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Midconvexity for finite sets
Journal of Inequalities and Applications volume 2013, Article number: 42 (2013)
Abstract
Let X be a finite subset of a real vector space. We study Jensentype convexity on subsets of X. In particular for subsets of X, we introduce the definition of Xmidconvex sets. We show that such a notion corresponds well to the classical notion of a convex set. Moreover, we prove that a function Xmidconvex set is a midconvex hull of all its extremal points. Other analogues of some classical results are also given. At the end we present an algorithmic approach to finding the midconvex hull of a given set.
MSC:26A51, 26B25.
1 Introduction
Thanks to the growing role of computers, one can recently notice an increasing interest in discrete (finite) sets in many parts of mathematics. In particular, one has to mention the recent development of convexity on subsets of {\mathbb{Z}}^{N}[1].
Therefore, there appears a natural need to introduce and develop (mid)convexity on finite sets. Since we are going to study finite subsets of a real vector space, we may restrict ourselves to finite subsets of {\mathbb{R}}^{N}. Since the middle point of an interval is one of the most important geometrical notions, we start our investigations with the simplest and earliest version of convexitymidconvexity.
The main difference between our approach and that of [1–3] is that we restrict our considerations to X, that is, we treat X as a space. Hence, our notions have relative character since we study convexity on the subsets of X. One can say that we investigate midconvexity on a restricted domain.
Now, we briefly describe the contents of the paper. First of all, we propose a proper definition of a midconvex set for finite sets. By ‘proper’ we mean here that we are looking for the notion which has possibly many properties analogous to those of convex sets in {\mathbb{R}}^{N}. We consider two different definitions of midconvexity. The first one, which is intuitively very similar to the classical definition, occurs to have weak properties. The other one, which we call function midconvexity, at a first glance seems a little bit artificial, but it corresponds well to the classical definition. It is based on the observation that in {\mathbb{R}}^{N} a closed set V is convex if and only if it is the preimage of the set (\mathrm{\infty},0] for some convex function f:{\mathbb{R}}^{N}\to \mathbb{R}.
Next, we study the properties of function midconvexity. In particular in Section 3, we prove an analogue of the theorem stating that a compact convex set in {\mathbb{R}}^{N} is the convex hull of its extremal points. We introduce also the notion of a convex function on a finite domain. We prove that the maximum of such a convex function is achieved at an extreme point of its domain.
2 Midconvexity on finite sets
In the whole paper, we assume that X is a finite subset of {\mathbb{R}}^{N}. By {\mathbb{N}}_{0} we denote the set of nonnegative integers and by {\mathbb{R}}_{+} the set of nonnegative real numbers. We adapt the classical definitions of a midconvex set and a midconvex function to our setting [4].
Definition 2.1 We say that W\subset X is Xmidconvex if
Evidently, X and a singleton are Xmidconvex sets. It is also obvious that the intersection of Xmidconvex sets is Xmidconvex.
Definition 2.2 We say that a function f:W\to \mathbb{R}, where W\subset X, is Xmidconvex (shortly midconvex) if W is Xmidconvex and
We will need some of their properties with respect to the following definition of a midconvex set.
Proposition 2.3 LetW\subset Xbe an Xmidconvex set, letc>0and letf:W\to \mathbb{R}, g:W\to \mathbb{R}be Xmidconvex functions. Thenmax(f,g)and cf are Xmidconvex functions.
Proof It is obvious that cf is Xmidconvex. We show that max(f,g) is Xmidconvex. Consider arbitrary u,v\in W such that \frac{u+v}{2}\in W. Then we have
and
Whence we obtain
□
To explain the unintuitive behavior of Xmidconvexity, we need the notion of a discrete interval (for m\in {\mathbb{N}}_{0}):
One could expect that if W is Xmidconvex and a,b\in W are such that {[a,b]}_{m}\subset X, then {[a,b]}_{m}\subset W. It occurs that this is not the case.
Example 2.4 Let N=1, X=\{0,1,2,3\} and W=\{0,3\}. Then W is Xmidconvex and 0,3\in W, but {[0,3]}_{3}=X\not\subset W.
Therefore, we need another definition of a midconvex set.
Let us present the idea which, in our opinion, leads to the ‘right’ definition. The observation that a closed set A\subset {\mathbb{R}}^{N} is convex if and only if there exists a convex function f:{\mathbb{R}}^{N}\to \mathbb{R} such that A=\{x\in {\mathbb{R}}^{N}:f(x)\le 0\} (the part ‘if’ of the statement is obvious; for the converse, take as f the distance from A) leads us to the following definition.
Definition 2.5 We say that W\subset X is function Xmidconvex (shortly Xfmidconvex) if there exists an Xmidconvex function f:X\to \mathbb{R} such that
Clearly, f:X\to \mathbb{R}, f\equiv 0 is Xmidconvex. Hence, X is Xfmidconvex.
Observation 2.6 Every Xfmidconvex set is Xmidconvex.
Proof Let W be Xfmidconvex. Then W:=\{x\in X:f(x)\le 0\} for a certain Xmidconvex function f:X\to \mathbb{R}. Consider arbitrary {x}_{1},{x}_{2}\in W such that \frac{{x}_{1}+{x}_{2}}{2}\in X. Then we have
Hence \frac{{x}_{1}+{x}_{2}}{2}\in W. □
As one could expect, the implication converse to Observation 2.6 does not hold. To notice this (see Example 2.8), we will need the following result, which shows that fmidconvexity resembles classical convexity.
Theorem 2.7 Let W be Xfmidconvex. Then
Proof The proof follows a similar approach from [5].
Since W is Xfmidconvex, there exists a midconvex function f:X\to \mathbb{R} such that W=\{x\in X:f(x)\le 0\}. Suppose, for the proof by contradiction, that there exist a,b\in W, m\in \mathbb{N}, such that {[a,b]}_{m}\subset X but {[a,b]}_{m}\not\subset W. It means that there exists a k\in \{0,\dots ,m\} such that
i.e., that f({x}_{k})>0. Then
Let l\in \{0,\dots ,m\} be chosen such that f({x}_{l})=C. Two cases can occur.

1.
l\le \frac{m}{2}. Then {x}_{2l}\in {[a,b]}_{m}, and we obtain that
\begin{array}{rcl}C& =& f({x}_{l})=f(\frac{l}{m}a+(1\frac{l}{m})b)=f\left(\frac{\frac{2l}{m}a+(1\frac{2l}{m})b+b}{2}\right)\\ \le & \frac{f(\frac{2l}{m}a+(1\frac{2l}{m})b)+f(b)}{2}=\frac{f(\frac{2l}{m}a+(1\frac{2l}{m})b)}{2}=\frac{f({x}_{2l})}{2}.\end{array}
Whence we have f({x}_{2l})\ge 2C>C, a contradiction.

2.
l>\frac{m}{2}. Then {x}_{2lm}\in {[a,b]}_{m}, and by reasoning analogous to case 1, we obtain a contradiction.
□
Now, we show that the implication converse to that from Observation 2.6 does not hold.
Example 2.8 We continue considerations from Example 2.4. Let N=1, m=3, X=\{0,1,2,3\}, W=\{0,3\}. Then {[0,3]}_{m}=X, and hence {[0,3]}_{3}\not\subset W. W is an Xmidconvex set which is not Xfmidconvex. It shows also that an analogue of Theorem 2.7 for Xmidconvex set does not hold.
Proposition 2.9 Let{W}_{2}\subset {W}_{1}\subset Xand let{W}_{1}be Xfmidconvex and{W}_{2}be{W}_{1}fmidconvex. Then{W}_{2}is Xfmidconvex.
Proof There exist midconvex functions {f}_{1}:X\to \mathbb{R}, {f}_{2}:{W}_{1}\to \mathbb{R} such that {W}_{1}={f}_{1}^{1}((\mathrm{\infty},0]) and {W}_{2}={f}_{2}^{1}((\mathrm{\infty},0]). Multiplying any of those functions by a positive real number, we do not destroy the above property. Therefore, we may assume that
We define a function f:X\to {\mathbb{R}}_{+} as follows:
Evidently, {f}^{1}((\mathrm{\infty},0])={W}_{2}. We have to prove yet that f is Xmidconvex. Consider arbitrary {x}_{1},{x}_{2}\in X such that \frac{{x}_{1}+{x}_{2}}{2}\in X. We distinguish five cases.
1^{o}. {x}_{1},{x}_{2}\in X\setminus {W}_{1}, \frac{{x}_{1}+{x}_{2}}{2}\in X\setminus {W}_{1}. Then we have
2^{o}. {x}_{1},{x}_{2}\in X\setminus {W}_{1}, \frac{{x}_{1}+{x}_{2}}{2}\in {W}_{1}. Making use of (1), we get
3^{o}. {x}_{1},{x}_{2}\in {W}_{1}. By Observation 2.6, {W}_{1} is Xmidconvex and therefore \frac{{x}_{1}+{x}_{2}}{2}\in {W}_{1}. As a consequence, we have
4^{o}. {x}_{1}\in X\setminus {W}_{1}, {x}_{2}\in {W}_{1}, \frac{{x}_{1}+{x}_{2}}{2}\in X\setminus {W}_{1}. Then we have
5^{o}. {x}_{1}\in X\setminus {W}_{1}, {x}_{2}\in {W}_{1}, \frac{{x}_{1}+{x}_{2}}{2}\in {W}_{1}. Applying (1), we get
We have proved that f is Xmidconvex. □
Now, we define an Xfmidconvex hull of a given set A\subset X. We begin with important remarks.
Proposition 2.10 Let{W}_{1}, {W}_{2}be Xfmidconvex subsets of X. Then{W}_{1}\cap {W}_{2}is Xfmidconvex.
Proof There exist midconvex functions {f}_{1}:X\to \mathbb{R}, {f}_{2}:X\to \mathbb{R} such that {W}_{1}={f}_{1}^{1}((\mathrm{\infty},0]) and {W}_{2}={f}_{2}^{1}((\mathrm{\infty},0]). We define the function f:X\to {\mathbb{R}}_{+} as follows:
By Proposition 2.3 f is Xmidconvex. Furthermore, {f}^{1}((\mathrm{\infty},0])={W}_{1}\cap {W}_{2}. This finishes the proof. □
Proposition 2.11 Let{X}_{1}\subset {\mathbb{R}}^{{N}_{1}}, {X}_{2}\subset {\mathbb{R}}^{{N}_{2}}be finite sets and let{W}_{1}\subset Xbe{X}_{1}fmidconvex and{W}_{2}\subset {X}_{2}be{X}_{2}fmidconvex. Then{W}_{1}\times {W}_{2}is({X}_{1}\times {X}_{2})fmidconvex.
Proof There exist an {X}_{1}midconvex function {f}_{1}:{X}_{1}\to \mathbb{R} and an {X}_{2}midconvex function {f}_{2}:{X}_{2}\to \mathbb{R} such that
We define the function f:{X}_{1}\times {X}_{2}\to \mathbb{R} as follows:
We are going to show that f is ({X}_{1}\times {X}_{2})midconvex and that
Consider arbitrary ({x}_{1},{x}_{2}),({y}_{1},{y}_{2})\in {X}_{1}\times {X}_{2} and assume that \frac{({x}_{1},{x}_{2})+({y}_{1},{y}_{2})}{2}\in {X}_{1}\times {X}_{2}. Then \frac{{x}_{1}+{y}_{1}}{2}\in {X}_{1}, \frac{{x}_{2}+{y}_{2}}{2}\in {X}_{2} and
Whence we obtain
It means that f is ({X}_{1}\times {X}_{2})midconvex. Furthermore, we have
□
We are going to define now an Xmidconvex hull and an Xfmidconvex hull of a given set A\subset X. We follow the classical definition of a midconvex hull [4].
Definition 2.12 Let A\subset X.
The intersection of all Xmidconvex sets containing A is called an Xmidconvex hull of A and is denoted by {mconv}_{X}(A).
The intersection of all Xfmidconvex sets containing A is called an Xfmidconvex hull of A and is denoted by {fmconv}_{X}(A).
One can directly verify that {mconv}_{X}(A) is Xmidconvex. By Proposition 2.10, we obtain that the intersection of a finite family of Xfmidconvex sets is Xfmidconvex, and consequently, by the finiteness of X, we obtain that {fmconv}_{X}(A) is Xfmidconvex. Consequently, the Xfmidconvex hull of A is the smallest Xfmidconvex set containing A.
The example below illustrates Definition 2.12 and shows that an Xmidconvex hull and an Xfmidconvex hull of a given set are, in general, different.
Example 2.13 Let N=2, X=\{(1,0),(0,0),(0,1),(0,2),(1,1),(1,0),(1,1),(2,1)\} (see Figure 1). Clearly, W:=\{(1,0),(0,2),(1,1),(2,1)\} is Xmidconvex. Hence, {mconv}_{X}(W)=W. One can show (it follows directly from the results of the next section) that {fmconv}_{X}(W)=X.
3 Extremal points
We begin with the definition of an extremal point.
Definition 3.1 Let W\subset {\mathbb{R}}^{N}. A point a\in W is called an extremal point of W if
We denote the set of all extremal points of W by extW. As we know [6], by the KreinMilman theorem, a compact convex set in {\mathbb{R}}^{N} is the convex hull of its extremal points. However, its analogue does not hold for Xmidconvexityjust consider X=\{0,1,2,3\} which is not the Xmidconvex hull of the set W=\{0,3\} of its extremal points.
In this section, we prove a version of the classical KreinMilman theorem for function midconvexity. We begin with the following observation.
Observation 3.2 IfW\subset {\mathbb{R}}^{N}is nonempty and finite, then theextW\ne \mathrm{\varnothing}.
Proof Take a convex hull of W in {\mathbb{R}}^{N}. It is nonempty, convex and compact (because W is finite). Therefore, [6] it has an extremal point. □
We will need the notion of index.
Definition 3.3 Let W\subset X. We say that a sequence ({x}_{0},\dots ,{x}_{n})\subset W is a Jchain of length n in W if either n=0 or n\ge 1 and
We call {x}_{0} the beginning and {x}_{n} the end of the Jchain.
Condition (2) means that {x}_{k} is the middle of {x}_{k1} and a certain point in W (or, in another words, that the symmetric point to {x}_{k1} with respect to {x}_{k} belongs to W).
Proposition 3.4 Let\mathrm{\varnothing}\ne W\subset X. Then for eachx\in W, there exists a Jchain in W beginning in extW and ending at x.
Proof Fix arbitrarily x\in W and denote by Z the set of points z of W such that there exists a Jchain in W beginning at z and ending at x. Obviously x\in Z, and hence Z\ne \mathrm{\varnothing}. By Observation 3.2, there exists z\in extZ. Let ({y}_{0},\dots ,{y}_{n}), {y}_{0}=z, {y}_{n}=x, be the Jchain with beginning at z and end at x.
We show that z\in extW. Suppose, for an indirect proof, that z\notin extW. Since z\in W, there exist v,w\in W, v\ne w such that z=(v+w)/2. Clearly, (v,{y}_{0},\dots ,{y}_{n}) and (w,{y}_{0},\dots ,{y}_{n}) are Jchains starting at v and w, respectively, and with end at x. This implies that v,w\in Z, which contradicts the assumption that z\in extZ. □
Let W\subset X be nonempty. By Observation 3.2, extW\ne \mathrm{\varnothing}. For x\in W by {i}_{W}(x), we denote the length of the shortest Jchain in W beginning in extW and ending at x. It follows from Proposition 3.4 that such a Jchain exists.
We also put
{i}_{W}(x) is called the index of x and {i}_{W}the index of W.
Remark 3.5 It follows from the proof of Proposition 3.4 that if {i}_{W}(x)=n, where n\in \mathbb{N}, then there exists an element \overline{x}\in W such that {i}_{W}(\overline{x})=n1. Therefore, if {i}_{W}=n, n\in {\mathbb{N}}_{0}, then for each m\in \{0,\dots ,n\}, we can find x\in W such that {i}_{W}(x)=m. Hence, the set of indices of elements of W coincides with the set \{0,\dots ,n\}.
One can easily notice that {i}_{W}(x)=0 if and only if x\in extW.
We illustrate the above notations by a simple example.
Example 3.6 Let N=2, W=X=\{(1,1),(1,0),(1,1),(0,0),(1,1),(1,0),(1,1)\}. One can easily notice (see Figure 2) that
We have
Proposition 3.7 We assume that\mathrm{\varnothing}\ne W\subset X. Then
Proof
Let
Clearly, the sequence {\{{W}_{k}\}}_{k\in {\mathbb{N}}_{0}} is ascending. Since W is finite, there exists a k\in {\mathbb{N}}_{0} such that {W}_{k}={W}_{k+1}. Let k be the smallest integer with this property. Then
and hence
which implies that
We are going to show that {W}_{k}=W, which will complete the proof. Suppose that this is not the case. Let P:=W\setminus {W}_{k}. Let p be an extremal point of P. We will prove that p is an extremal point of W. Suppose that it is not the case. Then there exist q,r\in W, q\ne r such that
Since p is an extremal point of P either q\notin P or r\notin P. Assume that q\notin P. Then q\in {W}_{k} and since p+(pq)=r\in W, we obtain that p\in {W}_{k+1}. Hence, {W}_{k+1}\ne {W}_{k}, a contradiction. Thus p is an extremal point of W and hence p\in {W}_{0}\subset {W}_{k}, a contradiction. We have proved that {W}_{k}=W. □
Now, we prove the main result of this section.
Theorem 3.8 Let\mathrm{\varnothing}\ne W\subset Xbe an Xfmidconvex set. Then
Proof
Evidently, we have
Suppose that
There exists a midconvex function f:X\to {\mathbb{R}}_{+} such that
If follows from this equality and (3) that
Let a\in W be such that f(a)=C and that
By Propositions 3.4 and 3.7, there exists a finite Jchain in W starting in extW and ending at a. Let ({a}_{0},\dots ,{a}_{n}) be the shortest Jchain with these properties. Then n\ge 1, 2a{a}_{n1}\in W and since {i}_{W}({a}_{n1})=n1, we have that f({a}_{n1})<C. By midconvexity of f, we obtain
a contradiction. □
Theorem 3.9 Let\mathrm{\varnothing}\ne W\subset X. Iff:W\to \mathbb{R}is Xmidconvex, then the maximum of f is achieved at an extreme point of W.
Proof
Consider the case when
and denote
For an indirect proof, suppose that
Since {i}_{W}(x)=0 iff x\in extW, we have then
We can find an a\in W\setminus extW such that
We choose the shortest Jchain ({a}_{0},\dots ,{a}_{n}=a) in W starting in extW and ending at a. Then 2a{a}_{n1}\in W and since {i}_{W}({a}_{n1})=n1, we have that f({a}_{n1})<C. By midconvexity of f, we obtain now
a contradiction. □
4 Construction of Xfmidconvex hull
There is a problem to find a convenient procedure to determinate an Xfmidconvex hull for a given set W\subset X. In this section, we present an algorithm for finding {fmconv}_{X}(W), which can be easily implemented in high level programming languages.
We start with the following auxiliary result.
Proposition 4.1 Let\mathrm{\varnothing}\ne W\subset Xand\overline{x}\in X. The following conditions are equivalent:

1.
\overline{x}\notin {fmconv}_{X}(W),

2.
there exists an Xmidconvex functionf:X\to \mathbb{R}such thatf{}_{W}\equiv 0andf(\overline{x})=1.
Proof Observe first that multiplying a midconvex function by a positive real number, we do not destroy midconvexity.
We show that (2) ⇒ (1). Let f:X\to \mathbb{R} be an Xmidconvex function such that f{}_{W}\equiv 0 and f(\overline{x})=1. We put
Then W\subset {W}_{1}, \overline{x}\notin {W}_{1} and {W}_{1} is Xfmidconvex. By Proposition 2.10 we obtain that {W}_{1}\cap {fmconv}_{X}(W) is Xfmidconvex. It follows from Definition 2.12 that {fmconv}_{X}(W)\subset {W}_{1}. Thus \overline{x}\notin {fmconv}_{X}(W).
For the proof of the converse implication, assume that \overline{x}\notin {fmconv}_{X}(W). As we have noticed directly after Definition 2.12 {fmconv}_{X}(W) is Xfmidconvex. Hence, there exists an Xmidconvex function g:X\to \mathbb{R} such that
Then g(\overline{x})>0. We put
By Proposition 2.3 f is Xmidconvex. Furthermore, we have
and f(\overline{x})=1. □
Two examples of Xfmidconvex hull are presented in Figure 3. X consists of all dots, W consists of big black dots, an Xfmidconvex hull of W is represented by big dots. Notice that extW=W.
Proposition 4.1 gives the way to determine {fmconv}_{X}(A). Namely, by condition (2) we can eliminate from X, step by step, elements of the set X\setminus {fmconv}_{X}(A).
To encode the result included in Proposition 4.1, we introduce some additional notations. Let X=\{{x}_{1},\dots ,{x}_{n}\}. Let W\subset X be fixed. Without loss of generality, we may assume that the elements \{{x}_{1},\dots ,{x}_{n}\} are ordered in such a way that
We denote
Observe that m:=cardK(X) is finite as m\le {n}^{2}.
Let us now fix v\in X. We want to check if there exists an Xmidconvex function f:X\to \mathbb{R} such that f{}_{W}\equiv 0 and f(v)=1.
Such a function f has to fulfill the following condition:
We enumerate the elements of K(X) and write K(X) in the form
Proposition 4.2 Letf:X\to \mathbb{R}, and lety:={[f({x}_{1}),\dots ,f({x}_{n})]}^{T}. We put
where
and({y}^{i},{z}^{i})\in K(X), i\in \{1,\dots ,m\}.
Then f is midconvex iff
where{\mathbf{0}}_{m}denotes zero in{\mathbb{R}}^{m} (we will often omit the subscript m and write 0).
Proof Observe that the system of inequalities (5) is equivalent to the following one:
for all ({y}^{i},{z}^{i})\in K(X), i\in \{1,\dots ,m\}.
Let us now fix i\in \{1,\dots ,m\} and consider the inequality corresponding to the i th element of K(X) (the i th inequality of system (7))
It is equivalent to the following one:
where {[{a}_{ij}]}_{j=1,\dots ,n} is the i th row of the matrix {A}_{f}. This finishes the proof. □
Remark 4.3 To save time and memory in practical implementation, in Proposition 4.2 we take into account only one of the pairs (y,z) and (z,y), since the inequalities f(\frac{y+z}{2})\le \frac{f(y)+f(z)}{2} and f(\frac{z+y}{2})\le \frac{f(z)+f(y)}{2} are equivalent.
Proposition 4.4 Letf:X\to \mathbb{R}be such thatf{}_{W}\equiv 0. Let{A}_{f}be a matrix defined in Proposition 4.2. We put
Then the following conditions are equivalent:

1.
f is midconvex,

2.
{A}_{W}{y}_{W}\le \mathbf{0}.
Proof Let {A}_{f}={[{A}_{f}^{i}]}_{i=1,\dots ,n}, where {A}_{f}^{i} denotes the i th column of the matrix {A}_{f}. Then the matrix {A}_{f} can be represented as the following block matrix:
where B:={[{A}_{f}^{i}]}_{i=ncardW+1,\dots ,n}\in {M}_{m\times cardW}.
Now, we use condition (4). By the condition f{}_{W}\equiv 0, we have that f({x}_{i})=0 for {x}_{i}\in X, i\in \{ncardW+1,\dots ,n\}. Thus
which can be written as the following block vector:
By Proposition 4.2, f is midconvex iff {A}_{f}y\le \mathbf{0}, which is equivalent in our situation to the following inequality:
The above inequality is equivalent to the following one:
This completes the proof since B{\mathbf{0}}_{cardW}={\mathbf{0}}_{m}. □
We illustrate the above considerations by a simple example.
Example 4.5 Let N=1, X=\{0,1,2,3,4\}, W=\{0,3\}. Then K(X)=\{(0,2),(1,3),(2,0),(3,1)\}. In this case, the system of linear inequalities (6) contains the following objects:
If we additionally want to consider only functions such that f{}_{W}\equiv 0, we have to remove the two last columns and consider the solutions of {A}_{W}{y}_{W}\le \mathbf{0} for {y}_{W}={[f(1),f(2),f(4)]}^{T} and
Making use of Remark 4.3, we may replace the matrix {A}_{W} by
Further discussion of this example can be found in Example 4.6.
The main question is still unanswered: Does there exist an Xmidconvex function f:X\to \mathbb{R} such that f{}_{W}\equiv 0 and f(v)=1? To answer this question we use linear programming (shortly LP), which is an efficient technique for finding a solution of optimization problems: to minimize or maximize a linear objective function subject to linear equality/inequality constrains [7]. It can be expressed in a canonical form:
where y represents the vector of variables (to be determined). In our case b=\mathbf{0}. The system of inequalities Ax\le b denotes the constraints which specify a convex polygon, over which the objective function is to be optimized.
We still need to ‘insert’ additional information that f(v)=1 (which is equivalent to f(v)>0, as we can divide f by f(v)). In other words, we ask if we can find a function satisfying the previously defined systems of inequalities and such that f(v)>0. To do this, we put c:={[0,\dots ,\underset{i}{\underset{\u23df}{1}},\dots ,0]}^{T}, where i\in \{1,\dots ,n\} is such that v={x}_{i}, {x}_{i}\in X (according to (4)), and find a solution to the system of inequalities defined above.
Now, we are ready to present a full algorithm for finding an Xfmidconvex hull of W\subset X:
Example 4.6
We will continue the discussion of Example 4.5. By the previous considerations, we have the following LP problem:
Firstly, we check if 1 is in the Xfmidconvex hull of the set W. We put c={[1,0,0]}^{T}. It is easy to get the following solution: f(1)=0, f(2)=0 and f(4)=a, a\in \mathbb{R}, a\ge 0 (this means that f(4) can be arbitrary, e.g., f(4)=0) this solution is bounded, so 1\in {fmconv}_{X}(W). The same consideration shows that also 2\in {fmconv}_{X}(W). On the contrary, for 4 the value f(4) is unbounded so 4\notin {fmconv}_{X}(W). Finally, we obtain that {fmconv}_{X}(W)=\{0,1,2,3\}.
There are many methods for solving LP (checking whether a solution exists is just as difficult as finding a solution): simplex algorithm of Dantzig [8], crisscross algorithm [9], projective algorithm of Karmarkar [10], etc.
However, the method for finding an Xfmidconvex hull presented in this section is not sufficiently efficient in practice because for each point of investigated space X, we solve LP. Moreover, if we use a classical simplex method, we can perform really badly since the worstcase complexity of the simplex method has exponential time.
Sample implementation of the algorithm described in this section with nice graphical interface prepared in Java programming language is available at http://www.ii.uj.edu.pl/~misztalk.
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Acknowledgements
The first author was supported by National Centre of Science (Poland) Grant No. 2011/01/B/ST6/01887.
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JaT carried out the conception studies, participated in the theorems proving and drafted the manuscript. JóT participated in its design and coordination. KM performed the algorithm implementation and description. All authors read and approved the final manuscript.
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Tabor, J., Tabor, J. & Misztal, K. Midconvexity for finite sets. J Inequal Appl 2013, 42 (2013). https://doi.org/10.1186/1029242X201342
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DOI: https://doi.org/10.1186/1029242X201342
Keywords
 midconvex set
 midconvex function
 midconvex hull