text:
\section{Introduction and notation}

As usual, we write $\R^n$ to represent the $n$-dimensional
Euclidean space, and we denote by $\e_i$ the $i$-th canonical unit
vector. We set $(x,y)$ for the open segment with endpoints $x,y\in\R^n$.
The $n$-dimensional volume of a compact set
$K\subset\R^n$, i.e., its $n$-dimensional Lebesgue measure, is
denoted by $\vol(K)$, and as a discrete counterpart, we use $|A|$
to represent the cardinality of a finite subset $A\subset\R^n$.

Let $\Z^n$ be the integer lattice, i.e., the lattice of all points with
integral coordinates in $\R^n$.
We will denote by $\G{n}(\cdot)$ the lattice point enumerator for the integer lattice $\Z^n$, i.e., $\G{n}(M)=|M\cap\Z^n|$.
By $\floor{x}$ we denote the floor function of the real number $x$, i.e., the
greatest integer less than or equal to $x$.
Similarly, $\ceil{x}$ represents the ceiling function of $x$, namely, the least integer greater than or equal to $x$.

Finally, given a set $M\subset\R^n$, $\chi_{_M}$ represents the characteristic function of $M$ and, moreover, we denote by $\inter M$, $\bd M$ and $\cl M$ its interior, boundary and closure, respectively.
Furthermore, given $r>0$, $rM$ stands for the set $\{rm:\, m\in M\}$.

\medskip

Relating the volume of the Minkowski addition of two sets in terms of their volumes, one is led to the famous \emph{Brunn-Minkowski inequality} (for extensive survey articles on this and related  inequalities we refer the
reader to \cite{Brt,G}).
One form of it asserts that if $\lambda\in(0,1)$ and $K$ and $L$ are non-empty compact subsets of $\R^n$ then
\begin{equation}\label{e:BM}
\vol\bigl((1-\lambda)K+\lambda L\bigr)^{1/n}\geq
(1-\lambda)\vol(K)^{1/n}+\lambda\vol(L)^{1/n}.
\end{equation}
Here $+$ is used for the Minkowski sum, i.e.,
$A+B=\{a+b:\, a\in A, \, b\in B\}$ for any non-empty sets $A, B\subset\R^n$.
Moreover, from the homogeneity of the volume, \eqref{e:BM} is equivalent to
\begin{equation}\label{e:BM_additive}
\vol\bigl(K+L\bigr)^{1/n}\geq
\vol(K)^{1/n}+\vol(L)^{1/n}.
\end{equation}

Next we move to the discrete setting, i.e., we consider finite
sets of (integer) points which are not necessarily
full-dimensional unless indicated otherwise. It can easily be seen
that one cannot expect to obtain a Brunn-Minkowski type inequality for
the cardinality in the classical form. Indeed, simply taking
$A=\{0\}$ to be the origin and any finite set $B\subset\Z^n$, we get
\[
|A+B|^{1/n}<|A|^{1/n}+|B|^{1/n}.
\]
Therefore, a discrete Brunn-Minkowski type inequality should either have a
different structure or involve modifications of the sets. A first example
is the simple inequality
\begin{equation}\label{e:|A+B|>|A|+|B|-1}
|A+B|\geq|A|+|B|-1,
\end{equation}
for finite $A,B\subset\Z^n$ (see e.g. \cite[Chapter~2]{Tao}).

In \cite{GG}, Gardner and Gronchi obtained an engaging
discrete Brunn-Min\-kows\-ki inequality: they proved that if $A,B$ are finite
subsets of the integer lattice $\Z^n$, with dimension $\dim B=n$, then
\begin{equation}\label{e:B-M_ineq_GG}
|A+B|\geq \bigl|D_{|A|}^B+D_{|B|}^B\bigr|.
\end{equation}
Here $D_{|A|}^B, D_{|B|}^B$ are {\it $B$-initial segments}: for $m\in\N$,
$D_m^B$ is the set of the first $m$ points of $\Z^n_+$ in the so-called
``$B$-order'', which is a particular order defined on $\Z^n_+$ depending
only on the cardinality of $B$. For a proper definition and a deep study
of it we refer the reader to~\cite{GG}. As consequences of
\eqref{e:B-M_ineq_GG} they also get two additional nice discrete
Brunn-Minkowski type inequalities:
\begin{equation*}
|A+B|^{1/n}\geq|A|^{1/n}+\dfrac{1}{(n!)^{1/n}}\bigl(|B|-n\bigr)^{1/n}
\end{equation*}
and, if $|B|\leq|A|$, then
\[
|A+B|\geq|A|+(n-1)|B|+\bigl(|A|-n\bigr)^{(n-1)/n}\bigl(|B|-n\bigr)^{1/n}-\dfrac{n(n-1)}{2}.
\]
These inequalities improve previous results obtained by Ruzsa in
\cite{Ru1,Ru2}.

\smallskip

An alternative to getting a ``classical'' Brunn-Minkowski type inequality
might be to transform (one of) the sets involved in the problem. In this regard,
in \cite{HCIYN} an extension $\bar{A}$ of $A$, is defined for any non-empty and finite set $A\subset\Z^n$
(for $n=1$ then $\bar{A}=A\cup\{\max(A)+1\}$, whereas for $n>1$ the set $\bar{A}$ is obtained by first adding the maximal cardinality section of $A$, and then applying the corresponding extension to every section of the latter
new set; we refer to Section 2.1 in \cite{HCIYN} for its precise definition and properties).
Using this technique the following discrete counterpart for \eqref{e:BM_additive} was shown:
\begin{thm}[\cite{HCIYN}]\label{t:B-M_discrete_adding}
Let $A,B\subset\Z^n$ be finite, $A,B\neq\emptyset$. Then
\begin{equation}\label{e:B-M_discrete_adding}
\bigl|\bar{A}+B\bigr|^{1/n}\geq |A|^{1/n}+|B|^{1/n}.
\end{equation}
The inequality is sharp.
\end{thm}

Aiming to get a discrete version of \eqref{e:BM}, it is worth noting the following:
for any pair of non-empty finite sets $A, B\subset\R^n$, by \eqref{e:|A+B|>|A|+|B|-1} (and using that $|A|, |B|\geq1$), one has
\begin{equation*}
\begin{split}
\bigl|(1-\lambda)A+\lambda B\bigr|&\geq \bigl|(1-\lambda)A\bigr|+\bigl|\lambda B\bigr|-1 =|A|+|B|-1 \\
&\geq (1-\lambda)|A|+\lambda |B|\geq \bigr((1-\lambda)|A|^{1/n}+\lambda |B|^{1/n}\bigr)^n,
\end{split}
\end{equation*}
where the last inequality follows from the convexity of the function $t\mapsto t^n$, for $t\geq0$.
Nevertheless this inequality is meaningless from a geometric point of view: the point is that while the quantities $|A|, |B|$ on the right-hand side are reduced by the factors $(1-\lambda)$ and $\lambda$, the sets $(1-\lambda)A$ and $\lambda B$ on the left-hand side have the same cardinality as $A$ and $B$, respectively. A possible solution would be to involve a more natural way to ``count points'' according to dilatations, namely, using the lattice point enumerator $\G{n}$ (for compact subsets of $\R^n$) instead of the cardinality $|\cdot|$ (for finite subsets of $\R^n$).

Again, one cannot expect to obtain a Brunn-Minkowski type inequality for $\G{n}$ in the classical form \eqref{e:BM} (which, as we have mentioned before, would be a similar situation to what happens for $|\cdot|$ regarding a discrete version of \eqref{e:BM_additive}). Indeed, just by taking $\lambda=1/2$, $K=[0,m-\epsilon]^n$ and $L=[0,m+\epsilon/2]^n$, with $m\in\N$ and $0<\epsilon<1$, one gets
\begin{equation}\label{e:counter_BM_G}
\G{n}\left(\frac{K+L}{2}\right)^{1/n}=m<m+\frac{1}{2}=\frac{\G{n}(K)^{1/n}+\G{n}(L)^{1/n}}{2}.
\end{equation}
Thus, as in \eqref{e:B-M_discrete_adding}, an alternative to get such an inequality for the lattice point enumerator would be to consider a certain extension of $(1-\lambda)K + \lambda L$. So, we pose the following question:
\begin{question}\label{q: BM_G}
Given compact sets $K, L\subset\R^n$ containing points from $\Z^n$,  what is the ``best'' way to define a set $M_\lambda$ such that
$(1-\lambda)K + \lambda L \subset M_\lambda$  and
\begin{equation*}
\G{n}(M_\lambda)^{1/n}\geq (1-\lambda)\G{n}(K)^{1/n}+\lambda\G{n}(L)^{1/n}
\end{equation*}
holds for a given $\lambda\in(0,1)$?
\end{question}

\section{Main results}
Here we give an answer to Question \ref{q: BM_G}, which we view as a discrete counterpart of \eqref{e:BM}:
\begin{theorem}\label{t: BM_lattice_point_no_G(K)G(L)>0}
Let $\lambda \in (0,1)$ and let $K, L \subset \R^n$ be non-empty bounded sets.
Then
\begin{equation}\label{e: BM_lattice_point_no_G(K)G(L)>0}
\G{n}\bigl((1-\lambda)K + \lambda L + (-1,1)^n\bigr)^{1/n}\geq (1-\lambda)\G{n}(K)^{1/n}+\lambda\G{n}(L)^{1/n}.
\end{equation}
The inequality is sharp.
\end{theorem}

We point out the following: when dealing with arbitrary non-empty subsets  $K, L\subset\R^n$ (i.e., not necessarily bounded), from \eqref{e: BM_lattice_point_no_G(K)G(L)>0} we immediately get that
\begin{equation*}
\begin{split}
\G{n}\bigl(M_\lambda + (-1,1)^n\bigr)^{1/n}
&\geq \sup_{m \in \N}\G{n}\bigl((1-\lambda)K_m + \lambda L_m + (-1,1)^n\bigr)^{1/n}\\
&\geq \sup_{m \in \N}\left((1-\lambda)\G{n}(K_m)^{1/n}+\lambda\G{n}(L_m)^{1/n}\right)\\
&=(1-\lambda)\G{n}(K)^{1/n}+\lambda\G{n}(L)^{1/n}
\end{split}
\end{equation*}
for any $\lambda \in (0,1)$, where $M_\lambda=(1-\lambda)K + \lambda L$, $K_m=K\cap [-m, m]^n$ and $L_m=L\cap [-m, m]^n$. So, for the sake of simplicity, we will present here the results in the setting of bounded sets, although they also hold in full generality.

The core ingredient in the proof of Theorem \ref{t: BM_lattice_point_no_G(K)G(L)>0} is its functional analogue.
To introduce such a result, we present an analytical counterpart for functions of the Brunn-Minkowski inequality,
the so-called \emph{Borell-Brascamp-Lieb inequality}, originally proved in
\cite{Borell} and \cite{BL}.
To this end, we first recall the definition of the $p$-mean $\Media{p}{\cdot}{\cdot}{\lambda}$
of two non-negative numbers, where $p$ is a parameter varying in $\R\cup\{\pm\infty\}$ (for a general reference for $p$-means of non-negative numbers, we refer the reader to the classic text of Hardy, Littlewood, and P\'olya \cite{HaLiPo} and to the handbook \cite{Bu}):
we consider first the case $p\in\R$, with $p\neq0$: given $a,b>0$ we set
\[
\Media{p}{a}{b}{\lambda}=\bigl((1-\lambda)a^p+\lambda b^p\bigr)^{1/p}.
\]
For $p=0$, we write $\Media{0}{a}{b}{\lambda}=a^{1-\lambda}b^{\lambda}$.
To complete the picture, for $p=\pm \infty$ we set $\Media{\infty}{a}{b}{\lambda}=\max\{a,b\}$ and
$\Media{-\infty}{x}{y}{\lambda}=\min\{a,b\}$. Finally, if $ab=0$, we define $\Media{p}{a}{b}{\lambda}=0$ for all $p\in\R\cup\{\pm\infty\}$. Note that $\Media{p}{a}{b}{\lambda}=0$, if $ab=0$, is redundant for all $p\leq0$, however it is relevant for $p>0$.
The reason to modify in this way the definition of $p$-mean given in \cite{HaLiPo} is due to the classical statement of the Borell-Brascamp-Lieb inequality, which is
collected below. In fact, without such a modification, the thesis of
the latter result would not have mathematical interest.

The following theorem (see
also \cite{G} for a detailed presentation), as previously stated, can be regarded as the
functional counterpart of the Brunn-Minkowski inequality.
In fact, a straightforward proof of \eqref{e:BM} for compact sets $K, L\subset\R^n$ of positive volume
can be obtained by applying \eqref{e:BBL_means} to the characteristic functions $f=\chi_{_K}$, $g=\chi_{_L}$ and
$h=\chi_{_{(1-\lambda)K + \lambda L}}$ with $p=\infty$. Moreover, in the literature, the case $p=0$ is referred to
as the \emph{Pr\'ekopa-Leindler inequality}, which is a powerful tool,
in particular when dealing with \emph{log-concave functions}.
\begin{thm}[The Borell-Brascamp-Lieb inequality]\label{t:BBL_means}
Let $\lambda\in(0,1)$. Let $-1/n\leq p\leq\infty$ and let $f,g,h:\R^n\longrightarrow\R_{\geq0}$ be measurable
functions such that
\begin{equation*}
h((1-\lambda)x + \lambda y)\geq\Media{p}{f(x)}{g(y)}{\lambda}
\end{equation*}
for all $x,y\in\R^n$. Then
\begin{equation}\label{e:BBL_means}
\int_{\R^n}h(x)\,\dlat x\geq
\Media{\frac{p}{np+1}}{\int_{\R^n}f(x)\,\dlat x}{\int_{\R^n}g(x)\,\dlat x}{\lambda}.
\end{equation}
\end{thm}

To state our main result and henceforth, we will need the following notation:
for a function $\phi:\R^n\longrightarrow\R_{\geq0}$, we denote by $\phi^{\symbol}:\R^n\longrightarrow\R_{\geq0}$ the function given by
\[\phi^{\symbol}(z) = \sup_{u\in(-1,1)^n}\phi(z+u)  \quad \text{ for all } z\in\R^n.\]
Such an extension of the function $\phi$ is nothing but the \emph{Asplund product} $\star$
(also referred to as the \emph{sup-convolution}, which can be regarded as the functional analogue of the Minkowski sum in the setting of log-concave functions, for which we refer the reader to \cite[Section~9.5]{Sch2} and the references therein)
of the functions $\phi$ and $\chi_{(-1,1)^n}$. Indeed,
\begin{equation*}
\begin{split}
\phi^{\symbol}(z) &= \sup_{u\in(-1,1)^n}\phi(z+u) =\sup_{u \in \R^n} \phi (z+ u)  \chi_{(-1,1)^n}(-u)\\
&=\sup_{u_1+u_2=z} \phi (u_1)  \chi_{(-1,1)^n}(u_2) = \bigl(\phi \star \chi_{(-1,1)^n}\bigr)(z).
\end{split}
\end{equation*}

\smallskip

Our main result reads as follows:
\begin{theorem}\label{t: BBL discreta (caso p-media)}
Let $\lambda\in(0,1)$ and let $K, L \subset \R^n$ be non-empty bounded sets. Let $-1/n\leq p\leq\infty$ and let
$f,g,h:\R^n\longrightarrow \R_{\geq 0}$ be functions such that
\[h((1-\lambda)x + \lambda y)\geq\Media{p}{f(x)}{g(y)}{\lambda}\]
for all $x \in K$, $y \in L$. Then
\begin{equation}\label{e: BBL discreta (caso p-media)}
\sum_{z \in M\cap\Z^n} h^\symbol(z)\geq \Media{\frac{p}{np+1}}{\sum_{x \in K\cap\Z^n} f(x)}{\sum_{y \in L\cap\Z^n} g(y)}{\lambda},
\end{equation}
where $M = (1-\lambda)K+\lambda L + (-1,1)^n$.
\end{theorem}

Let $\mathcal{B}=\{v_1\dots,v_n\}$ be a basis of an
$n$-dimensional lattice $\Lambda\subset\R^n$.
Under the same assumptions of the above result, we may consider the auxiliary functions
$f_{\mathcal{B}}, g_{\mathcal{B}}, h_{\mathcal{B}}: \R^n \rightarrow \R_{\geq0}$ defined by
\[ f_{\mathcal{B}}(x) = f(\varphi(x)), \quad g_{\mathcal{B}}(x) = g(\varphi(x))\text{ and } \quad h_{\mathcal{B}}(x) = h(\varphi(x)),\]
for all $x\in \R^n$, where $\varphi:\R^n\rightarrow\R^n$ is the linear (bijective) map given by $\varphi(x)=\sum_{i = 1}^n x_i v_i$ for any $x=(x_1,\dots,x_n)\in\R^n$. Thus, as an immediate consequence of the previous result (applied to the functions $f_{\mathcal{B}}, g_{\mathcal{B}}$ and $h_{\mathcal{B}}$), we get that
\begin{equation*}\label{e: BBL discrete (lattices)}
\sum_{z \in M\cap\Lambda} h^{\symbol_{_\mathcal{B}}}(z) \geq \Media{\frac{p}{np+1}}{\sum_{x \in K\cap\Lambda} f(x)}{\sum_{y \in L\cap\Lambda} g(y)}{\lambda},
\end{equation*}
where $M = (1-\lambda)K+\lambda L + \varphi((-1,1)^n)$ and $h^{\symbol_{_\mathcal{B}}}(z) = \sup_{u\in \varphi((-1,1)^n)} h(z+u)$ for all $z\in\R^n$.
So, Theorem \ref{t: BBL discreta (caso p-media)} also holds in the setting of an $n$-dimensional lattice $\Lambda\subset\R^n$.

\smallskip

We would like to point out that Theorem \ref{t:BBL_means} admits an equivalent version for $p$-sums (instead of $p$-means). In this regard, in \cite{IYN} a discrete version of the Borell-Brascamp-Lieb inequality for $p$-sums was provided. However, in contrast to the continuous setting, where one may directly obtain Theorem \ref{t:BBL_means} from the corresponding version for $p$-sums (and vice versa) because
of the homogeneity of the volume, one cannot expect to derive in a similar way a discrete version of Theorem \ref{t:BBL_means} (like Theorem \ref{t: BBL discreta (caso p-media)}) from \cite[Theorem~2.1]{IYN}.

\smallskip

In order to prove Theorem \ref{t: BBL discreta (caso p-media)}, we state an auxiliary result that will allow us to get the one-dimensional case of the above-mentioned Brunn-Minkowski type inequality for the lattice point enumerator (cf. Question \ref{q: BM_G}).
\begin{lemma}\label{l: G(M)+Z(M)>= (1-lam)G(K)+lamG(L)}
Let $\lambda \in (0,1)$ and let $K, L, M \subset \R$ be non-empty sets such that $(1-\lambda)K + \lambda L \subset M$. If $M = \bigcup_{i=1}^s [a_i, b_i]$ is a finite union of pairwise disjoint compact intervals then
\begin{equation}\label{e: G(M)+Z(M)>= (1-lam)G(K)+lamG(L)}
\G{1}(M) + \fun(M) \geq (1-\lambda) \G{1}(K) + \lambda \G{1}(L),
\end{equation}
where $\fun(M)$ denotes the number of non-integer extreme points of $M$, namely
\[ \fun(M) = | \{ a_i \notin \Z: 1\leq i \leq s\} | + | \{ b_i \notin \Z: 1\leq i \leq s \} |. \]
\end{lemma}
\begin{proof}
We show the result by induction on the number of intervals $s$ of $M$. For the case $s=1$, i.e., when
$M = [a_1, b_1]$ is a (non-empty) compact interval, we have on the one hand that
$\G{1}(M) = \floor{b_1} - \ceil{a_1} +1$.
Moreover, denoting by $a = \inf K$, $b = \sup K$, $c = \inf L$ and $d = \sup L$, we clearly get
$\G{1}(K)\leq \G{1}([a,b]) = \floor{b} - \ceil{a} +1$ and $\G{1}(L) \leq \G{1}([c,d]) = \floor{d} - \ceil{c} +1$.
On the other hand, the inclusion $(1-\lambda)K + \lambda L \subset M$ implies that
\begin{equation*}
\floor{b_1} \geq b_1 - \chi_{_{\R\setminus\Z}}(b_1)
\geq (1-\lambda)\floor{b} + \lambda\floor{d} - \chi_{_{\R\setminus\Z}}(b_1)
\end{equation*}
and
\begin{equation*}
-\ceil{a_1} \geq -a_1 - \chi_{_{\R\setminus\Z}}(a_1)
\geq -(1-\lambda)\ceil{a} - \lambda \ceil{c} - \chi_{_{\R\setminus\Z}}(a_1),
\end{equation*}
and thus
\begin{equation*}
\floor{b_1} - \ceil{a_1} +1
\geq (1-\lambda)(\floor{b}-\ceil{a} + 1) + \lambda(\floor{d}-\ceil{c}+1)-\fun(M).
\end{equation*}
This, together with the above upper bounds for the lattice point enumerator $G_1$ of $K$ and $L$, yields
$\G{1}(M)\geq (1-\lambda)\G{1}(K) + \lambda \G{1}(L)-\fun(M)$, which shows the case $s=1$.

So, we suppose that the inequality is true for $s\geq1$ and assume that
$M = \bigcup_{i=1}^{s+1} [a_i, b_i]$, where $b_i < a_{i+1}$ for all $1\leq i\leq s$.

Denoting by $M_1 = [a_1, b_1]$ and $M_2 = \bigcup_{i=2}^{s+1} [a_i, b_i]$, we may assume, without loss of generality, that $M_1\cap ((1-\lambda)K + \lambda L)\neq\emptyset$ (otherwise, the result follows directly from the induction hypothesis applied to the sets $K$, $L$ and $M_2$). Hence, we may define $m = \sup \bigl(M_1\cap((1-\lambda)K + \lambda L)\bigr)$ and then, since $K$ and $L$ are bounded (because $(1-\lambda)K + \lambda L\subset M$), there exist $k \in \cl K$ and $l \in \cl L$ such that $(1-\lambda)k + \lambda l = m$.
Thus, considering the sets $K_1 = \{x \in K: x \leq k\}$, $K_2 = K \setminus K_1$, $L_1 = \{x \in L: x \leq l\}$ and $L_2 = L \setminus L_1$, we clearly have that $(1-\lambda)K_1 + \lambda L_1 \subset M_1$ and $(1-\lambda)K_2 + \lambda L_2 \subset M_2$. Therefore, applying the induction hypothesis (and taking into account that $M_1$ are $M_2$ are disjoint), we get
\begin{equation*}
\begin{split}
\G{1}(M) + \fun(M) &= \G{1}(M_1) + \G{1}(M_2) + \fun(M_1) + \fun(M_2) \\
&\geq (1-\lambda)\bigl(\G{1}(K_1)+\G{1}(K_2)\bigr) + \lambda\bigl(\G{1}(L_1)+\G{1}(L_2)\bigr)\\
&= (1-\lambda) \G{1}(K) + \lambda \G{1}(L),
\end{split}
\end{equation*}
as desired.
\end{proof}

\begin{remark}
One might think that if $(1-\lambda)K + \lambda L$ is itself a finite union of (pairwise disjoint) compact intervals, the set $M = (1-\lambda)K + \lambda L$ would yield a tighter inequality in
\eqref{e: G(M)+Z(M)>= (1-lam)G(K)+lamG(L)} than other greater sets (with respect to set inclusion). Nevertheless, this is not true in general: if we consider $K = [-2m, -1] \cup [1,2m]$ with $m \in \N$, $L = \{0\}$ and $\lambda = 1/2$, then for $M = (1-\lambda)K + \lambda L=[-m, -1/2] \cup [1/2, m]$ we get $\G{1}(M) + \fun(M) = 2m + 2$ whereas for $M' = [-m, m]$ we have
$\G{1}(M') + \fun(M') = 2m+1$.
\end{remark}

We notice that, as shown in \eqref{e:counter_BM_G}, the quantity $\fun(M)$ cannot be (in general) omitted.
However, we can rewrite \eqref{e: G(M)+Z(M)>= (1-lam)G(K)+lamG(L)} to provide an answer to Question \ref{q: BM_G} for $n=1$, which is the one-dimensional case of Theorem \ref{t: BM_lattice_point_no_G(K)G(L)>0}:
\begin{lemma}\label{l: 1-dim_BM_G}
Let $\lambda \in (0,1)$ and let $K, L \subset \R$ be non-empty bounded sets. Then
\begin{equation}\label{e: 1-dim_BM_G}
\G{1}\bigl((1-\lambda)K + \lambda L + (-1,1)\bigr) \geq (1-\lambda) \G{1}(K) + \lambda \G{1}(L).
\end{equation}
The inequality is sharp.
\end{lemma}
\begin{proof}
Let $M = \bigcup_{x \in(1-\lambda)K + \lambda L} [\floor{x}, \ceil{x}]$. Clearly, $M$ is a finite union of compact intervals (since $K$ and $L$ are bounded) containing $(1-\lambda)K + \lambda L$. From Lemma
\ref{l: G(M)+Z(M)>= (1-lam)G(K)+lamG(L)} we then obtain $\G{1}(M)+\fun(M)\geq (1-\lambda) \G{1}(K) + \lambda \G{1}(L)$, which, together with the facts that $\fun(M) = 0$ and
$M \cap \Z =\left((1-\lambda)K + \lambda L + (-1,1)\right) \cap \Z$, yields \eqref{e: 1-dim_BM_G}.

Finally, in order to show that equality may be attained, we consider $K = L = [0,m]$ with $m \in \N$, for which $\G{1}\bigl((1-\lambda)K + \lambda L+(-1,1)\bigr) = m+1 = (1-\lambda) \G{1}(K) + \lambda \G{1}(L)$ for all $\lambda\in(0,1)$.
\end{proof}

\begin{remark}\label{r:best_interval}
Since both sides of \eqref{e: 1-dim_BM_G} are invariant under translations by integers of the sets $K$ and $L$,
we may replace $(-1,1)$ (in \eqref{e: 1-dim_BM_G}) by any other interval $(m,m+2)$, with $m\in\Z$.

We note however that the solution to Question \ref{q: BM_G} provided by Lemma \ref{l: 1-dim_BM_G} (for $n=1$), via
$M=(1-\lambda)K + \lambda L+(-1,1)$, for all $\lambda\in(0,1)$, cannot be in general improved by means of any other interval strictly contained in $(-1,1)$.
Indeed, by considering $I=[-a,1)$, with $-1<-a<0$, and taking $K=[-1,0]$, $L=[-2,0]$ and
$\lambda\in(0,1)$ such that $\lambda+a<1$, we get that
\[
\begin{split}
\G{1}\bigl((1-\lambda)K + \lambda L + I\bigr)&=\G{1}\bigl([-1-\lambda-a,1)\bigr)=2\\
&<2+\lambda=(1-\lambda) \G{1}(K) + \lambda \G{1}(L).
\end{split}
\]
The case $I=(-1,a]$, for $0<a<1$, is completely analogous and thus, no interval smaller than $(-1,1)$ (with respect to set inclusion) can be taken into account.
\end{remark}

\medskip

Now we state some auxiliary results. The following lemma can be regarded as a discrete counterpart of the well-known \emph{Cavalieri Principle} (see \cite[Lemma~3.1]{IYN}).

\begin{lemma}[\cite{IYN}]\label{l: Fubini discreto}
Let $\Omega \subset \R^n$ be a finite set, let $f:\Omega\longrightarrow\R_{\geq 0}$ and suppose $f(\Omega)\subset\{k_0, k_1, \dots, k_r\}$ where $0=k_0<k_1<\dots<k_r$.
Then
\begin{equation*}\label{e: Fubini discreto}
\sum_{x\in\Omega} f(x) = \sum_{i=1}^r (k_i - k_{i-1})\Bigl|\{ x \in \Omega : f(x) \geq k_i \}\Bigr|
=\int_0^{\infty} \Bigl|\{ x \in \Omega : f(x) \geq t \}\Bigr| \, \dlat t.
\end{equation*}
\end{lemma}

\begin{corollary}\label{c: Fubini discreto (lattice point enumerator)}
Let $\Omega\subset\R^n$ be a bounded set, let $f:\R^n\longrightarrow\R_{\geq 0}$ and suppose $f(\Omega\cap\Z^n)\subset\{k_0, k_1, \dots, k_r\}$ where $0=k_0<k_1<\dots<k_r$.
Then
\begin{equation*}
\sum_{x\in\Omega\cap\Z^n} f(x) = \sum_{i=1}^r (k_i - k_{i-1})\G{n}\bigl(\{ x \in \Omega : f(x) \geq k_i \}\bigr).
\end{equation*}
\end{corollary}

\medskip

The following result yields the case $n = 1$ of Theorem \ref{t: BBL discreta (caso p-media)} and
will be used to derive \eqref{e: BBL discreta (caso p-media)}.
\begin{lemma}\label{l: BBL Paso1 (caso p-media) new}
Let $\lambda \in (0,1)$ and let $\Omega_1, \Omega_2 \subset \R$ be non-empty bounded sets. Let $-1\leq p\leq\infty$ and let $f,g,h:\R \longrightarrow \R_{\geq 0}$ be functions such that
\begin{equation*}
h((1-\lambda)x + \lambda y)\geq\Media{p}{f(x)}{g(y)}{\lambda}
\end{equation*}
for all $x\in\Omega_1$, $y\in\Omega_2$. Then
\begin{equation*}
\sum_{z \in \Omega \cap \Z}h^\symbol(z) \geq \Media{\frac{p}{p+1}}{\sum_{x \in \Omega_1 \cap \Z}f(x)}{\sum_{y \in \Omega_2\cap \Z}g(y)}{\lambda},
\end{equation*}
where $\Omega = (1-\lambda)\Omega_1 + \lambda \Omega_2 + (-1,1)$.
\end{lemma}
\begin{proof}
Clearly, we may assume that $\sum_{x \in \Omega_1 \cap \Z}f(x), \sum_{y \in \Omega_2 \cap \Z}g(y)>0$.
We consider the functions $F, G, H, H^\symbol:\R\longrightarrow\R_{\geq0}$ given by
\[ F(x) = \frac{f(x)}{a}, \quad G(y) = \frac{g(y)}{b}, \quad H(z) = \frac{h(z)}{c_p}, \quad H^\symbol(z)=  \frac{h^\symbol(z)}{c_p},\]
where
\[ a = \max_{x \in \Omega_1\cap\Z} f(x)>0, \quad b=\max_{y \in \Omega_2\cap\Z} g(y)>0 \quad \text{and} \quad c_p= \Media{p}{a}{b}{\lambda}>0.\]
Then \[\max_{x \in \Omega_1\cap\Z} F(x) = \max_{y \in \Omega_2\cap\Z} G(y)=1.\]

First, we show that, for any $x \in \Omega_1$, $y \in \Omega_2$, we have
\begin{equation}\label{e: H((1-lambda)x + lambda y) >= min(F(x), G(y)) new}
H((1-\lambda)x + \lambda y) \geq \min(F(x), G(y)).
\end{equation}
To this aim, it is enough to consider $x \in \Omega_1$, $y \in \Omega_2$ with $f(x)g(y)>0$. If $p\neq0$ and $p\neq\infty$, writing $\theta = \lambda b^p / c_p^p \in (0,1)$, we get
\begin{equation*}
\begin{split}
h((1-\lambda)x + \lambda y) &\geq \left((1-\lambda)f(x)^p + \lambda g(y)^p \right)^{1/p}\\
&= c_p \left( \frac{(1-\lambda)a^pF(x)^p + \lambda b^p G(y)^p}{c_p^p} \right)^{1/p}\\
&= c_p \bigl( (1-\theta)F(x)^p + \theta G(y)^p \bigr)^{1/p}\\
&\geq c_p \min\{F(x), G(y)\}.
\end{split}
\end{equation*}

For $p=0$, we have
\begin{equation*}
h((1-\lambda)x + \lambda y) \geq f(x)^{1-\lambda}g(y)^\lambda
= c_0F(x)^{1-\lambda}G(y)^\lambda\geq c_0 \min\{F(x), G(y)\}.
\end{equation*}

For $p=\infty$, $h((1-\lambda)x + \lambda y) \geq \max\{f(x), g(y)\} \geq c_{\infty} \min\{F(x), G(y)\}$ clearly holds. Therefore, we have shown \eqref{e: H((1-lambda)x + lambda y) >= min(F(x), G(y)) new}.

Next, note that the definition of $F$ and $G$ implies that the level sets
\[ \{ x \in \Omega_1: F(x) \geq t \}, \quad \{ y \in \Omega_2 : G(y) \geq t \} \]
are non-empty for any $t \in [0,1]$. Moreover, writing $\Omega_{\lambda}=(1-\lambda)\Omega_1 + \lambda\Omega_2$,
we deduce from \eqref{e: H((1-lambda)x + lambda y) >= min(F(x), G(y)) new} that
\[ \{ z \in \Omega_{\lambda} : H(z) \geq t \} \supset (1-\lambda)\{ x \in \Omega_1 : F(x) \geq t \} + \lambda\{ y \in \Omega_2 : G(y) \geq t \}\]
and thus, by Lemma \ref{l: 1-dim_BM_G}, we have
\begin{equation}\label{e: Inequality of level sets (1-dim) new}
\begin{split}
&\G{1}\bigl(\{ z \in \Omega_{\lambda}: H(z) \geq t \} + (-1,1)\bigr)\\
\geq (1-\lambda)&\G{1}\bigl(\{ x \in \Omega_1 : F(x) \geq t \}\bigr)
+ \lambda\G{1}\bigl(\{ y \in \Omega_2 : G(y) \geq t \}\bigr)
\end{split}
\end{equation}
for all $t \in [0,1]$.

Note that, since $H^\symbol(z+u) \geq H\bigl((z+u)-u\bigr)=H(z)$ for all $u \in (-1,1)$, we also have
\begin{equation}\label{e: level_sets_Hsymbol}
\{ z\in \Omega : H^\symbol(z)\geq t \} \supset \{ z \in \Omega_{\lambda} : H(z) \geq t \} + (-1,1).
\end{equation}

Finally, set $\{k_0,k_1,\dots,k_r\} \supset F\bigl(\Omega_1\cap\Z\bigr) \cup G\bigl(\Omega_2\cap\Z\bigr)\cup H^\symbol\big(\Omega\cap\Z\big)$, with $0=k_0<k_1<\dots<k_r$ where, for some $s\in\{1,\dots,r\}$,
\[k_s = \max_{x \in \Omega_1\cap\Z} F(x) = \max_{y \in \Omega_2\cap\Z} G(y)=1.\]
Then, by Corollary \ref{c: Fubini discreto (lattice point enumerator)}, \eqref{e: Inequality of level sets (1-dim) new} and \eqref{e: level_sets_Hsymbol}, we get
\begin{equation*}
\begin{split}
\sum_{z \in \Omega\cap\Z} h^\symbol(z) &= \sum_{z \in \Omega\cap\Z} c_p H^\symbol(z)
= c_p \sum_{i = 1}^r (k_i - k_{i-1}) \G{1}\bigl(\{ z \in \Omega : H^\symbol(z) \geq k_i \}\bigr)\\
&\geq c_p \sum_{i = 1}^s (k_i - k_{i-1}) \G{1}\bigl(\{ z \in \Omega : H^\symbol(z) \geq k_i \}\bigr)\\
&\geq c_p \sum_{i = 1}^s (k_i - k_{i-1}) \Bigl( (1-\lambda)\G{1}\bigl(\{ x \in \Omega_1: F(x) \geq k_i \}\bigr) \\
& \hspace*{3.6cm} +  \lambda\G{1}\bigl(\{ y \in \Omega_2: G(y) \geq k_i \}\bigr) \Bigr)\\
&= c_p \left( (1-\lambda)\sum_{x \in \Omega_1\cap \Z} F(x) + \lambda\sum_{y \in \Omega_2\cap \Z} G(y) \right)\\
&= c_p \left( \frac{1-\lambda}{a}\,\sum_{x \in\Omega_1\cap \Z} f(x) +
\frac{\lambda}{b}\,\sum_{y \in \Omega_2\cap \Z} g(y) \right)\\
&\geq \Media{\frac{p}{p+1}}{\sum_{x \in \Omega_1\cap \Z} f(x)}{\sum_{y \in\Omega_2\cap \Z} g(y)}{\lambda}.
\end{split}
\end{equation*}
If $p\neq0$ and $p\neq\infty$, the last inequality follows from the reverse H\"older inequality (see e.g. \cite[Theorem~1, page~178]{Bu}),
\[ a_1b_1 + a_2b_2 \geq \left(a_1^{-p} + a_2^{-p}\right)^{-1/p} \left( b_1^q + b_2^q \right)^{1/q},\]
where $q = p/(p + 1)$ is the H\"older conjugate of $(-p) \leq 1$, just by taking $a_1 = ((1-\lambda)^{1/p} a)^{-1}$, $a_2 = (\lambda^{1/p} b)^{-1}$, $b_1 = (1-\lambda)^{1/q}\sum_{x \in \Omega_1\cap\Z} f(x)$ and $b_2 = \lambda^{1/q}\sum_{y \in \Omega_2\cap\Z} g(y)$.

The case $p = 0$ follows from the Arithmetic-Geometric mean inequality, whereas the case $p=\infty$ is immediate,
since there the $(p/(p+1))$-mean coincides with the $1$-mean.
\end{proof}

We would like to introduce a few additional notations which will be used to prove Theorem \ref{t: BBL discreta (caso p-media)}: we write $M(t)=\{x\in \R^{n-1}: (x,t)\in M\}$ for the $(n-1)$-dimensional section at height $t\in \R$ (in the direction of $\e_n$) whereas $\pi_n(M)$ denotes the orthogonal projection of $M$ onto $\R\e_n$ (regarded as a subset of $\R$), namely $\pi_n(M) = \{t \in \R: M(t)\neq\emptyset \}$.

\begin{proof}[Proof of Theorem \ref{t: BBL discreta (caso p-media)}]
If $n=1$, the result follows immediately from Lemma \ref{l: BBL Paso1 (caso p-media) new}.
Now suppose that $n>1$ and assume that the theorem holds for dimension $n-1$. Let $t_K \in \pi_n(K)$, $t_L\in\pi_n(L)$ and set, for the sake of brevity, $t_\lambda=(1-\lambda)t_K+ \lambda t_L$. Moreover, we denote by $C_n=(-1,1)^n$, $C_{n-1}=(-1,1)^{n-1}\times\{0\}$,  $M_{n-1}=(1-\lambda)K(t_K) + \lambda L(t_L) + (-1,1)^{n-1}$ and $M_\lambda=(1-\lambda)K+ \lambda L$ (thus, $M=M_\lambda + (-1,1)^n$). Consider the functions $f_1, g_1, h_1:\R^{n-1}\longrightarrow\R_{\geq 0}$ given by
\[ f_1(x) = f(x,t_K), \quad g_1(x) = g(x,t_L),\]
\[ h_1(x) = h(x, t_\lambda)\]
for any $x\in \R^{n-1}$.
Since for all $x \in K(t_K)$, $y\in L(t_L)$ we have
\begin{equation*}\begin{split}
h_1((1-\lambda)x+\lambda y) &= h((1-\lambda)x+\lambda y,(1-\lambda)t_K+ \lambda t_L)\\
&\geq \Media{p}{f(x,t_K)}{g(y,t_L)}{\lambda} = \Media{p}{f_1(x)}{g_1(y)}{\lambda},
\end{split}\end{equation*}
we may assert that
\begin{equation*}
\sum_{z \in M_{n-1}\cap\Z^{n-1}} h_1^\symbol(z) \geq
\Media{\frac{p}{(n-1)p+1}}{\sum_{x \in K(t_K)\cap\Z^{n-1}} f_1(x)}{\sum_{y \in L(t_L)\cap\Z^{n-1}} g_1(y)}{\lambda}.
\end{equation*}
This, together with the fact that
\[ ((1-\lambda)K+\lambda L)((1-\lambda)t_K + \lambda t_L) \supset (1-\lambda) K(t_K) + \lambda L(t_L),\]
and hence $(M_\lambda+C_{n-1})(t_\lambda)\supset M_{n-1}$, yields, in terms of $f$, $g$ and $h$,
\begin{equation}\label{e: suma de secciones n-1 dimensionales (caso p-medias) new}
\begin{split}
&\sum_{z \in ((M_\lambda+C_{n-1})(t_\lambda))\cap\Z^{n-1}} h^{\symbol\symbol}(z,t_\lambda)\\
&\geq \Media{\frac{p}{(n-1)p+1}}{\sum_{x \in K(t_K)\cap\Z^{n-1}} f(x,t_K)}{\sum_{y \in L(t_L)\cap\Z^{n-1}} g(y,t_L)}{\lambda},
\end{split}
\end{equation}
where $h^{\symbol\symbol}:\R^n\longrightarrow\R_{\geq 0}$ is the function given by $h^{\symbol\symbol}(z) = \sup_{v\in C_{n-1}} h(z+v)$, for which we have $h^{\symbol\symbol}(x,t_\lambda) = h_1^\symbol(x)$ for all $x\in\R^{n-1}$.

Now, let $f_2, g_2, h_2:\R \longrightarrow \R_{\geq 0}$ be the functions defined by
\[ f_2(t) = \sum_{x \in K(t)\cap\Z^{n-1}} f(x,t),
\;\; g_2(t) = \sum_{y \in L(t)\cap\Z^{n-1}} g(y,t),\,\text{ and }\]
\[h_2(t) = \sum_{z \in ((M_\lambda+C_{n-1})(t))\cap\Z^{n-1}} h^{\symbol\symbol}(z,t). \]
Hence, \eqref{e: suma de secciones n-1 dimensionales (caso p-medias) new} yields, in terms of $f_2$, $g_2$ and $h_2$,
\[ h_2((1-\lambda)t_K + \lambda t_L) \geq \Media{\frac{p}{(n-1)p+1}}{f_2(t_K)}{g_2(t_L)}{\lambda} \]
for any $t_K\in\pi_n(K)$, $t_L\in\pi_n(L)$, and thus we may use Lemma \ref{l: BBL Paso1 (caso p-media) new} with the sets $\pi_n(K)$, $\pi_n(L)$ and the functions $f_2$, $g_2$ and $h_2$ to obtain
\[ \sum_{t \in \Omega\cap\Z} h_2^{\symbol}(t) \geq \Media{\frac{p}{np+1}}{\sum_{t_K \in \pi_n(K)\cap\Z} f_2(t_K)}{\sum_{t_L \in \pi_n(L)\cap\Z} g(t_L)}{\lambda}, \]
where $\Omega = (1-\lambda)\pi_n(K) + \lambda \pi_n(L) + (-1,1)$.
In the following we prove that $\sum_{t\in\Omega\cap\Z} h_2^\symbol(t) \leq \sum_{z \in M\cap\Z^n} h^\symbol(z)$, and hence the above inequality together with the relations
\[ \sum_{t_K\in\pi_n(K)\cap\Z} f_2(t_K) =  \sum_{x \in K\cap\Z^n} f(x), \]
\[ \sum_{t_L\in\pi_n(L)\cap\Z} g_2(t_L) = \sum_{y\in L\cap\Z^n} g(y), \]
shows the result.
Indeed, from the fact that $(u,-w)\in C_{n}$ for any $(u,0)\in C_{n-1}$ and $w\in(-1,1)$, we have $(M_\lambda+C_{n-1})(t+w)\subset M(t)$ for all $w\in(-1,1)$ and thus we get
\begin{equation*}
\begin{split}
\sum_{t\in\Omega\cap\Z} h_2^\symbol(t) &= \sum_{t\in\Omega\cap\Z} \, \sup_{w\in(-1,1)} h_2(t+w)\\
&= \sum_{t\in\Omega\cap\Z} \, \sup_{w\in(-1,1)} \sum_{x \in ((M_\lambda+C_{n-1})(t+w))\cap\Z^{n-1}} h^{\symbol\symbol}(x, t+w)\\
&\leq \sum_{t\in\Omega\cap\Z} \; \sum_{x \in M(t)\cap\Z^{n-1}} \sup_{w\in(-1,1)} h^{\symbol\symbol}(x,t+w)\\
&= \sum_{t\in\Omega\cap\Z} \; \sum_{x \in M(t)\cap\Z^{n-1}} \sup_{w\in(-1,1)} \sup_{v \in (-1,1)^{n-1}} h(x+v, t+w)\\
&= \sum_{z \in M\cap\Z^n} \sup_{u\in (-1,1)^n} h(z+u) = \sum_{z \in M\cap\Z^n} h^\symbol(z),
\end{split}
\end{equation*}
as claimed. This finishes the proof.
\end{proof}

\begin{corollary}\label{c: BM_lattice_point}
Let $\lambda \in (0,1)$ and let $K, L \subset \R^n$ be non-empty bounded sets. Let $-1/n\leq p\leq\infty$.
Then
\begin{equation}\label{e: BM_lattice_point_general}
\G{n}\bigl((1-\lambda)K + \lambda L + (-1,1)^n\bigr)\geq \Media{\frac{p}{np+1}}{\G{n}(K)}{\G{n}(L)}{\lambda}.
\end{equation}
The inequality is sharp.
\end{corollary}
\begin{proof}
The result is an immediate consequence of Theorem \ref{t: BBL discreta (caso p-media)}, just by taking $f=\chi_{_K}$, $g=\chi_{_L}$ and $h=\chi_{_{(1-\lambda)K+\lambda L}}$, for which we clearly have that $h^\symbol=\chi_{_{(1-\lambda)K+\lambda L+(-1,1)^n}}$.

\smallskip

Now, in order to show that the equality may be attained, we consider $K = L = [0,m]^n$ with $m \in \N$,
for which
$\G{n}\bigl((1-\lambda)K + \lambda L+(-1,1)^n\bigr)=\G{n}(K)=\G{n}(L)=(m+1)^n$.
\end{proof}

We notice that \eqref{e: BM_lattice_point_general} for $p=\infty$ yields \eqref{e: BM_lattice_point_no_G(K)G(L)>0}
for bounded sets $K,L\subset\R^n$ with $\G{n}(K)\G{n}(L)>0$. So, to prove Theorem
\ref{t: BM_lattice_point_no_G(K)G(L)>0}, it is enough to deal with the case in which (only) one of the sets, say $L$, has no integer points. To this aim, first we show the following auxiliary result:
\begin{theorem}\label{t: BM_lattice_point_[-(q-1)/q,(q-1)/q]}
Let $K, L \subset \R^n$ be bounded sets such that $\G{n}(K)\G{n}(L)>0$
and let $\alpha=m/q$ and $\beta=p/q$ with $m,p,q\in\N$ so that $\alpha + \beta\leq1$. Then
\begin{equation}\label{e: BM_lattice_point_[-(q-1)/q,(q-1)/q]}
\G{n}\left(\alpha K+\beta L+\left[-\frac{q-1}{q},\frac{q-1}{q}\right]^n\right)^{1/n}
\geq \alpha\G{n}(K)^{1/n} + \beta\G{n}(L)^{1/n}.
\end{equation}
\end{theorem}
The proof of Theorem \ref{t: BM_lattice_point_[-(q-1)/q,(q-1)/q]}
is ideologically similar to the proof we have provided for Corollary \ref{c: BM_lattice_point}. So only the sketch of the proof is presented here. The main step is to prove an analogue of Theorem \ref{t: BBL discreta (caso p-media)} and apply it to $f=\chi_{_K}$, $g=\chi_{_L}$ and $h=\chi_{_{\alpha K+\beta L}}$.  The main new ingredient in the proof of such an analogue is the one dimensional case, for which we require an adaptation of Lemma \ref{l: 1-dim_BM_G}, which is done below in Lemma \ref{l: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])}.

\begin{lemma}\label{l: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])}
Let $K, L \subset \R$ be non-empty bounded sets and let $\alpha=m/q$ and $\beta=p/q$ with $m,p,q\in\N$ so that $\alpha + \beta\leq1$. Then
\begin{equation}\label{e: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])}
\G{1}\left(\alpha K+\beta L+\left[-\frac{q-1}{q},\frac{q-1}{q}\right]\right) \geq \alpha\G{1}(K) + \beta\G{1}(L).
\end{equation}
\end{lemma}
\begin{proof}
First we notice that, for any $x,y\in\R$, we have
\begin{equation}\label{e:property_floor_(mx+py)/q}
\floor{\frac{mx}{q}+\frac{py}{q}+\frac{q-1}{q}}\geq\frac{m\floor{x}+p\floor{y}}{q}.
\end{equation}
Indeed, given $z\in\R$, from the fact that
\[\frac{z}{q}+\frac{q-1}{q}\geq\frac{\floor{z}}{q}+\frac{q-1}{q}=c+\frac{r}{q}+\frac{q-1}{q}
\]
for some $c,r\in\Z$ with $0\leq r\leq q-1$, we get that
\begin{equation*}\label{e:property_floor_x/q}
\floor{\frac{z}{q}+\frac{q-1}{q}}\geq\frac{\floor{z}}{q}
\end{equation*}
for any $z\in\R$. This now implies that
\begin{equation*}
\floor{\frac{mx}{q}+\frac{py}{q}+\frac{q-1}{q}}\geq\frac{\floor{mx+py}}{q}\geq\frac{m\floor{x}+p\floor{y}}{q},
\end{equation*}
which yields \eqref{e:property_floor_(mx+py)/q}.

\medskip

Next we show the following:
if $K, L, M \subset \R$ are non-empty sets with $\alpha K+\beta L\subset M$ and
such that $M + [-(q-1)/q,(q-1)/q] = \bigcup_{i=1}^s [a_i, b_i]$ is a finite union of (pairwise disjoint)
compact intervals then
\begin{equation}\label{e: G(M+[-(q-1)/q,(q-1)/q])>=}
\G{1}\bigl(M+[-(q-1)/q,(q-1)/q]\bigr) \geq \alpha\G{1}(K) + \beta\G{1}(L).
\end{equation}
We prove it by induction on the number of intervals $s$ of $M+[-(q-1)/q,(q-1)/q]$. For $s=1$, i.e., when $M +[-(q-1)/q,(q-1)/q]= [a_1, b_1]$ is a (non-empty) compact interval, we have on the one hand that
$$
\G{1}\bigl(M+[-(q-1)/q,(q-1)/q]\bigr) = \floor{b_1} - \ceil{a_1} +1.
$$
Moreover, denoting by $a = \inf K$, $b = \sup K$, $c = \inf L$ and $d = \sup L$, we clearly get
$\G{1}(K)\leq \G{1}([a,b]) = \floor{b} - \ceil{a} +1$ and $\G{1}(L) \leq \G{1}([c,d]) = \floor{d} - \ceil{c} +1$.
On the other hand, the inclusion $\alpha K + \beta L \subset M$ implies that
\begin{equation*}
a_1\leq\alpha a+\beta c-\frac{q-1}{q}\leq\alpha b+\beta d+\frac{q-1}{q}\leq b_1.
\end{equation*}
Altogether, and using \eqref{e:property_floor_(mx+py)/q} jointly with the facts that
$\ceil{x}=-\floor{-x}$ for any $x\in\R$, and $\alpha+\beta\leq1$, we obtain
\begin{equation*}
\begin{split}
\G{1}\left(M+\left[-\frac{q-1}{q},\frac{q-1}{q}\right]\right) &= \floor{b_1} -\ceil{a_1} + 1\\
&\geq  \floor{\alpha b+\beta d+\frac{q-1}{q}} - \ceil{\alpha a+\beta c-\frac{q-1}{q}} + 1\\
&\geq\alpha\floor{b}+\beta\floor{d}-\alpha\ceil{a}+\beta\ceil{c}+1\\
&\geq\alpha\G{1}([a,b])+\beta\G{1}([c,d])\\
&\geq\alpha\G{1}(K)+\beta\G{1}(L).
\end{split}
\end{equation*}

Thus, we suppose that \eqref{e: G(M+[-(q-1)/q,(q-1)/q])>=} is true for $s\geq1$ and assume that
$M+[-(q-1)/q,(q-1)/q] = \bigcup_{i=1}^{s+1} [a_i, b_i]$, where $b_i < a_{i+1}$ for all $1\leq i\leq s$.

Denoting by $M_1$ and $M_2$ the subsets of $M$ such that $M_1+[-(q-1)/q,(q-1)/q]= [a_1, b_1]$ and
$M_2+[-(q-1)/q,(q-1)/q] = \bigcup_{i=2}^{s+1} [a_i, b_i]$, we may assume, without loss of generality, that $M_1\cap (\alpha K + \beta L)\neq\emptyset$ (otherwise, the result follows from the induction hypothesis applied
to the sets $K$, $L$ and $M_2$). Hence, we may define $m = \sup \bigl(M_1\cap(\alpha K + \beta L)\bigr)$ and then, since $K$ and $L$ are bounded, there exist $k \in \cl K$ and $l \in \cl L$ such that $\alpha k +  \beta l = m$.
Thus, considering the sets $K_1 = \{x \in K: x \leq k\}$, $K_2 = K \setminus K_1$, $L_1 = \{x \in L: x \leq l\}$ and $L_2 = L \setminus L_1$, and taking into account that $m+[-(q-1)/q,(q-1)/q]\subset[a_1, b_1]$, we clearly have that $\alpha K_1 + \beta L_1 \subset M_1$ and $\alpha K_2 + \beta L_2 \subset M_2$. Therefore, applying the induction hypothesis (and taking into account that $M_1+[-(q-1)/q,(q-1)/q]$ and $M_2+[-(q-1)/q,(q-1)/q]$ are disjoint), we get
\begin{equation*}
\begin{split}
&\,\G{1}\bigl(M+[-(q-1)/q,(q-1)/q]\bigr) \\
&= \G{1}\bigl(M_1+[-(q-1)/q,(q-1)/q]\bigr) + \G{1}\bigl(M_2+[-(q-1)/q,(q-1)/q]\bigr)\\
&\geq \alpha\G{1}(K_1)+\beta\G{1}(L_1) + \alpha\G{1}(K_2)+\beta\G{1}(L_2)
= \alpha\G{1}(K) + \beta\G{1}(L),
\end{split}
\end{equation*}
which shows \eqref{e: G(M+[-(q-1)/q,(q-1)/q])>=}.

\smallskip

Next we prove \eqref{e: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])}. We observe that we may assume, without loss of generality, that $K$ and $L$ are compact. Indeed, otherwise, considering the compact sets $K'= K\cap\Z$ and $L'=L\cap\Z$, for which we have $\G{1}(K')=\G{1}(K)$, $\G{1}(L')=\G{1}(L)$ and, from the monotonicity of $\G{1}(\cdot)$,
$\G{1}\bigl(\alpha K+\beta L+[-(q-1)/q,(q-1)/q]\bigr)\geq\G{1}\bigl(\alpha K'+\beta L'+[-(q-1)/q,(q-1)/q]\bigr)$, we would get the result.
So, \eqref{e: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])} follows from applying \eqref{e: G(M+[-(q-1)/q,(q-1)/q])>=} with $M=\alpha K+\beta L$, because the fact that $M$ is compact implies that there exists a finite sequence $m_1,\dots,m_r$ such that $\{m_1,\dots,m_r\}+[-(q-1)/q,(q-1)/q]=M+[-(q-1)/q,(q-1)/q]$. This concludes the proof.
\end{proof}

We are now in a position to prove Theorem \ref{t: BM_lattice_point_no_G(K)G(L)>0}.

\begin{proof}[Proof of Theorem \ref{t: BM_lattice_point_no_G(K)G(L)>0}]
For the sake of brevity, we will again denote by $M_\lambda=(1-\lambda)K+\lambda L$.
By Corollary \ref{c: BM_lattice_point} for $p=\infty$, and the monotonicity of $\G{n}(\cdot)$, it is enough to
show the result in the case in which $K=K\cap\Z^n$ (for which, clearly, $\G{n}(K)>0$) and $L=\{x\}$ with $x=(x_1,\dots,x_n)\notin\Z^n$.

First we show the case in which $\lambda=p/q$, $p,q\in\N$, is a rational number. Then,
writing $px=z+y$, with $z\in\Z^n$ and $y=(y_1,\dots,y_n)\in[0,1)^n$, and using
Theorem \ref{t: BM_lattice_point_[-(q-1)/q,(q-1)/q]}, we have
\begin{equation*}
\begin{split}
\G{n}\bigl(M_\lambda +(-1,1)^n\bigr)^{1/n}
&=\G{n}\left(\frac{q-p}{q}K+\frac{1}{q}z +\prod_{i=1}^n\left(-1+\frac{y_i}{q},1+\frac{y_i}{q}\right)\right)^{1/n}\\
&\geq\G{n}\left(\frac{q-p}{q}K+\frac{1}{q}z +\left[-\frac{q-1}{q},\frac{q-1}{q}\right]^n\right)^{1/n}\\
&\geq (1-\lambda)\G{n}(K)^{1/n}+\frac{1}{q}>(1-\lambda)\G{n}(K)^{1/n},
\end{split}
\end{equation*}
as desired.

Next we prove the case of an irrational $\lambda\in\R\setminus\Q$.
Let $I$ be the (possibly empty) subset of $\{1,\dots,n\}$ defined in the following way: $i\in I$ if and only if $x_i=a_i+b_i/\lambda$ for some $a_i,b_i\in\Z$. We point out that such $a_i,b_i\in\Z$ are necessarily unique,
since $\lambda\in\R\setminus\Q$.
Hence we may then consider the point $x'=(x_1',\dots,x_n')$ given by $x_i'=b_i/\lambda$ if $i\in I$ and $x_i'=0$ otherwise, for all $i=1,\dots, n$.
First we notice that, since $\lambda x'$ is an integer point, we have
\begin{equation*}
\begin{split}
\G{n}\bigl(M_\lambda +(-1,1)^n\bigr)&=\G{n}\bigl((1-\lambda)K+\lambda(x-x')+\lambda x'+(-1,1)^n\bigr)\\
&=\G{n}\bigl((1-\lambda)K+\lambda(x-x')+(-1,1)^n\bigr).
\end{split}
\end{equation*}

Next, denoting by $x_0=x-x'$, we will show that there exists $\delta>0$ such that
\begin{equation}\label{e:lambda_irrat_mu}
\G{n}\bigl((1-\lambda)K+\lambda x_0+(-1,1)^n\bigr)\geq\G{n}\bigl((1-\mu)K+\mu x_0+(-1,1)^n\bigr)
\end{equation}
for all $\mu$ with $|\mu-\lambda|<\delta$.
Thus, taking a sequence $(r_m)_m\subset\Q\cap(0,1)$ with $\lim_{m\to\infty} r_m=\lambda$
and $|r_m-\lambda|<\delta$ for all $m\in\N$, we get, from the previous case, that
\begin{equation*}
\begin{split}
\G{n}\bigl((1-\lambda)K+\lambda x_0+(-1,1)^n\bigr)^{1/n}&\geq\G{n}\bigl((1-r_m)K+r_m x_0+(-1,1)^n\bigr)^{1/n}\\
&\geq(1-r_m)\G{n}(K)^{1/n}+r_m\G{n}(\{x_0\})^{1/n}\\
&\geq(1-r_m)\G{n}(K)^{1/n}
\end{split}
\end{equation*}
for all $m\in\N$, which yields the result.

To show \eqref{e:lambda_irrat_mu} we notice that, since $K$ is finite, $(1-\lambda)K+\lambda x_0+[-1,1]^n$ is a finite union of closed unit cubes and then, for any $\mu$, $(1-\mu)K+\mu x_0+[-1,1]^n$ is the union of the corresponding translates of the cubes that constitute $(1-\lambda)K+\lambda x_0+[-1,1]^n$.
Thus, there exists $\delta_1>0$ such that if
$z\in\Z^n$ satisfies that $z\notin(1-\lambda)K+\lambda x_0+[-1,1]^n$ then $z\notin(1-\mu)K+\mu x_0+(-1,1)^n$ for all $|\mu-\lambda|<\delta_1$.
Moreover, if $(1-\lambda)K+\lambda x_0+(-1,1)^n$ contains no integer boundary points, we may take $\delta=\delta_1$ and we are done.

So, we may assume that $\bd\bigl(\bigl((1-\lambda)K+\lambda x_0+(-1,1)^n\bigr)\cap\Z^n\bigr)\neq\emptyset$.
Let $z\in\Z^n$ be a boundary point of $(1-\lambda)K+\lambda x_0+(-1,1)^n$ and let $k\in K$.
On the one hand, if $z$ is in the boundary of the cube $(1-\lambda)k+\lambda x_0+(-1,1)^n$, there exist $i\in\{1,\dots,n\}$ and $\varepsilon\in\{-1,1\}$ such that $(1-\lambda)k_i+\lambda(x_i-x_i')+\varepsilon=z_i$, where $k_i, z_i\in\Z$ are the $i$-th components of $k,z$, respectively. This implies that $x_i-x_i'=k_i+(z_i-\varepsilon-k_i)/\lambda$ and then, from the definition of both $I$ and $x'$, we get that $i\in I$ and so $a_i=k_i+(z_i-\varepsilon-k_i)/\lambda$, which yields that $z_i-\varepsilon-k_i=0$ (because $\lambda$ is irrational) and thus $x_i-x_i'=a_i=k_i$. But then $(1-\mu)k_i+\mu(x_i-x_i')+\varepsilon=(1-\lambda)k_i+\lambda(x_i-x_i')+\varepsilon=z_i$ for all $\mu\in\R$,
and hence $z$ lies in the affine hull of a facet of the open cube $(1-\mu)k+\mu x_0+(-1,1)^n$, which implies that $z\notin(1-\mu)k+\mu x_0+(-1,1)^n$ for any $\mu$.
On the other hand, if $z$ is not in the boundary of the cube $(1-\lambda)k+\lambda x_0+(-1,1)^n$ (and so not contained in it either), there exists $\delta_{z,k}>0$ so that $z\notin(1-\mu)k+\mu x_0+(-1,1)^n$ for all $\mu$ with $|\mu-\lambda|<\delta_{z,k}$.

Since the number of integer points $z$ in the boundary of $(1-\lambda)K+\lambda x_0+(-1,1)^n$ is finite (and so is $K$), we may define $\delta_2=\min_{z,k}\delta_{z,k}$.
Altogether, \eqref{e:lambda_irrat_mu} holds for $\delta=\min\{\delta_1,\delta_2\}$.
\end{proof}

\smallskip

We conclude this section by showing that the classical Borell-Brascamp-Lieb inequality \eqref{e:BBL_means} can be obtained from the discrete version \eqref{e: BBL discreta (caso p-media)} under some mild assumptions on the functions involved:
\begin{theorem}
The discrete Borell-Brascamp-Lieb type inequality \eqref{e: BBL discreta (caso p-media)} implies the classical Borell-Brascamp-Lieb inequality \eqref{e:BBL_means}, provided that the functions $f,g$ are Riemann integrable and $h$ is upper semicontinuous.
\end{theorem}

Before proving this result, we notice that it is not possible to directly obtain any of the
discrete Brunn-Minkowski type inequalities stated so far in the paper
from the classical one \eqref{e:BM_additive}, by using the method of replacing the points by suitable
compact sets. As pointed out by Gardner and Gronchi in
\cite[pages~3996--3997]{GG}:

\medskip

\begin{center}
\begin{minipage}{0.9\textwidth}
``{\it it is worth remarking that the obvious idea of replacing the points
in the two finite sets by small congruent balls and applying the classical
Brunn-Minkowski inequality to the resulting compact sets is doomed to
failure. The fact that the sum of two congruent balls is a ball of twice
the radius introduces an extra factor of $1/2$ that renders the resulting
bound weaker than even the trivial bound (11) below}\,''.
\end{minipage}
\end{center}

\medskip

\noindent We clarify that (11) in \cite{GG} coincides with
\eqref{e:|A+B|>|A|+|B|-1} of the present paper.

\smallskip

In the following, for a function $\phi:\R^n\longrightarrow\R_{\geq0}$, we write $\phi^{\symbol_k}:\R^n\longrightarrow\R_{\geq0}$ to denote the function given by
$\phi^{\symbol_k}(z) = \sup_{u\in(-2^{-k},2^{-k})^n}\phi(z+u)$ for all $z\in\R^n$.
\begin{proof}
Consider  $f,g,h:\R^n\longrightarrow\R_{\geq0}$  which satisfy the conditions of
Theorem \ref{t:BBL_means}, i.e., $f,g,h$ are measurable such that
\[
h((1-\lambda)x + \lambda y)\geq\Media{p}{f(x)}{g(y)}{\lambda}
\]
holds for all $x,y\in\R^n$ and for some fixed  $\lambda\in(0,1)$ and  $-1/n\leq p\leq\infty$.
Let $m\in \N$ and let $K = [-m,m]^n$. We will first show that
\begin{equation}\label{e:BBL_over_K}
\int_K h(x) \dlat x \geq \Media{\frac{p}{np+1}}{ \int_K f(x) \dlat x}{ \int_K g(x) \dlat x}{\lambda},
\end{equation}
for which we may assume (multiplying by $\chi_{_K}$ if necessary) that $f$, $g$ and $h$ vanish outside $K$.

For each $k\in\N$, we define the functions $f_k, g_k, h_k:\R^n\rightarrow\R_{\geq0}$ given by
\[ f_k(x) = \inf_{z \in x + [0,2^{-k}]^n} f(z), \quad g_k(x) = \inf_{z\in x + [0,2^{-k}]^n} g(z)\]
and
\[h_k(x) = \inf_{z\in x + [0,2^{-k}]^n} h(z). \]
Writing for short $K_0=\inter K$, note that for any $x,y\in K_0$ we have
\begin{equation*}
\begin{split}
h_k((1-\lambda)x+\lambda y) &= \inf_{z \in (1-\lambda)x+\lambda y + [0,2^{-k}]^n} h(z)\\
&= \inf_{z \in (1-\lambda)(x+ [0,2^{-k}]^n)+\lambda (y+[0,2^{-k}]^n)} h(z)\\
&= \inf_{z_1 \in x+[0,2^{-k}]^n, z_2 \in y+[0,2^{-k}]^n} h((1-\lambda)z_1 + \lambda z_2)\\
&\geq \inf_{z_1 \in x+[0,2^{-k}]^n, z_2 \in y+[0,2^{-k}]^n} \Media{p}{f(z_1)}{g(z_2)}{\lambda}\\
&\geq \Media{p}{\inf_{z_1 \in x+[0,2^{-k}]^n} f(z_1)}{\inf_{z_2 \in y+[0,2^{-k}]^n} g(z_2)}{\lambda}\\
&= \Media{p}{f_k(x)}{g_k(y)}{\lambda},
\end{split}
\end{equation*}
and thus, we can use Theorem \ref{t: BBL discreta (caso p-media)} for $2^{-k}\Z^n$ to deduce that, for any $k\in \N$, we have
\begin{equation}\label{e:discBBL_lattice2-kZn}
2^{-kn}\sum_{z \in K\cap2^{-k}\Z^n} (h_k)^{\symbol_k}(z) \geq \Media{\frac{p}{np+1}}{2^{-kn}\sum_{x \in K_0\cap2^{-k}\Z^n} f_k(x)}{2^{-kn}\sum_{y \in K_0\cap2^{-k}\Z^n} g_k(y)}{\lambda},
\end{equation}
where, on the left-hand side, we have used that $K=[-m,m]^n$, thus
\[
\left(2^kK_0+(-1,1)^n\right)\cap \Z^n= 2^kK\cap \Z^n
\]
and
\[
\left(K_0+(-2^{-k},2^{-k})^n\right)\cap2^{-k}\Z^n= K\cap2^{-k}\Z^n.
\]

The level sets $\{x \in K: h(x)\geq t\}$ are closed, because $h$ is upper semicontinuous and $K$ is closed
(see \cite[Theorem~1.6]{RoWe}),
and then a standard straightforward computation shows that
\[\{x \in K: h(x)\geq t\}=\bigcap_{k=1}^\infty\left(\{x \in K: h(x)\geq t\}+(-2^{-k},2^{-k})^n\right).\]
Moreover, since $h$ vanishes outside $K$, we have
$\{x \in K: h(x)> t\}+(-2^{-k},2^{-k})^n\supset\{x \in K+[0,2^{-k}]^n: h^{\symbol_k}(x)> t\}$ for all $t>0$.
Thus, by Fubini's theorem and the monotone convergence theorem, we get
\begin{equation}\label{e: ints_h_h^symbol}
\begin{split}
\int_K h(x) \dlat x &= \int_0^\infty \vol\bigl(\{x \in K: h(x)\geq t\}\bigr) \dlat t\\
&= \int_0^\infty \vol\left(\bigcap_{k=1}^\infty\left(\{x \in K: h(x)\geq t\}+(-2^{-k},2^{-k})^n\right)\right) \dlat t\\
&= \int_0^\infty \lim_{k\rightarrow\infty} \vol\left(\{x \in K: h(x)\geq t\}+(-2^{-k},2^{-k})^n\right) \dlat t \\
&= \lim_{k\rightarrow\infty} \int_0^\infty \vol\left(\{x \in K: h(x)\geq t\}+(-2^{-k},2^{-k})^n\right) \dlat t \\
&\geq \lim_{k\rightarrow\infty}\int_0^\infty \vol\bigl(\{x \in K+[0,2^{-k}]^n: h^{\symbol_k}(x)> t\}\bigr) \dlat t \\
&=\lim_{k\rightarrow\infty}\int_{K+[0,2^{-k}]^n} h^{\symbol_k}(x) \dlat x.
\end{split}
\end{equation}

Now we show that, given $z\in\R^n$, $h^{\symbol_k}(x)\geq(h_k)^{\symbol_k}(z)$ for all $x\in z+[0,2^{-k}]^n$. Indeed, we have
\begin{equation*}
\begin{split}
h^{\symbol_k}(x)&=\sup_{u\in(-2^{-k},2^{-k})^n} h(x+u)\geq\sup_{u\in(-2^{-k},2^{-k})^n} \inf_{v\in[0,2^{-k}]^n}h(z+v+u)\\
&=\sup_{u\in(-2^{-k},2^{-k})^n} h_k(z+u)=(h_k)^{\symbol_k}(z).
\end{split}
\end{equation*}
This, together with \eqref{e: ints_h_h^symbol} and the fact that $K+[0,2^{-k}]^n=K\cap2^{-k}\Z^n + [0,2^{-k}]^n$, implies that
\begin{equation*}
\int_K h(x) \dlat x \geq \lim_{k\rightarrow\infty}\int_{K+[0,2^{-k}]^n} h^{\symbol_k}(x) \dlat x
\geq \lim_{k\rightarrow\infty} 2^{-kn} \sum_{z \in K\cap2^{-k}\Z^n} (h_k)^{\symbol_k}(z).
\end{equation*}
Furthermore, since $f$ is Riemann integrable and $2^{-kn}\sum_{x \in K_0\cap2^{-k}\Z^n} f_k(x)$ is a \textit{lower sum} of $f\cdot\chi_{_{(-m,m]^n}}$ for the partition $\{x + [0,2^{-k}]^n\subset K: x\in2^{-k}\Z^n\}$ of $K$,
it is clear that
\begin{equation*}
\lim_{k\rightarrow\infty} 2^{-kn}\sum_{x \in K_0\cap2^{-k}\Z^n} f_k(x) = \int_K f(x)\dlat x.
\end{equation*}

Here we observe that it was crucial to work with $K_0$ in order to get a lower sum of $f\cdot\chi_{_{(-m,m]^n}}$
for the above partition. We also point out the necessity of considering the characteristic function $\chi_{_{(-m,m]^n}}$ instead of $\chi_{_{[-m,m]^n}}$, which has no influence when computing the above integral: in this way, the function $f\cdot\chi_{_{(-m,m]^n}}$ vanishes on the points of the corresponding facets of the cube.

The same holds for the function $g$ and then, taking limits on both sides of \eqref{e:discBBL_lattice2-kZn},
we get \eqref{e:BBL_over_K}.
Since \eqref{e:BBL_over_K} is true for $K=[-m,m]^n$, for every $m\in\N$, the proof is now concluded because
\[ \int_{\R^n} \phi(x) \dlat x = \lim_{m\rightarrow\infty} \int_{[-m,m]^n} \phi(x)\dlat x,\]
for every measurable function $\phi:\R^n\rightarrow\R_{\geq0}$.
\end{proof}

It is well-known that a function is Riemann integrable if and only if it is continuous almost everywhere. Since the boundary of a convex set has null measure (and from the characterization of the upper semicontinuity in terms of the level sets) we get the following result, as a straightforward consequence of the previous one.
\begin{corollary}
The discrete Brunn-Minkowski type inequality \eqref{e: BM_lattice_point_no_G(K)G(L)>0} implies the classical Brunn-Minkowski inequality \eqref{e:BM} for bounded convex sets $K$ and $L$.
\end{corollary}
We notice the necessity of assuming convexity in the latter result: for any measurable sets $K, L \subset \R^n$
of positive volume, containing no rational point, one cannot expect to recover the Brunn-Minkowski inequality
\eqref{e:BM} with the above method of shrinking the lattice $\Z^n$, since $K, L$ have no point in $2^{-k}\Z^n$, for any $k\in\N$.

\section{Relations with other inequalities}

The multiplicative version of the Brunn-Minkowski inequality is, among all its equivalent forms, the one that is
naturally connected to the Pr\'ekopa-Leindler inequality (the case $p=0$ of Theorem \ref{t:BBL_means}).
It asserts that if $\lambda\in(0,1)$ and $K$ and $L$ are non-empty compact subsets of $\R^n$ then
\begin{equation}\label{e:BM_multip}
\vol\bigl((1-\lambda)K+\lambda L\bigr)\geq
\vol(K)^{1-\lambda}\vol(L)^{\lambda}.
\end{equation}

In the discrete setting, considering now non-empty bounded sets $K, L\subset\R^n$,
from \eqref{e: BM_lattice_point_general} for $p=0$ we get
\begin{equation}\label{e:disc_multip_BM}
\G{n}\bigl((1-\lambda)K+\lambda L+(-1,1)^n\bigr)\geq\G{n}(K)^{1-\lambda}\G{n}(L)^\lambda.
\end{equation}
Regarding other possible discrete versions of \eqref{e:BM_multip}, we have the following engaging and elegant result, shown very recently by Halikias, Klartag and Slomka in \cite{HKS} (see also \cite{KL}):
\begin{thm}[\cite{HKS}]\label{t:HKS}
Let $\lambda\in(0,1)$ and let $f,g,h,k:\Z^n\longrightarrow\R_{\geq0}$ be functions
such that
\[h(\floor{(1-\lambda)x+\lambda y})k(\ceil{\lambda x+(1-\lambda) y})\geq f(x)g(y)\]
for all $x,y\in\Z^n$, where $\floor{x}=(\floor{x_1},\dots,\floor{x_n})$ and
$\ceil{x}=(\ceil{x_1},\dots,\ceil{x_n})$. Then
\[\left(\sum_{x\in\Z^n}h(x)\right)\left(\sum_{x\in\Z^n}k(x)\right)\geq
\left(\sum_{x\in\Z^n}f(x)\right)\left(\sum_{x\in\Z^n}g(x)\right).\]
\end{thm}
As they observed, when applying the above result to the functions $f=\chi_{_K}$, $g=\chi_{_L}$, $h=\chi_{_{\frac{K+L}{2}+(-1,0]^n}}$
and $k=\chi_{_{\frac{K+L}{2}+[0,1)^n}}$, one has
\begin{equation*}
\G{n}\left(\frac{K+L}{2}+(-1,0]^n\right)\G{n}\left(\frac{K+L}{2}+[0,1)^n\right)\geq\G{n}(K)\G{n}(L),
\end{equation*}
which yields the discrete multiplicative Brunn-Minkowski type inequality
\begin{equation}\label{e:disc_multip_BM_HKS}
\G{n}\left(\frac{K+L}{2}+[0,1]^n\right)\geq\sqrt{\G{n}(K)\G{n}(L)}.
\end{equation}
We notice that the sole difference between \eqref{e:disc_multip_BM_HKS} and \eqref{e:disc_multip_BM} (for $\lambda=1/2$) is the necessity of adding either the closed cube of edge length $1$ or the open cube of edge length $2$, respectively. However, they are not comparable. Indeed, let $n=1$ and let $K=L=[-x,x]$ with $x\in\R_{\geq0}$. On the one hand,
for $x\in\Z$, we have $\G{1}\bigl((K+L)/2+[0,1]\bigr)=2x+2>2x+1=\G{1}\bigl((K+L)/2+(-1,1)\bigr)$.
On the other hand, for $x\notin\Z$, we get
$\G{1}\bigl((K+L)/2+[0,1]\bigr)=2\floor{x}+2<2\floor{x}+3=\G{1}\bigl((K+L)/2+(-1,1)\bigr)$.

\smallskip

As pointed out in Remark \ref{r:best_interval}, inequality \eqref{e: BM_lattice_point_no_G(K)G(L)>0} is in general not true (even for $\lambda=1/2$) by just adding the cube $[0,1)^n$ to the convex combination $(K+L)/2$.
However, this can be solved by just considering the closed cube $[0,1]^n$, i.e., we show that
inequality \eqref{e:disc_multip_BM_HKS} also admits a $(1/n)$-form:
\begin{theorem}\label{t: BM_lattice_point_1/2}
Let $K, L \subset \R^n$ be bounded sets such that $\G{n}(K)\G{n}(L)>0$. Then
\begin{equation}\label{e: BM_lattice_point_1/2}
\G{n}\left(\frac{K + L}{2} + [0,1]^n\right)^{1/n}\geq \frac{\G{n}(K)^{1/n}+\G{n}(L)^{1/n}}{2}.
\end{equation}
The inequality is sharp.
\end{theorem}

As in Theorem \ref{t: BM_lattice_point_[-(q-1)/q,(q-1)/q]}, we just have to prove the corresponding one-dimen\-sio\-nal case, collected in Lemma \ref{l: G(M+[0,1])>=1/2(G(K)+G(L))}.
Then, the proof of Theorem \ref{t: BM_lattice_point_1/2} is completed in a way similar to the proof of Theorem
\ref{t: BBL discreta (caso p-media)} with the particular functions $f=\chi_{_K}$, $g=\chi_{_L}$ and $h=\chi_{_{\frac{K+L}{2}}}$, and $\lambda=1/2$, replacing there the use of inequality \eqref{e: 1-dim_BM_G} by
\eqref{e: G((K+L)/2+[0,1])>=1/2(G(K)+G(L))}.

\smallskip

Finally, to show that equality may be attained, we consider $K=-L=[0,m]^n$ with $m\in\N$ odd, for which
we have $\G{n}\bigl((K + L)/2+[0,1]^n\bigr)=\G{n}(K)=\G{n}(L)=(m+1)^n$.

\begin{lemma}\label{l: G(M+[0,1])>=1/2(G(K)+G(L))}
Let $K, L \subset \R$ be non-empty bounded sets. Then
\begin{equation}\label{e: G((K+L)/2+[0,1])>=1/2(G(K)+G(L))}
\G{1}\left(\frac{K+L}{2}+[0,1]\right) \geq \frac{\G{1}(K) + \G{1}(L)}{2}.
\end{equation}
\end{lemma}
\begin{proof}
The proof is completely analogous to that of Lemma \ref{l: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])}, and thus
we include here just the slight differences.

First we notice that, for any $x,y\in\R$, we have
\begin{equation}\label{e:property_floor_(x+y)/2}
\floor{\frac{x+y}{2}}+\frac{1}{2}\geq\frac{\floor{x}+\floor{y}}{2}.
\end{equation}
Indeed, if $\floor{x}+\floor{y}$ is even then $\floor{(x+y)/2}=(\floor{x}+\floor{y})/2$ whereas if
$\floor{x}+\floor{y}$ is odd we get $\floor{(x+y)/2}\geq(\floor{x}+\floor{y}-1)/2$.

\smallskip

Now, we notice that if $M\supset(K+L)/2$ is such that
$M +[0,1]= [a_1, b_1]$ is a (non-empty) compact interval, we have on the one hand that
$\G{1}\bigl(M+[0,1]\bigr) = \floor{b_1} - \ceil{a_1} +1$.
Moreover, denoting by $a = \inf K$, $b = \sup K$, $c = \inf L$ and $d = \sup L$, we clearly get
$\G{1}(K)\leq \G{1}([a,b]) = \floor{b} - \ceil{a} +1$ and $\G{1}(L) \leq \G{1}([c,d]) = \floor{d} - \ceil{c} +1$.
On the other hand, the inclusion $(K + L)/2 \subset M$ implies that
\begin{equation*}
a_1\leq\frac{a+c}{2}\leq\frac{b+d}{2}\leq b_1-1.
\end{equation*}
Altogether, and using \eqref{e:property_floor_(x+y)/2} jointly with the fact that
$\ceil{x}=-\floor{-x}$ for any $x\in\R$, we obtain
\begin{equation*}
\begin{split}
\G{1}\bigl(M+[0,1]\bigr) &= \floor{b_1} -\ceil{a_1} + 1 \geq  \floor{\frac{b+d}{2}} - \ceil{\frac{a+c}{2}} + 2\\
&\geq\frac{\floor{b}+\floor{d}}{2}-\frac{\ceil{a}+\ceil{c}}{2}+1=\frac{\G{1}([a,b])+\G{1}([c,d])}{2}\\
&\geq\frac{\G{1}(K)+\G{1}(L)}{2}.
\end{split}
\end{equation*}
The proof is then completed in a way analogous to the proof of Lemma \ref{l: G(alpha K+beta L+[-(q-1)/q,(q-1)/q])}.
\end{proof}

We would like to point out that the corresponding version of Theorem \ref{t: BM_lattice_point_1/2} for an arbitrary $\lambda\in(0,1)$ is, in general, not true. Indeed, taking $K=[0,1]^n$, $L=[-5,6]^n$ and $\lambda=1/3$ one has that
\begin{equation*}
\begin{split}
\G{n}\bigl((1-\lambda)K+\lambda L+[0,1]^n\bigr)^{1/n}
&=\G{n}\left(\left[\frac{-5}{3},\frac{11}{3}\right]^n\right)^{1/n}=5\\
&<\frac{16}{3}=(1-\lambda)\G{n}(K)^{1/n}+\lambda\G{n}(L)^{1/n}.
\end{split}
\end{equation*}

\smallskip

We include here an open question that arose during our study:
\begin{question}\label{q:G{n}(K)G{n}(L)>0}
Regarding the statement of Theorem \ref{t: BM_lattice_point_1/2},
is the assumption $\G{n}(K)\G{n}(L)>0$ necessary?
\end{question}

We conclude the section by proving an inequality similar to \eqref{e:B-M_discrete_adding} but in the spirit of \eqref{e: BM_lattice_point_no_G(K)G(L)>0}, namely that one may add another (fixed) set to the Minkowski sum $A+B$ instead of considering the extension $\bar{A}$ of $A$. We show that an appropriate set to be taken into account in this respect is the lattice cube $\{0,1\}^n$, which also fits well with inequality \eqref{e: BM_lattice_point_1/2}.
\begin{theorem}\label{t:B-M_discrete_sum}
Let $A,B\subset\Z^n$ be finite, $A,B\neq\emptyset$. Then
\begin{equation}\label{e:B-M_discrete_sum}
\bigl|A+B+\{0,1\}^n\bigr|^{1/n}\geq |A|^{1/n}+|B|^{1/n}.
\end{equation}
The inequality is sharp.
\end{theorem}
The idea of the proof we present here goes back to \cite[Lemma~2.4]{GT}.
\begin{proof}
Taking into account the relation
\[A+B+\{0,1\}^n+[0,1]^n=A+[0,1]^n+B+[0,1]^n,\]
jointly with the fact that $\vol(X+[0,1]^n)=|X|$, for any non-empty and finite set $X\subset\Z^n$,
the result directly follows from the Brunn-Minkowski inequality \eqref{e:BM_additive}. Finally, to show that the inequality is sharp, we consider the lattice cubes $A=\{0,\dots,m_1\}^n$ and $B=\{0,\dots,m_2\}^n$, for $m_1, m_2\in\Z_{\geq0}$.
\end{proof}

We conclude the paper by comparing inequalities \eqref{e:B-M_discrete_adding} and \eqref{e:B-M_discrete_sum}; again, we refer the reader to Section 2.1 in \cite{HCIYN} for the precise definition of the extension $\bar{A}$ of a given non-empty finite set $A\subset\Z^n$.

Since $A+(\{0\}^{n-1}\times\{0,1\})$ contains at least the same amount of points that the union of $A$ and its maximal cardinality section, and taking into account that $\{0,1\}^n=\bigl(\{0,1\}^{n-1}\times\{0\}\bigr)+\bigl(\{0\}^{n-1}\times\{0,1\}\bigr)$,
from the definition of $\bar{A}$ is then immediate that $\bigl|\bar{A}\bigr|\leq\bigl|A+\{0,1\}^n\bigr|$.
However, inequalities \eqref{e:B-M_discrete_adding} and \eqref{e:B-M_discrete_sum} are not comparable.
Indeed, if we consider on the one hand $A=\{(0,0), (1,0), (2,0), (1,1)\}$ and $B=\{0,1\}^2$, we have
$\bigl|\bar{A}\bigr|=10=\bigl|A+\{0,1\}^2\bigr|$ (with $\bar{A}\neq A+\{0,1\}^2$) and $\bigl|\bar{A}+B\bigr|=20 > 18 = \bigl|A+\{0,1\}^2+B\bigr|$ (see Figure \ref{fig: bar{A}+B vs A+C+B}).
On the other hand, for $A=\{(0,0), (0,1), (1,1), (4,1)\}$ and $B=\{0,1\}^2$ we obtain
$\bigl|\bar{A}+B\bigr|=21 < 24 = \bigl|A+\{0,1\}^2+B\bigr|$.

\smallskip

\begin{figure}[h]\label{fig: bar{A}+B vs A+C+B}
\centering
\begin{tikzpicture}[A/.style ={thick,color=black}, Amp/.style ={fill=none}, AmppB/.style ={color=black}, scale=0.8]
\draw (-0.5,0) -- (4.5,0);
\draw (0,-0.5) -- (0,3.5);
\filldraw[A] (0,0) circle (3pt);
\filldraw[A] (1,1) circle (3pt);
\filldraw[A] (1,0) circle (3pt);
\filldraw[A] (2,0) circle (3pt);
\filldraw[Amp] (0,2) circle (3pt);
\filldraw[Amp] (1,2) circle (3pt);
\filldraw[Amp] (2,2) circle (3pt);
\filldraw[Amp] (2,1) circle (3pt);
\filldraw[Amp] (3,2) circle (3pt);
\filldraw[Amp] (3,0) circle (3pt);
\filldraw[AmppB] (0,1) circle (2pt);
\filldraw[AmppB] (0,3) circle (2pt);
\filldraw[AmppB] (1,3) circle (2pt);
\filldraw[AmppB] (2,3) circle (2pt);
\filldraw[AmppB] (3,3) circle (2pt);
\filldraw[AmppB] (4,3) circle (2pt);
\filldraw[AmppB] (4,2) circle (2pt);
\filldraw[AmppB] (3,1) circle (2pt);
\filldraw[AmppB] (4,1) circle (2pt);
\filldraw[AmppB] (4,0) circle (2pt);
\end{tikzpicture}
\hspace{1.5cm}
\begin{tikzpicture}[A/.style ={thick,color=black}, Amp/.style ={fill=none}, AmppB/.style ={color=black}, scale=0.8]
\draw (-0.5,0) -- (4.5,0);
\draw (0,-0.5) -- (0,3.5);
\filldraw[A] (0,0) circle (3pt);
\filldraw[A] (1,1) circle (3pt);
\filldraw[A] (1,0) circle (3pt);
\filldraw[A] (2,0) circle (3pt);
\filldraw[Amp] (0,1) circle (3pt);
\filldraw[Amp] (1,2) circle (3pt);
\filldraw[Amp] (2,2) circle (3pt);
\filldraw[Amp] (2,1) circle (3pt);
\filldraw[Amp] (3,1) circle (3pt);
\filldraw[Amp] (3,0) circle (3pt);
\filldraw[AmppB] (0,2) circle (2pt);
\filldraw[AmppB] (1,3) circle (2pt);
\filldraw[AmppB] (2,3) circle (2pt);
\filldraw[AmppB] (3,3) circle (2pt);
\filldraw[AmppB] (4,2) circle (2pt);
\filldraw[AmppB] (3,2) circle (2pt);
\filldraw[AmppB] (4,1) circle (2pt);
\filldraw[AmppB] (4,0) circle (2pt);
\end{tikzpicture}
\caption{Left: the sets $A$ (thick points), $\bar{A}$ (thick and hollow points) and $\bar{A}+B$. Right:
the sets $A$ (thick points), $A+\{0,1\}^2$ (thick and hollow points) and $A+\{0,1\}^2+B$.}
\end{figure}

\medskip

\noindent {\it Acknowledgements.} We are very grateful to the anonymous referee for
her/his very helpful comments and remarks which have allowed us to improve the presentation of this work; we specially thank the referee for bringing to our attention the simpler and more elegant proof (than the one we originally did) of
Theorem \ref{t:B-M_discrete_sum}.
We would like to thank M. A. Hern\'andez Cifre for her very valuable suggestions and the careful reading of this paper.
We also thank D. Halikias, B. Klartag and B. A. Slomka for bringing to our attention Theorem \ref{t:HKS}
and J. Li for pointing out the possibility of considering the statement of Theorem \ref{t:B-M_discrete_sum}.

redpajama_set_name: RedPajamaArXiv


This week, SIWI's Dr. Alejandro Jiménez is in Addis Ababa attending a High Level Panel organized by Sanitation and Water For All (SWA).
During a World Water Week workshop, young water professionals drafted an action plan to address WASH issues in slums. We now invite you to contribute!
On World Toilet Day November 19, Marianne Kjellén of SIWI will participate in the seminar "Sanitation for improved lives of women and children", organised by Sida and WaterAid. Watch the webcast here!
SIWI has produced five thematic films and background papers to share insight on the centrality of water in building resilient future societies. Watch the films and download the papers.


Kris Moran is most comfortable behind the camera.
She attributes this to being camera shy, but, during the few hours I spend with her for this story, I come to realize that it's more because she is eternally preoccupied with capturing the perfectly framed shot. As NYLON's video team sets her up on a stool for the interview, Moran asks for trays to sit on so she can look taller in relation to the shot, gets up to move the tables behind her, and adjusts the camera angle so you can see more of the space she's showcasing. "If you show me anything, I will comment on it," she says when someone alerts us that we're low on time, "so just don't show me anything anymore!" It's that laser-focused attention to the most minuscule detail that has led Moran to become the similarly aesthetically obsessed director Wes Anderson's go-to set decorator.
If you've seen films like Moonrise Kingdom, The Darjeeling Limited, and The Life Aquatic with Steve Zissou, as well as films like Noah Baumbach's While We're Young and The Meyerowitz Stories, or indie darling Begin Again, you know what I am talking about: the lovingly selected retro mementos and odd trinkets that add nostalgic warmth to every shot, the perfect shade of blue wallpaper or carpeting centering its characters, and the plaid and floral upholstery that looks borrowed from your grandmother's basement in the best possible way. It's that incredible ability to make kitsch look modern, and clutter appear visually exciting that has made Moran the film industry's best secret weapon.
Moran has a signature style, and so she is not looking to reinvent the wheel with each new project (and she has many: most recently, she worked on Barry Jenkins' upcoming James Baldwin adaptation). So when four-time James Beard Award nominee, chef Floyd Cardoz, who had loved Moran's work on The Darjeeling Limited, cold-called her to design his new restaurant, The Bombay Bread Bar, she wasn't sure if she would take the job. "In the beginning, I thought that they just saw a Wes Anderson-themed restaurant, and I was kind of like... [sighs dramatically]," she tells me. "To make something that's derivative of him, I didn't feel very comfortable with it. It's kind of like selling out on Wes's thing, you know? It's not my thing. Like, I work for Wes." But as soon as she got to know Cardoz, Moran realized that the film was just his starting vision and not something that he wanted to actually imitate. "The whole idea and concept of the restaurant changed the more I got to know Floyd," she says. "I got on board and just sort of tried to make this train for him."
As Moran started interacting with the Indian-American chef, Cardoz inspired her with stories about his heritage and childhood, as well as his dishes that serve as love letters to India—the same ones beloved by judges of Season 3 Top Chef Masters (which, spoiler alert, he won). These inspirations can be seen in, for example, Moran's choice of vibrant oilcloth table coverings, one of Floyd's strongest visual memories: "Those tablecloths was something that Floyd was excited about from his childhood. That's what his auntie had at her table, and it was what all the cafes had when he was a kid in India." When Moran suggested they cover the bathroom wall with a collage of vintage Indian matchbox prints, Floyd was excited, distinctly remembering the designs from his youth. "Before we started, I would be walking around with paint chips, and Floyd would come in before work time for him, and he would just talk about his mother and his family house in India," she says. "I sort of got to hear his heartstrings about India, and he showed me pictures, so I just felt like this is his house, and I kind of couldn't help but really weave it in along with all his other hopes and dreams. I felt like I had a responsibility to represent this properly."
Considering herself only "a small cog in this giant machine," Moran wanted to incorporate other artists into the design process. She recruited Maria Qamar and Moshtari Hilal, amongst others, for the many art pieces adorning the space, like a floor-to-ceiling mural featuring a crying woman and man on a fire engine-red wall; a hallway featuring 80 stylized pink portraits, including one of Floyd, that Moran arranged over a chartreuse green wallpaper; and a tiger painted over an open wood-fire oven. Moran says that one of her favorite parts was discovering up-and-coming talent. "The younger Indian women artists that I looked around for—they are so incredible. It seems trifling for me to talk about them in this way, but, from the outside looking in, because I haven't noticed it before, there are so many of them. And they are doing this amazing, strong, feminist, outspoken work, and they're just beautiful and strong and embracing their qualities…," she says. "That was a fun journey for me to embark on."
Moran says, aside from the quick turnaround—she had less than a week to design the restaurant before it opened to the public—the process was not unlike designing a film set. "You have the character, which is Floyd, and then the subplot, which was that he wanted people to enjoy the food—he's like, 'I wanna have a place where people can feel good and be happy and wanna come and enjoy themselves and enjoy the food'—and then you have this restaurant structure, which is the set. Films are escapism, this restaurant is like escapism." Moran says. "It's like at the end of a Wes movie—you always have that amazing feeling, you get the song, you get the slo-mo, and you feel really good. That is what I was trying to bring here."
Take a tour of The Bombay Bread Bar with Moran, above, and learn more about the design details in the gallery, below.
Moral relied heavily on a saturated color palette of blues, reds, and pinks and mismatched prints. She commissioned an artist to create a portrait of Shah Jahan, who famously built the Taj Mahal, and hung portraits of men in suits, topped with animal heads, as nods to India's calling cards that "photographers" employed in the 1800s.


Essay On Sardar Patel | Speech On Sardar Patel : Sardar Patel's full name was Sardar Vallabhbhai Jhaverbhai Patel. He was one of the most important person involved in the struggle for independance. He was born on 31st October 1875 at Nadiad though his native place was at Karamsad. His family was not educated. He also initially passed matriculation very late at the age of 22 years. However he wanted to study law and he went to England for pursuing it. With great determination and hardwork he topped his college too. And after few years he emerged as a successful lawyer. He was married prior to matriculation at a tender age of 18 years. His wife however died due to cancer in January 1909. He received the news when he was busy working out a case. But he was a strong man and he didn't tell anybody about it until the case was over successfully.
The role of Sardar Patel in the history of India's independence is very important. He was the first Deputy Prime Minister and Home Minister of India. He did not become Prime Minister only because Gandhiji wanted Nehru to become the Prime Minister due to some reason. Sardarji became the first Home Minister of nascent India. Britishers left India after two centuries of struggle against them. Many people gave their lives and India got divided too. There were now many small kingdoms. Sardar Patel travelled to whole of India to convince all the 100 kings to join their kingdoms with the Independent Union of India. His most important role was the integration of 562 princely states under Indian Union. Within a year he redrew the boundaries of the India Map with every princely state merging the union of India and thereby making part of the political stream of life that was prosperous with harmony and cultural unity.
He was truly the Iron Man of India in its true meaning. He received the highest civilian award of Bharat Ratna for his immense contribution and 31st October is celebrated as Sardar Jayanti. There are lots of organizations and memorials after his name.
This iron man of India started to collapse in 1950 slowly due to blood in cough. After 2nd November in the same year he was confined to bed. He breathed his last on 15th December 1950. He will always be remembered for his brave and strong personality. Mr.Narendra Modi also stated in one of his addressing that India would have been much different if he was the Prime Minister then.
Ek Bharat Shreshtha Bharat is an initiative to boost cultural connect between different parts of the country by enhancing interaction among people living in different states.
Ek Bharat Shreshtha Bharat is a scheme, which is in the process to be launched as a new and effective scheme by the government of India. The Prime Minister of India, Shri Narendra Modi, has announced on Rashtriya Ekta Divas (31st of October 2015, the birth anniversary of Sardar Vallabhbhai Patel) to launch this new initiative. The aim of this scheme is to boost unity in India through the existing cultural connections among various parts of the country. It also aims to improve interaction between Indian people who are living at different places all through the country. This initiative will connect people to people which will really enhance unity in India.
The Prime Minister of India, Shri Narendra Modi, has announced on Rashtriya Ekta Divas address on 31st October in 2015 to launch a new initiative known as EK Bharat Shreshtha Bharat. Government of India is in process to launch this new initiative in order to boost the unity among people of India through the existing cultural connections between different parts of country. There is a plan to connect one state to any other state of the country every year on a reciprocal basis. In which, one state can popularize the rich heritage of another state such as Haryana state can connect to the Tamil Nadu state and can popularize its rich heritages through various events like literary events, book festivals, food festivals, song festivals, tours of Haryana people to Tamil Nadu, etc in its own state. In this way, a state would connect to any other state of the country every year and promote its heritages.
PM Narendra Modi has announced that EK Bharat Shreshtha Bharat is a new initiative to be launched in the near future in order to enhance unity in India. This initiative is in the process to be launched and aimed to connect people to people all through the country. This scheme was announced by the PM on Rashtriya Ekta Divas address, 31st of October 2015 (birth anniversary of Sardar Vallabhbhai Patel) to be launched. It is to initiate the cultural connections between different parts of the country. According to this initiative, one state of the country would connect to the other and promote each other's heritage in other state.
Through this scheme, people of a state will get proper knowledge of tradition, culture, and other practices of other states which will enhance the understanding and bonding among people thereby enhance and strengthen the unity and integrity of India. In order to make this scheme more effective 'EK Bharat Shreshtha Bharat' contest has been started to know the views and ideas of different people all through the country. Citizens of the country are invited to submit their views, ideas and suggestions (on government website) to structure the programme so that it can be effective in its all dimensions.
Indian Prime Minister, Narendra Modi, has talked about a new scheme called as Ek Bharat Shreshtha Bharat on 31st of October 2015 (birth anniversary of Sardar Vallabhbhai Patel) in his speech. This is an initiative in the process to be launched in near future. It is an effort by the Indian government to strengthen the unity and harmony all over the country. It is a programme aimed to connect people to people all through the country. India is a country which is the best example of "Unity in Diversity". This programme is also an initiative to improve the quality of unity in India. He also said in his 'Mann Ki Baat' programme that Ek Bharat Shreshtha Bharat scheme is to make India "One India Supreme India".
This solid initiative of enhancing the peace and harmony need to uphold the law and order in order to maintain the same. In order to make this programme more effective, PM has requested the views, ideas and suggestions of common public on a government portal 'MyGov.in'. He has also requested to suggest the structure, logo and ways to increase public participation in this programme. There are various creative minds hidden in the crowd who can give better suggestions to connect people of India to unity and harmony to make it Ek Bharat Shreshtha Bharat. There is a plan to make it a specific and prestigious scheme to connect people in easy manner to achieve the main goal of enriching the culture of unity and harmony in the country.
Ek Bharat Shreshtha Bharat scheme has got inspiration from the life of a great person and a freedom fighter of India named as Sardar Vallabhbhai Patel.
There is a plan to launch this scheme very soon in which, any one state of this country will connect to another state every year and promote each other's heritage like culture, tradition, language, etc.
There has been made committees to work effectively on the scheme in consultation with the states.
This scheme has aimed to get started with the unique partnership of two states for one year in which cultural and student exchanges would be followed.
Students of both states will go to each other's state in order to get knowledge about the culture, tradition and language.
In order to make this scheme en effective one, EK Bharat Shreshtha Bharat contest has been launched on 29th of November 2015. According to this contest, government of India needs views, ideas and better suggestions about this scheme by the common people to make it effective. People can submit their views, ideas and suggestions on 'mygov.in' till the last date of submission, 10th of December 2015.
PM Narendra Modi has talked about a scheme on the occasion of Rashtriya Ekta Divas, 31st of October 2015 (the birth anniversary of Sardar Vallabhbhai Patel). The name of the scheme is EK Bharat Shreshtha Bharat which is in the process to be launched in near future aiming to enhance and enrich the culture and tradition of country. Prime Minister has announced that Government of India will launch a new initiative to boost the cultural connections between different parts of the country. It is to connect people to people by enhancing interaction among people of different states.
According to this scheme, one state of the country will connect to other state on annual basis and both popularize the rich heritage of each other through the events like song festivals, food festivals, literary events, book festivals, tour and travel, etc. Next year, both states will connect to other two states aiming the same goal. This is the way to make people know about the culture, traditions and practices of various states all through the country. This will enhance the proper understanding and bonding among people as well as strengthen the unity and integrity in India.
It is said by the PM that the original and innovative ideas which are explained in detail will be more preferred and score higher. Details can be submitted in PDF format. First (Rs. 100,000/-), second (Rs. 75,000/-) and third (Rs. 50,000/-) cash prizes (along with a certificate) are also declared by the government in order to attract more participants towards this scheme. Citizens can submit the details about views and ideas till the last date of submission, 10th of December 2015.


text:
Later this month Starsailor play one of the XFM Manchester Access All Area April gigs which see the likes of Starsailor, The Zutons, Embrace and Goldfrapp in intimate venues across the city. Starsailor will return to Jabez Clegg to perform an acoustic set, their last appearance at Manchester was a sold out show at the 4000 capacity Apollo. Bizarrely the last time Designer Magazine saw Starsailor they were laughing outright at Colleen McLoughlin at the Kanye West gig at the MEN Arena where Ms Rooney to-be spent most of her time trying to get noticed by everyone in the building with her Agent Provocateur shopping bags.
Q: If you remember Designer Magazine did your first ever interview back at the Roadhouse when you were supporting Angelica?
A: Oh yeah, bringing back memories there
Q: We saw you at Jabez Clegg last October. Was that the first time you'd done shows as small as that since pre-NME tour in the early days?
A: Yeah. The idea was to get back to our roots and get warmed up for the theatre tour at the end of the month. Jabez Clegg was a great little gig. It was always a meeting point before going on to the Academy or the MDH for a gig and it has a good vibe. Obviously playing in Manchester any gig we have a great time and it's such a great city that there's a good band on at least once a week so when you're playing at the Apollo you're one of many good bands on that month, whereas that Jabez Clegg gig was a one-off and I think everyone that was able to be a part of it felt kind of special.
Q: I know you've done the Arena with the Charlatans and venues such as the Apollo as headliners. Do you lose something in cavernous venues like that?
A: It can do. It's really weird being the support band for gigs like that because when you get to that size you get more of the passing fan, if you know what I mean, like people who peruse the copy of City Life and think I quite like one of their songs I might head down to that. Whereas when you play the smaller venues tickets get bought up pretty quickly by the hardcore fans who know every word and sing every note. They're always the best kind of gigs.
Q: How quickly did the Jabez gig sell out last time?
A: It wasn't quite in half an hour, but it was sold old the day they went on sale
Q: How did it feel do comeback this time for the new album, a time where you have to have a habit or be connected to Pete Doherty in some way?
A: It's getting harder and harder, but we're still really grateful that we've still got our hand in so to speak. There's still enough people interested in the band that we can keep at it really. I'm one of these people that has more long-term goals as well. It's hard to compare yourselves to phenomenon's like Pete Doherty and the recent thing with the Arctic Monkeys because who knows what's going to happen to these people and these bands in years to come. I'd rather look back and compare in 10 or 20 years on what i've achieved.
The Arctic Monkey's are a great band and they deserve their success I think. But I think it's hard to compete with bands like that when there is so much hype around them. I think our time will come again and we've just got to keep our hand in and be ready for it when it comes and then look back in 10 or 20 years and think at this period in time we were the biggest band around, this time maybe not so quite so big and hopefully have a colourful career.
Q: When you think about Starsailor it is just about the music and that's strange when you consider that even Keane have gone through brand management to say what clothes makes the Keane brand look right
A: I think it's getting harder and harder to penetrate the british public in a big way because there seems to be a fascination and obsession with new things. Whereas elsewhere in the world there's more of a respect for experience. Having said that we've still got a lot of fans here so we can't have too many complaints. We can still fill the Apollo, but we're ambitious so we'd rather be playing the City Of Manchester stadium or down the road at Anfield, I won't mention that other place (laughs)
Q: Surely looking back and having a number 1 record in France as well must be one of the highlights?
A: It was unbelievable. It came out of the blue. It seemed like Silence Is Easy had, excuse the pun, sailed it's course. It seemed like it was starting to die down and we were ready to do the next album and suddenly this remix became this phenomenon. We took a few months out to promote the album in France and it came back to life and in the end with the remix being such a big success the album did just as well as "Love Is Here" in some respects.
In the top ten that week there was Mariah Carey and Eric Pridz's "Call On Me" with his dancers. It was really surreal because it stayed there for a few weeks. You put it in perspective and it sold half a million singles. I don't think Sugababes have sold that.
Also, it was kinda random because it was this strange remix that Jacques Lu Cont did, so it was more surreal and slightly ridiculous than if we'd got to number 1 with our own version cos that would have just been unbelievable.
Q: How was it recording your last album in LA?
A: The weather was a big factor. I think it definitely affects your mood when you can go outside after stressful times in the studio and sit out in the sunshine instead of rainy Manchester or in deepest Wales. Of course the weekends were amazing because you'd be going up to Tim Burgess' house for barbecues.
It's an inspiring place as well because it's like from the sublime to the ridiculous. Amazing people and beautiful scenery and you go out on Sunset strip and they're the most plastic empty people you could ever meet. It's a fascinating place.
Q: You can't imagine Starsailor going out of the North, never mind LA. Why did you decide to work with Rob Schnapf (Beck / Elliott Smith producer) on this album?
A: It's funny you should talk about the north. I love his stuff he did with Elliott Smith and The Vines and I went to a gig at the Limelight in Belfast that the Nine Black Alps were supporting the 22-20s. I remember Rob said he was working with the Nine Black Alps at the time so I asked them what he was like and what he was like to work with. They said he was amazing and had about 150 different guitars so that was me sold. So we've got Nine Black Alps to thank in some small way for us ending up working with Rob.
Q: Lyrically this album has changed somewhat from the first two albums as well?
A: That's through necessity because i've exhausted a lot of my personal life, especially on the second album with getting married and having a kid. Obviously when you get married and have a kid you get quite contented and cosy in your personal life and it's not necessary what people want to hear on a record. But you switch on the TV and there's all this crazy stuff happening in the world and it seems like every week there's a new global tragedy. It was a more slightly political album, hopefully with a small p. It's by no means an American Idiot or anything like that.
Q: The thing that Starsailor have always done is reverse the syndrome that most bands do. You've actually got more energetic with each album rather than the reverse
A: A lot of is through playing live as well. We made our name playing little places like the Social in London where there's not much room to swing a cat in, so an acoustic guitar and piano would suffice. You get famous and thrust upon a festival stage and you want to excite your audience as well as instilling emotion in them. There's nothing better than a big rousing version of "Good Souls" with a lot of electric guitar or "Silence Is Easy" with the amps on 11.
Watching other bands you just end up getting better and better. Like the Futureheads are a great live band and they really get the audience involved. It feels really communal and I feel they're a vastly underrated band. And obviously watching U2 they're the masters of live experience.
Q: The last tour you had a member of the Charlatans on tour, this one you had Richard from Echoboy joining you. Are you ever going to get a permanent 5th member of the band?
A: I don't think so because I do a lot of it myself in the studio. It's just really for the live shows to free me up to being a frontman instead of worrying about if I snap a string or hit the right pedal. I think you need someone getting the audience going and trying to be Ziggy Stardust (laughs).
We've known Richard cos he's always been around. He's done a few remixes for us and we remembered his band the Hybirds a few years back so we knew he could play instruments as well. He'd been playing bass for Spiritualised for a bit and I think that tour had just finished so luckily we managed to get him in time.
Q: So what can we expect from the XFM Access All Areas acoustic gig you're doing this month?
A: It will be tracks from the new album, but there will definitely be room for some of the old classics as well. We wouldn't want people going away disappointed because they haven't heard "Good Souls" or "Silence Is Easy". A Sugababes cover might also be in by popular demand
"On The Outside" is out now
Starsailor play Jabez Clegg on April 25th
For more info on Starsailor
www.starsailor.net
For more info on the XFM gigs and how to win tickets
www.xfmmanchester.co.uk
Click here to leave your Starsailor comments on the Message Board

redpajama_set_name: RedPajamaCommonCrawl


text:
О Ричарде Плантагенете, герцоге Глостере, см. Ричард III

Принц Ричард, герцог Глостерский (; род. ) — член британской королевской семьи, внук короля Георга V, двоюродный брат королевы Елизаветы II.

Биография
Крещён в Виндзорском замке, среди восприемников была тётка новорождённого королева Елизавета, жена Георга VI. В условиях Второй мировой войны место крестин (как и всех событий, связанных с королевской семьёй) было засекречено, и в газетах было написано, что племянника короля крестили в одной из часовен страны.

Наследовал титул герцога Глостерского от своего отца принца Генри в 1974 году (его мать Алиса, герцогиня Глостерская умерла в возрасте почти 103 лет в 2004 году). На сентябрь 2022 года является 30-м в порядке британского престолонаследия. По прямой мужской линии принц Ричард — старший в роду Виндзоров, потомков королевы Виктории и Альберта Саксен-Кобург-Готского. Таким образом, если бы в Великобритании при этой же правящей династии действовал салический закон, не допускающий женского наследования, он был бы королём.

В 1960-е годы Ричард занимался архитектурой и думал специализироваться в этой области, однако гибель в авиационной катастрофе старшего брата Уильяма (1941—1972) сделала его наследником герцогства и связала с ним известные обязательства перед короной (необходимость представлять Елизавету II в различных организациях, наносить визиты и т. п.). Герцог интересуется архитектурой и сохранностью исторических памятников и в настоящее время. Он является высочайшим патроном исторического общества имени Ричарда III (своего тёзки и также герцога Глостерского).

Резиденция — Кенсингтонский дворец в Лондоне.

Брак и дети
8 июля 1972 года женился на датчанке Биргитте ван Дёрс (фамилия по отцу — Хенриксен, родилась в 1946 году).

У пары трое детей:
 Александр, наследник герцогства (родился в 1974 году) носит титул графа Ольстерского
 Ксан (Зан, родился 12 марта 2007), лорд Куллоден
 леди Козима Роуз Александра Виндзор (родилась 20 мая 2010).
 леди Давина (родилась в 1977 году)
 Сенна Кауваи Льюис (родилась 22 июня 2010)
 Тейн Махута Льюис (родился 25 мая 2012)
 леди Роуз (родилась в 1980 году)
 Лайла Беатрикс Кристабель Гилман (родилась 30 мая 2010)
 Руфус Фредерик Монтегю Гилман (родился 2 ноября 2012).

Дети и внуки герцога Глостерского входят в порядок наследования британского престола, однако не имеют должностей Короны и не титулуются королевскими высочествами.

Награды

Блазон
Как член королевской семьи Ричард Глостерский имеет личный герб, основанный на гербе монарха Соединенного Королевства.

Четверочастный щит: в первом и четвёртом поле герб Англии — три золотых леопарда с лазоревым вооружением в червлёном поле, во втором поле герб Шотландии — в золотом поле с червлёной двойной внутренней каймой, проросшей лилиями червлёный восстающий лев с лазоревым вооружением, в третьем поле герб Ирландии — золотая арфа с серебряными струнами в лазоревом поле. Поверх щита серебряное титло о пяти концах, три из которых поочерёдно обременены крестом святого Георгия, а два — червлённым леопардом с лазуревым вооружением и языком.

Щит окружает лента ордена Подвязки.

Щитодержатели: справа — британский, коронованный открытой короной внуков суверена, лев c серебряным титлом (как в щите) на шее; слева — шотландский единорог с короной внуков суверена и серебряным титлом (как в щите) на шее.

Щит коронован короной внуков суверена с шапкой пэра внутри.

Нашлемник: золотой, коронованный открытой короной внуков суверена, леопард с серебряным титлом (как в щите) на шее, стоящий на короне внуков суверена.

Ссылки
 Profile at www.royal.gov.uk

Виндзоры
Графы Ольстер
Герцоги Глостер
Выпускники Кембриджского университета
Выпускники Колледжа Магдалины (Кембридж)

redpajama_set_name: RedPajamaWikipedia


Disney-Marvel's "Doctor Strange" opened this weekend to $86 million in 33 territories.
That's a very strong showing for the film starring Benedict Cumberbatch, even by Marvel standards, as the international debut is 49 percent higher than "Ant-Man," 37 percent more than "Guardians of the Galaxy" and 23 percent ahead of "Captain America: Winter Soldier," according to Walt Disney Studios' weekend estimates.
The film opens in an additional 787 IMAX screens globally next week, including North America and China, marking the widest IMAX release of all time and the first time a film will play in more than 1,000 locations.
Based in New York City's Greenwich Village, Doctor Strange must act as an intermediary between the real world and what lies beyond, utilizing a vast array of metaphysical abilities and artifacts to protect the Marvel cinematic universe.
It made $11.1 million in the U.K., $5.7 million in France and $4.9 million in Australia, followed by Germany, Mexico, Taiwan, Hong Kong, Indonesia and more.
Directed by Scott Derrickson ("Sinister"), "Doctor Strange" also stars Chiwetel Ejiofor, Rachel McAdams, Michael Stuhlbarg, Mads Mikkelsen, Tilda Swinton, Benedict Wong, Amy Landecker and Scott Adkins.


Trail begins along Kalmiopsis Rim Trail #1124 near Cold Springs and runs down a serpentine ridge to Emily Cabin area, where it follows an old mining road downstream to Copper Creek along the Little Chetco River. It crosses the river (ford required) near the Emily Cabin private inholding and then follows it downstream. It climbs over and down a steep hill to Madstone Cabin site on the main Chetco Wild and Scenic River between Madstone and Fresno Creeks. This trail is within the Kalmiopsis Wilderness. Likewise, Bailey Mountain Trail #1109 from the Little Chetco River up to Bailey Mountain may not be adequately maintained; it is best to check conditions prior to your visit to the area.
There is a small seasonal spring along the trail a short distance west of Kalmiopsis Rim Trail #1124. Near the confluence of Ditch Creek, there is a spur trail going upstream along the Little Chetco River to some nice camping spots and historic cabin sites.
This trail is located in an area that has burned in the Biscuit (2002) and Chetco Bar (2017) Fires. Although Forest Service trail crews and partner groups have been repairing and maintaining many of the affected trails following the fires, visitors should take into consideration (because of the substantially increased risk of fire-killed snags falling) that travel and camping may be more difficult and require greater concern for safety.


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