Directed discrete midpoint convexity

Abstract

For continuous functions, midpoint convexity characterizes convex functions. By considering discrete versions of midpoint convexity, several types of discrete convexities of functions, including integral convexity, L$^\natural$-convexity and global/local discrete midpoint convexity, have been studied. We propose a new type of discrete midpoint convexity that lies between L$^\natural$-convexity and integral convexity and is independent of global/local discrete midpoint convexity. The new convexity, named DDM-convexity, has nice properties satisfied by L$^\natural$-convexity and global/local discrete midpoint convexity. DDM-convex functions are stable under scaling, satisfy the so-called parallelogram inequality and a proximity theorem with the same small proximity bound as that for L$^{\natural }$-convex functions. Several characterizations of DDM-convexity are given and algorithms for DDM-convex function minimization are developed. We also propose DDM-convexity in continuous variables and give proximity theorems on these functions.

1 Introduction

For a continuous function f defined on a convex set $S \subseteq {\mathbb {R}}^n$, it was proved by Jensen [12] that midpoint convexity defined by

$$\begin{aligned} f(x) + f(y) \ge 2 f\left( \frac{x+y}{2}\right) \qquad (\forall x,y \in S) \end{aligned}$$

is equivalent to the inequality defining convex functions

$$\begin{aligned} \alpha f(x) + (1-\alpha )f(y) \ge f(\alpha x + (1-\alpha ) y) \qquad (\forall x,y \in S;\; \forall \alpha \in [0,1]). \end{aligned}$$

By capturing the concept of midpoint convexity, several types of ‘discrete’ midpoint convexities for functions defined on the integer lattice ${\mathbb {Z}}^n$ have been proposed.

A weak version of ‘discrete’ midpoint convexity is obtained by replacing $f((x+y)/2)$ by the smallest value of a linear extension of f among the integer points neighboring $(x+y)/2$. More precisely, for any point $x \in {\mathbb {R}}^n$, we consider its integer neighborhood

$$\begin{aligned} N(x) = \{ z \in {\mathbb {Z}}^n \mid |z_i-x_i| < 1 \;(i=1,,\ldots ,n)\} \end{aligned}$$

and the set $\Lambda (x)$ of all coefficients $(\lambda _z \mid z \in N(x))$ for convex combinations indexed by N(x). For a function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$, we define the local convex envelope ${\tilde{f}}$ of f by

$$\begin{aligned} {\tilde{f}}(x) = \min \left\{ \sum _{z \in N(x)}\lambda _z f(z) \mid \sum _{z \in N(x)}\lambda _z z = x,\; (\lambda _z) \in \Lambda (x)\right\} \qquad (x \in {\mathbb {R}}^n). \end{aligned}$$

We say that f satisfies weak discrete midpoint convexity if the following inequality holds

$$\begin{aligned} f(x) + f(y) \ge 2 {\tilde{f}} \left( \frac{x+y}{2}\right) \end{aligned}$$

(1.1)

for all $x,y \in {\mathbb {Z}}^n$. On the other hand, f is said to be integrally convex [3] if ${\tilde{f}}$ is convex on ${\mathbb {R}}^n$. Characterizations of integral convexity by using weak discrete midpoint convexity have been discussed in [3, 17, 18]. The simplest characterization, Theorem A.1 in [18], says that f is integrally convex if and only if f satisfies (1.1) for all $x,y \in \mathop {\mathrm{dom}}\limits f$ with¹$\Vert x-y\Vert _{\infty } \ge 2$, where the effective domain $\mathop {\mathrm{dom}}\limits f$ of f is defined by

$$\begin{aligned} \mathop {\mathrm{dom}}\limits f = \{ x \in {\mathbb {Z}}^n \mid f(x) < +\infty \}. \end{aligned}$$

The class of integrally convex functions establishes a general framework of discrete convex functions, including separable convex, L$^\natural$-convex, M$^\natural$-convex, L$^\natural _2$-convex, M$^\natural _2$-convex functions [21], BS-convex and UJ-convex functions [4], and globally/locally discrete midpoint convex functions [18]. The concept of integral convexity is used in formulating discrete fixed point theorems [8, 9, 29], designing algorithms for discrete systems of nonlinear equations [14, 28], and guaranteeing the existence of a pure strategy equilibrium in finite symmetric games [10].

A strong version of ‘discrete’ midpoint convexity is obtained by replacing $f((x+y)/2)$ by the average of the values of f at two integer points obtained by rounding-up and rounding-down of all components of $(x+y)/2$. More precisely, for a function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$, we say that f satisfies discrete midpoint convexity if it has

$$\begin{aligned} f(x) + f(y) \ge f\left( \left\lceil \frac{x+y}{2} \right\rceil \right) + f\left( \left\lfloor \frac{x+y}{2} \right\rfloor \right) \end{aligned}$$

(1.2)

for all $x,y \in {\mathbb {Z}}^n$, where $\lceil \cdot \rceil$ and $\lfloor \cdot \rfloor$ denote the integer vectors obtained by rounding up and rounding down all components of a given real vector, respectively. It is known that discrete midpoint convexity characterizes the class of L$^\natural$-convex functions [5, 20] which play important roles in both theoretical and practical aspects. L$^\natural$-convex functions are applied to several fields, including auction theory [15, 25], image processing [13], inventory theory [2, 27, 30] and scheduling [1]. Since discrete midpoint convexity (1.2) obviously implies weak discrete midpoint convexity (1.1), L$^\natural$-convex functions forms a subclass of integrally convex functions.

Moriguchi et al. [18] classified discrete convex functions between L$^\natural$-convex and integrally convex functions in terms of discrete midpoint convexity with $\ell _\infty$-distance requirements, and proposed two new classes of discrete convex functions, namely, globally/locally discrete midpoint convex functions. A function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ is said to be globally discrete midpoint convex if (1.2) holds for any pair $(x,y) \in {\mathbb {Z}}^n \times {\mathbb {Z}}^n$ with $\Vert x-y\Vert _\infty \ge 2$. A set $S \subseteq {\mathbb {Z}}^n$ is called a discrete midpoint convex set if its indicator function $\delta _S$ defined by

$$\begin{aligned} \delta _S(x)= {\left\{ \begin{array}{ll} 0 &{} (x\in S)\\ +\infty &{} (x\not \in S) \end{array}\right. } \qquad (x \in {\mathbb {Z}}^n) \end{aligned}$$

is globally discrete midpoint convex, that is, if

$$\begin{aligned} x, y\in S \text{ with } \Vert x-y\Vert _\infty \ge 2 \;\Rightarrow \; \left\lceil \frac{x+y}{2} \right\rceil ,\; \left\lfloor \frac{x+y}{2} \right\rfloor \in S. \end{aligned}$$

A function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ is said to be locally discrete midpoint convex if $\mathop {\mathrm{dom}}\limits f$ is a discrete midpoint convex set and (1.2) holds for any pair $(x,y) \in {\mathbb {Z}}^n \times {\mathbb {Z}}^n$ with $\Vert x-y\Vert _\infty = 2$. It is shown in [18] that the following inclusion relations among function classes hold:

$$\begin{aligned}&\{ \mathrm{L}^{\natural }\text{-convex } \} \ \subsetneqq \ \{ \text{ globally } \text{ discrete } \text{ midpoint } \text{ convex } \} \\&\subsetneqq \ \{ \text{ locally } \text{ discrete } \text{ midpoint } \text{ convex } \} \ \subsetneqq \ \{ \text{ integrally } \text{ convex } \}, \end{aligned}$$

and globally/locally discrete midpoint convex functions inherit nice features from L$^\natural$-convex functions, that is, for a globally/locally discrete midpoint convex f and a positive integer $\alpha$,

• the scaled function $f^\alpha$ defined by $f^\alpha (x) = f(\alpha x)\; (x \in {\mathbb {Z}}^n)$ belongs to the same class, that is, global/local discrete midpoint convexity is closed with respect to scaling operations,

• a proximity theorem with the same proximity distance with L$^\natural$-convexity holds, that is, given an $x^\alpha$ with $f(x^\alpha ) \le f(x^\alpha + \alpha d)$ for all $d \in \{-1,0,+1\}^n$, there exists a minimizer $x^*$ of f with $\Vert x^\alpha -x ^*\Vert _\infty \le n(\alpha -1)$,

• when f has a minimizer, a steepest descent algorithm for the minimization of f is developed such that the number of local minimizations in the neighborhood of $\ell _\infty$-distance 2 (the 2-neighborhood minimizations) is bounded by the shortest $\ell _\infty$-distance from a given initial feasible point to a minimizer of f, and

• when $\mathop {\mathrm{dom}}\limits f$ is bounded and $K_\infty$ denotes the $\ell _\infty$-size of $\mathop {\mathrm{dom}}\limits f$, a scaling algorithm minimizing f with $O(n \log _2 K_\infty$) calls of the 2-neighborhood minimization is developed.

This paper, strongly motivated by [18], proposes a new type of discrete midpoint convexity between L$^\natural$-convexity and integral convexity, but it is independent of global/local discrete midpoint convexity with respect to inclusion relation. We name the new convexity directed discrete midpoint convexity (DDM-convexity) which forms the following classification

$$\begin{aligned} \{ {\mathrm{L}}^{\natural}\text{-convex} \} \ \subsetneqq \ \{ \text{DDM-convex} \} \ \subsetneqq \ \{ \text{integrally} \text{ convex} \}. \end{aligned}$$

The same features as mentioned above are satisfied by DDM-convexity. The merits of DDM-convexity relative to global/local discrete midpoint convexity are the following properties:

• DDM-convexity is closed with respect to individual sign inversion of variables, that is, for a DDM-convex function f and $\tau _i \in \{+1,-1\}\;(i=1,\ldots ,n)$, $f(\tau _1 x_1,\ldots ,\tau _n x_n)$ is also DDM-convex (see Proposition 4 (3)). Neither L$^\natural$-convexity nor global nor local discrete midpoint convexity has this property, while integral convexity is closed with respect to individual sign inversion of variables.

• For a quadratic function $f(x) = x^\top Q x$ with a symmetric matrix $Q = [q_{ij}]$, DDM-convexity is characterized by the diagonal dominance with nonnegative diagonals of Q:

  $$\begin{aligned} q_{ii} \ge \sum _{j:j \ne i}|q_{ij}| \qquad (\forall i =1,\ldots ,n) \end{aligned}$$

  (see Theorem 9). While L$^\natural$-convexity is characterized by the combination of diagonal dominance with nonnegative diagonals and nonpositivity of all off-diagonal components of Q, global/local discrete midpoint convexity is independent of the diagonal dominance with nonnegative diagonals.

• A function $g : {\mathbb {Z}}\rightarrow {\mathbb {R}}\cup \{+\infty \}$ is said to be discrete convex if $g(t-1)+g(t+1) \ge 2g(t)$ for all $t \in {\mathbb {Z}}$. For univariate discrete convex functions $\xi _{i}, \varphi _{ij}, \psi _{ij} : {\mathbb {Z}}\rightarrow {\mathbb {R}}\cup \{+\infty \}\; (i=1,\ldots ,n; j \in \{1,\ldots ,n\} {\setminus } \{i\}$), a 2-separable convex function [7] defined as a function represented as

  $$\begin{aligned} f(x)=\sum _{i=1}^n \xi _i(x_i)+ \sum _{i,j:j\not =i}\varphi _{ij} (x_i-x_j)+ \sum _{i,j:j\not =i} \psi _{ij}(x_i+x_j)\qquad (x\in {\mathbb {Z}}^n). \end{aligned}$$

  The class of DDM-convex functions includes all 2-separable convex functions (see Theorem 4). It is known that if all $\psi _{ij}$ are identically zero, then f is L$^\natural$-convex, whereas there exists a 2-separable convex function not contained in the class of globally/locally discrete midpoint convex functions.

• A steepest descent algorithm for the minimization of DDM-convex functions requires only the 1-neighborhood minimization in contrast to the 2-neighborhood minimization (see Sect. 8.1).

In the next section, we give the definition of DDM-convexity and basic properties of DDM-convex functions. In Sect. 3, we discuss a relationship between DDM-convexity and known discrete convexities. For globally/locally discrete midpoint convex functions, Moriguchi et al. [18] revealed a useful property, which is expressed by the so-called parallelogram inequality. We show that a similar parallelogram inequality holds for DDM-convex functions in Sect. 4. Sections 5 and 6 are devoted to characterizations and operations for DDM-convexity. We prove a proximity theorem for DDM-convex functions in Sect. 7, while in Sect. 8 we propose a steepest descent algorithm and a scaling algorithm for DDM-convex function minimization. In Sect. 9, we define DDM-convex functions in continuous variables and give proximity theorems for such functions.

2 Directed discrete midpoint convexity

We give the definition of directed discrete midpoint convexity and show its basic properties.

For an ordered pair (x, y) of $x,y\in {\mathbb {Z}}^n$, we define $\mu (x,y) \in {\mathbb {Z}}^n$ by

$$\begin{aligned} \mu (x,y)_i= {\left\{ \begin{array}{ll} \displaystyle \left\lceil \frac{x_i+y_i}{2} \right\rceil &{} (x_i\ge y_i), \\ \displaystyle \left\lfloor \frac{x_i+y_i}{2} \right\rfloor &{} (x_i<y_i). \end{array}\right. } \end{aligned}$$

That is, each component $\mu (x,y)_i$ of $\mu (x,y)$ is defined by rounding up or rounding down $\frac{x_i+y_i}{2}$ to the integer in the direction of $x_i-y_i$. It is easy to show the next characterization of $\mu (x,y)$ and $\mu (y,x)$.

Proposition 1

For $x,y,p,q\in {\mathbb {Z}}^n$, $p=\mu (x,y)$ and $q=\mu (y,x)$ hold if and only if the following conditions (a)$\sim$(c) hold:

• (a) $p+q = x+y$,

• (b) $\Vert p-q\Vert _\infty \le 1$, and

• (c) for each $i=1,\dots ,n$, if $x_i\ge y_i$, then $p_i\ge q_i$; otherwise $p_i\le q_i$.

For every $a,b\in {\mathbb {R}}^n$, let us denote the n-dimensional vector $(a_1b_1,\dots ,a_nb_n)$ by $a\odot b$. The next proposition gives fundamental properties of $\mu (\cdot ,\cdot )$.

Proposition 2

For every $x,y,d\in {\mathbb {Z}}^n$, the following properties hold.

• (1) $x+y=\mu (x,y)+\mu (y,x)$.

• (2) If $x\ge y$, then $\mu (x,y)=\lceil \frac{x+y}{2}\rceil$ and $\mu (y,x)= \lfloor \frac{x+y}{2}\rfloor$.

• (3) If $\Vert x-y\Vert _\infty \le 1$, then $\mu (x,y)=x$ and $\mu (y,x)=y$.

• (4) $\mu (x+d,y+d)=\mu (x,y)+d$.

• (5) For any permutation $\sigma$ of $(1,\dots ,n)$,

  $$\begin{aligned} \mu ((x_{\sigma (1)},\dots ,x_{\sigma (n)}), (y_{\sigma (1)},\dots ,y_{\sigma (n)}))= (\mu (x,y)_{\sigma (1)},\dots ,\mu (x,y)_{\sigma (n)}). \end{aligned}$$

• (6) For any $\tau \in \{+1,-1\}^n$, $\mu (\tau \odot x,\tau \odot y)=\tau \odot \mu (x,y)$.

Proof

Properties (1)–(5) are obvious by the definition of $\mu (\cdot ,\cdot )$. Let us show (6). If $\tau _i=+1$,

$$\begin{aligned} \mu (\tau \odot x,\tau \odot y)_i= {\left\{ \begin{array}{ll} \left\lceil \frac{x_i+y_i}{2}\right\rceil =\mu (x,y)_i&{}\ (x_i\ge y_i)\\ \left\lfloor \frac{x_i+y_i}{2}\right\rfloor =\mu (x,y)_i&{}\ (x_i<y_i) \end{array}\right. } \end{aligned}$$

holds, and if $\tau _i=-1$,

$$\begin{aligned} \mu (\tau \odot x,\tau \odot y)_i= {\left\{ \begin{array}{ll} \left\lfloor \frac{-x_i-y_i}{2}\right\rfloor =-\left\lceil \frac{x_i+y_i}{2}\right\rceil =-\mu (x,y)_i&{}\ (x_i \ge y_i)\\ \left\lceil \frac{-x_i-y_i}{2}\right\rceil =-\left\lfloor \frac{x_i+y_i}{2}\right\rfloor =-\mu (x,y)_i&{}\ (x_i < y_i) \end{array}\right. } \end{aligned}$$

holds. $\square$

By using the introduced $\mu (\cdot ,\cdot )$, we propose new classes of functions and sets. We say that a function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ satisfies directed discrete midpoint convexity (DDM-convexity) or is a directed discrete midpoint convex function (DDM-convex function) if

$$\begin{aligned} f(x)+f(y)\ge f(\mu (x, y))+f(\mu (y, x)) \end{aligned}$$

(2.1)

for all $x,y \in {\mathbb {Z}}^n$. We call $S \subseteq {\mathbb {Z}}^n$ a directed discrete midpoint convex set (DDM-convex set) if its indicator function $\delta _S$ is DDM-convex, that is, if

$$\begin{aligned} x, y\in S\Rightarrow \mu (x, y),\; \mu (y, x)\in S \end{aligned}$$

holds.

The next propositions are direct consequences of Proposition 2 and the definition (2.1).

Proposition 3

The following statements hold:

• (1) Any function defined on $\{0, 1\}^n$ is a DDM-convex function.

• (2) Any subset of $\{0, 1\}^n$ is a DDM-convex set.

• (3) For a DDM-convex function f, its effective domain $\mathop {\mathrm{dom}}\limits f$ and the set $\mathop {\mathrm{argmin}}\limits f$ of minimizers of f are DDM-convex sets, where $\mathop {\mathrm{argmin}}\limits f$ is defined by

  $$\begin{aligned} \mathop {\mathrm{argmin}}\limits f = \{ x \in {\mathbb {Z}}^n \mid f(x) \le f(z)\; (\forall z \in {\mathbb {Z}}^n)\}. \end{aligned}$$

Proposition 4

Let $f, f_1, f_2:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be DDM-convex functions.

• (1) For any $d\in {\mathbb {Z}}^n$, $g(x)=f(x+d)$ is a DDM-convex function.

• (2) For any permutation $\sigma$ of $(1, \dots , n)$, $g(x)=f(x_{\sigma (1)}, \dots , x_{\sigma (n)})$ is a DDM-convex function.

• (3) For any $\tau \in \{+1, -1\}^n$, $g(x)=f(\tau \odot x)$ is a DDM-convex function.

• (4) For any $a_1, a_2\ge 0$, $g(x)=a_1f_1(x)+a_2f_2(x)$ is a DDM-convex function. In particular, the sum of DDM-convex functions is DDM-convex.

Proposition 5

Let $S, S_1, S_2\subseteq {\mathbb {Z}}^n$ be DDM-convex sets.

• (1) For any $d\in {\mathbb {Z}}^n$, $T=\{x+d \mid x\in S\}$ is a DDM-convex set.

• (2) For any permutation $\sigma$ of $(1, \dots , n)$, $T=\{(x_{\sigma (1)},\dots ,x_{\sigma (n)}) \mid (x_1,\dots ,x_n) \in S \}$ is a DDM-convex set.

• (3) For any $\tau \in \{+1, -1\}^n$, $T=\{\tau \odot x \mid x\in S\}$ is a DDM-convex set.

• (4) $T=S_1\cap S_2$ is a DDM-convex set.

3 Relationships with known discrete convexities

We discuss relationships between DDM-convexity and known discrete convexities, including integral convexity, L$^\natural$-convexity, global/local discrete midpoint convexity and 2-separable convexity.

As mentioned in Sect. 1, the class of integrally convex functions is characterized by weak discrete midpoint convexity (1.1). Since DDM-convexity (2.1) trivially implies (1.1), any DDM-convex function is integrally convex. Therefore, DDM-convex functions inherit many properties of integrally convex functions. We introduce a good property of integrally convex functions as well as DDM-convex functions, box-barrier property.

Theorem 1

(Box-barrier property [17, Theorem 2.6]) Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an integrally convex function, and let $p\in ({\mathbb {Z}}\cup \{-\infty \})^n$ and $q\in ({\mathbb {Z}}\cup \{+\infty \})^n$ with $p\le q$. Define

$$\begin{aligned} S&=\{x\in {\mathbb {Z}}^n \mid p_i<x_i<q_i\ (i=1, \dots , n)\}, \\ W^+_i&=\{x\in {\mathbb {Z}}^n \mid x_i=q_i, p_j\le x_j\le q_j\ (j\not =i)\}\quad (i=1, \dots , n),\\ W^-_i&=\{x\in {\mathbb {Z}}^n \mid x_i=p_i, p_j\le x_j\le q_j\ (j\not =i)\}\quad (i=1, \dots , n), \\ W&=\bigcup _{i=1}^n (W^+_i\cup W^-_i), \end{aligned}$$

and ${\hat{x}}\in S\cap \mathop {\mathrm{dom}}\limits f$. If $f({\hat{x}})\le f(y)$ for all $y\in W$, then $f({\hat{x}})\le f(z)$ for all $z\in {\mathbb {Z}}^n{\setminus } S$.

By setting $p = {\hat{x}} - \varvec{1}$ and $q = {\hat{x}} + \varvec{1}$ where $\varvec{1}$ denotes the vector of all ones, box-barrier property implies the minimality criterion of integrally convex functions.

Theorem 2

([3, Proposition 3.1]; see also [21, Theorem 3.21]) Let $f: {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an integrally convex function and ${\hat{x}} \in \mathop {\mathrm{dom}}\limits f$. Then ${\hat{x}}$ is a minimizer of f if and only if $f({\hat{x}}) \le f({\hat{x}} + d)$ for all $d \in \{ -1, 0, +1 \}^{n}$.

As a special case of Theorem 2, we have the minimality criterion of DDM-convex functions.

Corollary 1

Let $f: {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a DDM-convex function and ${\hat{x}} \in \mathop {\mathrm{dom}}\limits f$. Then ${\hat{x}}$ is a minimizer of f if and only if $f({\hat{x}}) \le f({\hat{x}} + d)$ for all $d \in \{ -1, 0, +1 \}^{n}$.

We next discuss the relationship between L$^\natural$-convexity and DDM-convexity. L$^\natural$-convex functions are originally defined by translation-submodularity:

$$\begin{aligned} f(x)+f(y)\ge f((x-\alpha \varvec{1})\vee y)+f(x\wedge (y+\alpha \varvec{1})) \end{aligned}$$

(3.1)

for all $x,y \in {\mathbb {Z}}^n$ and nonnegative integer $\alpha$, where $p \vee q$ and $p \wedge q$ denote the componentwise maximum and minimum of the vectors p and q, respectively. Translation-submodularity is a generalization of submodularity:

$$\begin{aligned} f(x)+f(y)\ge f(x \vee y)+f(x \wedge y). \end{aligned}$$

(3.2)

L$^\natural$-convexity has several equivalent characterizations as below.

Theorem 3

([3, Corollary 5.2.2], [5, Theorem 3], [21, Theorem 7.7] ) For a function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$, the following properties are equivalent:

• (1) f is $\text {L}^\natural$-convex, that is, (3.1) holds for all $x,y \in {\mathbb {Z}}^n$ and nonnegative integer $\alpha$.

• (2) f satisfies discrete midpoint convexity (1.2) for all $x,y \in {\mathbb {Z}}^n$.

• (3) f is integrally convex and submodular.

• (4) For every $x, y\in {\mathbb {Z}}^n$ with $x \not \ge y$ and $A={{\,\mathrm{\mathrm {argmax}}\,}}_i\{y_i-x_i\}$,

  $$\begin{aligned} f(x)+f(y)\ge f(x+\varvec{1}_A)+f(y-\varvec{1}_A), \end{aligned}$$

  where the i-component of $\varvec{1}_A$ is 1 if $i \in A$; otherwise 0.

Theorem 3 yields the next property.

Proposition 6

Any L$^\natural$-convex function is DDM-convex.

Proof

Let $f : {\mathbb {Z}}\rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an L$^\natural$-convex function. We arbitrarily fix $x,y \in \mathop {\mathrm{dom}}\limits f$ and show that (2.1) holds for x and y. By Proposition 2 (3), (2.1) holds if $\Vert x-y\Vert _\infty \le 1$. Suppose that $m = \Vert x-y\Vert _\infty \ge 2$ and $\Vert y-x\Vert _\infty =\max _i \{y_i-x_i\}$ (by exchanging the role of x and y if necessary). Let $A = {{\,\mathrm{\mathrm {argmax}}\,}}_i\{y_i - x_i\}$, $p=x+\varvec{1}_A$ and $q=y-\varvec{1}_A$. The vectors p and q satisfy the properties (a) and (c) of Proposition 1 and $\Vert p-q\Vert _\infty \le m$. Furthermore, by Theorem 3 (4), we have

$$\begin{aligned} f(x) + f(y) \ge f(p) + f(q). \end{aligned}$$

(3.3)

If $\Vert p-q\Vert _\infty =m$, by applying the above process for the pair (q, p) again, we obtain p and q with $\Vert p-q\Vert _\infty < m$ preserving (a), (c) of Proposition 1 and (3.3). By repeating this argument, we finally obtain p and q having (a)$\sim$(c) of Proposition 1 and (3.3), which means that (2.1) holds for x and y. $\square$

A set $S \subseteq {\mathbb {Z}}^n$ is called an L$^\natural$-convex set if its indicator function $\delta _S$ is L$^\natural$-convex.

Corollary 2

Any L$^\natural$-convex set is DDM-convex.

Example 1

([18, Remark 1]) The class of L$^\natural$-convex functions is a proper subclass of DDM-convex functions. For example,

$$\begin{aligned} S=\{(1, 0), (0, 1)\} \end{aligned}$$

is a DDM-convex set, but for $x=(1, 0), y=(0, 1)$, $\lceil \frac{x+y}{2}\rceil =(1, 1)\not \in S$ and $\lfloor \frac{x+y}{2}\rfloor =(0, 0)\not \in S$, which means that S is not L$^\natural$-convex. $\square$

Example 2

([18, Remark 2]) A set $S\subseteq {\mathbb {Z}}^n$ is said to be $L^\natural _2$-convex set if it is the Minkowski sum of two L$^\natural$-convex sets. DDM-convexity and L$^\natural _2$-convexity are mutually independent. For example,

$$\begin{aligned} \{(1, 0), (0, 1)\} \end{aligned}$$

is a DDM-convex set, but is not L$^\natural _2$-convex. On the other hand,

$$\begin{aligned} S=\{(0, 0, 0, 0), (0, 1, 1, 0), (1, 1, 0, 0), (1, 2, 1, 0)\} \end{aligned}$$

is the Minkowski sum of two L$^\natural$-convex sets $S_1 = \{\{(0, 0, 0, 0), (0, 1, 1, 0)\}$ and $S_2 = \{(0, 0, 0, 0),(1, 1, 0, 0)\}$, but S is not DDM-convex because for $x=(0, 0, 0, 0)$, $y=(1, 2, 1, 0)$, $\mu (x, y)=(0, 1, 0, 0)\not \in S$ and $\mu (y, x)=(1, 1, 1, 0)\not \in S$. $\square$

We next discuss the independence between global/local discrete midpoint convexity and DDM-convexity by showing the independence between discrete midpoint convex sets and DDM-convex sets.

Example 3

It is easy to show that the set S defined by

$$\begin{aligned} S=\{(0, 0, 0), (1, 1, 0), (1, 0, -1), (2, 1, -1)\} \end{aligned}$$

is discrete midpoint convex, but S is not DDM-convex because for $x=(0, 0, 0)$ and $y=(2, 1, -1)$, we have $\mu (x, y)=(1, 0, 0)\not \in S$ and $\mu (y, x)=(1, 1, -1)\not \in S$.

On the other hand,

$$\begin{aligned} T=\{(0, 0, 0), (1, 0, 0), (1, 1, 1), (2, 1, 1), (1, 1, -1), (2, 1, -1), (1, 1, 0), (2, 1, 0)\} \end{aligned}$$

is DDM-convex. However, T is not discrete midpoint convex, and moreover, for any $(\tau _1, \tau _2, \tau _3)\in \{-1, +1\}^3$, the modified set

$$\begin{aligned} \tau \odot T = \{(\tau _1x_1, \tau _2x_2, \tau _3x_3) \mid (x_1, x_2, x_3)\in T\} \end{aligned}$$

is not discrete midpoint convex while it is DDM-convex by Proposition 5 (3). The reason is as follows. Since T is symmetric on the third component, we can assume $\tau _3=+1$.

• In the case where $\tau = (\pm 1, +1, +1)$, for $x=(0, 0, 0) = \tau \odot (0,0,0)$ and $y=(\pm 2, 1, -1)=\tau \odot (2,1,-1)$, $\lfloor (x+y)/2 \rfloor = (\pm 1, 0, -1)\not \in \tau \odot T$.

• In the case where $\tau = (\pm 1, -1, +1)$, for $x = (0, 0, 0) = \tau \odot (0,0,0)$ and $y = (\pm 2, -1, 1)=\tau \odot (2,1,1)$, $\lceil (x+y)/2 \rceil = (\pm 1, 0, 1)\not \in \tau \odot T$. $\square$

We finally show that 2-separable convex functions are DDM-convex. Let $\xi _{i}, \varphi _{ij}, \psi _{ij} : {\mathbb {Z}}\rightarrow {\mathbb {R}}\cup \{+\infty \}\; (i=1,\ldots ,n; j \in \{1,\ldots ,n\} {\setminus } \{i\}$) be univariate discrete convex functions. A 2-separable convex function [7] is defined as a function represented as

$$\begin{aligned} f(x)=\sum _{i=1}^n \xi _i(x_i)+ \sum _{i,j:j\not =i}\varphi _{ij} (x_i-x_j)+ \sum _{i,j:j\not =i} \psi _{ij}(x_i+x_j)\qquad (x\in {\mathbb {Z}}^n). \end{aligned}$$

(3.4)

It is known that the function g defined by

$$\begin{aligned} g(x)=\sum _{i=1}^n \xi _i(x_i)+ \sum _{i,j:j\not =i}\varphi _{ij} (x_i-x_j) \qquad (x\in {\mathbb {Z}}^n) \end{aligned}$$

is L$^\natural$-convex [21, Proposition 7.9]. By Propositions 4 (4) and 6, it is enough to show that each $\psi _{ij}$ is DDM-convex in order to prove DDM-convexity of 2-separable convex function

$$\begin{aligned} f(x)=g(x)+ \sum _{i,j:j\not =i} \psi _{ij}(x_i+x_j)\qquad (x\in {\mathbb {Z}}^n). \end{aligned}$$

Lemma 1

For a univariate discrete convex function $\psi :{\mathbb {Z}}\rightarrow {\mathbb {R}}\cup \{+\infty \}$, $f:{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ defined by

$$\begin{aligned} f(x)=\psi (x_1+x_2)\qquad (x\in {\mathbb {Z}}^n) \end{aligned}$$

is DDM-convex.

Proof

For every $x,y \in {\mathbb {Z}}^n$, we show that

$$\begin{aligned} \psi (x_1{+}x_2) + \psi (y_1{+}y_2) \ge \psi (\mu (x,y)_1{+}\mu (x,y)_2) + \psi (\mu (y,x)_1{+}\mu (y,x)_2). \end{aligned}$$

(3.5)

Suppose that $\min \{x_1{+}x_2, y_1{+}y_2\}=x_1{+}x_2$ and $\max \{x_1{+}x_2, y_1{+}y_2\}=y_1{+}y_2$ without loss of generality. By convexity of $\psi$, for every $a,b,p,q \in {\mathbb {Z}}$ such that (i) $a+b=p+q$, (ii) $a \le p \le b$ and (iii) $a \le q \le b$, we have $\psi (a)+\psi (b)\ge \psi (p)+\psi (q)$. Thus, it is enough to show that

$$\begin{aligned}&(x_1+x_2)+(y_1+y_2)=(\mu (x, y)_1+\mu (x, y)_2)+(\mu (y, x)_1+\mu (y, x)_2), \end{aligned}$$

(3.6)

$$\begin{aligned}&x_1+x_2\le \mu (x, y)_1+\mu (x, y)_2\le y_1+y_2, \end{aligned}$$

(3.7)

$$\begin{aligned}&x_1+x_2\le \mu (y, x)_1+\mu (y, x)_2\le y_1+y_2. \end{aligned}$$

(3.8)

Obviously, (3.6) holds by

$$\begin{aligned} x_1+y_1=\mu (x, y)_1+\mu (y, x)_1, \quad x_2+y_2=\mu (x, y)_2+\mu (y, x)_2. \end{aligned}$$

To show (3.7) and (3.8) under $x_1+x_2 \le y_1+y_2$, we consider the following three cases separately: Case 1: $x_1\le y_1$ and $x_2\le y_2$, Case 2: $x_1>y_1$ and $x_2<y_2$, Case 3: $x_1<y_1$ and $x_2>y_2$.

Case 1 ($x_1\le y_1$ and $x_2\le y_2$). In this case, we have $\mu (x, y)_1=\lfloor \frac{x_1+y_1}{2}\rfloor$, $\mu (x, y)_2=\lfloor \frac{x_2+y_2}{2}\rfloor$, $\mu (y, x)_1=\lceil \frac{x_1+y_1}{2}\rceil$ and $\mu (y, x)_2=\lceil \frac{x_2+y_2}{2}\rceil$, which imply

$$\begin{aligned} x_1\le \mu (x, y)_1\le \mu (y, x)_1\le y_1, \quad x_2\le \mu (x, y)_2\le \mu (y, x)_2\le y_2. \end{aligned}$$

Conditions (3.7) and (3.8) are direct consequences of the above inequalities.

Case 2 ($x_1>y_1$ and $x_2<y_2$). In this case, under condition $x_1+x_2 \le y_1+y_2$, $c_1=x_1-y_1$ and $c_2=y_2-x_2$ satisfy $c_2\ge c_1>0$. By the following calculations:

$$\begin{aligned} \mu (x, y)_1=&\left\lceil \frac{x_1+y_1}{2}\right\rceil =x_1-\left\lfloor \frac{c_1}{2}\right\rfloor =y_1+\left\lceil \frac{c_1}{2}\right\rceil , \\ \mu (x, y)_2=&\left\lfloor \frac{x_2+y_2}{2}\right\rfloor =x_2+\left\lfloor \frac{c_2}{2}\right\rfloor =y_2-\left\lceil \frac{c_2}{2}\right\rceil , \\ \mu (y, x)_1=&\left\lfloor \frac{x_1+y_1}{2}\right\rfloor =x_1-\left\lceil \frac{c_1}{2}\right\rceil =y_1+\left\lfloor \frac{c_1}{2}\right\rfloor , \\ \mu (y, x)_2=&\left\lceil \frac{x_2+y_2}{2}\right\rceil =x_2+\left\lceil \frac{c_2}{2}\right\rceil =y_2-\left\lfloor \frac{c_2}{2}\right\rfloor , \end{aligned}$$

we have

$$\begin{aligned} \mu (x, y)_1{+}\mu (x, y)_2 =&x_1+x_2+ \left( \left\lfloor \frac{c_2}{2}\right\rfloor {-}\left\lfloor \frac{c_1}{2}\right\rfloor \right) =y_1+y_2- \left( \left\lceil \frac{c_2}{2}\right\rceil {-}\left\lceil \frac{c_1}{2}\right\rceil \right) , \\ \mu (y, x)_1{+}\mu (y, x)_2 =&x_1+x_2+ \left( \left\lceil \frac{c_2}{2}\right\rceil {-}\left\lceil \frac{c_1}{2}\right\rceil \right) =y_1+y_2- \left( \left\lfloor \frac{c_2}{2}\right\rfloor {-}\left\lfloor \frac{c_1}{2}\right\rfloor \right) . \end{aligned}$$

Conditions (3.7) and (3.8) follow from $\left\lfloor \frac{c_2}{2}\right\rfloor -\left\lfloor \frac{c_1}{2}\right\rfloor \ge 0$ and $\left\lceil \frac{c_2}{2}\right\rceil -\left\lceil \frac{c_1}{2}\right\rceil \ge 0$.

Case 3 ($x_1<y_1$ and $x_2>y_2$). In this case, we can show (3.7) and (3.8) in the same way as Case 2. $\square$

By Proposition 4 (4), Proposition 6 and Lemma 1, we have the next property.

Theorem 4

Any 2-separable convex function is DDM-convex.

Hence, 2-separable convex function is integrally convex.

4 Parallelogram inequality

Parallelogram inequality was originally proposed in [18] for globally/locally discrete midpoint convex functions. By borrowing arguments from [18], we show that DDM-convex sets/functions have similar properties.

For every pair $(x, y)\in {\mathbb {Z}}^n\times {\mathbb {Z}}^n$ with $\Vert y-x\Vert _\infty =m$, we consider sets defined by

$$\begin{aligned} A_k=\{i \mid y_i-x_i\ge k\}, \quad B_k=\{i \mid y_i-x_i\le -k\} \qquad (k=1, \dots , m), \end{aligned}$$

(4.1)

for which $A_1\supseteq A_2\supseteq \cdots \supseteq A_m$, $B_1\supseteq B_2\supseteq \cdots \supseteq B_m$, $A_1\cap B_1=\emptyset$ and $A_m\cup B_m\not =\emptyset$.

We first show the following property of DDM-convex sets.

Theorem 5

Let $S\subseteq {\mathbb {Z}}^n$ be a DDM-convex set, $x, y\in S$ with $\Vert y-x\Vert _\infty =m$, and $J\subseteq \{1, 2, \dots , m\}$. If $\{A_k\}$ and $\{B_k\}$ are defined by (4.1) and $d=\sum _{k\in J}(\varvec{1}_{A_k}-\varvec{1}_{B_k})$, we have $x+d\in S$ and $y-d\in S$.

To show this theorem, it is enough to verify

$$\begin{aligned} x+\sum _{k\in J}(\varvec{1}_{A_k}-\varvec{1}_{B_k})\in S \qquad (\forall J \subseteq \{1, 2, \dots , m\}). \end{aligned}$$

(4.2)

We first show that the decomposition $\sum _{k=1}^m (\varvec{1}_{A_k} - \varvec{1}_{B_k})$ of $y-x$ can be constructed by using the operation $\mu (\cdot ,\cdot )$.

For every $x\in {\mathbb {Z}}^n$, let us consider multiset D(x) of vectors by the following recursive formula:

$$\begin{aligned} D(x)= {\left\{ \begin{array}{ll} \emptyset &{} (x=\varvec{0}), \\ \{x\} &{} (\Vert x\Vert _\infty =1), \\ \{\mu (x, \varvec{0}), \mu (\varvec{0}, x)\} &{} (\Vert x\Vert _\infty =2), \\ D(\mu (x, \varvec{0}))\cup D(\mu (\varvec{0}, x)) &{} (\Vert x\Vert _\infty \ge 3), \end{array}\right. } \end{aligned}$$

(4.3)

where $\varvec{0}$ denotes the n-dimensional zero vector. We give several propositions.

Proposition 7

If a multiset $\{d^k\in \{-1, 0, 1\}^n{\setminus }\{\varvec{0}\} \mid k=1, \dots , m\}$ satisfies

$$\begin{aligned} 1\ge d^1_i \ge d^2_i \ge \cdots \ge d^m_i \ge 0 \; \text{ or } \; -1 \le d^1_i \le d^2_i \le \cdots \le d^m_i \le 0 \end{aligned}$$

(4.4)

for each $i\in \{1, \dots , n\}$, then

$$\begin{aligned} D\left( \sum _{k = 1}^m d^k\right) =\{d^k \mid k = 1,\ldots , m \}. \end{aligned}$$

(4.5)

Proof

We prove the assertion by induction on m. The assertion obviously holds if $m \le 1$.

Suppose that $m=2$. By (4.4), we have

$$\begin{aligned} \Vert d^{1}{+}d^{2}\Vert _\infty = 2, \quad \mu (d^{1}{+}d^{2}, \varvec{0})=d^{1}, \quad \mu (\varvec{0}, d^{1}{+}d^{2})=d^{2}, \end{aligned}$$

which, together with (4.3), imply the assertion.

Suppose that $m \ge 3$. Let $K=\{1, 2, \dots , m\}$, $K^\mathrm {O}=\{k \in K \mid k \text{ is } \text{ odd }\}$ and $K^\mathrm {E}=\{k \in K \mid k \text{ is } \text{ even }\}$. By induction hypothesis together with $|K^\mathrm {O}|, |K^\mathrm {E}|<|K|$, we obtain

$$\begin{aligned} D\left( \sum _{k\in K^\mathrm {O}}d^k\right) =\{d^k \mid k\in K^\mathrm {O}\}, \quad D\left( \sum _{k\in K^\mathrm {E}}d^k\right) =\{d^k \mid k\in K^\mathrm {E}\}. \end{aligned}$$

(4.6)

Furthermore, the claim below guarantees that

$$\begin{aligned} \mu \left( \sum _{k\in K}d^k, \varvec{0}\right) =\sum _{k\in K^\mathrm {O}}d^k, \quad \mu \left( \varvec{0}, \sum _{k\in K}d^k\right) =\sum _{k\in K^\mathrm {E}}d^k. \end{aligned}$$

(4.7)

By combining (4.3), (4.6) and (4.7), we have

$$\begin{aligned} D\left( \sum _{k\in K}d^k\right)&=D\left( \mu \left( \sum _{k\in K}d^k, \varvec{0}\right) \right) \cup D\left( \mu \left( \varvec{0}, \sum _{k\in K}d^k\right) \right) \\&=D\left( \sum _{k\in K^\mathrm {O}}d^k\right) \cup D\left( \sum _{k\in K^\mathrm {E}}d^k\right) \\&=\{d^k \mid k\in K^\mathrm {O}\}\cup \{d^k \mid k\in K^\mathrm {E}\}\\&=\{d^k \mid k\in K\}. \end{aligned}$$

Claim

(i) $\mu \left( \sum _{k\in K}d^k, \varvec{0}\right) =\sum _{k\in K^\mathrm {O}}d^k$, and (ii) $\mu \left( \varvec{0}, \sum _{k\in K}d^k\right) =\sum _{k\in K^\mathrm {E}}d^k$.

Proof

We show (i) (and can show (ii) in the same way). Let us fix $i \in \{1,\ldots ,n\}$. Assume that $\left( \sum _{k\in K}d^k\right) _i=l > 0$. Since $d^{1}_i=\cdots =d^{l}_i=1$ and $d^{{l+1}}_i=\cdots =d^{m}_i=0$ by (4.4), we have

$$\begin{aligned} \mu \left( \sum _{k\in K}d^k, \varvec{0}\right) _i =\left\lceil \frac{l}{2}\right\rceil =|\{1, \dots , l\} \cap \{k \mid k \text{ is } \text{ odd }\}| =\left( \sum _{k\in K^\mathrm {O}}d^k\right) _i. \end{aligned}$$

In the case where $\left( \sum _{k\in K}d^k\right) _i=-l < 0$, we have $d^{1}_i=\cdots =d^{l}_i=-1$ and $d^{{l+1}}_i=\cdots =d^{m}_i=0$ by (4.4), and hence

$$\begin{aligned} \mu \left( \sum _{k\in K}d^k, \varvec{0}\right) _i =\left\lfloor \frac{-l}{2}\right\rfloor =-\left\lceil \frac{l}{2}\right\rceil =-|\{1, \dots , l\}\cap \{k \mid k \text{ is } \text{ odd }\}| =\left( \sum _{k\in K^\mathrm {O}}d^k\right) _i. \end{aligned}$$

If $\left( \sum _{k\in K}d^k\right) _i=0$, by (4.4), we have $d^{1}_i=\cdots =d^{m}_i=0$ and

$$\begin{aligned} \mu \left( \sum _{k\in K}d^k, \varvec{0}\right) _i =0 =\left( \sum _{k\in K^\mathrm {O}}d^k\right) _i. \end{aligned}$$

Thus, (i) holds. (End of the proof of Claim). $\square$

Proposition 8

$D(y-x)=\{\varvec{1}_{A_k}-\varvec{1}_{B_k} \mid k=1, \dots , m\}$.

Proof

By the construction (4.1), $y-x=\sum _{k=1}^m(\varvec{1}_{A_k}-\varvec{1}_{B_k})$. By defining $d^k=\varvec{1}_{A_k}-\varvec{1}_{B_k}\ (k=1, \dots , m)$, condition (4.4) of Proposition 7 holds. The assertion is an immediate consequence of (4.5). $\square$

By Proposition 8, (4.2) can be rewritten as

$$\begin{aligned} x+\sum \{d \mid d\in E\}\in S \qquad (\forall E \subseteq D(y-x)). \end{aligned}$$

(4.8)

Therefore, Theorem 5 can be shown by the following proposition.

Proposition 9

For a DDM-convex set S and $x,y \in S$, (4.8) holds.

Proof

We prove (4.8) for x, y by induction on $\Vert y-x\Vert _\infty$. If $\Vert y-x\Vert _\infty \le 1$, (4.8) trivially holds. If $\Vert y-x\Vert _\infty =2$, then $D(y-x)=\{\mu (y{-}x, \varvec{0}), \ \mu (\varvec{0}, y{-}x)\}$. Since S is DDM-convex, we have

$$\begin{aligned} x+\mu (y{-}x, \varvec{0})&=\mu (y, x)\in S, \\ x+\mu (\varvec{0}, y{-}x)&=\mu (x, y)\in S, \end{aligned}$$

which guarantee that (4.8) holds.

Suppose that $\Vert y-x\Vert _\infty \ge 3$, and (4.8) holds for every $x'', y''\in S$ with $\Vert x''-y''\Vert _\infty <\Vert x-y\Vert _\infty$. We fix $E\subseteq D(y-x)$ arbitrarily. Let $x'=\mu (x, y)$ and $y'=\mu (y, x)$. Then we have $y' - x = y - x' = \mu (y-x,\varvec{0})$. By DDM-convexity of S, $x'$ and $y'$ also belong to S. By Proposition 2 (4) and the assumption $\Vert y-x\Vert _\infty \ge 3$, we have

$$\begin{aligned} \Vert y'-x\Vert _\infty = \Vert y-x'\Vert _\infty = \Vert \mu (y-x, \varvec{0})\Vert _\infty < \Vert y-x\Vert _\infty . \end{aligned}$$

By induction hypothesis, (4.8) holds for $(x, y')$ and $(x', y)$, and furthermore, by the equality $D(y'-x)=D(y-x')=D(\mu (y-x, \varvec{0}))$, we have

$$\begin{aligned} \begin{aligned}&u=x+\sum \{d \mid d\in E\cap D(\mu (y-x, \varvec{0}))\}\in S, \\&v=x'+\sum \{d \mid d\in E\cap D(\mu (y-x, \varvec{0}))\}\in S. \end{aligned} \end{aligned}$$

(4.9)

We also have $v-u = x'-x = \mu (\varvec{0}, y-x)$ and

$$\begin{aligned} \Vert v-u\Vert _\infty =\Vert x'-x\Vert _\infty =\Vert \mu (\varvec{0}, y-x)\Vert _\infty <\Vert y-x\Vert _\infty , \end{aligned}$$

which, together with the induction hypothesis, guarantee that (4.8) holds for (u, v). Moreover, by (4.3), $D(y-x) = D(\mu (y-x, \varvec{0}))\cup D(\mu (\varvec{0}, y-x))$ includes E, and hence, $D(v-u)=D(\mu (\varvec{0}, y-x))$ includes $E{\setminus } D(\mu (y-x, \varvec{0}))$. Thus, we have

$$\begin{aligned} w=u+\sum \{d \mid d\in E{\setminus } D(\mu (y-x, \varvec{0}))\}\in S. \end{aligned}$$

(4.10)

By (4.9) and (4.10), we obtain

$$\begin{aligned} w=x+\sum \{d \mid d\in E\}\in S, \end{aligned}$$

which implies (4.8). $\square$

We denote by DDMC(k) and by DDMC($\ge$ $k$) the classes of functions $f : {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ that satisfy DDM-convexity (2.1) for all $x,y \in {\mathbb {Z}}^n$ with $\Vert x-y\Vert _\infty = k$ and for all $x,y \in {\mathbb {Z}}^n$ with $\Vert x-y\Vert _\infty \ge k$, respectively. Before presenting parallelogram inequality for DDM-convex functions, we give a useful property.

Theorem 6

Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a function in DDMC(2) such that $\mathop {\mathrm{dom}}\limits f$ is DDM-convex. For $x\in \mathop {\mathrm{dom}}\limits f$ and $y\in {\mathbb {Z}}^n$ with $\Vert y-x\Vert _\infty =m$, and for any partition (I, J) of $\{1, \dots , m\}$, we consider

$$\begin{aligned} d_1=\sum _{i\in I}(\varvec{1}_{A_i}-\varvec{1}_{B_i}), \quad d_2=\sum _{j\in J}(\varvec{1}_{A_j}-\varvec{1}_{B_j}), \end{aligned}$$

where $A_k, B_k\ (k=1, \dots , m)$ are the sets defined by (4.1). Then we have

$$\begin{aligned} f(x)+f(x+d_1+d_2)\ge f(x+d_1)+f(x+d_2). \end{aligned}$$

(4.11)

Proof

We note that $y = x+d_1+d_2$. If $y\not \in \mathop {\mathrm{dom}}\limits f$, by $f(y)=+\infty$, (4.11) trivially holds. In the sequel, we assume that $y \in \mathop {\mathrm{dom}}\limits f$. Let I be denoted by $\{i_1,i_2\ldots ,i_{|I|}\}$ $(i_1< i_2< \cdots < i_{|I|})$ and J by $\{j_1,j_2\ldots ,j_{|J|}\}$ $(j_1< j_2< \cdots < j_{|J|})$. For every $k = 1,\ldots ,|I|$ and $l = 1,\ldots ,|J|$, we denote $d_1^k = \varvec{1}_{A_{i_k}}-\varvec{1}_{B_{i_k}}$ and, similarly, $d_2^l = \varvec{1}_{A_{j_l}}-\varvec{1}_{B_{j_l}}$. For every $k=0,1, \dots , |I|$ and for every $l=0,1, \dots , |J|$, define

$$\begin{aligned} x(k, l)=x+\sum _{i=1}^{k}d_1^ i + \sum _{j=1}^l d_2^j. \end{aligned}$$

By Theorem 5, for every k, l, we have $x(k, l) \in \mathop {\mathrm{dom}}\limits f$. We note that (4.11) is equivalent to

$$\begin{aligned} f(x(0, 0)) + f(x(|I|, |J|)) \ge f(x(|I|, 0)) + f(x(0, |J|)). \end{aligned}$$

(4.12)

Fix $k \in \{1,\ldots ,|I|\}$ and $l \in \{1,\ldots , |J|\}$. According to whether $i_k > j_l$ or $i_k < j_l$, either

$$\begin{aligned} A_{i_k} \subseteq A_{j_l},\quad B_{i_k} \subseteq B_{j_l},\quad A_{j_l} \cap B_{j_l} = \emptyset ,\quad A_{i_k} \cup B_{i_k} \ne \emptyset \end{aligned}$$

or

$$\begin{aligned} A_{j_l} \subseteq A_{i_k},\quad B_{j_l} \subseteq B_{i_k},\quad A_{i_k} \cap B_{i_k} = \emptyset ,\quad A_{j_l} \cup A_{j_l} \ne \emptyset . \end{aligned}$$

Thus, in the case where $i_k > j_l$, we have

$$\begin{aligned} (d_1^k+d_2^l)_p= {\left\{ \begin{array}{ll} 1 &{} (p\in A_{j_l}{\setminus } A_{i_k}), \\ 2 &{} (p\in A_{i_k}), \\ -2 &{} (p\in B_{i_k}), \\ -1 &{} (p\in B_{j_l}{\setminus } B_{i_k}), \\ 0 &{} (p\not \in (A_{j_l}\cup B_{j_l})), \end{array}\right. } \end{aligned}$$

which implies $\Vert d_1^k+d_2^l\Vert _\infty = 2$, $\mu (d_1^k+d_2^l, \varvec{0})=\varvec{1}_{A_{j_l}}-\varvec{1}_{B_{j_l}}=d^l_2$ and $\mu (\varvec{0}, d_1^k+d_2^l)=\varvec{1}_{A_{i_k}}-\varvec{1}_{B_{i_k}}=d^k_1$. Similarly, in the case where $i_k < j_l$, we have $\Vert d_1^k+d_2^l\Vert _\infty = 2$, $\mu (d_1^k+d_2^l, \varvec{0})=d^k_1$, $\mu (\varvec{0}, d_1^k+d_2^l)=d^l_2$. In both cases, since $\Vert x(k, l)-x(k-1, l-1)\Vert _\infty =\Vert d_1^k+d_2^l\Vert _\infty =2$ and $f \in$ DDMC(2), we obtain

$$\begin{aligned} \begin{aligned}&f(x(k, l))+f(x(k-1, l-1)) \\&\quad \ge f(\mu (x(k, l), x(k-1, l-1)))+f(\mu (x(k-1, l-1), x(k, l))). \end{aligned} \end{aligned}$$

(4.13)

On the other hand, the facts

$$\begin{aligned} \mu (x(k, l), x(k-1, l-1))&=x(k-1, l-1)+\mu (d_1^k+d_2^l, \varvec{0}), \\ \mu (x(k-1, l-1), x(k, l))&=x(k-1, l-1)+\mu (\varvec{0}, d_1^k+d_2^l), \end{aligned}$$

together with $\{\mu (d_1^k+d_2^l, \varvec{0}),\mu (\varvec{0}, d_1^k+d_2^l)\} = \{d_1^k, d_2^l\}$ in the both cases where $i_k > j_l$ and $i_k < j_l$, yield the right-hand side of (4.13) is equal to $f(x(k, l-1))+f(x(k-1, l))$. Therefore, we obtain

$$\begin{aligned} f(x(k, l)))+f(x(k-1, l-1)) \ge f(x(k, l-1))+f(x(k-1, l)). \end{aligned}$$

By adding the above inequalities for (k, l) with $1\le k\le |I|$ and $1\le l\le |J|$, we obtain (4.12). We emphasize that all the terms that are canceled in this addition of inequalities are finite valued because $x(k,l) \in \mathop {\mathrm{dom}}\limits f$ for all (k, l) with $0\le k\le |I|$ and $0\le l\le |J|$. $\square$

The next theorem is an immediate consequence of Theorem 6, because a DDM-convex function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ belongs to DDMC(2) and $\mathop {\mathrm{dom}}\limits f$ is DDM-convex.

Theorem 7

Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a DDM-convex function. For every $x, y\in \mathop {\mathrm{dom}}\limits f$, let $\{A_k \mid k=1,\ldots ,m\}$ and $\{B_k \mid k=1,\ldots ,m\}$ be the families defined by (4.1), where $m = \Vert y-x\Vert _\infty$. For any subset $J \subseteq \{1,\ldots ,m\}$, let $d = \sum _{k\in J}(\varvec{1}_{A_k}-\varvec{1}_{B_k})$. Then we have

$$\begin{aligned} f(x)+f(y)\ge f(x+d)+f(y-d). \end{aligned}$$

(4.14)

We call the inequality (4.14) parallelogram inequality of DDM-convex functions.

5 Characterizations

In this section, we give several equivalent conditions of DDM-convexity and a simple characterization of quadratic DDM-convex functions.

For every pair $(x,y) \in {\mathbb {Z}}^n \times {\mathbb {Z}}^n$, we recall that the families $\{A_k \mid k=1,\ldots ,m\}$ and $\{B_k \mid k=1,\ldots ,m\}$ are defined by

$$\begin{aligned} A_k=\{i \mid y_i-x_i\ge k\}, \quad B_k=\{i \mid y_i-x_i\le -k\} \qquad (k=1, \dots , m), \end{aligned}$$

in (4.1), where $m = \Vert y-x\Vert _\infty$.

Theorem 8

For a function $f:{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$, the following properties are equivalent to each other.

• (1) f is DDM-convex, i.e., $f\in$ DDMC($\ge$ $1$).

• (2) $\mathop {\mathrm{dom}}\limits f$ is DDM-convex and $f\in$ DDMC(2).

• (3) f satisfies parallelogram inequalities for every $x, y\in \mathop {\mathrm{dom}}\limits f$ and for any subset $J \subseteq \{1,\ldots ,m\}$ where $m = \Vert y-x\Vert _\infty$.

• (4) For every $x, y\in \mathop {\mathrm{dom}}\limits f$, we have

  $$\begin{aligned} f(x)+f(y)\ge f(x+\varvec{1}_{A_m}-\varvec{1}_{B_m})+f(y-\varvec{1}_{A_m}+\varvec{1}_{B_m}), \end{aligned}$$

  (5.1)

  where $m = \Vert y-x\Vert _\infty$, and the sets $A_m$ and $B_m$ are defined by (4.1).

• (5) For every $x, y\in \mathop {\mathrm{dom}}\limits f$ and for any $\alpha \in \{1, \dots , m\}$, by defining $d=\sum _{k=1}^\alpha (\varvec{1}_{A_k}-\varvec{1}_{B_k})$,we have

  $$\begin{aligned} f(x)+f(y)\ge f(x+d)+f(y-d), \end{aligned}$$

  where $m = \Vert y-x\Vert _\infty$, and the families $\{A_k \mid k=1,\ldots ,m\}$ and $\{B_k \mid k=1,\ldots ,m\}$ are defined by (4.1).

Proof

The implication $(1) \Rightarrow (2)$ is obvious and the implications from (1) to (3), (4) and (5) follow from Theorem 7. To prove opposite implications, we show $f \in$ DDMC(m) for every $m\ge 1$ from (4), and show [$(5) \Rightarrow (4)$], [$(3) \Rightarrow (4)$] and [$(2) \Rightarrow (3)$].

(4)$\Rightarrow$(1): Since $f \in$ DDMC(1) always holds from Proposition 2, we assume $m\ge 2$ and fix $x,y \in \mathop {\mathrm{dom}}\limits f$ with $\Vert x-y\Vert _\infty =m$, arbitrarily. Let denote $p=x+\varvec{1}_{A_m}-\varvec{1}_{B_m}$ and $q=y-\varvec{1}_{A_m}+\varvec{1}_{B_m}$ in (5.1). Vectors p and q satisfy conditions (a) and (c) of Proposition 1, $\Vert p-q\Vert _\infty <m$ and $p,q \in \mathop {\mathrm{dom}}\limits f$. By repeating (5.1) for (p, q) until $\Vert p-q\Vert _\infty \le 1$, the final p and q satisfy all conditions of Proposition 1, and hence, $p=\mu (x, y)$ and $q=\mu (y, x)$ are satisfied. This shows $f \in$ DDMC(m) holds.

(5)$\Rightarrow$(4): By setting $\alpha = m-1$ in (5), we obtain

$$\begin{aligned} x+d=y-\varvec{1}_{A_m}+\varvec{1}_{B_m}, \quad y-d=x+\varvec{1}_{A_m}-\varvec{1}_{B_m}, \end{aligned}$$

and hence, (4).

(3)$\Rightarrow$(4): By setting $J=\{1, \dots , m-1\}$ in (3), we obtain

$$\begin{aligned} x+d=y-\varvec{1}_{A_m}+\varvec{1}_{B_m}, \quad y-d=x+\varvec{1}_{A_m}-\varvec{1}_{B_m}, \end{aligned}$$

and hence, (4).

(2)$\Rightarrow$(3): Property (2) and Theorem 6 imply (3). $\square$

Remark 1

Theorem 8 may be regarded as a generalization of Theorem 3.

Property (4) of Theorem 8 corresponds to (4) of Theorem 3. For an L$^\natural$-convex function f, the following inequalities

$$\begin{aligned} f(x) + f(y) \ge f(x + \varvec{1}_{A_m}) + f(y - \varvec{1}_{A_m}) \end{aligned}$$

and

$$\begin{aligned} f(x) + f(y) \ge f(x - \varvec{1}_{B_m}) + f(y + \varvec{1}_{B_m}) \end{aligned}$$

hold. These inequalities imply (5.1). However, for a DDM-convex function f, the two parts on $A_m$ and $B_m$ must be combined to a single inequality (5.1). This demonstrates the difference between L$^\natural$-convexity and DDM-convexity.

Property (5) of Theorem 8 corresponds to (1) of Theorem 3. In (5) of Theorem 8, two vectors $x+d=x+\sum _{k=1}^\alpha (\varvec{1}_{A_k}-\varvec{1}_{B_k})$ and $y-d=y-\sum _{k=1}^\alpha (\varvec{1}_{A_k}-\varvec{1}_{B_k})$ are expressed by

$$\begin{aligned} (x+d)_i= {\left\{ \begin{array}{ll} x_i-\alpha &{} (x_i-y_i\ge \alpha ), \\ x_i+\alpha &{} (y_i-x_i\ge \alpha ), \\ y_i &{} (|x_i-y_i|<\alpha ), \end{array}\right. } \quad (y-d)_i= {\left\{ \begin{array}{ll} y_i+\alpha &{} (x_i-y_i\ge \alpha ), \\ y_i-\alpha &{} (y_i-x_i\ge \alpha ), \\ x_i &{} (|x_i-y_i|< \alpha ). \end{array}\right. } \end{aligned}$$

On the other hand, in (3.1), two vectors $(x-\alpha \varvec{1})\vee y$ and $x\wedge (y+\alpha \varvec{1})$ can be rewritten as

$$\begin{aligned} \left( (x-\alpha \varvec{1})\vee y\right) _i&= {\left\{ \begin{array}{ll} x_i-\alpha &{} (x_i-y_i\ge \alpha ), \\ y_i &{} (x_i-y_i<\alpha ), \end{array}\right. } \\ \left( x\wedge (y+\alpha \varvec{1}) \right) _i&= {\left\{ \begin{array}{ll} y_i+\alpha &{} (x_i-y_i\ge \alpha ), \\ x_i &{} (x_i-y_i<\alpha ). \end{array}\right. } \end{aligned}$$

In the same way as the relation between (4) of Theorem 8 and (4) of Theorem 3, two separate operations for L$^\natural$-convex functions must be executed simultaneously for DDM-convex functions. $\square$

For a quadratic function $f(x) = x^\top Q x\;(x\in {\mathbb {Z}}^n)$ with a symmetric matrix $Q = [q_{ij}]$, we show that f is DDM-convex if and only if Q is diagonally dominant with nonnegative diagonals:

$$\begin{aligned} q_{ii} \ge \sum _{j:j \ne i}|q_{ij}| \qquad (\forall i =1,\ldots ,n). \end{aligned}$$

(5.2)

For each $p\in {\mathbb {R}}$, let $p^+=\max \{p, 0\}$ and $p^-=\max \{-p, 0\}$. Note that $|p|=p^{+} + p^{-}$. Quadratic function $f(x)=x^\top Q x$ can be written as²

$$\begin{aligned} f(x)&=\sum _{i, j=1}^n q_{ij}x_ix_j \nonumber \\&=\sum _{i=1}^n\left( q_{ii}{-}\sum _{j:j\not =i}|q_{ij}|\right) x_i^2 +\frac{1}{2}\sum _{i,j:j\not =i}(q_{ij}^+(x_i{+}x_j)^2+q_{ij}^-(x_i{-}x_j)^2). \end{aligned}$$

(5.3)

In (5.3), the condition (5.2) of Q implies the nonnegativity of coefficients of $x_i^2$. Thus, if Q is diagonally dominant with nonnegative diagonals, then f is 2-separable convex, and hence, DDM-convex by Theorem 4. By proving the opposite implication, we obtain the following property.

Theorem 9

For a quadratic function $f(x) = x^\top Q x\; (x \in {\mathbb {Z}}^n)$ with a symmetric matrix $Q = [q_{ij}]$, f is DDM-convex if and only if Q is diagonally dominant with nonnegative diagonals.

Proof

It is enough to show that if f is DDM-convex, then Q is diagonally dominant with nonnegative diagonals. For each $i\in \{1, \dots , n\}$, define $z^i\in {\mathbb {Z}}^n$ by

$$\begin{aligned} z^i_j= {\left\{ \begin{array}{ll} 2 &{} \ (j=i), \\ -1 &{} \ (j\not =i \text{ and } q_{ij}\ge 0), \\ 1 &{} \ (j\not =i \text{ and } q_{ij}<0). \end{array}\right. } \end{aligned}$$

By DDM-convexity of f, the inequality

$$\begin{aligned} f(z^i)+f(\varvec{0})\ge f(\mu (z^i, \varvec{0}))+f(\mu (\varvec{0}, z^i)) \end{aligned}$$

must hold. Since $\mu (z^i, \varvec{0})=z^i-\varvec{1}_i$ and $\mu (\varvec{0}, z^i)=\varvec{1}_i$, we have

$$\begin{aligned} (z^i)^\top Q z^i \ge (z^i-\varvec{1}_i)^\top Q(z^i-\varvec{1}_i)+\varvec{1}_i^\top Q\varvec{1}_i, \end{aligned}$$

which implies

$$\begin{aligned} 0&\le (z^i)^\top Q \varvec{1}_i -\varvec{1}_i^\top Q \varvec{1}_i \\&=q_{ii}-\sum _{j:j\not =i\wedge q_{ji}\ge 0} q_{ji}+\sum _{j:j\not =i\wedge q_{ji}<0} q_{ji} \\&=q_{ii}-\sum _{j:j\not =i} |q_{ji}|. \end{aligned}$$

By $Q^\top = Q$, we obtain the diagonal dominance with nonnegative diagonals of Q. $\square$

The minimizers of DDM-convex functions are DDM-convex sets, while the minimizers of L$^\natural$-convex functions are L$^\natural$-convex sets. The class of L$^\natural$-convex functions has a characterization in terms of minimizers. For a function $f : {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ and $p \in {\mathbb {R}}^n$, we denote by $f-p$ the function given by

$$\begin{aligned} (f-p)(x) = f(x) - \sum _{i=1}^n p_ix_i \qquad (\forall x \in {\mathbb {Z}}^n). \end{aligned}$$

Theorem 10

([21, 23]) Under some regularity condition, a function $f : {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ is L$^\natural$-convex if and only if $\mathop {\mathrm{argmin}}\limits (f-p)$ is an L$^\natural$-convex set for every $p \in {\mathbb {R}}^n$ with $\inf (f-p) >-\infty$.

Unfortunately, the class of DDM-convex functions does not have a similar characterization.

Example 4

Let us consider the function $f : {\mathbb {Z}}^3 \rightarrow {\mathbb {R}}\cup \{+\infty \}$ given by

$$\begin{aligned}&f(0,0,0) = 0,\quad f(1,0,1) = f(1,1,0) = 1, \\&f(1,0,0) = f(1,1,1) = 2,\quad f(2,1,1) = 3, \\&f(x) = +\infty \quad (\text{otherwise}). \end{aligned}$$

The function f is not DDM-convex because

$$\begin{aligned} f(0,0,0) + f(2,1,1) = 3 < 4 = f(1,0,0) + f(1,1,1), \end{aligned}$$

while $\mathop {\mathrm{dom}}\limits f$ is a DDM-convex set (in fact, an L$^\natural$-convex set). Furthermore, $\mathop {\mathrm{argmin}}\limits (f-p)$ is a DDM-convex set for every $p \in {\mathbb {R}}^3$ as follows. There exists no $p \in {\mathbb {R}}^3$ such that $\{(0,0,0),(2,1,1)\} \subseteq \mathop {\mathrm{argmin}}\limits (f-p)$, because we have $0 \le 1-p_1-p_2$ from $(f-p)(0,0,0) \le (f-p)(1,1,0)$, and $2-p_1-p_2 \le 0$ from $(f-p)(2,1,1) \le (f-p)(1,0,1)$. For any $p \in {\mathbb {R}}^3$ and for any $x,y \in \mathop {\mathrm{argmin}}\limits (f-p)$, this fact implies that $\Vert x-y\Vert _\infty \le 1$ must hold, and hence, $\mathop {\mathrm{argmin}}\limits (f-p)$ is a DDM-convex set. We note that this example also shows that a similar characterization does not hold for the classes of globally/locally discrete midpoint convex functions. $\square$

6 Operations

We discuss several operations for discrete convex functions, including scaling operations [17, 18, 21], restriction, projection, direct sum and convolution operations [16, 21, 22].

6.1 Scaling operations

Scaling operations are useful techniques for designing efficient algorithms in discrete optimization. It is shown in [18] that global/local discrete midpoint convexity, including L$^\natural$-convexity, is closed under scaling operations. We show that DDM-convexity is also closed under scaling operations.

Given a function $f:{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ and a positive integer $\alpha$, the $\alpha$-scaling of f is the function $f^\alpha$ defined by

$$\begin{aligned} f^\alpha (x)=f(\alpha x)\qquad (x\in {\mathbb {Z}}^n). \end{aligned}$$

We also define the $\alpha$-scaling $S^\alpha$ of a set $S \subseteq {\mathbb {Z}}^n$ by

$$\begin{aligned} S^\alpha =\{x\in {\mathbb {Z}}^n \mid \alpha x\in S\}. \end{aligned}$$

Theorem 11

Given a DDM-convex function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ and a positive integer $\alpha$, the scaled function $f^\alpha :{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ is also DDM-convex.

Proof

By the equivalence between (1) and (4) of Theorem 8, it is sufficient to show that

$$\begin{aligned} f^\alpha (x)+f^\alpha (y) \ge f^\alpha (x+\varvec{1}_{A_m}-\varvec{1}_{B_m})+f^\alpha (y-\varvec{1}_{A_m}+\varvec{1}_{B_m}) \end{aligned}$$

for every $x, y\in {\mathbb {Z}}^n$ with $\Vert x-y\Vert _\infty = m$ and for families $\{A_k \mid k=1,\ldots ,m\}$ and $\{B_k \mid k=1,\ldots ,m\}$ defined by (4.1). The above inequality is written as

$$\begin{aligned} f(\alpha x)+f(\alpha y) \ge f(\alpha (x+\varvec{1}_{A_m}-\varvec{1}_{B_m}))+f(\alpha (y-\varvec{1}_{A_m}+\varvec{1}_{B_m})). \end{aligned}$$

(6.1)

For $\alpha x$ and $\alpha y$ with $\Vert \alpha x-\alpha y\Vert _\infty =\alpha m$, by defining

$$\begin{aligned} A_l^\alpha =\{i \mid \alpha y_i-\alpha x_i\ge l\}, \quad B_l^\alpha =\{i \mid \alpha y_i-\alpha x_i\le -l\}\qquad (l=1, \dots , \alpha m), \end{aligned}$$

we have $\alpha y-\alpha x=\sum _{l=1}^{\alpha m}(\varvec{1}_{A_l^\alpha }-\varvec{1}_{B_l^\alpha })$ and

$$\begin{aligned} A_k=A_{\alpha (k-1)+j}^\alpha , \quad B_k=B_{\alpha (k-1)+j}^\alpha \qquad (k=1, \dots , m;\; j=1, \dots , \alpha ). \end{aligned}$$

By (5) of Theorem 8 for

$$\begin{aligned} d =\sum _{l=1}^{\alpha (m-1)} (\varvec{1}_{A^\alpha _l}-\varvec{1}_{B^\alpha _l}), \quad d' =\sum _{l=\alpha (m-1)+1}^{\alpha m} (\varvec{1}_{A^\alpha _l}- \varvec{1}_{B^\alpha _l})=\alpha (\varvec{1}_{A_m}-\varvec{1}_{B_m}) \end{aligned}$$

we have

$$\begin{aligned} f(\alpha x)+f(\alpha y)&\ge f(\alpha x+d)+f(\alpha y-d)\\&=f(\alpha y-d')+f(\alpha x+d')\\&=f(\alpha (y-\varvec{1}_{A_m}+\varvec{1}_{B_m})) +f(\alpha (x+\varvec{1}_{A_m}-\varvec{1}_{B_m})), \end{aligned}$$

that is, (6.1). $\square$

Corollary 3

For a DDM-convex set $S\subseteq {\mathbb {Z}}^n$ and a positive integer $\alpha$, the $\alpha$-scaled set $S^\alpha$ is also DDM-convex.

6.2 Restrictions

For a function $f:{\mathbb {Z}}^{n+m}\rightarrow {\mathbb {R}}\cup \{+\infty \}$, the restriction of f on ${\mathbb {Z}}^n$ is the function g defined by

$$\begin{aligned} g(x)=f(x, \varvec{0}) \qquad (x\in {\mathbb {Z}}^n). \end{aligned}$$

For a set $S\subseteq {\mathbb {Z}}^{n+m}$, the restriction of S on ${\mathbb {Z}}^n$ is also defined by

$$\begin{aligned} T=\{x\in {\mathbb {Z}}^n \mid (x, \varvec{0})\in S\}. \end{aligned}$$

Obviously, the following properties hold.

Proposition 10

For a DDM-convex function, its restrictions are also DDM-convex.

Proposition 11

For a DDM-convex set, its restrictions are also DDM-convex.

6.3 Projections

For a function $f:{\mathbb {Z}}^{n+m}\rightarrow {\mathbb {R}}\cup \{+\infty \}$, the projection of f to ${\mathbb {Z}}^n$ is the function defined by

$$\begin{aligned} g(x)=\inf \{f(x, y) \mid y\in {\mathbb {Z}}^{m}\}\qquad (x\in {\mathbb {Z}}^n), \end{aligned}$$

(6.2)

where we assume that $g(x)>-\infty$ for all $x \in {\mathbb {Z}}^n$. For a set $S\subseteq {\mathbb {Z}}^{n+m}$, the projection of S to ${\mathbb {Z}}^n$ is also defined by

$$\begin{aligned} T=\{x\in {\mathbb {Z}}^n \mid \exists y\in {\mathbb {Z}}^m : (x, y)\in S\}. \end{aligned}$$

In the same way as the proof for globally discrete midpoint convex functions in [16, Theorem 3.5] we can show the following property.

Proposition 12

For a DDM-convex function, its projections are DDM-convex.

Proof

Let g be the projection defined by (6.2) of a DDM-convex function f. For every $x^{(1)}, x^{(2)} \in \mathop {\mathrm{dom}}\limits g$ and every $\varepsilon >0$, by the definition of the projection, there exist $y^{(1)}, y^{(2)}\in {\mathbb {Z}}^m$ with $g(x^{(i)})\ge f(x^{(i)}, y^{(i)})-\varepsilon$ for $i=1, 2$. Thus, we have

$$\begin{aligned} g(x^{(1)})+g(x^{(2)})\ge f(x^{(1)}, y^{(1)})+f(x^{(2)}, y^{(2)})-2\varepsilon . \end{aligned}$$

(6.3)

By DDM-convexity of f and the definition of the projection, we have

$$\begin{aligned}&f(x^{(1)}, y^{(1)})+f(x^{(2)}, y^{(2)})\nonumber \\&\quad \ge f(\mu ((x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)})))+ f(\mu ((x^{(2)}, y^{(2)}), (x^{(1)}, y^{(1)}))) \nonumber \\&\quad \ge g(\mu (x^{(1)}, x^{(2)}))+g(\mu (x^{(2)}, x^{(1)})). \end{aligned}$$

(6.4)

By (6.3) and (6.4), we obtain

$$\begin{aligned} g(x^{(1)})+g(x^{(2)})\ge g(\mu (x^{(1)}, x^{(2)}))+g(\mu (x^{(2)}, x^{(1)}))-2\varepsilon \end{aligned}$$

for any $\varepsilon >0$, which guarantees DDM-convexity of g. $\square$

Corollary 4

For a DDM-convex set, its projections are also DDM-convex.

6.4 Direct sums

For two functions $f_1:{\mathbb {Z}}^{n_1}\rightarrow {\mathbb {R}}\cup \{+\infty \}$ and $f_2:{\mathbb {Z}}^{n_2}\rightarrow {\mathbb {R}}\cup \{+\infty \}$, the direct sum of $f_1$ and $f_2$ is the function $f_1\oplus f_2:{\mathbb {Z}}^{n_1+n_2}\rightarrow {\mathbb {R}}\cup \{+\infty \}$ defined by

$$\begin{aligned} (f_1\oplus f_2)(x, y)=f_1(x)+f_2(y) \qquad (x\in {\mathbb {Z}}^{n_1},\; y\in {\mathbb {Z}}^{n_2}). \end{aligned}$$

For two sets $S_1\subseteq {\mathbb {Z}}^{n_1}$ and $S_2\subseteq {\mathbb {Z}}^{n_2}$, the direct sum of $S_1$ and $S_2$ is defined by

$$\begin{aligned} S_1\oplus S_2=\{(x, y) \mid x\in S_1,\; y\in S_2\}. \end{aligned}$$

DDM-convexity is closed under direct sums as below.

Proposition 13

For two DDM-convex functions $f_1:{\mathbb {Z}}^{n_1}\rightarrow {\mathbb {R}}\cup \{+\infty \}$ and $f_2:{\mathbb {Z}}^{n_2}\rightarrow {\mathbb {R}}\cup \{+\infty \}$, the direct sum $f_1\oplus f_2$ is also DDM-convex.

Proof

For every $x^{(1)}, y^{(1)}\in {\mathbb {Z}}^{n_1}$ and $x^{(2)}, y^{(2)}\in {\mathbb {Z}}^{n_2}$, it follows from DDM-convexity of $f_1$ and $f_2$ that

$$\begin{aligned}&f_1(x^{(1)})+f_1(y^{(1)})\ge f_1(\mu (x^{(1)}, y^{(1)}))+f_1(\mu (y^{(1)}, x^{(1)})), \end{aligned}$$

(6.5)

$$\begin{aligned}&f_2(x^{(2)})+f_2(y^{(2)})\ge f_2(\mu (x^{(2)}, y^{(2)}))+f_2(\mu (y^{(2)}, x^{(2)})). \end{aligned}$$

(6.6)

By the following relations

$$\begin{aligned}&\mu ((x^{(1)}, x^{(2)}), (y^{(1)}, y^{(2)})) =(\mu (x^{(1)}, y^{(1)}), \mu (x^{(2)}, y^{(2)})), \\&\mu ((y^{(1)}, y^{(2)}), (x^{(1)}, x^{(2)})) =(\mu (y^{(1)}, x^{(1)}), \mu (y^{(2)}, x^{(2)})), \end{aligned}$$

and by (6.5) and (6.6), we obtain

$$\begin{aligned}&(f_1\oplus f_2)(x^{(1)}, x^{(2)})+(f_1\oplus f_2)(y^{(1)}, y^{(2)})\\&\quad =[f_1(x^{(1)})+f_2(x^{(2)})]+[f_1(y^{(1)})+f_2(y^{(2)})]\\&\quad \ge [f_1(\mu (x^{(1)}, y^{(1)}))+f_2(\mu (x^{(2)}, y^{(2)}))]+[f_1(\mu (y^{(1)}, x^{(1)}))+f_2(\mu (y^{(2)}, x^{(2)}))]\\&\quad =(f_1\oplus f_2)( (\mu (x^{(1)}, y^{(1)}), \mu (x^{(2)}, y^{(2)})) )+ (f_1\oplus f_2)( (\mu (y^{(1)}, x^{(1)}), \mu (y^{(2)}, x^{(2)})) )\\&\quad =(f_1\oplus f_2)( \mu ((x^{(1)}, x^{(2)}), (y^{(1)}, y^{(2)})) )+ (f_1\oplus f_2)( \mu ((y^{(1)}, y^{(2)}), (x^{(1)}, x^{(2)})) ), \end{aligned}$$

which says DDM-convexity of $f_1 \oplus f_2$. $\square$

Corollary 5

For two DDM-convex sets $S_1\subseteq {\mathbb {Z}}^{n_1}$ and $S_2\subseteq {\mathbb {Z}}^{n_2}$, $S_1\oplus S_2$ is also DDM-convex.

6.5 Convolutions

For two functions $f_1, f_2:{\mathbb {Z}}^{n}\rightarrow {\mathbb {R}}\cup \{+\infty \}$, the convolution $f_1\square f_2$ is the function defined by

$$\begin{aligned} (f_1\square f_2)(x)=\inf \{f_1(y)+f_2(z) \mid x=y+z,\; \ y, z\in {\mathbb {Z}}^n\} \qquad (x\in {\mathbb {Z}}^n), \end{aligned}$$

where we assume $(f_1\square f_2)(x)>-\infty$ for every $x \in {\mathbb {Z}}^n$. For two sets $S_1, S_2\subseteq {\mathbb {Z}}^n$, the Minkowski sum $S_1 + S_2$ defined by

$$\begin{aligned} S_1+S_2=\{x+y \mid x\in S_1, y\in S_2 \} \end{aligned}$$

corresponds to the convolution of indicator functions $\delta _{S_1}$ and $\delta _{S_2}$. The next example shows that the Minkowski sum of two DDM-convex sets may not be DDM-convex, and hence, DDM-convexity is not closed under the convolutions.

Example 5

We borrow the example in [16, Example 4.2], which shows that L$^\natural$-convexity and global/local discrete midpoint convexity may not be closed under the convolutions.

Let $S_1=\{(0, 0, 0), (1, 1, 0)\}$ and $S_2=\{(0, 0, 0), (0, 1, 1)\}$ which are DDM-convex. The Minkowski sum of $S_1$ and $S_2$

$$\begin{aligned} S_1+S_2=\{(0, 0, 0), (0, 1, 1), (1, 1, 0), (1, 2, 1)\} \end{aligned}$$

is not DDM-convex, because for $x=(0, 0, 0)$ and $y=(1, 2, 1)$, we have $\mu (x, y)=(0, 1, 0)\not \in S_1+S_2$ and $\mu (y, x)=(1, 1, 1)\not \in S_1+S_2$. $\square$

It is known that the convolution of an L$^\natural$-convex function and a separable convex function is also L$^\natural$-convex, where a separable convex function $\varphi : {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ is given by

$$\begin{aligned} \varphi (x) = \sum _{i=1}^n \varphi _i(x_i) \qquad (x \in {\mathbb {Z}}^n) \end{aligned}$$

for univariate discrete convex functions $\varphi _i (i=1,\ldots ,n)$. By the same arguments of [16, Proposition 4.7], this can be extended to DDM-convexity.

Proposition 14

The convolution of a DDM-convex function and a separable convex function is also DDM-convex.

Proof

Let $f :{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a DDM-convex function, $\varphi :{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ a separable convex function represented as $\sum _{i=1}^n \varphi _i$ and let $g=f\square \varphi$. For every $x^{(1)}, x^{(2)}\in \mathop {\mathrm{dom}}\limits g$ and $\varepsilon >0$, by the definition of convolutions, there exist $y^{(i)}, z^{(i)}\; (i=1, 2)$ such that

$$\begin{aligned} g(x^{(i)})\ge f(y^{(i)})+\varphi (z^{(i)})-\varepsilon , \ \ x^{(i)}=y^{(i)}+z^{(i)} \qquad (i=1, 2). \end{aligned}$$

(6.7)

It follows from DDM-convexity of f that

$$\begin{aligned} f(y^{(1)})+f(y^{(2)})\ge f(\mu (y^{(1)}, y^{(2)}))+f(\mu (y^{(2)}, y^{(1)})). \end{aligned}$$

(6.8)

Let

$$\begin{aligned} z^\prime =\mu (x^{(1)}, x^{(2)})-\mu (y^{(1)}, y^{(2)}), \quad z^{\prime \prime }=\mu (x^{(2)}, x^{(1)})-\mu (y^{(2)}, y^{(1)}). \end{aligned}$$

By $g = f\square \varphi$, we have

$$\begin{aligned}&f(\mu (y^{(1)}, y^{(2)}))+\varphi (z^\prime )\ge g(\mu (x^{(1)}, x^{(2)})), \end{aligned}$$

(6.9)

$$\begin{aligned}&f(\mu (y^{(2)}, y^{(1)}))+\varphi (z^{\prime \prime })\ge g(\mu (x^{(2)}, x^{(1)})). \end{aligned}$$

(6.10)

The claim, below, states that

$$\begin{aligned} \varphi _i(z_i^{(1)})+\varphi _i(z_i^{(2)})\ge \varphi _i(z_i^\prime )+\varphi _i(z_i^{\prime \prime }) \end{aligned}$$

(6.11)

for each $i=1, \dots , n$. Thus, we have

$$\begin{aligned} \varphi (z^{(1)})+\varphi (z^{(2)})\ge \varphi (z^{\prime })+\varphi (z^{\prime \prime }). \end{aligned}$$

(6.12)

By summing up (6.7), (6.8), (6.9), (6.10) and (6.12), we obtain

$$\begin{aligned} g(x^{(1)})+g(x^{(2)})\ge g(\mu (x^{(1)}, x^{(2)}))+g(\mu (x^{(2)}, x^{(1)}))-2\varepsilon \end{aligned}$$

for any $\varepsilon >0$, which guarantees DDM-convexity of g.

Claim

(6.11) holds.

Proof

Since $\varphi _i$ is univariate discrete convex, for every $a, b\in {\mathbb {Z}}$ with $a\le b$ and for every $p, q\in {\mathbb {Z}}$ such that (i) $a+b=p+q$, (ii) $a \le p \le b$ and (iii) $a \le q \le b$, we have $\varphi _i(a)+\varphi _i(b) \ge \varphi _i(p)+\varphi _i(q)$. Thus, it is enough to show the following relations:

$$\begin{aligned}&z_i^{(1)}+z_i^{(2)}=z_i^\prime +z_i^{\prime \prime }, \end{aligned}$$

(6.13)

$$\begin{aligned}&\min \{z_i^{(1)}, z_i^{(2)}\} \le z_i^\prime \le \max \{z_i^{(1)}, z_i^{(2)}\}, \end{aligned}$$

(6.14)

$$\begin{aligned}&\min \{z_i^{(1)}, z_i^{(2)}\} \le z_i^{\prime \prime } \le \max \{z_i^{(1)}, z_i^{(2)}\}. \end{aligned}$$

(6.15)

Condition (6.13) follows from

$$\begin{aligned} \mu (x^{(1)}, x^{(2)})_i + \mu (x^{(2)}, x^{(1)})_i&= x_i^{(1)} + x_i^{(2)}, \\ \mu (y^{(1)}, y^{(2)})_i + \mu (y^{(2)}, y^{(1)})_i&= y_i^{(1)} + y_i^{(2)}, \end{aligned}$$

and

$$\begin{aligned} z_i^\prime +z_i^{\prime \prime }&=(\mu (x^{(1)}, x^{(2)})_i-\mu (y^{(1)}, y^{(2)})_i ) +( \mu (x^{(2)}, x^{(1)})_i-\mu (y^{(2)}, y^{(1)})_i )\\&=( \mu (x^{(1)}, x^{(2)})_i+\mu (x^{(2)}, x^{(1)})_i ) -( \mu (y^{(1)}, y^{(2)})_i+\mu (y^{(2)}, y^{(1)})_i )\\&=(x_i^{(1)}+x_i^{(2)})-(y_i^{(1)}+y_i^{(2)}) =(x_i^{(1)}-y_i^{(1)})+(x_i^{(2)}-y_i^{(2)}) \\&=z_i^{(1)}+z_i^{(2)}. \end{aligned}$$

To show (6.14) and (6.15), we consider the following cases: Case 1: $x^{(1)}_i\ge x^{(2)}_i$ and $y^{(1)}_i\ge y^{(2)}_i$, Case 2: $x^{(1)}_i< x^{(2)}_i$ and $y^{(1)}_i< y^{(2)}_i$, Case 3: $x^{(1)}_i\ge x^{(2)}_i$ and $y^{(1)}_i< y^{(2)}_i$, Case 4: $x^{(1)}_i< x^{(2)}_i$ and $y^{(1)}_i\ge y^{(2)}_i$.

Case 1 ($x^{(1)}_i\ge x^{(2)}_i$ and $y^{(1)}_i\ge y^{(2)}_i$). In this case, we have

$$\begin{aligned} z^\prime _i=\left\lceil \frac{x^{(1)}_i{+}x^{(2)}_i}{2} \right\rceil - \left\lceil \frac{y^{(1)}_i{+}y^{(2)}_i}{2} \right\rceil , \quad z^{\prime \prime }_i=\left\lfloor \frac{x^{(1)}_i{+}x^{(2)}_i}{2} \right\rfloor - \left\lfloor \frac{y^{(1)}_i{+}y^{(2)}_i}{2} \right\rfloor . \end{aligned}$$

By substituting $z_i^{(1)}{+}z_i^{(2)}=(x_i^{(1)}{+}x_i^{(2)})-(y_i^{(1)}{+}y_i^{(2)})$ into

$$\begin{aligned} \min \{z_i^{(1)}, z_i^{(2)}\} \le \frac{z_i^{(1)}{+}z_i^{(2)}}{2} \le \max \{z_i^{(1)}, z_i^{(2)}\}, \end{aligned}$$

we obtain

$$\begin{aligned} \min \{z_i^{(1)}, z_i^{(2)}\} + \frac{y_i^{(1)}{+}y_i^{(2)}}{2} \le \frac{x_i^{(1)}{+}x_i^{(2)}}{2} \le \max \{z_i^{(1)}, z_i^{(2)}\} + \frac{y_i^{(1)}{+}y_i^{(2)}}{2}. \end{aligned}$$

In the above inequalities, we round up and round down every terms, to obtain

$$\begin{aligned} \min \{z_i^{(1)}, z_i^{(2)}\}+\left\lceil \frac{y_i^{(1)}{+}y_i^{(2)}}{2}\right\rceil&\le \left\lceil \frac{x_i^{(1)}{+}x_i^{(2)}}{2}\right\rceil \le \max \{z_i^{(1)}, z_i^{(2)}\}+\left\lceil \frac{y_i^{(1)}{+}y_i^{(2)}}{2}\right\rceil , \\ \min \{z_i^{(1)}, z_i^{(2)}\}+\left\lfloor \frac{y_i^{(1)}{+}y_i^{(2)}}{2}\right\rfloor&\le \left\lfloor \frac{x_i^{(1)}{+}x_i^{(2)}}{2}\right\rfloor \le \max \{z_i^{(1)}, z_i^{(2)}\}+\left\lfloor \frac{y_i^{(1)}{+}y_i^{(2)}}{2}\right\rfloor . \end{aligned}$$

Thus, (6.14) and (6.15) are satisfied.

Case 2 ($x^{(1)}_i< x^{(2)}_i$ and $y^{(1)}_i< y^{(2)}_i$). In the same way as Case 1, we can show (6.14) and (6.15).

Case 3 ($x^{(1)}_i\ge x^{(2)}_i$ and $y^{(1)}_i< y^{(2)}_i$). In this case, we have

$$\begin{aligned} z^\prime _i=\left\lceil \frac{x^{(1)}_i{+}x^{(2)}_i}{2} \right\rceil - \left\lfloor \frac{y^{(1)}_i{+}y^{(2)}_i}{2} \right\rfloor , \quad z^{\prime \prime }_i=\left\lfloor \frac{x^{(1)}_i{+}x^{(2)}_i}{2} \right\rfloor - \left\lceil \frac{y^{(1)}_i{+}y^{(2)}_i}{2} \right\rceil . \end{aligned}$$

Moreover, we obtain

$$\begin{aligned} z_i^{(1)}=x_i^{(1)}-y_i^{(1)}&\ge \left\lceil \frac{x^{(1)}_i{+}x^{(2)}_i}{2} \right\rceil - \left\lfloor \frac{y^{(1)}_i{+}y^{(2)}_i}{2} \right\rfloor = z^{\prime }_i \\&\ge \left\lfloor \frac{x^{(1)}_i{+}x^{(2)}_i}{2} \right\rfloor - \left\lceil \frac{y^{(1)}_i{+}y^{(2)}_i}{2}\right\rceil = z^{\prime \prime }_i \\&\ge x_i^{(2)}-y_i^{(2)} =z_i^{(2)}, \end{aligned}$$

and hence, (6.14) and (6.15).

Case 4: ($x^{(1)}_i< x^{(2)}_i$ and $y^{(1)}_i\ge y^{(2)}_i$). In the same way as Case 3, we can show (6.14) and (6.15). (End of the proof of Claim). $\square$

Corollary 6

Minkowski sum of a DDM-convex set and an integral box is also DDM-convex, where an integral box is the set defined by $\{ x \in {\mathbb {Z}}^n \mid a \le x \le b\}$ for some $a \in ({\mathbb {Z}}\cup \{-\infty \})^n$ and $b \in ({\mathbb {Z}}\cup \{+\infty \})^n$ with $a \le b$.

7 Proximity theorems

For a function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ and a positive integer $\alpha$, a proximity theorem estimates the distance between a given local minimizer $x^\circ$ of the $\alpha$-scaled function $f^\alpha$ and a minimizer $x^*$ of f. For instance, the following proximity theorems for L$^\natural$-convex functions and globally/locally discrete midpoint convex functions are known.

Theorem 12

([11, 21]) Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an L$^\natural$-convex function, $\alpha$ be a positive integer and $x^\circ \in \mathop {\mathrm{dom}}\limits f$. If $f(x^\circ )\le f(x^\circ +\alpha d)$ for all $d\in \{0, +1\}^n \cup \{0,-1\}^n$, then there exists $x^*\in \mathop {\mathrm{argmin}}\limits f$ with $\Vert x^\circ -x^*\Vert _\infty \le n(\alpha -1)$.

Theorem 13

([18]) Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a globally/locally discrete midpoint convex function, $\alpha$ be a positive integer and $x^\circ \in \mathop {\mathrm{dom}}\limits f$. If $f(x^\circ )\le f(x^\circ +\alpha d)$ for all $d\in \{-1, 0, +1\}^n$, then there exists $x^*\in \mathop {\mathrm{argmin}}\limits f$ with $\Vert x^\circ -x^*\Vert _\infty \le n(\alpha -1)$.

In the same way as the arguments in [18], we can show the following proximity theorem for DDM-convex functions.

Theorem 14

Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a DDM-convex function, $\alpha$ be a positive integer and $x^\circ \in \mathop {\mathrm{dom}}\limits f$. If $f(x^\circ )\le f(x^\circ +\alpha d)$ for all $d\in \{-1, 0, +1\}^n$, then there exists $x^*\in \mathop {\mathrm{argmin}}\limits f$ with $\Vert x^\circ -x^*\Vert _\infty \le n(\alpha -1)$.

We note that $f^\alpha$ is also DDM-convex by Theorem 11 and $x^\circ$ corresponds to a minimizer $\varvec{0}$ of $f^\alpha (y) = f(x^\circ +\alpha y)$ by Corollary 1. We emphasize that the bound $n(\alpha -1)$ for DDM-convex functions is the same as that for L$^\natural$-convex functions and globally/locally discrete midpoint convex functions.

To prove Theorem 14, we assume $x^\circ = \varvec{0}$ without loss of generality. Let $S=\{x\in {\mathbb {Z}}^n \mid \Vert x\Vert _\infty \le n(\alpha -1)\}$, $W=\{x\in {\mathbb {Z}}^n \mid \Vert x\Vert _\infty =n(\alpha -1)+1\}$ and let $\gamma =\mathop {\mathrm{argmin}}\limits \{f(x) \mid x\in S\}$. We show that

$$\begin{aligned} f(y)\ge \gamma \qquad (\forall y\in W). \end{aligned}$$

(7.1)

Then Theorem 1 (box-barrier property) implies that $f(z)\ge \gamma$ for all $z\in {\mathbb {Z}}^n$.

Fix $y =(y_1,\ldots ,y_n)\in W$, and let $\Vert y\Vert _\infty =m (=n(\alpha -1)+1)$. By using

$$\begin{aligned} A_k=\{i \mid y_i\ge k\}, \quad B_k=\{i \mid y_i\le -k\}\qquad (k=1, \dots , m), \end{aligned}$$

we can write y as

$$\begin{aligned} y=\sum _{k=1}^m (\varvec{1}_{A_k}-\varvec{1}_{B_k}), \end{aligned}$$

where $A_1\supseteq \cdots \supseteq A_m$, $B_1\supseteq \cdots \supseteq B_m$, $A_1\cap B_1=\emptyset$ and $A_m\cup B_m\not =\emptyset$.

Lemma 2

There exists some $k_0\in \{1, \dots , m-\alpha +1\}$ with $(A_{k_0}, B_{k_0})=(A_{k_0+j}, B_{k_0+j})$ for $j=1, \ldots , \alpha -1$.

Proof

By $A_m \cup B_m \ne \emptyset$, we may assume $A_m\not =\emptyset$. Let $s=|A_1|$ and $(a_k, b_k)=(|A_k|, |B_k|+s)$ for $k=1, \dots , m$. Since $A_1\supseteq \cdots \supseteq A_m \ne \emptyset$, $B_1\supseteq \cdots \supseteq B_m$ and $A_1\cap B_1=\emptyset$, we have $n-s \ge |B_1|$, $s= a_1\ge \cdots \ge a_m \ge 1$ and $n \ge b_1 \ge \cdots \ge b_m\ge s$. Therefore, $(s, n) \ge (a_1, b_1) \ge \cdots \ge (a_m, b_m) \ge (1, s)$. Because $m=n(\alpha -1)+1$ and the length of a strictly decreasing chain connecting (s, n) to (1, s) in ${\mathbb {Z}}^2$ is bounded by n, there exists a constant subsequence of length $\ge \alpha$ in the sequence $\{(a_k,b_k)\}_{k=1,\ldots ,m}$ by the pigeonhole principle. Hence the assertion holds. $\square$

By using $k_0$ in Lemma 2, we define a subset J of $\{1, \dots , m\}$ by $J=\{k_0, \dots , k_0+\alpha -1\}$. By the parallelogram inequality (4.14) in Theorem 7, where $d_0=\varvec{1}_{A_{k_0}}-\varvec{1}_{B_{k_0}}$ and $d = \sum _{j\in J}(\varvec{1}_{A_j}-\varvec{1}_{B_j})=\alpha d_0$, we obtain

$$\begin{aligned} f(\varvec{0})+f(y)\ge f(\alpha d_0) + f(y - \alpha d_0). \end{aligned}$$

By the assumption, we have $f(\alpha d_0)\ge f(x^\circ ) = f(\varvec{0})$. We also have $y - \alpha d_0 \in S$ because

$$\begin{aligned} \Vert y - \alpha d_0\Vert _\infty =m-\alpha =(n-1)(\alpha -1)\le n(\alpha -1). \end{aligned}$$

By the definition of $\gamma$, $f(y - \alpha d_0)\ge \gamma$ must hold. Therefore,

$$\begin{aligned} f(y) \ge f(y - \alpha d_0)+[f(\alpha d_0)-f(\varvec{0})] \ge \gamma +0 = \gamma , \end{aligned}$$

which implies (7.1), completing the proof of Theorem 14.

8 Minimization algorithms

In this section, we propose two algorithms for DDM-convex function minimization.

8.1 The 1-neighborhood steepest descent algorithm

We first propose a variant of steepest descent algorithm for DDM-convex function minimization problem.

Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a DDM-convex function with $\mathop {\mathrm{argmin}}\limits f\not =\emptyset$. We suppose that an initial point

$$\begin{aligned} x^{(0)}\in \mathop {\mathrm{dom}}\limits f {\setminus } \mathop {\mathrm{argmin}}\limits f \end{aligned}$$

is given. Let L denote the minimum $l_\infty$-distance between $x^{(0)}$ and a minimizer of f, that is, L is defined by

$$\begin{aligned} L=\min \{\Vert x-x^{(0)}\Vert _\infty \mid x\in \mathop {\mathrm{argmin}}\limits f\}. \end{aligned}$$

For all $k=0, 1, \dots , L$ we define sets $S_k$ by

$$\begin{aligned} S_k=S_k(x^{(0)})=\{x\in {\mathbb {Z}}^n \mid \Vert x-x^{(0)}\Vert _\infty \le k\}. \end{aligned}$$

The idea of our algorithm is to generate a sequence of minimizers in $S_k$ for $k=1,\ldots ,L$. The next proposition guarantees that consecutive minimizers can be chosen to be close to each other.

Proposition 15

For each $k = 1,\ldots ,L$ and for any $x^{(k-1)}\in \mathop {\mathrm{argmin}}\limits \{f(x) \mid x\in S_{k-1}\}$, there exists $x^{(k)}\in \mathop {\mathrm{argmin}}\limits \{f(x) \mid x\in S_k \}$ with $\Vert x^{(k)}-x^{(k-1)}\Vert _\infty \le 1$.

Proof

If $k=1$, the assertion is obvious. Suppose that $k \ge 2$ and y is any point in $S_k$. By (2.1) for $x^{(k-1)}$ and y, we have

$$\begin{aligned} f(x^{(k-1)})+f(y)\ge f(\mu (x^{(k-1)}, y))+f(\mu (y, x^{(k-1)})). \end{aligned}$$

(8.1)

Since $x^{(k-1)}, y\in S_k$ and $S_k$ is a DDM-convex set, we also have

$$\begin{aligned} \mu (y, x^{(k-1)})\in S_k. \end{aligned}$$

(8.2)

Next, we show

$$\begin{aligned} \mu (x^{(k-1)}, y)\in S_{k-1}. \end{aligned}$$

(8.3)

To show this we arbitrarily fix $i\in \{1, \dots , n\}$, and consider the two cases: Case 1: $x^{(k-1)}_i-y_i=l\ (l\ge 1)$ and Case 2: $x^{(k-1)}_i-y_i=-l\ (l\ge 1)$.

Case 1 ($x^{(k-1)}_i-y_i=l\ (l\ge 1)$). In this case, $\mu (x^{(k-1)}, y)_i=x^{(k-1)}_i-\left\lfloor \frac{l}{2}\right\rfloor$ and $\left\lfloor \frac{l}{2}\right\rfloor \le l-1$. Thus, we have

$$\begin{aligned} x_i^{(0)}+(k-1)&\ge x_i^{(k-1)} \\&\ge \mu (x^{(k-1)}, y)_i =x^{(k-1)}_i-\left\lfloor \frac{l}{2}\right\rfloor \\&\ge x^{(k-1)}_i-(l-1)=y_i+1\ge x_i^{(0)}-(k-1). \end{aligned}$$

Case 2 ($x^{(k-1)}_i-y_i=-l\ (l\ge 1)$). In this case, $\mu (x^{(k-1)}, y)_i=x^{(k-1)}_i+\left\lfloor \frac{l}{2}\right\rfloor$ and $\left\lfloor \frac{l}{2}\right\rfloor \le l-1$. Thus, we have

$$\begin{aligned} x_i^{(0)}-(k-1)&\le x^{(k-1)}_i \\&\le \mu (x^{(k-1)}, y)_i=x^{(k-1)}_i+\left\lfloor \frac{l}{2}\right\rfloor \\&\le x^{(k-1)}_i+(l-1)=y_i-1\le x_i^{(0)}+(k-1). \end{aligned}$$

By the above arguments, (8.3) holds.

Let $y^*$ be a point y in $\mathop {\mathrm{argmin}}\limits \{f(x) \mid x\in S_k\}$ minimizing $\Vert y-x^{(k-1)}\Vert _\infty$. To prove $\Vert y^*-x^{(k-1)}\Vert _\infty \le 1$ by contradiction, suppose that $\Vert y^*-x^{(k-1)}\Vert _\infty \ge 2$, which yields $\Vert y^*-x^{(k-1)}\Vert _\infty >\Vert \mu (y^*, x^{(k-1)})-x^{(k-1)}\Vert _\infty$. Since $\mu (y^*, x^{(k-1)}) \in S_k$ by (8.2), this implies $f(y^*)<f(\mu (y^*, x^{(k-1)}))$. Moreover, by (8.3), we have $f(x^{(k-1)})\le f(\mu (x^{(k-1)}, y^*))$. These two inequalities contradict (8.1) for $x^{(k-1)}$ and $y^*$. Hence $\Vert y^*-x^{(k-1)}\Vert _\infty \le 1$ must hold. $\square$

By Proposition 15, it seems be natural to assume that we can find a minimizer of f within the 1-neighborhood $N_1(x)$ of x defined by

$$\begin{aligned} N_1(x) = \{ y \in {\mathbb {Z}}^n \mid \Vert z - x\Vert _\infty \le 1\}. \end{aligned}$$

With the use of a 1-neighborhood minimization oracle, which finds a point minimizing f in $N_1(x)$ for any $x \in \mathop {\mathrm{dom}}\limits f$, our algorithm can be described as below.

The 1-neighborhood steepest descent algorithm

• D0: Find $x^{(0)} \in \mathop {\mathrm{dom}}\limits f$, and set $k:= 1$.

• D1: Find $x^{(k)}$ that minimizes f in $N_{1}(x^{(k-1)})$.

• D2: If $f(x^{(k)}) = f(x^{(k-1)})$, then output $x^{(k-1)}$ and stop.

• D3: Set $k := k+1$, and go to D1.

Theorem 15

For a DDM-convex function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ with $\mathop {\mathrm{argmin}}\limits f\not =\emptyset$, the 1-neighborhood steepest descent algorithm finds a minimizer of f exactly in $(L{+}1)$ iterations, that is, exactly in $(L{+}1)$ calls of the 1-neighborhood minimization oracles.

Proof

By Proposition 15, the sequence $\{x^{(k)}\}$ generated by the 1-neighborhood steepest descent algorithm satisfy

$$\begin{aligned} x^{(k)}\in \mathop {\mathrm{argmin}}\limits \{f(x) \mid x\in S_k\} \qquad (k=1, 2, \dots ). \end{aligned}$$

(8.4)

Claim

If $f(x^{(k)})=f(x^{(k-1)})$ at Step D2, then $x^{(k-1)}\in \mathop {\mathrm{argmin}}\limits f$.

Proof

For any $d\in \{+1, 0, -1\}^n$, $x^{(k-1)}+d$ belongs to $S_{k}$, and hence, $f(x^{(k)})\le f(x^{(k-1)}+d)$ by (8.4). Therefore, if $f(x^{(k)})=f(x^{(k-1)})$, then $f(x^{(k-1)})\le f(x^{(k-1)}+d)$ for any d. Corollary 1 in Sect. 3 guarantees $x^{(k-1)}\in \mathop {\mathrm{argmin}}\limits f$. (End of the proof of Claim).

By the definition of L, $x^{(k)} \ne x^{(k-1)}$ if $k \le L$, and $x^{(L)}=x^{(L+1)}$. Therefore, our algorithm stops in $(L+1)$ iterations. $\square$

Remark 2

Theorem 15 says that the sequence of points generated by the 1-neighborhood steepest descent algorithm is bounded by the $\ell _\infty$-distance between an initial point and the nearest minimizer. Similar facts are pointed out for L$^\natural$-convex function minimization [13, 24, 26] and globally/locally discrete midpoint convex function minimization [18]. $\square$

Remark 3

Let F(n) denote the number of function evaluations in the 1-neighborhood minimization oracle. Since any function defined on $\{0,1\}^n$ is DDM-convex, the 1-neighborhood minimization problem is NP-hard. In almost cases, F(n) seems to be $\Theta (3^n)$ by a brute-force calculation, because $|N_1(\cdot )| = 3^n$. Fortunately, for L$^\natural$-convex functions, F(n) is bounded by a polynomial in n. Another hopeful case is a fixed parameter tractable case, that is, the case where there exists some parameter k such that the number of function evaluations F(n, k) in n and k is bounded by a polynomial p(n) in n times any function g(k) in k (see the next remark). $\square$

Remark 4

Let us consider the following problem:

$$\begin{aligned} \begin{array}{|lr} \text{ minimize } &{} \frac{1}{2} x^\top Q x + c^\top x \\ \text{ subject } \text{ to } &{} x \in {\mathbb {Z}}^n, \end{array} \end{aligned}$$

where a symmetric matrix $Q \in {\mathbb {R}}^{n \times n}$ is nonsingular and diagonally dominant with nonnegative diagonals, and $c \in {\mathbb {R}}^n$. Since Q is nonsingular, the (convex) continuous relaxation problem has a unique minimizer $-Q^{-1}c$. Furthermore, because the objective function is 2-separable convex, it follows from Theorem 18 in the next section that there exists an optimal solution in the box:

$$\begin{aligned} B = \{ x \in {\mathbb {Z}}^n \mid -Q^{-1}c - n\varvec{1}\le x \le -Q^{-1}c + n\varvec{1}\}. \end{aligned}$$

Therefore, the 1-neighborhood steepest descent algorithm with an initial point $\lfloor -Q^{-1}c \rfloor$ find an optimal solution in O(n) iterations. Furthermore, if $Q = [q_{ij}]$ is $(2k+1)$-diagonal, that is,

$$\begin{aligned} q_{ij} = 0 \qquad (i=1,\ldots ,n;\; j:|j-i| > k), \end{aligned}$$

then $F(n,k) = O(n 3^{k+1})$ as in [6]. $\square$

8.2 Scaling algorithm

In the same way as the scaling algorithm for minimization of globally/locally discrete midpoint convex functions in [18], the scaling property (Theorem 11) and the proximity theorem (Theorem 14) enable us to design a scaling algorithm for the minimization of DDM-convex functions with bounded effective domains.

Let $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ be a DDM-convex function with bounded effective domain. We suppose that $K_\infty =\max \{\Vert x-y\Vert _\infty \mid x, y\in \mathop {\mathrm{dom}}\limits f\} \; (K_\infty <+\infty )$ and an initial point $x \in \mathop {\mathrm{dom}}\limits f$ are given. Our algorithm can be described as follows.

Scaling algorithm for DDM-convex functions

• S0: Let $x \in \mathop {\mathrm{dom}}\limits f$ and $\alpha := 2^{\lceil \log _{2} (K_{\infty }+1) \rceil }$.

• S1: Find a vector y that minimizes $f^{\alpha }(y) = f(x{+}\alpha y)$ subject to $\Vert y \Vert _{\infty } \le n$ (by the 1-neighborhood steepest descent algorithm), and set $x:= x+ \alpha y$.

• S2: If $\alpha = 1$, then stop (x is a minimizer of f).

• S3: Set $\alpha :=\alpha /2$, and go to S1.

Theorem 16

For a DDM-convex function $f:{\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ with bounded effective domain, the scaling algorithm finds a minimizer of f in $O(n\log _2 K_\infty )$ calls of the 1-neighborhood minimization oracles.

Proof

The correctness of the algorithm can be shown by induction on $\alpha$. If $\alpha = 2^{\lceil \log _{2} (K_{\infty }+1) \rceil }$, then x is a unique point of $\mathop {\mathrm{dom}}\limits f^\alpha$ because $\alpha =2^{\lceil \log _2 (K_\infty +1)\rceil }>K_\infty$, that is, a minimizer of $f^\alpha$. Let $x^{2\alpha }$ denote the point x at the beginning of S1 for $\alpha$ and assume that $x^{2\alpha }$ is a minimizer of $f^{2\alpha }$. The function $f^\alpha (y)=f(x^{2\alpha }+\alpha y)$ is DDM-convex by Theorem 11. Let $y^\alpha =\mathop {\mathrm{argmin}}\limits \{f^\alpha (y) \mid \Vert y\Vert _\infty \le n\}$ and $x^\alpha =x^{2\alpha }+\alpha y^\alpha$. Theorem 14 guarantees that $x^\alpha$ is a minimizer of $f^\alpha$ because of $x^{2\alpha } \in \mathop {\mathrm{argmin}}\limits f^{2\alpha }$. At the termination of the algorithm, we have $\alpha =1$ and $f^\alpha =f$. The output of the algorithm, which is computed by the 1-neighborhood steepest descent algorithm, satisfies the condition of Corollary 1, and hence, the output is indeed a minimizer of f.

The time complexity of the algorithm can be analyzed as follows: by Theorem 15, S1 terminates in O(n) calls of the 1-neighborhood minimization oracles in each iteration. The number of iterations is $O(\log _2 K_\infty )$. Hence, the assertion holds. $\square$

9 DDM-convex functions in continuous variables

In [19], proximity theorems between L$^\natural$-convex functions and their continuous relaxations are proposed. We extend these results to DDM-convexity.

It is known that the continuous version of L$^\natural$-convexity can naturally be defined by using translation-submodularity (3.1). In this section, we define DDM-convexity in continuous variables in a different way. We call a continuous convex function $F : {\mathbb {R}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ a directed discrete midpoint convex function in continuous variables (${\mathbb {R}}$-DDM-convex function) if for any positive integer $\alpha$, the function $f^{1/\alpha } : {\mathbb {Z}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ defined by

$$\begin{aligned} f^{1/\alpha }(x)=F\left( \frac{x}{\alpha }\right) \qquad (x\in {\mathbb {Z}}^n) \end{aligned}$$

(9.1)

is DDM-convex. We denote by f the DDM-convex function $f^{1/1}$ which is nothing but the restriction of F to ${\mathbb {Z}}^n$.

An example of an ${\mathbb {R}}$-DDM-convex function is a continuous 2-separable convex function F which is defined as

$$\begin{aligned} F(x)=\sum _{i=1}^n \xi _i(x_i)+ \sum _{i,j:j\not =i}\varphi _{ij} (x_i-x_j)+ \sum _{i,j:j\not =i} \psi _{ij}(x_i+x_j)\qquad (x\in {\mathbb {R}}^n) \end{aligned}$$

for univariate continuous convex functions $\xi _{i}, \varphi _{ij}, \psi _{ij} : {\mathbb {R}}\rightarrow {\mathbb {R}}\cup \{+\infty \}\; (i=1,\ldots ,n; j \in \{1,\ldots ,n\} {\setminus } \{i\}$) as below. The restriction f of F to ${\mathbb {Z}}^n$ is trivially a 2-separable convex function on ${\mathbb {Z}}^n$ defined by (3.4). Furthermore, the function $F^{1/\alpha }:{\mathbb {R}}^n \rightarrow {\mathbb {R}}\cup \{+\infty \}$ defined by

$$\begin{aligned} F^{1/\alpha }(x) = F\left( \frac{x}{\alpha }\right) \qquad (x \in {\mathbb {R}}^n) \end{aligned}$$

is also a continuous 2-separable convex function, and hence, the restriction $f^{1/\alpha }$ of $F^{1/\alpha }$ to ${\mathbb {Z}}^n$ is also a 2-separable convex function on ${\mathbb {Z}}^n$.

We have the following proximity theorems between an ${\mathbb {R}}$-DDM-convex function F and its restriction f to ${\mathbb {Z}}^n$.

Theorem 17

Let $F:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an ${\mathbb {R}}$-DDM-convex function. For each $x^* \in \mathop {\mathrm{argmin}}\limits f\;(= \mathop {\mathrm{argmin}}\limits f^{1/1})$, there exists ${\overline{x}}\in \mathop {\mathrm{argmin}}\limits F$ with $\Vert x^*-{\overline{x}}\Vert _\infty \le n$.

Proof

Since f is DDM-convex, by Corollary 1 in Sect. 3, we have

$$\begin{aligned} f(x^*)\le f(x^*+d) \qquad (\forall d\in \{-1, 0, +1\}^n). \end{aligned}$$

Thus, for every integer $\alpha \ge 2$, by $f(x)=f^{1/\alpha }(\alpha x)\; (x\in {\mathbb {Z}}^n)$, we have

$$\begin{aligned} f^{1/\alpha }(\alpha x^*)\le f^{1/\alpha }(\alpha x^*+\alpha d)\qquad (\forall d\in \{-1, 0, +1\}^n). \end{aligned}$$

By Theorem 14 for $f^{1/\alpha }$, there exists $x^{1/\alpha } \in \mathop {\mathrm{argmin}}\limits f^{1/\alpha }$ with

$$\begin{aligned} \alpha x^*-(\alpha -1)n\varvec{1}\le x^{1/\alpha } \le \alpha x^*+(\alpha -1)n\varvec{1}. \end{aligned}$$

(9.2)

By dividing all terms in (9.2) by $\alpha$, we obtain

$$\begin{aligned} x^*-n\varvec{1}\le x^*-\frac{\alpha -1}{\alpha }n\varvec{1}\le \frac{x^{1/\alpha }}{\alpha } \le x^*+\frac{\alpha -1}{\alpha }n\varvec{1}\le x^*+n\varvec{1}. \end{aligned}$$

Let $B=\{x\in {\mathbb {R}}^n \mid x^*-n\varvec{1}\le x \le x^*+n\varvec{1}\}$. For each integer $k\ge 1$, considering $\alpha _k=2^k$ and $x^{1/\alpha _k}\in \mathop {\mathrm{argmin}}\limits f^{1/\alpha _k}$, we have $\frac{x^{1/\alpha _k}}{\alpha _k}\in B$. Since B is compact, there exists a subsequence $\{\frac{x^{1/\alpha _{k_i}}}{\alpha _{k_i}}\}$ with

$$\begin{aligned} \lim _{i\rightarrow \infty }\frac{x^{1/\alpha _{k_i}}}{\alpha _{k_i}}={\overline{x}}\in B. \end{aligned}$$

Since F is continuous, we have

$$\begin{aligned} \lim _{i\rightarrow \infty }F\left( \frac{x^{1/\alpha _{k_i}}}{\alpha _{k_i}}\right) =F\left( \lim _{i\rightarrow \infty }\frac{x^{1/\alpha _{k_i}}}{\alpha _{k_i}}\right) =F({\overline{x}}). \end{aligned}$$

Since $x^{1/\alpha _{k_i}}\in \mathop {\mathrm{dom}}\limits f^{1/\alpha _{k_{i+1}}}$ holds for each i by the definition (9.1), we have $F(\frac{x^{1/\alpha _{k_1}}}{\alpha _{k_1}}) \ge F(\frac{x^{1/\alpha _{k_2}}}{\alpha _{k_2}}) \ge \cdots \ge F(\frac{x^{1/\alpha _{k_i}}}{\alpha _{k_i}}) \ge \cdots$ which together with $x^{1/2^{k_i}}\in \mathop {\mathrm{argmin}}\limits f^{1/2^{k_i}}$ for all i, guarantees that

$$\begin{aligned} F({\overline{x}} )\le F\left( \frac{x^{1/2^{k_i}}}{2^{k_i}}\right) =\min f^{1/2^{k_i}} \qquad (i=1,2,\ldots ). \end{aligned}$$

(9.3)

We finally show $F({\overline{x}})=\min F$, that is, ${\overline{x}} \in \mathop {\mathrm{argmin}}\limits F$. Suppose to the contrary that there exists $x'$ with $F(x') < F({\overline{x}})$. Let $\varepsilon = F({\overline{x}})-F(x') > 0$. By the continuity of F, there exists $\delta _{\varepsilon }$ such that

$$\begin{aligned} \forall y \in {\mathbb {R}}^n,\;\Vert x'-y\Vert _\infty<\delta _{\varepsilon } \;\Rightarrow \; |F(x')-F(y)|<\varepsilon . \end{aligned}$$

(9.4)

Because there exist $N\in \{k_i \mid i=1, 2, \dots \}$ and $y \in {\mathbb {R}}^n$ such that $2^N y \in {\mathbb {Z}}^n$ and $\Vert x'-y\Vert _\infty <\delta _{\varepsilon }$, by (9.4), we have

$$\begin{aligned} \min f^{1/2^N} \le F(y) <F(x')+\varepsilon = F({\overline{x}}), \end{aligned}$$

which contradicts (9.3). Therefore, ${\overline{x}}$ must be a minimizer of F. $\square$

If F has a unique minimizer, the converse of Theorem 17 also holds.

Theorem 18

Let $F:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an ${\mathbb {R}}$- DDM-convex function. If F has a unique minimizer ${\overline{x}}$, there exists $x^*\in \mathop {\mathrm{argmin}}\limits f$ with $\Vert x^*-{\overline{x}}\Vert _\infty \le n$.

Proof

Let ${\overline{x}}$ be a unique minimizer of F. If f has a minimizer $x^*$, then $\Vert x^*-{\overline{x}}\Vert _\infty \le n$ must hold by Theorem 17. Thus, it is enough to show that f has a minimizer.

Suppose to the contrary that f has no minimizer and let $B=\{x\in {\mathbb {R}}^n \mid {\overline{x}}-n\varvec{1}\le x \le {\overline{x}}+n\varvec{1}\}$. Then, there exists $y \in \mathop {\mathrm{dom}}\limits f {\setminus } B$ such that $f(y) < f(x)$ for all $x \in \mathop {\mathrm{dom}}\limits f \cap B$. Let $\ell = \Vert y-{\overline{x}}\Vert _\infty$ and $B' = \{x\in {\mathbb {R}}^n \mid {\overline{x}}-\ell \varvec{1}\le x \le {\overline{x}}+\ell \varvec{1}\}$. Note that $\ell > n$ and $B' \supset B$. Let us consider the restriction G of F to $B'$ defined by

$$\begin{aligned} G(x) = {\left\{ \begin{array}{ll} F(x) &{} (x \in B') \\ +\infty &{} (x \not \in B') \end{array}\right. } \qquad (x \in {\mathbb {R}}^n). \end{aligned}$$

Obviously, G is ${\mathbb {R}}$-DDM-convex and ${\overline{x}}$ is a unique minimizer of G. In particular, the restriction g of G to ${\mathbb {Z}}^n$ is DDM-convex and has a minimizer z since $B'$ is bounded. This point z does not belong to B since $y \not \in B$ and $f(y) < f(x)$ for all $x \in \mathop {\mathrm{dom}}\limits f \cap B$. However, this contradicts Theorem 17 for G and g. $\square$

If F has a bounded effective domain, a similar statement holds. Let ${\overline{K}}_{\infty }=\sup \{\Vert x-y\Vert _\infty \mid x, y\in \mathop {\mathrm{dom}}\limits F\}$.

Theorem 19

Let $F:{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ be an ${\mathbb {R}}$-DDM-convex function. If $\mathop {\mathrm{dom}}\limits F$ is bounded (i.e., ${\overline{K}}_{\infty } < \infty$), for each ${\overline{x}} \in \mathop {\mathrm{argmin}}\limits F$, there exists $x^*\in \mathop {\mathrm{argmin}}\limits f$ with $\Vert x^*-{\overline{x}}\Vert _\infty \le n$.

Proof

If $\mathop {\mathrm{dom}}\limits f=\mathop {\mathrm{argmin}}\limits f$, the assertion holds. In the sequel, we assume that $\mathop {\mathrm{dom}}\limits f\not = \mathop {\mathrm{argmin}}\limits f$. We fix a minimizer ${\overline{x}}$ of F, arbitrarily. For a sufficiently small $\varepsilon >0$, let us consider functions $F_\varepsilon :{\mathbb {R}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ and $f_\varepsilon :{\mathbb {Z}}^n\rightarrow {\mathbb {R}}\cup \{+\infty \}$ defined by

$$\begin{aligned}&F_\varepsilon (x)=F(x) +\varepsilon \sum _{i=1}^n(x_i-{\overline{x}}_i)^2 \qquad (x\in {\mathbb {R}}^n), \\&f_\varepsilon (x)=F_\varepsilon (x) \qquad (x\in {\mathbb {Z}}^n). \end{aligned}$$

Function $F_\varepsilon$ has the unique minimizer ${\overline{x}}$ and satisfies the conditions of Theorem 18, by Proposition 4 (4), because $f_\varepsilon ^{1/\alpha }$ defined by (9.1) for $F_\varepsilon$ is the sum of $f^{1/\alpha }$ and a separable convex function which are DDM-convex. Thus, by Theorem 18 for $F_\varepsilon$ and $f_\varepsilon$, there exists $x^{\varepsilon }\in \mathop {\mathrm{argmin}}\limits f_\varepsilon$ with $\Vert x^{\varepsilon }-{\overline{x}}\Vert _\infty \le n$.

Let $\beta =\min \{f(x) \mid x\in \mathop {\mathrm{dom}}\limits f{\setminus } \mathop {\mathrm{argmin}}\limits f\} > 0$. Note that $\beta$ is well-defined by boundedness of $\mathop {\mathrm{dom}}\limits f$. We show that if $\varepsilon <(\beta -\min f)/(n{\overline{K}}_{\infty }^2)$, then $x^{\varepsilon } \in \mathop {\mathrm{argmin}}\limits f$. For any $x\in \mathop {\mathrm{argmin}}\limits f$, by $f_\varepsilon (x^{\varepsilon })\le f_\varepsilon (x)$, we have

$$\begin{aligned} f(x^{\varepsilon })\le f(x) + \varepsilon \sum _{i=1}^n \{(x_i-{\overline{x}}_i)^2-(x^{\varepsilon }_i-{\overline{x}}_i)^2\}. \end{aligned}$$

As $f(x^{\varepsilon })\ge f(x)$, $\sum _{i=1}^n \{(x_i-{\overline{x}}_i)^2-(x^{\varepsilon }_i-{\overline{x}}_i)^2\} \ge 0$ must hold. Therefore, we obtain

$$\begin{aligned} f(x^{\varepsilon })&< f(x)+\frac{\beta -\min f}{n{\overline{K}}_{\infty }^2} \sum _{i=1}^n \{(x_i-{\overline{x}}_i)^2-(x^{\varepsilon }_i-{\overline{x}}_i)^2\}\\&\le f(x)+\beta -\min f = \beta , \end{aligned}$$

which says $x^{\varepsilon }\in \mathop {\mathrm{argmin}}\limits f$. $\square$

Remark 5

There is a convex function which is not ${\mathbb {R}}$-DDM-convex. For example, for a positive definite matrix $Q = \left[ {\begin{matrix} 5 &{} 2 \\ 2 &{} 1 \end{matrix}} \right]$, the function

$$\begin{aligned} F(x) = x^\top Q x \qquad (x \in {\mathbb {R}}^2) \end{aligned}$$

is convex, but the restriction f of F to ${\mathbb {Z}}^2$ is not DDM-convex by Theorem 9, and hence, F is not ${\mathbb {R}}$-DDM-convex. $\square$

Remark 6

The convex extension of a DDM-convex function may not be ${\mathbb {R}}$-DDM-convex. For example,

$$\begin{aligned} S = \{(1,0,0), (0,1,0), (0,0,1)\} \end{aligned}$$

is a DDM-convex set, and hence, its indicator function $f = \delta _S$ is DDM-convex. We denote the convex hull of S by ${\overline{S}}$. Then the convex extension F of f is expressed by

$$\begin{aligned} F(x) = {\left\{ \begin{array}{ll} 0 &{} (x \in {\overline{S}}) \\ +\infty &{} (x \not \in {\overline{S}}) \end{array}\right. } \qquad (x \in {\mathbb {R}}^3), \end{aligned}$$

and $f^{1/2}$ is given by

$$\begin{aligned} f^{1/2}(x) = {\left\{ \begin{array}{ll} 0 &{} (x \in T) \\ +\infty &{} (x \not \in T) \end{array}\right. } \qquad (x \in {\mathbb {Z}}^3), \end{aligned}$$

where $T = \{(2,0,0),(1,1,0),(0,2,0),(0,1,1),(0,0,2),(1,0,1)\}$. The function $f^{1/2}$ is not DDM-convex, because for $x = (2,0,0)$ and $y = (0,1,1)$, we have $\mu (x,y) = (1,0,0) \not \in T$. $\square$