Proof By Fubini’s theorem, for any $p>0,$ we have

$$ \begin{align*} &p\int_0^{\infty}r^{p-1}\mathrm{vol}_n(K\cap(re_n+K))dr\\& \quad = p\int_0^{\infty}r^{p-1}\int_{P_{e_n^\perp}(K)}\mathrm{vol}_1(K\cap(re_n+K)\cap(y+\langle e_n\rangle))dydr\\& \quad = p\int_0^{\infty}r^{p-1}\int_{P_{e_n^\perp}(K)}\max\{\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))-r,0\}dydr\\& \quad = \int_{P_{e_n^\perp}(K)}\int_0^{\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))}pr^{p-1}(\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))-r)drdy\\& \quad = \int_{P_{e_n^\perp}(K)}\left((\mathrm{vol}_1(K\cap (y+\langle e_n\rangle)))^{p+1}-\frac{p}{p+1}(\mathrm{vol}_1(K\cap (y+\langle e_n\rangle)))^{p+1}\right)dy\\& \quad = \frac{1}{p+1}\int_{P_{e_n^\perp}(K)}\left(\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))\right)^{p+1}dy, \end{align*} $$

which proves the first equality. Besides, we also have that

$$ \begin{align*} 2^p\int_{S_{e_n}(K)}|\langle x,e_n\rangle|^pdx&=2^p\int_{P_{e_n^\perp}(K)}\int_{-\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}}^{\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}}|t|^pdtdy\\& = 2^{p+1}\int_{P_{e_n^\perp}(K)}\int_{0}^{\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}}t^pdtdy\\& = \frac{1}{p+1}\int_{P_{e_n^\perp}(K)}(\mathrm{vol}_1(K\cap(y+\langle e_n\rangle)))^{p+1}dy, \end{align*} $$

which proves the second equality.

Proof of Theorem 1.1

Let $f: P_{e_n^\perp }(K)\to [0,\infty )$ be the function given by

$$ \begin{align*}f(y)=\mathrm{vol}_1(K\cap(y+\langle e_n\rangle)), \end{align*} $$

which is concave by Brunn’s principle [Reference Brazitikos, Giannopoulos, Valettas and Vritsiou9, Theorem 1.2.2]. Then, by Berwald’s inequality (Theorem 2.3) applied on $P_{e_n^\perp }(K)$ with $p=1$ and $q=n+1,$ we have that

$$ \begin{align*}\left(\frac{{{2n}\choose{n-1}}}{\mathrm{vol}_{n-1}(P_{e_n^\perp}(K))}\int_{P_{e_n^\perp}(K)} f^{n+1}(y)dy\right)^{\frac{1}{n+1}}\leq\frac{n}{\mathrm{vol}_{n-1}(P_{e_n^\perp}(K))}\int_{P_{e_n^\perp}(K)} f(y)dy. \end{align*} $$

Equivalently, taking into account that $\frac {1}{n}{{2n}\choose {n-1}}=\frac {1}{n+1}{{2n}\choose {n}}$ and that, by Fubini’s theorem, ${\int _{P_{e_n^\perp }(K)} f(y)dy=\mathrm {vol}_n(K)}$ , we obtain

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}\frac{1}{n+1}\int_{P_{e_n^\perp}(K)} f^{n+1}(y)dy\leq\frac{(\mathrm{vol}_n(K))^{n+1}}{(\mathrm{vol}_{n-1}(P_{e_n^\perp}(K)))^n}. \end{align*} $$

By Lemma 3.1, this inequality is equivalent to

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}n\int_0^\infty r^{n-1}\mathrm{vol}_n(K\cap(re_n+K))dr\leq\frac{(\mathrm{vol}_n(K))^{n+1}}{(\mathrm{vol}_{n-1}(P_{e_n^\perp}(K)))^n}.\\[-41pt] \end{align*} $$

Proof By Theorem 1.1 we have that for any $U\in O(n),$

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}n\int_0^\infty r^{n-1}\mathrm{vol}_n(U(K)\cap(re_n+U(K)))dr\leq\frac{(\mathrm{vol}_n(U(K)))^{n+1}}{(\mathrm{vol}_{n-1}(P_{e_n^\perp}(U(K))))^n}. \end{align*} $$

Equivalently, taking into account that $P_{e_n^\perp }(U(K))=P_{(U^t(e_n))^\perp }(K)$ , $\mathrm {vol}_n(U(K)\cap (re_n+U(K)))=\mathrm {vol}_n(K\cap (rU^t(e_n)+K))$ , and that $\mathrm {vol}_n(U(K))=\mathrm {vol}_n(K)$ ,

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}n\int_0^\infty r^{n-1}\mathrm{vol}_n(K\cap(rU^t(e_n)+K))dr\leq\frac{(\mathrm{vol}_n(K))^{n+1}}{(\mathrm{vol}_{n-1}(P_{(U^t(e_n))^\perp}(K)))^n}. \end{align*} $$

Therefore, since for every $\theta \in S^{n-1}$ there exists $U\in O(n)$ such that $U^t(e_n)=\theta $ , we have that for every $\theta \in S^{n-1}$ ,

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}n\int_0^\infty r^{n-1}\mathrm{vol}_n(K\cap(r\theta+K))dr\leq\frac{(\mathrm{vol}_n(K))^{n+1}}{(\mathrm{vol}_{n-1}(P_{\theta^\perp}(K)))^n}. \end{align*} $$

Taking into account (1.6) and that $\rho _{\Pi ^*K}(\theta )=\frac {1}{\mathrm {vol}_{n-1}(P_{\theta ^\perp }(K))}$ for every $\theta \in S^{n-1}$ , we have

$$ \begin{align*}{{2n}\choose{n}}^{\frac{1}{n}}\rho_{K_n(g_K)}(\theta)\leq n\mathrm{vol}_n(K)\rho_{\Pi^*(K)}(\theta)\quad\forall\theta\in S^{n-1}, \end{align*} $$

which is equivalent to the inclusion relation

$$ \begin{align*}{{2n}\choose{n}}^{\frac{1}{n}}K_n(g_K)\subseteq n\mathrm{vol}_n(K)\Pi^*(K) \end{align*} $$

given by (1.7). Taking volumes and using (2.15), we obtain the result.

Proof On the one hand, if $P_{e_n^\perp }(S_{e_n}(K)+C_{n-1})=\emptyset $ then $\mu (S_{e_n}(K)+C_{n-1})=0$ and the inequality is trivial. Let us assume that $P_{e_n^\perp }(S_{e_n}(K)+C_{n-1})\neq \emptyset $ . Notice also that since $C_{n-1}$ is contained in the hyperplane $e_n^\perp $ , we have that $S_{e_n}(C_{n-1})=C_{n-1}$ . Therefore, by (2.6), we have that

$$ \begin{align*}S_{e_n}(K)+C_{n-1}=S_{e_n}(K)+S_{e_n}(C_{n-1})\subseteq S_{e_n}(K+C_{n-1}). \end{align*} $$

Since $\mu (S_{e_n}(L))=\mu (L)$ for every bounded convex set $L\subseteq \mathbb {R}^n$ , we have that

$$ \begin{align*}\mu(S_{e_n}(K)+C_{n-1})\leq\mu(S_{e_n}(K+C_{n-1}))=\mu(K+C_{n-1}). \end{align*} $$

Proof First of all, notice that if $P_{e_n^\perp }(K)\cap \mathbb {Z}^n=\emptyset $ , then $K\cap \mathbb {Z}^n=\emptyset $ . Thus, ${G_n(K)=0}$ , $G_{n-1}(P_{e_n^\perp }(K))=0$ and $\mu (K)=0$ and both inequalities hold. Let us assume that $P_{e_n^\perp }(K)\cap \mathbb {Z}^n\neq \emptyset $ . Since K is bounded and convex, for every $y\in P_{e_n^\perp }(K)\cap \mathbb {Z}^n$ , if $G_1(K\cap (y+\langle e_n\rangle ))=k(y)$ , then, if $k(y)\geq 1$ , we have $K\cap (y+\langle e_n\rangle )$ contains a segment of length $k(y)-1$ (see figure 1 below) and then

$$ \begin{align*}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))\geq G_1(K\cap(y+\langle e_n\rangle))-1, \end{align*} $$

while if $k(y)=0$ , then the latter inequality is trivial. Furthermore, independently of whether $k(y)=0$ or $k(y)\geq 1$ , there exists a segment of length $k(y)+1$ containing $k(y)+2$ points in $\mathbb {Z}^n$ , which contains (see figure 1 below) $K\cap (y+\langle e_n\rangle )$ . Then,

$$ \begin{align*}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))\leq G_1(K\cap(y+\langle e_n\rangle))+1. \end{align*} $$

Therefore, summing in $y\in P_{e_n^\perp }(K)\cap \mathbb {Z}^n$ , from the definition (1.8) of $\mu ,$ we obtain that

$$ \begin{align*} \mu(K)=&\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))\\\geq &\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n} G_1(K\cap(y+\langle e_n\rangle))-G_{n-1}(P_{e_n}^\perp(K)\cap\mathbb{Z}^n) \end{align*} $$

and

$$ \begin{align*} \mu(K)=&\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))\\\leq &\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n} G_1(K\cap(y+\langle e_n\rangle))+G_{n-1}(P_{e_n}^\perp(K)\cap\mathbb{Z}^n). \end{align*} $$

Since ${\sum _{y\in P_{e_n^\perp }(K)\cap \mathbb {Z}^n} G_1(K\cap (y+\langle e_n\rangle ))=G_n(K)}$ , we obtain that

$$ \begin{align*}G_n(K)-G_{n-1}(P_{e_n^\perp}(K))\leq\mu(K)\leq G_n(K)+G_{n-1}(P_{e_n^\perp}(K)). \end{align*} $$

Finally, taking into account (2.1) for any bounded set M containing the origin, we obtain that for any convex body $K\subseteq \mathbb {R}^n$ ,

$$ \begin{align*}\lim_{r\to\infty}\frac{G_{n-1}(P_{e_n^\perp}(rK+M))}{r^n}=\lim_{r\to\infty}\frac{1}{r}\frac{G_{n-1}(rP_{e_n^\perp}(K)+P_{e_n^\perp}(M))}{r^{n-1}}=0 \end{align*} $$

and then

$$ \begin{align*}\lim_{r\to\infty}\frac{\mu(rK+M)}{r^n}=\lim_{r\to\infty}\frac{G_n(rK+M)}{r^n}=\mathrm{vol}_n(K). \end{align*} $$

Proof First of all, notice that if $P_{e_n^\perp }(K)\cap \mathbb {Z}^n=\emptyset $ , then $\mu (K)=0$ , and therefore, for every $r\geq 0$ , we have that $\mu (K\cap (re_n+K))=0$ . In such case, we also have $P_{e_n^\perp }(S_{e_n}(K))\cap \mathbb {Z}^n=\emptyset $ and $\mu (S_{e_n}(K))=0$ . Thus, all the identities are trivial. Let us assume that $P_{e_n^\perp }(K)\cap \mathbb {Z}^n\neq \emptyset $ . From the definition of $\mu $ given in (1.8), we have that for any $r\geq 0,$

$$ \begin{align*} \mu(K\cap(re_n+K))&=\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\mathrm{vol}_1(K\cap(re_n+K)\cap(y+\langle e_n\rangle))\\&=\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\max\{\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))-r,0\}. \end{align*} $$

Then, for any $p>0,$ we have

$$ \begin{align*} p &\int_0^{\infty}r^{p-1}\mu(K\cap(re_n+K))dr\\&\quad =p\int_0^{\infty}r^{p-1}\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\max\{\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))-r,0\}dr\\&\quad =\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\int_0^{\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))}pr^{p-1}(\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))-r)dr\\&\quad =\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\left((\mathrm{vol}_1(K\cap (y+\langle e_n\rangle)))^{p+1}-\frac{p}{p+1}(\mathrm{vol}_1(K\cap (y+\langle e_n\rangle)))^{p+1}\right)\\&\quad =\frac{1}{p+1}\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\left(\mathrm{vol}_1(K\cap (y+\langle e_n\rangle))\right)^{p+1}, \end{align*} $$

which proves the first identity. Finally, notice that

$$ \begin{align*} 2^p\int_{S_{e_n}(K)}|\langle x,e_n\rangle|^pd\mu(x)&=2^p\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\int_{-\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}}^{\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}}|t|^pdt\\&=2^{p+1}\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\int_0^{\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}}t^pdt\\&=\frac{1}{p+1}\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}(\mathrm{vol}_1(K\cap(y+\langle e_n\rangle)))^{p+1}, \end{align*} $$

which proves the second identity.

Proof of Theorem 1.2

Let $K\subseteq \mathbb {R}^n$ be a convex body satisfying

$$ \begin{align*}\max_{y\in e_n^\perp}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))=\mathrm{vol}_1(K\cap\langle e_n\rangle),\end{align*} $$

and let

$$ \begin{align*} K_1&=S_{e_n}(K)\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle\geq0\}\\&=\left\{(y,t)\in\mathbb{R}^{n-1}\times\mathbb{R}\,:\,y\in P_{e_n^\perp}(K),\,0\leq t\leq\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}\right\}. \end{align*} $$

Notice that $K_1$ is the hypograph of the concave function $f:P_{e_n^\perp }(K)\to [0,\infty )$ given by

$$ \begin{align*}f(y)=\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))}{2}. \end{align*} $$

By Theorem 2.4 applied on $P_{e_n^\perp }(K)$ with $p=1$ and $q=n$ , we have

$$ \begin{align*} &\left(\frac{{{2n}\choose{n-1}}}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}f^{n+1}(y)\right)^{\frac{1}{n+1}}\\& \quad \leq \frac{n}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{y\in (P_{e_n^\perp}(K)+C_{n-1})\cap\mathbb{Z}^n}f^{\diamond}(y). \end{align*} $$

Equivalently, taking into account that $\frac {1}{n}{{2n}\choose {n-1}}=\frac {1}{n+1}{{2n}\choose {n}}$ , we obtain that

$$ \begin{align*} &\frac{{{2n}\choose{n}}}{n^n}\sum_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}\frac{\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))^{n+1}}{n+1}\\& \quad \leq \frac{1}{G_{n-1}(P_{e_n^\perp}(K))^n}\left(\sum_{y\in (P_{e_n^\perp}(K)+C_{n-1})\cap\mathbb{Z}^n}2f^{\diamond}(y)\right)^{n+1}. \end{align*} $$

On the one hand, by Lemma 4.3,

$$ \begin{align*}\frac{1}{n+1}\sum_{y\in P_{e_n^\perp}(K)}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))^{n+1}=n\int_0^{\infty}r^{n-1}\mu(K\cap(re_n+K))dr. \end{align*} $$

On the other hand, notice that the hypograph of $f^\diamond $ is the closure of

$$ \begin{align*} L_1&:=K_1+C_{n-1}=S_{e_n}(K)\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle\geq0\}+C_{n-1}\\&=(S_{e_n}(K)+C_{n-1})\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle\geq0\}, \end{align*} $$

and then

$$ \begin{align*}\sum_{y\in (P_{e_n^\perp}(K)+C_{n-1})\cap\mathbb{Z}^n}2f^{\diamond}(y)=2\mu(L_1)=\mu(S_{e_n}(K)+C_{n-1}). \end{align*} $$

Therefore,

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}n\int_0^\infty r^{n-1}\mu(K\cap(re_n+K))dr\leq\frac{(\mu(S_{e_n}(K)+C_{n-1}))^{n+1}}{(G_{n-1}(P_{e_n^\perp}(K)))^n}.\\[-41pt] \end{align*} $$

Proof of Corollary 1.5

Let $K\subseteq \mathbb {R}^n$ be a convex body satisfying ${\max _{y\in e_n^\perp }\mathrm {vol}_1(K\cap (y+\langle e_n\rangle ))}=\mathrm {vol}_1(K\cap \langle e_n\rangle )$ . By Theorem 1.2, we have that

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}n\int_0^\infty r^{n-1}\mu(K\cap(re_n+K))dr\leq\frac{(\mu(S_{e_n}(K)+C_{n-1}))^{n+1}}{(G_{n-1}(P_{e_n^\perp}(K)))^n}. \end{align*} $$

On the one hand, by Lemma 4.2, we have that

$$ \begin{align*} \mu(S_{e_n}(K)+C_{n-1})&\leq G_n(S_{e_n}(K)+C_{n-1})+G_{n-1}(P_{e_n^\perp}(S_{e_n}(K)+C_{n-1}))\\&=G_n(S_{e_n}(K)+C_{n-1})+G_{n-1}(P_{e_n^\perp}(K)+C_{n-1}). \end{align*} $$

On the other hand, by (2.11), we have that $K\cap (re_n+K)=\emptyset $ if $r>\rho _{K-K}(e_n)$ . Therefore, using Lemma 4.2 and the fact that $K\cap (re_n+K)\subseteq K,$ we have

$$ \begin{align*} &n\int_0^\infty r^{n-1}\mu(K\cap(re_n+K))dr=n\int_0^{\rho_{K-K}(e_n)} r^{n-1}\mu(K\cap(re_n+K))dr\\& \quad \geq n\int_0^{\rho_{K-K}(e_n)} r^{n-1}(G_n(K\cap(re_n+K))-G_{n-1}(P_{e_n^\perp}(K\cap(re_n+K))))dr\\& \quad \geq n\int_0^{\rho_{K-K}(e_n)} r^{n-1}(G_n(K\cap(re_n+K))-G_{n-1}(P_{e_n^\perp}(K)))dr\\& \quad =n\int_0^{\rho_{K-K}(e_n)} r^{n-1}G_n(K\cap(re_n+K))dr-\rho_{K-K}^n(e_n)G_{n-1}(P_{e_n^\perp}(K))\\& \quad =n\int_0^{\infty} r^{n-1}G_n(K\cap(re_n+K))dr-\rho_{K-K}^n(e_n)G_{n-1}(P_{e_n^\perp}(K)). \end{align*} $$

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body. Since the inequality we want to prove is invariant by translations, we can assume without loss of generality that ${\max _{y\in P_{e_n^\perp }(K)}=\mathrm {vol}_1(K\cap (y+\langle e_n\rangle ))=\mathrm {vol}_1(K\cap \langle e_n\rangle )}$ . Then, for any $\lambda>0,$ we have that

$$ \begin{align*}\max_{y\in P_{e_n^\perp}(\lambda K)}\mathrm{vol}_1(\lambda K\cap(y+\langle e_n\rangle))=\mathrm{vol}_1(\lambda K\cap\langle e_n\rangle). \end{align*} $$

Thus, by Theorem 1.2 and Lemma 4.1, for any $\lambda>0$ ,

$$ \begin{align*} \frac{{{2n}\choose{n}}}{n^n}\int_0^\infty nr^{n-1}\mu(\lambda K\cap(re_n+\lambda K))dr\leq &\frac{(\mu(S_{e_n}(\lambda K)+C_{n-1}))^{n+1}}{(G_{n-1}(P_{e_n^\perp}(\lambda K)))^n}\\ \leq &\frac{\mu(\lambda K+C_{n-1})^{n+1}}{G_{n-1}(P_{e_n^\perp}(\lambda K))^n}. \end{align*} $$

By Lemma 4.2, we have that for every $r>0,$

$$ \begin{align*}\mu(\lambda K\cap(re_n+\lambda K))\geq G_n(\lambda K\cap(re_n+\lambda K))-G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K))) \end{align*} $$

and that

$$ \begin{align*}\mu(\lambda K+C_{n-1})\leq G_n(\lambda K+C_{n-1})+G_{n-1}(\lambda P_{e_n^\perp}(K)+C_{n-1}). \end{align*} $$

Then, we obtain that

$$ \begin{align*} &\frac{{{2n}\choose{n}}}{n^n}\int_0^\infty nr^{n-1}(G_n(\lambda K\cap(re_n+\lambda K))-G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K))))dr\\ \leq &\frac{(G_n(\lambda K+C_{n-1})+G_{n-1}(\lambda P_{e_n^\perp}(K)+C_{n-1}))^{n+1}}{G_{n-1}(\lambda P_{e_n^\perp}(K))^n}. \end{align*} $$

Therefore, dividing the latter inequality by $\lambda ^{2n}=\frac {\lambda ^{n(n+1)}}{\lambda ^{n(n-1)}}$ , we obtain that for any $\lambda>0$ , the quantity

(4.1) $$ \begin{align} \frac{{{2n}\choose{n}}}{n^n}\frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}(G_n(\lambda K\cap(re_n+\lambda K))-G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K))))dr\nonumber\\ \end{align} $$

is bounded above by

$$ \begin{align*}\frac{(G_n(\lambda K+C_{n-1})+G_{n-1}(\lambda P_{e_n^\perp}(K)+C_{n-1}))^{n+1}}{\lambda^{2n}G_{n-1}(\lambda P_{e_n^\perp}(K))^n}, \end{align*} $$

which equals

(4.2) $$ \begin{align} \frac{\left(\frac{G_n(\lambda K+C_{n-1})}{\lambda^n}+\frac{G_{n-1}(\lambda P_{e_n^\perp}(K)+C_{n-1})}{\lambda^n}\right)^{n+1}}{\left(\frac{G_{n-1}(\lambda P_{e_n^\perp}(K))}{\lambda^{n-1}}\right)^n}. \end{align} $$

On the one hand, we are going to prove that

(4.3) $$ \begin{align} &\lim_{\lambda\to\infty}\frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}G_n(\lambda K\cap(re_n+\lambda K))dr \nonumber\\& \quad =\frac{{{2n}\choose{n}}}{n^n}\int_0^\infty nr^{n-1}\mathrm{vol}_n( K\cap(re_n+ K))dr \end{align} $$

and that

(4.4) $$ \begin{align} \lim_{\lambda\to\infty}\frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K)))dr=0. \end{align} $$

As a consequence, we will obtain that (4.1) converges, as $\lambda \to \infty $ , to

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}\int_0^\infty nr^{n-1}\mathrm{vol}_n( K\cap(re_n+ K))dr. \end{align*} $$

On the other hand, we have that, by (2.2),

$$ \begin{align*}\lim_{\lambda\to\infty}\frac{G_{n-1}(\lambda P_{e_n^\perp}(K))}{\lambda^{n-1}}=\mathrm{vol}_{n-1}(P_{e_n^\perp}(K)) \end{align*} $$

and, by (2.1) (with $M=C_{n-1}$ ),

$$ \begin{align*} &\lim_{\lambda\to\infty}\left(\frac{G_n(\lambda K+C_{n-1})}{\lambda^n}+\frac{G_{n-1}(\lambda P_{e_n^\perp}(K)+C_{n-1})}{\lambda^n}\right)\\& \quad =\mathrm{vol}_n(K)+0\cdot\mathrm{vol}_{n-1}(P_{e_n^\perp}(K))=\mathrm{vol}_n(K). \end{align*} $$

Therefore, (4.2) converges to $\frac {\mathrm {vol}_n(K)^{n+1}}{\mathrm {vol}_{n-1}(P_{e_n^\perp }(K))^n}$ as $\lambda \to \infty $ . As a consequence, we will obtain

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}\int_0^\infty nr^{n-1}\mathrm{vol}_n( K\cap(re_n+ K))dr\leq\frac{\mathrm{vol}_n(K)^{n+1}}{\mathrm{vol}_{n-1}(P_{e_n^\perp}(K))^n}. \end{align*} $$

Let us prove (4.3): By (2.11), $\lambda K\cap (re_n+\lambda K)=\emptyset $ for every $r>\rho _{\lambda (K-K)}(e_n)$ . Therefore, changing variables $r=\lambda s$ , for any $\lambda>0,$ we have

$$ \begin{align*} &\frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}G_n(\lambda K\cap(re_n+\lambda K))dr=\\& \quad =\frac{1}{\lambda^{2n}}\int_0^{\rho_{\lambda(K-K)}(e_n)} nr^{n-1}G_n\left(\lambda\left( K\cap\left(\frac{r}{\lambda} e_n+ K\right)\right)\right)dr\\& \quad =\int_0^{\rho_{K-K}(e_n)} ns^{n-1}\frac{G_n(\lambda(K\cap(s e_n+ K)))}{\lambda^n}ds. \end{align*} $$

For any $s\in [0,\rho _{K-K}(e_n)],$ we have by (2.2) that

$$ \begin{align*}\lim_{\lambda\to\infty}\frac{G_n(\lambda(K\cap(s e_n+ K)))}{\lambda^n}=\mathrm{vol}_n(K\cap (se_n+K)). \end{align*} $$

Also by (2.2),

$$ \begin{align*}\lim_{\lambda\to\infty}\frac{G_n(\lambda K)}{\lambda^n}=\mathrm{vol}_n(K). \end{align*} $$

Thus, given $\varepsilon _0>0,$ there exists $\lambda _0>0$ such that if $\lambda>\lambda _0$

$$ \begin{align*}ns^{n-1}\frac{G_n(\lambda(K\cap(s e_n+ K)))}{\lambda^n}\leq ns^{n-1}\frac{G_n(\lambda K)}{\lambda^n}\leq ns^{n-1}(\mathrm{vol}_n(K)+\varepsilon_0). \end{align*} $$

The function $f(s)=ns^{n-1}(\mathrm {vol}_n(K)+\varepsilon _0)$ is integrable in $[0,\rho _{K-K}(e_n)]$ . Thus, by the dominated convergence theorem and using again that, by (2.11), $K\cap (se_n+ K)=\emptyset $ if $s>\rho _{K-K}(e_n)$ , we obtain

$$ \begin{align*} &\lim_{\lambda\to\infty}\frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}G_n(\lambda K\cap(re_n+\lambda K))dr\\& \quad =\int_0^{\rho_{K-K}(e_n)}ns^{n-1}\mathrm{vol}_n(K\cap(se_n+ K))ds\\& \quad =\int_0^{\infty}ns^{n-1}\mathrm{vol}_n(K\cap(se_n+ K))ds, \end{align*} $$

which, renaming s as r, proves (4.3).

Let us now prove (4.4): Again, by (2.11), we have that $\lambda K\cap (re_n+\lambda K)=\emptyset $ if $r>\rho _{\lambda (K-K)}(e_n)$ and then, for any $\lambda>0,$ we have

$$ \begin{align*} 0&\leq \frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K)))dr\\& =\frac{1}{\lambda^{2n}}\int_0^{\rho_{\lambda(K-K)}(e_n)} nr^{n-1}G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K)))dr\\& \leq \frac{G_{n-1}(\lambda P_{e_n^\perp}(K))}{\lambda^{2n}}\int_0^{\rho_{\lambda(K-K)}(e_n)}nr^{n-1}dr=\frac{G_{n-1}(\lambda P_{e_n^\perp}(K))\rho_{\lambda(K-K)}^n(e_n)}{\lambda^{2n}}\\& =\frac{1}{\lambda}\frac{G_{n-1}(\lambda P_{e_n^\perp}(K))}{\lambda^{n-1}}\rho_{K-K}^n(e_n). \end{align*} $$

Since by (2.2),

$$ \begin{align*}\lim_{\lambda\to\infty}\frac{1}{\lambda}\frac{G_{n-1}(\lambda P_{e_n^\perp}(K))}{\lambda^{n-1}}\rho_{K-K}^n(e_n)=0\cdot\mathrm{vol}_{n-1}(P_{e_n^\perp}(K))\rho_{K-K}^n(e_n)=0, \end{align*} $$

we have that

$$ \begin{align*}\lim_{\lambda\to\infty}\frac{1}{\lambda^{2n}}\int_0^\infty nr^{n-1}G_{n-1}(P_{e_n^\perp}(\lambda K\cap(re_n+\lambda K)))dr=0. \end{align*} $$

Now that we have proved (4.3) and (4.4), the proof is complete.

Proof Let $f:[0,1]\to \mathbb {R}$ be the function $f(x)=p(1-x)^{n-1}x^{p-1}$ , which is Riemann-integrable in $[0,1]$ with

$$ \begin{align*}\int_0^1 f(x)dx={{n-1+p}\choose{n-1}}^{-1}. \end{align*} $$

Since the norm of the partition of the interval $[0,1]$ given by $\mathcal {P}_x=\left \{0,\frac {1}{\lfloor x\rfloor },\frac {2}{\lfloor x\rfloor },\dots ,1\right \}$ tends to $0$ as x tends to $\infty $ , and

$$ \begin{align*}\sum_{k=0}^{\lfloor x\rfloor}\frac{p}{\lfloor x\rfloor}\left(1-\frac{k}{\lfloor x\rfloor}\right)^{n-1}\left(\frac{k}{\lfloor x\rfloor}\right)^{p-1}=\sum_{k=0}^{\lfloor x\rfloor-1}\frac{p}{\lfloor x\rfloor}\left(1-\frac{k}{\lfloor x\rfloor}\right)^{n-1}\left(\frac{k}{\lfloor x\rfloor}\right)^{p-1} \end{align*} $$

is a Riemann sum of f associated with $\mathcal {P}_x$ , we have that

$$ \begin{align*}\lim_{x\to\infty}\sum_{k=0}^{\lfloor x\rfloor}\frac{p}{\lfloor x\rfloor}\left(1-\frac{k}{\lfloor x\rfloor}\right)^{n-1}\left(\frac{k}{\lfloor x\rfloor}\right)^{p-1}={{n-1+p}\choose{n-1}}^{-1}. \end{align*} $$

Moreover, as ${\lim _{x\to \infty }\frac {\lfloor x\rfloor }{x}=1}$ , we obtain that

$$ \begin{align*}\lim_{x\to\infty} B_{x}(p)=\lim_{x\to\infty}\sum_{k=0}^{\lfloor x\rfloor}\frac{p}{x}\left(1-\frac{k}{x}\right)^{n-1}\left(\frac{k}{x}\right)^{p-1}={{n-1+p}\choose{n-1}}^{-1}.\\[-41pt] \end{align*} $$

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in P_{e_n^\perp }(K)$ satisfying the hypotheses (H) and let $f:[0,\infty )\to \mathbb {R}$ be the function given by

$$ \begin{align*}f(r)=G_{n-1}(S_{e_n}(K)\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle=r\}). \end{align*} $$

By Remark 5.2, f satisfies that $f(r)\leq \tilde {f}(r)$ for every $r\geq 0$ . Besides, since $M\geq 1$ , we have that $f(1)\geq 1$ . For any $p\geq 1$ , let also $h_p:(0,\infty )\to [0,\infty )$ be the function given by

(5.4) $$ \begin{align} h_p(x)=x^pB_x(p)=\sum_{k=0}^{\lfloor x\rfloor}p\left(1-\frac{k}{x}\right)^{n-1}k^{p-1}. \end{align} $$

Our purpose is to prove the existence of some $m_0(p)\geq M$ such that $m_0(p)>1$ and

$$ \begin{align*}h_p(m_0(p))=\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^\infty pk^{p-1}\tilde{f}(k). \end{align*} $$

Notice that for any $p\geq 1$ , $h_p$ is clearly continuous at any $x_0\in (0,\infty )\setminus \mathbb {N}$ . If $x_0=k_0\in \mathbb {N}$ then

$$ \begin{align*} \lim_{x\to x_0^-}h_p(x)&=\lim_{x\to k_0^-}\sum_{k=0}^{k_0-1}p\left(1-\frac{k}{x}\right)^{n-1}k^{p-1}=\sum_{k=0}^{k_0-1}p\left(1-\frac{k}{k_0}\right)^{n-1}k^{p-1}\\&=\sum_{k=0}^{k_0}p\left(1-\frac{k}{k_0}\right)^{n-1}k^{p-1}=h_p(k_0) \end{align*} $$

and

$$ \begin{align*} \lim_{x\to x_0^+}h_p(x)&=\lim_{x\to k_0^+}\sum_{k=0}^{k_0}p\left(1-\frac{k}{x}\right)^{n-1}k^{p-1}=\sum_{k=0}^{k_0}p\left(1-\frac{k}{k_0}\right)^{n-1}k^{p-1}\\&=h_p(k_0). \end{align*} $$

Thus, for any $p\geq 1$ , $h_p$ is continuous at $x_0=k_0$ and then $h_p$ is continuous on $(0,\infty )$ . Besides, since for every $p\geq 1,$ the function $h_p$ is defined as $h_p(x)=x^pB_x(p)$ for every $x\in (0,\infty )$ , we have that for any $p\geq 1,$ Lemma 5.1 implies that

(5.5) $$ \begin{align} \lim_{x\to\infty}h_p(x)=\infty. \end{align} $$

Let us now distinguish two cases, depending on whether $p>1$ or $p=1$ : Assume first that $p>1$ . Since $B_x(p)=0$ for every $x\in (0,1]$ , we have that for every $x\in (0,1],$

(5.6) $$ \begin{align} h_p(x)=0. \end{align} $$

Since, by definition, M is a non-negative integer and, by b), we are assuming that $M\geq 1$ , we have that $\tilde {f}(1)\geq f(1)\geq 1>0$ and then

$$ \begin{align*}\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}pk^{p-1}\tilde{f}(k)\geq\frac{p\tilde{f}(1)}{G_{n-1}(P_{e_n^\perp}(K))}>0. \end{align*} $$

By the continuity of $h_p$ on $(0,\infty )$ , (5.6) and (5.5), there exists $m_0(p)>1$ such that

$$ \begin{align*}h_p(m_0(p))=\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}pk^{p-1}\tilde{f}(k)>0. \end{align*} $$

Let us assume now that $p=1$ . Then, $B_x(1)=\frac {1}{x}$ for every $x\in (0,1]$ and, since for every $x\in (0,\infty ),$ the function $h_1$ is defined as $h_1(x)=xB_x(1)$ , we have that for every $x\in (0,1],$

(5.7) $$ \begin{align} h_1(x)=xB_x(1)=1. \end{align} $$

Since by b), we are assuming that $M\geq 1$ ,

$$ \begin{align*}\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}pk^{p-1}\tilde{f}(k)=\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}\tilde{f}(k)\geq\frac{\tilde{f}(0)+\tilde{f}(1)}{G_{n-1}(P_{e_n^\perp}(K))}>1, \end{align*} $$

where we have used that $\tilde {f}(1)\geq 1$ and that, from the definition of $\tilde {f}$ ,

$$ \begin{align*}\tilde{f}(0)=G_{n-1}(P_{e_n^\perp}(K)+C_{n-1})\geq G_{n-1}(P_{e_n^\perp}(K)). \end{align*} $$

By the continuity of $h_1$ on $(0,\infty )$ , (5.7) and (5.5), there exists $m_0(1)>1$ such that

$$ \begin{align*}h_1(m_0(1))=\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}pk^{p-1}\tilde{f}(k)>0. \end{align*} $$

In both cases, since $m_0(p)>1$ , such $m_0(p)$ satisfies that $B_{m_0(p)}(p)>0$ and

$$ \begin{align*}m_0(p)^p=\frac{\sum_{k=0}^{\infty}pk^{p-1}\tilde{f}(k)}{G_{n-1}(P_{e_n^\perp}(K))B_{m_0(p)}(p)}. \end{align*} $$

Let us now see that $m_0(p)\geq M$ . Assume that $m_0(p)<M$ . By a), we are assuming

$$ \begin{align*}\max_{y\in P_{e_n^\perp}(K)\cap\mathbb{Z}^n}G_1(S_{e_n}(K)\cap(y+\langle e_n\rangle))=G_1(S_{e_n}(K)\cap\langle e_n\rangle). \end{align*} $$

Thus, from the definition of ${M=\max _{x\in S_{e_n}(K)\cap \mathbb {Z}^n}\langle x,e_n\rangle }$ , together with the hypotheses (H), we have that $Me_n\in S_{e_n}(K)\cap \{x\in \mathbb {R}^n\,:\langle x,e_n\rangle \geq 0\}$ . Consequently, the convex hull of $P_{e_n^\perp }(K)$ and the point $m_0(p)e_n$ is contained in $S_{e_n}(K)\cap \{x\in \mathbb {R}^n\,:\langle x,e_n\rangle \geq 0\}$ . Therefore, for every $0\leq k\leq \lfloor m_0(p)\rfloor $ ,

$$ \begin{align*}S_{e_n}(K)\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle=k\}\supseteq\left(1-\frac{k}{m_0(p)}\right)P_{e_n^\perp}(K)\times\{k\}. \end{align*} $$

Thus,

$$ \begin{align*}(S_{e_n}(K)+C_{n-1})\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle=k\}\supseteq\left(\left(1-\frac{k}{m_0(p)}\right)P_{e_n}(K)+C_{n-1}\right)\times\{k\}. \end{align*} $$

We obtain, as a consequence of the discrete Brunn–Minkowski inequality (Theorem 2.2), that

(5.8) $$ \begin{align} \tilde{f}^{\frac{1}{n-1}}(k)&\geq \left(1-\frac{k}{m_0(p)}\right)G_{n-1}(P_{e_n^\perp}(K))^{\frac{1}{n-1}}+\frac{k}{m_0(p)}G_{n-1}(\{0\})^{\frac{1}{n-1}} \nonumber\\&\geq \left(1-\frac{k}{m_0(p)}\right)G_{n-1}(P_{e_n^\perp}(K))^{\frac{1}{n-1}}. \end{align} $$

Since $\lfloor m_0(p)\rfloor \leq m_0(p)<M$ and

$$ \begin{align*}\tilde{f}^{\frac{1}{n-1}}(M)\geq f^{\frac{1}{n-1}}(M)\geq1>0, \end{align*} $$

we obtain that

$$ \begin{align*}\sum_{k=0}^{M}pk^{p-1}\tilde{f}(k)>\sum_{k=0}^{\lfloor m_0(p)\rfloor}pk^{p-1}\tilde{f}(k) \end{align*} $$

and then, as a consequence of (5.8) and using (5.4), that

$$ \begin{align*} m_0(p)^p&=\frac{\sum_{k=0}^{\infty}pk^{p-1}\tilde{f}(k)}{G_{n-1}(P_{e_n^\perp}(K))B_{m_0(p)}(p)}\geq\frac{\sum_{k=0}^{M}pk^{p-1}\tilde{f}(k)}{G_{n-1}(P_{e_n^\perp}(K))B_{m_0(p)}(p)}\\&>\frac{\sum_{k=0}^{\lfloor m_0(p)\rfloor}pk^{p-1}\tilde{f}(k)}{G_{n-1}(P_{e_n^\perp}(K))B_{m_0(p)}(p)}\geq\frac{\sum_{k=0}^{\lfloor m_0(p)\rfloor}pk^{p-1}\left(1-\frac{k}{m_0(p)}\right)^{n-1}}{B_{m_0(p)}(p)}\\&=m_0(p)^p, \end{align*} $$

which is a contradiction. Therefore, $m_0(p)\geq M$ .

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in P_{e_n^\perp }(K)$ satisfying the hypotheses (H) and let $f,\tilde {f}$ , and $g_p$ be the functions defined in the statement. Let us denote, for every $r\geq 0$ ,

$$ \begin{align*}D_r=S_{e_n}(K)\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle=r\}, \end{align*} $$

and

$$ \begin{align*}\tilde{D}_r=(S_{e_n}(K)+C_{n-1})\cap\{x\in\mathbb{R}^n\,:\,\langle x,e_n\rangle=r\}=D_r+C_{n-1}. \end{align*} $$

Therefore, $f(r)=G_{n-1}(D_r)$ and $\tilde {f}(r)=G_{n-1}(\tilde {D}_r)$ for every $r\geq 0$ .

Notice that, as mentioned in Remark 5.2,

$$ \begin{align*}\text{supp}(f)\subseteq\text{supp}(\tilde{f})\subseteq\left[0,\frac{1}{2}\max_{y\in P_{e_n^\perp}}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))\right] \end{align*} $$

and then, for every ${r>\max _{y\in P_{e_n^\perp }(K)}\frac {1}{2}\mathrm {vol}_1(K\cap (y+\langle e_n\rangle ))}$ , we have

$$ \begin{align*}0=\tilde{f}(r)\leq g_p(r). \end{align*} $$

Therefore, we can define

$$ \begin{align*}\tilde{r}_0(p)=\inf\{r\geq0\,:\,\tilde{f}(r)\leq g_p(r)\}\in[0,\infty). \end{align*} $$

From the definition of $\tilde {r}_0(p)$ , we trivially have that for every $0\leq r<\tilde {r}_0(p),$

$$ \begin{align*}\tilde{f}(r)> g_p(r). \end{align*} $$

In particular, for every $r=k\in \mathbb {N}\cup \{0\}$ with $0\leq k<\tilde {r}_0(p)$ , we have that

$$ \begin{align*}\tilde{f}(k)>g_p(k). \end{align*} $$

Notice also that M is defined in (H) as ${M=\max _{x\in S_{e_n}(K)\cap \mathbb {Z}^n}\langle x, e_n\rangle }$ and, as mentioned in Remark 5.2, $\text {supp}(f)\subseteq [0, M+1)$ . Then, for every $k\in \mathbb {N}$ such that $k>M,$ we have

$$ \begin{align*}g_p(k)\geq f(k)=0. \end{align*} $$

Therefore, if $\tilde {r}_0(p)>M$ , then we can take $r_0(p)=\tilde {r}_0(p)$ . Let us assume that $\tilde {r}_0(p)\leq M$ .

Let us assume first that $\tilde {r}_0(p)=M\in \mathbb {N}$ . If $f(M)\leq g_p(M)$ , then we can take $r_0(p)=\tilde {r}_0(p)$ . If, on the contrary, $f(M)>g_p(M)$ , we have, by Remark 5.2,

$$ \begin{align*}\tilde{f}(M)\geq f(M)>g_p(M) \end{align*} $$

and we can take any $r_0(p)>\tilde {r}_0(p)=M$ .

Let us assume now that $\tilde {r}_0(p)<M$ . In such case, let us take $r_0(p)=\tilde {r}_0(p)$ and let us see that for every $k\in \mathbb {N}$ with $r_0(p)\leq k\leq M,$ we have that $g_p(k)\geq f(k)$ . Notice that for every $0\leq r\leq M$ , the set $\tilde {D}_r$ is a convex open (in the topology induced in $\mathbb {R}^{n-1}\times \{r\}$ by the standard topology in $\mathbb {R}^n$ ) bounded set, and that $P_{e_n^\perp }(\tilde {D}_{r_1})\supseteq P_{e_n^\perp }(\tilde {D}_{r_2})$ if $r_1\leq r_2$ . Moreover, for any $0\leq r_1<\max _{y\in P_{e_n^\perp }(K)}\frac {1}{2}\mathrm {vol}_1(K\cap (y+\langle e_n\rangle )),$ we have ${\bigcup _{r_2>r_1}P_{e_n^\perp }(\tilde {D}_{r_2})=P_{e_n^\perp }(\tilde {D}_{r_1})}$ . Therefore, the function $\tilde {f}$ is continuous from the right at every $0\leq r_1<\max _{y\in P_{e_n^\perp }(K)}\frac {1}{2}\mathrm {vol}_1(K\cap (y+\langle e_n\rangle ))$ and, from the definition of $\tilde {r}_0(p)$ as an infimum, we obtain that if $r_0(p)=\tilde {r}_0(p)<\max _{y\in P_{e_n^\perp }(K)}\frac {1}{2}\mathrm {vol}_1(K\cap (y+\langle e_n\rangle ))$ , then

(5.11) $$ \begin{align} \tilde{f}(r_0(p))\leq g_p(r_0(p)) \end{align} $$

and

$$ \begin{align*}r_0(p)=\min\{r\geq0\,:\,\tilde{f}(r)\leq g_p(r)\}. \end{align*} $$

Notice that, as explained in Remark 5.2,

$$ \begin{align*}[0,M]\subseteq\text{supp}(f)\subseteq\text{supp}(\tilde{f})\subseteq\left[0,\frac{1}{2}\max_{y\in P_{e_n^\perp}}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))\right]. \end{align*} $$

Thus, $M\leq \max _{y\in P_{e_n^\perp }(K)}\frac {1}{2}\mathrm {vol}_1(K\cap (y+\langle e_n\rangle ))$ and, since $r_0(p)<M$ , then $r_0(p)$ satisfies (5.11). If $r_0(p)< r\leq M$ , then, taking $\lambda =\frac {r_0(p)}{r}\in [0,1)$ , we have, by the convexity of $S_{e_n}(K)+C_{n-1}$ , that

$$ \begin{align*}\tilde{D}_{r_0(p)}\supseteq (1-\lambda)\tilde{D}_0+\lambda \tilde{D}_r=(1-\lambda)D_0+\lambda D_r+C_{n-1}. \end{align*} $$

By the discrete Brunn–Minkowski inequality (Theorem 2.2), we have

$$ \begin{align*} \tilde{f}^{\frac{1}{n-1}}(r_0(p))&\geq G_{n-1}((1-\lambda)D_0+\lambda D_r+C_{n-1})^{\frac{1}{n-1}}\\&\geq (1-\lambda)f^{\frac{1}{n-1}}(0)+\lambda f^{\frac{1}{n-1}}(r). \end{align*} $$

Taking into account that $r_0(p)<r\leq M\leq m_0(p)$ ,

$$ \begin{align*} g_p^{\frac{1}{n-1}}(r_0(p))&=\left(1-\frac{r_0(p)}{m_0(p)}\right)G_{n-1}(P_{e_n^\perp}(K))^{\frac{1}{n-1}}\\&=(1-\lambda)G_{n-1}(P_{e_n^\perp}(K))^{\frac{1}{n-1}}+\lambda\left(1-\frac{r}{m_0}\right)G_{n-1}(P_{e_n^\perp}(K))^{\frac{1}{n-1}}\\&=(1-\lambda)f^{\frac{1}{n-1}}(0)+\lambda g_p^{\frac{1}{n-1}}(r). \end{align*} $$

Thus, by (5.11), we have that for every $r_0(p)<r\leq M,$

$$ \begin{align*}(1-\lambda)f^{\frac{1}{n-1}}(0)+\lambda g_p^{\frac{1}{n-1}}(r)=g_p^{\frac{1}{n-1}}(r_0(p))\geq \tilde{f}^{\frac{1}{n-1}}(r_0(p))\geq (1-\lambda)f^{\frac{1}{n-1}}(0)+\lambda f^{\frac{1}{n-1}}(r). \end{align*} $$

Therefore, for every $r_0(p)<r\leq M$ ,

$$ \begin{align*}g_p(r)\geq f(r). \end{align*} $$

By (5.11), this inequality also holds for $r=r_0(p)$ . Therefore, if $\tilde {r}_0(p)=r_0(p)<M$ , for every $k\in \mathbb {N}$ such that $r_0(p)\leq k\leq M$ , we have that

$$ \begin{align*}g_p(k)\geq f(k).\\[-33pt] \end{align*} $$

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in P_{e_n^\perp }(K)$ satisfying the hypotheses (H) and let $p\geq 1$ . By Lemma 5.3 and Remark 5.4, there exists $m_0(p)>1$ such that

$$ \begin{align*}m_0(p)^pB_{m_0(p)}(p)=\frac{1}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^\infty pk^{p-1}\tilde{f}(k)>0. \end{align*} $$

Let also $g_p$ be the function defined in (5.1), given by

$$ \begin{align*}g_p(r)=\begin{cases} \left(1-\frac{r}{m_0(p)}\right)^{n-1}G_{n-1}(P_{e_n^\perp}(K)) &\text{ if }0\leq r\leq m_0(p)\\0&\text{ if }r>m_0(p),\\ \end{cases} \end{align*} $$

which, from the definition of $B_{m_0(p)}(q)$ for $q>0$ , satisfies (5.9). That is, for every $q>0,$

$$ \begin{align*}\sum_{k=0}^\infty qk^{q-1}g_p(k)=m_0(p)^qB_{m_0(p)}(q)G_{n-1}(P_{e_n^\perp}(K)). \end{align*} $$

Therefore, taking $q=p,$

$$ \begin{align*}\sum_{k=0}^{\infty}k^{p-1}g_p(k)=\sum_{k=0}^{\lfloor m_0(p)\rfloor}k^{p-1}g_p(k)=\sum_{k=0}^\infty k^{p-1}\tilde{f}(k). \end{align*} $$

Let now $q>p$ . By Lemma 5.5, there exists $r_0(p)$ such that $\tilde {f}(k)\geq g_p(k)$ for every $k\in \mathbb {N}\cup \{0\}$ with $0\leq k< r_0(p)$ and $g_p(k)\geq f(k)$ for every $k\in \mathbb {N}$ with $k\geq r_0(p)$ . Taking into account that $f(k)\leq \tilde {f}(k)$ for every $k\geq 0$ , and understanding the first sum as $0$ if $r_0(p)=0$ , we have

$$ \begin{align*} &\sum_{k=0}^{\lceil r_0(p)\rceil-1}k^{q-1}\left(\tilde{f}(k)-g_p(k)\right)-\sum_{k=\lceil r_0(p)\rceil}^\infty k^{q-1}\left(g_p(k)-f(k)\right)\\&\quad =\sum_{k=0}^{\lceil r_0(p)\rceil-1}k^{q-p}k^{p-1}\left(\tilde{f}(k)-g_p(k)\right)-\sum_{k=\lceil r_0(p)\rceil}^\infty k^{q-p}k^{p-1}\left(g_p(k)-f(k)\right)\\&\quad \leq {r_0(p)}^{q-p}\sum_{k=0}^{\lceil r_0(p)\rceil-1}k^{p-1}\left(\tilde{f}(k)-g_p(k)\right)-{r_0(p)}^{q-p}\sum_{k=\lceil r_0(p)\rceil}^\infty k^{p-1}\left(g_p(k)-f(k)\right)\\&\quad ={r_0(p)}^{q-p}\sum_{k=0}^{\lceil r_0(p)\rceil-1}k^{p-1}\left(\tilde{f}(k)-g_p(k)\right)+{r_0(p)}^{q-p}\sum_{k=\lceil r_0(p)\rceil}^\infty k^{p-1}\left(f(k)-g_p(k)\right)\\&\quad \leq {r_0(p)}^{q-p}\sum_{k=0}^{\lceil r_0(p)\rceil-1}k^{p-1}\left(\tilde{f}(k)-g_p(k)\right)+{r_0(p)}^{q-p}\sum_{k=\lceil r_0(p)\rceil}^\infty k^{p-1}\left(\tilde{f}(k)-g_p(k)\right)\\&\quad ={r_0}^{q-p}\sum_{k=0}^{\infty}k^{p-1}\left(\tilde{f}(k)-g_p(k)\right)=0. \end{align*} $$

Therefore, we have that

$$ \begin{align*}\sum_{k=0}^{\lceil r_0(p)\rceil-1}k^{q-1}\tilde{f}(k)+\sum_{k=\lceil r_0(p)\rceil}^\infty k^{q-1}f(k)\leq\sum_{k=0}^\infty k^{q-1}g_p(k) \end{align*} $$

and then, since $\tilde {f}(k)\geq f(k)$ for every $k\geq 0$ ,

$$ \begin{align*}\sum_{k=0}^{\infty}k^{q-1}f(k)\leq\sum_{k=0}^\infty k^{q-1}g_p(k). \end{align*} $$

Consequently, since $m_0(p)$ satisfies (5.10), $g_p$ satisfies (5.9), and $B_{m_0(p)}(q)>0$ for every $q\geq 1$ , as mentioned in Remark 5.4,

$$ \begin{align*} \left(\frac{B_{m_0(p)}(q)^{-1}}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}qk^{q-1}f(k)\right)^{\frac{1}{q}}&\leq \left(\frac{B_{m_0(p)}(q)^{-1}}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}qk^{q-1}g_p(k)\right)^{\frac{1}{q}}=m_0(p)\\&=\left(\frac{B_{m_0(p)}(p)^{-1}\sum_{k=0}^\infty pk^{p-1}\tilde{f}(k)}{G_{n-1}(P_{e_n^\perp}(K))}\right)^{\frac{1}{p}}. \end{align*} $$

Proof of Theorem 1.7

Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in P_{e_n^\perp }(K)$ such that ${\max _{y\in P_{e_n^\perp }(K)\cap \mathbb {Z}^n}G_1(S_{e_n}(K)\cap (y+\langle e_n\rangle ))=G_1(S_{e_n}(K)\cap \langle e_n\rangle )}$ and let ${M=\max _{x\in S_{e_n}(K)\cap \mathbb {Z}^n}\langle x,e_n\rangle }$ . If $M=0$ then, as mentioned in Remark 1.9, any value of $m_0>1$ gives the inequality, even though $m_0$ is not defined by Lemma 5.3. Let us assume that $M\geq 1$ . Therefore, K satisfies the hypotheses (H). Let $p=1$ , $f,\tilde {f}:[0,\infty )\to \mathbb {N}\cup \{0\}$ be defined as in Theorem 5.6 by

• • $f(r)=G_{n-1}(S_{e_n}(K)\cap \{x\in \mathbb {R}^n\,:\,\langle x,e_n\rangle =r\})$ ,

• • $\tilde {f}(r)=G_{n-1}((S_{e_n}(K)+C_{n-1})\cap \{x\in \mathbb {R}^n\,:\,\langle x,e_n\rangle =r\})$ ,

and let $m_0=m_0(1)>1$ be the number given by Lemma 5.3. Applying Theorem 5.6 with $p=1$ and $q=n+1,$ we obtain that

$$ \begin{align*}\left(\frac{B_{m_0}(n+1)^{-1}}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}(n+1)k^{n}f(k)\right)^{\frac{1}{n+1}}\leq \frac{B_{m_0}(1)^{-1}}{G_{n-1}(P_{e_n^\perp}(K))}\sum_{k=0}^{\infty}\tilde{f}(k). \end{align*} $$

Equivalently,

$$ \begin{align*}B_{m_0}(n+1)^{-1}\sum_{k=0}^{\infty}(n+1)k^{n}f(k)\leq \frac{\left(B_{m_0}(1)^{-1}\sum_{k=0}^{\infty}\tilde{f}(k)\right)^{n+1}}{(G_{n-1}(P_{e_n^\perp}(K)))^n}. \end{align*} $$

From (2.5), $S_{e_n}(K)$ is symmetric with respect to the hyperplane $\{x\in \mathbb {R}^n\,:\,\langle x, e_n\rangle =0\}$ and then, from the definition of f and taking into account that for any $y\in \mathbb {Z}^{n-1}$ and $k\in \mathbb {N,}$ we have $(y,k)\neq (y,-k)$ and that the term corresponding to $k=0$ in the following sum is $0$ , we obtain

$$ \begin{align*}\sum_{k=0}^{\infty}k^{n}f(k)=\frac{1}{2}\sum_{x\in S_{e_n}(K)\cap\mathbb{Z}^n}|\langle x,e_n\rangle|^n=\frac{1}{2}\int_{S_{e_n}(K)}|\langle x,e_n\rangle|^ndG_n(x). \end{align*} $$

In the same way, since $S_{e_n}(K)+C_{n-1}$ is also symmetric with respect to the hyperplane $\{x\in \mathbb {R}^n\,:\,\langle x, e_n\rangle =0\}$ and taking into account that for any $y\in \mathbb {Z}^{n-1}$ and $k\in \mathbb {N,}$ we have $(y,k)\neq (y,-k)$ , but $(y,0)=(y,-0)$ and the term corresponding to the following sum is not 0, we obtain

$$ \begin{align*}\sum_{k=0}^{\infty}\tilde{f}(k)=\frac{1}{2}\left(G_n(S_{e_n}(K)+C_{n-1})+G_{n-1}(P_{e_n^\perp}(K)+C_{n-1})\right). \end{align*} $$

Therefore,

$$ \begin{align*} &\frac{(n+1)B_{m_0}(n+1)^{-1}}{B_{m_0}(1)^{-(n+1)}}2^n\int_{S_{e_n}(K)}|\langle x,e_n\rangle|^ndG_n(x)\leq\\& \quad \leq \frac{\left(G_n(S_{e_n}(K)+C_{n-1})+G_{n-1}(P_{e_n^\perp}(K)+C_{n-1})\right)^{n+1}}{(G_{n-1}(P_{e_n^\perp}(K)))^n}. \end{align*} $$

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body. Since the inequality we want to prove is invariant under translations, we can assume, without loss of generality, that

$$ \begin{align*}\max_{y\in P_{e_n^\perp}(K)}\mathrm{vol}_1(K\cap(y+\langle e_n\rangle))=\mathrm{vol}_1(K\cap\langle e_n\rangle). \end{align*} $$

Therefore, for any $\lambda>0,$

$$ \begin{align*}\max_{y\in P_{e_n^\perp}(\lambda K)}\mathrm{vol}_1(\lambda K\cap(y+\langle e_n\rangle))=\mathrm{vol}_1(\lambda K\cap\langle e_n\rangle). \end{align*} $$

Since for every $y\in P_{e_n^\perp }(\lambda K),$ the segment $S_{e_n}(\lambda K)\cap (y+\langle e_n\rangle )$ is centered at y, we have

$$ \begin{align*}\max_{y\in P_{e_n^\perp}(\lambda K)}G_1(S_{e_n}(\lambda K)\cap(y+\langle e_n\rangle))=G_1(S_{e_n}(\lambda K)\cap\langle e_n\rangle). \end{align*} $$

In particular,

$$ \begin{align*}\max_{y\in P_{e_n^\perp}(\lambda K)\cap\mathbb{Z}^n}G_1(S_{e_n}(\lambda K)\cap(y+\langle e_n\rangle))=G_1(S_{e_n}(\lambda K)\cap\langle e_n\rangle). \end{align*} $$

By Theorem 1.7, there exists ${m_{0,\lambda }\geq \max _{x\in S_{e_n}(\lambda K)\cap \mathbb {Z}^n}\langle x,e_n\rangle }$ such that

$$ \begin{align*} &\frac{(n+1)B_{m_{0,\lambda}}(n+1)}{B_{m_{0,\lambda}}(1)^{-(n+1)}}2^n\int_{S_{e_n}(\lambda K)}|\langle x,e_n\rangle|^ndG_n(x)\leq\\ \leq &\frac{\left(G_n(S_{e_n}(\lambda K)+C_{n-1})+G_{n-1}(P_{e_n^\perp}(\lambda K)+C_{n-1})\right)^{n+1}}{(G_{n-1}(P_{e_n^\perp}(\lambda K)))^n}. \end{align*} $$

Therefore, taking into account that $S_{e_n}(\lambda K)=\lambda S_{e_n}(K)$ for any $\lambda>0$ and dividing both sides of the inequality by $\lambda ^{2n}=\frac {\lambda ^{n(n+1)}}{\lambda ^{n(n-1)}}$ , we have that for every $\lambda>0,$

$$ \begin{align*} &\frac{(n+1)B_{m_{0,\lambda}}(n+1)}{B_{m_{0,\lambda}}(1)^{-(n+1)}}2^n\frac{1}{\lambda^n}\int_{\lambda S_{e_n}(K)}|\langle \frac{x}{\lambda},e_n\rangle|^ndG_n(x)\leq\\ \leq &\frac{\left(\frac{G_n(\lambda S_{e_n}(K)+C_{n-1})}{\lambda^n}+\frac{1}{\lambda}\frac{G_{n-1}(P_{e_n^\perp}(\lambda K)+C_{n-1})}{\lambda^{n-1}}\right)^{n+1}}{\left(\frac{G_{n-1}(\lambda P_{e_n^\perp}(K))}{\lambda^{n-1}}\right)^n}. \end{align*} $$

Taking the limit as $\lambda \to \infty $ and taking into account that ${\lim _{\lambda \to \infty }\max _{x\in S_{e_n}(\lambda K)\cap \mathbb {Z}^n}\langle x,e_n\rangle =\infty }$ , and therefore ${\lim _{\lambda \to \infty }m_{0,\lambda }=\infty }$ , we obtain, using Lemma 5.1,

$$ \begin{align*}\frac{(n+1){{2n}\choose{n-1}}}{n^{n+1}}2^n\int_{S_{e_n}(K)}|\langle x,e_n\rangle|^ndx\leq\frac{(\mathrm{vol}_n(S_{e_n}(K))+0)^{n+1}}{(\mathrm{vol}_{n-1}(P_{e_n^\perp}(K)))^n}=\frac{(\mathrm{vol}_n(K))^{n+1}}{(\mathrm{vol}_{n-1}(P_{e_n^\perp}(K)))^n}, \end{align*} $$

where we have also used (2.1) and (2.3), since the function $|\langle x,e_n\rangle |^n$ is Riemann-integrable on $S_{e_n}(K)$ .

Equivalently,

$$ \begin{align*}\frac{{{2n}\choose{n}}}{n^n}2^n\int_{S_{e_n}(K)}|\langle x,e_n\rangle|^ndx\leq\frac{(\mathrm{vol}_n(K))^{n+1}}{(\mathrm{vol}_{n-1}(P_{e_n^\perp}(K)))^n}. \end{align*} $$

Proof First of all, let us define $\tilde {K}_p(\tilde {g}_{K+C_n})$ as the following set:

$$ \begin{align*} \tilde{K}_p(\tilde{g}_{K+C_n})&=\left\{x\in\mathbb{R}^n\,:\,p\int_0^\infty r^{p-1}\tilde{g}_{K+C_n}(rx)dr\geq \tilde{g}_K(0)\right\}\\&=\left\{x\in\mathbb{R}^n\,:\,p\int_0^\infty r^{p-1}G_n((K+C_n)\cap(rx+K+C_n))dr\geq G_n(K)\right\}. \end{align*} $$

Notice that since the right-hand side in the inequality defining this set is $\tilde {g}_K(0)$ rather than $\tilde {g}_{K+C_n}(0)$ , this set is not $K_p(\tilde {g}_{K+C_n})$ but a dilation of it. Indeed, for any $x\in \mathbb {R}^n$ and any $\lambda>0,$

$$ \begin{align*}p\int_0^\infty r^{p-1}\tilde{g}_{K+C_n}(\lambda r x)dr=\frac{p}{\lambda^p}\int_0^\infty s^{p-1}\tilde{g}_{K+C_n}(sx)ds \end{align*} $$

and then, if $0<\lambda \leq 1$ ,

$$ \begin{align*}p\int_0^\infty r^{p-1}\tilde{g}_{K+C_n}(\lambda r x)dr\geq p\int_0^\infty r^{p-1}\tilde{g}_{K+C_n}(rx)dr. \end{align*} $$

Thus, if $x\in \tilde {K}_p(\tilde {g}_{K+C_n})$ , we have that also $\lambda x\in \tilde {K}_p(\tilde {g}_{K+C_n})$ . Therefore, $\tilde {K}_p(\tilde {g}_{K+C_n})$ is a star set with $0$ as a center. Moreover, as in (2.13), for any $\theta \in S^{n-1}$ ,

$$ \begin{align*} \rho_{\tilde{K}_p(\tilde{g}_{K+C_n})}(\theta)&=\sup\left\{\lambda>0\,:\,p\int_0^\infty r^{p-1}\tilde{g}_{K+C_n}(\lambda r\theta)dr\geq \tilde{g}_K(0)\right\}\\&=\sup\left\{\lambda>0\,:\,\frac{p}{\lambda^p}\int_0^\infty s^{p-1}\tilde{g}_{K+C_n}(s\theta)ds\geq \tilde{g}_K(0)\right\}\\&=\left(\frac{p}{\tilde{g}_K(0)}\int_0^\infty s^{p-1}\tilde{g}_{K+C_n}(s\theta)ds\right)^{\frac{1}{p}}\\&=\left(\frac{p}{G_n(K)}\int_0^\infty r^{p-1}G_n((K+C_n)\cap(r\theta+K+C_n))dr\right)^{\frac{1}{p}}\\&=\left(\frac{G_n(K+C_n)}{G_n(K)}\right)^{\frac{1}{p}}\rho_{K_p(\tilde{g}_{K+C_n})}(\theta). \end{align*} $$

Therefore, $\tilde {K}_p(\tilde {g}_{K+C_n})$ is the dilation of $K_p(\tilde {g}_{K+C_n})$ in the right-hand side of (6.2):

$$ \begin{align*}\tilde{K}_p(\tilde{g}_{K+C_n})=\left(\frac{G_n(K+C_n)}{G_n(K)}\right)^{\frac{1}{p}}K_p(\tilde{g}_{K+C_n}). \end{align*} $$

Let $x,y\in K_p(\tilde {g}_K)$ , $\lambda ,\mu \in [0,1]$ such that $\lambda +\mu =1$ , and $\gamma =\frac {1}{p}$ . Let us define the functions:

• • $h(t)=\tilde {g}_{K+C_n}(t^\gamma (\lambda x+\mu y))$ , $t> 0$ ,

• • $w(r)=\tilde {g}_K(r^\gamma x)$ , $r> 0$ ,

• • $g(s)=\tilde {g}_K(s^\gamma y)$ , $s>0$ .

Denoting by $M_{-\gamma }^\lambda (r,s)$ for any $r,s>0$ the number

$$ \begin{align*}M_{-\gamma}^\lambda(r,s)=(\lambda r^{-\gamma}+\mu s^{-\gamma})^{-\frac{1}{\gamma}}, \end{align*} $$

our purpose is to show that for any $r,s>0,$ we have that

(6.3) $$ \begin{align} h(M_{-\gamma}^\lambda(r,s))\geq w(r)^{\frac{\lambda s^\gamma}{\lambda s^\gamma+\mu r^\gamma}}g(s)^{\frac{\mu r^\gamma}{\lambda s^\gamma+\mu r^\gamma}} \end{align} $$

in order to apply [Reference Brazitikos, Giannopoulos, Valettas and Vritsiou9, Theorem 2.5.4] and obtain

(6.4) $$ \begin{align} \int_0^\infty h(t)dt\geq M_{-\gamma}^\lambda\left(\int_0^\infty w(r)dr,\int_0^\infty g(s)ds\right). \end{align} $$

If $K\cap (r^\gamma x+K)=\emptyset $ or $K\cap (s^\gamma y+K)=\emptyset $ then $w(r)=0$ or $g(s)=0$ and inequality (6.3) is trivial. Otherwise, calling

• • $\lambda _1=\frac {\lambda s^\gamma }{\lambda s^\gamma +\mu r^\gamma }$

• • $\mu _1=\frac {\mu r^\gamma }{\lambda s^\gamma +\mu r^\gamma }$

we have that $\lambda _1+\mu _1=1$ and, since K is convex,

$$ \begin{align*} &(K+C_n)\cap(\lambda_1r^\gamma x+\mu_1s^\gamma y+K+C_n)\supseteq\\& \quad \supseteq \lambda_1(K\cap(r^\gamma x+K))+\mu_1(K\cap(s^\gamma y+K))+C_n \end{align*} $$

and then, by the discrete Brunn–Minkowski inequality (Theorem 2.2),

$$ \begin{align*} &G_n((K+C_n)\cap(\lambda_1r^\gamma x+\mu_1s^\gamma y+K+C_n))\geq\\&\quad \geq G_n(\lambda_1(K\cap(r^\gamma x+K))+\mu_1(K\cap(s^\gamma y+K))+C_n)\\&\quad \geq \left(\lambda_1G_n^{\frac{1}{n}}(K\cap(r^\gamma x+K))+\mu_1G_n^{\frac{1}{n}}(K\cap(s^\gamma y+K))\right)^n\\&\quad \geq G_n^{\lambda_1}(K\cap(r^\gamma x+K))G_n^{\mu_1}(K\cap(s^\gamma y+K)). \end{align*} $$

Therefore,

$$ \begin{align*} h(M_{-\gamma}^\lambda(r,s))&=\tilde{g}_{K+C_n}\left(\frac{1}{\lambda r^{-\gamma}+\mu s^{-\gamma}}(\lambda x+\mu y)\right)\\&=\tilde{g}_{K+C_n}\left(\frac{\lambda s^\gamma}{\lambda s^\gamma+\mu r^\gamma}r^\gamma x+\frac{\mu r^\gamma}{\lambda s^\gamma+\mu r^\gamma}s^\gamma y\right)\\&=\tilde{g}_{K+C_n}\left(\lambda_1r^\gamma x+\mu_1s^\gamma y\right)\\&\geq \tilde{g}_K^{\lambda_1}(r^\gamma x)\tilde{g}_K^{\mu_1}(s^\gamma y)\\&=\tilde{g}_K(r^\gamma x)^{\frac{\lambda s^\gamma}{\lambda s^\gamma+\mu r^\gamma}}\tilde{g}_K(s^\gamma y)^{\frac{\mu r^\gamma}{\lambda s^\gamma+\mu r^\gamma}}\\&=w(r)^{\frac{\lambda s^\gamma}{\lambda s^\gamma+\mu r^\gamma}}g(s)^{\frac{\mu r^\gamma}{\lambda s^\gamma+\mu r^\gamma}}. \end{align*} $$

Thus, by [Reference Brazitikos, Giannopoulos, Valettas and Vritsiou9, Theorem 2.5.4] we have (6.4). Equivalently, taking into account that $-\gamma <0$ ,

$$ \begin{align*} \left(\int_0^\infty \tilde{g}_{K+C_n}(t^\gamma(\lambda x+\mu y))dt\right)^{-\gamma}&\leq \lambda\left(\int_0^\infty \tilde{g}_K(r^\gamma x)dr\right)^{-\gamma}\\& \quad +\mu\left(\int_0^\infty \tilde{g}_K(s^\gamma y)ds\right)^{-\gamma}. \end{align*} $$

Changing variables, using that $\gamma =\frac {1}{p}$ , and taking into account that $x,y\in K_p(\tilde {g}_K)$ ,

$$ \begin{align*} \left(\int_0^\infty pt^{p-1}\tilde{g}_{K+C_n}(t(\lambda x+\mu y))dt\right)^{-\frac{1}{p}}&\leq \lambda\left(\int_0^\infty pr^{p-1}\tilde{g}_K(r x)dr\right)^{-\frac{1}{p}}\\&\quad +\mu\left(\int_0^\infty ps^{p-1}\tilde{g}_K(s y)ds\right)^{-\frac{1}{p}}\\&\leq \lambda G_n(K)^{-\frac{1}{p}}+\mu G_n(K)^{-\frac{1}{p}}=G_n(K)^{-\frac{1}{p}}. \end{align*} $$

Therefore,

$$ \begin{align*}\int_0^\infty pt^{p-1}\tilde{g}_{K+C_n}(t(\lambda x+\mu y))dt\geq G_n(K) \end{align*} $$

and

$$ \begin{align*}\lambda x+\mu y\in \tilde{K}_p(\tilde{g}_{K+C_n})=\left(\frac{G_n(K+C_n)}{G_n(K)}\right)^{\frac{1}{p}}K_p(\tilde{g}_{K+C_n}).\\[-41pt] \end{align*} $$

Proof Integrating in polar coordinates, we have that

$$ \begin{align*} &\int_{K_{n+p}(\tilde{g}_K)}h(x)dx=n\mathrm{vol}_n(B_2^n)\int_{S^{n-1}}\int_0^{\rho_{K_{n+p}(\tilde{g}_K)}(\theta)}r^{n-1}h(r\theta)drd\sigma(\theta)\\&\quad =n\mathrm{vol}_n(B_2^n)\int_{S^{n-1}}\int_0^{\rho_{K_{n+p}(\tilde{g}_K)}(\theta)}r^{n+p-1}h(\theta)drd\sigma(\theta)\\&\quad =\frac{n}{n+p}\mathrm{vol}_n(B_2^n)\int_{S^{n-1}}\rho_{K_{n+p}(\tilde{g}_K)}^{n+p}(\theta)h(\theta)d\sigma(\theta)\\&\quad =\frac{n\mathrm{vol}_n(B_2^n)}{G_n(K)}\int_{S^{n-1}}\int_0^\infty r^{n+p-1}G_n(K\cap(r\theta+K))h(\theta)drd\sigma(\theta)\\&\quad =\frac{n\mathrm{vol}_n(B_2^n)}{G_n(K)}\int_{S^{n-1}}\int_0^\infty r^{n-1}G_n(K\cap(r\theta+K))h(r\theta)drd\sigma(\theta)\\&\quad =\frac{1}{G_n(K)}\int_{\mathbb{R}^n}h(x)G_n(K\cap(x+K))dx. \end{align*} $$

Notice that if $h(x)=1$ , which is homogeneous of degree $0$ , we have that

$$ \begin{align*} \mathrm{vol}_n(K_n(\tilde{g}_K))&=\frac{1}{G_n(K)}\int_{\mathbb{R}^n}G_n(K\cap(x+K))dx\\&=\frac{1}{G_n(K)}\int_{\mathbb{R}^n}\sum_{y\in \mathbb{Z}^n}\chi_{K}(y)\chi_{x+K}(y)dx\\&=\frac{1}{G_n(K)}\sum_{y\in \mathbb{Z}^n}\int_{\mathbb{R}^n}\chi_{K}(y)\chi_{y-K}(x)dx\\&=\frac{1}{G_n(K)}\sum_{y\in \mathbb{Z}^n}\chi_{K}(y)\mathrm{vol}_n(K)=\mathrm{vol}_n(K).\\[-44pt] \end{align*} $$

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in K$ , $\theta \in S^{n-1}$ and $p>0$ . Let $m_{p,\theta }$ and $g_{p,\theta }$ be defined as in the statement. First of all, notice that since $0\in K,$ we have that $0\in \text {int}(K+C_n)$ and then $m_{p,\theta }>0$ . Notice also that for every convex set L, every $\theta \in S^{n-1}$ , and every $0\leq r_1<r_2$ , we have that $L\cap (r_1\theta +L)\supseteq L\cap (r_2\theta +L)$ and then the functions $\tilde {g}_K(r\theta )$ and $\tilde {g}_{K+C_n}(r\theta )$ are decreasing in $r\in [0,\infty )$ . Moreover, since $K\subseteq K+C_n$ , we have that $K\cap (x+K)\subseteq (K+C_n)\cap (x+K+C_n)$ for every $x\in \mathbb {R}^n$ and then $\tilde {g}_K(x)\leq \tilde {g}_{K+C_n}(x)$ for every $x\in \mathbb {R}^n$ . Furthermore, since K is a compact convex set, $\tilde {g}_K(r\theta )$ is continuous from the left in $r\in [0,\infty )$ and since $K+C_n$ is an open bounded convex set, $\tilde {g}_{K+C_n}(r\theta )$ is continuous from the right in $r\in [0,\infty )$ . Let us call

(6.5) $$ \begin{align} M_\theta=\max\{r\geq0\,:\, \tilde{g}_{K}(r\theta)\geq 1\} \end{align} $$

and

$$ \begin{align*}\tilde{M}_\theta=\sup\{r\geq0\,:\, \tilde{g}_{K+C_n}(r\theta)\geq 1\} \end{align*} $$

and notice that, necessarily, $m_{p,\theta }\geq M_\theta $ . Otherwise, if $m_{p,\theta }<M_\theta \leq \tilde {M}_\theta $ then $K\cap (m_{p,\theta }\theta +K)\neq \emptyset $ and for every $0\leq r\leq m_{p,\theta }$ , we have that

$$ \begin{align*}(K+C_n)\cap\left(r\theta+K+C_n\right)\supseteq\left(1-\frac{r}{m_{p,\theta}}\right)K+\frac{r}{m_{p,\theta}}(K\cap(m_{p,\theta}\theta+K))+C_n \end{align*} $$

and then, by the discrete Brunn–Minkowski inequality (Theorem 2.2),

$$ \begin{align*} \tilde{g}_{K+C_n}^{\frac{1}{n}}(r\theta)&\geq G_n\left(\left(1-\frac{r}{m_{p,\theta}}\right)K+\frac{r}{m_{p,\theta}}(K\cap(m_{p,\theta}\theta+K))+C_n\right)^{\frac{1}{n}}\\&\geq \left(1-\frac{r}{m_{p,\theta}}\right)G_n(K)^{\frac{1}{n}}+\frac{r}{m_{p,\theta}}G_n(K\cap(m_{p,\theta}\theta+K))^{\frac{1}{n}}\\&\geq \left(1-\frac{r}{m_{p,\theta}}\right)G_n(K)^{\frac{1}{n}}. \end{align*} $$

Therefore, for every $0\leq r\leq m_{p,\theta }$ , we have that

$$ \begin{align*}\tilde{g}_{K+C_n}(r\theta)\geq g_{p,\theta}(r) \end{align*} $$

and then, taking into account that $\tilde {g}_{K+C_n}(r\theta )\geq 1$ for every $0\leq r<\tilde {M}_\theta $ ,

$$ \begin{align*} m_{p,\theta}^p&=\frac{{{n+p}\choose{n}}}{G_n(K)}\int_0^\infty pr^{p-1}\tilde{g}_{K+C_n}(r\theta)dr\\&=\frac{{{n+p}\choose{n}}}{G_n(K)}\int_0^{\tilde{M}_\theta} pr^{p-1}\tilde{g}_{K+C_n}(r\theta)dr\\&>\frac{{{n+p}\choose{n}}}{G_n(K)}\int_0^{m_{p,\theta}} pr^{p-1}\tilde{g}_{K+C_n}(r\theta)dr\\&\geq \frac{{{n+p}\choose{n}}}{G_n(K)}\int_0^{m_{p,\theta}} pr^{p-1}g_{p,\theta}(r)dr\\&={{n+p}\choose{n}}\int_0^{m_{p,\theta}} pr^{p-1}\left(1-\frac{r}{m_{p,\theta}}\right)dr=m_{p,\theta}^p, \end{align*} $$

which is a contradiction.

Since we trivially have that for any $r>\tilde {M}_\theta $ ,

$$ \begin{align*}0=\tilde{g}_{K+C_n}(r\theta)\leq g_{p,\theta}(r), \end{align*} $$

we can define

$$ \begin{align*}r_0(p,\theta)=\inf\{r\geq0\,:\tilde{g}_{K+C_n}(r\theta)\leq g_{p,\theta}(r)\}<\infty. \end{align*} $$

From the definition of $r_0(p,\theta )$ as an infimum, we trivially have that for every $0\leq r<r_0(p,\theta ),$

$$ \begin{align*}\tilde{g}_{K+C_n}(r\theta)> g_{p,\theta}(r). \end{align*} $$

Besides, since $\tilde {g}_{K+C_n}(r\theta )$ is continuous from the right on $[0,\infty )$ and $\tilde {g}_K(x)\leq \tilde {g}_{K+C_n}(x)$ for every $x\in \mathbb {R}^n$ , from the definition of $r_0(p,\theta )$ as an infimum, we obtain

(6.6) $$ \begin{align} \tilde{g}_{K}(r_0(p,\theta)\theta)\leq\tilde{g}_{K+C_n}(r_0(p,\theta)\theta)\leq g_{p,\theta}(r_0(p,\theta)). \end{align} $$

and $r_0(p,\theta )$ is a minimum.

Notice that also, from the definition of $M_\theta $ , for every $r>M_\theta $ , we have that

$$ \begin{align*}g_{p,\theta}(r)\geq \tilde{g}_K(r\theta)=0. \end{align*} $$

Therefore, if $r_0(p,\theta )> M_\theta ,$ the theorem is proved. Thus, we will assume that $r_0(p,\theta )\leq M_\theta $ .

If $r_0(p,\theta )= M_\theta $ then, by (6.6), we have that

$$ \begin{align*}\tilde{g}_{K}(M_\theta\theta)\leq\tilde{g}_{K+C_n}(M_\theta\theta)\leq g_{p,\theta}(M_\theta) \end{align*} $$

and the theorem is proved.

Let us assume, then, that $r_0(p,\theta )<M_\theta $ and let us prove that if $r_0(p,\theta )< r\leq M_\theta $ then $g_{p,\theta }(r)\geq \tilde {g}_K(r\theta )$ .

If $r_0(p,\theta )<r\leq M_\theta $ then, taking $\lambda =\frac {r_0(p,\theta )}{r}\in [0,1),$ we have

$$ \begin{align*}(K+C_n)\cap(r_0(p,\theta)\theta+K+C_n)\supseteq (1-\lambda) K+\lambda (K\cap(r\theta+K))+C_n \end{align*} $$

and then by the discrete Brunn–Minkowski inequality (Theorem 2.2),

$$ \begin{align*} \tilde{g}_{K+C_n}^{\frac{1}{n}}(r_0(p,\theta)\theta)&\geq G_n\left((1-\lambda) K+\lambda (K\cap(r\theta+K)+C_n)\right)^{\frac{1}{n}}\\&\geq (1-\lambda)G_n(K)^{\frac{1}{n}}+\lambda G_n(K\cap(r\theta+K))^{\frac{1}{n}}\\&=(1-\lambda)\tilde{g}_K^{\frac{1}{n}}(0)+\lambda \tilde{g}_K^{\frac{1}{n}}(r\theta). \end{align*} $$

Taking into account that $r_0(p,\theta )<r\leq M_\theta \leq m_{p,\theta }$

$$ \begin{align*} g_{p,\theta}^{\frac{1}{n}}(r_0(p,\theta))&=\left(1-\frac{r_0(p,\theta)}{m_{p,\theta}}\right)G_n(K)^{\frac{1}{n}}\\&=(1-\lambda)G_n(K)^{\frac{1}{n}}+\lambda\left(1-\frac{r}{m_{p,\theta}}\right)G_n(K)^{\frac{1}{n}}\\&=(1-\lambda)\tilde{g}_K^{\frac{1}{n}}(0)+\lambda g_{p,\theta}^{\frac{1}{n}}(r). \end{align*} $$

Thus, by (6.6), if $r_0(p,\theta )<r\leq M_\theta $

$$ \begin{align*} (1-\lambda)\tilde{g}_K^{\frac{1}{n}}(0)+\lambda g_{p,\theta}^{\frac{1}{n}}(r)&=g_{p,\theta}^{\frac{1}{n}}(r_0(p,\theta))\geq\tilde{g}_{K+C_n}^{\frac{1}{n}}(r_0(p,\theta)\theta)\\&\geq (1-\lambda)\tilde{g}_K^{\frac{1}{n}}(0)+\lambda \tilde{g}_K^{\frac{1}{n}}(r\theta) \end{align*} $$

and then

$$ \begin{align*}g_{p,\theta}(r)\geq \tilde{g}_K(r\theta).\\[-33pt] \end{align*} $$

Proof of Theorem 6.2

Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in K$ . Let $\theta \in S^{n-1}$ . Fix $p>0$ and let, as in Lemma 6.6, $m_{p,\theta }$ be defined as

$$ \begin{align*}m_{p,\theta}=\left(\frac{{{n+p}\choose{n}}}{G_n(K)}\int_0^\infty pr^{p-1}\tilde{g}_{K+C_n}(r\theta)dr\right)^{\frac{1}{p}} \end{align*} $$

and $g_{p,\theta }:[0,\infty )\to [0,\infty )$ be the function given by

$$ \begin{align*}g_{p,\theta}(r)=\begin{cases} \left(1-\frac{r}{m_{p,\theta}}\right)^nG_n(K)&\text{ if }0\leq r\leq m_{p,\theta}\\ 0 &\text{ otherwise}. \end{cases} \end{align*} $$

Notice that, changing variables $r=m_{p,\theta }s$ , for every $q>0,$ we have that

$$ \begin{align*} \frac{{{n+q}\choose{n}}}{G_n(K)}\int_0^\infty qr^{q-1}g_{p,\theta}(r)dr&=\frac{{{n+q}\choose{n}}}{G_n(K)}\int_0^{m_{p,\theta}} qr^{q-1}\left(1-\frac{r}{m_{p,\theta}}\right)^nG_n(K)dr\\&={{n+q}\choose{n}}m_{p,\theta}^q\int_0^1 qs^{q-1}(1-s)^nds\\&=m_{p,\theta}^q. \end{align*} $$

In particular, taking $q=p$ , we have that

$$ \begin{align*}\int_0^\infty r^{p-1}g_{p,\theta}(r)dr=\int_0^\infty r^{p-1}\tilde{g}_{K+C_n}(r\theta)dr. \end{align*} $$

Let $q>p$ . Taking $r_0(p,\theta )$ provided by Lemma 6.6, we have

$$ \begin{align*} &\int_0^{r_0(p,\theta)}r^{q-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr-\int_{r_0(p,\theta)}^\infty r^{q-1}\left(g_{p,\theta}(r)-\tilde{g}_K(r\theta)\right)dr\\& \quad =\int_0^{r_0(p,\theta)}r^{q-p}r^{p-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr\\& \qquad -\int_{r_0(p,\theta)}^\infty r^{q-p}r^{p-1}\left(g_{p,\theta}(r)-\tilde{g}_K(r\theta)\right)dr\\& \quad \leq r_0(p,\theta)^{q-p}\int_0^{r_0(p,\theta)}r^{p-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr\\& \qquad -r_0(p,\theta)^{q-p}\int_{r_0}^\infty r^{p-1}\left(g_{p,\theta}(r)-\tilde{g}_K(r\theta)\right)dr\\& \quad =r_0(p,\theta)^{q-p}\int_0^{r_0(p,\theta)}r^{p-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr\\& \qquad +r_0(p,\theta)^{q-p}\int_{r_0(p,\theta)}^\infty r^{p-1}\left(\tilde{g}_K(r\theta)-g_{p,\theta}(r)\right)dr\\& \quad \leq r_0(p,\theta)^{q-p}\int_0^{r_0(p,\theta)}r^{p-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr\\& \qquad +r_0(p,\theta)^{q-p}\int_{r_0(p,\theta)}^\infty r^{p-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr\\& \quad =r_0(p,\theta)^{q-p}\int_0^{\infty}r^{p-1}\left(\tilde{g}_{K+C_n}(r\theta)-g_{p,\theta}(r)\right)dr=0. \end{align*} $$

Therefore, we have that

$$ \begin{align*}\int_0^{r_0(p,\theta)}r^{q-1}\tilde{g}_{K+C_n}(r\theta)dr+\int_{r_0(p,\theta)}^\infty r^{q-1}\tilde{g}_{K}(r\theta)dr\leq\int_0^\infty r^{q-1}g_{p,\theta}(r)dr \end{align*} $$

and then, since $\tilde {g}_K(x)\leq \tilde {g}_{K+C_n}(x)$ for every $x\in \mathbb {R}^n$ ,

$$ \begin{align*}\int_{0}^\infty r^{q-1}\tilde{g}_{K}(r\theta)dr\leq\int_0^\infty r^{q-1}g_{p,\theta}(r)dr. \end{align*} $$

Consequently,

$$ \begin{align*} {{n+q}\choose{n}}^{\frac{1}{q}}\rho_{K_q(\tilde{g}_K)}(\theta)&=\left(\frac{{{n+q}\choose{n}}}{G_n(K)}\int_0^\infty qr^{q-1}\tilde{g}_{K}(r\theta)dr\right)^{\frac{1}{q}}\\&\leq\left(\frac{{{n+q}\choose{n}}}{G_n(K)}\int_0^\infty qr^{q-1}g_{p,\theta}(r)dr\right)^{\frac{1}{q}}=m_{p,\theta}\\&=\left(\frac{{{n+p}\choose{n}}}{G_n(K)}\int_0^\infty pr^{p-1}\tilde{g}_{K+C_n}(r\theta)dr\right)^{\frac{1}{p}}\\&=\left(\frac{{{n+p}\choose{n}}G_n(K+C_n)}{G_n(K)}\right)^{\frac{1}{p}}\rho_{K_p(\tilde{g}_{K+C_n})}(\theta). \end{align*} $$

Since this is true for every $\theta \in S^{n-1}$ , we have the inclusion relation stated in the theorem.

Proof Let $K\subseteq \mathbb {R}^n$ be a convex body with $0\in K$ , $\theta \in S^{n-1}$ , and $p>0$ . We have seen in Theorem 6.2 that for any $0<p<q,$ we have that

$$ \begin{align*}{{n+q}\choose{n}}^{\frac{1}{q}}\rho_{K_q(\tilde{g}_K)}(\theta)\leq\left(\frac{{{n+p}\choose{n}}G_n(K+C_n)}{G_n(K)}\right)^{\frac{1}{p}}\rho_{K_p(\tilde{g}_{K+C_n})}(\theta). \end{align*} $$

Taking limits as $q\to \infty $ we have that ${\lim _{q\to \infty }{{n+q}\choose {n}}^{\frac {1}{q}}=1}$ . Besides, calling as in (6.5), $M_\theta =\max \{r\geq 0\,:\, \tilde {g}_{K}(r\theta )\geq 1\}$ , if we consider the measure $d\nu (r)=\tilde {g}_K(r\theta )dr$ on the interval $[0,M_\theta ]$ , denote by $L^q(\nu )$ the corresponding Lebesgue space on $([0,M_\theta ],d \nu )$ , and take $h(r)=r$ , then

$$ \begin{align*}M_\theta=\Vert h\Vert_\infty=\lim_{q\to\infty}\Vert h\Vert_{L^{q-1}(\nu)}=\lim_{q\to\infty}\left(\int_0^{M_\theta}r^{q-1}\tilde{g}_K(r\theta)dr\right)^{\frac{1}{q-1}}. \end{align*} $$

Therefore,

$$ \begin{align*} \lim_{q\to\infty}\rho_{K_q(\tilde{g}_K)}&=\lim_{q\to\infty}\left(\frac{q}{G_n(K)}\int_0^{\infty}r^{q-1}\tilde{g}_K(r\theta)dr\right)^{\frac{1}{q}}\\&=\lim_{q\to\infty}\left(\frac{q}{G_n(K)}\int_0^{M_\theta}r^{q-1}\tilde{g}_K(r\theta)dr\right)^{\frac{1}{q}}\\&=\lim_{q\to\infty}\frac{q^{\frac{1}{q}}}{G_n(K)^{\frac{1}{q}}}\left(\int_0^{M_\theta}r^{q-1}\tilde{g}_K(r\theta)dr\right)^{\frac{1}{q-1}\frac{q-1}{q}}\\&=M_\theta. \end{align*} $$

Taking into account Remark 6.7, we have

$$ \begin{align*}\rho_{(K\cap \mathbb{Z}^n)-K}(\theta)\leq\left(\frac{{{n+p}\choose{n}}G_n(K+C_n)}{G_n(K)}\right)^{\frac{1}{p}}\rho_{K_p(\tilde{g}_{K+C_n})}(\theta). \end{align*} $$

Since this is true for every $\theta \in S^{n-1}$ , we obtain the result.