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This article is cited in 4 scientific papers (total in 4 papers) Uniformly and locally convex asymmetric spaces I. G. Tsar'kovab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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     § 1. IntroductionIn the present paper we consider linear spaces with asymmetric norm ${\|\cdot|}$, which are more general than classical normed linear spaces. An asymmetric norm on a linear space $X$ is defined by the following axioms: 1) $\|\alpha x|=\alpha\| x|$ for all $\alpha\geqslant 0$, $x\in X$; 2) $\| x+y|\leqslant \| x |+\| y|$ for all $x,y\in X$; 3) $\|x|\geqslant 0$ for all $x\in X$, and 3a) $\|x|= 0$ $\Longleftrightarrow $ $x=0$. An asymmetric norm is generated by the Minkowski functional of some (generally speaking, asymmetric) set containing the origin in its kernel. In parallel with asymmetric norms ${\|\cdot|}$ it is often convenient to consider the symmetrization norms defined by $\|x\|:=\max\{\|x|,\|-x|\}$ $(x\in X)$. In general, a space with an asymmetric norm (an asymmetric space) satisfies only the $T_1$ separation axiom (that is, for any points $a, b \in X$, one can find neighbourhoods $O(a)$ and $O(b)$ of these points such that $a\notin O(b)$ and $b\notin O(a)$), and can fail to be Hausdorff. We will also consider asymmetric seminorms ${\|\cdot|}$, which are defined by axioms 1)–3), while axiom 3a) is replaced by the condition $\|x|= 0=\|-x|$ $\Longrightarrow$ $x=0$. An asymmetric space $(X,{\|\cdot|})$ equipped with an asymmetric seminorm will simply be called a seminormed space. The case of seminormed spaces will always be explicitly mentioned; otherwise a space will be assumed to be equipped with an asymmetric norm. Among the asymmetric spaces, we will consider the subclass of uniformly convex spaces. The main purpose of this paper is to find a definition of a uniformly convex space for which most properties of uniformly convex spaces (in the normed linear space setting) can be carried over to essentially asymmetric spaces (a space is said to be essentially asymmetric if its asymmetric norm is not equivalent to the symmetrization norm). Given a linear asymmetric normed space or a seminormed space $X=(X,{\|\cdot|})$, $B(x,r)=\{y\in X\mid \|y-x|\leqslant r \} $ and $\mathring{B}(x,r) =\{y\in X\mid \|y-x|< r\}$ are, respectively, the ‘closed’ and open balls of radius $r$ with centre $x$. Note that the ‘closed’ ball $B(x,r)$ can fail to be closed relative to the topology generated by the subbase of open balls. The sphere of radius $r$ with centre $x$ is defined by $S(x,r) = \{y\in X\mid \|y-x |=r\}$. The unit sphere in $X$ is denoted by $S=S(0,1)$, and the dual unit sphere in the dual space $X^*$ of $X$ is denoted by $S^*=S^*(0,1)$. Given an arbitrary subset $M$ of an asymmetric normed space $X$ or a seminormed space $X $, the distance of a point $y\in X$ to the set $M$ is defined by $\varrho(y,M) := \inf_{z\in M}\|z-y|$. The set of nearest points in $M$ to some $x\in X$ is defined by $P_Mx :=\{y\in M\mid { }{\|y-x|=\varrho(x,M)\}}$. Given $x\in X $ and $\delta>0$, we also consider the metric $\delta$-projections $P_M^\delta x := \{y\in M\mid \|y-x|\leqslant\varrho(x,M)+\delta\}=M\cap B(x,\varrho(x,M)+\delta)$ and $\mathring{P}_M^\delta x := \{y\in M\mid \|y-x|<\varrho(x,M)+\delta\}=M\cap \mathring{B}(x,\varrho(x,M)+\delta)$. For more on asymmetric spaces, see the survey [1]. Various approximative properties in asymmetric spaces were discussed in [2]–[12]. The novelty of our paper is in the application of the new definition of a uniformly convex asymmetric normed space to classical problems of geometric approximation theory. The paper is organized as follows. In § 2 we obtain conditions under which the intersection of any nested family of nonempty convex closed sets is nonempty (Theorem 2). Then we present conditions under which the set of points of approximative uniqueness for an arbitrary nonempty closed subset of an essentially asymmetric uniformly convex space is dense in this space (Theorem 3). Problems of the existence and uniqueness of Chebyshev centres in asymmetric uniformly convex spaces are discussed in § 4 (see Theorem 7). In the concluding section, § 5, we consider the relationships between $\gamma$-suns, suns and the existence of best approximation (Theorems 8 and 9); we also find a sufficient condition for the radial $\delta$-solarity of a set in terms of the convergence to zero of its modulus of uniform approximative continuity (Theorem 10). Given a uniformly convex symmetric norm $\Psi(\,\cdot\,{,}\,\cdot\,)\colon \mathbb{R}^2\to \mathbb{R}$, and $1<p,q<\infty$, the space $L_{p,q,\Psi}(\Omega)$ is defined as the space of equivalence classes of measurable functions $f\colon \Omega\to \mathbb{R}$ with finite $L_p(\Omega)$- and $L_q(\Omega)$-norms. It is equipped with the asymmetric norm $\|f|$ defined by $\Psi(\|f_+\|_p,\|f_-\|_q)$, where $\|\varphi\|_\theta:=\|\varphi\|_{L_\theta(\Omega)}$; see [12]). This space, which serves as a model example in the study of asymmetric uniformly convex spaces, originates from the classical $L_p$-spaces, which are frequently useful in classical approximation of functions (and, in particular, in problems of approximation by trigonometric polynomials). Numerous general applications of asymmetric norms include numerical mathematics, economics, geometrical optics and differential equations, in which asymmetric convex positively homogeneous functionals (asymmetric norms, in our language) are useful. Asymmetric norms seem to date back to Minkowski (the Minkowski functional; see [13]). M. Krein was the first to use asymmetric norms in infinite-dimensional analysis (see [14]); he also gave them the name of asymmetric norms in 1938 (see [14]). Krein, as a single author and with co-authors, used asymmetric norms when dealing with extremal problems related to the Markov moment problem. The importance of sublinear functionals for several problems in convex analysis and calculus was pointed out by König (see [15] and [16]). Asymmetric distances also appear naturally in approximation theory (in particular, in problems of best one-sided approximations). In this regard, Babenko [17] noted that approximations in spaces with asymmetric norms serve as a ‘bridge’ between best approximations and best one-sided approximations (in this connection we mention the papers [18] and [19] by Dolzhenko and Sevast’yanov and the books [20] (§ I.9.E) by Collatz and Krabs and [1] by Cobzaş). From the point of view of approximation theory it seems interesting to apply the machinery of asymmetric norms to problems of complexity analysis in computer science (see [21]). In geometric approximation theory and convex analysis, asymmetric distances were considered by Borodin [22], [23], Ivanov and Lopushanski [24], [25], Alimov [4]–[6], and by this author, on his own [7]–[12] and in collaboration with Alimov [26], [27]. Let us now give the definition of a uniformly convex asymmetric space. We set Definition 1. An asymmetric space $X=(X,{\|\cdot|})$ is called uniformly convex if for all $\varepsilon>0$ and $ a\in (0,1]$ there exists $\delta>0$ such that for all $f,g \in X$, $\|f|=\|g|=1$, the condition $\Delta(a)<\delta$ implies that $f \in B(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. It follows from this definition that for any $\varepsilon>0$ there exists $\delta>0$ such that, for all $f,g \in X$, $\|f|=\|g|=1$, the condition $\|(f+g)/2|\geqslant 1-\delta$ implies that $\|f -\mu g|\leqslant\varepsilon$ for some $\mu\in [1-\varepsilon,1]$. Remark 1. Note that Indeed, by the triangle inequality $\|f-a_0g|\leqslant \|f- ag|+(a-a_0)\|g|$, which gives Remark 1 implies that if in the definition of uniform convexity we choose $\delta>0$ for some $\varepsilon>0$ and $a_0\in (0,1]$, then the same $\delta>0$ also works for the same $\varepsilon>0$ and all $a\in [a_0,1]$. So, in order to show that $X$ is uniformly convex, it suffices to verify that for any $\varepsilon>0$ and arbitrarily small $a>0$ (for example, for $0<a\ll\varepsilon$), there exists $\delta=\delta(a)>0$ such that, for all $f,g\in X$, $\|f|=\|g|=1$, the condition $\Delta(a)<\delta$ implies that $f \in B(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. Remark 2. Note that if $X=(X,{\|\cdot|})$ is uniformly convex, then so is the mirror asymmetric space $X^-=(X,{\|-\cdot|})$. Definition 2. An asymmetric space $X=(X,{\|\cdot|})$ is called left-uniformly convex if for all $\varepsilon>0$ and $ a\in (0,1]$ there exists $\delta>0$ such that for all $f,g\in X$, $\|f|=\|g|=1$, the condition $\Delta(a)<\delta$ implies that $g \in B(\mu f,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. Remark 3. Definitions 1 and 2 can be given not only for normed linear spaces $X$, but also for seminormed spaces $X$. § 2. Nested bounded convex closed setsDefinition 3. We say that $\{x_n\} \subset X$ is a Cauchy sequence (an inverse Cauchy sequence) if for each $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that $\|x_m-x_n|< \varepsilon$ (respectively, $\|x_n-x_m|<\varepsilon)$ for all $m\geqslant n\geqslant N$. An asymmetric space $X=(X,{\|\cdot|})$ is called right-complete (left-complete) if for each Cauchy sequence $\{x_n\}\subset X$ there exists a point $x\in X$ such that $\|x-x_n|\to 0$ (respectively, $\|x_n-x|\to 0)$ as $n\to\infty$. A right-complete space will be simply called a complete space. An asymmetric space $X=(X,{\|\cdot|})$ is called inversely right-complete (inversely left-complete) if for each inverse Cauchy sequence $\{x_n\}\subset X$ there exists a point $x\in X$ such that $\|x\,{-}\,x_n| \to 0$ (respectively, $\|x_n\,{-}\,x| \to 0)$ as $n\to\infty$. An inversely right-complete space will simply be referred to as an inversely-complete space. A right-complete space will be called a complete space. Remark 4. Note that if $X=(X,\|\cdot|)$ is right-complete (left-complete), then the mirror asymmetric space $X^-=(X,{\|-\cdot|})$ is inversely left-complete (respectively, inversely right-complete). Remark 5. A set $N$ in an asymmetric space $X=(X,{\|\cdot|})$ is closed (relative to the topology generated by open balls) if $y\in N$ whenever there exists a sequence $\{y_n\}\subset N$ such that $\|y_n-y|\to 0$ as $n\to\infty$. Remark 6. Definition 3 can also be introduced not only for normed linear spaces $X$, but also for seminormed spaces $X$. Remark 7. The unit ball $B(0,1)$ of an asymmetric space $X=(X,{\|\cdot|})$ is closed relative to right-convergence (this convergence corresponds to the topology generated by the left-open balls), that is, $y\in B(0,1)$ whenever there exists a sequence $\{y_n\}\subset B(0,1)$ such that $\|y-y_n|\to 0$ as $n\to\infty$. Indeed, if $\|y|>1$, then there exists a bounded linear functional $y^*\in S^*$ such that $y^*(y)=\alpha>1\geqslant y^*(z)$ for all $z\in B(0,1)$ (see [3]). Then $\|y-y_n|\geqslant y^*(y-y_n)\geqslant \alpha -1\nrightarrow 0$ as $n\to\infty$, which is impossible. Definition 4. An asymmetric space $X=(X,{\|\cdot|})$ is called locally uniformly convex if, for all $g\in S$, $\varepsilon>0$ and $ a\in (0,1]$, there exists $\delta>0$ such that for any $f\in S$ the condition $\|f-ag|+a\|g|-\|f|<\delta$ implies that $\|f-g|<\varepsilon$. If $\|g-f|<\varepsilon$ under the above conditions, then we say that $X$ is a left-locally uniformly convex space. Remark 8. Note that if a linear space $X{=}\,(X,{\|\cdot|})$ is locally uniformly convex, then so is the mirror asymmetric space $X^- = (X,{\|-\cdot|})$. For ordinary normed spaces the above definition of locally uniformly convex spaces is equivalent to several well-known characteristic properties of such spaces. The following result is Corollary 1 in [12], proved there for asymmetric normed spaces. Corollary 1. Any uniformly convex linear space with asymmetric seminorm is a locally uniformly convex asymmetric space. Proof. There exists $\mu\in [1-\varepsilon,1]$ such that $f,g\in B(\mu g,\varepsilon)$. Hence and the required result follows. The following result is Theorem 1 in [12]. Theorem A. Let $X=(X,{\|\cdot|})$ be an asymmetric locally uniformly convex space, and let $x\in S$ and $x^*\in S^*$ be vectors such that $x^*(x)=1$. Also let $\{x_n\}\subset M:=\{z\in X\mid x^*(z)\geqslant 1\}$, and let $\|x_n|\to 1$ as $n\to\infty$. Then $\|x_n-x|\to 0$ as $n\to\infty$. As usual, a space is Hausdorff (or satisfies the $T_2$ separation axiom) if any two distinct points $a,b\in X$ have disjoint neighbourhoods. We also note that an asymmetric space $(X,{\|\cdot|})$ is Hausdorff if and only if so is the mirror space $(X,{\|-\cdot|})$. The following remark is frequently useful. Remark 9. Let $X$ be a locally uniformly convex space with $T_2$ separation axiom (respectively, $T_1$ separation axiom), and let $\{x_n\}\subset B(0,1)$ and $\|x_n-x|\to 0$ (respectively, $\|x-x_n|\to 0$) as $n\to\infty$. By the asymmetric version of the Hahn-Banach theorem (see [3], Theorem 4.3) there exists a bounded linear functional $x^*\in S^*$ such that $x^*(x)=\|x|$. Assume that $x^*(x_n)\to 1$. Then $x\in S$. To prove the claim in Remark 9 we consider the Hausdorff case first. Since the sets $\{z\in X\mid x^*(z)\geqslant c\}$ are closed for all $c\in \mathbb{R}$, it suffices to show that $x\in B(0,1)$. Assume on the contrary that $\|x|>1$. Let $x_0=x/\|x|$. The linear functional $x^*\in S^*$ separates the ball $B(0,1)$ and the point $x_0$. Hence $\{x'_n:= x_n/x^*(x_n)\}$ is a minimizing sequence for the origin in $ M:=\{z\in X\mid x^*(z)\geqslant 1\}$. Now Theorem A implies that $\| x'_n- x_0|\to 0$ as $n\to\infty$. In the triangle with vertices $0$, $x$ and $x'_n$ the line segments $[x_n,x]$ and $[x'_n,x_0]$ intersect at some point $y_n$, $n\in \mathbb{N}$. Hence $\|y_n-x|\leqslant \|x_n-x|\to 0$ and $\|y_n-x_0|\leqslant \|x'_n-x_0|\to 0$ as $n\to\infty$, which is impossible for Hausdorff topological spaces. This contradiction completes the proof in the first case. In the second case (for $T_1$ spaces), at the last step we have $\|x -y_n|\leqslant \|x-x_n|\to 0$ and $\|y_n-x_0|\leqslant \|x'_n-x_0|\to 0$ as $n\to\infty$, which cannot be the case for $T_1$ topological spaces. Proposition 1. Let $X=(X,{\|\cdot|})$ be a Hausdorff asymmetric normed space and let $N$, $\varnothing\ne N\subset X$, be such that $N\subset B(x_n,\varepsilon_n)$ for all $n\in \mathbb{N}$, for some sequence $\{x_n\}\subset X$ and some nonnegative null sequence $\{\varepsilon_n\}$. Then $N$ is a singleton. Proof. Assume on the contrary that $N$ contains two distinct points $a,b\in N$. It can be assumed without loss of generality that $N=[a,b] $. Then $\|a-x_n| \to 0$ and $\|b-x_n|\to 0$, that is, the sequence $\{x_n\}$ right-converges to two distinct limit points, which is impossible in the Hausdorff mirror space $X=(X,{\|-\cdot|})$. Proposition 1 is proved. Theorem 1. Let $X=(X,{\|\cdot|})$ be a right-complete uniformly convex asymmetric space and let $x^*$ be a norm-one linear functional. Then the origin has a unique nearest point in the set $M:=\{z\in X\mid x^*(z)\geqslant 1\}$ and, moreover, if $\{x_n \}\subset M$ and $\|x_n|\to 1=\varrho(0,M)$ as $n\to\infty$, then $\|x_n -x|\to 0$ as $n\to\infty$. Proof. Let $\varepsilon\in (0,1)$ and let $\varepsilon_n:={\varepsilon}/{2^{n+2}}$. Since $X$ is uniformly convex, for any $\varepsilon\in (0,1)$ there exists $\delta=\delta(\varepsilon)\in (0,{\varepsilon}/{8})$ such that for all $f,g\in S$ the condition $0\leqslant \|f-\frac{1}{2}g|+\frac{1}{2}\|g|-\|f|<\delta(\varepsilon)$ implies that $f,g\in B(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$.
We construct the sequence $\{f_i\}_{i=0}^{\infty}\subset S$ recursively. 1. Let $\delta_1:=\delta(\varepsilon_1)$. There exist $f_0,f_1\in S$ such that $x^*(f_1),x^*(f_0)\geqslant 1-{\delta_1}/{100}$. Setting $g=f_0$ and $f=(f_1+g)/2$, we have $x^*(f)\geqslant 1-{\delta_1}/{100}$ and $\|f-\frac{1}{2}g|+\frac{1}{2}\|g|=\frac{1}{2}\|f_1|+\frac{1}{2}\|g|=1=\|f_1|$. Let $\beta_1\geqslant 1$ be such that $\beta_1 f\in S$. Then $1\geqslant x^*(\beta_1 f)=\beta_1 (1-{\delta_1}/{100})$, which implies that $1\leqslant\beta_1\leqslant {1}/(1-\delta_1/100)$. Hence $0\leqslant\beta_1-1\leqslant ({\delta_1}/{100})/(1-{\delta_1}/{100})<{\varepsilon_1}/{4}$ and
$$
\begin{equation*}
1=\|\beta_1 f|\leqslant \biggl\|\beta_1 f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g| \leqslant \biggl\| f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|+(\beta_1-1)\|f| \leqslant1+\frac{{\delta_1}/{100}}{1-{\delta_1}/{100}}.
\end{equation*}
\notag
$$
Consequently,
$$
\begin{equation*}
\biggl\|\beta_1 f-\frac{1}{2}g\biggr| +\frac{1}{2}\|g|-1\leqslant\frac{{\delta_1}/{100}}{1-{\delta_1}/{100}}.
\end{equation*}
\notag
$$
Therefore, $\beta_1 f,g\in B(\widehat{\mu}_1 g,\varepsilon_1)$ for some $\widehat{\mu}_1\in [1-\varepsilon_1,1]$. Since $[f,(\widehat{\mu}_1/\beta_1)g]$ is the midline of the triangle with vertices $f_1$, $g$ and $(2{\widehat{\mu}_1}/{\beta_1}-1)g$, we obtain
$$
\begin{equation*}
\biggl\|f-\frac{\widehat{\mu}_1}{\beta_1}g\biggr| \leqslant \frac{\varepsilon_1}{\beta_1}\quad\text{and} \quad \|f_1-\mu_1g|\leqslant 2 \frac{\varepsilon_1}{\beta_1} \leqslant 4\varepsilon_1, \quad\text{where } \mu_1=2\frac{\widehat{\mu}_1}{\beta_1}-1.
\end{equation*}
\notag
$$
Hence $f_1\in B(\mu_1 g,4 \varepsilon_1)$.
2. Let $\delta_k:=\min_{m=0,\dots,k}\delta(\varepsilon_{m+1})$. Assume that we have already constructed points $f_0,\ldots,f_{n-1}\in S$ such that
$$
\begin{equation*}
x^*(f_k)\geqslant 1-\frac{\delta_{k+1}}{100}, \qquad k=0,\dots,n-1.
\end{equation*}
\notag
$$
There exists a point $ f_n\in S$ such that $x^*(f_{n}) \geqslant 1-{\delta_{n+1}}/{100}\geqslant 1-{\delta_{n}}/{100}$. Setting $g=f_{n-1}$ and $f=(f_{n}+g)/2$ we have $x^*(f)\geqslant 1-{\delta_n}/{100}$ and $\|f-\frac{1}{2}g|+\frac{1}{2}\|g|=\frac{1}{2}\|f_n|+\frac{1}{2}\|g|=1=\|f_n|$. Let $\beta_{n}\geqslant 1$ be such that $\beta_{n} f\in S$. Then $1\geqslant x^*(\beta_{n} f)=\beta_{n} (1-{\delta_{n}}/{100})$, and therefore $1\leqslant\beta_{n}\leqslant {1}/(1-{\delta_{n}}/{100})$. Hence $0\leqslant\beta_{n}-1\leqslant ({\delta_{n}}/{100})/(1-\delta_{n}/100)<{\varepsilon_n}/{4}$ and
$$
\begin{equation*}
1=\|\beta_{n} f|\leqslant \biggl\|\beta_n f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g| \leqslant \biggl\| f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|+(\beta_{n}-1)\|f| =1+\frac{{\delta_{n}}/{100}}{1-{\delta_{n}}/{100}}.
\end{equation*}
\notag
$$
Consequently,
$$
\begin{equation*}
\biggl\|\beta_n f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|-1\leqslant\frac{{\delta_n}/{100}}{1-{\delta_n}/{100}}.
\end{equation*}
\notag
$$
Therefore, $\beta_{n} f,g\in B(\widehat{\mu}_{n} g,\varepsilon_{n})$ for some $\widehat{\mu}_{n}\in [1-\varepsilon_{n},1]$. Thus,
$$
\begin{equation*}
\biggl\|f-\frac{\widehat{\mu}_n}{\beta_n}g\biggr| \leqslant \frac{\varepsilon_n}{\beta_n}\quad\text{and} \quad\|f_n-\mu_ng|\leqslant 2 \frac{\varepsilon_n}{\beta_n} \leqslant 4\varepsilon_n, \quad\text{where } \mu_n=2\frac{\widehat{\mu}_n}{\beta_n}-1.
\end{equation*}
\notag
$$
Hence $f_{n}\in B(\mu_{n} g,4 \varepsilon_n)$.
3. Note that at each step we can select $\varepsilon_{n+1}$ sufficiently close to zero so that the product $\prod_{k=1}^\infty\mu_k$ is convergent. In this case, of course, $P_m:=\prod_{k=m+1}^\infty\mu_k\to 1$ as ${m\to\infty}$. Setting $\widehat{f}_k:=P_kf_k$, $k\in \mathbb{N}$, we have
$$
\begin{equation*}
\|\widehat{f}_{k+1}-\widehat{f}_k|=P_{k+1}\|f_{k+1}-\mu_{k+1} f_k|\leqslant \|f_{k+1}-\mu_{k+1} f_k|, \qquad k\in \mathbb{N}.
\end{equation*}
\notag
$$
It follows that
$$
\begin{equation*}
\begin{gathered} \, 1\geqslant \|\widehat{f}_k|\geqslant x^*(\widehat{f}_k)=P_k x^*( {f}_k)\geqslant P_k \biggl(1-\frac{\delta_{n}}{100}\biggr)\to 1, \qquad k\to\infty, \\ \sum_{k= m}^{n}\|\widehat{f}_{k+1}-\widehat{f}_k |\leqslant 4\sum_{k= m}^{n}\frac{\varepsilon}{2^{k+2}}\to 0, \qquad n\to\infty. \end{gathered}
\end{equation*}
\notag
$$
So $\{\widehat{f}_n\}$ is a Cauchy sequence in the complete space $X$. Therefore, there exists $x\in X$ such that $\|x -\widehat{f}_n|\to 0$ as $n\to\infty$. Now, $x\in S$ and $x^*(x)=1 $ by Remark 9.
We claim that $\{g\in S\mid x^*(g)=1\}=\{x\}$. Indeed, if there exists $g\in S$ such that $x^*(g)=1$, then the line segment $[g,x]\subset B(0,1)\cap M$ lies in the sphere $S$ and consists of nearest points in $M$ to the origin. By uniform convexity there exists a sequence $\{\mu_n\}\subset [0,1]$ such that $\lim_{n\to\infty}\mu_n=1$ and $\|g-\mu_n x|\to 0$ as $n\to\infty$. Hence
$$
\begin{equation*}
\|g- x|\leqslant \|g-\mu_n x|+\|\mu_n x-x|=\|g-\mu_n x|+(1-\mu_n)\|-x|\to 0, \qquad n\to\infty.
\end{equation*}
\notag
$$
Therefore, $g=x$.
By Corollary 1 and Theorem A, if $\{x_n\}\subset M:=\{z\in X\mid x^*(z)\geqslant 1\}$ and $\|x_n|\to 1$ as $n\to\infty$, then $\|x_n-x|\to 0$ as $n\to\infty$. Theorem 1 is proved. Corollary 2. Let $N$ be a nonempty closed convex subset of an asymmetric uniformly convex right-complete space $X=(X,{\|\cdot|})$. Then each $x\in X$ has a unique point $y\in N$ in $N$ (that is, $\|y-x|=\varrho(x,N)$) and, moreover, if $\{y_n\}\subset N$ is a minimizing sequence in $N$ for $x$ (that is, $\|y_n-x|\to \varrho(x,N) $ as $n\to\infty$), then $\|y_n-y|\to 0$ as $n\to\infty$. Proof. It can be assumed without loss of generality that $x\in X\setminus N$, $x=0$ and $\varrho(x,N)=1$. By the asymmetric Hahn-Banach theorem (see [3], Theorem 4.3, there exists a bounded linear functional $x^*\in S^*$ separating the ball $\mathring{B}(0,1)$ and the set $N$, that is, $N\subset M:=\{y\in X\mid x^*(y)\geqslant 1\}$. By Theorem 2 there exists a unique point $y\in M$ such that $\|y-x|=\varrho(0,M)$ and, moreover, $\|y_n-y|\to 0$ as $n\to\infty$ for any sequence $\{y_n\}\subset N\subset M$ such that $\|y_n-x|\to \varrho(x,M)=1=\varrho(x,N)$ as ${n\to\infty}$. Since $N$ is closed, $y$ lies in $N$. This proves Corollary 2. Remark 10. The existence of a nearest point in $M $ (in Theorem 1) and in $N$ (in Corollary 2) can also be verified in seminormed spaces. In this case, a nearest point is unique if the unit sphere $S$ does not contain nondegenerate line segments. For asymmetric normed spaces this result easily follows from local uniform convexity, which, in turn, is secured by Corollary 1. The following result is Corollary 2 in [12]. Theorem B. Let $N$ be a nonempty closed convex subset of a Hausdorff asymmetric uniformly convex left-complete space $X=(X,{\|\cdot|})$. Then each $x\in X$ has a unique nearest point $y\in N$ in $N$ (that is, $\|y-x|=\varrho(x,N)$) and, moreover, if $\{y_n\}\subset N$ is a minimizing sequence in $N$ for $x$ (that is, $\|y_n- x|\to \varrho(x,N)$ as $n\to\infty$), then $\|y_n-y|\to 0$ as $n\to\infty$. Theorem 2. Let $X=(X,{\|\cdot|})$ be a Hausdorff left-complete uniformly convex asymmetric space, and let $\{N_k\}$ be a nested sequence of nonempty convex closed bounded sets. Then the intersection $N:=\bigcap_{k=1}^\infty N_k$ is nonempty. Proof. It can be assumed without loss of generality that $0\notin N_1$. By assumption the sequence $\{\varrho_k:=\varrho(0,N_k)\}$ is bounded and monotone nondecreasing, and therefore it converges to a finite limit $\varrho$. It can be assumed without loss of generality that $\varrho=1$. By Corollary 2 in [12] a nearest point in $N_k$ to the origin exists and is unique.
Let $\varepsilon\in (0,1)$ and let $\varepsilon_n:={\varepsilon}/{2^{n+2}}$. Since $X$ is uniformly convex, for all $\varepsilon\in (0,1)$ there exists $\delta=\delta(\varepsilon)\in (0,{\varepsilon}/{8})$ such that for any $f,g\in S$ the condition $0\leqslant \|f-\frac{1}{2}g|+\frac{1}{2}\|g|-\|f|<\delta(\varepsilon)$ implies that $f,g\in B(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. We can choose a subsequence $\{N_{k_n}\}_{n=0}^\infty$ such that $\varrho_{k_n}\geqslant 1-{\delta(\varepsilon_n/2)}/{200}$. For convenience we denote this subsequence and the corresponding subsequence $\{\varrho_{k_n}\}$ by $\{N_n\}$ and $\{\varrho_n\}$ again. For each $n\in \mathbb{Z}_+$ there exists a bounded linear functional $\{x^*_n\}\subset S^*$ that separates the ball $B(0,1-{\delta(\varepsilon_n/2)}/{100})$ and the set $N_n$. We construct the sequence $\{f_n\}_{n=0}^{\infty}\subset S$ recursively. 1. Let $\delta_1:=\delta(\varepsilon_1/2)$ and let $\psi_0$ and $\psi_1$ be nearest points to the origin in the sets $N_0$ and $N_1$, respectively, that is, $ \|\psi_0|=\varrho_0$ and $\|\psi_1|=\varrho_1$. Setting $f_0:=\varrho_0^{-1}\psi_0$ and $f_1:=\varrho_1^{-1}\psi_1\in S$, we have $x^*_0(f_1),x^*_0(f_0)\geqslant 1-{\delta_1}/{100}$. We also put $g=f_0$ and $f=(f_1+g)/{2}$. Then we have
$$
\begin{equation*}
x^*_0(f)\geqslant 1-\frac{\delta_1}{100}\quad\text{and} \quad \biggl\|f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|=\frac{1}{2}\|f_1|+\frac{1}{2}\|g|=1=\|f_1|.
\end{equation*}
\notag
$$
Let $\beta_1\geqslant 1$ and $\beta_1 f\in S$. Then $1\geqslant x^*_0(\beta_1 f)=\beta_1 (1-{\delta_1}/{100})$, and so $1\leqslant\beta_1\leqslant {1}/(1-{\delta_1}/{100})$. Therefore, $0\leqslant\beta_1-1\leqslant (\delta_1/100)/(1-\delta_1/100)<\varepsilon_1/4$ and
$$
\begin{equation*}
1=\|\beta_1 f|\leqslant \biggl\|\beta_1 f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g| \leqslant \biggl\| f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|+(\beta_1-1)\|f| \leqslant 1+\frac{{\delta_1}/{100}}{1-{\delta_1}/{100}}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\biggl\|\beta_1 f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|-1\leqslant\frac{\delta_1/100}{1-\delta_1/100}.
\end{equation*}
\notag
$$
Consequently, $\beta_1 f,g\in B(\widehat{\mu}_1 g,\varepsilon_1)$ for some $\widehat{\mu}_1\in [1-\varepsilon_1/2,1]$. Since $[f,(\widehat{\mu}_1/\beta_1)g]$ is the midline of the triangle with vertices $f_1$, $g$ and $(2{\widehat{\mu}_1}/{\beta_1}-1)g$, we obtain
$$
\begin{equation*}
\biggl\|f-\frac{\widehat{\mu}_1}{\beta_1}g\biggr| \leqslant \frac{\varepsilon_1}{\beta_1}\quad\text{and} \quad\|f_1-\mu_1g|\leqslant 2 \frac{\varepsilon_1}{\beta_1} \leqslant 4\varepsilon_1, \quad\text{where } \mu_1=2\frac{\widehat{\mu}_1}{\beta_1}-1\in [1-\varepsilon_1,1],
\end{equation*}
\notag
$$
Hence $f_1\in B(\mu_1 g,4 \varepsilon_1)$. As a result,
$$
\begin{equation*}
\|f_1-(1-\varepsilon_1)g|\leqslant \|f_1-\mu_1g|+\|\mu_1g-(1-\varepsilon_1)g|\leqslant 4 \varepsilon_1 + \varepsilon_1=5\varepsilon_1,
\end{equation*}
\notag
$$
and $f_1\in B(\mu_1 g, 5\varepsilon_1)$, where we take $\mu_1=1-\varepsilon_1$. Therefore,
$$
\begin{equation*}
\psi_1\in B(\mu_1\varrho_1 g,5 \varepsilon_1\varrho_1)= B\biggl(\frac{\mu_1\varrho_1}{\varrho_0} \psi_0,5 \varepsilon_1\varrho_1\biggr).
\end{equation*}
\notag
$$
2. Let $\delta_k:=\min_{m=0,\dots,k}\delta(\varepsilon_{m+1}/2)$, $k=0,\dots,n$. Assume that
$$
\begin{equation*}
f_0,\ldots,f_{n-1}\in S \ \ \text{are such that}\ \ x^*(f_k)\geqslant 1-\frac{\delta_{k+1}}{100}, \quad k=0,\dots, n-1.
\end{equation*}
\notag
$$
Let $\psi_n$ be a nearest point to the origin in the set $N_n$, that is, $ \|\psi_n|=\varrho_n$. Setting $f_n:=\varrho_n^{-1}\psi_n \in S$ we have
$$
\begin{equation*}
x^*_n(f_{n}) \geqslant 1-\frac{\delta_{n+1}}{100}\geqslant 1-\frac{\delta_{n}}{100}.
\end{equation*}
\notag
$$
We also set $g=f_{n-1}$ and $f=(f_{n}+g)/2$. Then
$$
\begin{equation*}
x^*_{n-1}(f)\geqslant 1-\frac{\delta_n}{100}\quad\text{and} \quad \biggl\|f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g| =\frac{1}{2}\|f_n|+\frac{1}{2}\|g|=1=\|f_n|.
\end{equation*}
\notag
$$
Let $\beta_{n}\geqslant 1$ be such that $\beta_{n} f\in S$. Then $1\geqslant x^*(\beta_{n} f)=\beta_{n} (1-{\delta_{n}}/{100})$, and therefore $1\leqslant\beta_{n}\leqslant {1}/(1-{\delta_{n}}/{100})$. Hence ${0\leqslant\beta_{n}-1\leqslant ({\delta_{n}}/{100})/(1-{\delta_{n}}/{100})<\varepsilon_n/{4}}$ and
$$
\begin{equation*}
1=\|\beta_{n} f|\leqslant \biggl\|\beta_n f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g| \leqslant \biggl\| f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|+(\beta_{n}-1)\|f| \leqslant 1+\frac{\delta_{n}/100}{1-\delta_{n}/100}.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\biggl\|\beta_n f-\frac{1}{2}g\biggr|+\frac{1}{2}\|g|-1 \leqslant\frac{\delta_n/100}{1-\delta_n/100}.
\end{equation*}
\notag
$$
Therefore, $\beta_{n} f,g\in B(\widehat{\mu}_{n} g,\varepsilon_{n})$ for some $\widehat{\mu}_{n}\in [1-\varepsilon_{n}/2,1]$. Hence
$$
\begin{equation*}
\biggl\|f-\frac{\widehat{\mu}_n}{\beta_n}g\biggr| \leqslant \frac{\varepsilon_n}{\beta_n}\ \ \text{and} \ \ \|f_n-\mu_ng|\leqslant 2 \frac{\varepsilon_n}{\beta_n} \leqslant 4\varepsilon_n, \quad\text{where } \mu_n=2\frac{\widehat{\mu}_n}{\beta_n}-1\in [1-\varepsilon_{n},1].
\end{equation*}
\notag
$$
As a result, $f_{n}\in B(\mu_{n} g,4 \varepsilon_n)$. Proceeding as at step 1 and taking the number $1-\varepsilon_n$ as $\mu_n$ we see that $f_{n}\in B(\mu_{n} g,5\varepsilon_n)$. Therefore,
$$
\begin{equation*}
\psi_n\in B(\mu_n\varrho_n g,5 \varepsilon_n\varrho_n)= B\biggl(\frac{\mu_n\varrho_n}{\varrho_{n-1}} \psi_{n-1},5 \varepsilon_n\varrho_n\biggr).
\end{equation*}
\notag
$$
Note that by the choice of $\varrho_n$ the quantity ${\mu_n\varrho_n}/{\varrho_{n-1}} $ is smaller than 1 (here we recall that $\mu_{n}=1-\varepsilon_n$).
3. Note that at each step we can take $\varepsilon_{n+1}$ to be sufficiently close to 0 so that the product $\prod_{k=1}^\infty\mu_k$ is convergent; of course, in this case we have $P_m:={\prod_{k=m+1}^\infty{\mu_k\varrho_k}/(\varrho_{k-1})\to1}$ as ${m\to\infty}$. Setting ${\widehat{\psi}_k:=P_k\psi_k}$, $k\in \mathbb{N}$, we obtain
$$
\begin{equation*}
\|\widehat{\psi}_{k+1}-\widehat{\psi}_k| =P_{k+1}\biggl\|\psi_{k+1}-\frac{\mu_{k+1}\varrho_{k+1}}{\varrho_{k}} \psi_k\biggr| \leqslant \biggl\|\psi_{k+1}-\frac{\mu_{k+1}\varrho_{k+1}}{\varrho_{k}} \psi_k\biggr|, \qquad k\in \mathbb{N}.
\end{equation*}
\notag
$$
It follows that
$$
\begin{equation*}
\sum_{k= m}^{n}\|\widehat{\psi}_{k+1}-\widehat{\psi}_k |\leqslant 5\sum_{k= m}^{n}\frac{\varepsilon}{2^{k+2}}\to 0 \quad\text{as } n\to\infty.
\end{equation*}
\notag
$$
So $\{\widehat{\psi}_n\}$ is a Cauchy sequence in the left-complete space $X$. Therefore, there exists a point $x\in X$ such that $\|\widehat{\psi}_n -x|\to 0$ as $n\to\infty$. Hence $\|{\psi}_n -x \|\leqslant \|{\psi}_n-P_n{\psi}_n|+\|P_n{\psi}_n-x|= (1-P_n)\|{\psi}_n|+\|\widehat{\psi}_n-x|\to 0$ as $n\to \infty$.
The set $N:=\bigcap_{k=1}^\infty N_k$ is closed, so $x\in N$. Consequently, $N$ is nonempty. Theorem 2 is proved. Remark 11. If, in addition to the hypotheses of Theorem 2 regarding the space $X $, we assume that the unit ‘closed’ ball of $X$ is closed (in this case $X$ is a fortiori a Hausdorff space; see [2]), then $\psi_n$ (the nearest point to the origin in the set $N_n$) tends to $x$ as $n\to\infty$, which is a nearest point to the origin in $N:=\bigcap_{k=1}^\infty N_k$. Of course, the same holds not only for the origin, but also for any point $y\in X$, that is, $\|P_{N_n} y-P_Ny|\to 0$ as $n\to\infty$. A similar result about nearest points is valid if the above additional property is replaced by the condition that $\varrho(y,N_n)\to \varrho(y, N)$ as ${n\to\infty}$. The same result also holds if at least one set in this family is left-strongly closed (see Definition 5 below). § 3. Points of existence and points of approximative uniquenessLemma 1. Let $X$ be a uniformly convex asymmetric seminormed space, let ${\Delta\in (0,1)}$, $M\subset X$, $g_0\in X\setminus M$ and $r=\varrho(g_0,M)>0$. Then for any $\varepsilon\in (0,1)$ there exists $\delta_0\in (0,\varepsilon/8)$ such that if $\widetilde{f}_0\in M$, $\|\widetilde{f}_0-g_0|<r+\delta_0$, and $f\in M$, $\|f-g_1|<\varrho(g_1,M)+\delta_0$, where $g^1:=(\widetilde{f}_0-g_0)/\|\widetilde{f}_0-g_0|$ and $g_1:=g_0+\Delta g^1$, then $f\in B(\mu(\widetilde{f}_0-g_0)+g_0,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. Proof. It can be assumed without loss of generality that $g_0=0$ and $r=1$. Since $X$ is uniformly convex, for any $\varepsilon\in (0,1)$ and $\Delta\in (0,1)$ there exists $\delta=\delta(\varepsilon,\Delta)>0$ such that if $f_0,f_1\in S$ and $\|f_1-\Delta f_0|+\Delta\|f_0|-\|f_1|<\delta$, then $\|f_1-\mu f_0|<\varepsilon$ for some $\mu\in [1-\varepsilon,1]$. Let $\delta_0:=\frac{1}{2}\min\{\delta(\varepsilon/2,\Delta),\varepsilon/8\}$. Consider an arbitrary point $\widetilde{f}_0\in M$ such that $\|\widetilde{f}_0|<1+\delta_0$. Setting $g^1:=\widetilde{f}_0/\|\widetilde{f}_0|$ and $g_1:=g_0+\Delta g^1=\Delta g^1$, we have
$$
\begin{equation*}
\varrho(g_1,M)\leqslant \|\widetilde{f}_0-g_1|=\|g^1\cdot \|\widetilde{f}_0|-\Delta g^1|\leqslant \|g^1(\|\widetilde{f}_0|-\Delta)| <1+\delta_0-\Delta
\end{equation*}
\notag
$$
and $1-\Delta=\varrho(0,M)-\|g_1|\leqslant \varrho(g_1,M)$. Given an arbitrary $f\in M$, we have
$$
\begin{equation*}
\|f-g_1|<\varrho(g_1,M)+\delta_0\leqslant 1-\Delta+2\delta_0\leqslant 1-\Delta +\delta\biggl(\frac\varepsilon2,\Delta\biggr).
\end{equation*}
\notag
$$
Hence $\|f-g_1|+\|g_1|\leqslant 1-\Delta +\delta(\varepsilon/2,\Delta)+\Delta= 1 +\delta(\varepsilon/2,\Delta)$. As a result, $\|f\,{-}\,g_1|+\|g_1|-\|f|<\delta(\varepsilon/2,\Delta)$, and therefore $f\in B(\mu' g^1,\varepsilon/2)$ for some $\mu'\in [1-\varepsilon/2,1]$. We have
$$
\begin{equation*}
1\geqslant\frac{\mu'}{\|\widetilde{f}_0|}\geqslant\frac{1-\varepsilon/2}{1+\delta_0}\geqslant \frac{1-\varepsilon/2}{1+\varepsilon/8}\geqslant 1-\varepsilon,
\end{equation*}
\notag
$$
so setting $\mu={\mu'}/{\|\widetilde{f}_0|}$ we find that $f\in B(\mu'{\widetilde{f}_0}/{\|\widetilde{f}_0|},\varepsilon/2)=B(\mu {\widetilde{f}_0} , \varepsilon/2)$. Lemma 1 is proved. Definition 5. A subset $M$ of an asymmetric seminormed space $X =(X,{\|\cdot|})$ is called left-strongly closed if $x\in M$ and $\|x_n-x\|:=\max\{\|x_n-x|,\|x-x_n|\}\to 0$ as $n\to\infty$, whenever $\{x_n\}\subset M$ and $\|x_n-x|\to 0$ as $n\to\infty$. Lemma 2. Let $X$ be a left-complete uniformly convex asymmetric seminormed space with closed unit ball and $M\subset X$ be a left-strongly closed nonempty subset. Then any neighbourhood of an arbitrary point $x\in X\setminus M$ contains an existence point for $M$ (that is, a point that has a nearest point in $M$) at which the distance function $\varrho(\cdot,M)$ is continuous. Proof. By assumption the ball $B(0,1)$ is closed, hence any ‘closed’ ball is too. It can be assumed without loss of generality that $\varrho(x,M)=1$. We claim that any neighbourhood $O_\sigma(x)$ of $x$ contains a required existence point $v_0\in X\setminus M$. Let $\Delta\in (0,\min\{1/3,\sigma/3\})$. We construct sequences $\{y_n\}\subset M$ and $\{x_n\}\subset X$ recursively. For sufficiently small $\varepsilon\in (0,1)$ we set $\{\varepsilon_n={\varepsilon}/{2^n}\}$, $\delta_0=\varepsilon$ and $\sigma_0=\Delta/2$. We also let $x_0=x$.
1. We use Lemma 1. Let $\delta_1\in (0,\varepsilon_1/8)$, let $y_1\in M$ satisfy $\|y_1-x|<\varrho(x,M)+\delta_1$, and let $x_1\in [x,y_1]$ satisfying $\|x_1-x|=\Delta_1:=\Delta/2$ (in this case $\|y_1-x_1|<\varrho(x_1,M)+\delta_1$) be such that, for any point $f\in M$ such that $\|f-x_1|<\varrho(x_1,M)+3\delta_1$ we have $f\in B(\mu_1(y_1-x)+x,\varepsilon_1)$ for some $\mu_1\in [1-\varepsilon_1,1]$. Next, let $\sigma_1\in (0,\sigma)$, $\sigma_1<\delta_0/2$, be such that $\varrho(x_1,M)\leqslant \varrho(z,M)+\delta_0/2$ for all $z\in B(x_1,\sigma_1)$. 2. Suppose that the points $\{y_k\}_{k=1}^{n}\subset M$, $\{x_k\}_{k=1}^{n}\subset X$ and numbers $\{\delta_k\}_{k=1}^{n}$, $\{\mu_k\}_{k=1}^{n}$, $\{\sigma_k\}_{k=1}^{n}$, $n\geqslant 2$, have already been constructed so that, for all $k $,
$$
\begin{equation*}
\begin{gathered} \, \|y_{k}-x_{k-1}|<\varrho(x_{k-1},M)+\delta_{k}, \\ x_{k}\in [x_{k-1},y_{k}] \ \ \text{and}\ \ \|x_{k}-x_{k-1}|=\Delta_{k}:=\frac{1}{2}\min\{\Delta_{k-1},\sigma_{k-1},\delta_{k-1}\} \end{gathered}
\end{equation*}
\notag
$$
(in this case $\|y_{k}-x_{k}|<\varrho(x_{k},M)+\delta_{k}$) and, in addition, for an arbitrary point $f\in M$, $\|f-x_{k}|<\varrho(x_{k},M)+3\delta_{k}$ the inclusion $f\in B(\mu_{k}(y_{k}-x_{k-1})+x_{k-1},\varepsilon_{k})$ holds for some $\mu_{k}\in [1-\varepsilon_{k},1]$. Then $y_{k+1}\in B(\mu_{k}(y_{k}-x_{k-1})+x_{k-1},\varepsilon_{k})$, $k<n$. We also have $\sigma_{k}<\delta_{k-1}/2$, and, moreover, $\varrho(x_{k},M)\leqslant \varrho(z,M)+\delta_{k-1}/2$ for any $z\in B(x_{k},\sigma_{k})$.
By Lemma 1 there exist a number $\delta_{n+1}\in (0,\min_{k=1,n}\delta_k)$, a point $y_{n+1}\in M$ such that $\|y_{n+1}-x_{n}|<\varrho(x_{n},M)+\delta_{n+1}$, and a point
$$
\begin{equation*}
x_{n+1}\in [x_{n},y_{n+1}]\colon \|x_{n+1}-x_{n}|=\Delta_{n+1}:=\frac{1}{2}\min\{\Delta_{n},\sigma_n,\delta_{n}\}
\end{equation*}
\notag
$$
(in this case $\|y_{n+1}-x_{n+1}|<\varrho(x_{n+1},M)+\delta_{n+1}$) such that for an arbitrary point ${f\in M}$ such that $\|f-x_{n+1}|<\varrho(x_{n+1},M)+3\delta_{n+1}$ the inclusion $f\in B(\mu_{n+1}(y_{n+1}- x_{n})+x_{n},\varepsilon_{n+1})$ holds for some $\mu_{n+1}\in [1-\varepsilon_{n+1},1]$. Moreover, $y_{n+1}\in B(\mu_{n}(y_{n}-x_{n-1})+x_{n-1},\varepsilon_{n })$. Let $\sigma_{n+1}\in (0,\min_{k=1,n}\sigma_k)$, $\sigma_{n+1}<\delta_{n}/2$, be a number such that $\varrho(x_{n+1},M)\leqslant \varrho(z,M)+\delta_{n}/2$ for any $z\in B(x_{n+1},\sigma_{n+1})$.
3. We have
$$
\begin{equation*}
\begin{aligned} \, &\biggl\|\biggl(\prod_{k\geqslant n+1}\mu_k\biggr)y_{n+1}-\biggl(\prod_{k\geqslant n}\mu_k\biggr)y_n\biggr|= \biggl(\prod_{k\geqslant n+1}\mu_k \biggr)\|y_{n+1}-\mu_n y_n| \\ &\qquad\leqslant \|y_{n+1}-(\mu_{n}(y_{n}-x_{n-1})+x_{n-1})| +\|\mu_{n}(y_{n}-x_{n-1})+x_{n-1}-\mu_{n}y_{n}| \\ &\qquad\leqslant \varepsilon_{n}+\|(1-\mu_n)x_{n-1}|\leqslant 2 \varepsilon_{n}. \end{aligned}
\end{equation*}
\notag
$$
Hence $\{\widehat{y}_n:=(\prod_{k\geqslant n}\mu_k)y_n\}$ is a Cauchy sequence. Since $X$ is left-complete, there exists a point $y\in X$ such that $\|\widehat{y}_n-y|\to 0$ as $n\to \infty$. We have
$$
\begin{equation*}
\|{y}_n-y|\leqslant \|{y}_n-\widehat{y}_n|+\|\widehat{y}_n-y|= \biggl(1-\prod_{k\geqslant n}\mu_k \biggr)\|y_n|+\|\widehat{y}_n-y|\to 0, \qquad n\to\infty.
\end{equation*}
\notag
$$
This implies that $y\in M$ because $M$ is closed.
By construction $\{x_n\}$ is a Cauchy sequence, and so there exists a point $x^0\in X$ such that $\|x_n-x^0|\to 0$ as $n\to\infty$. Consequently,
$$
\begin{equation*}
\varrho(x^0,M)\leqslant\|y_n-x^0|\leqslant \|y_n-x_n|+\|x_n-x^0|=\varrho(x_n,M)+\delta_n+\|x_n-x^0|
\end{equation*}
\notag
$$
for all $n\in \mathbb{N}$. Since $x^0\in B(x_n,\sigma_n)$ (here we have used the fact that the ball $ B(x_n,\sigma_n)$ is closed and $\{x_k\}_{k\geqslant n+1}\subset B(x_n,\sigma_n)$), we see that $\varrho(x^0,M)+{\delta_n}/{2}\geqslant \varrho(x_n,M)$, and therefore $\|y_n-x^0|,\varrho(x_n,M)\to \varrho(x^0,M)$ as $n\to\infty$. By assumption $M$ is a left-strongly closed set, hence $\|y-y_n|\to 0$ as $n\to\infty$. Therefore,
$$
\begin{equation*}
\varrho(x^0,M)\leqslant \|y-x^0|\leqslant \|y-y_n|+ \|y_n-x^0|\to \varrho(x^0,M), \qquad n\to\infty,
\end{equation*}
\notag
$$
that is, $y\in M$ is a nearest point to $x^0$ in $M$. Lemma 2 is proved. Theorem 3. Let $X=(X,{\|\cdot|})$ be a complete (or complete with respect to the symmetrization norm) locally uniformly convex asymmetric space, $M$ be a nonempty closed set such that the set $E$ of points of existence for $M$ is dense with respect to the asymmetric norm of $X$ (respectively, with respect to the symmetrization norm). Then the set of points of approximative uniqueness for $M$ is of second category with respect to the asymmetric norm of $X$ (respectively, with respect to the symmetrization norm). Proof. Let $x\in E$ be such that $\varrho(x,M)>0$. It can be assumed without loss of generality that $\varrho(x,M)=1$ and $x=0$. By assumption the space $X=(X,{\|\cdot|})$ is locally uniformly convex. Hence for an arbitrary nearest point $y\in P_Mx$ to $x$ and arbitrary $\varepsilon>0$ and $ \Delta\in (0,1]$ there exists $\delta=\delta(\varepsilon,\Delta)\in (0,1)$ such that, for any $f\in S$ and $\Delta'\in [\Delta/4,1]$, the condition $\|f-\Delta' y|+\Delta'\|y|-\|f|\leqslant 3\delta$ implies that $\|f-y|<\varepsilon$. Moreover, here it can be assumed that $\delta<\varepsilon$. Let $z\in [x,y]$, $\|z-x|=\Delta$. Then for any point $w\in O^{\mathrm{sym}}(z)=O^{\mathrm{sym}}_\delta(z):=\{t\mid \|z-t\|<\delta\}$,
$$
\begin{equation*}
\|y-w|\leqslant \|z-w|+\|y-z|\leqslant \delta+1-\Delta,
\end{equation*}
\notag
$$
which yields $\varrho(w,M)\leqslant \delta+1-\Delta$. Correspondingly, if $s\in M$, $\|s-w|\leqslant \varrho(w,M)+\delta$, then
$$
\begin{equation*}
\begin{aligned} \, \|s-x| &\leqslant \|s-z|+\|z-x|\leqslant \|s-w|+\|w-z\|+\|z-x| \\ &\leqslant \varrho(w,M)+\delta+\delta +\Delta\leqslant \delta+1-\Delta+2\delta +\Delta=1+3\delta. \end{aligned}
\end{equation*}
\notag
$$
Note that then $1\leqslant \|s|\leqslant 1+3\delta$. Setting $s':=(s-x)/\|s-x|+x=s/\|s|$ we have $\|s-s'|\leqslant {3\delta}/(1+3\delta)\|s|\leqslant 3\delta$. Let $z_1\in [x,z]$ be a point such that the line segments $[z_1,s_1]$ and $[z,s]$ are parallel. The triangles $\triangle xz_1s_1$ and $\triangle xz s $ are similar with magnification ratio ${\|s-x|}/{\|s_1-x|}\leqslant 1$, hence which implies that $\|s_1-y|<\varepsilon$. Thus,
Given points $x\in E$, where $\varrho(x,M)>0$, and $y\in P_Mx$ and numbers $\varepsilon\in (0,1)$ and $\Delta\in (0,\|y-x|)$, consider the set $A(x,y,\varepsilon,\Delta)$ defined as the union of the neighbourhoods $O^{\mathrm{sym}}(z)$. The set $A(x,y,\varepsilon,\Delta)$ is open with respect to the symmetrization norm. The closure of $B(\varepsilon,\Delta):=\bigcup_{x,y}A(x,y,\varepsilon,\Delta)$ in $X$ with respect to the symmetrization norm contains $E$. It is easily seen that $\bigcap_{\varepsilon,\Delta}B(\varepsilon,\Delta)$ is a set of second category with respect to the asymmetric norm of $X$ (or with respect to the symmetrization norm), and by construction it consists of points of approximative uniqueness. This proves Theorem 3. Theorem 4. Let $X=(X,{\|\cdot|})$ be a complete asymmetric space (or a left-complete uniformly convex $T_2$ asymmetric space), and let $\{A_k\}$ be a sequence of closed sets without interior points with respect to the asymmetric norm (respectively, with respect to the symmetrization norm $\|\cdot\|:=\max\{{\|\cdot|},{\|-\cdot|}\})$. Then the complement of the set $A:=\bigcup_{k=1}^\infty A_k$ is dense in $X$ with respect to the asymmetric norm of $X$ (respectively, with respect to the symmetrization norm of $X$). Proof. The arguments for the symmetrization norm are the same as for the asymmetric one. The only difference is that at the end of the proof we use the fact that, under the hypotheses of the theorem, the intersection of any nested bounded sequence of nonempty closed convex sets is nonempty (see Theorem 2). So we give a proof only for an asymmetric norm. Let $B(x,R)\supset B(y,r)$. Then $\|y-x|+r\leqslant R$, that is, $\|y-x|\leqslant R-r$.
Let $B (x_0,r_0)$ be a ball in $X$. We construct recursively a nested sequence of balls
$$
\begin{equation*}
\{B(x_n,r_n)\}\subset B(x_0,r_0) \quad\text{such that }B(x_n,r_n)\cap A_n\neq \varnothing \text{ for all }n\in \mathbb{N}.
\end{equation*}
\notag
$$
As a result, $\|x_m-x_n|\leqslant r_n-r_m$ for all $m\geqslant n$, which implies that $\{x_n\}$ is a Cauchy sequence. Since $X$ is complete, there exists a point $x\in X$ such that $\|x-x_m|\to 0$ as $m\to\infty$. Hence
$$
\begin{equation*}
\|x-x_n|\leqslant \|x-x_m|+\|x_m-x_n|\to r_n-\lim_{m\to\infty}r_m\leqslant r_n, \qquad n\leqslant m\to\infty.
\end{equation*}
\notag
$$
As a result, $x\in B(x_n,r_n)$. Therefore, $x\notin A_n$ for all $n\in \mathbb{N}$, which implies the required result. Theorem 4 is proved. Corollary 3. Assume that each nested sequence of nonempty closed convex sets in $X=(X,{\|\cdot|})$ has a nonempty intersection. Then $X$ is complete with respect to the symmetrization norm. Proof. Note that any ball $B_{\mathrm{sym}}(x ,r ):=\{y\in X\mid \|y-x\|\leqslant r\}$ is a closed set. It is known that if an arbitrary nested sequence of balls $\{B_{\mathrm{sym}}(x_n,r_n)\}$ has a nonempty intersection, then the normed (with respect to the symmetrization norm) linear space is complete. Now the required result follows.
§ 4. Existence and uniqueness of Chebyshev centres for bounded setsIn this section we consider the classical problems of the existence and uniqueness of Chebyshev centres for asymmetric spaces. For a survey of similar problems in normed spaces, see [26]. Definition 6. Let $X=(X,{\|\cdot|})$ be an asymmetric space. A set $M\subset X$ is said to be bounded if $M\subset B(z,C)$ for some $z\in X$, $C>0$. A set $M$ is left-bounded if $M\subset B^-(w,C):=\{y\in X\mid \|w-y|\leqslant C'\}$ for some $w\in X$ and $C'>0$. A set $M$ is said to be bibounded if it is both bounded and left-bounded. Note that $M$ is bibounded if and only if $M$ is bounded with respect to the symmetrization norm. Theorem 5. Let $X=(X,{\|\cdot|})$ be a complete uniformly convex asymmetric space and let $M \subset X$ be a bibounded set. Given $C > 0$, consider the nonempty set $M_C\,{:=}\,\{z \in X\,|\, B(z,C) \supset M\}$. Then $M$ has a left-relative Chebyshev centre with respect to $ \overline{M_C}$, that is, there exists a point $x\in \overline{M_C}$ such that $r(M,M_C):=\inf_{z\in M_C}\sup_{y\in M }\|z-y|$ is equal to $\sup_{y\in M }\|x-y|$. Proof. It can be assumed without loss of generality that $r(M,M_C)=1$ and $0\in M$. Since $X$ is uniformly convex, for any $\varepsilon\in (0,1)$ there exists $\delta=\delta(\varepsilon)\in (0,{\varepsilon}/{8})$ such that for all $f,g\in S$ the condition $0\leqslant \|f-\frac{1}{2}g|+\frac{1}{2}\|g|-\|f|<\delta(\varepsilon)$ implies that $f,g\in B(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. We set $\varepsilon_n:={\varepsilon}/{2^n}$, where $\varepsilon\in (0,1)$ and ${n\in \mathbb{N}}$. Consider the nonincreasing number sequence $\delta_n:=\min_{k=1,\dots,n}\{\delta(\varepsilon_k/2)\}$, $n\in \mathbb{N}$.
We construct the sequence $\{x_n\}_{n=1}^{\infty}\subset M_C$ using recursion. We take $x_1\in M_C$ such that $M\subset B^-(x_1,1+{\delta_1}/{10^4}) $, and for $n>1$ we take $x_n\in M_C$ such that $M\subset B^-(x_n,1+{\delta_n}/{10^4}) $. There exists a point $y\in M$ such that $\|(x_n+x_{n-1})/2-y|\geqslant 1-{\delta_n}/{10^4}$. Setting $f_{n}:=x_{n}-y$ and $f_{n-1}:=x_{n-1}-y$, we have
$$
\begin{equation*}
\frac{x_n+x_{n-1}}2-y=\frac{f_n+f_{n-1}}2, \qquad \|f_{n}|\leqslant 1+\frac{\delta_n}{10^4}\quad\text{and} \quad \|f_{n-1}|\leqslant 1+\frac{\delta_{n-1}}{10^4}.
\end{equation*}
\notag
$$
Next,
$$
\begin{equation*}
\frac{1}{2}\|f_{n}|+\frac{1}{2}\|f_{n-1}|\geqslant \biggl\|\frac{f_n+f_{n-1}}2 \biggr|\geqslant 1-\frac{\delta_n}{10^4},
\end{equation*}
\notag
$$
so that
$$
\begin{equation*}
\|f_{n}|\geqslant 1-\frac{2\delta_n}{10^4}-\frac{\delta_{n-1}}{10^4}>1-\frac{\delta_{n-1}}{10^3}, \qquad \|f_{n-1}|\geqslant 1-\frac{2\delta_n}{10^4}-\frac{\delta_{n}}{10^4}>1-\frac{ \delta_{n-1}}{10^3}.
\end{equation*}
\notag
$$
Let $\alpha_n$ and $\alpha_{n-1}$ be numbers such that $f'_{n}:=\alpha_nf_n\in S$ and $f'_{n-1}:=\alpha_{n-1}f_{n-1}\in S$. Since
$$
\begin{equation*}
1-\frac{\delta_{n-1}}{10^3}<\|f_n|,\|f_{n-1}|<1+\frac{\delta_{n-1}}{10^3},
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
\alpha_{n-1},\alpha_{n}\in \biggl[\biggl(1+\frac{\delta_{n-1}}{10^3}\biggr)^{-1}, \biggl(1-\frac{\delta_{n-1}}{10^3}\biggr)^{-1}\biggr].
\end{equation*}
\notag
$$
If $\alpha_n\geqslant \alpha_{n-1}$, then
$$
\begin{equation*}
\begin{aligned} \, &\biggl\|\frac{f'_{n}+f'_{n-1}}{2}\biggr| =\biggl\|\alpha_{n-1}\frac{f_{n}+f_{n-1}}{2} +\frac{\alpha_n-\alpha_{n-1}}{2}f_n\biggr| \\ &\qquad\geqslant \alpha_{n-1}\biggl\|\frac{f_{n}+f_{n-1}}{2}\biggr| -\frac{(1-{\delta_{n-1}}/10^3)^{-1}-(1+{\delta_{n-1}}/10^3)^{-1}}{2} \biggl(1+\frac{\delta_n}{10^4}\biggr) \\ &\qquad \geqslant \biggl(1+\frac{\delta_{n-1}}{10^3}\biggr)^{-1} \biggl(1-\frac{\delta_n}{10^4}\biggr) -\frac{{2\delta_{n-1}}/{10^3}}{2(1-{\delta_{n-1}}/{10^3})(1+{\delta_{n-1}}/{10^3})} \biggl(1+\frac{\delta_n}{10^4}\biggr) \\ &\qquad \geqslant\biggl(1-\frac{\delta_{n-1}}{10^3}\biggr)\biggl(1-\frac{\delta_n}{10^4}\biggr) -\frac{{\delta_{n-1}}/{10^3}}{(1-{\delta_{n-1}}/{10^3})} \\ &\qquad \geqslant 1-\frac{2 \delta_{n-1}}{10^3}-\frac{\delta_{n-1}}{10^3} \biggl(1+2\frac{\delta_{n-1}}{10^3}\biggr) \geqslant 1-\frac{\delta_{n-1}}{250}. \end{aligned}
\end{equation*}
\notag
$$
In the case when $\alpha_{n-1}\geqslant \alpha_n$ we obtain
$$
\begin{equation*}
\begin{aligned} \, &\biggl\|\frac{f'_{n}+f'_{n-1}}{2}\biggr|=\biggl\|\alpha_{n}\frac{f_{n}+f_{n-1}}{2} +\frac{\alpha_{n-1}-\alpha_n}{2}f_{n-1}\biggr| \\ &\qquad \geqslant \alpha_{n}\biggl\|\frac{f_{n}+f_{n-1}}{2}\biggr| -\frac{(1-{\delta_{n-1}}/{10^3})^{-1}-(1+{\delta_{n-1}}/{10^3})^{-1}}{2} \biggl(1+\frac{\delta_{n-1}}{10^4}\biggr) \\ &\qquad \geqslant\biggl(1-\frac{\delta_{n-1}}{10^3}\biggr)\biggl(1-\frac{\delta_n}{10^4}\biggr) -\frac{{2\delta_{n-1}}/{10^3}}{2(1-{\delta_{n-1}}/{10^3})(1+{\delta_{n-1}}/{10^3})} \biggl(1+\frac{\delta_{n-1}}{10^4}\biggr) \\ &\qquad\geqslant 1-\frac{2 \delta_{n-1}}{10^3}-\frac{\delta_{n-1}}{10^3} \biggl(1+2\frac{\delta_{n-1}}{10^3}\biggr) \geqslant 1-\frac{\delta_{n-1}}{250}. \end{aligned}
\end{equation*}
\notag
$$
We have $\|(f'_{n}+f'_{n-1})/2-\frac{1}{2}f'_{n-1}|+\frac{1}{2}\|f'_n|=1$. Hence, if $\beta_n\geqslant 1$ is such that $\beta_n(f'_{n}+f'_{n-1})/2\in S$, then
$$
\begin{equation*}
0\leqslant \beta_n-1\leqslant \frac{{\delta_n}/{250}}{1-{\delta_n}/{250}}<\frac{\varepsilon_n}{4}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \biggl\|\beta_n\frac{f'_{n}+f'_{n-1}}{2}-\frac{1}{2}f'_{n-1}\biggr| +\frac{1}{2}\|f'_n|-1 &\leqslant\biggl\|(\beta_n-1)\frac{f'_{n}+f'_{n-1}}{2}\biggr| \\ &\leqslant (\beta_n-1)\biggl(1+\frac{\delta_{n-1}}{10^4}\biggr)<\frac{1}{2}\delta_n. \end{aligned}
\end{equation*}
\notag
$$
Since $X$ is uniformly convex, there exists a number $\widehat{\mu}_{n-1}\in[1- \varepsilon_{n-1}/2,1]$ such that $\beta_{n}(f'_n+f'_{n-1})/2,f'_{n-1}\in B(\widehat{\mu}_{n-1}f'_{n-1},\varepsilon_{n-1})$. Therefore,
$$
\begin{equation*}
\begin{gathered} \, \biggl\|\frac{f'_n+f'_{n-1}}{2}-\frac{\widehat{\mu}_{n-1}}{\beta_n}f'_{n-1}\biggr| \leqslant \frac{\varepsilon_{n-1}}{\beta_n}\quad\text{and} \quad \|f'_n-\mu_{n-1}f'_{n-1}|\leqslant 2 \frac{\varepsilon_{n-1}}{\beta_n} \leqslant 4\varepsilon_{n-1}, \\ \text{where } \mu_{n-1}=2\frac{\widehat{\mu}_{n-1}}{\beta_n}-1\geqslant 1-\varepsilon_{n-1}, \end{gathered}
\end{equation*}
\notag
$$
which implies that $f'_{n}\in B(\mu_{n-1} f'_{n-1},4 \varepsilon_{n-1})$. Hence $\|f'_n-\mu_{n-1} f'_{n-1}|\leqslant 4\varepsilon_{n-1}$, that is,
$$
\begin{equation*}
f_n\in B\biggl(\frac{\mu_{n-1}}{\alpha_n} f'_{n-1},\frac{4\varepsilon_{n-1}}{\alpha_n}\biggr)= B\biggl(\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n} f_{n-1},\frac{4\varepsilon_{n-1}}{\alpha_n}\biggr).
\end{equation*}
\notag
$$
Proceeding as in the proof of Theorem 2 and taking $\mu_{n-1}=1-\varepsilon_{n-1}$ we see that ${\mu_{n-1}\alpha_{n-1}}/{\alpha_n} <1$ and
$$
\begin{equation*}
\begin{aligned} \, 0 &\leqslant 1-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n} =\frac{\alpha_n-(1-\varepsilon_{n-1})\alpha_{n-1}}{\alpha_n} \\ &\leqslant\frac{(1-{\delta_{n-1}}/{10^3})^{-1}-(1-\varepsilon_{n-1})(1+{\delta_{n-1}}/{10^3})^{-1}} {(1+{\delta_{n-1}}/{10^3})^{-1}} \\ &=\frac{{2\delta_{n-1}}/{10^3}}{1-{\delta_{n-1}}/{10^3}}+\varepsilon_{n-1}<2\varepsilon_{n-1}. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, \frac{5\varepsilon_{n-1}}{\alpha_n} &\geqslant \biggl\|f_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n} f_{n-1}\biggr| =\biggl\|x_n-y-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}(x_{n-1}-y)\biggr| \\ &=\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}- \biggl(y-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}y\biggr)\biggr| \\ &\geqslant \biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr| \biggl(1-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}\biggr)\|y| \\ &\geqslant\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr| -\biggl(1-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}\biggr)d, \end{aligned}
\end{equation*}
\notag
$$
where $d:=\sup_{z\in M}\|z|<+\infty$. Hence
$$
\begin{equation*}
\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}|\leqslant \varepsilon_{n-1}\biggl(\frac{5}{\alpha_n}+2d\biggr) \leqslant (8+2d)\varepsilon_{n-1}.
\end{equation*}
\notag
$$
We set $P_n:=(\prod_{k\geqslant n}{\mu_{k}\alpha_{k}}/{\alpha_{k+1}})$ and $\widehat{x}_n:=P_nx_n$, $n\in \mathbb{N}$. Then
$$
\begin{equation*}
\|\widehat{x}_{n}-\widehat{x}_{n-1}|=P_n\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr| \leqslant (8+2d)\varepsilon_{n-1}, \qquad n\in \mathbb{N},
\end{equation*}
\notag
$$
and therefore $\{\widehat{x}_n\}$ is a Cauchy sequence. Since $X$ is complete, there exists a point $x\in X$ (in the actual fact, $x$ lies in the closure of $M_C$) such that $\|x-\widehat{x}_n|\to 0$ as $n\to \infty$, and therefore $\|x-x_n|\leqslant \|x- \widehat{x}_n|+\|\widehat{x}_n-{x}_n|=\|x-\widehat{x}_n|+(1-P_n)\|-x_n|\leqslant \|x-\widehat{x}_n|+(1-P_n)(d+C)\to 0$ as $n\to \infty$. Moreover, $\|x-z|\leqslant \|x-x_n|+\|x_n-z|\leqslant \|x-x_n|+ 1+{\delta_{n-1}}/{10^4}\to 1=r(M,M_C)$ for all $z\in M$. Theorem 5 is proved. Remark 12. A similar analysis proves the existence of both left- and right-relative Chebyshev centres if $X=(X,{\|\cdot|})$ is a complete uniformly convex asymmetric space, $M $ is a bounded subset of $X$ with respect to the symmetrization norm, and $A\subset X$ is a bounded closed set with respect to the symmetrization norm. In other words, for such $M \subset X$ and $A\subset X$ there exists a relative Chebyshev centre $x\in A$ (a left-relative Chebyshev centre $x'\in A)$, that is, Definition 7. A seminormed space is said to be inversely uniformly convex (or inversely right-uniformly convex) if for all $\varepsilon\in (0,1)$ there exists $\delta=\delta(\varepsilon)\in (0,{\varepsilon}/{8})$ such that for all $f,g\in S$ the condition $0\leqslant \|f-\frac{1}{2}g|+\frac{1}{2}\|g|-\|f|<\delta(\varepsilon)$ implies that $f \in B^-(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. The definition of an inversely left-uniformly convex space is similar; the only difference is that the last conclusion is replaced by the following one: $g \in B^-(\mu f,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. Note that if $(X,{\|\cdot|})$ is an inversely right-uniformly convex (inversely left-uniformly convex) linear space, then so is the mirror space $(X,{\|-\cdot|})$. Theorem 6. Let $X=(X,{\|\cdot|})$ be a left-complete inversely uniformly convex asymmetric seminormed space, and let $M\subset X$ be a right-bounded set. Then $M$ has a right-Chebyshev centre in $X$, that is, there exists a point $x\in X$ such that Proof. The proof is similar to the proof of Theorem 5, the only difference is that the sequence $\{x_n\}_{n=1}^{\infty} $ is taken in $X$, rather than in $M_C$.
It can be assumed without loss of generality that $r(M,X)=1$ and ${0\in M}$. Since $X$ is inversely uniformly convex, for any $\varepsilon\in (0,1)$ there exists a number ${\delta=\delta(\varepsilon)\in (0,{\varepsilon}/{8})}$ such that for all $f,g\in S$ the condition $0\leqslant \|f-\frac{1}{2}g|+{\frac{1}{2}\|g|-\|f|<\delta(\varepsilon)}$ implies that $f \in B^-(\mu g,\varepsilon)$ for some $\mu\in [1-\varepsilon,1]$. We set $\varepsilon_n:={ }{\varepsilon}/{2^n}$, where $\varepsilon\in (0,1)$ and $n\in \mathbb{N}$, and choose a nonincreasing number sequence $\delta_n:=\min_{k=1,\dots,n}\{\delta(\varepsilon_k/2)\}$, $n\in \mathbb{N}$. We construct the sequence $\{x_n\}_{n=1}^{\infty}\subset M_C$ recursively. We consider $x_1\in X$ such that $M\subset B(x_1,1+{\delta_1}/{10^4}) $ and then select $x_n\in X$ so that $M\subset B(x_n,1+{\delta_n}/{10^4}) $ for $n>1$. There exists a point $y\in M$ such that $\|y-(x_n+x_{n-1})/2|\geqslant 1-{\delta_n}/{10^4}$. Setting $f_{n}:=y -x_{n}$ and $f_{n-1}:=y -x_{n-1}$ we have
$$
\begin{equation*}
y-\frac{x_n+x_{n-1}}2=\frac{f_n+f_{n-1}}2, \qquad \|f_{n}|\leqslant 1+\frac{\delta_n}{10^4}\quad\text{and} \quad \|f_{n-1}|\leqslant 1+\frac{\delta_{n-1}}{10^4}.
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\frac{1}{2}\|f_{n}|+\frac{1}{2}\|f_{n-1}|\geqslant \biggl\|\frac{f_n+f_{n-1}}2 \biggr|\geqslant 1-\frac{\delta_n}{10^4},
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
\|f_{n}|\geqslant 1-\frac{2\delta_n}{10^4}-\frac{\delta_{n-1}}{10^4}>1-\frac{\delta_{n-1}}{10^3}\quad\text{and} \quad \|f_{n-1}|\geqslant 1-\frac{2\delta_n}{10^4}-\frac{\delta_{n}}{10^4}>1-\frac{\delta_{n-1}}{10^3}.
\end{equation*}
\notag
$$
Let $\alpha_n$ and $\alpha_{n-1}$ be such that $f'_{n}:=\alpha_nf_n\in S$ and $f'_{n-1}:=\alpha_{n-1}f_{n-1}\in S$. Then
$$
\begin{equation*}
1-\frac{\delta_{n-1}}{10^3}<\|f_n|,\|f_{n-1}|<1+\frac{\delta_{n-1}}{10^3},
\end{equation*}
\notag
$$
and so
$$
\begin{equation*}
\alpha_{n-1},\alpha_{n}\in \biggl[\biggl(1+\frac{\delta_{n-1}}{10^3}\biggr)^{-1}, \biggl(1-\frac{\delta_{n-1}}{10^3}\biggr)^{-1}\biggr].
\end{equation*}
\notag
$$
Arguing as in the proof of Theorem 5 we can show that
$$
\begin{equation*}
\biggl\|\frac{f'_{n}+f'_{n-1}}{2}\biggr| \geqslant 1-\frac{\delta_{n-1}}{250}.
\end{equation*}
\notag
$$
We have $\|(f'_{n}+f'_{n-1})/2-f'_{n-1}/2|+\|f'_n|/2=1$, and so, if $\beta_n\geqslant 1$ is such that $\beta_n(f'_{n}+f'_{n-1})/2\in S$, then
$$
\begin{equation*}
0\leqslant \beta_n-1\leqslant \frac{{\delta_n}/{250}}{1-{\delta_n}/{250}} <\frac{\varepsilon_n}{4}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \biggl\|\beta_n\frac{f'_{n}+f'_{n-1}}{2}-\frac{1}{2}f'_{n-1}\biggr|+\frac{1}{2}\|f'_n|-1 &\leqslant \biggl\|(\beta_n-1)\frac{f'_{n}+f'_{n-1}}{2}\biggr| \\ &\leqslant (\beta_n-1)\biggl(1+\frac{\delta_{n-1}}{10^4}\biggr)<\frac{1}{2}\delta_n. \end{aligned}
\end{equation*}
\notag
$$
Since $X$ is inversely uniformly convex, there exists a number $\widehat{\mu}_{n-1}\in[1-\varepsilon_{n-1}/2,1]$ such that $\beta_{n} (f'_n+f'_{n-1})/2 \in B^-(\widehat{\mu}_{n-1}f'_{n-1},\varepsilon_{n-1})$. Hence
$$
\begin{equation*}
\biggl\|\frac{\widehat{\mu}_{n-1}}{\beta_n}f'_{n-1}-\frac{f'_n+f'_{n-1}}{2}\biggr|\leqslant \frac{\varepsilon_{n-1}}{\beta_n}\quad\text{and} \quad \|\mu_{n-1}f'_{n-1}-f'_n|\leqslant 2 \frac{\varepsilon_{n-1}}{\beta_n} \leqslant 4\varepsilon_{n-1},
\end{equation*}
\notag
$$
where Thus, $f'_{n}\in B^-(\mu_{n-1} f'_{n-1},4 \varepsilon_{n-1})$. As a result, $\|\mu_{n-1} f'_{n-1}-f'_n|\leqslant 4\varepsilon_{n-1}$, that is,
$$
\begin{equation*}
f_n\in B^-\biggl(\frac{\mu_{n-1}}{\alpha_n} f'_{n-1},\frac{4\varepsilon_{n-1}}{\alpha_n}\biggr) = B^-\biggl(\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n} f_{n-1},\frac{4\varepsilon_{n-1}}{\alpha_n}\biggr).
\end{equation*}
\notag
$$
Proceeding as in the proof of Theorem 2 and taking $\mu_{n-1}=1-\varepsilon_{n-1}$ we see that ${\mu_{n-1}\alpha_{n-1}}/{\alpha_n} <1$ and
$$
\begin{equation*}
\begin{aligned} \, 0 &\leqslant 1-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n} =\frac{\alpha_n-(1-\varepsilon_{n-1})\alpha_{n-1}}{\alpha_n} \\ &\leqslant\frac{(1-{\delta_{n-1}}/{10^3})^{-1} -(1-\varepsilon_{n-1})(1+{\delta_{n-1}}/{10^3})^{-1}}{(1+{\delta_{n-1}}/{10^3})^{-1}} \\ &=\frac{{2\delta_{n-1}}/{10^3}}{1-{\delta_{n-1}}/{10^3}}+\varepsilon_{n-1}<2\varepsilon_{n-1}. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, \frac{5\varepsilon_{n-1}}{\alpha_n} &\geqslant \biggl\|\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n} f_{n-1}-f_n\biggr| =\biggl\|\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}(y -x_{n-1})-(y- x_n)\biggr| \\ &=\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha-_n}x_{n-1}- \biggl(y-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}y\biggr)\biggr| \\ &\geqslant \biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr| -\biggl(1-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}\biggr)\|y| \\ &\geqslant\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr| -\biggl(1-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}\biggr)d, \end{aligned}
\end{equation*}
\notag
$$
where $d= \sup_{z\in M}\|z|<+\infty$. Hence
$$
\begin{equation*}
\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr|\leqslant \varepsilon_{n-1}\biggl(\frac5\alpha_n+2d\biggr)\leqslant (8+2d)\varepsilon_{n-1}.
\end{equation*}
\notag
$$
Setting $P_n:=\prod_{k\geqslant n} ({\mu_{k}\alpha_{k}}/{\alpha_{k+1}})$ and $\widehat{x}_n:=P_nx_n$, $n\in \mathbb{N}$, we have
$$
\begin{equation*}
\|\widehat{x}_{n}-\widehat{x}_{n-1}| =P_n\biggl\|x_n-\frac{\mu_{n-1}\alpha_{n-1}}{\alpha_n}x_{n-1}\biggr|\leqslant (8+2d)\varepsilon_{n-1},\qquad n\in \mathbb{N},
\end{equation*}
\notag
$$
and therefore $\{\widehat{x}_n\}$ is a Cauchy sequence. Since $X$ is left-complete, there exists a point $x\in X$ such that $\|\widehat{x}_n- x|\to 0$ as $n\to \infty$, and so $\|x_n- x|\leqslant \|x_n- \widehat{x}_n|+\|\widehat{x}_n- {x}|= (1-P_n)\|x_n|+\|\widehat{x}_n- {x}| \to 0$ as $n\to \infty$. Moreover, $\| z-x|\leqslant \|z- x_n|+\|x_n-x|\leqslant \|x_n-x|+ 1+{\delta_{n-1}}/{10^4}\to 1=r(M,X)$ for all $z\in M$. Hence $x$ is a right-Chebyshev centre in $X$ for $M$. Theorem 6 is proved. Remark 13. A result similar to in Theorem 6 also holds for left-Chebyshev centres. Namely, let $X=(X,{\|\cdot|})$ be a left-complete inversely uniformly convex asymmetric seminormed space and $M\subset X$ be a left-bounded set. Then there exists a left-Chebyshev centre for $M$ in $X$, that is, there exists a point $x\in X$ such that Definition 8. A set $M\subset X$ is called left-compact if it is compact relative to the topology generated by the subbase of left-open balls. Theorem 7. Let $X=(X,{\|\cdot|})$ be a complete locally uniformly convex asymmetric space, $M\subset X$ be a left-compact set and $x$ be a left-Chebyshev centre of $M$. Then $\|x-x_n|\to 0$ as $n\to\infty$ whenever $\{x_n\}\in X$ is a sequence such that Proof. Assume the contrary. Passing to a subsequence if necessary we can assume that $\|x-x_n|> \delta$, $n\in \mathbb{N}$, for some $\delta>0$. It can also be assumed without loss of generality that $r=1$. Since $X$ is locally uniformly convex, for all ${g\in S}$ and ${\varepsilon\in(0,1)}$ there exists $\delta=\delta(\varepsilon)\in (0,{\varepsilon}/{8})$ such that for any $f\in S$ the condition $0\leqslant \|f-\frac{1}{2}g|+\frac{1}{2}\|g|-\|f|<\delta(\varepsilon)$ implies that $\|f-g|<\varepsilon$. We set $\varepsilon_n:={\varepsilon}/{2^n}$, $n\in \mathbb{N}$, for some $\varepsilon\in (0,1)$. Consider the nonincreasing number sequence $\delta_n:=\min_{k=1,\dots,n}\{\delta(\varepsilon_k/2)\}$, $n\in \mathbb{N}$. Passing to a subsequence again we can assume that $M\subset B^-(x_n,1+{\delta_n}/{10^4}) $ for all $n $. The left-norm is upper semicontinuous with respect to the topology generated by it, hence the function $\|x-\cdot|$ assumes its maximum on any nonempty left-compact set. Therefore, there exists a point ${y_n\in M}$ such that $\|(x_n+x)/2-y_n|\geqslant 1$. Since $M$ is left-compact, there exists $y\in M$ such that $\|y-y_{n_k}|\to 0$ as $k\to\infty$. Moreover, this subsequence contains a new subsequence such that $\|y-y_{n'_k}|\leqslant {\delta_k}/{10^4}$. For convenience we also denote this subsequence by $\{y_k\}$. Then
Setting $f_{n}:=x_{n}-y$ and $f:=x-y$, we have $(x_n+x)/2-y=(f_n+f)/2 $, $\|f_{n}|\leqslant 1+{\delta_n}/{10^4}$ and $\|f|\leqslant 1$. Let $\alpha_n$ and $\alpha $ be numbers such that $f'_{n}:=\alpha_nf_n\in S$ and $f' :=\alpha f \in S$. We also take $\beta_n\geqslant 1$ such that $\beta_n(f'_{n}+f')/{2}\in S$. Arguing as in the proof of the previous theorem we obtain
$$
\begin{equation*}
0\leqslant \beta_n-1\leqslant \frac{{\delta_n}/{250}}{1-{\delta_n}/{250}}<\frac{\varepsilon_n}{4}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \biggl\|\beta_n\frac{f'_{n}+f'}{2}-\frac{1}{2}f'_n\biggr|+\frac{1}{2}\|f'_n|-1 &\leqslant \biggl\|(\beta_n-1)\frac{f'_{n}+f'}{2}\biggr| \\ &\leqslant (\beta_n-1)\biggl(1+\frac{\delta_{n-1}}{10^4}\biggr)<\frac{1}{2}\delta_n. \end{aligned}
\end{equation*}
\notag
$$
Now, since $X$ is locally uniformly convex, we show as in the proof of Theorem 5 that $\|f'- f'_n|\leqslant 4\varepsilon_{n-1}$, that is, $f\in B(({1}/{\alpha}) f'_n,{4\varepsilon_{n-1}}/{\alpha})= B(({\alpha_{n}}/{\alpha})f_n,{4\varepsilon_{n}}/{\alpha})$. Hence
$$
\begin{equation*}
\biggl\|\frac{\alpha_{n}}{\alpha} f-\frac{\alpha_{n}}{\alpha} f_n\biggr|\leqslant \biggl\|\frac{\alpha_{n}}{\alpha} f-f\biggr|+ \biggl\|f-\frac{\alpha_{n}}{\alpha} f_n\biggr| \leqslant\biggl|\frac{\alpha_{n}}{\alpha}-1\biggr|\|f\|+\frac{4\varepsilon_{n}}{\alpha}\to 0, \qquad n\to\infty,
\end{equation*}
\notag
$$
and therefore
$$
\begin{equation*}
\|x-x_n|=\|f-f_n |=\frac{\alpha}{\alpha_{n}} \biggl\|\frac{\alpha_{n}}{\alpha} f-\frac{\alpha_{n}}{\alpha} f_n\biggr| \to 0, \qquad n\to\infty,
\end{equation*}
\notag
$$
which contradicts the original assumption. This proves Theorem 7. § 5. Properties of sunsDefinition 9. Let $\varnothing \ne M\subset X$. We say that $x\in X\setminus M$ is a solar point for $M$ if there exists a point $y\in P_Mx\ne \varnothing$ (called a luminosity point) such that $y\in P_M((1-\lambda)y+\lambda x)$ for all $\lambda\geqslant 0$ (geometrically, this means that there is a ‘solar’ ray emanating from $y$ and passing through $x$ such that $y$ is a nearest point in $M$ to any point on the ray). We say that $x\in X\setminus M$ is a strict solar point if $P_Mx\ne\varnothing$ and each $y\in P_Mx$ is a luminosity point for $x$. A set $M$ is a sun (a strict sun) with respect to a set $K\subset X\setminus M$ if any point in $K$ is a solar (strict solar) point for $M$. For $K= X\setminus M$ we say in this case that $M$ is a sun (a strict sun, respectively). Definition 10. A set $M$ of an asymmetric seminormed space $X=(X,{\|\cdot|})$ is called a $\gamma$-sun if for any $\delta>0$ the ball $B(x_0,r_0-\delta)$, $r_0=\varrho(x_0,M)$, can be put in some ball $B(x,R)$ of arbitrarily large radius $R$ which is disjoint from $M$. Theorem 8. Let $X=(X,{\|\cdot|})$ be a locally uniformly convex asymmetric space. Then any proximinal $\gamma$-sun $M\subset X$ in $X$ is a Chebyshev sun. Proof. First we show that $M$ is a strict sun. Consider arbitrary points ${x\in X\setminus M}$ and $y\in P_Mx$. It can be assumed without loss of generality that ${\varrho(x,M)=1}$ and $x=0$. We claim that $y$ is a nearest point in $M$ to any point on the ray $\ell:=\{y+t(x-y)\mid t\geqslant 0\}$. Let $z\in \ell$ and $\|y-z|=R>1$. Consider sufficiently small $\delta\in (0,1/2)$. We set $k=(R-\delta)/(1-\delta)$, $a:=(1-\delta)/(R-\delta)$ ($\geqslant a_0:={1}/(2R+1)$), $g':=y$ and $y':=y(1-\delta)$. Let us construct a function ${\varepsilon=\varepsilon(\delta)>0}$ that tends to zero as $\delta\to 0$ and such that for all $\varphi\in S(0,1)$ the condition $\|\varphi-ag'|+\|ag'|-1\leqslant \delta$ implies that $\|\varphi-g'|<\varepsilon(\delta)$. Since $M$ is a $\gamma$-sun, there exist $z_\delta\in X$ and $R'>R+3\delta$ such that $B(z_\delta,R')\cap M=\varnothing$ and $B(z_\delta,R')\supset B(0,1-\delta)$. Recall that the condition $B(u_1,r_1)\subset B(u_2,r_2)$ is equivalent to $\|u_1-u_2|\leqslant r_2-r_1$. Let $w$ be the point of intersection of $S(0,1-\delta)$ and the ray $\{ t(0-z_\delta)\mid t\geqslant 0\}$. We have $B(z_\delta,R')\supset B(z_\delta,\|w-z_\delta|)\supset B(0,1-\delta)$, and therefore $ B(z_\delta,\|w-z_\delta|)\cap M=\varnothing$. Let $z'\in [z_\delta,w]$ be a point such that $R-\delta=\|w-z'|$. In this case $B(z_\delta,\|w-z_\delta|)\supset B(z',\|w-z'|)\supset B(0,1-\delta)$ and $\|-z'|=R-1$. Let $y',y''\in [0,y]$ be such that $\|y'|=1-\delta$ and $y''\in S(z',R-\delta)$. In this case $y''\in [y',y]$.
Let $f':={\omega}/(1-\delta)$, $f'':=ag'+(1-a)f'$ and $\theta:=\|f''|<1$. Assume that, for some null sequence $\delta=\delta_n$, $n\in \mathbb{N}$, we have $\theta_n\leqslant \theta_0<1$, $n\in \mathbb{N}$, for the corresponding quantities $\theta=\theta_n$ (constructed from the $\delta_n$). Then $B(f'',1-\theta_0)\subset B(0,1)$, and therefore
$$
\begin{equation*}
B\bigl(ay'+(1-a)w,(1-\theta_0)(1-\delta)\bigr)=B\bigl(f''(1-\delta),(1-\theta_0)(1-\delta)\bigr) \subset B(0,1-\delta).
\end{equation*}
\notag
$$
A homothety with centre $w$ and ratio $k$ yields $B(y',k(1-\theta_0)(1-\delta))\subset B(z',R-\delta)$. However, this inclusion is impossible when $k(1-\theta_0)(1-\delta)>\delta$, since in this case the ball $B(z',R-\delta)$ does not meet $M$, and therefore does not contain the point $y$.
Consider the case when $\theta_n\to 1$ as $n\to \infty$. In this case we have $\|\theta_n^{-1}f'' -ag'|+\|ag'|-1\leqslant\|\theta_n^{-1}f''-f''|+\|f'' -ag'|+\|ag'|-1=(\theta_n^{-1}-1)\|f''|\leqslant \theta_n^{-1}-1$. Hence $\|\theta_n^{-1}f''-g'|< \varepsilon(\theta_n^{-1}-1)=:\varepsilon_n\to 0$ as $n\to \infty$, and so
$$
\begin{equation*}
(1-a)\|f'-g'|=\|\theta_ng'-g'|+\|f''-\theta_ng'|\leqslant (1-\theta_n)\|-g'|+\varepsilon_n\theta_n\to 0, \qquad n\to\infty.
\end{equation*}
\notag
$$
Consequently, for the sequences $y'=y'_n$ and $w=w_n$, which are constructed from the $\delta_n$, we have $\|y-y'_n|\to 0$ and $\|w_n-y|\to 0$ as $n\to\infty$. Next, for the sequence $z'=z'_n $, which is also constructed from the $\delta_n$, we have $\|z-z'_n|\to 0$ as $n\to\infty$. Hence $\|q-z'_n|\leqslant \|q-z|+\|z-z'_n|\leqslant R-\sigma+\|z-z'_n|\to R-\sigma$ for any $\sigma\in (0,R)$ and $q$ such that $\|q-z|\leqslant R-\sigma$. Therefore, for all $n$ exceeding some number, $q$ lies in $\mathring B(z'_n,R-\delta_n)$ and does not lie in $M$. Hence the ball $B(z,R)$ supports the set $M$, and $y$ is a nearest point to $z$ in $M$. Since $R$ is arbitrary, it follows that $y$ is a nearest point in $M$ to any point on the ray $\ell $. This proves Theorem 8. Definition 11. Let $X=(X,{\|\cdot|})$ be an asymmetric seminormed space and let $M\subset X$. We say that $M$ is left-approximatively closed at $x\in X$ if for each minimizing sequence $\{y_n\}\subset M$ for $x$ (so that $\|y_n-x|\to\varrho(x,M)$ as $n\to\infty$) that converges to some $y\in X$ (that is, $\|y_n-y| \to 0$ as $n\to\infty)$ we have $y\in M$ and ${\|y-x|=\varrho(x,M)}$. We say that $M$ is left-approximatively closed if it is left-approximatively closed at any point $x\in X$. The definition of a right-approximatively closed set is similar. Remark 14. Each left-approximatively closed set is closed. Conversely, if a nonempty subset $M$ of an asymmetric normed space is closed and the ball $B(0,1)$ is closed, then $M$ is approximatively closed (that is, $M$ is both left- and right-approximatively closed). Theorem 9. Let $X=(X,{\|\cdot|})$ be a left-complete uniformly convex asymmetric seminormed space and $M\subset X$ be a left-approximatively closed $\gamma$-sun. Then $M$ is a set of existence. Proof. Let $x\in X\setminus M$. It can be assumed without loss of generality that $\varrho(x,M)= 1$ and $x=0$. For any sequence $\{\delta_n\}\subset (0,+\infty)$ there exists a sequence $\{R_n\}\subset (1,+\infty)$ such that $\delta_n\to 0$ and $R_n\to +\infty$ as $n\to\infty$, and $B(x_n,R_n)\supset B(0,1-\delta_n)$ for $n\in \mathbb{N}$. We can assume that $R_n=\|-x_n|+1-\delta_n$, $n\in \mathbb{N}$.
By Lemma 1, for the sequence $\{\varepsilon_n={\varepsilon}/{2^n}\}$ $(\varepsilon\in (0,1))$ there exists a null sequence $\{\delta_n\}$ such that $\delta_n\in (0,\varepsilon_n/16)$ for $n\in \mathbb{N}$, and for an arbitrary point $f_{n-1}\in M$ such that $\|f_{n-1}|<1+\delta_{n-1}$ and any $f \in M$, $\|f |<1+\delta_{n}$, we have $f\in B(\mu_{n-1} f_{n-1},\varepsilon_n)$ for some $\mu_{n-1}\in[1-\varepsilon_{n-1},1]$. Here, as $\Delta$ and $\delta_0$ from Lemma 1 we take, respectively, $\Delta_n={\|-x_n|}/{R_n}$ and $2\delta_n$. Completing our recursive construction of the sequence $\{f_n\}$, we take $f_n\in M$ such that $\|f_n |<1+\delta_{n}$. In this case $f_n\in B(\mu_{n-1} f_{n-1},\varepsilon_{n-1})$. We have
$$
\begin{equation*}
\biggl\|\biggl(\prod_{k\geqslant n+1}\mu_k\biggr)f_{n+1}-\biggl(\prod_{k\geqslant n}\mu_k\biggr)f_n\biggr|= \biggl(\prod_{k\geqslant n+1}\mu_k\biggr)\|f_{n+1}-\mu_n f_n|\leqslant \varepsilon_{n+1}.
\end{equation*}
\notag
$$
Hence $\{\widehat{f}_n:=(\prod_{k\geqslant n}\mu_k)f_n\}$ is a Cauchy sequence in $X$ and, next, since $X$ is left-complete, there exists a point $y\in X$ such that $\|\widehat{f}_n-y|\to 0$ as $n\to \infty$. As a result,
$$
\begin{equation*}
\|{f}_n-y|\leqslant \|{f}_n-\widehat{f}_n|+\|\widehat{f}_n-y|=\biggl(1-\biggl(\prod_{k\geqslant n}\mu_k\biggr)\biggr)\|f_n|+\|\widehat{f}_n-y|\to 0, \qquad n\to\infty,
\end{equation*}
\notag
$$
and, moreover, $y\in M$ and $\varrho(0,M)=\|y|$ since $M$ is approximatively closed. So $y$ is a nearest point to $x$ in $M$. Now the conclusions of the theorem follows since $x$ is arbitrary. Theorem 9 is proved. Definition 12. Let $\varepsilon\geqslant 0 $ and $M\subset X$. We say that $\varphi\colon X\to M$ is an additive (multiplicative) $ \varepsilon $-selection (of the operator of near-best approximation) if, for all $x\in X $ we have ($ \varphi(x)\in P_M^{\varepsilon\varrho(x,M)} x$, respectively). An $ \varepsilon $-selection on $E\subset X$ is defined similarly. Let $\Xi(x,\varepsilon,\tau)=\Xi_M(x,\varepsilon,\tau) = \{\varphi\colon B(x,\tau) \to M \mid \|\varphi(y)-y| \leqslant \varrho(y,M)+\varepsilon \}$ be the class of additive $\varepsilon$-selections onto a set $M$ defined on the ball $B(x,\tau)$. The modulus of uniform approximative continuity (at a point $x)$, where $x\in X$, is defined for $\tau>0$ by
$$
\begin{equation*}
uc_x(\tau)=\inf\biggl\{ \frac{\omega_x(\varphi,\tau)}{\varrho(x,M)}+\frac{\varepsilon}{\tau}\biggm| \varepsilon>0,\,\varphi\in \Xi(x,\varepsilon,\tau) \biggr\},
\end{equation*}
\notag
$$
and the oscillation of a mapping $\varphi$ at a point $x$ with step $\tau$ is defined by $\omega_x(\varphi,\tau)=\sup_{\|x-y|\leqslant \tau}\|\varphi(y)-\varphi(x)|$. We recall the definition of a $\delta$-sun (see [27]). Definition 13. Let $M$ be a nonempty subset of an asymmetric seminormed space $X=(X,{\|\cdot|}))$. A point $x_0\in X$ such that $\varrho(x_0,M)>0$ is called a $\delta$-solar point for $M$ if there exists a sequence of points $\{x_n\}$, $\|x_0-x_n|\to 0$ as $n\to\infty$, such that A set $M$ is called a $\delta$-sun if any $x \in X $ satisfying $\varrho(x ,M)>0$ is a $\delta$-solar point for $M$. Definition 14. Set A point $x\in X\setminus M$ is a radial $\delta$-solar point for $M$ if there exists a sequence $\{\tau_n\}$ such that $\tau_n\to 0+$ as $n\to\infty$ and $\delta_x^r(\tau_n) = o(1)$ as $n\to\infty$. Theorem 10. Let $M$ be a nonempty subset of an asymmetric seminormed space $X=(X,{\|\cdot|})$. Assume that $uc_x(\tau)=o(1)$ as $\tau\to0+$. Then $\delta_x^r(\tau)=o(1)$ as ${\tau\to0+}$, that is, $x\in X\setminus M$ is a radial $\delta$-solar point for $M$. Proof. It can be assumed without loss of generality that $\varrho(x,M)=1$ and $\tau\in (0,1)$. Indeed, the condition that $uc_x(\tau)=o(1) $ as $\tau\to0+$ implies that there exists a map $\varphi=\varphi_{x,\varepsilon,2\tau}\in \Xi(x,\varepsilon,2\tau)$ such that $\omega_x(\varphi,2\tau)=o(1) $ and $\varepsilon=o(\tau)$ as $\tau\to0+$. Let
$$
\begin{equation*}
y=\varphi(x), \qquad v=x+\tau\frac{x-y}{\|y-x|}, \qquad v'=\varphi(v)\quad\text{and} \quad v''=x+\tau\frac{x-v'}{\|y-x|}.
\end{equation*}
\notag
$$
Then $\|v'-y|=\|\varphi(v)-\varphi(x)|\leqslant \omega_x(\varphi,\tau)=o(1) $,
$$
\begin{equation*}
\biggl|\frac{\|v'-x|}{\|y-x|}-1\biggr|=\biggl|\frac{\|v'-x|-\|y-x|}{\|y-x|}\biggr| \leqslant\biggl|\frac{v'-x}{\|y-x|}-\frac{y-x}{\|y-x|}\biggr|=\frac{\|v'-y|}{\|y-x|}=o(1)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\|v-v''|=\tau\biggl\|\frac{x-v'}{\|y-x|}-\frac{x-y}{\|y-x|}\biggr\|=o(\tau), \qquad \tau\to0+.
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\varrho(v,M)\leqslant \|v'-v|\leqslant \varrho(v,M)+\varepsilon=\varrho(v,M)+o(\tau)
\end{equation*}
\notag
$$
we have the asymptotic equality Next,
$$
\begin{equation*}
\|v'-v''|-\|v' -v| \leqslant \|v-v'' | =o(\tau)\quad\text{and} \quad \varrho(v,M)-\varrho(v'',M) \leqslant \|v''-v \| =o(\tau),
\end{equation*}
\notag
$$
hence
$$
\begin{equation*}
\begin{aligned} \, \|v'-x|&=\|v'-v''|-\|x-v''|=\|v'-v''|-\tau\frac{\|v'-x|}{\|y-x|} =\|v'-v''|-\tau(1+o(1)) \\ &= \varrho(v,M)+o(\tau)-\tau+o(\tau)\leqslant \varrho(v'',M)+o(\tau)-\tau. \end{aligned}
\end{equation*}
\notag
$$
Now,
$$
\begin{equation*}
\varrho(v'',M)-\|v'-x|\leqslant \|v'-v''|-\|v'-x|=\|x-v''|=\tau\frac{\|x-v'|}{\|y-x|}=\tau(1+o(1)),
\end{equation*}
\notag
$$
and so we have
$$
\begin{equation*}
\frac{\varrho(v'',M)-\|v'-x|}{\tau}\to 1, \qquad \tau\to 0+.
\end{equation*}
\notag
$$
Hence $\delta_x^r(\tau)\to 0$, that is, $x$ is a radial $\delta$-solar point for $M$. This proves Theorem 10. Remark 15. If there exists a sequence $\{\tau_n\}$ such that $\tau_n \to 0+$ as $n \to \infty$ and $uc_x(\tau_n)=o(1) $ as $n\to \infty$, then there exists a sequence $\{\tau'_n\} $ such that $\tau'_n\to 0+$ as ${n\to\infty}$ and $\delta_x^r(\tau'_n)=o(1)$ as $n\to\infty$. Hence $x\in X\setminus M$ is a radial $\delta$-solar point for $M$. Remark 16. Since $\varrho(v'',M)\leqslant \varrho(x,M)+\|x-v''|=\varrho(x,M)+\tau$, we have and so the radial $\delta$-solarity of $M$ implies that $M$ is a $\delta$-sun. Remark 17. Alimov proved that in a left-complete asymmetric normed space any $\delta$-sun with continuous distance function is a $\gamma$-sun. It follows from this result and Theorem 9 that in a left-complete uniformly convex asymmetric space any left-approximatively closed $\delta$-sun with continuous distance function is an existence set, which, moreover, is a Chebyshev sun in view of Theorem 8. 
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\Bibitem{Tsa22} |
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