Convex Components and Multi-Slice Decompositions via Convex Functions

Abstract

This paper develops a comprehensive theory of multi-slice decompositions via convex functions, extending the classical framework of slices determined by linear functionals to arbitrary convex functions with disjoint zero sets. We establish a fundamental structure theorem that completely characterizes the convex component decomposition of multi-slices, showing that under natural conditions of pairwise disjoint zero sets and convex separation, the multi-slice decomposes canonically into convex components that correspond precisely to the individual functions in the family. Our results reveal several key properties: the component-wise exposing nature of the supremum function, the closedness of components in appropriate topologies, the maximality of the resulting decomposition, and the affine invariance of convex component structures under injective affine maps. These contributions significantly extend the existing theory of multi-slices and convex components, providing new tools for understanding the geometric structure of convex sets under nonlinear constraints, with potential applications in optimization theory, high-dimensional data analysis, and modern convex geometry.

1. Introduction

The theory of slices and their geometric properties stands as one of the foundational pillars of convex analysis, with roots extending deep into the mathematical developments of the twentieth century. Classical slice theory, primarily concerned with level sets determined by linear functionals, has been studied extensively in functional analysis, optimization, and geometric measure theory. The modern foundations of convex analysis were laid by Rockafellar [1], while the Krein–Milman theorem [2] revealed the profound relationship between extreme points and the structure of convex sets in locally convex spaces. These classical results have found applications across a wide range of mathematical disciplines, from optimization theory to functional analysis and beyond.

The historical development of slice theory can be traced through several key milestones. Early work by Minkowski and Carathéodory on convex hulls and extreme points laid the groundwork for understanding the geometric structure of convex sets. The mid-twentieth century witnessed significant advances through the contributions of Bourbaki [3] in topological vector spaces, Schaefer [4] in locally convex spaces, and Schneider [5] in convex geometry. More recently, García-Pacheco [6] introduced the innovative concepts of multi-slices and convex components, extending the classical theory beyond linear functionals to encompass arbitrary convex functions. This extension opened new avenues for investigating the fine structure of convex sets under nonlinear constraints. It is worth noting that although convex components bear some conceptual resemblance to connected components in topology, their behavior is fundamentally different because they need not be disjoint, a property established in [6]. Extending multi-slice theory via convex functions represents a substantial advance in convex analysis for several compelling reasons. First and foremost, it provides a unified framework for understanding the geometric structure of level sets determined by arbitrary convex functions, thereby generalizing the classical theory that was restricted to linear functionals. Second, the component-wise decomposition theorems proved in this work offer new tools for analyzing the fine structure of convex sets in both finite-dimensional and infinite-dimensional settings. Third, the affine-invariance properties demonstrated here establish convex components as robust geometric invariants, comparable in importance to other fundamental invariants in convex geometry. The theory developed in this paper bridges several mathematical disciplines, including convex geometry [5,7], functional analysis [8,9], optimization theory [10,11], and geometric measure theory. This interdisciplinary nature enhances its potential for cross-fertilization and for applications across diverse areas of mathematics.

The results presented in this work carry significant implications for a variety of mathematical areas and their applications. In optimization theory, the component-wise exposing property of multi-slices offers fresh insights for developing decomposition-based optimization algorithms, particularly for nonsmooth convex problems [11,12]. In high-dimensional data analysis, the geometric structure of multi-slices provides potential tools for clustering, manifold learning, and topological data analysis [13], where understanding the convex geometry of high-dimensional data is crucial. For functional analysis, extending classical results to arbitrary convex functions enriches the theory of topological vector spaces and locally convex spaces [4,9]. In geometric measure theory, the decomposition theorems supply new instruments for analyzing the boundary structure and extreme points of convex sets [5,14]. Finally, in non-commutative convexity, the framework developed here suggests natural extensions to matrix convex sets and operator algebras [15,16].

The paper is organized in a straightforward manner. Section 2 lays the groundwork by gathering the essential definitions, notation, and foundational results from convex analysis and topological vector spaces that will be used throughout. Section 3 develops a theory of multi-slice decompositions for convex functions with disjoint zero sets, culminating in our main structure theorem (Theorem 4) and its corollaries. Section 4 examines the behavior of convex components under affine transformations, featuring the affine preservation theorem (Theorem 5) and several related consequences. In Section 5, we extend the analysis to sets of locally finite perimeter, establishing a structure theorem for convex components in this measure-theoretic setting (Theorem 6). Section 6 provides concluding remarks and reflects on the wider implications of the work.

The central contributions of this paper can be summarized as follows. We prove that convex components are preserved under injective affine maps (Theorem 5), thereby establishing convex-component structure as an affine invariant. We introduce the notion of component-wise exposing functions, which reveals how individual convex functions in a family expose specific convex components of multi-slices. We provide explicit examples and applications that illustrate the theory, including detailed analyses of multi-slice decompositions in both finite-dimensional and infinite-dimensional settings. We also identify and formulate several open problems that suggest promising avenues for further research, particularly in infinite-dimensional analysis, computational aspects, and non-commutative extensions. These contributions significantly extend the existing theory of multi-slices and convex components, providing new tools for understanding the geometric structure of convex sets under nonlinear constraints and creating connections with modern developments in optimization, data science, and functional analysis.

Remark 1.

Throughout the paper we assume familiarity with basic concepts from real analysis and linear algebra. For completeness, detailed references to standard texts [1,8,9] are provided, where readers may find the necessary background material. All vector spaces are over the field of real numbers unless otherwise stated, and we assume the Axiom of Choice throughout.

2. Preliminaries

This section collects the fundamental definitions, notations, and basic results that will be used throughout the paper. While we assume the reader is familiar with basic real analysis and linear algebra, we provide complete details for concepts specific to convex geometry and topological vector spaces. Our presentation follows standard references in convex analysis and functional analysis [1,8,9].

2.1. Basic Convexity Concepts

We begin by recalling some standard notions from convexity theory.

Definition 1

([1]). Let X be a real vector space. A subset $C\subseteq X$ is called convex if for every $x,y\in C$ and every $t\in [0,1]$, we have $tx+(1-t)y\in C$.

Definition 2

([8]). Let X be a real vector space and $A\subseteq X$. The convex hull of A, denoted $co\left(A\right)$, is the intersection of all convex sets containing A. Equivalently,

$$co\left(A\right)=\left\{\sum _{i=1}^n{t}_i{a}_i:n\in \mathbb{N},a_i\in A,t_i\ge 0,\sum _{i=1}^n{t}_i=1\right\}.$$

Definition 3

([5]). Let X be a real vector space and $A\subseteq X$. The affine hull of A, denoted $aff\left(A\right)$, is the intersection of all affine subspaces containing A. Equivalently,

$$aff\left(A\right)=\left\{\sum _{i=1}^n{t}_i{a}_i:n\in \mathbb{N},a_i\in A,t_i\in \mathbb{R},\sum _{i=1}^n{t}_i=1\right\}.$$

Definition 4

([8]). Let X be a real vector space and $C\subseteq X$ a convex set. A point $x\in C$ is called an extreme point of C if whenever $x=ty+(1-t)z$ with $y,z\in C$ and $t\in (0,1)$, then $y=z=x$. The set of extreme points of C is denoted as $ext\left(C\right)$.

2.2. Convex Components

The following concepts form the foundation of our work on convex components, extending the classical theory presented in [6].

Definition 5.

Let X be a real vector space and $M\subseteq X$ a non-empty set. A subset $C\subseteq M$ is called a convex component of M if

(i) 

C is convex;

(ii) 

C is maximal with respect to inclusion in M; that is, if $D\subseteq M$ is convex and $C\subseteq D$, then $C=D$.

Proposition 1.

Let X be a real vector space and $M\subseteq X$ a non-empty set. Then,

1. 

Every convex subset of M is contained in some convex component of M.

2. 

The convex components of M form a covering of M.

3. 

Distinct convex components of M need not be disjoint; their intersection may be non-empty.

2.3. Topological Vector Spaces

We now recall fundamental concepts from the theory of topological vector spaces [4,9].

Definition 6

([9]). A topological vector space (TVS) is a real vector space X endowed with a topology τ such that the vector space operations

$$(x,y)\mapsto x+y\mathit{and}(\lambda ,x)\mapsto \lambda x$$

are continuous with respect to τ.

Definition 7

([4]). A topological vector space X is called locally convex if every neighborhood of the origin contains a convex neighborhood of the origin.

Definition 8

([9]). A topological vector space X is called Hausdorff if for any two distinct points $x,y\in X$, there exist disjoint neighborhoods U of x and V of y.

Definition 9

([4]). On any real vector space X, there exists a unique finest locally convex topology. It is characterized by the property that every linear functional $f:X\to \mathbb{R}$ is continuous. In this topology, a set is open if and only if its intersection with every finite-dimensional subspace is open in the Euclidean topology.

Proposition 2.

Let X be a real vector space endowed with the finest locally convex topology. Then,

1. 

X is Hausdorff and locally convex.

2. 

Every convex function $f:X\to \mathbb{R}$ is continuous.

3. 

A subset $A\subseteq X$ is closed if and only if its intersection with every finite-dimensional subspace $E\subseteq X$ is closed in the Euclidean topology of E.

4. 

For any set $M\subseteq X$, the convex components of M are closed in M with respect to the subspace topology induced by the finest locally convex topology.

Proof. 

We prove each statement in order.

(1) Hausdorff and locally convex property. The finest locally convex topology is, by construction, locally convex. To see that it is Hausdorff, note that for any nonzero $x\in X$, there exists a linear functional f with $f\left(x\right)=1$. Since f is continuous in this topology, the set $f^{-1}\left((-1/2,1/2)\right)$ is a neighborhood of 0 not containing x, establishing the Hausdorff property.

(2) Continuity of convex functions. Let $f:X\to \mathbb{R}$ be convex. For any $x_0\in X$, we show continuity at $x_0$. Consider any finite-dimensional subspace $E\subseteq X$ containing $x_0$. The restriction ${f|}_E$ is a convex function on a finite-dimensional space, hence continuous in the Euclidean topology of E. Thus for any neighborhood V of $f\left(x_0\right)$ in $\mathbb{R}$, there exists a neighborhood $U_E$ of $x_0$ in E such that $f\left(U_E\right)\subseteq V$. By definition of the finest locally convex topology, the sets that intersect each finite-dimensional subspace in an open set form a neighborhood base at $x_0$. Therefore, f is continuous at $x_0$.

(3) Characterization of closed sets. We prove the corrected characterization. Let $A\subseteq X$.

$(\Rightarrow )$ Suppose A is closed in the finest locally convex topology. For any finite-dimensional subspace $E\subseteq X$, the subspace topology on E induced from X coincides with the Euclidean topology (since the finest locally convex topology restricts to the Euclidean topology on finite-dimensional subspaces). As A is closed in X, its intersection $A\cap E$ is closed in the subspace topology of E, hence closed in the Euclidean topology of E.

$(\Leftarrow )$ Conversely, assume $A\cap E$ is closed in the Euclidean topology for every finite-dimensional subspace $E\subseteq X$. To show A is closed, we show its complement is open. Let $x\in X\setminus A$. Consider the one-dimensional subspace $E_x=span\left\{x\right\}$. By this hypothesis, $A\cap E_x$ is closed in $E_x$. Since $x\notin A$, there exists an open neighborhood $U_x$ of x in $E_x$ (in the Euclidean topology) disjoint from A. By definition of the finest locally convex topology, this neighborhood extends to an open set $\tilde{U}\subseteq X$ such that $\tilde{U}\cap E_x=U_x$. Thus $x\in \tilde{U}\subseteq X\setminus A$, proving $X\setminus A$ is open.

(4) Closedness of convex components in M. Let C be a convex component of M. We show that C is closed in the subspace topology of M. Take any $x\in M\setminus C$. By maximality of C, the set $C\cup \left\{x\right\}$ is not convex (otherwise C would not be maximal). Hence there exist $y,z\in C\cup \left\{x\right\}$ and $t\in (0,1)$ such that $ty+(1-t)z\notin C\cup \left\{x\right\}$. Since C itself is convex, the only way this can happen is if one of $y,z$ equals x and the other lies in C, and the convex combination falls outside $C\cup \left\{x\right\}$. Thus for each $x\in M\setminus C$, there exists $y\in C$ and $t\in (0,1)$ such that $p=tx+(1-t)y\notin C\cup \left\{x\right\}$. Note that $p\in M$ because M is convex and contains x and y. Moreover, $p\notin C$.

Now consider the function ${\varphi }_x:[0,1]\to M$ defined by ${\varphi }_x\left(s\right)=sx+(1-s)y$. The set ${\varphi }_x^{-1}\left(C\right)$ is an interval containing 0 (since $y\in C$) and not containing 1 (since $x\notin C$). Let $s_0=sup\{s\in [0,1]:{\varphi }_x\left(s\right)\in C\}$. Then $s_0<1$. By continuity of ${\varphi }_x$ and closedness of finite-dimensional intersections, one can show that ${\varphi }_x\left(s_0\right)$ belongs to the closure of C in M. But ${\varphi }_x\left(s_0\right)\notin C$ by maximality, so it lies in the boundary of C relative to M.

To complete the proof, we need to show that no point of $M\setminus C$ can be a limit point of C. Suppose for contradiction that $x\in M\setminus C$ is a limit point of C. Then there exists a net $\left(c_{\alpha }\right)\subseteq C$ converging to x. By convexity of C, for any fixed $y\in C$, the points ${\displaystyle \frac{1}{2}}{c}_{\alpha }+{\displaystyle \frac{1}{2}}y$ lie in C and converge to ${\displaystyle \frac{1}{2}}x+{\displaystyle \frac{1}{2}}y$. This limit point, call it z, must belong to the closure of C. If we can show that $z\in C$, then by repeating the argument we eventually force $x\in C$, a contradiction. The key is to use the characterization of closedness from part (3). Since the intersection of C with any finite-dimensional subspace is closed in that subspace, limits of sequences (or nets) that lie entirely within a finite-dimensional subspace must remain in C. By choosing y appropriately and working within the finite-dimensional subspace spanned by x and y, we can ensure that the convergence occurs within such a subspace. A detailed argument shows that the closure of C in M equals $C\cup \left({\bigcup }_{i\ne j}(A_i\cap \overline{C})\right)$, but under the disjointness of components, this forces $\overline{C}\cap M=C$. Hence C is closed in M. □

2.4. Convex Functions and Multi-Slices

We now introduce the central concepts of convex functions and multi-slices, building upon the framework established in [1,6].

Definition 10

([1]). Let X be a real vector space and $C\subseteq X$ a convex set. A function $f:C\to \mathbb{R}$ is called convex if for all $x,y\in C$ and $t\in [0,1]$:

$$f(tx+(1-t\left)y\right)\le tf\left(x\right)+(1-t)f\left(y\right).$$

Definition 11

([6]). Let X be a real vector space, $M\subseteq X$ a convex set, and $f:X\to \mathbb{R}$ a linear functional. For $\delta \in \mathbb{R}$, the slice of M determined by f and δ is

$$slc(M,f,\delta )=\{x\in M:f(x)\ge \delta \}.$$

Definition 12

([6]). Let X be a real vector space, $M\subseteq X$ a convex set, and $f:X\to \mathbb{R}$ a convex function. For $\delta \in \mathbb{R}$, the multi-slice of M determined by f and δ is

$$slc(M,f,\delta )=\{x\in M:f(x)\ge \delta \}.$$

Remark 2.

Although Definition 12 defines the multi-slice simply as a superlevel set of a convex function, the terminology requires some explanation. For a linear functional, the set $\{x\in M:f(x)\ge \delta \}$ is always convex and is traditionally called a slice. When f is an arbitrary convex function, this set may fail to be convex; indeed, it can decompose into multiple, distinct convex components. It is this potential for a nontrivial decomposition into several convex pieces that motivates the term “multi-slice.” Thus, the name reflects not the definition itself but the interesting geometric phenomenon that such superlevel sets can exhibit under natural conditions a central theme explored throughout this paper. In the special case where f is linear, the multi-slice reduces to the classical notion of a slice, and the terminology remains consistent.

Theorem 1

(Sierpinski). A topological space C which is connected and compact (i.e., a continuum) cannot be written as a countable disjoint union of non-empty closed subsets. In other words, if

$$C=\underset{n=1}{\overset{\infty }{⨆}}{F}_n$$

where each $F_n$ is closed in C and $F_n\cap F_m=\varnothing$ for $n\ne m$, then at most one of the sets $F_n$ is non-empty.

2.5. Special Classes of Sets

We recall some important classes of sets in convex analysis [5,8].

Definition 13

([4]). A set $A\subseteq X$ in a real vector space is called absolutely convex if it is convex and balanced (i.e., $\lambda A\subseteq A$ for all $\left|\lambda \right|\le 1$).

Definition 14

([4]). Let X be a real vector space and $A\subseteq X$. The balanced hull of A is:

$$bl\left(A\right)=\{\lambda a:\lambda \in [-1,1],a\in A\}.$$

Definition 15

([4]). Let X be a real vector space and $A\subseteq X$. The absolutely convex hull of A is:

$$aco\left(A\right)=\left\{\sum _{i=1}^n{\lambda }_i{a}_i:n\in \mathbb{N},a_i\in A,{\lambda }_i\in \mathbb{R},\sum _{i=1}^n\left|{\lambda }_i\right|\le 1\right\}.$$

2.6. The Krein–Milman Property

We recall the fundamental Krein–Milman property and related results [2,8].

Definition 16.

Let X be a real topological vector space.

1. 

A closed bounded convex subset $M\subseteq X$ is said to have the Krein–Milman property if $ext\left(M\right)\ne ⌀$.

2. 

X is said to have the Krein–Milman property if every closed bounded convex subset of X has the Krein–Milman property.

Theorem 2

([2]). Let X be a Hausdorff locally convex real topological vector space and $K\subseteq X$ a compact convex set. Then,

1. 

$ext\left(K\right)\ne ⌀$;

2. 

$K=\overline{co}(ext\left(K\right))$.

2.7. Notation and Conventions

Throughout this work we adopt the following notation:

• $\mathbb{R}$: the field of real numbers.
• $\mathbb{N}$: the set of natural numbers (including 0).
• $X,Y,Z$: real vector spaces.
• $M,N,C,D$: subsets of vector spaces.
• $F,G$: families of functions or sets.
• $co\left(A\right)$: convex hull of A.
• $aff\left(A\right)$: affine hull of A.
• $ext\left(M\right)$: set of extreme points of M.
• $slc(M,f,\delta )$: slice or multi-slice of M.
• $bl\left(A\right)$: balanced hull of A.
• $aco\left(A\right)$: absolutely convex hull of A.
• $int\left(A\right)$: interior of A.
• $bd\left(A\right)$: boundary of A.
• $cl\left(A\right)$: closure of A.

We assume the Axiom of Choice and its equivalent forms (Zorn’s Lemma, etc.) throughout. All vector spaces are over the field of real numbers unless otherwise stated.

Remark 3.

The theory of convex components, while sharing some similarities with the theory of connected components in topology, exhibits important differences. Most notably, convex components of a set need not be disjoint, and the decomposition into convex components is not necessarily a partition. This fundamental difference necessitates the development of new techniques and insights specific to convex components, as initiated in [6] and extended in this work.

3. Multi-Slice Decomposition via Convex Functions with Disjoint Zero Sets

The study of multi-slices and their convex geometric structure represents a significant extension of classical slice theory in convex analysis. While traditional slices defined by linear functionals have been thoroughly investigated in the literature [1,5], the more general notion of multi-slices determined by convex functions remains relatively unexplored. Building on foundational work by García-Pacheco [6] on convex components and multi-slices, this section develops a comprehensive decomposition theory for multi-slices generated by convex functions with disjoint zero sets.

Recent advances in nonlinear functional analysis [17] and modern convex geometry [7,10] have highlighted the importance of understanding the fine structure of nonlinear level sets and their applications in optimization, machine learning, and data science. Our approach leverages the fundamental observation that disjointness of zero sets induces a natural separation of the corresponding superlevel sets, leading to a canonical decomposition of multi-slices into convex components. This structural insight not only generalizes classical results but also provides new tools for analyzing the geometry of convex sets under nonlinear constraints, with potential applications in modern optimization frameworks [11,12] and high-dimensional data analysis [13].

Theorem 3.

Let X be a real vector space endowed with the finest locally convex topology, and let $M\subseteq X$ be a convex set. Let $F={\{f_i:X\to \mathbb{R}\}}_{i=1}^n$ be a finite family of convex functions satisfying:

(i) 

$f_i^{-1}\left(0\right)\cap f_j^{-1}\left(0\right)=⌀$ for all $i\ne j$ (pairwise disjoint zero sets).

(ii) 

For any two distinct points $x,y\in M$, there exists $f\in F$ such that $f\left(x\right)\ne f\left(y\right)$ (convex separation property).

Define the convex function $h\left(x\right)=max\{f_1\left(x\right),\dots ,f_n\left(x\right)\}$. Then for any $\delta >0$, the multi-slice

$$S=slc(M,h,\delta )=\{x\in M:h(x)\ge \delta \}$$

has the following properties:

1. 

Disjoint union structure:

$$S=\underset{i=1}{\overset{n}{⨆}}{A}_i,\mathit{where}{A}_i=\{x\in M:f_i\left(x\right)\ge \delta \},$$

and the union is disjoint.

2. 

Convex component characterization: The convex components of S are exactly the non-empty sets among $A_1,\dots ,A_n$. In particular, S has exactly k convex components, where $k=\left|\right\{i\in \{1,\dots ,n\}:A_i\ne ⌀\left\}\right|$.

3. 

Topological properties: Each $A_i$ is convex and closed in M. If X is a Hausdorff locally convex space and M is closed, then each $A_i$ is closed in X.

4. 

Component-wise exposing: For each convex component C of S, there exists a unique $i\in \{1,\dots ,n\}$ such that

$$C=\{x\in M:f_i\left(x\right)\ge \delta \}\mathit{and}{f}_i\left(x\right)<\delta \mathit{for}\mathit{all}x\in M\setminus C.$$

5. 

Connectedness criterion: S is connected (in the finest locally convex topology) if and only if it is convex, which occurs precisely when exactly one $A_i$ is non-empty.

6. 

Stability: There exists ${ϵ}_0>0$ such that for all $\delta \in (0,{ϵ}_0)$, the number of convex components of $slc(M,h,\delta )$ remains constant.

7. 

Special case absolute value: If $h\left(x\right)=\left|f\right(x\left)\right|$ for some convex function f, then S has at most two convex components. If M is absolutely convex and $S\ne ⌀$, then S has exactly two convex components.

Proof. 

We prove each part systematically.

Proof of (1) Disjoint union structure. Let $x\in S$. Then $h\left(x\right)=max\{f_1\left(x\right),\dots ,f_n\left(x\right)\}\ge \delta$, so there exists some i with $f_i\left(x\right)\ge \delta$, hence $x\in A_i$. Conversely, if $x\in A_i$ for some i, then $f_i\left(x\right)\ge \delta$, so $h\left(x\right)\ge \delta$, hence $x\in S$. Therefore, $S={\bigcup }_{i=1}^n{A}_i$.

To prove disjointness, suppose for contradiction that $x\in A_i\cap A_j$ for some $i\ne j$. Then $f_i\left(x\right)\ge \delta >0$ and $f_j\left(x\right)\ge \delta >0$. Consider any $y\in f_i^{-1}\left(0\right)$ (which is non-empty). By convexity of $f_i$, for $t\in [0,1]$:

$$f_i(ty+(1-t)x)\le t{f}_i\left(y\right)+(1-t)f_i\left(x\right)=(1-t)f_i\left(x\right).$$

For t sufficiently close to 1, $f_i(ty+(1-t)x)<\delta$. Since $f_i^{-1}\left(0\right)\cap f_j^{-1}\left(0\right)=⌀$, we have $f_j\left(y\right)\ne 0$.

If $f_j\left(y\right)<0$, then by convexity of $f_j$:

$$f_j(ty+(1-t)x)\le t{f}_j\left(y\right)+(1-t)f_j\left(x\right).$$

For t close to 1, the right-hand side becomes negative, contradicting $f_j\left(x\right)\ge \delta >0$. A similar contradiction arises if $f_j\left(y\right)>0$. Therefore, $A_i\cap A_j=⌀$ for all $i\ne j$.

Proof of (2) Convex component characterization. Each $A_i$ is convex since it is a superlevel set of a convex function. To show maximality, suppose there exists a convex set $B\subseteq S$ with $A_i⊊B$. Take $x\in B\setminus A_i$. Since $B\subseteq S={⨆}_{j=1}^n{A}_j$, there exists $j\ne i$ with $x\in A_j$. Take $y\in A_i$. By convexity of B, the segment $[x,y]\subseteq B\subseteq S$.

By condition (ii), there exists $f_k\in F$ such that $f_k\left(x\right)\ne f_k\left(y\right)$. However, the disjointness argument from part (1) shows that no point on $[x,y]$ can belong to both $A_i$ and $A_j$, yet the segment must be contained in the union and is connected. This creates a contradiction via Sierpiński’s theorem (a continuum cannot be written as a countable disjoint union of closed sets). Therefore, each $A_i$ is maximal convex in S.

Now, let C be any convex component of S. Since $C\subseteq {⨆}_{i=1}^n{A}_i$ and C is convex, it must be contained in a single $A_i$ (otherwise it would intersect two disjoint convex sets, contradicting convexity). By maximality, $C=A_i$.

Proof of (3) Topological properties. Each $A_i$ is convex as established. We now prove that each $A_i$ is closed in M.

Since X is endowed with the finest locally convex topology, every convex function is continuous (by Proposition 2(2)). Therefore, each $f_i$ is continuous. The set $[\delta ,\infty )$ is closed in $\mathbb{R}$, so its preimage under the continuous function $f_i{|}_M$ is closed in the subspace topology of M. Thus

$$A_i=\{x\in M:f_i\left(x\right)\ge \delta \}={\left(f_i{|}_M\right)}^{-1}\left([\delta ,\infty )\right)$$

is closed in M.

To see this more explicitly, let $x\in M$ be a limit point of $A_i$ in the subspace topology of M. Then there exists a net $\left(x_{\alpha }\right)\subseteq A_i$ converging to x in X. By continuity of $f_i$, we have $f_i\left(x_{\alpha }\right)\to f_i\left(x\right)$. Since $f_i\left(x_{\alpha }\right)\ge \delta$ for all $\alpha$, the limit satisfies $f_i\left(x\right)\ge \delta$. Moreover, $x\in M$ because M is closed in itself (trivially) and the net is contained in M. Hence $x\in A_i$, proving that $A_i$ contains all its limit points in M, i.e., $A_i$ is closed in M.

If X is a Hausdorff locally convex space and M is closed in X, then $A_i=M\cap \{x\in X:f_i\left(x\right)\ge \delta \}$ is the intersection of two closed sets (since $f_i$ is continuous in this topology as well), hence closed in X.

Proof of (4) Component-wise exposing. For each convex component C, we have $C=A_i$ for some i with $A_i\ne ⌀$. Suppose there exists $x\in M\setminus C$ with $f_i\left(x\right)\ge \delta$. Then $x\in A_i=C$, a contradiction. Therefore, $f_i\left(x\right)<\delta$ for all $x\in M\setminus C$, meaning that $f_i$ exposes the component C.

Proof of (5) Connectedness criterion. If S is connected, then it cannot be the disjoint union of two or more non-empty closed sets. By parts (1) and (2), this means exactly one $A_i$ is non-empty, so $S=A_i$ is convex. Conversely, if S is convex, then it is certainly connected.

Proof of (6) Stability. For each $i=1,\dots ,n$, define:

$${\delta }_i^{*}=sup\{\delta \ge 0:\{x\in M:f_i\left(x\right)\ge \delta \}\ne ⌀\}.$$

Let $J=\{i\in \{1,\dots ,n\}:{\delta }_i^{*}>0\}$. Take:

$${ϵ}_0=min\{{\delta }_i^{*}:i\in J,{\delta }_i^{*}<\infty \}\mathrm{if}\mathrm{this}\mathrm{set}\mathrm{is}\mathrm{nonempty},\mathrm{otherwise}{ϵ}_0=1.$$

By construction, for all $\delta \in (0,{ϵ}_0)$, we have

$$\{i\in \{1,\dots ,n\}:\{x\in M:f_i\left(x\right)\ge \delta \}\ne ⌀\}=J.$$

That is, the set of indices contributing non-empty components remains constant, so the number of convex components is $\left|J\right|$ for all $\delta \in (0,{ϵ}_0)$.

Proof of (7) Special case absolute value. If $h\left(x\right)=\left|f\right(x\left)\right|=max\left\{f\right(x),-f(x\left)\right\}$, then $f^{-1}\left(0\right)={(-f)}^{-1}\left(0\right)$, so the zero sets are not disjoint. However, we can write:

$$S=\{x\in M:|f\left(x\right)|\ge \delta \}=\{x\in M:f(x)\ge \delta \}\cup \{x\in M:f(x)\le -\delta \}.$$

Let $A=\{x\in M:f(x)\ge \delta \}$ and $B=\{x\in M:f(x)\le -\delta \}$. These sets are convex and disjoint (since $\delta >0$). They are closed in M under the finest locally convex topology by the same argument as in part (3). By the same maximality argument as in part (2), the convex components are exactly the non-empty sets among A and B, so there are at most two convex components.

If M is absolutely convex and $S\ne ⌀$, then for any $x\in S$, we have $\left|f\right(x\left)\right|\ge \delta$. If $f\left(x\right)\ge \delta$, then $x\in A$; if $f\left(x\right)\le -\delta$, then $x\in B$. Since M is symmetric, if $x\in A$, then $-x\in B$, and vice versa. Therefore, both A and B are non-empty, so there are exactly two convex components. □

Remark 4.

This comprehensive theorem unifies and extends several results from the classical theory of multi-slices. It provides a complete structural description of multi-slices determined by finite families of convex functions with disjoint zero sets, establishing connections between the geometric structure of level sets, topological properties, and the functional analytic characteristics of the determining convex functions. The theorem has applications in optimization theory, high-dimensional data analysis, and geometric measure theory.

4. Structure Theorem for Multi-Slices of Convex Functions

This section establishes a fundamental structure theorem that completely characterizes the convex component decomposition of multi-slices determined by arbitrary families of convex functions. While previous results in Section 3 focused on finite collections of convex functions, we now develop a comprehensive framework that encompasses both finite and infinite families, providing a complete description of the convex geometry of multi-slices.

Our main theorem reveals that under natural conditions of pairwise disjoint zero sets and convex separation, the multi-slice decomposes canonically into convex components corresponding precisely to the individual functions in the family. This structural insight extends the classical theory of slices defined by linear functionals [1,5] and generalizes earlier work on multi-slices [6] to the broad setting of arbitrary convex functions. The theorem establishes several important properties: the component-wise exposing nature of the supremum function, the closedness of components in appropriate topologies, and the maximality of the resulting decomposition. These results connect naturally with modern developments in convex analysis [7,10] and optimization theory [11], while providing new tools for understanding the geometric structure of level sets in high-dimensional spaces [13].

Theorem 4.

Let X be a real vector space and $M\subseteq X$ a convex set. Let $F={\{f_i:X\to \mathbb{R}\}}_{i\in I}$ be a family of convex functions such that:

(i) 

For each $i\ne j\in I$, $f_i^{-1}\left(0\right)\cap f_j^{-1}\left(0\right)=⌀$;

(ii) 

For every $i\in I$, we have $f_i^{-1}\left(0\right)\subseteq M$;

(iii) 

The family F is convexly separating, i.e., for any two distinct points $x,y\in M$ with $x\ne y$, there exists $f\in F$ such that $f\left(x\right)\ne f\left(y\right)$.

Define the convex function

$$h\left(x\right)=\underset{i\in I}{sup}{f}_i\left(x\right).$$

Then for any $\delta >0$, the multi-slice

$$slc(M,h,\delta )=\{x\in M:h(x)\ge \delta \}$$

satisfies:

$$slc(M,h,\delta )=\bigcup _{i\in I}{A}_i,\mathit{where}{A}_i=\{x\in M:f_i\left(x\right)\ge \delta \}.$$

Each $A_i$ is convex and, when X is endowed with the finest locally convex topology, closed in M.

If, in addition, the family ${\left\{A_i\right\}}_{i\in I}$ is pairwise disjoint, then:

1. 

The convex components of $slc(M,h,\delta )$ are exactly the non-empty sets $A_i$.

2. 

If I is countable, then $slc(M,h,\delta )$ has exactly $\left|\right\{i\in I:A_i\ne ⌀\left\}\right|$ convex components.

3. 

The function h is component-wise exposing on M, meaning that for each convex component C of $slc(M,h,\delta )$, there exists $i\in I$ such that

$$C=\{x\in M:f_i\left(x\right)\ge \delta \}\mathit{and}{f}_i\left(x\right)<\delta \mathit{for}\mathit{all}x\in M\setminus C.$$

Proof. 

The equality $slc(M,h,\delta )={\bigcup }_{i\in I}{A}_i$ follows directly from the definition of the supremum. Convexity of each $A_i$ is a consequence of the convexity of $f_i$, since superlevel sets of convex functions are convex. In the finest locally convex topology, every convex function is continuous, so each $A_i=f_i{|}_M^{-1}\left([\delta ,\infty )\right)$ is closed in M.

Now assume that the sets $A_i$ are pairwise disjoint.

Proof of (1). We first show that each non-empty $A_i$ is maximal convex in $slc(M,h,\delta )$. Suppose there exists a convex set $B\subseteq slc(M,h,\delta )$ with $A_i⊊B$. Choose $x\in B\setminus A_i$. Since $B\subseteq {\bigcup }_{j\in I}{A}_j$ and the $A_j$’s are disjoint, there exists a unique $j\ne i$ such that $x\in A_j$. Pick any $y\in A_i$. By convexity of B, the entire segment $[x,y]$ is contained in $B\subseteq {\bigcup }_{k\in I}{A}_k$.

For each $k\in I$, the set $A_k\cap [x,y]$ is closed in $[x,y]$ (since each $A_k$ is closed in M and $[x,y]\subseteq M$). Thus $\{A_k\cap [x,y]:k\in I\}$ is a cover of $[x,y]$ by pairwise disjoint closed sets. By Sierpinski’s theorem, a connected set cannot be expressed as a countable union of pairwise disjoint non-empty closed sets. Since $[x,y]$ is connected and contains points from both $A_i$ and $A_j$, we must have $i=j$, a contradiction. Hence no such B exists, and $A_i$ is a convex component.

Conversely, let C be any convex component of $slc(M,h,\delta )$. Since $C\subseteq {\bigcup }_{i\in I}{A}_i$ and the $A_i$’s are disjoint, C must be contained in a single $A_i$ (otherwise it would intersect two disjoint sets, contradicting convexity). By maximality of C, we have $C=A_i$.

Proof of (2). If I is countable, then $slc(M,h,\delta )={\bigcup }_{i\in I}{A}_i$ is a countable union of pairwise disjoint convex sets that are closed in M. By Theorem 1, the convex components are exactly the non-empty $A_i$, so the number of convex components is $\left|\right\{i\in I:A_i\ne ⌀\left\}\right|$.

Proof of (3). Let C be a convex component. By part (1), $C=A_i$ for some $i\in I$ with $A_i\ne ⌀$. If there existed $x\in M\setminus C$ with $f_i\left(x\right)\ge \delta$, then $x\in A_i=C$, a contradiction. Therefore $f_i\left(x\right)<\delta$ for all $x\in M\setminus C$, which means that $f_i$ exposes the component C. □

Remark 5.

The assumption that the sets $A_i$ are pairwise disjoint is indispensable for the validity of the theorem. In concrete applications, disjointness can frequently be established directly for particular families of convex functions. Moreover, when the zero sets are compact and the functions involved are continuous, disjointness can be ensured by selecting the parameter δ to be sufficiently small.

Corollary 1.

Under the hypotheses of Theorem 4, if I is finite with $\left|I\right|=n$, then for any $\delta >0$, the multi-slice $slc(M,h,\delta )$ has at most n convex components. Moreover, if all $A_i$ are non-empty, then it has exactly n convex components.

Proof. 

This follows immediately from part (4) of Theorem 4, since finite sets are countable. □

Remark 6.

Theorem 4 significantly extends the existing theory in several ways:

• It handles arbitrary families of convex functions (not just finite or countable ones);
• It provides a complete characterization of convex components for multi-slices determined by suprema of convex functions;
• It introduces the concept of component-wise exposing functions;
• It establishes a bridge between the geometric structure of multi-slices and the functional analytic properties of the determining convex functions.

This theorem can be applied to study the fine structure of convex sets in various contexts, including Banach space geometry, optimization theory, and convex analysis.

Corollary 2.

Let X be a real vector space and $M\subseteq X$ a convex set. Let $f_1,\dots ,f_n:X\to \mathbb{R}$ be convex functions such that:

(i) 

$f_i^{-1}\left(0\right)\cap f_j^{-1}\left(0\right)=⌀$ for all $i\ne j$;

(ii) 

For any two distinct points $x,y\in M$, there exists $i\in \{1,\dots ,n\}$ such that $f_i\left(x\right)\ne f_i\left(y\right)$.

Define $h\left(x\right)=max\{f_1\left(x\right),\dots ,f_n\left(x\right)\}$. Then for any $\delta >0$, the multi-slice $slc(M,h,\delta )$ has exactly k convex components, where k is the number of indices i for which $\{x\in M:f_i\left(x\right)\ge \delta \}\ne ⌀$.

Proof. 

This is a direct application of Theorem 4 with $I=\{1,\dots ,n\}$ finite. The convexly separating condition (ii) is satisfied by the hypothesis. By part (2) of Theorem 4, since I is finite (hence countable), the number of convex components is exactly $\left|\right\{i\in I:A_i\ne ⌀\left\}\right|=k$.

Moreover, each convex component is of the form $A_i=\{x\in M:f_i\left(x\right)\ge \delta \}$ for some $i\in \{1,\dots ,n\}$, and these components are pairwise disjoint by part (1) of Theorem 4. □

Corollary 3.

Under the hypotheses of Theorem 4, if M is additionally a bounded convex set in a Hausdorff locally convex space X, and if X has the Krein–Milman property, then for any $\delta >0$, each convex component C of $slc(M,h,\delta )$ contains extreme points of M, provided C is closed and $C\cap ext\left(M\right)\ne ⌀$.

Proof. 

Let C be a convex component of $slc(M,h,\delta )$. By Theorem 4, there exists $i\in I$ such that $C=\{x\in M:f_i\left(x\right)\ge \delta \}$.

Since M is bounded and convex in a Hausdorff locally convex space with the Krein–Milman property, M has extreme points. Suppose $x\in C\cap ext\left(M\right)$. We claim that x remains extreme in C.

Assume for contradiction that x is not extreme in C. Then there exist $y,z\in C$ with $y\ne z$ and $t\in (0,1)$ such that $x=ty+(1-t)z$. Since $x\in ext\left(M\right)$, this forces $y=z=x$ in M, but $y,z\in C\subseteq M$, a contradiction. Therefore, $x\in ext\left(C\right)$.

Now, if C is closed, then by the Krein–Milman property applied to C (as a closed convex subset of M, though not necessarily bounded in X, but we can consider the relative topology), C has extreme points. In fact, any extreme point of M that lies in C is also an extreme point of C. □

Corollary 4.

Let X be a real vector space and $M\subseteq X$ a convex set. Let $F={\left\{f_i\right\}}_{i=1}^n$ be a finite family satisfying the conditions of Theorem 4. Then there exists ${ϵ}_0>0$ such that for all $\delta \in (0,{ϵ}_0)$, the number of convex components of $slc(M,h,\delta )$ is constant.

Proof. 

For each $i=1,\dots ,n$, define:

$${\delta }_i^{*}=sup\{\delta \ge 0:\{x\in M:f_i\left(x\right)\ge \delta \}\ne ⌀\}.$$

Note that ${\delta }_i^{*}\ge 0$ and ${\delta }_i^{*}$ could be $+\infty$ if $f_i$ is unbounded above on M.

Let $J=\{i\in \{1,\dots ,n\}:{\delta }_i^{*}>0\}$ be the indices for which $f_i$ attains values strictly greater than 0 on M. For $i\notin J$, we have $f_i\left(x\right)\le 0$ for all $x\in M$, so these indices do not contribute to multi-slices for $\delta >0$.

Now define:

$${ϵ}_0=\left\{\begin{array}{cc}\min \{{\delta }_i^{*}:i\in J,{\delta }_i^{*}<\infty \},\hfill & \mathrm{if}\mathrm{there}\mathrm{exists}i\in J\mathrm{with}{\delta }_i^{*}<\infty ;\hfill \\ 1,\hfill & \mathrm{if}{\delta }_i^{*}=\infty \mathrm{for}\mathrm{all}i\in J\mathrm{and}J\ne ⌀;\hfill \\ 1,\hfill & \mathrm{if}J=⌀.\hfill \end{array}\right.$$

We verify that this definition is meaningful and that ${ϵ}_0>0$:

• If there exists $i\in J$ with ${\delta }_i^{*}<\infty$, then the minimum is taken over a non-empty finite set of positive numbers, hence ${ϵ}_0>0$.
• If ${\delta }_i^{*}=\infty$ for all $i\in J$ and $J\ne ⌀$, then all relevant functions are unbounded above on M. In this case, any $\delta >0$ will yield non-empty superlevel sets for all $i\in J$. The choice ${ϵ}_0=1$ is arbitrary but serves as a convenient positive threshold; any positive number would work equally well.
• If $J=⌀$, then no function attains positive values on M, so $slc(M,h,\delta )=⌀$ for all $\delta >0$. The choice ${ϵ}_0=1$ is again arbitrary but harmless.

By construction, for all $\delta \in (0,{ϵ}_0)$, we have

$$\{i\in \{1,\dots ,n\}:\{x\in M:f_i\left(x\right)\ge \delta \}\ne ⌀\}=J.$$

That is, the set of indices contributing non-empty components remains constant for $\delta \in (0,{ϵ}_0)$. Note that when ${\delta }_i^{*}=\infty$ for all $i\in J$, the condition $\delta <{ϵ}_0=1$ is sufficient but not necessary; any finite upper bound would work, and we simply choose ${ϵ}_0=1$ as a convenient positive number.

By Corollary 2, the number of convex components is exactly $\left|J\right|$ for all $\delta \in (0,{ϵ}_0)$. This number is constant on this interval, establishing the result. □

Remark 7.

The choice ${ϵ}_0=1$ in the cases where all relevant ${\delta }_i^{*}$ are infinite is indeed arbitrary, but it serves the purpose of selecting some positive threshold below which the behavior is stable. Any positive number would work equally well; the important point is that such an ${ϵ}_0$ exists. In practice, one could simply take ${ϵ}_0=\infty$ in these cases, meaning the stability holds for all $\delta >0$. However, to maintain a uniform statement with ${ϵ}_0$ finite, we adopt the convention ${ϵ}_0=1$ as a harmless normalization.

Corollary 5.

Let X be a Hausdorff locally convex real topological vector space and $M\subseteq X$ a closed convex set with non-empty interior. Let $F={\left\{f_i\right\}}_{i\in I}$ satisfy the conditions of Theorem 4, with each $f_i$ continuous. Then for any $\delta >0$, the convex components of $slc(M,h,\delta )$ are separated by open sets in X.

Proof. 

Let C and D be two distinct convex components of $slc(M,h,\delta )$. By Theorem 4, there exist $i\ne j\in I$ such that:

$$C=\{x\in M:f_i\left(x\right)\ge \delta \},D=\{x\in M:f_j\left(x\right)\ge \delta \}.$$

Since $f_i$ and $f_j$ are continuous and M is closed, both C and D are closed in X.

Consider the function $\varphi \left(x\right)=f_i\left(x\right)-f_j\left(x\right)$. This is continuous since $f_i$ and $f_j$ are continuous. Note that for $x\in C$, we have $f_i\left(x\right)\ge \delta$ and $f_j\left(x\right)<\delta$ (by the disjointness property and Theorem 4(3)), so $\varphi \left(x\right)>0$. Similarly, for $x\in D$, we have $\varphi \left(x\right)<0$.

Define:

$$U=\{x\in X:\varphi (x)>0\},V=\{x\in X:\varphi (x)<0\}.$$

These are open sets (by continuity of $\varphi$) that separate C and D, with $C\subseteq U$ and $D\subseteq V$, and $U\cap V=⌀$.

Therefore, the convex components are separated by open sets in X. □

Corollary 6.

Let X be a Hausdorff locally convex real topological vector space and $M\subseteq X$ a closed convex set with non-empty interior. Let $F={\left\{f_i\right\}}_{i=1}^n$ be continuous convex functions satisfying the conditions of Theorem 4. If $h\left(x\right)=max\{f_1\left(x\right),\dots ,f_n\left(x\right)\}$ is such that $slc(M,h,0)=bd\left(M\right)$, then the boundary of M decomposes into at most n convex components.

Proof. 

Since $slc(M,h,0)=bd\left(M\right)$ and h is continuous (as the maximum of finitely many continuous functions), we have

$$bd\left(M\right)=\{x\in M:h\left(x\right)\ge 0\}=\bigcup _{i=1}^n\{x\in M:f_i\left(x\right)\ge 0\}.$$

By Theorem 4, this is a disjoint union of convex sets that are closed in M (hence in $bd\left(M\right)$ with the subspace topology). By Theorem 2.10 of [6], since this is a finite disjoint union of convex sets closed in $bd\left(M\right)$, these are exactly the convex components of $bd\left(M\right)$.

Therefore, the boundary of M has at most n convex components. □

Example 1.

Let $X={\mathbb{R}}^2$ and let $M\subseteq X$ be the closed unit square:

$$M=[-1,1]\times [-1,1].$$

Partition M into four disjoint convex sets:

$$\begin{array}{cc}\hfill M_1& =\left\{\right(x,y)\in M:x\ge 0,y\ge 0,x+y\ge 1\},\hfill \\ \hfill M_2& =\left\{\right(x,y)\in M:x\le 0,y\ge 0,-x+y\ge 1\},\hfill \\ \hfill M_3& =\left\{\right(x,y)\in M:x\le 0,y\le 0,-x-y\ge 1\},\hfill \\ \hfill M_4& =\left\{\right(x,y)\in M:x\ge 0,y\le 0,x-y\ge 1\}.\hfill \end{array}$$

Geometrically, $M_1$ is the right-top triangle of the square, $M_2$ the left-top triangle, $M_3$ the left-bottom triangle, and $M_4$ the right-bottom triangle. These four triangles are pairwise disjoint, convex, and their union is the set of points in M satisfying $\left|x\right|+\left|y\right|\ge 1$.

Define four convex functions $f_1,f_2,f_3,f_4:{\mathbb{R}}^2\to \mathbb{R}$ by:

$$\begin{array}{cc}\hfill f_1(x,y)& =min\{x,y,x+y-1\},\hfill \\ \hfill f_2(x,y)& =min\{-x,y,-x+y-1\},\hfill \\ \hfill f_3(x,y)& =min\{-x,-y,-x-y-1\},\hfill \\ \hfill f_4(x,y)& =min\{x,-y,x-y-1\}.\hfill \end{array}$$

Each $f_i$ is convex (as the pointwise minimum of affine functions). For $(x,y)\in M_i$, we have $f_i(x,y)\ge 0$, with equality on the boundary of $M_i$. For $(x,y)\notin M_i$, we have $f_i(x,y)<0$.

Set $F=\{f_1,f_2,f_3,f_4\}$ and define

$$h(x,y)=max\{f_1(x,y),f_2(x,y),f_3(x,y),f_4(x,y)\}.$$

For $\delta =0$, consider the multi-slice

$$S=slc(M,h,0)=\left\{\right(x,y)\in M:h(x,y)\ge 0\}.$$

Proof. 

We verify that the hypotheses of Theorem 4 are satisfied and examine the structure of S.

Step 1: Disjointness of the zero sets. For each i, the zero set $f_i^{-1}\left(0\right)$ is precisely the boundary of $M_i$. Specifically:

$$\begin{array}{cc}\hfill f_1^{-1}\left(0\right)& =\{(x,y)\in {\mathbb{R}}^2:min\{x,y,x+y-1\}=0\}\hfill \\ \hfill & =\left\{\right(x,0):x\ge 1\}\cup \left\{\right(0,y):y\ge 1\}\cup \left\{\right(x,1-x):0\le x\le 1\},\hfill \\ \hfill f_2^{-1}\left(0\right)& =\{(x,y)\in {\mathbb{R}}^2:min\{-x,y,-x+y-1\}=0\}\hfill \\ \hfill & =\left\{\right(-x,0):x\ge 1\}\cup \left\{\right(0,y):y\ge 1\}\cup \left\{\right(-x,1+x):-1\le x\le 0\},\hfill \\ \hfill f_3^{-1}\left(0\right)& =\{(x,y)\in {\mathbb{R}}^2:min\{-x,-y,-x-y-1\}=0\}\hfill \\ \hfill & =\left\{\right(-x,0):x\ge 1\}\cup \left\{\right(0,-y):y\ge 1\}\cup \left\{\right(-x,-1-x):-1\le x\le 0\},\hfill \\ \hfill f_4^{-1}\left(0\right)& =\{(x,y)\in {\mathbb{R}}^2:min\{x,-y,x-y-1\}=0\}\hfill \\ \hfill & =\left\{\right(x,0):x\ge 1\}\cup \left\{\right(0,-y):y\ge 1\}\cup \left\{\right(x,x-1):0\le x\le 1\}.\hfill \end{array}$$

These zero sets are pairwise disjoint. Indeed, each consists of three rays/segments that lie outside the square M except for the boundaries of the triangles, and these boundaries do not intersect each other.

Step 2: Convex separation property. Take any two distinct points $p,q\in M$. If p and q lie in different triangles $M_i$ and $M_j$, then $f_i\left(p\right)\ge 0$ while $f_i\left(q\right)<0$, so $f_i\left(p\right)\ne f_i\left(q\right)$. If p and q lie in the same triangle $M_i$, then they are distinguished by the affine functions defining $f_i$. Thus the family F separates points of M.

Step 3: Description of the multi-slice. For each i, define

$$A_i=\{x\in M:f_i\left(x\right)\ge 0\}=M_i.$$

These sets are exactly the four triangles. They are convex, pairwise disjoint, and $S=A_1\cup A_2\cup A_3\cup A_4$.

Step 4: Convex components. We claim that each $A_i$ is a convex component of S. Clearly each $A_i$ is convex. To see maximality, suppose $B\subseteq S$ is convex with $A_i⊊B$. Then there exists $x\in B\setminus A_i$. Since the $A_j$’s are disjoint and cover S, we have $x\in A_j$ for some $j\ne i$. Choose any $y\in A_i$. By convexity of B, the entire segment $[x,y]$ lies in $B\subseteq S$. But this segment must cross the region where all $f_k<0$ (which lies outside S), a contradiction. Hence no such B exists, and each $A_i$ is maximal convex in S.

Thus the convex components of S are exactly $A_1,A_2,A_3,A_4$.

Step 5: Topological properties. In the finest locally convex topology on ${\mathbb{R}}^2$, each $A_i$ is closed in M (since it is the intersection of M with the closed set $f_i^{-1}\left([0,\infty )\right)$, and $f_i$ is continuous in this topology). Moreover, each $A_i$ is closed in ${\mathbb{R}}^2$ because it is a compact subset of ${\mathbb{R}}^2$.

Step 6: Component-wise exposing. For each convex component $A_i$, the corresponding function $f_i$ exposes it: for any $x\in M\setminus A_i$, we have $f_i\left(x\right)<0$, while for $x\in A_i$, $f_i\left(x\right)\ge 0$.

Step 7: Geometric interpretation. The multi-slice $S=\left\{\right(x,y)\in M:|x|+|y|\ge 1\}$ is the part of the square outside the diamond $\left|x\right|+\left|y\right|<1$. This set decomposes naturally into four convex components the four corner triangles each exposed by one of the functions $f_i$. The functions $f_i$ are constructed so that their superlevel sets are exactly these triangles.

Thus this example perfectly illustrates Theorem 4: a finite family of convex functions with pairwise disjoint zero sets and the convex separation property yields a multi-slice that decomposes into convex components corresponding precisely to the individual functions to illustrate the example, see the Figure 1. □

5. Convex Component Preservation Under Affine Maps

This section investigates the behavior of convex components under affine transformations, establishing fundamental invariance properties that significantly enhance the utility of convex component analysis in geometric contexts. A central question in convex geometry concerns which structural properties remain invariant under affine maps, which play a crucial role in simplifying geometric problems through appropriate coordinate changes [5,7]. We prove that when an affine map is injective on a given set, it preserves the complete convex component structure, mapping each convex component bijectively to a convex component of the image. This preservation theorem extends beyond mere convexity preservation, a classical result in convex analysis [1], to encompass the finer granularity of maximal convex subsets. Our results demonstrate that convex components constitute an affine invariant, providing a powerful tool for analyzing geometric structures across different coordinate systems and ambient spaces. The theorem finds particular strength in applications involving linear isomorphisms, translations, and affine retractions, with implications for optimization problems where affine transformations are employed to simplify constraint structures [10,11]. Furthermore, these invariance properties establish convex components as robust geometric invariants, comparable in significance to other fundamental affine invariants in convex geometry [5].

Theorem 5.

Let X and Y be real vector spaces, and let $T:X\to Y$ be an affine map (i.e., $T\left(x\right)=L\left(x\right)+b$ where $L:X\to Y$ is linear and $b\in Y$). Let $M\subseteq X$ be a non-empty set and let ${\left\{C_i\right\}}_{i\in I}$ be the convex components of M.

If T is injective on M (i.e., ${T|}_M$ is one-to-one), then the convex components of $T\left(M\right)$ are exactly ${\left\{T\left(C_i\right)\right\}}_{i\in I}$. Moreover, if T is affine and bijective on M (i.e., ${T|}_M$ is a bijection onto $T\left(M\right)$), then the convex component structure is preserved in the strong sense that for any convex set $D\subseteq M$, D is a convex component of M if and only if $T\left(D\right)$ is a convex component of $T\left(M\right)$.

Proof. 

We prove the theorem in several steps, carefully noting where hypotheses are used.

Step 1: Preliminary observations. Affine maps preserve convexity: if $C\subseteq X$ is convex, then $T\left(C\right)\subseteq Y$ is convex. This follows directly from the definition of convexity and the affine property of T.

If T is injective on M, then T preserves disjointness: if $C_i\cap C_j=⌀$, then $T\left(C_i\right)\cap T\left(C_j\right)=⌀$.

Step 2: Each $T\left(C_i\right)$ is convex in $T\left(M\right)$. Since each $C_i$ is convex and T is affine, $T\left(C_i\right)$ is convex in Y. As $T\left(C_i\right)\subseteq T\left(M\right)$, it is also convex in $T\left(M\right)$.

Step 3: Each $T\left(C_i\right)$ is maximal convex in $T\left(M\right)$. Suppose, for contradiction, that for some $i\in I$, $T\left(C_i\right)$ is not maximal convex in $T\left(M\right)$. Then there exists a convex set $D\subseteq T\left(M\right)$ such that $T\left(C_i\right)⊊D$. Define

$$E=T^{-1}\left(D\right)\cap M.$$

We claim that E is convex. To see this, take $x_1,x_2\in E$ and $t\in [0,1]$. Since $x_1,x_2\in M$, we need to show that $t{x}_1+(1-t)x_2\in E$, i.e., that $T(t{x}_1+(1-t)x_2)\in D$ and that $t{x}_1+(1-t)x_2\in M$.

The first condition follows from the convexity of D:

$$T(t{x}_1+(1-t)x_2)=tT\left(x_1\right)+(1-t)T\left(x_2\right)\in D,$$

since $T\left(x_1\right),T\left(x_2\right)\in D$.

The second condition that $t{x}_1+(1-t)x_2\in M$ does not follow automatically, because M is not assumed to be convex. However, we do not actually need E to be convex as a subset of X; we only need to show that $C_i⊊E$ leads to a contradiction with the maximality of $C_i$ as a convex component of M. To obtain this contradiction, we proceed differently.

Since D is convex and contains $T\left(C_i\right)$, for any $y\in C_i$ and any $z\in T^{-1}\left(D\right)\cap M\setminus C_i$, the segment $\left[T\right(y),T(z\left)\right]$ lies in D. By injectivity of T on M, the preimage of this segment is the segment $[y,z]$ (since T is affine and injective, it maps line segments to line segments bijectively). This preimage segment must lie in $T^{-1}\left(D\right)$. However, points on this segment may or may not belong to M.

Now, crucially, we use the fact that $C_i$ is a convex component of M. Consider any $x\in T^{-1}\left(D\right)\cap M\setminus C_i$. Since $C_i$ is maximal convex in M, the set $co(C_i\cup \left\{x\right\})$ (the convex hull of $C_i$ and x) cannot be entirely contained in M; otherwise, $C_i$ would not be maximal. Therefore, there exist points in the convex hull that lie outside M. By the injectivity and affinity of T, these points map to points in $co(T\left(C_i\right)\cup \left\{T\left(x\right)\right\})$ that lie outside $T\left(M\right)$. But this contradicts the fact that D is convex and contains both $T\left(C_i\right)$ and $T\left(x\right)$, since then the entire convex hull of $T\left(C_i\right)\cup \left\{T\left(x\right)\right\}$ would be contained in $D\subseteq T\left(M\right)$.

Thus, no such D exists, and each $T\left(C_i\right)$ is maximal convex in $T\left(M\right)$.

Step 4: The sets ${\left\{T\left(C_i\right)\right\}}_{i\in I}$ cover $T\left(M\right)$. Since ${\left\{C_i\right\}}_{i\in I}$ covers M and $T\left(M\right)={\bigcup }_{i\in I}T\left(C_i\right)$, the family ${\left\{T\left(C_i\right)\right\}}_{i\in I}$ covers $T\left(M\right)$.

Step 5: The sets ${\left\{T\left(C_i\right)\right\}}_{i\in I}$ are the convex components of $T\left(M\right)$. From Steps 2–4, we have that ${\left\{T\left(C_i\right)\right\}}_{i\in I}$ is a family of convex sets that are maximal in $T\left(M\right)$ and cover $T\left(M\right)$. By the definition of convex components, these must be exactly the convex components of $T\left(M\right)$.

Step 6: Strong preservation under affine bijections. Now assume ${T|}_M$ is a bijection onto $T\left(M\right)$. We need to show that for any convex set $D\subseteq M$, D is a convex component of M if and only if $T\left(D\right)$ is a convex component of $T\left(M\right)$.

The forward direction (⇒) was proved in Steps 2–5.

For the reverse direction (⇐), suppose $T\left(D\right)$ is a convex component of $T\left(M\right)$. Since ${T|}_M$ is a bijection, we have $D=T^{-1}\left(T\left(D\right)\right)\cap M$. We need to show D is a convex component of M.

First, D is convex: for $x_1,x_2\in D$ and $t\in [0,1]$, we have $T\left(x_1\right),T\left(x_2\right)\in T\left(D\right)$, and since $T\left(D\right)$ is convex:

$$T(t{x}_1+(1-t)x_2)=tT\left(x_1\right)+(1-t)T\left(x_2\right)\in T\left(D\right).$$

By injectivity, $t{x}_1+(1-t)x_2\in D$, so D is convex.

Next, D is maximal convex in M. Suppose there exists a convex set $E\subseteq M$ with $D⊊E$. Then $T\left(E\right)$ is convex (since T is affine and preserves convexity) and $T\left(D\right)⊊T\left(E\right)\subseteq T\left(M\right)$, contradicting the maximality of $T\left(D\right)$ as a convex component of $T\left(M\right)$. Therefore, D is a convex component of M.

This completes the proof of the strong preservation property. □

Remark 8.

The key subtlety in Step 3 is that we cannot simply claim that $T^{-1}\left(D\right)\cap M$ is convex, since M is not assumed convex. Instead, we use the maximality of $C_i$ in M and the properties of affine injective maps to derive a contradiction. The argument relies on the fact that if $T\left(C_i\right)⊊D$ with D convex, then taking any $x\in T^{-1}\left(D\right)\cap M\setminus C_i$, the convex hull of $C_i\cup \left\{x\right\}$ cannot be entirely contained in M (otherwise $C_i$ would not be maximal). This forces points outside M, which map via T to points outside $T\left(M\right)$, contradicting the convexity of D and its containment in $T\left(M\right)$.

Corollary 7.

Let X and Y be real vector spaces and let $T:X\to Y$ be a linear isomorphism. Then for any non-empty set $M\subseteq X$, the convex components of $T\left(M\right)$ are exactly

$$\left\{T\right(C):C\mathit{is}a\mathit{convex}\mathit{component}\mathit{of}M\}.$$

Moreover, the convex component lattice of M is isomorphic to the convex component lattice of $T\left(M\right)$.

Proof. 

Since T is a linear isomorphism, it is affine and bijective. The result follows immediately from Theorem 5. The lattice isomorphism follows from the fact that T establishes a bijection between convex subsets of M and convex subsets of $T\left(M\right)$ that preserves the inclusion relation. □

Corollary 8.

Let X be a real vector space and $M\subseteq X$ a non-empty set. For any $v\in X$, the convex components of $M+v$ are exactly $\{C+v:C\mathit{is}a\mathit{convex}\mathit{component}\mathit{of}M\}$.

Proof. 

The translation map $T\left(x\right)=x+v$ is affine and bijective. Apply Theorem 5. □

Proposition 3.

Let X and Y be real vector spaces, $M\subseteq X$ non-empty, and $T:X\to Y$ an affine map injective on M. Then the number of convex components of M equals the number of convex components of $T\left(M\right)$. In particular, if M has finitely many convex components, so does $T\left(M\right)$, and if M has infinitely many convex components, so does $T\left(M\right)$.

Proof. 

By Theorem 5, the convex components of $T\left(M\right)$ are exactly ${\left\{T\left(C_i\right)\right\}}_{i\in I}$, where ${\left\{C_i\right\}}_{i\in I}$ are the convex components of M. Since T is injective on M, the index set I is the same for both families. Therefore, the number of convex components is preserved. □

Remark 9.

Theorem 5 shows that the convex component structure is an affine invariant. This provides a powerful tool for analyzing convex components: we can apply affine transformations to simplify the geometry while preserving the essential convex structure. For example, we can translate sets to move them to more convenient positions, or apply linear isomorphisms to transform them into canonical forms, without changing their convex component decomposition.

Remark 10.

The injectivity condition in Theorem 5 is necessary. Consider $T:{\mathbb{R}}^2\to \mathbb{R}$ defined by $T(x,y)=x$ (projection onto the x-axis). Let $M=\left\{\right(0,0),(1,0),(0,1\left)\right\}$. The convex components of M are the three singleton sets. However, $T\left(M\right)=\{0,1\}$, whose convex components are the two singleton sets $\left\{0\right\}$ and $\left\{1\right\}$. Thus the convex component structure is not preserved under non-injective affine maps.

Corollary 9.

Let X and Y be real vector spaces, and let $L:X\to Y$ be an injective linear map. Let $M\subseteq X$ be a non-empty set with convex components ${\left\{C_i\right\}}_{i\in I}$. Then the convex components of $L\left(M\right)$ are exactly ${\left\{L\left(C_i\right)\right\}}_{i\in I}$. Moreover, if L is a linear isomorphism, then for any convex set $D\subseteq M$, D is a convex component of M if and only if $L\left(D\right)$ is a convex component of $L\left(M\right)$.

Proof. 

Since L is linear, it is affine (with $b=0$). The injectivity of L ensures that ${L|}_M$ is injective. Therefore, by Theorem 5, the convex components of $L\left(M\right)$ are exactly ${\left\{L\left(C_i\right)\right\}}_{i\in I}$.

If L is a linear isomorphism, then ${L|}_M$ is bijective onto $L\left(M\right)$, so the strong preservation property from Theorem 5 applies: D is a convex component of M if and only if $L\left(D\right)$ is a convex component of $L\left(M\right)$. □

Corollary 10.

Let $X,Y$ be real vector spaces, $T:X\to Y$ an affine map injective on $M\subseteq X$, and $f:Y\to \mathbb{R}$ a convex function. Define $g:X\to \mathbb{R}$ by $g\left(x\right)=f\left(T\right(x\left)\right)$. Then for any $\delta \in \mathbb{R}$, the convex components of the multi-slice $slc(M,g,\delta )$ are in bijection with the convex components of $slc\left(T\right(M),f,\delta )$. Specifically, if ${\left\{C_i\right\}}_{i\in I}$ are the convex components of $slc(M,g,\delta )$, then ${\left\{T\left(C_i\right)\right\}}_{i\in I}$ are the convex components of $slc\left(T\right(M),f,\delta )$.

Proof. 

First, note that g is convex since it is the composition of a convex function with an affine map. Indeed, for $x_1,x_2\in X$ and $t\in [0,1]$, we have

$$\begin{array}{ccc}\hfill g(t{x}_1+(1-t)x_2)& =& f\left(T(t{x}_1+(1-t)x_2)\right)=f(tT\left(x_1\right)+(1-t)T\left(x_2\right))\hfill \\ & \le & tf\left(T\left(x_1\right)\right)+(1-t)f\left(T\left(x_2\right)\right)=tg\left(x_1\right)+(1-t)g\left(x_2\right).\hfill \end{array}$$

Now observe that

$$\begin{array}{ccc}\hfill slc(M,g,\delta )& =& \{x\in M:g(x)\ge \delta \}=\{x\in M:f(T\left(x\right))\ge \delta \}\hfill \\ & =& T^{-1}(slc(T\left(M\right),f,\delta ))\cap M.\hfill \end{array}$$

Since T is injective on M, the restriction ${T|}_{slc(M,g,\delta )}$ is a bijection onto $slc\left(T\right(M),f,\delta )$. Moreover, T is affine, so by Theorem 5, the convex components are preserved: if ${\left\{C_i\right\}}_{i\in I}$ are the convex components of $slc(M,g,\delta )$, then ${\left\{T\left(C_i\right)\right\}}_{i\in I}$ are the convex components of $slc\left(T\right(M),f,\delta )$. □

Corollary 11.

Let X be a finite-dimensional real vector space and $M\subseteq X$ a non-empty set. Let Y be any real vector space with $dimY\ge dimX$, and let $L:X\to Y$ be an injective linear map. Then the convex component structure of M is isomorphic to the convex component structure of $L\left(M\right)$. In particular, the number and combinatorial structure of convex components are independent of the ambient space dimension, as long as the dimension is at least $dim(spanM)$.

Proof. 

Since $dimY\ge dimX$, there exist injective linear maps from X to Y. Let $L:X\to Y$ be such an injective linear map. Then ${L|}_M$ is injective, so by Corollary 9, the convex components of $L\left(M\right)$ are exactly ${\left\{L\left(C_i\right)\right\}}_{i\in I}$, where ${\left\{C_i\right\}}_{i\in I}$ are the convex components of M.

The isomorphism of convex component structures means that

• The number of convex components is the same;
• The inclusion relations between convex components are preserved;
• The lattice of convex subsets generated by the components is isomorphic.

All these properties follow from the fact that L establishes a bijection between convex subsets of M and convex subsets of $L\left(M\right)$ that preserves the inclusion relation.

The condition $dimY\ge dim\left(\mathrm{span}M\right)$ ensures that we can embed M injectively into Y, and the convex component structure depends only on the affine geometry of M, not on the ambient space. □

Corollary 12.

Let $X,Y$ be real vector spaces, $T:X\to Y$ an affine map injective on $M\subseteq X$. If the convex components of M are pairwise disjoint, then the convex components of $T\left(M\right)$ are also pairwise disjoint. Conversely, if T is affine and bijective on M and the convex components of $T\left(M\right)$ are pairwise disjoint, then the convex components of M are pairwise disjoint.

Proof. 

(⇒) Suppose the convex components ${\left\{C_i\right\}}_{i\in I}$ of M are pairwise disjoint. By Theorem 5, the convex components of $T\left(M\right)$ are ${\left\{T\left(C_i\right)\right\}}_{i\in I}$. Since T is injective on M, if $C_i\cap C_j=⌀$ for $i\ne j$, then $T\left(C_i\right)\cap T\left(C_j\right)=⌀$. Therefore, the convex components of $T\left(M\right)$ are pairwise disjoint.

(⇐) Now assume T is affine and bijective on M, and the convex components ${\left\{D_j\right\}}_{j\in J}$ of $T\left(M\right)$ are pairwise disjoint. By Theorem 5, the convex components of M are ${\{T^{-1}\left(D_j\right)\cap M\}}_{j\in J}$. Since T is injective, if $D_i\cap D_j=⌀$ for $i\ne j$, then $T^{-1}\left(D_i\right)\cap T^{-1}\left(D_j\right)\cap M=⌀$. Therefore, the convex components of M are pairwise disjoint. □

Corollary 13.

Let $X,Y$ be real vector spaces, and let $M\subseteq X$ be a non-empty set. Let $T:X\to Y$ be an affine map that is bijective on M (i.e., ${T|}_M$ is a bijection onto $T\left(M\right)$). Then,

(i) 

If M is convex, then $x\in M$ is an extreme point of M if and only if $\left\{x\right\}$ is a convex component of M.

(ii) 

Under the same hypothesis that M is convex, $T(ext(M\left)\right)=ext\left(T\right(M\left)\right)$.

Proof. 

We first clarify the relationship between extreme points and convex components. For a general set M (not necessarily convex), a singleton $\left\{x\right\}$ may be a convex component without x being an extreme point, simply because the definition of extreme point requires the set to be convex. The equivalence holds precisely when M is convex.

Proof of (i). Assume M is convex.

$(\Rightarrow )$ Suppose $x\in ext\left(M\right)$. We claim that $\left\{x\right\}$ is a convex component of M. Clearly $\left\{x\right\}$ is convex. To show maximality, assume there exists a convex set $D\subseteq M$ with $\left\{x\right\}\subseteq D$. Take any $y\in D$ with $y\ne x$. Since M is convex, the segment $[x,y]$ lies in M. But then x would be an interior point of this segment (unless $y=x$), contradicting that x is an extreme point of M. Therefore, no such y exists, and $D=\left\{x\right\}$. Hence $\left\{x\right\}$ is a convex component.

$(\Leftarrow )$ Conversely, suppose $\left\{x\right\}$ is a convex component of M. If x were not an extreme point of M, then there exist distinct $y,z\in M$ and $t\in (0,1)$ such that $x=ty+(1-t)z$. Since M is convex, the entire segment $[y,z]$ lies in M. But then $\left\{x\right\}$ would be properly contained in the convex set $[y,z]\subseteq M$, contradicting the maximality of $\left\{x\right\}$ as a convex component. Hence $x\in ext\left(M\right)$.

Proof of (ii). Now assume M is convex. Since T is affine and bijective on M, $T\left(M\right)$ is also convex (affine maps preserve convexity). By part (i), for any $x\in M$, $x\in ext\left(M\right)$ if and only if $\left\{x\right\}$ is a convex component of M. By Theorem 5, $\left\{x\right\}$ is a convex component of M if and only if $T\left(\right\{x\left\}\right)=\left\{T\right(x\left)\right\}$ is a convex component of $T\left(M\right)$. Applying part (i) to $T\left(M\right)$, we have that $\left\{T\right(x\left)\right\}$ is a convex component of $T\left(M\right)$ if and only if $T\left(x\right)\in ext\left(T\right(M\left)\right)$. Chaining these equivalences yields $x\in \mathrm{ext}\left(M\right)$ if and only if $T\left(x\right)\in \mathrm{ext}\left(T\right(M\left)\right)$. Bijectivity of T on M then gives $T\left(\mathrm{ext}\right(M\left)\right)=\mathrm{ext}\left(T\right(M\left)\right)$. □

Remark 11.

The hypothesis that M is convex is essential for the equivalence between extreme points and singleton convex components. In the absence of convexity, a singleton may be a convex component without the point being extreme in any meaningful sense, as the definition of extreme point presupposes a convex ambient set. Example 1 illustrates this: in the cross-shaped set, the origin is not an extreme point (it lies in the interior of segments), yet it forms a singleton convex component. This underscores the importance of the convexity assumption in the corollary.

Corollary 14.

Let X be a real vector space and $M\subseteq X$ a non-empty set. Let $P:X\to X$ be an affine retraction onto an affine subspace $Y\subseteq X$ (i.e., $P^2=P$ and $P\left(X\right)=Y$). If ${P|}_M$ is injective, then the convex components of $P\left(M\right)$ are exactly $\left\{P\right(C):C\mathrm{is}\mathrm{a}\mathrm{convex}\mathrm{component}\mathrm{of}M\}$. Moreover, if $M\subseteq Y$, then the convex component structures of M and $P\left(M\right)$ coincide.

Proof. 

Since P is affine and ${P|}_M$ is injective, we can apply Theorem 5 directly. The convex components of $P\left(M\right)$ are ${\left\{P\left(C_i\right)\right\}}_{i\in I}$, where ${\left\{C_i\right\}}_{i\in I}$ are the convex components of M.

If $M\subseteq Y$, then ${P|}_M$ is the identity map (since P is a retraction onto Y), so $P\left(M\right)=M$ and the convex component structures are identical. □

Corollary 15.

Let $X,Y$ be real vector spaces, and let $M\subseteq X$, $N\subseteq Y$ be non-empty sets that are both convex. Let ${\left\{C_i\right\}}_{i\in I}$ and ${\left\{D_j\right\}}_{j\in J}$ be the convex components of M and N respectively. Then the convex components of $M\times N\subseteq X\times Y$ are exactly ${\{C_i\times D_j\}}_{(i,j)\in I\times J}$.

Proof. 

We first note that the assumption that M and N are convex is essential. For non-convex sets, the convex components of a product need not decompose as products of convex components, as illustrated in the remark following this proof.

Step 1: Preliminary observations. Since M and N are convex, their Cartesian product $M\times N$ is convex. For any convex sets $C\subseteq M$ and $D\subseteq N$, the product $C\times D$ is convex in $X\times Y$. Moreover, the projection maps ${\pi }_X:X\times Y\to X$ and ${\pi }_Y:X\times Y\to Y$ are linear (hence affine) and preserve convexity: if $E\subseteq M\times N$ is convex, then ${\pi }_X\left(E\right)\subseteq M$ and ${\pi }_Y\left(E\right)\subseteq N$ are convex.

Step 2: Each $C_i\times D_j$ is convex in $M\times N$. This follows directly from the convexity of $C_i$ and $D_j$.

Step 3: Maximality of $C_i\times D_j$. Suppose $E\subseteq M\times N$ is convex and contains $C_i\times D_j$. We claim that $E=C_i\times D_j$.

Consider the projections. Since E is convex, ${\pi }_X\left(E\right)$ is convex in M and contains ${\pi }_X(C_i\times D_j)=C_i$. By maximality of $C_i$ as a convex component of M, we must have ${\pi }_X\left(E\right)=C_i$. Similarly, ${\pi }_Y\left(E\right)=D_j$.

Now take any $(x,y)\in E$. Then $x\in {\pi }_X\left(E\right)=C_i$ and $y\in {\pi }_Y\left(E\right)=D_j$, so $(x,y)\in C_i\times D_j$. Hence $E\subseteq C_i\times D_j$, and together with the assumption $C_i\times D_j\subseteq E$, we obtain $E=C_i\times D_j$. Thus each $C_i\times D_j$ is maximal convex in $M\times N$.

Step 4: Coverage and disjointness. Since ${\left\{C_i\right\}}_{i\in I}$ covers M and ${\left\{D_j\right\}}_{j\in J}$ covers N, the family ${\{C_i\times D_j\}}_{(i,j)\in I\times J}$ covers $M\times N$. Moreover, if $(i,j)\ne (i^{\prime },j^{\prime })$, then either $i\ne i^{\prime }$ or $j\ne j^{\prime }$. In the first case, $C_i\cap C_{i^{\prime }}=⌀$ implies $(C_i\times D_j)\cap (C_{i^{\prime }}\times D_{j^{\prime }})=⌀$; in the second case, $D_j\cap D_{j^{\prime }}=⌀$ similarly implies disjointness. Hence the sets are pairwise disjoint.

Step 5: Conclusion. The family ${\{C_i\times D_j\}}_{(i,j)\in I\times J}$ is a pairwise disjoint collection of convex sets that are maximal in $M\times N$ and cover $M\times N$. By definition, these are precisely the convex components of $M\times N$. □

Remark 12.

The assumption that M and N are convex is indispensable. For a counterexample when convexity fails, take $X=Y=\mathbb{R}$ and let $M=\{0,1\}$ (two points) and $N=\left\{0\right\}$. The convex components of M are $\left\{0\right\}$ and $\left\{1\right\}$; the convex component of N is $\left\{0\right\}$. The product $M\times N=\left\{\right(0,0),(1,0\left)\right\}$ has convex components $\left\{\right(0,0\left)\right\}$ and $\left\{\right(1,0\left)\right\}$, which are indeed of the form $\left\{0\right\}\times \left\{0\right\}$ and $\left\{1\right\}\times \left\{0\right\}$. However, if we take $M=\{0,1\}$ and $N=\{0,1\}$, the product has four points, and its convex components are the singletons, which again decompose as products. A more subtle counterexample would require a non-convex set where a convex subset of the product is not a product set for instance; take $M=\{(0,0),(1,1)\}\subset {\mathbb{R}}^2$ (not convex) and $N=\left\{0\right\}\subset \mathbb{R}$. Then $M\times N\subset {\mathbb{R}}^3$ consists of two points, and its convex components are the singletons, which are not products of convex components of M and N because M itself does not have a product structure. This illustrates why the convexity of M and N is necessary for the decomposition to hold in the simple product form.

Corollary 16

(Product Decomposition for Convex Sets). Let $X,Y$ be real vector spaces, and let $M\subseteq X$, $N\subseteq Y$ be non-empty convex sets. Let ${\left\{C_i\right\}}_{i\in I}$ and ${\left\{D_j\right\}}_{j\in J}$ be the convex components of M and N respectively. Then,

(i) 

Every convex component of $M\times N$ is of the form $C_i\times D_j$ for some $i\in I$, $j\in J$.

(ii) 

Conversely, every such product $C_i\times D_j$ is a convex component of $M\times N$.

(iii) 

The convex components of $M\times N$ are pairwise disjoint and form a covering of $M\times N$.

Remark 13.

These corollaries demonstrate the power and versatility of Theorem 5. They show that convex component structure is preserved under various natural operations: linear maps, multi-slice formations, dimension changes, extreme point characterizations, retractions, and products. This makes convex components a robust and useful invariant in convex geometry and analysis.

Example 2.

Consider the cross–shaped set in ${\mathbb{R}}^2$ given by

$$M=\left([-2,2]\times \left\{0\right\}\right)\cup \left(\left\{0\right\}\times [-2,2]\right),$$

which consists of a horizontal and a vertical line segment intersecting at the origin. Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be the affine map defined by

$$T(x,y)=(2x+1,y+x+3).$$

We shall verify that T preserves the convex-component structure of M, thus illustrating Theorem 5.

Proof. 

The verification proceeds in several steps.

Step 1: Convex components of M. Write the two segments as

$$H=[-2,2]\times \left\{0\right\},V=\left\{0\right\}\times [-2,2].$$

Both are convex, but they are not maximal convex subsets of M because each contains the origin, and the origin can be separated from the rest. A careful inspection shows that the maximal convex subsets of M are the following five sets:

$$\begin{array}{cccc}\hfill C_1& =[-2,0)\times \left\{0\right\},\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}{C}_2& =(0,2]\times \left\{0\right\},\hfill \\ \hfill C_3& =\left\{0\right\}\times [-2,0),\hfill & \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}{C}_4& =\left\{0\right\}\times (0,2],\hfill \\ \hfill C_5& =\left\{\right(0,0\left)\right\}.\hfill \end{array}$$

Indeed, any convex subset of M must be contained either in H or in V (or be a single point). The four open half-segments $C_1,\dots ,C_4$ are convex and cannot be enlarged without losing convexity or leaving M; the isolated intersection point $\left\{\right(0,0\left)\right\}$ is also a convex component. Hence M has exactly five convex components.

Step 2: Properties of the map T. The map can be written as

$$T(x,y)=\left(\begin{array}{cc}2& 0\\ 1& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}1\\ 3\end{array}\right),$$

so it is affine. Its linear part has determinant $2\ne 0$; consequently T is a bijection of ${\mathbb{R}}^2$ and, in particular, injective on M.

Step 3: Image of M under T. Applying T to the horizontal segment H gives

$$T\left(H\right)=\left\{T\right(x,0):x\in [-2,2\left]\right\}=\left\{\right(2x+1,x+3):x\in [-2,2\left]\right\},$$

which is the straight-line segment joining $(-3,1)$ to $(5,5)$. The vertical segment V becomes

$$T\left(V\right)=\left\{T\right(0,y):y\in [-2,2\left]\right\}=\left\{\right(1,y+3):y\in [-2,2\left]\right\},$$

the vertical segment from $(1,1)$ to $(1,5)$. Thus

$$T\left(M\right)=T\left(H\right)\cup T\left(V\right)$$

is again a cross-shaped set, now slanted and translated; the two image segments intersect at $T(0,0)=(1,3)$.

Step 4: Images of the convex components. By direct computation,

$$\begin{array}{cc}\hfill T\left(C_1\right)& =\left\{\right(2t+1,t+3):t\in [-2,0\left)\right\},\hfill \\ \hfill T\left(C_2\right)& =\left\{\right(2t+1,t+3):t\in (0,2\left]\right\},\hfill \\ \hfill T\left(C_3\right)& =\left\{\right(1,y+3):y\in [-2,0\left)\right\},\hfill \\ \hfill T\left(C_4\right)& =\left\{\right(1,y+3):y\in (0,2\left]\right\},\hfill \\ \hfill T\left(C_5\right)& =\left\{\right(1,3\left)\right\}.\hfill \end{array}$$

Each $T\left(C_i\right)$ is convex (the affine image of a convex set) and, by the same maximality argument used for the $C_i$, it is a maximal convex subset of $T\left(M\right)$. Hence the family $\{T\left(C_1\right),\dots ,T\left(C_5\right)\}$ is precisely the set of convex components of $T\left(M\right)$.

Step 5: Verification of Theorem 5. Because T is affine and injective on M, the theorem guarantees that the convex components of $T\left(M\right)$ are exactly the images of the convex components of M. This is exactly what we have exhibited. Moreover, since T is globally bijective, the “strong preservation” part also holds: a subset $D\subseteq M$ is a convex component of M if and only if $T\left(D\right)$ is a convex component of $T\left(M\right)$.

Step 6: Geometric interpretation. The transformation T consists of a scaling in the x-direction by a factor 2, a shear that adds the x-coordinate to the y-coordinate, and finally a translation by $(1,3)$. The original orthogonal cross becomes a slanted cross, yet the five convex components (the four open arms and the centre point) are faithfully carried to the five convex components of the image. Figure 2 illustrates the two sets. □

Remark 14.

The example shows that Theorem 5 applies even when convex components are not disjoint (here they all meet at the centre point). The crucial hypothesis is the injectivity of the affine map on the set, which guarantees that the component structure is carried over exactly. It also illustrates that convex components may be relatively open within the ambient set, and this topological property is likewise preserved by affine transformations.

Corollary 17.

For the affine map $T(x,y)=(2x+1,y+x+3)$ and the set M defined above, the lattices of convex subsets of M and of $T\left(M\right)$ are isomorphic. In particular, the convex components correspond bijectively.

Proof. 

Since T is an affine bijection of the whole plane, it restricts to a bijection ${T|}_M:M\to T\left(M\right)$. By Theorem 5 (strong preservation property), a subset $D\subseteq M$ is convex if and only if $T\left(D\right)\subseteq T\left(M\right)$ is convex, and inclusion relations are preserved. Hence the map $D\mapsto T\left(D\right)$ is an isomorphism between the lattices of convex subsets. Restricting this isomorphism to the convex components yields the desired bijection. □

6. Convex Components of Sets with Locally Finite Perimeter

The interplay between convex geometry and geometric measure theory offers a powerful framework for studying sets that lack smooth boundaries. In this section, we generalize the idea of convex components to sets of locally finite perimeter, a concept extensively developed by De Giorgi and later systematized in the foundational monographs [15,18]. The theory of sets of locally finite perimeter, also known as Caccioppoli sets, provides a robust measure-theoretic notion of boundary that coincides with the topological boundary for sufficiently regular sets [5,19]. We prove a structure theorem that connects the convex component decomposition with the reduced boundary in the sense of De Giorgi, revealing how the geometry of convex functions interacts with the measure-theoretic boundary. This connection builds upon classical results in geometric measure theory [20,21] and aligns with recent research on the interplay between convexity and variational problems, as discussed in works such as [11,12,18] which explore applications in optimization and partial differential equations.

Theorem 6.

Let $E\subset {\mathbb{R}}^n$ be a convex set with nonempty interior and locally Lipschitz boundary. Let $\mathcal{F}={\{f_i:{\mathbb{R}}^n\to \mathbb{R}\}}_{i=1}^k$ be a family of convex functions satisfying:

(i) 

The zero sets $f_i^{-1}\left(0\right)$ are pairwise disjoint and contained in E.

(ii) 

Each $f_i$ is continuously differentiable in a neighborhood of $\partial E\cap f_i^{-1}\left(0\right)$ and satisfies

$$\nabla f_i\left(x\right)\ne 0\mathit{for}\mathit{all}x\in f_i^{-1}\left(0\right).$$

Define the convex function

$$h\left(x\right)=\underset{1\le i\le k}{max}{f}_i\left(x\right).$$

Then, for every sufficiently small $\delta >0$, the set

$$E_{\delta }=\{x\in E:h\left(x\right)\ge \delta \}$$

has the following properties:

1. 

$E_{\delta }$ decomposes into a finite disjoint union of convex sets:

$$E_{\delta }=\underset{i=1}{\overset{m}{⨆}}{A}_i,m\le k,$$

where $A_i=\{x\in E:f_i\left(x\right)\ge \delta \}\ne ⌀$. These are precisely the convex components of $E_{\delta }$.

2. 

Each $A_i$ has locally finite perimeter and its reduced boundary satisfies, up to an ${\mathcal{H}}^{n-1}$-null set,

$${\partial }^{*}{A}_i\subset \partial E\cup f_i^{-1}\left(\delta \right).$$

3. 

If, in addition, $\partial E$ is $C^1$ and each $\nabla f_i\left(x\right)·{\nu }_E\left(x\right)\ne 0$ for $x\in \partial E\cap f_i^{-1}\left(0\right)$, then the perimeter measure decomposes additively:

$$P(E_{\delta };·)=\sum _{i=1}^{m}P(A_i;·).$$

Proof. 

We provide detailed proof, carefully justifying each step with appropriate hypotheses.

• Step 1: Decomposition of $E_{\delta }$

For each $i=1,\dots ,k$, set

$$A_i=\{x\in E:f_i\left(x\right)\ge \delta \}.$$

Since $h={max}_i{f}_i$, we have $E_{\delta }={\bigcup }_{i=1}^k{A}_i$.

• Disjointness of the components

We show that for sufficiently small $\delta >0$, the sets $A_i$ are pairwise disjoint. Suppose, for contradiction, that there exist $i\ne j$ and ${\delta }_n\to 0^{+}$ with points $x_n\in A_i\cap A_j$ for each n. Then $f_i\left(x_n\right)\ge {\delta }_n>0$ and $f_j\left(x_n\right)\ge {\delta }_n>0$.

Take any $y\in f_i^{-1}\left(0\right)$ (which is nonempty by hypothesis). By convexity of $f_i$,

$$f_i(ty+(1-t)x_n)\le t{f}_i\left(y\right)+(1-t)f_i\left(x_n\right)=(1-t)f_i\left(x_n\right).$$

For t sufficiently close to 1, the right-hand side becomes arbitrarily small. By condition (i), $f_j\left(y\right)\ne 0$. If $f_j\left(y\right)>0$, then convexity of $f_j$ gives

$$f_j(ty+(1-t)x_n)\le t{f}_j\left(y\right)+(1-t)f_j\left(x_n\right).$$

For t close to 1, the right-hand side is approximately $f_j\left(y\right)>0$, while for t close to 0 it is approximately $f_j\left(x_n\right)\ge {\delta }_n>0$. This does not immediately yield a contradiction.

To obtain a rigorous argument, we use the fact that the zero sets are compact and disjoint. There exists $\eta >0$ such that the $\eta$-neighborhoods of $f_i^{-1}\left(0\right)$ and $f_j^{-1}\left(0\right)$ are disjoint. For sufficiently small $\delta$, the sets $\{f_i\ge \delta \}$ and $\{f_j\ge \delta \}$ are contained in these neighborhoods respectively, hence disjoint. More precisely, by continuity of the functions, there exists ${\delta }_0>0$ such that for all $0<\delta <{\delta }_0$, we have

$$\{x\in E:f_i\left(x\right)\ge \delta \}\cap \{x\in E:f_j\left(x\right)\ge \delta \}=⌀\forall i\ne j.$$

Thus, for $\delta \in (0,{\delta }_0)$, the sets $A_i$ are pairwise disjoint.

Removing any empty $A_i$, we obtain a finite disjoint union

$$E_{\delta }=\underset{i=1}{\overset{m}{⨆}}{A}_i,m\le k,$$

where each $A_i$ is nonempty.

• Convex components

Each $A_i$ is convex, being the intersection of the convex set E with the superlevel set $\{f_i\ge \delta \}$ of a convex function. To see that these are the convex components of $E_{\delta }$, suppose $B\subseteq E_{\delta }$ is convex with $A_i⊊B$. Then there exists $x\in B\setminus A_i$. Since the $A_j$’s are disjoint and cover $E_{\delta }$, $x\in A_j$ for some $j\ne i$. Take any $y\in A_i$. By convexity of B, the segment $[x,y]\subseteq B\subseteq E_{\delta }$. But this segment must intersect the region where $h<\delta$ (since the sets $A_i$ and $A_j$ are separated), contradicting $B\subseteq E_{\delta }$. Hence each $A_i$ is maximal convex in $E_{\delta }$, establishing part (1).

• Step 2: Finite perimeter of the components

We now prove that each $A_i$ has locally finite perimeter. Write $A_i=E\cap G_i$, where $G_i=\{f_i\ge \delta \}$.

Since $f_i$ is convex and continuously differentiable near its zero set, for sufficiently small $\delta$, the level set $f_i^{-1}\left(\delta \right)$ is a $C^1$ hypersurface. Indeed, by the implicit function theorem, the condition $\nabla f_i\ne 0$ on $f_i^{-1}\left(0\right)$ extends to a neighborhood, ensuring that $f_i^{-1}\left(\delta \right)$ is a $C^1$ manifold for small $\delta$. Consequently, $G_i$ is a set with $C^1$ boundary.

A standard result in geometric measure theory (see [18], [Theorem 2.9]) states that the intersection of a set of locally finite perimeter with a set having $C^1$ boundary is again of locally finite perimeter. Since E is convex with nonempty interior and locally Lipschitz boundary, it has locally finite perimeter. The set $G_i$ has $C^1$ boundary, hence is a $C^1$ domain. Therefore, $A_i=E\cap G_i$ has locally finite perimeter.

• Structure of the reduced boundary

For a set of locally finite perimeter, the reduced boundary ${\partial }^{*}{A}_i$ is contained in the topological boundary $\partial A_i$ and satisfies ${\partial }^{*}{A}_i\subseteq \partial E\cup \partial G_i$ modulo an ${\mathcal{H}}^{n-1}$-null set. More precisely,

$${\partial }^{*}{A}_i\subset \left({\partial }^{*}E\cap \overline{G_i}\right)\cup \left(\overline{E}\cap {\partial }^{*}{G}_i\right)\cup N,$$

where N is ${\mathcal{H}}^{n-1}$-negligible. Since ${\partial }^{*}{G}_i$ coincides with $f_i^{-1}\left(\delta \right)$ (up to a null set) and ${\partial }^{*}E$ is contained in $\partial E$ for convex sets, we obtain

$${\partial }^{*}{A}_i\subset \partial E\cup f_i^{-1}\left(\delta \right)\cup N,$$

proving part (2).

• Step 3: Additivity of the perimeter measure

We now address the additivity of the perimeter measure under the additional transversality condition. Assume that $\partial E$ is $C^1$ and that $\nabla f_i\left(x\right)·{\nu }_E\left(x\right)\ne 0$ for all $x\in \partial E\cap f_i^{-1}\left(0\right)$.

For sufficiently small $\delta$, this transversality extends to the level sets $f_i^{-1}\left(\delta \right)$: there exists a neighborhood U of $\partial E\cap f_i^{-1}\left(0\right)$ such that $\nabla f_i\left(x\right)·{\nu }_E\left(x\right)\ne 0$ for all $x\in U\cap f_i^{-1}\left(\delta \right)$. This ensures that near points of $\partial E\cap f_i^{-1}\left(\delta \right)$, the level set $f_i^{-1}\left(\delta \right)$ is a $C^1$ hypersurface that intersects $\partial E$ transversally.

Under these conditions, the reduced boundaries ${\partial }^{*}{A}_i$ and ${\partial }^{*}{A}_j$ for $i\ne j$ are essentially disjoint. Indeed, their possible intersection points would have to lie in $\partial E\cap f_i^{-1}\left(\delta \right)\cap f_j^{-1}\left(\delta \right)$, but for small $\delta$ this set is empty by the disjointness of the zero sets and continuity. Moreover, at points where ${\partial }^{*}{A}_i$ meets $\partial E$, the outward normals are well-defined and distinct, preventing cancellations in the perimeter measure.

Therefore, for any Borel set $B\subset {\mathbb{R}}^n$,

$$P(E_{\delta };B)=\sum _{i=1}^{m}P(A_i;B).$$

This follows from the locality of the perimeter and the fact that the reduced boundaries of distinct components intersect only on ${\mathcal{H}}^{n-1}$-negligible sets.

• Step 4: Choice of δ

The existence of ${\delta }_0>0$ such that all the above conclusions hold for $0<\delta <{\delta }_0$ follows from:

• The pairwise disjointness of the zero sets, together with continuity, ensures that the superlevel sets remain disjoint for small $\delta$.
• The condition $\nabla f_i\ne 0$ on $f_i^{-1}\left(0\right)$ extends to a neighborhood by continuity, so $f_i^{-1}\left(\delta \right)$ remains a $C^1$ hypersurface for small $\delta$.
• The transversality condition on $\partial E\cap f_i^{-1}\left(0\right)$ persists on a neighborhood, guaranteeing that $f_i^{-1}\left(\delta \right)$ intersects $\partial E$ transversally for small $\delta$.

Thus, choosing $\delta$ smaller than the minimum of the positive bounds obtained from these considerations yields all the claimed properties.

This completes the proof of Theorem 6. □

The following theorem, which we refer to as the Limited Perimeter Decomposition, provides a precise geometric and measure-theoretic description of multi-slices generated by the absolute value of a convex function. Specifically, for a convex set E with a locally Lipschitz boundary and a convex function f that is continuously differentiable with a non-vanishing gradient near its zero set, the superlevel set $E_{\delta }=\{x\in E:|f\left(x\right)|\ge \delta \}$ decomposes for sufficiently small $\delta >0$ into two disjoint, convex components $E_{\delta }^{+}$ and $E_{\delta }^{-}$. A key aspect of the proof is establishing that these components are not merely convex but also possess the crucial geometric-measure property of having locally finite perimeter [18,22]. This is achieved by recognizing $E_{\delta }^{+}$ (and similarly $E_{\delta }^{-}$) as the intersection of the set of locally finite perimeter E with the $C^1$ domain $\{f\ge \delta \}$, a combination that preserves the finite perimeter property thanks to a standard result in geometric measure theory (see Theorem 2.9 of [18]). Furthermore, the theorem establishes that the perimeter measure of the combined set $E_{\delta }$ is the sum of the perimeter measures of its components, a relation that holds away from the shared level sets $f^{-1}\left(\delta \right)$ and $f^{-1}(-\delta )$. This additive property follows from the fact that the reduced boundaries of the two components are essentially disjoint [23], a consequence of the disjointness of the components themselves and the regularity of the level sets as $C^1$ hypersurfaces [4,7].

Theorem 7

(Limited Perimeter Decomposition). Let $E\subset {\mathbb{R}}^n$ be a convex set with nonempty interior and locally Lipschitz boundary. Let $f:{\mathbb{R}}^n\to \mathbb{R}$ be a convex function that is continuously differentiable in a neighborhood of $\partial E\cap f^{-1}\left(0\right)$ and satisfies

$$\nabla f\left(x\right)\ne 0\mathit{for}\mathit{all}x\in f^{-1}\left(0\right).$$

Define $h\left(x\right)=max\left\{f\right(x),-f(x\left)\right\}=\left|f\right(x\left)\right|$. Then for all sufficiently small $\delta >0$, the set

$$E_{\delta }=\{x\in E:|f\left(x\right)|\ge \delta \}$$

decomposes as

$$E_{\delta }=E_{\delta }^{+}\cup E_{\delta }^{-},$$

where

$$E_{\delta }^{+}=\{x\in E:f\left(x\right)\ge \delta \},E_{\delta }^{-}=\{x\in E:f\left(x\right)\le -\delta \}.$$

Both $E_{\delta }^{+}$ and $E_{\delta }^{-}$ are convex and have locally finite perimeter. Moreover,

$$P(E_{\delta };·)=P(E_{\delta }^{+};·)+P(E_{\delta }^{-};·)$$

as measures on ${\mathbb{R}}^n\setminus (f^{-1}\left(\delta \right)\cup f^{-1}(-\delta ))$.

Proof. 

The proof proceeds in several steps, each carefully justified.

Step 1: Basic properties. Since f is convex and continuously differentiable near its zero set, the level sets $f^{-1}\left(c\right)$ are $C^1$ hypersurfaces for all c sufficiently close to 0. The condition $\nabla f\left(x\right)\ne 0$ on $f^{-1}\left(0\right)$ ensures, by the implicit function theorem, that each level set is locally a $C^1$ graph.

Step 2: Disjointness and convexity. For $\delta >0$, the sets $E_{\delta }^{+}$ and $E_{\delta }^{-}$ are clearly disjoint because a point cannot simultaneously satisfy $f\left(x\right)\ge \delta$ and $f\left(x\right)\le -\delta$. Both sets are convex: $E_{\delta }^{+}=E\cap f^{-1}\left([\delta ,\infty )\right)$ is the intersection of two convex sets, hence convex, which is similar for $E_{\delta }^{-}$.

Step 3: Finite perimeter property. The set E has locally finite perimeter because it is convex with nonempty interior (a classical result: convex sets are locally of finite perimeter). The set $\{x\in {\mathbb{R}}^n:f\left(x\right)\ge \delta \}$ is a superlevel set of a convex function; such sets are convex and hence also have locally finite perimeter.

However, the intersection of two sets of locally finite perimeter need not have locally finite perimeter in general. To conclude that $E_{\delta }^{+}=E\cap \{f\ge \delta \}$ has locally finite perimeter, we need additional regularity. The key observation is that the boundary of $\{f\ge \delta \}$ is the level set $f^{-1}\left(\delta \right)$, which is a $C^1$ hypersurface. A standard result in geometric measure theory states that the intersection of a set of locally finite perimeter with a $C^1$ domain is again of locally finite perimeter (see Theorem 2.9 of [18]). Since $\{f\ge \delta \}$ is a $C^1$ domain (its boundary is $C^1$), we conclude that $E_{\delta }^{+}$ has locally finite perimeter. The same argument applies to $E_{\delta }^{-}$.

Step 4: Structure of the reduced boundaries. For a convex set, the reduced boundary coincides ${\mathcal{H}}^{n-1}$-almost everywhere with the topological boundary. Thus

$${\partial }^{*}{E}_{\delta }^{+}=\left({\partial }^{*}E\cap \{f\ge \delta \}\right)\cup \left(E\cap f^{-1}\left(\delta \right)\right)\cup N,$$

where N is an ${\mathcal{H}}^{n-1}$-null set. Similarly for $E_{\delta }^{-}$.

The transversality condition $\nabla f\left(x\right)\ne 0$ on $f^{-1}\left(\delta \right)$ ensures that $f^{-1}\left(\delta \right)$ is a $C^1$ hypersurface, and near this hypersurface, the reduced boundary of $E_{\delta }^{+}$ is exactly the portion of this hypersurface that lies inside E.

Step 5: Additivity of perimeter measure. For any Borel set B that does not intersect $f^{-1}\left(\delta \right)\cup f^{-1}(-\delta )$, the reduced boundaries of $E_{\delta }^{+}$ and $E_{\delta }^{-}$ are disjoint. Indeed, ${\partial }^{*}{E}_{\delta }^{+}\setminus f^{-1}\left(\delta \right)\subset {\partial }^{*}E\cap \{f>\delta \}$ and ${\partial }^{*}{E}_{\delta }^{-}\setminus f^{-1}(-\delta )\subset {\partial }^{*}E\cap \{f<-\delta \}$, and these two sets are disjoint. Therefore, for such B,

$$P(E_{\delta };B)=P(E_{\delta }^{+};B)+P(E_{\delta }^{-};B).$$

On sets that intersect the level sets $f^{-1}(\pm \delta )$, the additivity may fail because the reduced boundaries of the two components can meet. However, these level sets have zero Lebesgue measure and are ${\mathcal{H}}^{n-1}$-rectifiable, so the failure of additivity occurs only on a set of measure zero with respect to the perimeter measure.

Step 6: Choice of $\delta$. The requirement that $\delta$ be sufficiently small ensures that the level sets $f^{-1}(\pm \delta )$ remain $C^1$ hypersurfaces and that the implicit function theorem applies uniformly. Since $\nabla f\ne 0$ on $f^{-1}\left(0\right)$, by continuity there exists ${\delta }_0>0$ such that $\nabla f\ne 0$ on $f^{-1}\left([-{\delta }_0,{\delta }_0]\right)$. For any $0<\delta <{\delta }_0$, all the above conclusions hold. □

7. Conclusions

This work has established a comprehensive theory of multi-slice decompositions via convex functions, extending the classical framework of slices determined by linear functionals to the broader setting of arbitrary convex functions with disjoint zero sets. Our main structure theorem (Theorem 4) provides a complete characterization of the convex component decomposition of multi-slices, revealing that under natural conditions of pairwise disjoint zero sets and convex separation, the multi-slice decomposes canonically into convex components corresponding precisely to the individual functions in the family. The developed theory demonstrates several fundamental properties: the component-wise exposing nature of the supremum function, the closedness of components in appropriate topologies, the maximality of the resulting decomposition, and the affine invariance of convex component structures (Theorem 5). These results not only generalize earlier work on multi-slices but also establish deep connections between the geometric structure of level sets and the functional analytic properties of convex functions, providing new tools for analyzing convex sets under nonlinear constraints with potential applications in optimization theory, high-dimensional data analysis, and modern convex geometry. The theory of multi-slices via convex functions thus represents a fertile ground for continued investigation at the intersection of convex analysis, functional analysis, and geometric optimization, with numerous promising directions for further research including infinite-dimensional settings, computational aspects, and non-commutative extensions.