| Sbornik: Mathematics |
|
|
|
|
|
Autopolar conic bodies and polyhedra M. S. Makarovab, V. Yu. Protasovc a Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia b Lomonosov Moscow State University, Moscow, Russia; c Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
Bibliographic databases:
    § 1. Introduction1.1. Antinorms and conic polyhedraThe word ‘antinorm’ is understood in various senses. Following most of the literature we define it as a ‘concave norm’, that is, a positively homogeneous concave nonnegative function. It is shown easily that such a function cannot be defined on the whole of $\mathbb{R}^d$ unless it is identical zero. Therefore, an antinorm is usually restricted to a cone. Brief historical remarks and examples of applications are presented below, in § 1.2. Level sets of an antinorm are convex and unbounded. Such ‘antiballs’ are referred to as conic bodies; in particular, conic polyhedra are conic bodies bounded by several hyperplanes. Most geometric properties of these objects are similar to what we have in the classical convex analysis. There are, however, some significant differences, one of which, the self-duality phenomenon, is the subject of this paper. We consider a cone $K \subset \mathbb{R}^d$ with apex at the origin. It is assumed to be closed, solid (with nonempty interior) and pointed (containing no straight lines). As usual, faces of a cone are its intersections with planes of support. The apex is a face of dimension $0$, a face of dimension $1$ is an edge, and a face of dimension $d-1$ is a facet. A cone is polyhedral if it is defined by a system of several linear inequalities. An interior ray is a ray going from the origin and passing through an interior point. The dual cone is defined in the usual way by $K^*= \bigl\{\boldsymbol{y} \in \mathbb R^d\colon \inf_{\boldsymbol{x} \in K} (\boldsymbol{y}, \boldsymbol{x}) \geqslant 0\bigr\}$. Definition 1. An antinorm on a cone $K$ is a concave, homogeneous, nonnegative function $f\colon K\to \mathbb{R}_+$ which is positive somewhere. Homogeneity means that $f(\lambda\boldsymbol{x})=\lambda f(\boldsymbol{x})$, $\lambda \in \mathbb{R}$. Since the cone $K$ is pointed, $\lambda$ must be nonnegative unless $\boldsymbol{x} =0$, otherwise $\lambda\boldsymbol{x} \notin K$. An antinorm can vanish only on the boundary of $K$. It may not be continuous, but all of its points of discontinuity are on the boundary. Surprisingly enough, there are examples of antinorms that do not have a continuous extension from the interior of $K$ to the boundary. However, such an extension always exists and is unique for polyhedral cones, in particular, for $K = \mathbb{R}^d_+$; see [27]. The set $G=\{\boldsymbol{x} \in K\colon f(\boldsymbol{x}) \geqslant R\}$ is a ball of radius $R$ of the antinorm $f$. Since it is defined by a reverse inequality ($f(\boldsymbol{x}) \geqslant R$ instead of $f(\boldsymbol{x})\leqslant R$), it should rather be called an ‘antiball’, but we use simplified terminology. If $\boldsymbol{x} \in G$, then $\lambda \boldsymbol{x} \in G$ for all $\lambda \geqslant 1$, and therefore a ball of an antinorm is never bounded. We usually deal with unit balls, when $R=1$. The set $S = \partial G$ is a sphere of the antinorm. Definition 2. Given a cone $K$, a conic body $G$ is a closed convex proper subset of $K$ such that for every interior ray $\ell \subset K$ the set $G\cap \ell$ is a ray. A ball of an antinorm is a conic body. Conversely, every conic body is a unit ball of the antinorm $f_G(\boldsymbol{x})=\sup\{\lambda>0\colon \lambda^{-1}\boldsymbol{x} \in G\}$, which is an analogue of the standard Minkowski functional. A finite intersection of conic bodies is a conic body, and the same is true for their Minkowski sum. However, this is in general not true for the convex hull. This operation is replaced by the positive convex hull $\operatorname{co}_{+} (X)=\operatorname{co}(X)+K$, where $\operatorname{co}(\,{\cdot}\,)$ denotes the standard convex hull. For every $X\subset K$, the closure of $\operatorname{co}_{+} (X)$ is either a conic body or the whole cone $K$. In particular, the positive convex hull of a finite set not containing the origin is a conic polytope. Definition 3. A conic polyhedron $G$ is the set $P = \{\boldsymbol{x}\in K\colon (\boldsymbol{a}_j,\boldsymbol{x})\geqslant1, {j=1,\dots,n}\}$, where the $\boldsymbol{a}_j$ are nonzero elements of $K^*$. Thus, a conic polyhedron is a conic body bounded by several hyperplanes. The corresponding antinorm is piecewise linear: $f_P(\boldsymbol{x})=\min_{i} (\boldsymbol{a}_i, \boldsymbol{x})$. In general, a conic polyhedron may not be a conic polytope. Nevertheless, if the cone $K$ is polyhedral, in particular, if $K = \mathbb{R}^d_+$, then these two notions coincide. For the sake of simplicity, in what follows we assume that $K=\mathbb{R}^d_+$. Since each antinorm on $\mathbb{R}^d_+$ has a unique continuous extension, we consider only continuous antinorms. As we see, the main facts of the classical convex analysis have analogues for antinorms and conic bodies/polyhedra, with the replacement of $\sup$ by $\inf$ and $\operatorname{co}(\,{\cdot}\,)$ by $\operatorname{co}_+(\,{\cdot}\,)$. The same can be said of duality theory but with one exception: while in the classical convex analysis there is a unique autopolar set (the unit Euclidean ball), the family of autopolar conic bodies and polyhedra in $\mathbb{R}^d_+$ is quite rich and diverse. We give formal definitions and preliminary facts in § 2, and then in § 3 we formulate the main results. Theorem 1 will allow us to construct infinitely many families of self-dual antinorms and autopolar conic bodies. In case $d=2$ this gives a complete classification of autopolar sets (Theorem 3) while for $d=3$ there is a counterexample (Theorem 4) showing that not all such sets are produced by means of Theorem 1. A complete classification of self-dual antinorms and autopolar sets is formulated as an open problem in § 6. Throughout the paper we denote vectors by bold characters and their coordinates and other scalars by ordinary characters, so that $\boldsymbol{x} = (x_1,\dots,x_d)$. The Euclidean norm is denoted by $|\boldsymbol{x}|$. By a projection we always mean the orthogonal projection of $\mathbb{R}^d$ onto a proper affine subspace. The preimage under this operator will be referred to as the orthogonal extension. So the orthogonal extension of a set $X \subset \mathbb{R}^d$ is $\{\boldsymbol{x} + \boldsymbol{h}\colon \boldsymbol{x} \in X,\,\boldsymbol{h}\in X^{\perp}\}$, where $X^{\perp}$ denotes the orthogonal complement to the linear span of $X$. 1.2. Applications and related worksTo the best of our knowledge, Definition 1 originated with Merikoski [18], and the first results on antinorms were obtained in [19] and [20]. Matrix antinorms on a positive semidefinite cone were studied in [2] and [3]; some authors considered the Minkowski antinorm $f(A) = (\operatorname{det}A)^{1/d}$ and the Schatten $q$-antinorms $f(A) = (\operatorname{tr}A)^{1/q}$, $q \in (-\infty, 1]$; see also [4] for extensions to von Neumann algebras and [17] on antinorms on Radon curves. Independently and later, antinorms were defined in [24] for an analysis of infinite products of random matrices. Antinorms have been applied to linear dynamical systems of the form $\dot{\boldsymbol{x}}=A(t)\boldsymbol{x}$, $t\geqslant 0$ (continuous time), and $\boldsymbol{x}_{k+1} = A(k)\boldsymbol{x}_k$, $k\geqslant 0$ (discrete time), where the matrix $A(\,{\cdot}\,)$ is chosen independently from a fixed compact set for each value of the argument. If the system is asymptotically stable, that is, all of its trajectories $\boldsymbol{x}_k$ tend to zero as $k\to \infty$, then it possesses a convex Lyapunov function (a Lyapunov norm). This norm is the main tool in the proof of stability [13], [21], [14]. However, for other types of stability a Lyapunov norm may not exist. This occurs, for instance, for almost sure stability, when almost all trajectories tend to zero [5], [7]. Nevertheless, in this case a Lyapunov function can be constructed as a concave homogeneous function, that is, an antinorm [12], [26], [25]. The same can be said on the stabilizability problem, when the existence of at least one stable trajectory is proved using a suitable antinorm [1], [6], [8]–[10], [29]. Since a nontrivial antinorm cannot be defined on the whole space, this is applied to systems with an invariant cone, in particular, positive systems. Dual norms and antinorms appear naturally in considerations of the dual systems defined by transposed matrices [16], [23], [27]. For applications of antinorms to convex trigonometry in optimal control, see [15]. Remark 1. Since a positive homogeneous function cannot be concave on the whole space $\mathbb{R}^d$, antinorms are usually defined on convex cones. Some works consider collections of antinorms on partitions of $\mathbb{R}^d$ into convex cones. An equivalent definition uses the Minkowski functionals of star-shaped sets [22], [28]. § 2. Duality2.1. Dual antinorms and polar conic bodiesDefinition 4. For an antinorm $f$ on $\mathbb{R}^d_+$, its dual antinorm is $f^{*}(\boldsymbol{y})=\inf_{f(\boldsymbol{x}) \geqslant 1}(\boldsymbol{y}, \boldsymbol{x})$. We also write $f^{*}(\boldsymbol{y})=\inf_{\boldsymbol{x} \in \mathbb R^d_+}(\boldsymbol{y}, \boldsymbol{x})/f(\boldsymbol{x})$, assuming that the infimum is taken over the points such that $f(\boldsymbol{x})\ne 0$. Definition 5. For a conic body $G\mkern-1mu\!\subset\!\mkern-1mu \mathbb{R}^d_+$, its polar is ${G^*\mkern-1mu\!=\!\mkern-1mu \bigl\{\mkern-1mu\boldsymbol{y}\mkern-1mu\!\in\!\mkern-1mu \mathbb R^d_+\colon \mkern-2mu\inf_{\boldsymbol{x} \in G}(\mkern-1mu\boldsymbol{y}, \boldsymbol{x}\mkern-1mu)\mkern-1mu\!\geqslant\!\mkern-1mu 1\mkern-2mu\bigr\}}$. It follows easily that if an antinorm $f$ has a unit ball $G$, then the dual antinorm $f^*$ has the unit ball $G^*$. The main properties of duality are very similar to those in the classical convex analysis:
The first and third properties follow directly from the definition. For the proof of the second (on the bipolar and double dual), see [11]. Formally, one can define the polar of an arbitrary subset $M\subset \mathbb{R}^d_+$ by the same formula:
$$
\begin{equation*}
M^*=\Bigl\{\boldsymbol{y} \in \mathbb R^d_+\colon \inf_{\boldsymbol{x} \in M}(\boldsymbol{x}, \boldsymbol{y}) \geqslant 1\Bigr\}.
\end{equation*}
\notag
$$
If the closure of $M$ does not contain the origin, then $M^*$ is nonempty. Let $G$ denote the closure of $\operatorname{co}_+(M)$. Then $G$ is a conic body and $M^* = G^*$. Thus, the polar of an arbitrary set is the polar of the closure of its positive convex hull (which is a conic body). In particular, for every set $M$ separated from the origin we have $M^{**}=\overline{\operatorname{co}_+(M)}$. Example 1. The polar of a point $\boldsymbol{a} \in \mathbb{R}^d_+$, $\boldsymbol{a} \ne 0$, is the same as the polar of the conic polyhedron $\operatorname{co}_+(\boldsymbol{x}) =\{\boldsymbol{x} \in \mathbb{R}^d_+\colon x_i \geqslant a_i,\,i = 1,\dots, d\}$, and it is equal to the conic polyhedron $\boldsymbol{a}^*=\{\boldsymbol{y} \in \mathbb{R}^d_+\colon (\boldsymbol{a}, \boldsymbol{y}) \geqslant 1\}$. This polyhedron is obtained by cutting off the corner of the positive orthant $\mathbb{R}^d_+$ by the hyperplane $H$ orthogonal to $\boldsymbol{a}$ and passing through the point $\boldsymbol{a}/|\boldsymbol{a}|^2$ (reciprocal to the point $\boldsymbol{a}$). Sometimes we consider this hyperplane as the polar of $\boldsymbol{a}$. The polar of a conic polyhedron $Q$ with vertices $\boldsymbol{a}_1,\dots,\boldsymbol{a}_n$ (in which case $Q= \operatorname{co}_+ \{\boldsymbol{a}_1,\dots,\boldsymbol{a}_n\}$) is the conic polyhedron $Q^*$ defined by the system of inequalities $(\boldsymbol{a}_i, \boldsymbol{x}) \geqslant 1$, $i=1,\dots,n$. We see that duality theory is very similar to what we have in the classical case, with two exceptions. First, the map $f\mapsto f^*$ can be discontinuous [27]. The second exception is the self-duality issue addressed in the next subsection. 2.2. Self-dualityAn antinorm $f$ on $\mathbb{R}^d_+$ is self-dual if $f^* = f$; a conic body $G$ is autopolar if $G^* = G$. The self-duality of an antinorm is equivalent to the autopolarity of its unit ball $G$. This means that for an arbitrary point $A \in \partial G$ its polar is a hyperplane of support for $G$, and vice versa (Figure 1). Recall that in convex analysis self-duality is very rare. To avoid a confusion we denote the (classical) polar of a set $Q\subset \mathbb{R}^d$ by $\widehat Q$. Thus, $\widehat Q=\{\boldsymbol{y} \in \mathbb{R}^d\colon\sup_{\boldsymbol{x} \in Q} (\boldsymbol{y}, \boldsymbol{x}) \leqslant 1\}$. The following fact is well known, and we include its proof for the convenience of the reader. Fact. The Euclidean unit ball $B=\{\boldsymbol{x} \in \mathbb{R}^d\colon |\boldsymbol{x}|\leqslant 1\}$ is the unique subset of $\mathbb{R}^d$ such that $\widehat B=B$. Indeed, if a set $Q$ coincides with $\widehat Q$, then $Q \subset B$. Otherwise, $Q$ contains a point $\boldsymbol{y}$ such that $(\boldsymbol{y}, \boldsymbol{y}) > 1$ and therefore $\boldsymbol{y} \notin \widehat Q$. On the other hand the inclusion ${Q \subset B}$ implies that $\widehat B \subset \widehat Q$, and so $B \subset Q$ because both $Q$ and $B$ are autopolar. Hence ${Q=B}$. Thus, apart from the unit Euclidean ball, there are no autopolar sets in $\mathbb{R}^d$. One, therefore, could expect that there are no autopolar conic bodes in $\mathbb{R}^d_+$ either. However, the situation is totally opposite: there is a lot of autopolar conic bodies, including polyhedra. This phenomenon was observed for the first time for $d=2$ in [27]. For general $d \geqslant 3$ the orthogonal extension (preimage under the projection) of an autopolar plane body is also autopolar. Are there other examples? The following family of smooth autopolar conic bodies (with smooth strictly convex boundary) was found in [27]. We describe it in terms of the corresponding antinorms. Example 2. For an arbitrary collection of nonnegative numbers $\{p_i\}_{i=1}^d$ such that $\sum_{i=1}^d p_i = 1$ the function (for $p_i=0$ we set $({x_i}/{\sqrt{p_i}})^{p_i}=1$), is a continuous self-dual antinorm. For example, if $p_1=p_2 = 1/2$, then we obtain the self-dual antinorm $f(x_1, x_2) = \sqrt{2x_1x_2}$ in $\mathbb{R}^2_+$. The corresponding autopolar ball $G$ is bounded by the hyperbola. The function (1) can be expressed in a simpler form as $f(\boldsymbol{x}) = Cx_1^{p_1}\cdots x_d^{p_d}$, where $C$ is a constant depending on $p_i$. There was a mistake in the computation of this constant in [27], and now we correct it. In Proposition 3 we show that the self-duality of (1), along with infinitely many other smooth self-dual antinorms, follows from the general result of Theorem 1. This has been the only known example of a nontrivial smooth self-affine antinorm. Theorem 1 in § 3 provides a rich variety of other examples. The problem of the existence of nontrivial autopolar conic polyhedra in dimensions $d\geqslant 3$ was stated in [27] and positively solved in [16]. For all $d\geqslant 3$ and $n\geqslant 1$ there exists an autopolar conic polyhedron with $n$ vertices in $\mathbb{R}^d_+$. We consider an example for $d=3$ and $n=1$. Example 3. In the positive orthant in $\mathbb{R}^3$ we take an arbitrary point $A_1$ on the $OX_3$-axis such that $OA_1 > 1$ (Figure 2). We denote by $\alpha_1$ the hyperplane polar to $A_1$. It is parallel to $OX_1X_2$ and passes through the point $A_1' \in OX_3$ such that $OA_1'={1}/{OA_1}$. We denote by $\ell_2$ the ray of intersection of $\alpha_1$ and the sector $OX_1X_3$. Fix an arbitrary point $A_2 \in \ell_2$ and denote by $\alpha_2$ its polar (a hyperplane). Let it intersect $\ell_2$ at a point $A_2'$. Let $\ell_3$ and $\ell_1$ denote the rays of intersection of $\alpha_2$ with $\alpha_1$ and $OX_2X_3$, respectively. Note that $A_1$ is the endpoint of $\ell_1$. Finally, consider the point $A_3$ on $\ell_3$ such that all three angles $A_1A_3O$, $A_2A_3O$ and $A_1A_3A_2$ are right. Such a point exists and is unique since the ray $\ell_3$ is a perpendicular to the plane of the triangle $A_1A_2O$ erected at its orthocentre. We denote the plane of the triangle $A_1A_2A_3$ by $\alpha_3$. Then the conic polyhedron $P=\operatorname{co}_+\{A_1, A_2, A_3\}$ is autopolar. Indeed, it is bounded (apart from the coordinate planes) by the three hyperplanes $\alpha_1$, $\alpha_2$ and $\alpha_3$, and each plane $\alpha_i$ is polar to $A_i$, $i = 1, 2, 3$; see [16] for the details. The polyhedron $P$ has three vertices $A_1$, $A_2$ and $A_3$, three facets $A_1A_2A_3$, $\ell_2A_2A_3\ell_3$ and $\ell_1A_1A_3\ell_3$ and three facets lying on coordinate subspaces. No autopolar polyhedra different from those presented in [16] have been known. Theorem 1 produces an infinite variety of autopolar bodies, both smooth and polyhedral ones. Before formulating it, we observe several general properties of autopolar objects. Proof. The definition of the dual antinorm implies that if ${f\geqslant g}$, then ${f^*\leqslant g^*}$ and therefore ${f\leqslant g}$. The proposition is proved. Proof. Let $f$ be a self-dual antinorm generated by an autopolar body $G$. We denote by $A$ the point in $G$ closest to the origin and by $\boldsymbol{a}$ the vector $OA$. The analogue to Young’s inequality (see above) yields $(\boldsymbol{a}, \boldsymbol{a}) \geqslant f(\boldsymbol{a})f^*(\boldsymbol{a})$. On the other hand ${\boldsymbol{a} \in \partial G}$, and therefore $f(\boldsymbol{a}) = 1$, so that $f^*(\boldsymbol{a}) =1$. Thus, $(\boldsymbol{a}, \boldsymbol{a}) \geqslant 1$. The hyperplane $H$ passing through $A$ and orthogonal to $\boldsymbol{a}$ is a plane of support for $G$. Therefore, its polar $H^*$, which is the point $(1/|\boldsymbol{a}|^2)A$, lies on the unit sphere of the antinorm $f^* = f$. Hence the length of $\boldsymbol{a}/|\boldsymbol{a}|^2$ is at least $1$, and therefore $|\boldsymbol{a}| \leqslant 1$, so that $|\boldsymbol{a}| =1$.
The proposition is proved. Example 4. For the unit ball of the smooth antinorm (1) from Example 2 the point closest to the origin is $\boldsymbol{a} = (\sqrt{p_1},\dots,\sqrt{p_d})$. Clearly, $|\boldsymbol{a}| = 1$. For the conic polyhedron $P$ in Example 3 the point closest to the origin is $\boldsymbol{a} = A_3$, and $|\boldsymbol{a}| = 1$. Indeed, since the plane $A_1A_2A_3$ is the polar of $A_3$, it follows that $\boldsymbol{a} = |OA_3| = 1$. Now Proposition 2 implies that $A_3$ is the closest point. Applying Proposition 2 to the unit ball of a self-dual antinorm we see that $|\boldsymbol{x}| \geqslant 1$ whenever $f(\boldsymbol{x}) \geqslant 1$. Therefore, $f(\boldsymbol{x}) \geqslant |\boldsymbol{x}|$ and this inequality is strict unless $\boldsymbol{x}$ is collinear to $\boldsymbol{a}$. Thus, we obtain the following result. Corollary 1. Every self-dual antinorm does not exceed the Euclidean norm and there is a unique point where both of them are equal to $1$. § 3. The main resultsWe begin with Theorem 1, which asserts that every self-dual antinorm can be extended (lifted) from a suitable hyperplane to the whole cone $\mathbb{R}^d_+$. This lifting is not unique, and every $(d-1)$-dimensional self-dual antinorm can produce infinitely many $d$-dimensional ones. This gives us a large variety of autopolar conic bodies, smooth and polyhedral alike, constructed inductively. Then we prove Theorem 2, which describes the structure of these antinorms. For $d=2$ this method produces all autopolar sets (Theorem 3), while for $d=3$ there is an autopolar polyhedron not obtained in this way (Example 9). We call a hyperplane $V \subset \mathbb{R}^d$ admissible if it has the form ${V\!=\!\{\boldsymbol{x} \!\in\! \mathbb R^d\colon x_i\!=\!\mu x_j\}}$, where $i$ and $j$ are two different indices from $\{1,\dots,d\}$ and $\mu \geqslant 0$ is arbitrary. Thus, an admissible hyperplane contains $d-2$ coordinate axes and intersect $\mathbb{R}^d_{+}$ in an $(d-1)$-dimensional orthant $\mathbb{R}^{d-1}_+$, which we denote by $K'$. 3.1. Lifting theoremsConsider an arbitrary admissible hyperplane $V$ that cuts the orthant $\mathbb{R}^d_+$ into two parts $K_1$ and $K_2$. We set $K'= \mathbb{R}^d_+ \cap V$. Let $\varphi$ be an antinorm on $K'$, $G' \subset K'$ be its unit ball and $\Phi$ be its orthogonal extension to $\mathbb{R}^d_+$ defined by $\Phi(\boldsymbol{x}) = \varphi(\boldsymbol{x}')$, $\boldsymbol{x}\in \mathbb{R}^d_+$, where $\boldsymbol{x}'$ is the orthogonal projection of $\boldsymbol{x}$ onto $V$. Thus, $\varphi$ is defined on $K'$, while $\Phi$ is defined on $\mathbb{R}^d_+$. The unit ball of $\Phi$ is the intersection of $\mathbb{R}^d_+$ with the right cylinder based on $G'$. Theorem 1. Let an admissible hyperplane $V$ cut $\mathbb{R}^d_+$ into closed parts $K_1$ and $K_2$ and let $\varphi$ be a self-dual antinorm on $K'$. Let $f_1\colon K_1\to \mathbb{R}_+$ be an arbitrary antinorm that coincides with $\varphi$ on $K'$ and does not exceed $\Phi$ on $K_1$. Then the function $f(\boldsymbol{x})$ defined on $\mathbb{R}^d_+$ by is a self-dual antinorm on $\mathbb{R}^d_+$. Thus, a self-dual antinorm $\varphi$ on $\mathbb{R}^{d-1}_+$ produces infinitely many antinorms of this type on $\mathbb{R}^d_+$. Namely, consider $\varphi$ on an arbitrary admissible hyperplane $V$, and take an arbitrary antinorm $f_1$ not exceeding the orthogonal extension of $\varphi$ on the part $K_1$. Then we obtain a self-dual antinorm by formula (2). The structure of this antinorm is clarified by Theorem 2 below. Let an admissible hyperplane $V$ cut the positive orthant $K=\mathbb{R}^d_+$ into two closed parts $K_1$ and $K_2$, and let $K'= K\cap V$. Suppose functions $f_1$ and $f_2$ are defined on $K_1$ and $K_2$, respectively, and coincide on $K'$; then their concatenation $f = f_1\cup f_2$ is defined on $\mathbb{R}^d_+$ by
$$
\begin{equation}
f(\boldsymbol{x})= \begin{cases} f_1(\boldsymbol{x}),&\boldsymbol{x} \in K_1, \\ f_2(\boldsymbol{x}),&\boldsymbol{x} \in K_2 . \end{cases}
\end{equation}
\tag{3}
$$
Theorem 2. Under the assumptions of Theorem 1 consider the function Then (1) $f_2|_{V}=\varphi$ and $f_2\leqslant \Phi$ on $K_2$; (2) $f_1(\boldsymbol{x})=\inf_{\boldsymbol{y} \in K_2}(\boldsymbol{x}, \boldsymbol{y})/f_2(\boldsymbol{y})$. Thus, $f_1$ and $f_2$ are obtained from each other by the same formula (4). Theorem 1 asserts that the concatenation $f = f_1 \cup f_2$ is a self-dual antinorm on $\mathbb{R}^d_+$. Remark 2. The proofs of Theorems 1 and 2 are presented in § 5. We will see that the admissibility of the plane $V$ is used only once: the projection of an arbitrary point $\boldsymbol{x} \in \mathbb{R}^d_+$ onto $V$ lies in $K'$. We are not aware whether it is possible to generalize Theorem 1, maybe in a slightly different form, to an arbitrary hyperplane $V$ intersecting the interior of $\mathbb{R}^d_+$. Example 9 in § 6 (see Figure 8) shows that the answer can be affirmative. 3.2. Geometric formulationsA reformulation of Theorems 1 and 2 in terms of conic bodies is straightforward. For an admissible hyperplane $V$ that cuts $\mathbb{R}^d_+$ into two parts $K_1$ and $K_2$ (for the sake of simplicity, we assume that both have a nonempty interior) set $K'=\mathbb{R}^d_+ \cap V$. Corollary 2. Let $V$ be an admissible hyperplane and $G'$ be a $(d-1)$-dimensional autopolar conic body in $K'$. Then for an arbitrary conic body $G_1 \subset K_1$ whose intersection with $V$ and projection onto $V$ coincide with $G'$, the set $G=G_1\cup G_2$, where is an autopolar conic body in $G$. Below we use the notation from Corollary 2: the conic body $G_2 \subset K_2$ is obtained from $G_1$ by formula (5), and $G = G_1\cup G_2$. Recall that an orthogonal extension of a set $M\subset V$ is the preimage of $M$ under the orthogonal projection $\mathbb{R}^d \to V$. Corollary 3. If $G'$ is autopolar in $K'$ and the orthogonal extension of every ${(d\!-\!2)}$-dimensional plane of support of $G'$ is a $(d-1)$-dimensional plane of support of $G_1$, then $G$ is an autopolar conic body. If all these planes of support are tangent to $S_1 = \partial G_1$ and $S_1$ is smooth and strictly convex, then so is $S$. In this way we can produce smooth and strictly convex autopolar surfaces. Now we turn to conic polyhedra. Corollary 4. If $G_1 \subset K_1$ is an arbitrary conic polyhedron, its facet $G' = G_1\cap V$ is autopolar in $K'$ and all dihedral angles adjacent to this facade do not exceed $90^{\circ}$, then $G$ is an autopolar polyhedron. Let $G'$ be defined by a system of $n$ linear inequalities $(\boldsymbol{a}_i, \boldsymbol{x}) \geqslant 1$, $\boldsymbol{x} \in K'$, where all the $\boldsymbol{a}_i$ belong to $ K'$. We add arbitrary linear inequalities $(\boldsymbol{a}_j, \boldsymbol{x}) \geqslant 1$ for $j\in n+1,\dots,n+m$ and $\boldsymbol{x} \in \mathbb{R}^d_+$ such that $\boldsymbol{a}_{j} \in K_2$ and the hyperplanes $(\boldsymbol{a}_j, \boldsymbol{x}) = 1$ do not intersect the relative interior of $K'$. Then the conic polyhedron $G_1=\{\boldsymbol{x} \in K_1\colon (\boldsymbol{a}_i, \boldsymbol{x}) \geqslant 1,\,i=1,\dots,n+m\}$ satisfies the assumptions of Corollary 4. Conversely, every conic polyhedron $G_1$ from Corollary 4 is obtained in this way. Each of them produces an autopolar polyhedron in $\mathbb{R}^d_+$. 3.3. Self-duality in low dimensionsTheorem 1 provides a method of an inductive construction of autopolar sets, beginning with dimension one. The case $d=1$. This is a trivial case: every antinorm on $K=\mathbb{R}_+$ is a linear function $f(x) = kx$, $k> 0$ (we do not use bold letters here since we deal with scalars). Then $f^*(y)={y}/{k}$, and so the only self-dual antinorm is $f(x)=x$. The case $d=2$. Theorem 1 produces infinitely many self-dual antinorms in $\mathbb{R}^2_+$. Moreover, it actually gives their complete classification. In this case $V$ is a straight line $\{t\boldsymbol{a}\colon t\in \mathbb{R}\}$, where $\boldsymbol{a} \in \mathbb{R}^2_+$ and $|\boldsymbol{a}| = 1$, and, correspondingly, $K'$ is the ray $\{t\boldsymbol{a}\colon t \geqslant 0\}$. Theorem 1 suggests the following construction. Algorithm (constructing self-dual antinorms in $\mathbb{R}^2_+$). Take an arbitrary suitable unit vector $\boldsymbol{a}\in\mathbb{R}^2_+$ and set $\varphi(t\boldsymbol{a})=t$, $t\geqslant 0$. This is a self-dual antinorm on $K'$. Its orthogonal extension to $\mathbb{R}^2_+$ is $\Phi(\boldsymbol{x}) = (\boldsymbol{a}, \boldsymbol{x})$. Then choose an arbitrary antinorm $f_1$ on $K_1$ (the sector between $V$ and the $OX_1$-axis) such that $f_1(\boldsymbol{a}) = 1$ and $f_1(\boldsymbol{x}) \leqslant (\boldsymbol{x}, \boldsymbol{a})$, $\boldsymbol{x} \in K_1$. Geometrically, this means that the unit ball $G_1 \subset K_1$ is separated from the origin by the perpendicular to $V$ erected at the point $\boldsymbol{a}$. Equivalently, the tangent to $S_1$ at the point $\boldsymbol{a}$ makes an angle of at most $90^{\circ}$ with the ray $K'$. Then $f_2$ is defined in $K_2$ by formula (4) and $f=f_1\cup f_2$ is a self-dual antinorm on $\mathbb{R}^2_+$. Theorem 3. Every self-dual antinorm on $\mathbb{R}^2_+$ can be obtained by means of the above algorithm. Proof. Let $f$ be an arbitrary self-dual antinorm on $\mathbb{R}^2_+$. We denote by $A$ the point on the conic sphere $S = \partial G$ closest to the origin $O$, and $\boldsymbol{a}$ is the vector $OA$. The line $\ell$ passing through $A$ and orthogonal to $\boldsymbol{a}$ separates $S$ from $O$, hence the values of $f$ on $\ell$ do not exceed $1$. By Proposition 2, $|\boldsymbol{a}| = 1$, hence for all $\boldsymbol{x} \in \ell$ we have $f(\boldsymbol{x}) \leqslant 1=(\boldsymbol{a}, \boldsymbol{x})$. Therefore, $f(\boldsymbol{x}) \leqslant (\boldsymbol{a}, \boldsymbol{x})$ for all $\boldsymbol{x} \in \mathbb{R}^2_{+}$ by homogeneity. The ray $K'$ cuts $\mathbb{R}^2_+$ into two angles $K_1$ and $K_2$, and the function $f_1 = f|_{K_1}$ satisfies all the assumptions of Theorem 1. Hence the function $\widetilde f = f_1 \cup f_2$, where $f_2(\boldsymbol{y})=\inf_{\boldsymbol{x} \in K_1} (\boldsymbol{y}, \boldsymbol{x})/{f(\boldsymbol{x})}$, is self-dual. On the other hand, for every $\boldsymbol{y} \in K_2$ we have
$$
\begin{equation*}
f(\boldsymbol{y})=f^*(\boldsymbol{y})=\inf_{\boldsymbol{x} \in \mathbb R^d_2} \frac{(\boldsymbol{y}, \boldsymbol{x})}{f(\boldsymbol{x})} \leqslant\inf_{\boldsymbol{x} \in K_1} \frac{(\boldsymbol{y}, \boldsymbol{x})}{f(\boldsymbol{x})}= \widetilde f(\boldsymbol{y}).
\end{equation*}
\notag
$$
Thus, $f\leqslant \widetilde f$. Since both these functions are self-dual, we obtain $f=\widetilde f$ (Proposition 1).
The theorem is proved. Corollary 5. Every autopolar conic polygon in $\mathbb{R}^2_+$ has the form $G=G_1\cup G_2$, where $K'$ is an arbitrary ray cutting $\mathbb{R}^2_+$ into two sectors, $G_1$ is an arbitrary conic polygon in $K_1$ whose angle at the vertex $\boldsymbol{a}$ is at most $90^{\circ}$ and $G_2$ is the polar transform of $G_1$ defined in (5). § 4. Examples and special casesWe consider several examples of application of Theorems 1 and 2 Example 5 (case of a right cylinder). In Theorem 1 we can always take $f_1 = \Phi$: this is the maximum possible $f_1$. In this case $G_1$ is a right cylinder with base $G'$. Let $V$ separate the $OX_{d-1}$- and $OX_d$-axes, and let $K_1$ contain the first of them. The function $f_2$ and the set $G_2$ are found by Theorem 2. The unit ball $G_2$ is an oblique cylinder with base $G'$ and element parallel to the $OX_d$-axis. The conic body $G$ formed by these two cylinders is autopolar. Example 6 (orthogonal extension). In the limit case the hyperplane $V$ coincides with one of the coordinate planes, for instance, $OX_1\dots X_{d-1}$. In this case $f_1$ must be equal to $\Phi$. Indeed, if $f_1(\boldsymbol{x}) < \Phi(\boldsymbol{x})$ for some $\boldsymbol{x}$, then $f_1(\boldsymbol{x}' + x_d\boldsymbol{e}_d) < \varphi(\boldsymbol{x}')$, where $\boldsymbol{e}_d$ is the $d$th basis vector and $\boldsymbol{x}'$ is the projection of $\boldsymbol{x}$ onto $V$. The concavity of the function $f_1$ implies that it becomes negative for large $x_d$, which is impossible. Thus, if $V$ coincides with a coordinate plane, then $f_1 = \Phi|_{K_1}$ and the only possible conic body $G_1$ is an orthogonal extension of $G'$. Hence the autopolar set $G$ produced by Theorem 1 is an orthogonal extension of $G'$. Example 7 (autopolar polygons). In view of Theorem 3 every autopolar conic polygon is constructed as follows. We take a point $A_0 \in \mathbb{R}^2_+$ such that $OA_0 = 1$ and choose an arbitrary conic polygon $P = A_0A_1\dots A_n$ inside the sector $A_0OX_1$ so that the angle of $P$ at the vertex $A_0$ is at most $90^{\circ}$. The polygon $P$ has two infinite sides: one beginning at $A_0$ parallel to $OA$ (we denote it by $a$) and the other beginning at $A_n$ parallel to $OX_1$. This is $G_1$. We denote by $a_i$ the line that is the polar of $A_i$. The lines $a,a_0,\dots,a_n$ (in this order) form the polygon $G_2$ in the sector $AOX_2$. The polygon $G=G_1\cup G_2$ is autopolar (Figure 3). Now we turn to the smooth antinorm from Example 2 and show that it can also be constructed using Theorem 1. Proposition 3. For any nonnegative numbers $p_1,\dots,p_{d}$ such that $\sum_{i=1}^d p_i = 1$, the self-dial antinorm (1) is obtained by Theorem 1 from the $(d-1)$-dimensional self-dual antinorm $\varphi$ corresponding to the numbers $p_1,\dots,p_{d-2},p_{d-1}+p_{d}$. Proof. We assume that all the $p_i$ are nonzero, otherwise we reduce the dimension. Let us show that $f$ is obtained by Theorem 1 from the hyperplane
$$
\begin{equation*}
V=\biggl\{\boldsymbol{x} \in \mathbb R^d\colon\frac{x_{d-1}}{\sqrt{p_{d-1}}}= \frac{x_d}{\sqrt{p_d}}\biggr\}
\end{equation*}
\notag
$$
and $f_1=f|_{K_1}$. We introduce the coordinate system in $V$ as follows: all the $OX_i$-axes, $i = 1, \dots, d-2$, are the same as in $\mathbb{R}^d$ and the $OX_0$-axis is $ (OX_{d-1}X_{d})\cap V$. We have
$$
\begin{equation*}
x_{d-1}=\frac{\sqrt{p_{d-1}}}{\sqrt{p_{d-1}+p_d}} x_0 \quad\text{and}\quad x_{d}=\frac{\sqrt{p_{d}}}{\sqrt{p_{d-1}+p_d}} x_0.
\end{equation*}
\notag
$$
Set $p_0 = p_{d-1}+p_{d}$ and consider the function
$$
\begin{equation*}
\varphi(\boldsymbol{x})= \prod_{i=0}^{d-2}\biggl(\frac{x_i}{\sqrt{p_i}}\biggr)^{p_i} \quad\text{on } K'.
\end{equation*}
\notag
$$
By the inductive assumption $\varphi$ is self-dual. Then the gradient
$$
\begin{equation*}
f'(\boldsymbol{x})=\biggl( \frac{p_1}{x_1}, \dots, \frac{p_d}{x_d} \biggr) f(\boldsymbol{x})
\end{equation*}
\notag
$$
belongs to $V$ whenever $\boldsymbol{x} \in V$. Indeed, if $\frac{x_{d-1}}{x_{d}}= \frac{\sqrt{p_{d-1}}}{\sqrt{p_d}}$, then
$$
\begin{equation*}
\frac{f_{x_{d-1}}}{f_{x_d}}= \frac{p_{d-1}}{x_{d-1}}f(\boldsymbol{x})\Bigm/\Bigl(\frac{p_d}{x_d}f(\boldsymbol{x})\Bigr)= \frac{\sqrt{p_{d-1}}}{\sqrt{p_d}}.
\end{equation*}
\notag
$$
Thus, the tangent plane to $S$ at every point $\boldsymbol{x} \in V\cap S$ is orthogonal to $S$. In view of Corollary 3 we have $f\leqslant \Phi$.
The proposition is proved. For $d=1$ the antinorm (1) becomes $f(x) = x$, which is self-dual. Then every $d$-dimensional antinorm of type (1) can be obtained from this one by $d$ successive applications of Proposition 3 with the preservation of self-duality by Theorem 1. Example 8. The three-dimensional autopolar conic polyhedron $P$ presented in Example 3 can be obtained by means of lifting from Theorem 1. In this case the hyperplane $V$ passes through the $OX_3$-axis and point $A_3$ (Figure 4). The intersection $P\cap V$ is a conic polygon $P'$ bounded by the line segment $A_1A_3$, the ray $A_1X_3$ and the ray $\ell$ going from $A_3$ along the line of intersection of $V$ with $\alpha_1$. The polyhedron $P'$ is autopolar. Indeed, the polar of $A_1$ is the plane $\alpha_1$, hence $P'$ is obtained by applying the algorithm in § 3.3 to the partition of $\mathbb{R}^2_+$ by the line $OA_3$. Then we pass from $P'$ to $P$ using Corollary 4. The assumptions of the corollary are satisfied. Indeed, $V$ is orthogonal to both $\alpha_1$ and $\alpha_3$, and therefore the dihedral angles adjacent to the edges $\ell$ and $A_1A_2$ are right. § 5. Proofs of the main theorems Proof of Theorem 2. For every pair of points $\boldsymbol{x}\in K_1, \, \boldsymbol{y} \in K'$ we have $(\boldsymbol{y} , \boldsymbol{x})=(\boldsymbol{y}, \boldsymbol{x}')$ and $f_1(\boldsymbol{x}) \leqslant \Phi(\boldsymbol{x}) = \varphi(\boldsymbol{x}')= f_1(\boldsymbol{x}')$. Therefore,
$$
\begin{equation*}
\frac{(\boldsymbol{y}, \boldsymbol{x})}{f_1(\boldsymbol{x})} \geqslant\frac{(\boldsymbol{y}, \boldsymbol{x}')}{f_1(\boldsymbol{x}')}.
\end{equation*}
\notag
$$
We take the infima of both parts over $\boldsymbol{x} \in K_1$. In the left-hand side we obtain $f_2(\boldsymbol{y})$, and the right-hand side becomes Thus, for every $\boldsymbol{y} \in K'$ we have $f_2(\boldsymbol{y}) \geqslant \varphi^*(\boldsymbol{y})$. On the other hand the infimum over the set $K_1$ does not exceed the infimum over $K'$, hence Thus, $f_2(\boldsymbol{y}) = \varphi^* (\boldsymbol{y})$, and therefore $f_2|_{V} = \varphi^* = \varphi$.
Now take an arbitrary $\boldsymbol{y} \in K_2$. For every $\boldsymbol{x} \in K_1$ we have
$$
\begin{equation*}
(\boldsymbol{y}, \boldsymbol{x})- (\boldsymbol{y}', \boldsymbol{x})=(\boldsymbol{y} - \boldsymbol{y}', \boldsymbol{x}) =(\boldsymbol{y} - \boldsymbol{y}', \boldsymbol{x}')+(\boldsymbol{y} - \boldsymbol{y}', \boldsymbol{x} - \boldsymbol{x}') \leqslant 0,
\end{equation*}
\notag
$$
because $(\boldsymbol{y} - \boldsymbol{y}', \boldsymbol{x}') = 0$ and $(\boldsymbol{y} - \boldsymbol{y}', \boldsymbol{x} - \boldsymbol{x}')\leqslant0$ since the vectors $\boldsymbol{y} - \boldsymbol{y}'$ and $\boldsymbol{x} - \boldsymbol{x}'$ have opposite directions. Thus, $(\boldsymbol{y}, \boldsymbol{x}) - (\boldsymbol{y}', \boldsymbol{x})\leqslant0$. Consequently,
$$
\begin{equation*}
\frac{(\boldsymbol{y}, \boldsymbol{x})}{f_1(\boldsymbol{x})} \leqslant \frac{(\boldsymbol{y}', \boldsymbol{x})}{f_1(\boldsymbol{x})}.
\end{equation*}
\notag
$$
Taking the infima of both parts over $\boldsymbol{x} \in K_1$, we conclude that
$$
\begin{equation*}
f_2(\boldsymbol{y}) \leqslant f_2(\boldsymbol{y}')=\varphi(\boldsymbol{y}'),
\end{equation*}
\notag
$$
which proves the inequality $f_2 \leqslant \Phi$ on $K_2$. The proof of part (1) is complete.
Taking an arbitrary $\boldsymbol{y} \in K_2$ we show that for every $\boldsymbol{u} \in K_2$ there exists $\boldsymbol{x} \in K_1$ for which ${(\boldsymbol{y}, \boldsymbol{x})}/{f_1(\boldsymbol{x})} \leqslant {(\boldsymbol{y}, \boldsymbol{u})}/{f_2(\boldsymbol{u})}$. This will imply that the infimum of the quantity $(\boldsymbol{y}, \boldsymbol{z})/f(\boldsymbol{z})$ over $\boldsymbol{z} \in \mathbb{R}^d_+$ (which is $f^*(\boldsymbol{y})$) is attained on $K_1$, that is, it is equal to
$$
\begin{equation*}
\inf_{\boldsymbol{z} \in K_1}\frac{(\boldsymbol{y}, \boldsymbol{z})}{f_1(\boldsymbol{z})}=f_2(\boldsymbol{y}).
\end{equation*}
\notag
$$
This proves the equality $f(\boldsymbol{y})=f^*(\boldsymbol{y})$ for $\boldsymbol{y} \in K_2$. To establish the existence of a required point $\boldsymbol{x}$ it suffices to take $\boldsymbol{x} = \boldsymbol{u}'$, which means that ${(\boldsymbol{y}, \boldsymbol{u}')}/{f_1(\boldsymbol{u}')} \leqslant {(\boldsymbol{y}, \boldsymbol{u})}/{f_2(\boldsymbol{u})}$. This follows from two inequalities:
By part (1) the function $f_2$ on $K_2$ possesses the same properties as $f_1$ on $K_1$. On the other hand $f^*|_{K_2} = f|_{K_2} = f_2$. Hence for arbitrary $\boldsymbol{x} \in K_1$ we can argue as above and show that
$$
\begin{equation*}
f_1^{**}(\boldsymbol{x})=\inf_{\boldsymbol{u} \in K_2}\frac{(\boldsymbol{x}, \boldsymbol{u})}{f^*(\boldsymbol{u})}= \inf_{\boldsymbol{u} \in K_2}\frac{(\boldsymbol{x}, \boldsymbol{u})}{f_2(\boldsymbol{u})}.
\end{equation*}
\notag
$$
Invoking now the equality $f^{**} = f$, we complete the proof of part (2).
Theorem 2 is proved. Proof of Theorem 1. In the proof of Theorem 2 we showed that By assertion (2) of Theorem 2 the function $f_1$ is determined by $f_2$ in the same way as $f_2$ is determined by $f_1$. Interchanging these functions we obtain $f^*|_{K_1} = f|_{K_1} = f_1$. Therefore, $f^* = f$. From the definition of the dual function it immediately follows that $f^*$ is an antinorm, hence $f$ is too.
The theorem is proved. § 6. A counterexample and open problemsWe present an example of a three-dimensional conic polyhedron that cannot be obtained by the lifting procedure from Theorem 1. This leaves open the problem of a complete classification of autopolar sets (Problem 1 below). We begin with auxiliary results. For every conic body $G$ the minimum distance to the origin is attained at a unique point $A \in G$. This is a simple consequence of convexity. If $G$ is autopolar, then $OG=1$ by Proposition 2. It turns out that if $G$ is obtained by lifting from an admissible hyperplane $V$, then $V$ passes through $A$. Proposition 4. Assume that an autopolar conic body $G\subset \mathbb{R}^d_+$ is obtained by lifting a $(d-1)$-dimensional conic body $G'$ from an admissible hyperplane $V$ (Theorem 1). Then $G$ and $G'$ share the point closest to the origin, and this point belongs to $V$. Proof. Let $A$ and $A'$ be the points closest to the origin on the conic bodies $G$ and $G'$, respectively. Since both bodies are autopolar, it follows that $OA=OA'=1$. On the other hand $A$ and $A'$ belong to $G$, and the uniqueness of the closest point yields $A=A'$.
The proposition is proved. Thus, if an autopolar body $G$ is obtained by Theorem 1, then the cutting hyperplane $V$ contains the point $A\in G$ closest to the origin. This reduces the search for this hyperplane for a given autopolar set to looking at $d(d-1)/2$ admissible hyperplanes passing through $A$. The second property that follows directly from Theorem 2 is the following analogue of Corollary 4. It gives a necessary condition for a conic polyhedron to be obtained by lifting. Proposition 5. If a conic polyhedron $P$ is obtained by Theorem 1, then the admissible hyperplane $V$ intersects $P$ in an autopolar $(d-1)$-dimensional polyhedron $P'$ and cuts $P$ into two polyhedra $P_1$ and $P_2$ with a common facet $P'$. This common facet makes non-obtuse dihedral angles with all the adjacent facets of $P_1$ and $P_2$. In particular, every facet of $P$ cut by $V$ is orthogonal to $V$. Now we are ready to present the promised counterexample. We use the construction from [16]. To simplify the notation, we sometimes replace a polar by its boundary. For example, the polar of a point $\boldsymbol{a}$ is the set $\{\boldsymbol{y} \in \mathbb{R}^{d}_+\colon (\boldsymbol{y}, \boldsymbol{a}) = 1\}$. Similarly, we by the polar of a line segment $[\boldsymbol{a}_1, \boldsymbol{a}_2]$ we mean the set $\{\boldsymbol{y} \in \mathbb{R}^{d}_+\colon {(\boldsymbol{y}, \boldsymbol{a}_i) = 1}, i=1,2\}$. Example 9. The following series of autopolar conic polytopes $\{P_n\}_{n \geqslant 3}$ was presented in [16]. For every $n\geqslant 3$ the polytope $P_n$ is constructed in $\mathbb{R}^3_{+}$ as follows. We take an arbitrary point $A_1$ on the $OX_3$-axis such that $|OA_1| > 1$. Then we take an arbitrary point $A_2$ on the intersection of the polar of $A_1$ with the hyperplane $OX_1X_3$ (Figure 5). Then the construction goes by recursion. Suppose the points $A_1,\dots,A_k$, $k\leqslant n-2$, have been constructed; then $A_{k+1}$ is the point on the polar of the line segment $A_{k-1}A_k$ such that $|OA_{k+1}| > 1$. Finally, at the last iteration $A_n$ is chosen on the polar of the segment $A_{n-2}A_{n-1}$ so that $|OA_{n}| = 1$. Then $P_n = \operatorname{co}_+\{A_1,\dots,A_n\}$ is an autopolar conic polyhedron, for which $A_n$ is the point closest to the origin. (See the proof in [16], Theorem 2.) Moreover, the construction is well defined for every choice of points $A_1,\dots,A_n$ on the corresponding polars satisfying the conditions on the lengths of the $OA_{k}$; see [16]. Thus, we have actually constructed a family of autopolar polyhedra . Figure 5 presents one of its representatives, the polyhedron $P_5$, Figure 6 below shows the same polyhedron without auxiliary lines. The polyhedron $P$ constructed in Example 3 (see Figure 2) also belongs to this series, and in fact, $P=P_3$. As we see, it can be obtained by Theorem 1. The same is true for $P_4$. It turns out, however, that $P_5$ is not obtained by this theorem, which gives the required counterexample. Theorem 4. The conic polyhedron $P_5$ cannot be constructed by the lifting procedure described in Theorem 1. Proof. If $P_5$ is constructed by lifting from an admissible hyperplane $V$ , then $V$ passes through the point in $P_5$ closest to the origin (Proposition 4). This closest point is $A_5$ [16]. Moreover, by Proposition 5 $V$ is orthogonal to all transversal (that is, cut by $V$ into two parts) facets of $P_5$. If $V$ passes through the $OX_3$-axis, then it is transversal (and therefore orthogonal) to the faces $A_1A_2A_5A_4$ and $A_2A_3A_5$ (Figure 7).
Therefore, $V$ is orthogonal to the edge $A_2A_5$, and therefore $A_2A_5$ is parallel to the coordinate $OX_1X_2$-plane. In this case the point $A_5$ lies on the plane $\alpha_1$, which is the polar of $A_1$. The latter contradicts to the construction of $P_5$. The cases when $V$ passes through the other coordinate axes are considered in the same way. The theorem is proved. Thus, not all $d$-dimensional autopolar conic polyhedra are obtained by lifting from $(d-1)$-dimensional ones. A natural question arises: how to classify all possible autopolar sets? Problem 1. What is the complete classification of autopolar conic bodies (polyhedra) in $\mathbb{R}^d_+$? A complete classification can be either a description of all autopolar conic polyhedra or an algorithm that can construct all of them. Theorem 3 solves this problem for $d=2$; for higher dimensions $d$ it remains open. A weaker version of Problem 1 is to generalize Theorem 1 to cover all autopolar conic bodies. For example, does there exist a lifting procedure from a nonadmissible hyperplane $V$, that is, one not containing $(d-2)$ coordinate axes? This can be formulated as follows. Problem 2. Is it true that every autopolar conic polyhedron $P\subset \mathbb{R}^d_+$ is cut by a (possibly nonadmissible) hyperplane $\widetilde V$ that is orthogonal to all transversal faces of $P$ and makes non-obtuse angles with all facets intersecting it? For the polyhedron $P_5$ such a hyperplane exists: this is the $OA_2A_4$-plane ( Figure 8). In the proof of Theorem 1 we essentially used the fact that the orthogonal projection of an arbitrary point $X\in \mathbb{R}^d_+$ to $V$ is contained in $V\subset \mathbb{R}^d_+$. This is true only for admissible hyperplanes $V$. AcknowledgementThe authors are grateful to L. V. Lokutsievskiy for his valuable comments and for useful discussions. 
Citation in format AMSBIB
\Bibitem{MakPro25} |
|
||||||||||||
|
Contact us: |
Terms of Use
|
Registration to the website |
Logotypes |
|