| Izvestiya: Mathematics |
|
|
|
|
|
Chebyshev sets composed of subspaces in asymmetric normed spaces A. R. Alimovabc , I. G. Tsar'kova a Lomonosov Moscow State University b Steklov Mathematical Institute of Russian Academy of Sciences, Moscow c Moscow Center for Fundamental and Applied Mathematics
§ 1. IntroductionWe study properties of Chebyshev sets composed of finitely or infinitely many planes (closed affine subspaces, possibly degenerated to points). The problem is considered in linear spaces with asymmetric norm, which include, as a particular case, the normed linear spaces. By definition, an asymmetric norm $\|\,{\cdot}\,\,|$ on a real linear space $X$ satisfies the following axioms: 1) $\|\alpha x|=\alpha\| x|$ for all $\alpha\geqslant 0$, $x\in X$; 2) $\|x+y|\leqslant \| x |+\| y|$ for all $x,y\in X$, and 3) $\|x|\geqslant 0$ for all $x\in X$, $\|x|= 0\Leftrightarrow x=0$. Below, we will always assume that the union of planes $M:=\bigcup _{i\in A} L_i$ is irreducible, that is, the set $L_j\setminus \bigcup _{i\in A, \ i\ne j} L_i$ is dense in $L_j$ for any $j\in A$ with respect to the symmetrization norm $\|x\|_{\mathrm{sym}}=\max\{\|x\,|,\|{-}x\,|\}$ As a corollary, if the union $M:=\bigcup _{i\in A} L_i$ is irreducible, then no plane from this union contains any other plane. The Baire category theorem implies that any at most countable union of planes $\bigcup _iL_i$, $L_i\not\subset L_j$, $i\ne j$, in a Banach space is irreducible. The problem of approximation by Chebyshev unions of planes is considered both in usual normed linear spaces and in linear spaces with asymmetric norm (see § 2 below). The present paper continues and extends our previous studies on approximation by unions of planes begun in [1] and [2] in the normed linear space setting. Aggregates composed of at most countably many planes are being actively investigated at present and have a number of applications, for example, in problems of signal recovery (in many cases, signals are known to lie on a union of planes) [3], representation of signals from sparse base signals (the “union of subspaces” model in compressive sensing) [4], and in problems of mathematical economics and related rank- and $\ell_0$-norm minimization problems [3]. Problems involving approximation by unions of planes are related, on the one hand, to approximation by ridge functions
$$
\begin{equation*}
\mathscr R_N:= \biggl\{\sum^N_{k=1}f_k((a_k,x)+c_k) \biggm| x\in \mathbb R^n,\, a_k\in \mathscr A\subset \mathbb R^n, \, c_k\in \mathscr C \subset \mathbb R^n\biggr\},
\end{equation*}
\notag
$$
where $ \mathscr A$ and $ \mathscr C$ are at most countable sets. Ridge approximation appeared first in mathematical physics, and then proved useful in computer tomography, neural network, learning theory, greedy algorithms, and other applied and theoretical problems (see, for example, [5]). In addition, approximations by unions of planes are also closely related to classical variants of $n$-term approximation problems. Problems of this kind were considered by S. B. Stechkin, B. S. Kashin, V. N. Temlyakov, R. DeVore, S. V. Konyagin, P. Petrushev, E. D. Livshits, and many others (see, for example, [6]). In the above problems one usually deals with finite, countable, and even uncountable unions of planes. As a possible application, we also mention problems involving approximation in spaces of $A$-finitely generated matrices (possibly infinite). A matrix $\mathbf A= (a_{ij})$ is called $A$-finite (where $A$ is a set of pairs of indexes) if $a_{ij}=0$ for $(i,j)\notin A$. A matrix is said to be $A$-finitely generated if it can be obtained from an $A$-finite matrix by an arbitrary permutation of finitely many rows and columns. Let $X$ be a space of matrices, and let $\|\,{\cdot}\,\|$ be a norm on $X$ (as such a norm, one may consider the Frobenius, Schatten, Grothendieck norms, the operator $\ell^p$-norm, etc.). The problem of approximation of an arbitrary matrix by $A$-finitely generated matrices is a problem of approximation by an at most countable union of subspaces in $(X,\|\,{\cdot}\,\|$). With the help of the results that follow below one can extend some of the previous results on Chebyshev sets (and, in particular, subspaces) in matrix spaces (see [7] and [8]). Below, we will mainly follow the definitions of the survey [9] and the book [10]. The main definitions are given in § 2. Definition 1.1. A set $M\subset X $ is a Chebyshev set if each point from $X$ has a unique nearest point in $M$. A Chebyshev sets is always closed. Chebyshev sets composed of at most countably many of approximatively compact sets have been recently studied by Tsar’kov [2], § 3 (in most cases, in uniformly convex spaces). In particular, he established the following result. Theorem 1.A. Let $X$ be a uniformly convex Banach space, $M\subset X$ be a Chebyshev set composed of at most countably many approximatively compact sets (in particular, planes). Then $M$ is a Chebyshev sun. If, in addition, $X$ is a smooth space, then $M$ is convex. The definition of a sun is given below in § 6. Remark 1.1. Note that in Theorem 1.A the set $M$ can be composed of a continuum number of planes. Below, we will mainly consider at most countable unions of planes. This constraint stems from the fact that, even in Hilbert spaces, the study of Chebyshev sets composed of a continuum (or a large number) of planes (and even straight lines) will inevitably lead to the well-known unsolved problem of whether any Chebyshev set in an (infinite-dimensional) Hilbert space is convex. Indeed, if a Hilbert space $H$ contains a non-convex Chebyshev set, then it contains a Chebyshev set $C$ in the form of a Klee cavern; that is, $C$ is the complement of a non-empty convex open bounded set; see [10], § 6.3. We set $G:=X\setminus C$, where we assume without loss of generality that $0\in G$. Consider the space $X=H\oplus \mathbb{R}$ and equip it with the Euclidean norm. We also set $N:=\{t(y\,{\oplus}\, \{1\}) \mid y\in G,\, t\in \mathbb{R}\}$. The set $M:=X\setminus N$ is a Chebyshev set in $X$. By construction, $M$ is composed of straight lines passing through $0$ (the number of such lines has cardinality of a continuum if the original space $X$ is separable). Another example of a Chebyshev set which is a disjoint continual union of planes (more precisely, points) is given by the famous Klee example of a discrete Chebyshev set ([10], § 7.4), which is constructed in the Banach space $\ell=\ell^1(\Gamma )$, where $|\Gamma|=\omega$ and $\omega$ is such that $\omega^{\aleph_{0}}=\omega$. In the present paper, we partially extend Theorem 1.A by considering, generally, a wider class of spaces $X$, but with more restrictions on the set – namely, we study the Chebyshev sets composed of an irreducible (finite, countable, or uncountable) union of planes $L_j$, that is, Note that in (1.1) (unlike Theorem 1.A) we do not assume that the planes $L_n$ should be approximatively compact. The assumption that the union is at most countable is not burdensome in applications; nevertheless, we extend the countable union setting to the more general case (see condition (4.2)). The irreduciblility of the union (1.1) implies that no plane from (1.1) contains any other plane from this union.The paper is organized as follows. The main definitions are given in § 2, where we also recall necessary facts from the theory of asymmetric normed spaces. In § 3, we give two auxiliary lemmas on the structure of Chebyshev sets composed of planes. These results are of independent interest. In the main Theorem 4.1 from § 4, we show, in particular, that if, in a Banach space, $M$ is a Chebyshev set (1.1) consisting of at least two planes, then $M$ is not $B$-connected (that is, its intersection with some closed ball is disconnected) and is not $\mathring{B}$-complete (see the definition in § 2). In Corollary 4.1, we show that, in the case of symmetrizable spaces (and, in particular, in the normed space setting), a subset of a reflexive $(\mathrm{CLUR})$-space set of the form (1.1) consisting of at least two planes is not a Chebyshev set. This partially extends Theorem 1.A and does not involve constraints in the form of uniform convexity or smoothness of a space (like in [2]). The picture becomes quite transparent in case of finitely many planes. In this case, by Theorem 5.1 from § 5, no finite union of planes (different from a single plane) is a Chebyshev set for any (asymmetric) norm on the space. Some applications of our results are given in § 6. § 2. Spaces with asymmetric norm. Main definitionsIn parallel with the usual normed linear spaces, we will consider spaces with asymmetric norm. The term asymmetric norm was coined by M. G. Krein [11] in 1938. The functional
$$
\begin{equation*}
\|x\|_{\mathrm{sym}}=\max\{\|x|,\|{-}x|\}, \qquad x\in X
\end{equation*}
\notag
$$
(the symmetrization norm), is a norm. The entities defined in terms of the symmetrization norm will be given the subscript $\mathrm{sym}$. For example, $B_\mathrm{sym}(x,r):=\{y\mid\|x-y\|_\mathrm{sym}\leqslant r\}$. An asymmetric normed space $X=(X,\|\,{\cdot}\,|)$ is called symmetrizable if the asymmetric norm $\|\,{\cdot}\,|$ is equivalent to the symmetrization norm, that is, there exists a number $K\geqslant 1$ such that
$$
\begin{equation*}
K^{-1}\|x\|_{\mathrm{sym}}\leqslant \|x|\leqslant \|x\|_{\mathrm{sym}} \quad \text{for all}\quad x\in X.
\end{equation*}
\notag
$$
The topology $\tau$ of an asymmetric space is defined by the subbase of open balls $\mathring{B}(x,r)=\{z\in X\mid \|y-x|<r\}$. In general, this topology satisfies only $T_1$ separation axiom, and may fail to be Hausdorff. An asymmetric space (with natural $T_1$-topology $\tau$) will be simply called an asymmetric space. For results and examples on separation axioms in asymmetric spaces, see [12]. The class of asymmetric spaces, which is an important and useful extension of normed linear spaces, has numerous applications in problems of approximation theory, variational calculus, computer science, and mathematical economics. The theory of asymmetric spaces and applications thereof is in active development at present. For example, in this regard, one considers problems in approximation theory and topology, location problems with asymmetric norms (Chebyshev centres and networks with respect to asymmetric norms are important in this setting), and also problems in principal component analysis (a popular dimension reduction tool). (For these problems and other applications, see, for example, [13], [14], § 2.) Asymmetric distances are also natural in many problems of geometric approximation theory (see, for example, [15]–[20]). Below $\mathring{B}(x,r)=\{y\in X \mid \|y-x| < r\}$ is the open ball with centre $x$ and radius $r$; $B(x,r)=\{y\in X \mid \|y-x|\leqslant r\}$ is the “closed” ball1 ball with centre $x$ and radius $r$; $S(x,r) =\{y\in X \mid \|y-x|=r\}$ is the sphere with centre $x$ and radius $r$. For brevity, we set $B:=B(0,1)$ (the unit ball), $S=S(0,1)$ (the unit sphere). Remark 2.1. A subset $N$ of an asymmetric space $X$ is (left) closed (relative to the topology $\tau$) if $y\in N$, whenever $(y_n)\subset N$, $\|y_n - y|\to 0$, $n\to\infty$. Definition 2.1. A set $N\subset X=(X,{\|\cdot|})$ is right-closed if $y\in N$, whenever $(y_n)\subset N$, $\|y-y_n|\to 0$, $n\to\infty$. Remark 2.2. The ball $B(0,1)$ of any asymmetric space $X=(X,\|\,{\cdot}\,|)$ is closed relative to the right convergence (relative to the topology generated by the left open balls), that is, $y\in B(0,1)$, whenever $(y_n)\subset B(0,1)$, $\|y-y_n|\to 0$, $n\to\infty$. Let $\varnothing \ne M\subset X$. The distance from a point $x\in X$ to a set $M$ is denoted by In symmetrizable spaces, the distance function $\rho(\,{\cdot}\,,M)$ is continuous; however, in the general case, it is only lower semi-continuous (see [21], p. 146). The set of nearest points from $M$ for a given $x$ is defined by $P_Mx$, that is,Definition 2.2. A set $M$ is $B$-connected if its intersection with any closed ball $B(x,r)$ is connected; $M$ is $\mathring{B}$-connected if its intersection with any open ball $\mathring{B}(x,r)$ is connected. Definition 2.3 (see [13]). A sequence $(x_n)\subset X$ is called a Cauchy sequence if, for any $\varepsilon>0$, there exists $N\in \mathbb{N}$ such that $\|x_m-x_n|< \varepsilon$ for all $m\geqslant n\geqslant N$. An asymmetric normed space $X=(X,\|\cdot|)$ is called right- (left-) complete if, for any Cauchy sequence $(x_n) \subset X$, there exists a point $x\in X$ such that $\|x-x_n|\to 0$ $(\|x_n-x|\to 0)$ as $n\to\infty$. Definition 2.4. A set $M$ is $\mathring{B}$-complete2 (see [22], [17]) if, for all $x\in X$ and $r>0$, § 3. Auxiliary resultsWe will require the following auxiliary result on the linearity and homogeneity of the metric projection along a subspace (for the normed space setting, see, for example, Proposition 1.6 in [10]). Proposition 3.1. Let $L$ be a (closed) subspace of an asymmetric normed linear space $X$. Then the metric projection operator is homogeneous and linear along $L$: In addition, if $P_Lx=\varnothing$, then $\lambda P_L(x)+y=\varnothing$. Proof. Let $x\in X$, $y\in L$, $\lambda>0$. We have (in the next-to-last equality, we used the assumption that $L$ is a subspace).
For $\lambda=0$, the result is trivial. Let $\lambda >0$. We have
$$
\begin{equation*}
\begin{aligned} \, &z \in P_M(\lambda x+y)\quad \Longleftrightarrow\quad \|z-(\lambda x+y)|= \rho(\lambda x+y,L) \\ &\qquad\Longleftrightarrow \quad \lambda \biggl\| \biggl(\frac {z-y}\lambda\biggr)-x\biggr| \stackrel{(3.1)}{=} \lambda \rho(x,L) \quad \Longleftrightarrow\quad \biggl\| \biggl( \frac {z-y}\lambda\biggr)-x\biggr| = \rho(x,L) \\ &\qquad\Longleftrightarrow\quad \frac{z-y}\lambda \in P_Lx \quad \Longleftrightarrow \quad z\in \lambda P_Lx + y. \end{aligned}
\end{equation*}
\notag
$$
In what follows, we will always assume that a countable set is infinite. The following two important auxiliary results will be required in the proof of the main Theorem 4.1. Lemma 3.1. Let $X$ be a left-complete asymmetric normed space, and let $M :=\bigcup_{i\in \mathbb{N}}L_i$ be a Chebyshev set which is an irreducible union of countably many planes $L_i$ in $X$. Next, let $\varnothing\ne U\subset X$ be open, $U\cap M \ne \varnothing$, and $A:=\{i \in \mathbb{N}\mid L_i\cap U\ne \varnothing\}$. Then 1) if $A$ is finite, then any plane $L_i$, $i\in A$, is a Chebyshev plane; 2) if $A$ is countable, then there exists an infinitely large $i_0\in A$ such that Proof. Let $A=\{n_1,\dots, n_\nu\}$, where $n_\nu$ is either finite or infinite. Assume that there exists a point $x_1\in L_{n_1}\cap U$ such that some neighbourhood $\mathscr O (x_1)\subset U$ of $x_1$ intersects only finitely many planes $L_i$, $i\geqslant n_1$, $i\in A$. Shrinking homothetically, if necessary, the neighbourhood $\mathscr O (x_1)$ and translating insignificantly the point $x_1$ in the plane $L_{n_1}$, it can be assumed that
$$
\begin{equation}
\mathscr O (x_1) \text{ does not intersect any of the planes } L_i, \qquad i\in A\setminus\{ n_1\}.
\end{equation}
\tag{3.3}
$$
For some $r>0$, we have $\mathring{B}(x_1,2r)\subset \mathscr O (x_1)$. It is clear that $\mathring{B}_\mathrm{sym} (x_1,r)\subset \mathring{B}(x_1,r)\subset \mathring{B}(x_1,2r)$. An application of the triangle inequality shows that, for each point from the ball $\mathring{B}_{\mathrm{sym} }(x_1,r)$, its (unique) nearest point from $M$ lies in $L_{n_1}$. Now $L_{n_1}$ is a Chebyshev plane by Proposition 3.1. A similar analysis shows that, in the case of a finite $A$, each of the planes $L_{n_2}, \dots, L_{n_\nu}$ is a Chebyshev plane.
We next assume that $A$ is infinite. By the above, the result of Lemma 3.1 in case 2) will be proved if, for each $N$, we will find an $i_N\in A$, $i_N>N$, and a point $x_N\in L_{n_N}\cap U$ such that some neighbourhood $\mathscr O (x_N)$ of this point intersects only a finite number of planes $L_{n_i}$, $i\ne N$. Correspondingly, if this assumption is not true, then there exists a number $\mu$ such that
$$
\begin{equation}
\begin{aligned} \, &\forall\, m\geqslant \mu \quad \forall\, x\in L_{n_m}\cap U \ \text{ any neighbourhood } \mathscr O (x)\subset U \text{ of } x \\ &\text{intersects infinitely many planes }L_{n_i}, \qquad i\ne m. \end{aligned}
\end{equation}
\tag{3.4}
$$
Next, let $N_1>\mu$, $N_1\in A$, $x_0\in L_{N_1}\cap U $, be arbitrary, and let $\mathscr O (x_0)\subset U$ be a neighbourhood of $x_0$. By the assumption, the irreducible set $L_{N_1} \setminus \bigcup _{i\ne N_1}L_i$ is dense in $L_{N_1}$. Hence there exists a point $s_1 \in L_{N_1}\cap \mathscr O (x_0)$ such that $s_1\notin L_i$, $i\ne N_1$. In addition, in view of (3.4) we can choose $\varepsilon_1>0$ such that $\mathring{B}(s_1,3\varepsilon_1)\subset \mathscr O (x_0)$, and
$$
\begin{equation}
\begin{aligned} \, &\text{the ball }\mathring{B}(s_1,3\varepsilon_1)\text{ intersects infinitely many planes }L_i,\ i > N_1, \\ &i\in A,\text{ and does not intersect any of the planes }L_1, \dots, L_{N_1-1}. \end{aligned}
\end{equation}
\tag{3.5}
$$
Similarly, for each arbitrarily large $N_2 >N_1$, for any plane $L_{N_2}$, $N_2>N_2$, $N_2\in A$, $L_{N_2}\cap \mathring{B}(s_1,3\varepsilon_1)\ne \varnothing$, and an arbitrary point $x_1\in L_{N_2}\cap \mathring{B}(s_1,\varepsilon_1)$, using the irreducibility and (3.5), we find a point $s_2\in L_{N_2}$ and a number $\varepsilon_2>0$ such that $s_2\notin L_i$ for each $i\ne N_2$, $\mathring{B}(s_2,3\varepsilon_2)\subset \mathring{B}(s_1,\varepsilon_1)$, and
$$
\begin{equation}
\begin{aligned} \, &\text{the ball }\mathring{B}(s_2,3\varepsilon_2)\text{ intersects infinitely many planes }L_i, \\ &i> N_2,\text{ and does not intersect any of the planes }L_1, \dots, L_{N_2-1}. \end{aligned}
\end{equation}
\tag{3.6}
$$
We next argue by induction. Assume that we have found a point $s_k$ and number $N_k >N_{k-1}$, $N_k\in A$. Proceeding as above, we find a point $s_{k+1}\in L_{N_{k+1}}\cap \mathring{B}(s_k,\varepsilon_k)$ and a ball $\mathring{B}(s_{k+1},3\varepsilon_{k+1})\subset \mathring{B}(s_k,\varepsilon _k)$ such that
$$
\begin{equation}
\begin{aligned} \, &\text{the ball }\mathring{B}(s_{k+1},3\varepsilon_{k+1})\text{ intersects an infinite number of planes } L_i, \\ &i> N_{k+1},\text{ and does not intersect any of the planes }L_1, \dots, L_{N_{k+1}-1}. \end{aligned}
\end{equation}
\tag{3.7}
$$
As a result, we get a sequence $(s_k)$ such that $\|s_m-s_{m-1}| < 3^{2-m}\varepsilon_1$ for each $m$. Since $X$ is left-complete, there exists a point $s$ such that $\|s_k-s|\to 0$, and, further, $s\in \mathring{B}(s_m, 3\varepsilon_m)$ for each $m$. Since $M$ is closed and $s_k\in M$ for each $k\in\mathbb{N}$, we have $s\in M$. However, by construction, $s$ does not lie in any of the planes $L_i$, which contradicts the closedness of $M$. Lemma 3.1 is proved. Remark 3.1. For right-complete spaces, an analogue of Lemma 3.1 is formulated and proved similarly under the natural additional assumption that the set $M$ is right-closed. (In Lemma 3.1, the Chebyshev property of the set implies that it is left-closed.) We omit further details. Lemma 3.2. Consider two cases. 1. Let $X$ be a left-complete asymmetric normed space in which the unit ball $B(0,1)$ is closed, and let $M :=\bigcup_{i\in \mathbb{N}}L_i$ be a Chebyshev set which is an irreducible union of countably many planes $L_i$ in $X$. 2. Let $X$ be a right-complete asymmetric normed space, and let $M :=\bigcup_{i\in \mathbb{N}}L_i$ be a right-closed Chebyshev set which is an irreducible union of countably many planes $L_i$ in $X$. In each of cases 1, 2, there exist a Chebyshev plane $L_{j_0}$ and a ball $B(x',r')$ which supports this plane at some point $s'\in L_{j_0}$ and such that 1) $\mathring{B}(x',r')\cap M \ne\varnothing$; 2) $s'$ is an isolated point of the intersection $B(x',r')\cap M $. As a corollary, the set $M$ is not $\mathring{B}$-complete and is not $B$-connected (that is, the intersection of $M$ with some closed ball is disconnected). Remark 3.2. Let $L_i$, $L_j$ be closed planes. Assume that Remark 3.3. A plane $L$ will be called regular if, for each point $x\notin M$, the point $x$ and the plane $L$ have some disjoint neighbourhoods. In assertion 1 of Lemma 3.2, the assumption that the unit ball is closed can be replaced by regularity of each of the planes $L_i$ composing the set $M$. Remark 3.4. If, under the hypotheses of Lemma 3.2, the Chebyshev set $M $ is an irreducible union of a finite number of planes $L_i$, $i=1,\dots, N$, $N> 1$, then assertions 1), 2) of Lemma 3.2 hold without the additional assumption that the space is complete. Proof of Lemma 3.2. Our plan is as follows. We will twice apply Remark 3.2, and then, by passing to the limit, construct a point from $M$ not lying in any of the planes $L_i$ that compose the set $M$, which will give us a contradiction.
So, assume on the contrary that By Lemma 3.1, there exists a Chebyshev plane $L_{n_0}$ from the family $\{L_i \mid i\in \mathbb{N}\}$ composing the set $M$.We choose a point $v\in M\setminus L_{n_0}$ (this can be done, because $M$ is an irreducible union of planes). Let $u$ be a unique nearest point from the Chebyshev plane $L_{n_0}$ for $v$. Let $r:=\|u-v|$. It is clear that $\mathring{B}(v,r)\cap (M\setminus L_{n_0}) \ne\varnothing$. By Remark 3.2, there is a point $s\in L_{n_0}\setminus \bigcup_{m\neq n_0}L_m$ (that is, $s\notin M\setminus L_{n_0}$) lying arbitrarily close to the point $u$ with respect to the norm $\|\,{\cdot}\,\|_{\mathrm{sym}}$ and
$$
\begin{equation}
\mathring{B}(w,r)\cap (M\setminus L_{n_0})\ne \varnothing, \quad \text{where } \ w:= v+s-u.
\end{equation}
\tag{3.10}
$$
The metric projection is linear along $L_{n_0}$ (see Proposition 3.1), and hence
$$
\begin{equation}
s\text{ is a unique nearest point from the Chebyshev plane }L_{n_0}\text{ for }w.
\end{equation}
\tag{3.11}
$$
We set
$$
\begin{equation*}
\alpha_0:=\inf\{\alpha\mid \mathring{B}(w_\alpha,\alpha r)\cap (M\setminus L_{n_0})\neq\varnothing\},
\end{equation*}
\notag
$$
where $w_\alpha := \alpha w + (1-\alpha)s $. From (3.10) we have $\alpha _0 <1$; note that $\alpha_0>0$ by (3.11).
The induction base. We set $s_0:=s$, $x_0:=w_{\alpha _0}$, $r_0 :=\alpha _0 r$, and, further, for $0<\varepsilon_0<(r-r_0)/10=(1-\alpha_0)r/10$, we define $\widehat{x}_0 = w_{\alpha_0 + \varepsilon_0}$, $\widehat{r}_0=r_0+\varepsilon_0$. Here, we assume that the number $\varepsilon_0$ is so small that the intersection $\mathring{B}(s_0,\varepsilon_0)\cap L_j$ is non-empty only for $j>n_0$. We next define
$$
\begin{equation}
T^0_{\varepsilon_0}:= \bigl(\mathring{B}(\widehat{x}_0,\widehat{r}_0)\setminus B(x_0,r_0)\bigr)\cap \mathring{B}\biggl(s_0, \frac{\varepsilon_0}{10^2}\biggr) .
\end{equation}
\tag{3.12}
$$
Let us show that Indeed, if $T^0_{\varepsilon_0}$ were empty, then the set $D^0:=(B(\widehat{x}_0,\widehat{r}_0)\setminus B(x_0,r_0))\cap \mathring{B}(s_0,\varepsilon_0/10^2)$ would also be empty, since otherwise there would exist a point $y\in D^0$ such that $y\in {B}(\widehat{x}_0,\widehat{r}_0)$, $r_0<\|y-x_0|$ and $\|y-s_0|<\varepsilon_0/10^2$. Hence there would also exist a point $y_0\in \mathring{B}(\widehat{x}_0,\widehat{r}_0)$ sufficiently close to $y$ and such that $r_0<\|y-x_0|$ and $\|y-s_0|<\varepsilon_0/10^2$. But then we would have $y_0\in T^0_{\varepsilon_0}$, which would contradict the assumption $T^0_{\varepsilon_0}=\varnothing$. So, $D^0=\varnothing$. Note that in this case the ball $B(x_0,r_0)$ is not a singleton, and its interior does not intersect $M\setminus L_{n_0}$ (by definition of the number $\alpha_0)$. Since $M$ is a Chebyshev set, we have $M\cap B(x_0,r_0) = \{s\}=\{s_0\}$. Therefore, the ball $B(\widehat{x}_0,\widehat{r}_0)$ intersects the set $M\setminus L_{n_0}$, ($\mathring{B}(\widehat{x}_0,\widehat{r}_0)\cap M\ne \varnothing$), and $s_0=s\in L_{n_0}$ is an isolated point of the intersection $M\cap B(\widehat{x}_0,\widehat{r}_0)$. But this is impossible, since by assumption (3.9) the set $M$ is $B$-connected. So, $T^0_{\varepsilon_0}$ is non-empty, that is, (3.13) holds.
We also note that $B$-connectedness of $M$ also implies that the set $T^0_{\varepsilon_0}$ intersects infinitely many planes $L_i$ with numbers $i>n_0$ from some countable index set $A_1 \subset \mathbb{N}\setminus\{1,\dots, n_0\}$. Now an appeal to Lemma 3.1 gives us a Chebyshev plane $L_{n_1}$, $n_1>n_0$, from the family $\{L_i \mid i\in A_1\}$ such that $L_{n_1}\cap T^0_{\varepsilon_0}\ne\varnothing$. Let $t_1\in L_{n_1}$ be a nearest point from the Chebyshev plane $L_{n_1}$ for $s_0$. Then $\|s_1-s_0|\leqslant \varepsilon_0/10^2$ by (3.12). As above, using Remark 3.2, we find a point $s_1\in L_{n_1}$ sufficiently close to the point $t_1$ (say, $\|t_1-s_1\|_\mathrm{sym}<\varepsilon_0/10^3$) such that $s_1\in L_{n_1}\setminus \bigcup_{m\neq n_1}L_m$, and
$$
\begin{equation}
\mathring{B}(w_1,r^1)\cap (M\setminus L_{n_1})\ne \varnothing, \quad \text{where } \ w_1:= s_0+t_1-s_1, \ \ r^1:=\|t_1-s_0|.
\end{equation}
\tag{3.14}
$$
By construction, $\|s_1-s_0| < (2/10^2)\varepsilon_0$.
Let, as above,
$$
\begin{equation*}
\alpha ^1_0:=\inf \{\alpha \in [0,1] \mid \mathring{B}(w_\alpha, \alpha r^1)\cap (M\setminus L_{n_1})\ne \varnothing\},
\end{equation*}
\notag
$$
where $w_\alpha := \alpha w _1 + (1-\alpha)s_1$. The number $\alpha_0$ is positive by Remark 3.2 since $M$ is closed. We choose a sufficiently small $0<\varepsilon_1<\varepsilon_0/10^2$ such that the ball $\mathring{B}(s_1,\varepsilon_1)$ intersects the planes $L_j$ only with numbers $j>n_1$ and $\varepsilon_1<(r^1-r_1)/10$, where $r_1:=\alpha^1_0 r^1$.
We set $\widehat{x} _1 = w_{\alpha ^1_0 + \varepsilon_1}$, $\widehat{r}_1=r_1+\varepsilon_1$, $x_1:=w_{\alpha ^1_0}$, and define
$$
\begin{equation}
T^1_{\varepsilon_1}:= \bigl(\mathring{B}(\widehat{x}_1, \widehat r_1)\setminus B(x_1,r_1)\bigr)\cap \mathring{B}\biggl(s_1, \frac{\varepsilon_1}{10^2}\biggr) .
\end{equation}
\tag{3.15}
$$
Similarly to (3.13) it can be verified that the set $T^1_{\varepsilon_1}$ is non-empty and intersects infinitely many planes $L_i$, $i\in A_2$, where the set $A_2\subset \mathbb{N}$ is countable, $i> n_1$ ($A_2$ cannot be finite, since $M$ is $B$-connected, and in this case, $B(x_1,r_1)\cap M$ would contain an isolated point $s_1$). By Lemma 3.1, there exists a Chebyshev plane $L_{n_2}$, $n_1>n_1$, from the family $\{L_i \mid i\in A_2\}$ such that $L_{n_2}\cap T^1_{\varepsilon_1}\ne \varnothing$.
Proceeding as above, we assume that we have constructed the corresponding Chebyshev planes $L_{n_0},\dots, L_{n_{k-1}}$, $n_0<n_1< \dots < n_{k-1}$, points $s_0, \dots, s_{k-1}$ from these planes, sets $T^j_{\varepsilon_j}$, $A_j$, $j=0, \dots, k-1$, points $x_0,\dots, x_{k-1}$, $\widehat{x}_0,\dots, \widehat{x}_{k-1}$, and numbers $\varepsilon_0, \dots, \varepsilon_{k-1}$. By the construction,
$$
\begin{equation}
\|s_j -s_{j-1}| \leqslant \frac2{10^2} \varepsilon_{j-1} ,\qquad 1\leqslant j < k-1.
\end{equation}
\tag{3.16}
$$
From estimate (3.16) it follows that the balls $B(s_j,\varepsilon_j)$ are nested.
We set
$$
\begin{equation}
T^k_{\varepsilon_k}:= (\mathring{B}(\widehat{x}_k,\widehat{r}_k)\setminus B(x_k,r_k))\cap \mathring{B}\biggl(s_k, \frac{\varepsilon_{k-1}}{10^2}\biggr).
\end{equation}
\tag{3.17}
$$
This set is non-empty, and, since $M$ is $B$-connected, it follows that $ T^k_{\varepsilon_k}$ intersects infinitely many planes $L_i$, $i\in A_k$, where $A_k\subset \mathbb{N}$ is countable, $i\ne n_0, \dots, n_{k-1}$. Next, we choose a sufficiently small $0<\varepsilon_k<\varepsilon_{k-1}/10^2$ such that the ball $B(s_k,\varepsilon_k)$ intersects the plane $L_j$ only with indexes $j>n_k$. By Lemma 3.1, there exists a Chebyshev plane $L_{n_k}$, $n_k>n_{k-1}$, from the family $\{L_i \mid i\in A_k\}$ such that $L_{n_k}\cap T^k_{\varepsilon_k}\ne\varnothing$.
Let, as above, $t_k\in L_{n_k}$ be a nearest point from the Chebyshev plane $L_{n_k}$ for $s_{k-1}$. Then $\|t_k-s_{k-1}|\leqslant \varepsilon_k/10^2$ by (3.17). Using Remark 3.2 and proceeding as above, we find a point $s_k\in L_{n_k}$ sufficiently close to the point $t_k$ (say, $\|t_k-s_k\|_\mathrm{sym}<\varepsilon_k/10^3$) such that $s_k\in L_{n_k}\setminus \bigcup_{m\neq n_k}L_m$, and
$$
\begin{equation}
\mathring{B}(w_k,r_k)\cap (M\setminus L_{n_k})\ne \varnothing, \quad \text{where } \ w_k:= s_{k-1}-t_k-s_k.
\end{equation}
\tag{3.18}
$$
By the triangle inequality, $\|s_k-s_{k-1}| < (2/10^2)\varepsilon_{k-1}$. As a result, the sequence $(s_k)$ is constructed by induction. By (3.16) and since the space is left-complete by the assumption, there exists a point $s$ such that $\|s_k-s|\to 0$, and $s\in M$, since $M$ is closed. In addition, $s\in B(s_k,\varepsilon_k)$ for all $k$, because by the assumption the ball $B(0,1)$ is closed. Further, since the balls $B(s_k,\varepsilon_k)$ are nested, we have $s\in \bigcap _k B(s_k,\varepsilon_k)$. However, by construction of the balls $B(s_k,\varepsilon_k)$, the point $s$ does not lie in any of the planes $L_i$, that is, $s\notin M$. This contradiction shows that the process of construction of the points $s_k$ cannot be infinite, and so it will terminate at some step $m$. As a corollary, in this case, $M$ is not $B$-connected (see Remark 3.4), which contradicts the original assumption (3.9).
Now let us assume that the space $X$ is right-complete and $M$ is right-closed (case 2 of the lemma). It is known that in this case the unit ball $B(0,1)$ is right-closed. By right-closedness, there exists a point $s$ such that $\|s-s_k|\to 0$; in addition, $s\in M$, since $M$ is right-closed. However, by construction $s$ does not lie in any of the planes $L_i$. This contradiction shows that the process of construction of the planes $s_k$ cannot be infinite, and, therefore, it terminates at some step $m$. As a corollary, in this case, $M$ is $B$-connected (see Remark 3.4), which contradicts the original assumption (3.9). In this case, there exists a ball $B(x,r)$ which supports the Chebyshev plane $L_{n_m}$ at a unique point $s_m$, $\mathring{B}(x,r)\cap M\ne\varnothing$, and $s_m$ is an isolated point of the intersection $B(x,r)\cap M$. That $M$ is not $\mathring{B}$-complete is clear (see (2.1)) from assertions 1, 2 of the lemma. The proof of this lemma shows that assumption (3.9) can be replaced by a weaker assumption, which is the negation of the following condition:
$$
\begin{equation}
\begin{aligned} \, &\text{there exist a Chebyshev plane }L_{k_0}\text{ and a ball }B(x',r') \\ &\text{which supports this plane at an isolated point }a \in L_{k_0}\subset M \\ &\text{of the intersection } M\cap B(x',r')\text{ such that }\mathring{B}(x',r')\cap M\ne \varnothing. \end{aligned}
\end{equation}
\tag{3.19}
$$
In this case, a contradiction is obtained similarly. Lemma 3.2 is proved. § 4. The main theoremWe first recall some definitions. Definition 4.1. A set $M$ is called right- (left-) approximatively compact if the conditions $(y_n)\subset M$, $\|y_n-x|\to \rho (x,M)$ (correspondingly, $\|x-y_n|\to \rho ^-(x,M)$) imply that there exists a subsequence $(y_{n_k})$, (left-) converging a point $\widehat y\in M$, that is, $\|y_{n_k}-\widehat y|\to 0$. The corresponding point $x$ is a point of right- (left-) approximative compactness for $M$. Remark 4.1. If $x$ is a point of right-approximative compactness for $M$, then the point $\widehat y$ from Definition 4.1 may fail to be a nearest point from $M$ for $x$ (the situation $\widehat y\notin P_Mx$ is, of course, impossible in the symmetrizable case); see also Remark 4.2 below. Correspondingly, to exclude the “improper” case $\widehat y\notin P_Mx$ for $x\in \operatorname{AC}(M)$, we introduce the following definition. Definition 4.2. A set $M$ is said to be regularly (right-) approximatively compact if the conditions $(y_n)\subset M$, $\|y_n-x|\to \rho (x,M)$, imply that there exist a point $\widehat y \in P_Mx$ and a subsequence $(y_{n_k})$ converging to a point $\widehat y$, that is, $\|y_{n_k}-\widehat y|\to 0$. The corresponding point $x$ is called a point of regular (right-) approximative compactness. A set $M\subset X$ is called regularly left-approximatively compact if the conditions $(y_n)\subset M$, $\|x-y_n|\to \rho ^-(x,M)$, imply that there exist a point $\widehat y \in P_M^-x$ and a subsequence $(y_{n_k})$ (left-) converging to $\widehat y$, that is, $\|y_{n_k}-\widehat y|\to 0$. Remark 4.2. If $x$ is a point of left-approximative compactness for $M$, then the inclusion $\widehat y\in P_M^-x$ always holds. As a corollary, a left-approximatively compact set is necessarily regularly left-approximatively compact. Remark 4.3. It is easily checked that if the unit ball $B(0,1)$ of the space $X$ is closed,3 then in $X$ Definition 4.3. A closed set $M\ne\varnothing$ is unimodal (or an $\mathrm{LG}$-set; see, for example, [9], § 3.3) if, for each $x\notin M$, each local minimum of the function $\Phi_x(y)=\|y-x|$, $y\in M$, is global. In other words, a closed set $M$ is unimodal if $y\in P_Mx$, whenever $y\in P_{M\cap V}x$, where $V:=B(y,\varepsilon)$. Remark 4.4. In any asymmetric space, the class of unimodal sets coincides with the class of closed $\mathring{B}$-complete sets (see Theorem 8 in [17]). Theorem 4.1. Consider two conditions. 1. Let $X$ be a left-complete asymmetric normed space in which the unit ball $B(0,1)$ is closed, and let $M :=\bigcup_{i\in \mathbb{N}}L_i$ be an irreducible union of countably many planes $L_i$ in $X$. 2. Let $X$ be a right-complete asymmetric normed space, and $M :=\bigcup_{i\in \mathbb{N}}L_i$ be a right-closed set which is an irreducible union of countably many planes $L_i$ in $X$. Next, assume that at least one of the following conditions is satisfied: 1) $M$ is $B$-connected; 2) $M$ is $\mathring{B}$-complete; 3) $M$ is right-regularly approximatively compact; 4) $M$ is unimodal. Then $M$ is not a Chebyshev set in $X$. Remark 4.5. In assertion 1 of Theorem 4.1, the assumption that the unit ball $B(0,1)$ is closed can be replaced by the requirement that the distance $\rho(\,{\cdot}\,,M)$ to $M$ is continuous. In addition, the closedness condition can be relaxed as follows: if a sequence from $B(0,1)$ converges, then one of its limit points lies in the ball. Proof of Theorem 4.1. The conclusion under the conditions 1), 2) was proved in Lemma 3.2. The result in case 4) follows from case 2) in view of Remark 4.4. Now let condition 3) be met. By Lemma 3.2, there exist a Chebyshev plane $L_{j_0}$ and a ball $B(x',r')$ which supports this plane at some point $s'\in L_{j_0}$ such that $\mathring{B}(x',r')\cap M \ne\varnothing$, and $s'$ is an isolated point of the intersection $B(x',r')\cap M $. We assume without loss of generality that $s'=0$, $\|x'\|=1$. We set
$$
\begin{equation*}
\gamma := \inf \{\alpha >0 \mid \mathring{B}(\alpha x',\alpha )\cap M \ne\varnothing\}.
\end{equation*}
\notag
$$
Since $0$ is an isolated point of the intersection $B(x',r')\cap M $, we have $\gamma >0$. Next, $\mathring{B}(x_\gamma, \gamma )\cap M=\varnothing$, 0 is a nearest point from $M$ for $x_\gamma$, and, by the definition of the number $\gamma$, there exists a minimizing sequence $(y_n)$ from $M$ for $x_\gamma $, $(y_n)\not\subset L_{j_0}$. Since $M$ is right-regularly approximatively compact, there exists a subsequence $(y_{n_k})$ converging to some point $y'$ from $M$. Next, $0$ is an isolated point of the intersection $B(x',r')\cap M $, and hence $y'\ne 0$. As a result, the point $x_\gamma$ has two nearest points $0$ and $y'$ from $M$, which contradicts the assumption that $M$ is a Chebyshev set. Theorem 4.1 is proved. Definition 4.4. Let $X$ be a symmetrizable asymmetric space. We recall that $X\in (\mathrm{CLUR})$ if $x\in S$, $y_n\in S$, $\|x+y_n|\to 2$ imply that $(y_n)$ has a converging subsequence. Note that any reflexive $(\mathrm{CLUR})$-space is a Efimov–Stechkin space (see, for example, § 2 in [9], [24], [25]). Corollary 4.1. Let $X$ be a symmetrizable asymmetric space, and $M :=\bigcup_{i\in \mathbb{N}}L_i$ be an irreducible union of an at most countable number of planes $L_i$ in $X$. Next, let at least one of the following conditions be met: 1) $M$ is $B$-connected; 2) $M$ is $\mathring{B}$-complete; 3) $M$ is right- (left-) approximatively compact; 4) $X$ is a reflexive $ (\mathrm{CLUR})$-space; 5) $M$ is unimodal. Then $M$ is not a Chebyshev set in $X$. Proof. Let condition 4) be met. By Lemma 8 in [26], the classes of $\mathring{B}$- and $B$-connected closed sets are equivalent in $(\mathrm{CLUR})$-spaces. Any reflexive $(\mathrm{CLUR})$-space is a Efimov–Stechkin space (see, for example, [9], § 2), and in Efimov–Stechkin spaces any Chebyshev set is $\mathring{B}$-connected (see, for example, Theorem 5.2 in [9]). Remark 4.6. Corollary 4.1 gives the following partial answer to the famous Efimov–Stechkin–Klee problem on convexity of Chebyshev sets: namely, in any Hilbert space (and, more generally, in any symmetrizable $ (\mathrm{CLUR})$-space) an at most countable union of planes is a Chebyshev set if and only if this union is a Chebyshev plane itself. Let us give some comments for possible generalizations of the above results to not necessarily countable union of subspaces. For a set ($L_\alpha $ is a plane), which is not necessary a countable or finite union of planes, consider the following condition:
$$
\begin{equation}
\begin{aligned} \, &\text{any plane from union (4.1) contains a dense} \\ &\text{(relative to the symmetrization norm) set of points,} \\ &\text{each of which has a neighbourhood that intersects} \\ &\text{an at most countable set of planes, and, in addition,} \\ &\text{the intersections of the these neighbourhoods with this plane } \\ &\text{are not contained in the union of other planes.} \end{aligned}
\end{equation}
\tag{4.2}
$$
It is easily seen that, under condition (4.2), the results of Lemmas 3.1 and 3.2 extend to the case, where $M$ is a not necessary countable or finite union of planes (see (4.1)). We thus have the following result. Theorem 4.2. Consider two conditions. 1. Let $X$ be a left-complete asymmetric normed space with closed unit ball $B(0,1)$, and $M $ be an irreducible union of arbitrarily many planes (4.1). 2. Let $X$ be a right-complete asymmetric normed space, and $M $ be a right-closed set (4.1) which is an irreducible union of arbitrarily many planes. Next, assume that condition (4.2) is met and at least one of the following conditions is satisfied: 1) $M$ is $B$-connected; 2) $M$ is $\mathring{B}$-complete; 3) $M$ is right-regularly approximatively compact; 4) $M$ is unimodal. Then $M$ is not a Chebyshev set in $X$. Remark 4.7. As above (see Remark 4.3), the condition that the unit ball $B(0,1)$ is closed in assertion 1 of Theorem 4.2 can be replaced by the requirement that the distance function $\rho(\,{\cdot}\,,M)$ onto the set $M$ be continuous. Corollary 4.2. Let $X$ be a complete symmetrizable asymmetric space, $M \subset X$ be an irreducible union of planes (4.1). Next, let condition (4.2) be met and at least one of the following conditions be satisfied: 1) $M$ is $B$-connected; 2) $M$ is $\mathring{B}$-complete; 3) $M$ is right-regularly approximatively compact; 4) $M$ is unimodal; 5) $X$ is a reflexive $ (\mathrm{CLUR})$-space; Then $M$ is not a Chebyshev set in $X$. § 5. Approximation by a union of a finite number of planesNow we assume that is an irreducible union of a finite number of planes $L_n$, and $L_n\not\subset L_m$ for $n\ne m$.The following Theorem 5.1 establishes that there does not exist a non-trivial Chebyshev set composed of a finite number of planes. Theorem 5.1. In an arbitrary asymmetric normed linear space, any finite union of planes $M=M_N$ from (5.1) is a Chebyshev set if and only if $M$ is a Chebyshev plane (in this case, $N=1)$. Proof. The following fact is immediate from Lemma 3.1: if a finite union of planes $M =M_N:= \bigcup _{1\leqslant n\leqslant N} L_n $ is a Chebyshev set, then any of the planes $L_n$ is a Chebyshev set. By Lemma 3.2 and Remark 3.4, there exist a Chebyshev plane $L_{j_0}$ and a ball $B(x',r')$ which supports this plane at some point $s'\in L_{j_0}$ such that $\mathring{B}(x',r')\cap M \ne\varnothing$, and $s'$ is an isolated point of the intersection $B(x',r')\cap M $. We assume without loss of generality that $s'=0$, $\|x'\|=1$. Let
$$
\begin{equation*}
\gamma := \inf \{\alpha >0 \mid \mathring{B}(\alpha x',\alpha )\cap M \ne\varnothing\}.
\end{equation*}
\notag
$$
Since $0$ is an isolated point of the intersection $B(x',r')\cap M $, we have $\gamma >0$. Next, $\mathring{B}(x_\gamma, \gamma )\cap M=\varnothing$, $0$ is a nearest point from $M$ for $x_\gamma $, and further, by the definition of the number $\gamma$, one of the planes $L_i$, $i\ne j_0$, supports the ball $B(x_\gamma, \gamma )$. By the above, $L_i$ is a Chebyshev plane, and hence the point $x_\gamma $ has a nearest point which lies in the sphere $S(x_\gamma, \gamma )$ since $L_i$ supports $B(x_\gamma, \gamma )$. However, $0$ is a nearest point from $M$ for $x_\gamma$, but this contradicts the condition that $M$ is a Chebyshev set. Theorem 5.1 is proved. § 6. Some applicationIn this section, we consider the problem of uniqueness and existence (or the lack thereof) for sets composed of planes. We will consider some classical asymmetric spaces. The solarity problem will also be studied. Some applications of the above results for classical normed linear spaces $C(Q)$, $L^1$, and $L^\infty$ can be found in [1]. A plane will be said to non-trivial if it is distinct from a point and the entire space. In particular, in [1] it was shown that that, in the space $L^1(\Omega,\sigma,\mu)$ with atomless $\sigma$-finite measure $\mu$, any at most countable irreducible union of non-trivial planes of finite dimension or codimension is not a Chebyshev set. It was also shown that if $L^\infty (\Omega,\sigma,\mu)$ is infinite-dimensional, then in this space any at most countable irreducible union of planes of finite dimension $\geqslant 2$ is not a Chebyshev set in $L^\infty (\Omega,\sigma,\mu)$. In [1], it was also verified that if a metrizable compact set $Q$ cannot be embedded in a circle, then any at most countable irreducible union of planes of finite dimension $\geqslant 2$ is not a Chebyshev set in $C(Q)$. All these results were obtained with the use of the analogue of Lemma 3.1 for normed spaces (see Lemma 3.1 in [1]). Asymmetric analogues of classical normed spaces $C(Q)$ and $L^p$ were extensively studied starting from Krein [11] (see also [27]–[31]). For example, the spaces $L^p=L^p (\Omega,\sigma,\mu)$, $1\leqslant p<\infty$, with $\sigma$-finite norm are equipped with the asymmetric norm
$$
\begin{equation}
\biggl(\int_\Omega |f_+/w_+ +f_-/w_-|^p\,d\mu \biggr)^{1/p},
\end{equation}
\tag{6.1}
$$
where $f_+(t):=\max\{f(t),0\}$, $f_-(t):=\min\{f(t),0\}$, $w_+$, $w_-$ are measurable weights such that $0<\alpha \leqslant w_\pm(t)\leqslant \beta <\infty$. The space $L^p$, $1<p<\infty$, with norm (6.1) is a complete symmetrizable $ (\mathrm{CLUR})$-space. The following result now follows from Corollaries 4.1 and 4.2. Theorem 6.1. No irreducible union of planes in $L^p$, $1<p<\infty$, with norm (6.1) is a Chebyshev set. Let $C_\psi(Q)$ be the space of real-valued continuous functions $ C(Q,\mathbb{R})$ on a Hausdorff compact set $Q$ equipped with the asymmetric norm
$$
\begin{equation}
\|x|_{\psi} = \|x|_{\psi_+,\psi_-}:=\max_{t\in Q}\biggl\{\frac{x_+ (t)}{\psi_+(t)},\, \frac{x_-(t)}{\psi_-(t)} \biggr\},
\end{equation}
\tag{6.2}
$$
where $\psi_+$, $\psi_-$ are fixed positive continuous functions. For this space, asymmetric analogues of the Chebyshev equioscillation theorem were obtained in [28], and some characterization properties of strict protosuns (Kolmogorov sets) were proposed in [32] in the spaces $C_\psi(Q)$ and in the more general spaces $C_{\psi,0}(Q)$ of functions that vanish at infinity. Let us briefly recall the well-known Haar problem (which, in the classical setting, calls for necessary and sufficient conditions for a system $\{\varphi _1,\dots, \varphi _n\}$ of continuous functions on a compact set $Q$ that its linear hull $L$ be a Chebyshev set in $C(Q)$. This problem was stated and solved by A. Haar: $L$ is Chebyshev plane if and only if each non-zero polynomial $c_1\varphi _1(t) + \dots + c_n\varphi _n(t)$ has at most $n- 1$ different zeros on $Q$. This condition is known as the Haar condition, and a subspace satisfying this condition is called a Haar subspace. It is easily checked that a system $\{\varphi _1,\dots, \varphi _n\}$ in the space $C_\psi(Q)$ is a Haar system if and only if, for all different points $t_1,\dots, t_n\in [a, b]$, the Haar determinant
$$
\begin{equation*}
\Delta (t_1,\dots, t_n):= \begin{vmatrix} \varphi _1(t_1) & \dots & \varphi _n(t_1) \\ \dots & \dots & \dots \\ \varphi _1(t_n) & \dots & \varphi _n(t_n) \end{vmatrix}
\end{equation*}
\notag
$$
is non-zero. Consequently, if $Q\subset \mathbb R^d$ is a compact set with non-empty interior, $d\geqslant 2$, then in $C_\psi(Q)$ any system $\{\varphi _1,\dots, \varphi _n\}$, $n\geqslant 2$, is not a Haar system (see also [27], Chap. IX, § 5). As a result, the corresponding subspace $L:=\operatorname{span}\{\varphi _1, \dots, \varphi _n\}$ is not a Chebyshev plane in $C_\psi(Q)$. For some extensions of this result, see Sevast’yanov [29]. The next result is now a consequence of Lemma 3.1 and the fact just mentioned. Theorem 6.2. If $Q\subset \mathbb R^n$ is a compact set with non-empty interior, $n \geqslant 2$, then, in the space $C_\psi(Q)$ with asymmetric norm (6.2), any at most countable irreducible union of finite-dimensional planes of dimension $\geqslant 2$ is not a Chebyshev set. Given $\varnothing \ne M\subset X$, we say that $x\in X\setminus M$ is a solar point if there exists a point $y\in P_Mx\ne \varnothing$ such that $ y\in P_M\bigl((1-\lambda)y+\lambda x\bigr)$ for each $\lambda\geqslant 0$. A set $M\subset X$ is a sun if each point $x\in X\setminus M$ is a solar point for $M$. The following result, which is new also in the setting of usual normed spaces, establishes the lack of solarity of Chebyshev sets only under structural constraints on the set. For other results on solarity of Chebyshev sets, see Chap. 10 in [10]. Theorem 6.3. Let $X$ be a left-complete asymmetric normed space with closed unit ball. Then no Chebyshev set in $X$ which is an irreducible union of at most countably many planes is a sun. For a proof of Theorem 6.3 it suffices to note that if $M$ is a Chebyshev sun (that is, a Chebyshev set which is a sun), then by Theorem 8 in [17] this set is unimodal. Now it remains to invoke assertion 1 of Theorem 4.1. Remark 6.1. A result similar to Theorem 6.3 for right-complete spaces and their right-closed subsets composed of at most countably many planes can be derived from assertion 2 of Theorem 4.1. The result for arbitrary sets of planes can be verified as above under condition (4.2). Remark 6.2. In any (not necessarily complete) space, the union of a finite number of non-equal straight lines is not a Chebyshev set. In particular, the union of two non-equal straight lines is not a Chebyshev set. We note without proof that, in the spaces $C(Q)$ and $C_\psi(Q)$, $\operatorname{card}Q\geqslant 2$, one can construct a sun composed of two different straight lines. Such an example can also be constructed in the spaces $\ell^1$ and $\ell^\infty$. However, such an example is impossible in the spaces $L^1$ with atomless $\sigma$-finite measure (see Theorem 3 in [16] and Remark 1 in [33]), because in such spaces any finite-dimensional sun is convex. In smooth spaces, the union of two non-equal lines (and any irreducible finite union of different finite-dimensional planes) cannot be a Chebyshev set and cannot be a sun. Remark 6.3. Above, we have noted that V. Klee constructed an example of a discrete Chebyshev set in the Banach space $\ell^1(\Gamma )$, where $|\Gamma|=\omega$ and $\omega$ such that $\omega^{\aleph_{0}}=\omega$. This set is an irreducible union of uncountably many $0$-dimensional planes (points). It is easily seen that in $\ell^1(\Gamma)$ one can construct a similar example of a Chebyshev sun which is an irreducible union of uncountably many $k$-dimensional planes. By Remark 6.1, this Chebyshev set is not a sun. AcknowledgementsThe authors are grateful to V. I. Berdyshev for useful discussions and to the referee for several remarks and suggestions, which improved this text. 
Citation in format AMSBIB
\Bibitem{AliTsa24} |
|
||||||||||||||||||||||||||
|
Contact us: |
Terms of Use
|
Registration to the website |
Logotypes |
|