Convex Sets and Minimal Sublinear Functions

TL;DR

This paper characterizes the support function of a specific convex set related to a given convex set with the origin in its interior, showing it is the minimal sublinear function defining that set.

Contribution

It establishes that the support function of a particular convex set is the smallest among all sublinear functions representing the original convex set.

Findings

Support function is the minimal among sublinear functions for the set

Characterization of convex sets via sublinear functions

Theoretical insight into convex analysis and support functions

Abstract

We show that, given a closed convex set $K$ containing the origin in its interior, the support function of the set $\{y\in K^*: \exists x\in K\mbox{ such that } \langle x,y \rangle =1\}$ is the pointwise smallest among all sublinear functions $\sigma$ such that $K=\{x: \sigma(x)\leq 1\}$.

Equations

$$K^{\ast }$$

$$\hat{K}$$

$${\sigma }_T(x)=\underset{y\in T}{\sup }⟨x,y⟩\text{mbox}forallx\in {\mathbb{R}}^n.$$

$${\gamma }_K(x)=\inf \{t>0|t^{-1}x\in K\}\text{mbox}forallx\in {\mathbb{R}}^n.$$

$${\rho }_K(x)\ge t\ge \underset{y\in K^{\ast }}{\sup }⟨x,y⟩\ge \underset{y\in \hat{K}}{\sup }⟨x,y⟩={\rho }_K(x),$$

$$\sigma (\bar{x}-{\delta }^{-1}x)=\underset{y\in K^{\ast }}{\sup }⟨\bar{x}-{\delta }^{-1}x,y⟩.$$

$$\sigma (x)\ge \underset{y\in K^{\ast }}{\inf }[\delta (1-⟨\bar{x},y⟩)+⟨x,y⟩],\forall \delta >0$$

$$\sigma (x)\ge \underset{\delta >0}{\sup }\underset{y\in K^{\ast }}{\inf }[\delta (1-⟨\bar{x},y⟩)+⟨x,y⟩].$$

$$\sigma (x)\ge \underset{\delta >0}{\sup }g(\delta )$$

$$x=f+\sum _{j=1}^m{r}^j{s}_j,x\in P\cap {\mathbb{Z}}^n,s\ge 0,\text{mbox}and{s}_j\in \mathbb{Z},j\in I,$$

$$x=f+\sum _{j=1}^m{r}^j{s}_j,x\in S,s\ge 0,$$

$$\sum _{j=1}^m\psi (r^j)s_j\ge 1$$

$$B_{\psi }=\{x\in {\mathbb{R}}^n|\psi (x-f)\le 1\}$$

$${\psi }_B(r)=\underset{i=1,\dots ,k}{\max }⟨a_i,r⟩,\forall r\in {\mathbb{R}}^n.$$

Full text

1 Introduction

The purpose of this note is to prove the following theorem. For $K\subseteq\mathbb{R}^{n}$, we use the notation

[TABLE]

The set $K^{*}$ is the polar of $K$. The set $\hat{K}$ is contained in the relative boundary of $K^{*}$. The polar $K^{*}$ is a convex set whereas $\hat{K}$ is not convex in general.

The support function of a nonempty set $T\subset\mathbb{R}^{n}$ is defined by

[TABLE]

It is straightforward to show that support functions are sublinear, that is they are convex and positively homogeneous (A function $f:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ is positively homogeneous if $f(tx)=tf(x)$ for every $x\in\mathbb{R}^{n}$ and $t>0$), and $\sigma_{T}=\sigma_{\overline{\mathrm{conv}}(T)}$ [5]. We will show that, if $K\subset\mathbb{R}^{n}$ is a closed convex set containing the origin in its interior, then $K=\{x\,|\,\sigma_{\hat{K}}(x)\leq 1\}$. The next theorem shows that $\sigma_{\hat{K}}$ is the smallest sublinear function with this property.

Theorem 1

Let $K\subset\mathbb{R}^{n}$ be a closed convex set containing the origin in its interior. If $\sigma:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ is a sublinear function such that $K=\{x\in\mathbb{R}^{n}\,|\,\sigma(x)\leq 1\}$, then $\sigma_{\hat{K}}(x)\leq\sigma(x)$ for all $x\in\mathbb{R}^{n}$.

In the remainder we define $\rho_{K}=\sigma_{\hat{K}}$.

Let $K\subset\mathbb{R}^{n}$ be a closed convex set containing the origin in its interior. A standard concept in convex analysis [5, 7] is that of gauge (sometimes called Minkowski function), which is the function $\gamma_{K}$ defined by

[TABLE]

By definition $\gamma_{K}$ is nonnegative. One can readily verify that $K=\{x\,|\,\gamma_{K}(x)\leq 1\}$. It is well known that $\gamma_{K}$ is the support function of $K^{*}$ (see [5] Proposition 3.2.4).

Given any sublinear function $\sigma$ such that $K=\{x\,|\,\sigma(x)\leq 1\}$, it follows from positive homogeneity that $\sigma(x)=\gamma_{K}(x)$ for every $x$ where $\sigma(x)>0$. Hence $\sigma(x)\leq\gamma_{K}(x)$ for all $x\in\mathbb{R}^{n}$. On the other hand, we prove in Theorem 1 that the sublinear function $\rho_{K}$ satisfies $\rho_{K}(x)\leq\sigma(x)$ for all $x\in\mathbb{R}^{n}$. In words, $\gamma_{K}$ is the largest sublinear function such that $K=\{x\in\mathbb{R}^{n}\,|\,\sigma(x)\leq 1\}$ and $\rho_{K}$ is the smallest.

Note that $\rho_{K}$ can take negative values, so in general it is different from the gauge $\gamma_{K}$. Indeed the recession cone of $K$, which is the set $\mathrm{rec}(K)=\{x\in K\,|\,tx\in K\mbox{ for all }t>0\}$, coincides with $\{x\in K\,|\,\sigma(x)\leq 0\}$ for every sublinear function $\sigma$ such that $K=\{x\,|\,\sigma(x)\leq 1\}$. In particular $\rho_{K}(x)$ can be negative for $x\in\mathrm{rec}(K)$. For example, let $K=\{x\in\mathbb{R}^{2}\,|\,x_{1}\leq 1,\,x_{2}\leq 1\}$. Then $K^{*}=\mathrm{conv}\{(0,0),(1,0),(0,1)\}$ and $\hat{K}=\mathrm{conv}\{(1,0),(0,1)\}$. Therefore, for every $x\in\mathbb{R}^{2}$, $\gamma_{K}(x)=\max\{0,x_{1},x_{2}\}$ and $\rho_{K}(x)=\max\{x_{1},x_{2}\}$. In particular, $\rho_{K}(x)<0$ for every $x$ such that $x_{1}<0$, $x_{2}<0$.

By Hörmander’s theorem [6], a sublinear function $\sigma:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ is the support function of a unique bounded closed convex set $C\subset\mathbb{R}^{n}$, say $\sigma=\sigma_{C}$. So the condition $K=\{x\in\mathbb{R}^{n}\,|\,\sigma(x)\leq 1\}$ says $K=C^{*}$. Thus Theorem 1 can be restated in its set version.

Theorem 2

Let $K\subset\mathbb{R}^{n}$ be a closed convex set containing the origin in its interior. If $C\subset\mathbb{R}^{n}$ is a bounded closed convex set such that $K=C^{*}$, then $\hat{K}\subset C$.

When $K$ is bounded, this theorem is trivial (the hypothesis $K=C^{*}$ becomes $K^{*}=C$. The conclusion $\hat{K}\subset C$ is obvious because one always has $\hat{K}\subset K^{*}$.) So the interesting case of Theorem 1 is when $K$ is unbounded.

We present the proof of Theorem 1 in Section 2. Theorem 1 has applications in integer programming. In particular it is used to establish the relationship between minimal inequalities and maximal lattice-free convex sets [1], [2]. We summarize these results in Section 3.

2 Proof of Theorem 1

We will need Straszewicz’s theorem [8] (see [7] Theorem 18.6). Given a closed convex set $C$, a point $x\in C$ is extreme if it cannot be written as a proper convex combination of two distinct points in $C$. A point $x\in C$ is exposed if there exists a supporting hyperplane $H$ for $C$ such that $H\cap C=\{x\}$. Clearly exposed points are extreme. We will denote by $\mathrm{ext}(C)$ the set of extreme points and $\mathrm{exp}(C)$ the set of exposed points of $C$.

Theorem 3

Given a closed convex set $C$, the set of exposed points of $C$ is a dense subset of the set of extreme points of $C$.

Let $K$ be a closed convex set containing the origin in its interior. Let $\sigma$ be a sublinear function such that $K=\{x\,|\,\sigma(x)\leq 1\}$. The boundary of $K$, denoted by $\mathrm{\bf bd}(K)$, is the set $\{x\in K\,|\,\sigma(x)=1\}$.

Lemma 4

For every $x\notin\mathrm{rec}(K)$, $\sigma(x)=\rho_{K}(x)=\sup_{y\in K^{*}}\langle x,y\rangle$.

Proof. Let $x\notin\mathrm{rec}(K)$. Then $t=\sigma(x)>0$. By positive homogeneity, $\sigma(t^{-1}x)=1$, hence $t^{-1}x\in\mathrm{\bf bd}(K)$. Since $K$ is closed and convex, there exists a supporting hyperplane for $K$ containing $t^{-1}x$. Since $0\in\mathrm{\bf int}(K)$, this implies that there exists $\bar{y}\in K^{*}$ such that $(t^{-1}x)\bar{y}=1$. In particular $\bar{y}\in\hat{K}$, hence by definition $\rho_{K}(x)\geq\langle x,\bar{y}\rangle=t$.

Furthermore, for any $y\in K^{*}$, $\langle t^{-1}x,y\rangle\leq 1$, hence $\langle x,y\rangle\leq t$, which implies $t\geq\sup_{y\in K^{*}}\langle x,y\rangle$. Thus

[TABLE]

where the last inequality holds since $\hat{K}\subset K^{*}$, hence equality holds throughout. $\square$

Corollary 5

$K=\{x\,|\,\rho_{K}(x)\leq 1\}$.

Lemma 6

Given an exposed point $\bar{y}$ of $K^{*}$ different from the origin, there exists $x\in K$ such that $\langle x,\bar{y}\rangle=1$ and $\langle x,y\rangle<1$ for all $y\in K^{*}$ distinct from $\bar{y}$.

Proof. If $\bar{y}\neq 0$ is an exposed point of $K^{*}$, then there exists a supporting hyperplane $H=\{y\,|\,\langle a,y\rangle=\beta\}$ such that $\langle a,\bar{y}\rangle=\beta$ and $\langle a,y\rangle<\beta$ for every $y\in K^{*}\setminus\{\bar{y}\}$. Since $0\in K^{*}$ and $\bar{y}\neq 0$, $\beta>0$. Thus the point $x=\beta^{-1}a\in K^{**}=K$ satisfies the statement (where $K=K^{**}$ holds because $0\in K$). $\square$

The next lemma states that $\hat{K}$ and $\hat{K}\cap\mathrm{exp}(K^{*})$ have the same support function.

Lemma 7

For every $x\in\mathbb{R}^{n}$, $\rho_{K}(x)=\sup_{y\in\hat{K}\cap\mathrm{exp}(K^{*})}\langle x,y\rangle$.

Proof. We first show that $\rho_{K}(x)=\sup_{y\in\hat{K}\cap\mathrm{ext}(K^{*})}\langle x,y\rangle$. Given $y\in\hat{K}$ we show that there exists an extreme point $y^{\prime}$ of $K^{*}$ in $\hat{K}$ such that $\langle x,y\rangle\leq\langle x,y^{\prime}\rangle$. Since $y\in\hat{K}$, there exists $\bar{x}\in K$ such that $\langle\bar{x},y\rangle=1$. The point $y$ is a convex combination of extreme points $y_{1},\ldots,y_{k}$ of $K^{*}$, and each $y_{i}$ satisfies $\langle\bar{x},y^{i}\rangle=1$. Thus $y^{1},\ldots,y^{k}\in\hat{K}$, and $\langle x,y^{i}\rangle\geq\langle x,y\rangle$ for at least one $i$.

By Straszewicz’s theorem (Theorem 3) the set of exposed points in $K^{*}$ is a dense subset of the extreme points of $K^{*}$. By Lemma 6, all exposed points of $K^{*}$ except the origin are in $\hat{K}$, hence $\mathrm{exp}(K^{*})\cap\hat{K}$ is dense in $\mathrm{ext}(K^{*})\cap\hat{K}$. Therefore $\rho_{K}(x)=\sup_{y\in\hat{K}\cap\mathrm{exp}(K^{*})}\langle x,y\rangle$. $\square$

A function $\sigma$ is subadditive if $\sigma(x_{1}+x_{2})\leq\sigma(x_{1})+\sigma(x_{2})$ for every $x_{1},x_{2}\in\mathbb{R}^{n}$. It is easy to show that $\sigma$ is sublinear if and only if it is subadditive and positively homogeneous.

Proof of Theorem 1. By Lemma 4, we only need to show $\sigma(x)\geq\rho_{K}(x)$ for points $x\in\mathrm{rec}(K)$. By Lemma 7 it is sufficient to show that, for every exposed point $\bar{y}$ of $K^{*}$ contained in $\hat{K}$, $\sigma(x)\geq\langle x,\bar{y}\rangle$.

Let $\bar{y}$ be an exposed point of $K^{*}$ in $\hat{K}$. By Lemma 6 there exists $\bar{x}\in K$ such that $\langle\bar{x},\bar{y}\rangle=1$ and $\langle\bar{x},y\rangle<1$ for all $y\in K^{*}$ distinct from $\bar{y}$. Note that $\bar{x}\in\mathrm{\bf bd}(K)$.

We observe that for all $\delta>0$, $\bar{x}-\delta^{-1}x\notin\mathrm{rec}(K)$. Indeed, since $x\in\mathrm{rec}(K)$, $\bar{x}+\delta^{-1}x\in K$. Hence $\bar{x}-\delta^{-1}x\notin\mathrm{\bf int}(K)$ because $\bar{x}\in\mathrm{\bf bd}(K)$. Since $0\in\mathrm{\bf int}(K)$ and $\bar{x}-\delta^{-1}x\notin\mathrm{\bf int}(K)$, then $\bar{x}-\delta^{-1}x\notin\mathrm{rec}(K)$. Thus by Lemma 4

[TABLE]

Since $\bar{x}\in\mathrm{\bf bd}(K)$, $\sigma(\bar{x})=1$. By subadditivity, $1=\sigma(\bar{x})\leq\sigma(\bar{x}-\delta^{-1}x)+\sigma(\delta^{-1}x).$ By positive homogeneity, the latter implies that $\sigma(x)\geq\delta-\delta\sigma(\bar{x}-\delta^{-1}x)$ for all $\delta>0$. By (1),

[TABLE]

hence

[TABLE]

Let $g(\delta)=\inf_{y\in K^{*}}\delta(1-\langle\bar{x},y\rangle)+\langle x,y\rangle$. Since $\bar{x}\in K$, $1-\langle\bar{x},y\rangle\geq 0$ for every $y\in K^{*}$. Hence $\delta(1-\langle\bar{x},y\rangle)+\langle x,y\rangle$ defines an increasing affine function of $\delta$ for each $y\in K^{*}$, therefore $g(\delta)$ is increasing and concave. Thus $\sup_{\delta>0}g(\delta)=\lim_{\delta\rightarrow\infty}g(\delta)$.

Since $0\in\mathrm{\bf int}(K)$, $K^{*}$ is compact. Hence, for every $\delta>0$ there exists $y(\delta)\in K^{*}$ such that $g(\delta)=\delta(1-\langle\bar{x},y(\delta)\rangle)+\langle x,y(\delta)\rangle$. Furthermore, there exists a sequence $(\delta_{i})_{i\in\mathbb{N}}$ such that $\lim_{i\rightarrow\infty}\delta_{i}=+\infty$ and the sequence $(y_{i})_{i\in\mathbb{N}}$ defined by $y_{i}=y(\delta_{i})$ converges, because in a compact set every sequence has a convergent subsequence. Let $y^{*}=\lim_{i\rightarrow\infty}y_{i}$.

We conclude the proof by showing that $\sigma(x)\geq\langle x,y^{*}\rangle$ and $y^{*}=\bar{y}$.

[TABLE]

where the last inequality follows from the fact that $\delta_{i}(1-\langle\bar{x},y_{i}\rangle)\geq 0$ for all $i\in\mathbb{N}$. Finally, since $\lim_{i\rightarrow\infty}\delta_{i}(1-\langle\bar{x},y_{i}\rangle)$ is bounded and $\lim_{i\rightarrow\infty}\delta_{i}=+\infty$, it follows that $\lim_{i\rightarrow\infty}(1-\langle\bar{x},y_{i}\rangle)=0$, hence $\langle\bar{x},y^{*}\rangle=1$. By our choice of $\bar{x}$, $\langle\bar{x},y\rangle<1$ for every $y\in K^{*}$ distinct from $\bar{y}$. Hence $y^{*}=\bar{y}$. $\Box$

3 An application to integer programming

In [1] and [2], Basu et al. apply Theorem 1 to cutting plane theory.

Consider a mixed integer linear program, and the optimal tableau of the linear programming relaxation. We select $n$ rows of the tableau, relative to $n$ basic integer variables $x_{1},\ldots,x_{n}$. Let $s_{1},\ldots,s_{m}$ denote the nonbasic variables. Let $f_{i}$ be the value of $x_{i}$ in the basic solution associated with the tableau, $i=1,\ldots,n$, and suppose $f\notin\mathbb{Z}^{n}$. The tableau restricted to these $n$ rows is of the form

[TABLE]

where $r^{j}\in\mathbb{R}^{n}$, $j=1,\ldots,m$, $I$ denotes the set of integer nonbasic variables, and $P$ is some full-dimensional rational polyhedron in $\mathbb{R}^{n}$, representing constraints on the basic variables (typically nonnegativity or bounds on the variables).

An important question in integer programming is to derive valid inequalities for (2), cutting off the current infeasible solution $x=f$, $s=0$. We consider a simplified model where the integrality conditions are relaxed on all nonbasic variables. So we study the following model, introduced by Johnson [4],

[TABLE]

where where $S=P\cap\mathbb{Z}^{n}$ and $f\in\mathrm{conv}(S)\setminus\mathbb{Z}^{n}$. Note that every inequality cutting off the point $(f,0)$ can be expressed in terms of the nonbasic variables $s$ only, and can therefore be written in the form $\sum_{j=1}^{m}\alpha_{j}s_{j}\geq 1$.

Basu et al. [2] study “general formulas” to generate such inequalities. By this, we mean functions $\psi\,:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that the inequality

[TABLE]

is valid for (3) for every choice of $m$ and vectors $r^{1},\ldots,r^{m}\in\mathbb{R}^{n}$. We refer to such functions $\psi$ as valid functions (with respect to $f$ and $S$). Since one is interested in the deepest inequalities cutting off $(f,0)$, one only needs to investigate (pointwise) minimal valid functions.

Given a sublinear function $\psi$ such that the set

[TABLE]

is $S$-free (i.e. $\mathrm{\bf int}(B_{\psi})\cap S=\emptyset$), it is easily shown that $\psi$ is a valid function. Indeed, since $\psi$ is sublinear, $B_{\psi}$ is a closed convex set with $f$ in its interior, thus, given any solution $(\bar{x},\bar{s})$ to (3), we have $\sum_{j=1}^{m}\psi(r^{j})\bar{s}_{j}\geq\psi(\sum_{j=1}^{m}r^{j}\bar{s}_{j})=\psi(\bar{x}-f)\geq 1,$ where the first inequality follows from sublinearity and the last one from the fact that $\bar{x}\notin\mathrm{\bf int}(B_{\psi})$.

On the other hand, Dey and Wolsey [3] show that, if $\psi$ is a minimal valid function, $\psi$ is sublinear and $B_{\psi}$ is an $S$-free convex set with $f$ in its interior.

Using Theorem 1, Basu et al. [2] prove that, if $\psi$ is a minimal valid function, then $B_{\psi}$ is a maximal $S$-free convex set. That is, $B_{\psi}$ is an inclusionwise maximal convex set such that $\mathrm{\bf int}(B_{\psi})\cap S=\emptyset$. Furthermore, they give a explicit formulas for all minimal valid functions.

In order to prove this result, they first show that maximal $S$-free convex sets are polyhedra. Therefore, a maximal $S$-free convex set $B\subseteq\mathbb{R}^{n}$ containing $f$ in its interior can be uniquely written in the form $B=\{x\in\mathbb{R}^{n}:\;\langle a_{i},x-f\rangle\leq 1,\;i=1,\ldots,k\}$. Thus, if we let $K:=\{x-f\,|\,x\in B\}$, Theorem 1 implies that the function $\psi_{B}:=\rho_{K}$ is the minimal sublinear function such that $B=\{x\in\mathbb{R}^{n}\,|\,\psi_{B}(x-f)\leq 1\}$. Note that, since $K^{*}=\mathrm{conv}\{0,a_{1},\ldots,a_{k}\}$, $\psi_{B}$ has the following simple form

[TABLE]

From the above, it is immediate that, if $B$ is a maximal $S$ free convex set, then the function $\psi_{B}$ is a minimal valid function.

The main use of Theorem 1 is in the proof of the converse statement, namely, that every minimal valid function is of the form $\psi_{B}$ for some maximal $S$-free convex set $B$ containing $f$ in its interior. The proof outline is as follows. Suppose $\psi$ is a minimal valid function. Thus $\psi$ is sublinear and $B_{\psi}$ is an $S$-free convex set. Let $K:=\{x-f\,|\,x\in B_{\psi}\}$.

$i)$ Since $\{x\in\mathbb{R}^{n}\,|\,\rho_{K}(x-f)\leq 1\}=B_{\psi}$, Theorem 1 implies that $\psi\geq\rho_{K}$.

$ii)$ Theorem [2]. There exists a maximal $S$-free convex set $B=\{x\in\mathbb{R}^{n}:\;\langle a_{i},x-f\rangle\leq 1,\;i=1,\ldots,k\}$ such that $a_{i}\in\overline{\mathrm{conv}}(\hat{K})$ for $i=1,\ldots,k$.

$iii)$ For every $r\in\mathbb{R}^{n}$, we have $\psi(r)\geq\rho_{K}(r)=\sup_{y\in\overline{\mathrm{conv}}(\hat{K})}\langle y,r\rangle\geq\max_{i=1,\ldots,k}\langle a_{i},r\rangle=\psi_{B}(r)$, where the first inequality follows from i) and the second from ii). Since $\psi_{B}$ is a valid function, it follows by the minimality of $\psi$ that $\psi=\psi_{B}$.