On Godbersen’s conjecture

Abstract

We provide a natural generalization of a geometric conjecture of Fáry and Rédei regarding the volume of the convex hull of $K \subset {\mathbb {R}}^n$, and its negative image $-K$. We show that it implies Godbersen’s conjecture regarding the mixed volumes of the convex bodies $K$ and $-K$. We then use the same type of reasoning to produce the currently best known upper bound for the mixed volumes $V(K[j], -K[n-j])$, which is not far from Godbersen’s conjectured bound. To this end we prove a certain functional inequality generalizing Colesanti’s difference function inequality.

Appendix

We give here another proof of Theorem 1.6. The proof is attained by essentially repeating the arguments from [12], but instead of considering $K$ and $-K$, we consider two general convex bodies $K$ and $L$. Moreover, we will be able to characterize the equality case in Theorem 1.5.

1.1 The Rogers–Shephard body

We consider the following $(2n+1)$-dimensional body (see Fig. 2), a special case of which, where $K=L$, plays a central role in [12].

$$\begin{aligned} G{(K,L)}:=\left\{ \left( x,y,\theta \right) \in {\mathbb {R}}^{n}\times {\mathbb {R}}^{n}\times {\mathbb {R}}\,|\,\theta \in \left[ 0,1\right] ,\quad x\in \theta K,\quad x+y\in (1-\theta ) L\right\} . \end{aligned}$$

Fig. 2

The section of $G{(K,L)}$ where $\theta =\theta _0,\,y=0$, is an intersection of two cylinders

The projection of $G{(K,L)}$ onto the $\left( n+1\right)$-dimensional subspace of points of the form $\left( 0,y,\theta \right)$ is denoted by $C(K,L) =\{ (0,y,\theta ) \ | \ \theta \in [0,1], \ y \in (1-\theta ) L - \theta K \}$. Equivalently,

$$\begin{aligned} C(K,L)=\{0\}\times {\mathrm{conv}}\left( L\times \left\{ 0\right\} ,\quad -K\times \left\{ 1\right\} \right) \subseteq \{0\}\times {\mathbb {R}}^{n+1}. \end{aligned}$$

(13)

When $K=L$, this is exactly the body $C(K)$ used in [12].

The main tool used by Rogers and Shephard in [12] for finding an upper bound for the volume of the difference body is the following theorem which we provide along with its proof, for the convenience of the reader.

Theorem 5.1

(Rogers and Shephard) Let $K\subset {\mathbb {R}}^n$ be a convex body, and let $H=K\cap E$ be a $j$-dimensional section of $K$, and $L$ the orthogonal projection of $K$ onto $E^\perp$. Then

$$\begin{aligned} \frac{j!\left( n-j\right) !}{n!}\mathrm{Vol}_{j}\left( H\right) \mathrm{Vol}_{n-j}\left( L\right) \le \mathrm{Vol}_{n}\left( K\right) . \end{aligned}$$

(14)

Moreover, equality holds if and only if for every direction $v\in E^\perp$, the intersection of $K$ with $E+{\mathbb {R}}^+v$ is obtained by taking the convex hull of $H$ and one more point.

Proof of Theorem 5.1

The proof consists of two arguments. The first one is that all of the quantities in (14) are invariant under a Schwarz symmetrization, so we may assume that all intersections of $K$ parallel to $H$ are (centered) dilates of a ball. That is, we may consider the body:

$$\begin{aligned} K^*=\left\{ (l,y)\in {\mathbb {R}}^{n-j}\times {\mathbb {R}}^j\quad |\quad l\in L, \ |y|\le \left( \frac{\mathrm{Vol}_j(K_l)}{\mathrm{Vol}_j(B_2^j)} \right) ^{1/j} \right\} , \end{aligned}$$

where $K_l=K\cap (E+l)$, for every $l\in L$. The second argument is that $H^*\vee L \subseteq K^*$, where $H^*$ denotes the section $K^* \cap E$. Then, a simple computation shows that the left-hand side of (14) equals $\mathrm{Vol}_n(H^*\vee L)$, and thus inequality (14) holds.

Assume now that equality holds in (14). First note that $H^*\vee L\subseteq K^*$, thus equality in volumes implies $K^*=H^*\vee L=H^*\vee \partial L$. Moreover, for every direction $v\in E^{\perp }$, let $[0,l] = L\cap ({\mathbb {R}}^+v)$, and for every $t\in [0,1]$, let $f_v(t)=\mathrm{Vol}_{j} (K_{tl})^{1/j}$. Note that $f_v$ is invariant under Schwartz symmetrization, hence for every $v\in E^{\perp }$, $f_v$ is linear, and $f_v(1)=0$. Since $K_{tl}$ contains $t K_{l}+(1-t)H$, by the Brunn–Minkowski inequality one has

$$\begin{aligned} f_v(t)=\mathrm{Vol}_{j}(K_{tl})^{1/j} \ge t \mathrm{Vol}_{j}(K_{l})^{1/j} + (1-t) \mathrm{Vol}_{j}(H)^{1/j} = (1-t)f_v(0). \end{aligned}$$

Thus, from the equality case in the Brunn–Minkowski inequality, $K_l$ is a homothety of $H$ of zero volume, i.e. it is a point. Moreover, for every $t \in [0,1]$, one has $K_{tl} = t K_{l}+(1-t)H$, and thus $K \cap (E+{\mathbb {R}}^+v) = H \vee K_l$, and the proof is complete. $\square$

1.2 An upper bound for the volume of $C(K,L)$

Theorem 5.2

For convex bodies $K,L\subseteq {\mathbb {R}}^n$, let $C(K,L)$ be as defined in (13). For every $\theta \in [0,1]$,

$$\begin{aligned} \mathrm{Vol}_{n+1}\left( C\left( K,L\right) \right) \le \frac{1}{n+1} \left( \frac{\mathrm{Vol}_{n}\left( K\right) \mathrm{Vol}_{n}\left( L\right) }{\mathrm{Vol}_n(\theta K\cap (1-\theta ) L) } \right) . \end{aligned}$$

In [12] this bound was obtained for the specific case $K=L$ and $\theta =\frac{1}{2}$.

Proof of Theorem 5.2

In order to estimate $\mathrm{Vol}_{n+1}\left( C\left( K,L\right) \right)$, we apply Theorem 5.1 for the body $G{(K,L)}$, and the $n$-dimensional affine subspace $E= \{ \theta = \theta _0, y=0 \}$. First, by Fubini’s Theorem,

$$\begin{aligned} \mathrm{Vol}_{2n+1}\left( G{(K,L)}\right)= & {} \int _{0}^{1}d\theta \int _{\theta K}\mathrm{Vol}_n ((1-\theta )L )dx\\= & {} \mathrm{Vol}_n\left( K\right) \mathrm{Vol}_n\left( L\right) \int _{0}^{1}\theta ^n \left( 1-\theta \right) ^n d\theta \\= & {} \mathrm{Vol}_n\left( K\right) \mathrm{Vol}_n\left( L\right) \frac{n!n!}{\left( 2n+1\right) !}. \end{aligned}$$

As in Theorem 5.1, set $H = G{(K,L)} \cap E$. Note that $\mathrm{Vol}_n(H) =\mathrm{Vol}_n(\theta _0 K\cap (1-\theta _0)L)$. As mentioned before, the projection of $G{(K,L)}$ on $E^{\perp }$ is exactly $C\left( K,L\right)$, and using Theorem 5.1 we get

$$\begin{aligned} \frac{n!\left( n+1\right) !}{\left( 2n+1\right) !} \mathrm{Vol}_{n+1}\left( C\left( K,L\right) \right) \mathrm{Vol}_{n}\left( \theta _0 K\cap (1-\theta _0)L \right) \le \mathrm{Vol}_{2n+1}\left( G{(K,L)}\right) . \end{aligned}$$

(15)

Plugging in the volume of $G{(K,L)}$ completes the proof of the theorem. $\square$

1.3 A second proof of Theorem 1.6

Using Theorem 5.1, this time for the body $C(K,L)$, and the volume bound from Theorem 5.2, we can give yet another proof of Theorem 1.6.

Proof of Theorem 1.6

Let $K,L\subseteq {\mathbb {R}}^n$ be two convex bodies such that $0\in K\cap L$, and set $\theta \in [0,1]$. We need to show that

$$\begin{aligned} \mathrm{Vol}(L \vee -K) \, \mathrm{Vol}( \theta K\cap (1-\theta ) L) \le \mathrm{Vol}(K)\mathrm{Vol}(L). \end{aligned}$$

Let $E$ be the 1-dimensional subspace of ${\mathbb {R}}^{n+1}$ given by $E = \{ x =0 \}$. The body $L \vee -K$ is the $n$-dimensional projection of $C(K,L)$ onto the subspace $E^{\perp }= \{(x,0) \, | \, x\in {\mathbb {R}}^n\}$. Since $0 \in K \cap L$, the section $H = E \cap C(K,L)$ is a unit segment. By Theorem 5.1,

$$\begin{aligned} \frac{1}{n+1}\mathrm{Vol}_{n}(-K\vee L) \le \mathrm{Vol}_{n+1}(C(K,L)). \end{aligned}$$

(16)

Combining this with the volume bound for $C(K,L)$ established in Theorem 5.2 above, we get the desired inequality. $\square$

1.4 The equality case in Theorem 1.5 and in Theorem 1.6

Here we characterize the equality cases in Theorems 1.5 and 1.6. We start with the former, and show that equality holds in (7) if and only if $K$ is a simplex with a vertex at the origin. Indeed, it is not hard to check that for the standard simplex $S$ one has

$$\begin{aligned} \mathrm{Vol}((1-\lambda )S\vee -\lambda S)= \sum _{k=0}^n \left( {\begin{array}{c}n\\ k\end{array}}\right) (1-\lambda )^k \lambda ^{n-k}\mathrm{Vol}(S) =\mathrm{Vol}(S). \end{aligned}$$

As for the other direction, assume that $\mathrm{Vol}((1-\lambda )K\vee -\lambda K)= \mathrm{Vol}(K)$. We wish to show that $K$ is a simplex with 0 as one of its vertices. Recall from the introduction that Theorem 1.6, for $(1-\lambda )K, \lambda K$, and $\theta _0 = \lambda$, immediately yields inequality (7). Combining (15) and (16) in this case yields

$$\begin{aligned} \mathrm{Vol}_n(-(1-\lambda )K\vee \lambda K)\le (n+1) \mathrm{Vol}_{n+1}(C((1-\lambda )K,\lambda K))\le \mathrm{Vol}_n(K). \end{aligned}$$

(17)

From the assumption $\mathrm{Vol}((1-\lambda )K\vee -\lambda K)=\mathrm{Vol}(K)$ it follows that both inequalities in (17) are in fact equalities. In particular, equality holds in (15) for the bodies $(1-\lambda )K, \lambda K$, and $\theta _0 = \lambda$. By the equality condition in Theorem 5.1, this implies in particular that sections of the body $G{((1-\lambda )K,\lambda K)}$ by affine subspaces of the form $\{(x,y,\theta )| x\in {\mathbb {R}}^n, y=y_0, \theta =\lambda \}$, for any $y_0\in \lambda (1-\lambda )K-\lambda (1-\lambda )K$ (which are given by $\lambda (1-\lambda )K \cap \lambda (1-\lambda )K -y_0$), are homothetic. Thus, $K$ must be a simplex by the following lemma due to Rogers and Shephard.

Lemma 4 from [11]. Let $K$ be a convex body in ${\mathbb {R}}^n$. If the intersections $K \cap (K+x)$ are homothetic for all $x \in K - K$, then $K$ is a simplex.

Finally, in order to show that 0 is a vertex of $K$, note that equality holds also in (16), for the bodies $(1-\lambda )K, \lambda K$. Hence, among all sections of $C((1-\lambda )K, \lambda K)$ parallel to $H = E \cap C((1-\lambda )K, \lambda K)$, $H$ is the only one with unit length. Since we assumed that $K$ contains the origin, one has $(1-\lambda ) K \cap - \lambda K=\{0\}$. This, together with the fact that $K$ is a simplex, means that 0 must be one of the vertices of $K$.

We turn now to show that equality in Theorem 1.6 holds if and only if $K$ and $L$ are simplices with a common vertex at the origin, and such that $(1-\theta )L = \theta K$. To this end, we shall make use of Theorem 1.9. Assume that for some given $\theta \in (0,1)$,

$$\begin{aligned} \mathrm{Vol}(L \vee -K) \, \mathrm{Vol}(\theta K\cap (1-\theta ) L) = \mathrm{Vol}(K)\mathrm{Vol}(L). \end{aligned}$$

(18)

Since the inclusion $\theta K\cap (1-\theta ) L\subseteq \left( K^\circ +L^\circ \right) ^\circ$ always holds, by Theorem 1.9 we then must have

$$\begin{aligned} \theta K\cap (1-\theta ) L = \left( K^\circ +L^\circ \right) ^\circ \end{aligned}$$

(as both are compact convex sets, inclusion together with equality of volumes implies equality of sets). We claim that this equality implies that $K$ and $L$ are homothetic. Indeed, we may rewrite the above equality as

$$\begin{aligned} \theta ^{-1} K^\circ \vee (1-\theta )^{-1} L^\circ = K^\circ +L^\circ , \end{aligned}$$

so that in particular

$$\begin{aligned} \theta ^{-1} K^\circ \subset K^\circ +L^\circ , \qquad (1-\theta )^{-1} L^\circ \subset K^\circ +L^\circ . \end{aligned}$$

Thus, $\theta ^{-1}h_{K^\circ }\le h_{K^\circ }+h_{L^\circ }$, and $(1-\theta )^{-1}h_{L^\circ }\le h_{K^\circ }+h_{L^\circ }$. Putting the two together one has

$$\begin{aligned} h_{L^\circ } = \frac{1-\theta }{\theta }h_{K^\circ }, \end{aligned}$$

or equivalently, $(1-\theta )L = \theta K$. This, together with (18) implies that equality in (7) holds for $K$. Thus, from the characterization of the equality case in Theorem 1.5, the body $K$ must be a simplex with a vertex at the origin. This completes the proof for the equality case in Theorem 1.6.

Remark 5.3

It is worthwhile to notice that in Theorem 1.9 there are more equality cases than in Theorem 1.5. Indeed, one may readily check that for positive $\lambda _1,\dots ,\lambda _n$, and the bodies $K=\mathrm{conv}\{ 0, e_1, \ldots , e_n\}$, $L = \mathrm{conv}\{ 0, \lambda _1e_1, \ldots , \lambda _n\}$ we have that $\mathrm{Vol}(K)\mathrm{Vol}(L) = \frac{1}{n!^2}\prod _{i=1}^n \lambda _i$, that

$$\begin{aligned} \mathrm{Vol}\left( \left( K^\circ +L^\circ \right) ^\circ \right) = \frac{1}{n!}\prod _{i=1}^n\frac{\lambda _i}{1+\lambda _i} \end{aligned}$$

and that

$$\begin{aligned} \mathrm{Vol}(K \vee -L)= \frac{1}{n!}\sum _{A\subset \{1,\ldots n\} }\prod _{i\in A} \lambda _i = \frac{1}{n!}\prod _{i=1}^n (1+ \lambda _i). \end{aligned}$$

The characterization of the equality case in Theorem 1.9 is thus left open.