_{\alpha}(M|A)\geq\frac{r_{\alpha}(M)}{n},$

then

$V_{\alpha}(M)\leq(1+n2^{d+1}\eta)V_{\alpha}(M|A).$

###### Proof.

We may assume that $o\in A$. We set $C=M|A$ and $B^{\alpha}=A\cap B^{d}$, and
as before we write $v^{(\alpha)}(\cdot)$ to denote the mixed volume in $A$.

Lemma 5.3 is equivalent to the inequality

$V(M[\alpha],B^{d}[d-\alpha])\leq(1+n2^{d+1}\eta)V(C[\alpha],B^{d}[d-\alpha]).$

Since $B^{d}$ is a zonoid, using Theorem 1.3 for the second and
$r_{\alpha}(M)\leq nr_{\alpha}(C)$ for the third inequality, we obtain that

$\displaystyle V(M[\alpha],B^{d}[d-\alpha])$ $\displaystyle\leq V((C+\eta
r_{\alpha}(M)B^{d})[\alpha],B^{d}[d-\alpha])$
$\displaystyle=V(C[\alpha],B^{d}[d-\alpha])+\sum_{j=1}^{\alpha}\binom{\alpha}{j}(\eta
r_{\alpha}(M))^{j}V(C[\alpha-j],B^{d}[j],B^{d}[d-\alpha])\allowdisplaybreaks$
$\displaystyle\leq
V(C[\alpha],B^{d}[d-\alpha])+\sum_{j=1}^{\alpha}\frac{\binom{\alpha}{j}}{\binom{d}{\alpha-j,j,d-\alpha}}(\eta
r_{\alpha}(M))^{j}\kappa_{j}V_{\alpha-j}(C)V_{d-\alpha}(B^{d})\allowdisplaybreaks$
$\displaystyle\leq
V(C[\alpha],B^{d}[d-\alpha])+V_{d-\alpha}(B^{d})\sum_{j=1}^{\alpha}\frac{\binom{\alpha}{j}^{2}}{\binom{d}{\alpha-j,j,d-\alpha}}(\eta
nr_{\alpha}(C))^{j}v^{(\alpha)}(C[\alpha-j],B^{\alpha}[j])\allowdisplaybreaks$
$\displaystyle=V(C[\alpha],B^{d}[d-\alpha])+\frac{V_{d-\alpha}(B^{d})}{\binom{d}{\alpha}}\sum_{j=1}^{\alpha}\binom{\alpha}{j}(\eta
n)^{j}v^{(\alpha)}(C[\alpha-j],r_{\alpha}(C)B^{\alpha}[j])\allowdisplaybreaks$
$\displaystyle\leq
V(C[\alpha],B^{d}[d-\alpha])+\frac{V_{d-\alpha}(B^{d})}{\binom{d}{\alpha}}V_{\alpha}(C)\left[(1+\eta
n)^{\alpha}-1\right],$

where for the final inequality we used that a translate of
$r_{\alpha}(C)B^{\alpha}$ is contained in $C$.

Since $n\eta<\frac{1}{\alpha}$ implies that
$(1+n\eta)^{\alpha}<1+2\alpha\,n\eta$, and

$\frac{V_{d-\alpha}(B^{d})}{\binom{d}{\alpha}}V_{\alpha}(C)=\frac{V_{d-\alpha}(B^{d})}{\kappa_{d-\alpha}}V(C[\alpha],B^{d}[d-\alpha])\leq
2^{\frac{d}{2}}V(C[\alpha],B^{d}[d-\alpha]),$

where also Lemma 5.2 was used, we finally get

$\displaystyle V(M[\alpha],B^{d}[d-\alpha])$ $\displaystyle\leq
V(C[\alpha],B^{d}[d-\alpha])\left(1+2^{\frac{d}{2}}\cdot 2\alpha n\eta\right)$
$\displaystyle\leq\left(1+n2^{d+1}\eta\right)V(C[\alpha],B^{d}[d-\alpha]),$

which proves the lemma. ∎

###### Proof of Theorem 5.1.

According to Proposition 4.5, possibly after translating $K_{1}$, there exists
a linear subspace $L_{1}\in\operatorname{G}(n,\alpha_{1})$ such that

$K_{1}\subset L_{1}+c\sqrt{\varepsilon}r_{\alpha_{1}}(K_{1})B^{n}\quad\mbox{
and }\quad r_{\alpha_{1}}(K_{1}|L_{1})\geq r_{\alpha_{1}}(K_{1})/n$ (61)

where $c=2^{\frac{n+2}{2}}n^{\frac{7}{2}}$. We set

$C_{1}=K_{1}|L_{1}\quad\mbox{ and }\quad B^{\alpha_{1}}=L_{1}\cap B^{n}.$

Note that (61) implies that

$K_{1}\subset C_{1}+c\sqrt{\varepsilon}r_{\alpha_{1}}(K_{1})B^{n}.$ (62)

For $\mathcal{Z}=(Z_{2}[\alpha_{2}],\ldots,Z_{{m}}[\alpha_{m}])$, we deduce
from (62) that

$\displaystyle V(K_{1}[\alpha_{1}],\mathcal{Z})$ $\displaystyle\leq
V((C_{1}+c\sqrt{\varepsilon}r_{\alpha_{1}}(K_{1})B^{n})[\alpha_{1}],\mathcal{Z})$
$\displaystyle=V(C_{1}[\alpha_{1}],\mathcal{Z})+\sum_{j=1}^{\alpha_{1}}\binom{\alpha_{1}}{j}(c\sqrt{\varepsilon}r_{\alpha_{1}}(K_{1}))^{j}V(C_{1}[\alpha_{1}-j],B^{n}[j],\mathcal{Z}).$
(63)

Next we provide several upper and lower bounds for the individual expressions
involved in (63). The conditions in Theorem 5.1 and
$1-\varepsilon\leq(1+\varepsilon)^{-1}$ yield

$(1-\varepsilon)V_{\alpha_{1}}(K_{1})\prod_{i=1}^{m}V_{\alpha_{i}}(Z_{i})\leq\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(K_{1}[\alpha_{1}],\mathcal{Z}).$
(64)

Setting $\beta=n-\alpha_{1}$, it follows from an application of [49, Theorem
5.3.1] that

$\displaystyle\binom{n}{\alpha_{1},\ldots,\alpha_{m}}V(C_{1}[\alpha_{1}],\mathcal{Z})$
$\displaystyle=V_{\alpha_{1}}(C_{1})\binom{\beta}{\alpha_{2},\ldots,\alpha_{m}}$
$\displaystyle\qquad\times
v^{(\beta)}\left((Z_{2}|L_{1}^{\bot})[\alpha_{1}],\ldots,(Z_{m}|L_{1}^{\bot})[\alpha_{m}]\right).$
(65)

An application of Theorem 1.3 yields that

$\binom{n}{\alpha_{1},\ldots,\alpha_{m}}\binom{\alpha_{1}}{j}V(C_{1}[\alpha_{1}-j],B^{n}[j],\mathcal{Z})\leq\kappa_{j}V_{\alpha_{1}-j}(C_{1})\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i}).$
(66)

In addition, using first (61), then Lemma 5.2 and finally the fact that a
translate of $r_{\alpha_{1}}(C_{1})B^{\alpha_{1}}$ lies in $C_{1}$, we get

$\displaystyle\binom{\alpha_{1}}{j}^{-1}r_{\alpha_{1}}(K_{1})^{j}\kappa_{j}V_{\alpha_{1}-j}(C_{1})$
$\displaystyle\leq n^{j}\cdot r_{\alpha_{1}}(C_{1})^{j}\cdot
v^{(\alpha_{1})}(C_{1}[\alpha_{1}-j],B^{\alpha_{1}}[j])$
$\displaystyle=n^{j}v^{(\alpha_{1})}(C_{1}[\alpha_{1}-j],r_{\alpha_{1}}(C_{1})B^{\alpha_{1}}[j])$
$\displaystyle\leq n^{j}V_{\alpha_{1}}(C_{1}).$ (67)

Combining (66) and (67), we get

$\binom{n}{\alpha_{1},\ldots,\alpha_{m}}r_{\alpha_{1}}(K_{1})^{j}V(C_{1}[\alpha_{1}-j],B^{n}[j],\mathcal{Z})\leq
n^{j}V_{\alpha_{1}}(C_{1})\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i}).$ (68)

Substituting (64), (65) and (68) into (63) and using
$V_{\alpha_{1}}(C_{1})\leq V_{\alpha_{1}}(K_{1})$ by Lemma 3.1, we finally
arrive at (with $\beta=n-\alpha_{1}$)

$\displaystyle(1-\varepsilon)\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i})$
$\displaystyle\leq\binom{\beta}{\alpha_{2},\ldots,\alpha_{m}}v^{(\beta)}\left((Z_{2}|L_{1}^{\bot})[\alpha_{2}],\ldots,(Z_{m}|L_{1}^{\bot})[\alpha_{m}]\right)$
$\displaystyle\quad+\left(\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i})\right)\sum_{j=1}^{\alpha_{1}}\binom{\alpha_{1}}{j}(cn\sqrt{\varepsilon})^{j}.$

Since $\varepsilon\leq(5cn^{2})^{-2}$, we have
$cn\sqrt{\varepsilon}<\frac{1}{n}$, and hence

$\sum_{j=1}^{\alpha_{1}}\binom{\alpha_{1}}{j}(cn\sqrt{\varepsilon})^{j}=\left(1+cn\sqrt{\varepsilon}\right)^{\alpha_{1}}-1\leq
2cn^{2}\sqrt{\varepsilon}.$

Using that $1-\varepsilon-2cn^{2}\sqrt{\varepsilon}\geq
1-2.5\,cn^{2}\sqrt{\varepsilon}$, we deduce that

$0<(1-2.5\,cn^{2}\sqrt{\varepsilon})\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i})\leq\binom{\beta}{\alpha_{2},\ldots,\alpha_{m}}v^{(\beta)}\left((Z_{2}|L_{1}^{\bot})[\alpha_{2}],\ldots,(Z_{m}|L_{1}^{\bot})[\alpha_{m}]\right).$
(69)

In particular, this shows that $V_{\alpha_{i}}(Z_{i}|L_{1}^{\bot})>0$ for
$i=2,\ldots,m$. Moreover, from (69), $V_{\alpha_{i}}(Z_{i}|L_{1}^{\bot})\leq
V_{\alpha_{i}}(Z_{i})$ for $i=2,\ldots,m$ and
$(1-2.5\,cn^{2}\sqrt{\varepsilon})^{-1}\leq 1+5\,cn^{2}\sqrt{\varepsilon}$ we
now obtain that

$\displaystyle 0<\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i}|L_{1}^{\bot})$
$\displaystyle\leq\prod_{i=2}^{m}V_{\alpha_{i}}(Z_{i})$ (70)
$\displaystyle\leq(1+5cn^{2}\sqrt{\varepsilon})\binom{\beta}{\alpha_{2},\ldots,\alpha_{m}}v^{(\beta)}\left((Z_{2}|L_{1}^{\bot})[\alpha_{2}],\ldots,(Z_{m}|L_{1}^{\bot})[\alpha_{m}]\right).$
(71)

We set $\widetilde{Z}_{i}=Z_{i}|L_{1}^{\bot}$ for $i=2,\ldots,m$. Based on
(70) and (71), we apply Proposition 4.5 in $L_{1}^{\bot}$ if $m\geq 3$ (and
hence $2\leq m-1\leq\alpha_{2}+\cdots+\alpha_{m}=n-\alpha_{1}$), and directly
apply Proposition 4.3 and Lemma 4.2 in $L_{1}^{\bot}$ if $m=2$. Thus we obtain
$\alpha_{i}$-dimensional linear subspaces $\widetilde{L}_{i}\subset
L_{1}^{\bot}$ for $i=2,\ldots,m$ such that

$r_{\alpha_{i}}(\widetilde{Z}_{i}|\widetilde{L}_{i})\geq
r_{\alpha_{i}}(\widetilde{Z}_{i})/n\mbox{ \ \ and \ \
}\widetilde{Z}_{i}\subset\widetilde{L}_{i}+2^{n+2}n^{7}\varepsilon^{\frac{1}{4}}r_{\alpha_{i}}(\widetilde{Z}_{i})\,(B^{n}\cap
L_{1}^{\bot}),$

where we used that $c\sqrt{5cn^{2}\sqrt{\varepsilon}}\leq
2^{n+2}n^{7}\varepsilon^{\frac{1}{4}}$.

We deduce from Lemma 5.3 that if $i=2,\ldots,m$, then

$V_{\alpha}(\widetilde{Z}_{i})\leq(1+n2^{d+1}2^{n+2}n^{7}\varepsilon^{\frac{1}{4}})V_{\alpha}(\widetilde{Z}_{i}|\widetilde{L}_{i})\leq(1+n^{8}2^{2n+3}\varepsilon^{\frac{1}{4}})V_{\alpha}(\widetilde{Z}_{i}|\widetilde{L}_{i}).$
(72)

On the other hand,

$V_{\alpha_{i}}(Z_{i})\leq(1+5cn^{2}\sqrt{\varepsilon})V_{\alpha_{i}}(\widetilde{Z}_{i}),\quad
i=2,\ldots,m,$ (73)

follows from (71), Theorem 1.3 and again from
$V_{\alpha_{j}}(\widetilde{Z}_{j})\leq V_{\alpha_{j}}(Z_{j})$ for
$j=2,\ldots,m$.

From $\varepsilon\leq(5cn^{2})^{-2}$, (72), (73) and
$\widetilde{Z}_{i}|\widetilde{L}_{i}=Z_{i}|\widetilde{L}_{i}$ for
$i=2,\ldots,m$ (note that $\widetilde{L}_{i}\subset L_{1}^{\bot}$) we deduce
that

$\displaystyle V_{\alpha_{i}}(Z_{i})$
$\displaystyle\leq(1+5cn^{2}\sqrt{\varepsilon})V_{\alpha_{i}}(\widetilde{Z}_{i})\leq(1+5cn^{2}\sqrt{\varepsilon})(1+n^{8}2^{2n+3}\varepsilon^{\frac{1}{4}})V_{\alpha}(\widetilde{Z}_{i}|\widetilde{L}_{i})$
$\displaystyle\leq(1+n^{8}2^{2n+5}\varepsilon^{\frac{1}{4}})V_{\alpha_{i}}(Z_{i}|\widetilde{L}_{i}).$

It follows from Proposition 4.3 that

$r_{\alpha_{i}+1}(Z_{i})\leq\sqrt{2^{n+2}n^{5}n^{8}2^{2n+5}\varepsilon^{\frac{1}{4}}}\cdot
r_{\alpha_{i}}(Z_{i})\leq n^{7}2^{2n+4}\varepsilon^{\frac{1}{8}}\cdot
r_{\alpha_{i}}(Z_{i}),$

and in turn, Lemma 4.2 (ii) yields that there exists a linear subspace
$L_{i}\in\operatorname{G}(n,\alpha_{i})$ such that

$Z_{i}\subset
L_{i}+n^{8}2^{2n+4}\varepsilon^{\frac{1}{8}}r_{\alpha_{i}}(Z_{i})\cdot
B^{n}\mbox{ \ \ and \ \ }r(Z_{i}|L_{i})\geq r_{\alpha_{i}}(Z_{i})/n>0.$ (74)

Given (61) and (74), we conclude

$[L_{1},\ldots,L_{m}]\geq 1-n^{5}\cdot
n^{8}2^{2n+4}\varepsilon^{\frac{1}{8}}=1-n^{13}2^{2n+4}\varepsilon^{\frac{1}{8}}$

by Lemma 4.6. ∎

Acknowledgement. K. Böröczky was supported by research grant NKFIH 132002.

Daniel Hug was supported by research grant HU 1874/5-1 (DFG).

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