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This article is cited in 4 scientific papers (total in 4 papers) I. G. Tsar'kovab a Lomonosov Moscow State University, Faculty of Mechanics and Mathematics b Moscow Center for Fundamental and Applied Mathematics
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2024, Volume 88, Issue 2, DOI: https://doi.org/10.4213/im9393 
Bibliographic databases:
    § 1. IntroductionSolution of special problems depends on minimization of some functionals under the assumption that the solution to this problem is unique or stable. These functionals will be considered on asymmetric normed linear spaces $ (X,\|\,{\cdot}\,|)$. Let $\varphi\colon X\to \mathbb{R}$ be a lower bounded functional and $\psi\colon X\to \mathbb{R}_+$ be a continuous functional. By $\theta$ we denote the pair $(\psi,\varphi)$. In most cases, it suffices to define the functional $\varphi$ on a given set $M$ and then extend somehow, if necessary, to the whole space $X$. For a set $M\subset X$ and $y\in X$, the $\theta$-metric function (the $\theta$-distance) to $M$ from $y$ is defined by
$$
\begin{equation*}
\varrho_\theta(y,M):= \inf_{z\in M}(\psi(z-y)+\varphi(z)).
\end{equation*}
\notag
$$
Sometimes, instead of the function $\varphi\colon X(M)\to \mathbb{R}$ we will consider a function $\varphi\colon X\times X\to \mathbb{R}$ or $\varphi\colon M\times X\to \mathbb{R}$. In this case, the $\theta$-distance to $M $ is defined by
$$
\begin{equation*}
\varrho_\theta(y)=\varrho_\theta(y,M):=\inf_{z\in M}(\psi(z-y)+\varphi(z,y)).
\end{equation*}
\notag
$$
The left $\theta$-distance to a set $M$ is defined similarly by
$$
\begin{equation*}
\varrho_\theta^-(y)=\varrho_\theta^-(y,M):=\inf_{z\in M}(\psi(y-z)+\varphi(z,y)).
\end{equation*}
\notag
$$
We also consider the usual metric function The set of all $\theta$-nearest points in $M$ for $x\in X$ is defined by
$$
\begin{equation*}
\theta\text{-}P_{M}x (= \theta\text{-}P_{M}(x)):= \{y\in M\mid \psi(y-x)+\varphi(y)=\varrho_\theta(x,M)\}.
\end{equation*}
\notag
$$
Given $\varepsilon\geqslant 0$, the set of all $\varepsilon$-nearest points in $M$ for $x$ is defined by
$$
\begin{equation*}
\theta\text{-}P_M^\varepsilon x(=\theta\text{-}P_M^\varepsilon (x))=\{y\in M\mid \psi(y-x)+\varphi(y)\leqslant\varrho_\theta(x,M)+\varepsilon\}.
\end{equation*}
\notag
$$
The mapping $\theta\text{-}P_M$ is called the $\theta$-metric projection to the set $M$. The ball and sphere with centre $x$ and radius $r\geqslant 0$ are defined by
$$
\begin{equation*}
B(x,r)=\{y\in X\mid \|y-x|\leqslant r\}\quad\text{and}\quad S(x,r)=\{y\in X\mid \|y-x|= r\}.
\end{equation*}
\notag
$$
For brevity, we set $B:=B(0,1)$, $S:=S(0,1)$ (the unit ball and the unit sphere, respectively). By $X^*$ we denote the space of all bounded linear functionals (see [1], [2]); the dual unit sphere is denoted as $S^*$. Similarly, the $\theta$-ball with centre $x$ of radius $r\in \mathbb{R}$ in the space $X$ is defined by $B_\theta(x,r)=\{y\in X\mid \psi(y-x)+\varphi(y)\leqslant r\}$ (respectively, $B_\theta(x,r)=\{y\in X\mid \psi(y- x)+ \varphi(y,x)\leqslant r\}$). The $\theta$-sphere with centre $x$ of radius $r\in \mathbb{R}$ is defined by $S_\theta(x,r)=\{y\in X \mid \psi(y-x)+\varphi(y)= r\}$ (respectively, $S_\theta(x,r)=\{y\in X\mid \psi(y-x)+\varphi(y,x)= r\}$). Below, $\theta\text{-}P_{M}x =\theta\text{-}P_{M}(x):= \{y\in M\mid \psi(y-x)+\varphi(y,x)=\varrho_\theta(x,M)\}$ is the set of $\theta$-nearest points in $M$ for $x\in X$. The set of all $\varepsilon$-nearest points is defined by $\theta$-$P_{M}^\varepsilon x= \theta$-$P_{M}^\varepsilon(x)=\{y\in M\mid \psi(y-x)+\varphi(y,x)\leqslant\varrho_\theta(x,M)+\varepsilon\}$. Given $K\subset X$, by $\operatorname{conv}K$ and $\operatorname{\overline{conv}}K$ we denote, respectively, the convex hull of $K$ and its closure. The main purpose of the present paper is to carry over $\delta$-solar-like properties from normed spaces to spaces with more general gauge functionals. We mention in this regard Moors [3], who obtained an estimate for $\varepsilon$-convexity of an approximating set $M$ in a Hilbert space $H$ under the condition that $\operatorname{diam} P_M x\leqslant \varepsilon$ for all $x\in H$. In this regard, we also mention interesting applications in the problem of non-unique solvability of non-linear differential equations obtained recently by Ricceri [4] with the help of minimization of functionals of special form. Some of these results were refined in [5]. For definitions and results on solar-like sets in normed spaces, see, for example, [5]–[17]. In Theorems 1–3, we obtain analogues of $\delta$- and $\gamma$-solar-like properties with a given lower estimate for the rate of variation of the (left- or right-) $\theta$-metric function. In these results, we estimate the distance form some given point to a point with larger distance of the $\theta$-metric function. In these results, the role of $\delta$-solarity is played by a lower estimate (in terms of a function $f$) for the rate of variation of the $\theta$-metric function. Theorem 3 deals with essentially asymmetric spaces, but for a narrower class of functions $f$ (here, we consider the class of monotone non-decreasing functions). In Theorems 4–7, we study conditions and geometric characteristics from which one can estimate from below the rate of variation of the $\theta$-metric function for various $\theta$. Several stability conditions for the $\theta$-metric projection are considered. In essence, in this way we obtain various extensions of $\delta$-solar properties, and in this sense, these results can be looked upon as inverse theorems of the above results. In Theorem 8, we obtain conditions under which a boundedly compact set with continuous $\theta$-metric projection is a $\theta$-sun (this concept is a direct analogue of the classical definition of a sun). We next give an example illustrating the potency of our results for establishing non-unique solvability in Dirichlet problems. In the remaining part of this section, we consider simple conditions for continuity of the $\theta$-metric function. The corresponding result is used in Theorem 1, which deals with symmetrizable asymmetric spaces. Here, it is worth noting that this continuity may fail to hold (as a rule) in a more general setting. Moreover, in essentially (non-symmetrizable) asymmetric spaces, the continuity of the usual metric projection may fail to imply that of the usual metric function. Definition 1. A set $E\subset X$ will be said to be $\theta$-bounded if there exist $r>0$ and a compact set $K$ such that $\bigcup_{x\in K}B_\theta(x,r)$ contains $E$. A set is $\theta$-boundedly compact if its intersection with any set $\bigcup_{x\in K}B_\theta(x,r)$, where $K$ is some compact set, is compact in $X$. Note that, for all $x,y\in X$,
$$
\begin{equation*}
\varrho_\theta(y)-\varrho_\theta(x)\geqslant \inf_{z\in M}(\psi(z-y)-\psi(z-x)).
\end{equation*}
\notag
$$
Indeed, for each $\varepsilon>0$, there exists $z\in M$ such that $\varrho_\theta(y)\geqslant \psi(z-y)+\varphi(z)-\varepsilon$. Hence
$$
\begin{equation*}
\varrho_\theta(y)-\varrho_\theta(x)\geqslant \psi(z-y) +\varphi(z) -\varepsilon-(\psi(z-x)+\varphi(z))=\psi(z-y)-\psi(z-x)-\varepsilon,
\end{equation*}
\notag
$$
which implies the required inequality. Similarly, for each $\varepsilon>0$, there exists $z\in M$ such that $\varrho_\theta(x)\geqslant \psi(z-x)+\varphi(z)-\varepsilon$. Hence
$$
\begin{equation*}
\varrho_\theta(y)-\varrho_\theta(x)\leqslant \psi(z-y)+\varphi(z) -(\psi(z-x)+\varphi(z)-\varepsilon)=\psi(z-y)-\psi(z-x)+\varepsilon.
\end{equation*}
\notag
$$
Hence, for all $x,y\in X$,
$$
\begin{equation*}
\varrho_\theta(y)-\varrho_\theta(x)\leqslant \sup_{z\in M}(\psi(z-y)-\psi(z-x)).
\end{equation*}
\notag
$$
As a result, we have the inequality
$$
\begin{equation*}
|\varrho_\theta(y)-\varrho_\theta(x)|\leqslant \sup_{z\in M}|\psi(z-y)-\psi(z-x)|.
\end{equation*}
\notag
$$
An asymmetric space $(X,\|\,{\cdot}\,|)$ is called symmetrizable if the asymmetric norm $\|\,{\cdot}\,|$ is equivalent to the symmetrization norm
$$
\begin{equation*}
\|\,{\cdot}\,\|:=\max\{\|\,{\cdot}\,|,\|-\,{\cdot}\,|\},
\end{equation*}
\notag
$$
that is, for some $A>0$, for all $x\in X$. We also note that, for all fixed $z,x\in X$,
$$
\begin{equation*}
|\psi(z-y)-\psi(z-x)|\to 0 \quad\text{if}\quad \|x-y\|\to 0.
\end{equation*}
\notag
$$
As a result, the function $\varrho_\theta(\,{\cdot}\,,M)$ is continuous on any symmetrizable $X$ if either $M\subset X$ is $\theta$-boundedly compact or the function $\psi$ is uniformly continuous on any $\theta$-bounded subset of $X$. This follows from the above inequality if $X$ is symmetrizable. The next result strengtheners this assertion. Lemma 1. Let $(X,\|\,{\cdot}\,|)$ be a symmetrizable asymmetric space. Assume either that $M\subset X$ is $\theta$-boundedly compact, and both $\psi\colon X\to \mathbb{R}$ and $\varphi\colon X\times X\to \mathbb{R}$ are continuous on $X$, or $\psi$ is uniformly continuous on the difference of any $\theta$-bounded set and a compact subspace of $X$, and, for each $\theta$-bounded $N\subset M$, there exists an infinitely small function $\omega_N(\varepsilon)=o(1)$ as $\varepsilon\to 0+$ such that Then the function $\varrho_\theta(\,{\cdot}\,,M)$ is continuous on $X$. Proof. In each of the above cases, the proof is reduced to the situation, where, for each compact set $K\subset X$ and for each $\theta$-bounded set $N\subset M$, the function $\varphi$ is uniformly continuous on $N\times K$, and $\psi$ is uniformly continuous on $N-K$.
Let us first show that the function $\varrho_\theta(\,{\cdot}\,,M)$ is bounded on $K$. Assume on the contrary that there exists a sequence $\{x_n\}\subset K$ such that $\varrho_\theta(x_n)\to +\infty$. It can be assumed without loss of generality that the sequence $\{x_n\}$ converges to some point $x\in K$. Consider an arbitrary point $z\in M$. Each of the sequences $|\psi(z-x_n)-\psi(z-x)|$, $|\varphi(z,x_n)-\varphi(z,x)|$ tends to zero as $\|x_n-x\|\to 0$ as $n\to\infty$, and hence the right-hand side in the inequality $\varrho_\theta(x_n)\leqslant \psi(z-x_n)+\varphi(z,x_n)$ is bounded, which contradicts the construction of the sequence $\{x_n\}$. Let the compact set $K_0$ be defined as the union of an arbitrary sequence $\{x_n\}$ converging to $x$ and its limit point $x$. We set $R := \sup_{z\in K_0} \varrho_\theta(z,M)$, and define $N:=M\cap \bigl( \bigcup_{z\in K_0}B_\theta(z,R+1)\bigr)$. For arbitrary $\varepsilon\in (0,1)$, there exists a point $z \in N$ such that $\varrho_\theta(x,M) \geqslant \psi(z-x)+\varphi(z,x)-\varepsilon$. Hence $\varrho_\theta(x_n,M)-\varrho_\theta(x,M) \leqslant (\psi(z-x_n)+\varphi(z,x_n))-(\psi(z-x)+\varphi(z,x) - \varepsilon) \to \varepsilon$ as $n\to\infty$. Similarly, there exists a point $w_n\in N$ such that $\varrho_\theta(x_n,M)\geqslant \psi(w_n- x_n)+ \varphi(w_n,x_n)-\varepsilon$, $n\in \mathbb{N}$. We have $\varrho_\theta(x_n,M)-\varrho_\theta(x,M)\geqslant (\psi(w_n-x_n)+\varphi(w_n,x_n) - \varepsilon)-(\psi(w_n-x)+\varphi(w_n,x))\to -\varepsilon$ as $n\to\infty$. Since $\varepsilon$ is arbitrary, we have $\varrho_\theta(x_n,M)\to\varrho_\theta(x,M)$ as $n\to\infty$. Since the sequence $\{x_n\}$ is arbitrary, the function $\varrho_\theta(\,{\cdot}\,,M)$ is continuous on $X$. This proves Lemma 1. § 2. Approximative properties of sets as a function of the rate of variation of the $\theta$-metric functionFor brevity, for $x\in X$, we set $r(x)=\varrho_\theta(x,M)$. From a lower estimate for the maximal rate of variation $\varlimsup_{\Delta x\to 0}(r(x+\Delta x)-r(x))/\|\Delta x|$ of the metric $\theta$-function $r(x)$, we will obtain approximative and geometric properties of the approximating set $M$; the other way round, in the next section, from approximative and geometric properties of an approximating set, we will obtain estimates for the rate of variation of the metric $\theta$-function. To estimate from below the maximal rate of variation of the metric function, we will use a continuous function
$$
\begin{equation*}
f\colon [a,+\infty)\to \mathbb{R},\quad f\ \text{is positive on } (a,+\infty),\quad a\geqslant 0.
\end{equation*}
\notag
$$
We will assume that, for each $x\in X$ with $r(x)>a:=\inf_{x\in X}\varrho_\theta(x,M)$, we have
$$
\begin{equation*}
\varlimsup_{\Delta x\to 0}\frac{r(x+\Delta x)-r(x)}{\|\Delta x|}\geqslant f(r(x)).
\end{equation*}
\notag
$$
Note that, for normed spaces $X$, this condition with $f\equiv 1$ and $a=0$ means in the case $\psi(t)\equiv \|t\|$ and $\varphi\equiv 0$ the so-called $\delta$-solarity of the set $M$; in complete spaces, this condition implies that $M$ is almost convex. An account of this is given in the surveys [5]–[8], [10]. Let $a_0>a$ be sufficiently close to $a$ (in what follows, we will assume that $a_0\to a$). Then, for each $R\geqslant a_0$, there exists $\delta_0>0$ such that $\inf_{[a_0,R]}f> 12\mu$, where $\mu=\omega(f,\delta_0)$ (here $\omega(f,\,{\cdot}\,)$ is the modulus of continuity of the function $f$ on $[a,R]$). For all $x\in X$ such that $r(x)>a_0$ and sufficiently small numbers $\Delta,\Delta_1>0$, let $\delta(x)= \delta(x,\Delta,\Delta_1)$ be the supremum of all $\delta\in (0,\min\{\delta_0,\Delta\})$ such that there exist a vector $\Delta x$, $\|\Delta x|=\delta$, and a number $\lambda\in (0,\mu)$ satisfying and, in addition, $r(x+\Delta x)-r(x)<\Delta_1$.Let us first assume that $X$ is symmetrizable, $M \subset X$, and the function $r(\,{\cdot}\,)=\varrho_\theta(\,{\cdot}\,,M)$ is continuous on $X$ (see Lemma 1). Assume that, for some point $y\in X$, there exist a vector $\Delta x$ and number $\lambda\in (0,\mu)$ such that
$$
\begin{equation*}
\|\Delta x|\in \biggl(\frac{\delta(y)}2,\delta(y)\biggr] \quad\text{and}\quad \Delta_1>r(y+\Delta x)-r(y)> (f(r(y))- \lambda)\|\Delta x|.
\end{equation*}
\notag
$$
One can find a neighbourhood $O_\eta(y):=\{z\in X\mid \|z-y|<\eta\}$ such that for any $\mu'\in (0, (\mu-\lambda)/3)$ and any point $x'\in O_\eta(y)$. In addition, for the vector $\Delta x':=y+\Delta x-x'$, we have
$$
\begin{equation*}
\begin{aligned} \, &\Delta_1 >r(x'+\Delta x')-r(x')=r(y+\Delta x)-r(x')> r(y+\Delta x)-r(y)-\mu' \|\Delta x| \\ &\ \geqslant (f(r(y))- (\lambda+ \mu'))\|\Delta x |\geqslant (f(r(x'))- (\lambda+2\mu'))\|\Delta x |=(f(r(x'))- \lambda')\|\Delta x|, \end{aligned}
\end{equation*}
\notag
$$
where $\lambda':=\lambda+2\mu'\in (0,\mu)$. Here, for sufficiently small $\eta$, it can be assumed that $\|\Delta x'|>\|\Delta x|/2$. Theorem 1. Let $X$ be a symmetrizable complete (relative to the symmetrization norm) space, $M\subset X$, the function $r(\,{\cdot}\,)=\varrho_\theta(\,{\cdot}\,,M)$ be continuous on $X$, and let for any point $x\in X$ such that $r(x):=\varrho_\theta(x,M)>a$. Then, for each number $R>a$ and each $x\in X$ such that $R>r(x)\geqslant a$, Proof. Let $\Delta,\Delta_1>0$ be arbitrary small numbers. We argue by induction. Consider an arbitrary point $x\in X$ such that $r(x)\geqslant a_0>a$ and define $x_0=x$.
$1^\circ$. Let us find a vector $\Delta x_1$ and a number $\lambda_1\in (0,\mu)$ such that $\|\Delta x_1|\in (\delta(x_0)/2,\delta(x_0)]$ and $\Delta_1>r(x_0+\Delta x_1)-r(x_0)> (f(r(x_0))-\lambda_1)\|\Delta x_1|\geqslant \mu\|\Delta x_1|$. $2^\circ$. Assume that the point $x_n=x_{n-1}+\Delta x_n$ is constructed. Let us find a vector $\Delta x_{n+1}$ and a number $\lambda_{n+1}\in (0,\mu)$ such that $\|\Delta x_{n+1}|\in (\delta(x_n)/2,\delta(x_n)]$ and $\Delta_1>r(x_n+\Delta x_{n+1})-r(x_n)> (f(r(x_n))-\lambda_{n+1})\|\Delta x_{n+1}|\geqslant \mu\|\Delta x_{n+1}|$. $3^\circ$. We claim that there exists $N\in \mathbb{N}$ such that $r(x_N)> R$. Indeed, if $r(x_n)\leqslant R$ for all $n\in \mathbb{N}$, then $\sum_{n=1}^{\infty}\|\Delta x_{n }|\leqslant (1/\mu)\sum_{n=1}^{\infty}(r(x_n)-r(x_{n-1}))\leqslant R/\mu$. Hence the sequence $\{x_n\}$ converges to some point $y$. Here, $r(y)\leqslant R$. There exist a vector $\Delta x$ and a number $\lambda\in (0,\mu)$ such that $\|\Delta x|\in (\delta(y)/2,\delta(y)]$ and $\Delta_1>r(y+\Delta x)-r(y)> (f(r(y))- \lambda)\|\Delta x | $. Therefore, in some neighbourhood $O_\eta(y)$, for some $\lambda'\in (0,\mu)$, we have
$$
\begin{equation*}
\Delta_1>r(x'+\Delta x')-r(x')> (f(r(x'))-\lambda')\|\Delta x|.
\end{equation*}
\notag
$$
Let $C_R>0$ be such that $f<C_R$ on $[a,R+\mu\Delta]$. Consider an arbitrary $\lambda''\in (\lambda',\mu)$. We will assume that $\eta$ is so small that $\|\Delta x'|<\|\Delta x|+\eta_0$ for some $\eta_0>0$ satisfying $\eta_0 C_R<(\lambda''-\lambda')\|\Delta x|$. We have
$$
\begin{equation*}
(f(r(x'))-\lambda')\|\Delta x| >(f(r(x'))-\lambda'')(\|\Delta x| +\eta_0)>(f(r(x'))-\lambda'')\|\Delta x'|.
\end{equation*}
\notag
$$
For sufficiently small $\eta>0$ (see the last paragraph before the theorem), we have $x_n\in O_\eta(y)$ starting from some $n_0\in \mathbb{N}$, and hence, for $x'=x_n$ we have $\delta(x_n)\geqslant \|\Delta x'|$, and, therefore, $\|\Delta x_n|\geqslant \delta(x_n)/2\geqslant\|\Delta x'|/2>\|\Delta x|/4$, which is impossible because the series $\sum_{n=1}^{\infty}\|\Delta x_{n }|$ is convergent. Hence there exists $N\in \mathbb{N}$ such that $r(x_N)> R$. It can be assumed that $r(x_N)\in [R,R+\Delta_1]$. $4^\circ$. So, we have
$$
\begin{equation*}
\frac{\Delta r_i}{f(r_i)-\mu}\geqslant \|\Delta x_i|,
\end{equation*}
\notag
$$
where $\Delta r_i=r(x_i)-r(x_{i-1})$, $r_i=r(x_i)$, $i=1,\dots, N$. Hence
$$
\begin{equation*}
\sum_{i=1}^{N}\frac{\Delta r_i}{f(r_i)-\mu}\geqslant \sum_{i=1}^{N}\|\Delta x_i|\geqslant \|x_N-x_0|=\|x_N-x|.
\end{equation*}
\notag
$$
We claim that, from any $\varepsilon>0$ one can find a small $\Delta_1>0$ such that
$$
\begin{equation*}
\int_{r(x)}^R\frac{dr}{f(r)-\mu}+\varepsilon\geqslant \|x_N-x| .
\end{equation*}
\notag
$$
This follows from the fact that, for a continuous function on a closed interval, its Riemann sums uniformly approximate the integral over this interval as the diameter of the partition tends to zero. In our case, the diameter of the partition $\max_{i=1,\dots,N} \Delta r_i< \Delta_1$ tends to zero as $ \Delta_1\to 0$. Here, it can be assumed that $r(x_{N-1})\leqslant R<R+\Delta_1$. The function $r(\,{\cdot}\,)$ is continuous, and hence there exists a point $x_R\in [x_{N-1},x_{N}]$ such that $r(x_R)=R$. Hence
$$
\begin{equation*}
\int_{r(x)}^R\frac{dr}{f(r)-\mu}+\varepsilon\geqslant \inf\{\|x_R-x|\mid r(x_R)=R \}-\Delta.
\end{equation*}
\notag
$$
The quantity $\mu$ can be made arbitrarily small by reducing $\delta_0$, and hence since $\varepsilon$ is arbitrary, we have
$$
\begin{equation*}
\inf\{\|x_R-x|\mid r(x_R)=R \}\leqslant \int_{r(x)}^R\frac{dr}{f(r)}.
\end{equation*}
\notag
$$
Making $a_0\to a$, we get the required estimate also in the case $r(x)\geqslant a$. Theorem 1 is proved. Remark 1. The proof of Theorem 1 shows that, in the above process of construction of elements $x_n$, we can reach, after a finite number of steps, a point $x_N$ such that $R+\Delta_1\geqslant r(x_N)>R>r(x)\geqslant a$. Remark 2. The case $f>0$ on $(a,b]$ can be extended to the case $f\geqslant 0$ on $[a,b]$. Here, if the integral $\int_{r(x)}^{R}dr/f(r)$ is divergent, it is assumed to be equal to $+\infty$. Remark 3. Indeed, the proof of Theorem 1 shows that the conclusion of the theorem remains valid if one assumes the condition not for all points $x\in X$ such that $r(x):=\varrho_\theta(x,M)>a$, but only for the points from the ball $B(x,R+\varepsilon)$, where $\varepsilon>0$ is an arbitrary small fixed number. We next consider the case of asymmetric spaces with different conditions on the functions $\psi$ and $\varphi$. Let us first examine the case where $X$ is an asymmetric (not necessarily symmetrizable) space, $\varphi\colon X\to \mathbb{R}$, and in lieu of the continuity of the function $\psi$, we will assume that this function is left- (right-)uniformly upper semicontinuous, that is, for each $\varepsilon>0$, there exists $\delta>0$ such that $\psi(x)-\psi(x')\geqslant -\varepsilon/2$ for all $x,x'\in X$ such that $\|x-x'|<\delta$ ($\|x'-x|<\delta$). Assume first that the function $\psi$ is left-uniformly upper semicontinuous. Hence $\varrho_\theta(\,{\cdot}\,)$ is right-upper semicontinuous on $X$. Indeed, for each $\varepsilon>0$, there exists $z\in M$ such that $\varrho_\theta(x)-\varrho_\theta(x')\geqslant\psi(z-x)-\psi(z- x')-\varepsilon/2$. So, for an appropriate $\delta>0$, which is constructed from $\varepsilon/2$ and is such that, for all $x,x'\in X$ with $\|x'-x|<\delta$, we have $\psi(z-x)-\psi(z-x')-\varepsilon/2\geqslant -\varepsilon$. Remark 4. Here it is worth pointing out that the last estimate is derived from the already fixed point $z$, and so the condition that $\psi$ is left-uniform upper semicontinuity can be replaced by the left-upper semicontinuity at $z-x$. In particular, as $\psi$ one can consider any asymmetric norm $\|\,{\cdot}\,|$ (assuming that it is continuous). Note that continuity of $\|\,{\cdot}\,|$ is equivalent to saying that the ball $B(0,1)$ is closed (see [18]). We will also assume in addition that the function $f$ is (non-strictly) monotone increasing. In addition, for all $x\in X$ with $r(x)>a_0$ and sufficiently small $\Delta >0$, let $\delta(x)= \delta(x,\Delta )$ be the supremum of all $\delta\in (0,\min\{\delta_0,\Delta\})$ such that there exist a vector $\Delta x$, $\|\Delta x|=\delta$, and a number $\lambda\in (0,\mu)$ satisfying Let, for some point $y\in X$, there exist a vector $\Delta x$ and a number $\lambda\in (0,\mu)$ such that
$$
\begin{equation*}
\|\Delta x|\in\biggl(\frac{\delta(y)}2,\delta(y)\biggr] \quad\text{and}\quad r(y+\Delta x)-r(y)> (f(r(y))- \lambda)\|\Delta x |.
\end{equation*}
\notag
$$
We fix an arbitrary number $\lambda'\in (\lambda,\mu)$. For a sufficiently small $\eta>0$, an arbitrary point $x'\in O_\eta(y)$, and $\varepsilon=(\lambda'-\lambda)\|\Delta x|/3$, there exists $z\in M$ such that
$$
\begin{equation*}
r(y)-r(x')=\varrho_\theta(y)-\varrho_\theta(x')\geqslant\psi(z-y)-\psi(z-x')-\varepsilon> -2\varepsilon.
\end{equation*}
\notag
$$
For the vector $\Delta x':=y+\Delta x-x'$, we have
$$
\begin{equation*}
\begin{aligned} \, r(x'+\Delta x')-r(x') &=r(y+\Delta x)-r(x') \\ &=r(y+\Delta x)-r(y)+(r(y)-r(x'))\geqslant r(y+\Delta x)-r(y)-2\varepsilon. \end{aligned}
\end{equation*}
\notag
$$
Here, it can be assumed that $\eta$ is so small that $f(r(y))\geqslant f(r(x')) - \varepsilon$ (since $f(\,{\cdot}\,)$ is continuous and monotone increasing and since $r(\,{\cdot}\,)$ is right-upper semicontinuous). Hence
$$
\begin{equation*}
r(x'+\Delta x')-r(x')> (f(r(y))- \lambda)\|\Delta x |-3\varepsilon\geqslant (f(r(x'))- \lambda')\|\Delta x |.
\end{equation*}
\notag
$$
Theorem 2. Let $(X,\|\,{\cdot}\,|)$ be a complete symmetrizable space, $\varphi :X\to \mathbb{R}$, let a function $\psi$ be left-upper semicontinuous at each point from $X$, $f\uparrow$, and let at each point $x\in X$ such that $r(x):=\varrho_\theta(x,M)>a$. Then, for each $R>a$ and each $x\in X$ with $R>r(x)\geqslant a$, In addition, if the restriction of the function $r(\,{\cdot}\,)$ to any interval is continuous on this interval, then it can be assumed that $r(x_R)= R$. Proof. The proof is similar to that of Theorem 1. Let us use the inequality which was proved before this theorem. Here, $ \lambda'\in (0,\mu)$ is some number, $x'\in O_\eta(y)$ is an arbitrary point, and $\eta>0$ is a sufficiently small number.
We fix an arbitrary small number $\Delta>0$. Let, as before, $\delta(x)=\delta(x,\Delta)$. We argue by induction. Consider an arbitrary point $x\in X$ such that $r(x)\geqslant a_0>a$ and define $x_0=x$. $1^\circ$. Let us find a vector $\Delta x_1$ and a number $\lambda_1\in (0,\mu)$ such that $\|\Delta x_1|\in (\delta(x_0)/2,\delta(x_0)]$ and $r(x_0+\Delta x_1)-r(x_0)> (f(r(x_0))-\lambda_1)\|\Delta x_1|\geqslant \mu\|\Delta x_1|$. $2^\circ$. Assume that a point $x_n=x_{n-1}+\Delta x_n$ is constructed. Let us find a vector $\Delta x_{n+1}$ and a number $\lambda_{n+1}\in (0,\mu)$ such that $\|\Delta x_{n+1}|\in (\delta(x_n)/2,\delta(x_n)]$ and $r(x_n+\Delta x_{n+1})-r(x_n)> (f(r(x_n))-\lambda_{n+1})\|\Delta x_{n+1}|\geqslant \mu\| \Delta x_{n+1}|$. $3^\circ$. We claim that there exists $N\in \mathbb{N}$ such that $r(x_N)> R$. Indeed, if $r(x_n)\leqslant R$ for all $n\in \mathbb{N}$, then $\sum_{n=1}^{\infty}\|\Delta x_{n }|\leqslant \frac{1}{\mu}\sum_{n=1}^{\infty}(r(x_n)-r(x_{n-1}))\leqslant R/\mu$. Since $X$ is complete, the sequence $\{x_n\}$ converges to some point $y\in X$. There exist a vector $\Delta x$ and a number $\lambda'\in (0,\mu)$ such that $\|\Delta x|\in (\delta(y)/2,\delta(y)]$ and $r(x'+\Delta x')-r(x')> (f(r(x'))- \lambda')\|\Delta x | $. Proceeding as in the proof of Theorem 1, we can find a sufficiently small $\eta$ such that
$$
\begin{equation*}
(f(r(x'))-\lambda')\|\Delta x| >(f(r(x'))-\lambda'')\|\Delta x'|,
\end{equation*}
\notag
$$
for some $\lambda''\in (\lambda',\mu)$ and $\Delta x':=y+\Delta x-x'$. We will assume that $\eta<\|\Delta x|/2$.
Starting from some $n_0\in \mathbb{N}$ we have $(x'=)x_n\in O_\eta(y)$. In this case,
$$
\begin{equation*}
\|\Delta x'|\geqslant \|\Delta x |-\|x'-y|\geqslant \|\Delta x |-\eta>\frac{\|\Delta x|}2.
\end{equation*}
\notag
$$
For sufficiently small $\eta>0$ (see the last paragraph before the theorem) we have $x_n\in O_\eta(y)$ starting from some $n_0\in \mathbb{N}$, and hence for $x'=x_n$ we have $\delta(x_n)\geqslant \|\Delta x'|$. Therefore, $\|\Delta x_n|\geqslant \delta(x_n)/2\geqslant \|\Delta x'|/2>\|\Delta x|/4$, which is impossible since the series $\sum_{n=1}^{\infty}\|\Delta x_{n }|$ is convergent. Hence there exists $N\in \mathbb{N}$ such that $r(x_N)> R$.
$4^\circ$. So, we have
$$
\begin{equation*}
\frac{\Delta r_i}{f(r_i)-\mu}\geqslant \|\Delta x_i|,
\end{equation*}
\notag
$$
where $\Delta r_i=r(x_i)-r(x_{i-1})$, $i=1,\dots, N$. Hence
$$
\begin{equation*}
\sum_{i=1}^{N}\frac{\Delta r_i}{f(r_i)-\mu}\geqslant \sum_{i=1}^{N}\|\Delta x_i|\geqslant \|x_N-x_0|=\|x_N-x|.
\end{equation*}
\notag
$$
It can be assumed that $r(x_N)>R$ and $r(x_{N-1})\leqslant R$. If the restriction of the function $r(\,{\cdot}\,)$ to any interval is continuous on this interval, then there exists a point $x_R\in [x_{N-1},x_N]$ such that $r(x_R)=R$.
Since $f$ is monotone increasing, we have The quantity $\mu$ can be made arbitrarily small by reducing $\delta_0$, and hence since $\varepsilon$ is arbitrary, we have Making $a_0$ to $a$, we get the required estimate also in the case $r(x)\geqslant a$. If the restriction of the function $r(\,{\cdot}\,)$ to any interval is continuous on this interval, then the inequality $r(x_R)\geqslant R$ can be replaced by the equality $r(x_R)= R$. This proves the theorem.Remark 5. If the unit ball $B(0,1)$ is a closed set, then in view of Remark 4 one can take as $\psi$ a norm or a seminorm $\|\,{\cdot}\,|$. We also note that if $\psi$ is a norm $\|\,{\cdot}\,|$, then even in the non-symmetrizable setting, the restriction of the functions $\varrho_\theta(x,M)$ and $\psi(x)$ to any interval is continuous. Indeed, for $\|\,{\cdot}\,\|=\max\{\|\,{\cdot}\,|,\| - {\cdot|}\,\}$, we have the inequality $\varrho_\theta(x_1,M)\leqslant \|z-x_1|+\varphi(z)\leqslant \|z-x_2|+\varphi(z)+\|x_1-x_2\|$ for arbitrary $z\in M$, $x_1,x_2\in X$. Hence $\varrho_\theta(x_1,M)\leqslant \inf_{z\in M}(\|z-x_2|+\varphi(z))+\|x_1-x_2\|=\varrho_\theta(x_2,M)+\|x_1-x_2\|$. Similarly $\varrho_\theta(x_2,M)\leqslant \varrho_\theta(x_1,M)+\|x_1-x_2\|$. Hence $|\varrho_\theta(x_1,M)-\varrho_\theta(x_2,M)|\leqslant \|x_1-x_2\|$. It remains to note that on any interval the asymmetric norm $\|\,{\cdot}\,|$ is equivalent to the symmetrization norm $\|\,{\cdot}\,\|$. Hence in Theorem 2, for this $\psi$ it can be assumed that $r(x_R)=R$. Remark 6. As in Theorem 1, in the process of construction of elements $x_n$, we reach, after a finite number of steps, a point $x_N$ such that $R+\varepsilon>r(x_N)>R>r(x)\geqslant a$, where $\varepsilon>0$ is an arbitrarily small number. In the same way, the case $f>0$ on $(a,b]$ can be extended to the case $f\geqslant 0$ on $[a,b]$. Here, if the integral $\int_{r(x)}^{R}dr/f(r)$ is divergent, it is assumed to be equal to $+\infty$. Remark 7. If as $\varphi$ one takes a function defined on $X\times X$ and which is right-uniformly upper semicontinuous on $M\times X$ with respect to the second variable (that is, for each $\varepsilon>0$, there exists $\delta>0$ such that for all $z\in M$ and $x,x'\in X$ with $\|x'-x|<\delta$), and if the remaining conditions of Theorem 2 are met, then the conclusion of this theorem remains valid. In the actual fact, the condition on $\varphi$ can be relaxed even further on assuming that, for each $\varepsilon>0$, there exists $\delta>0$ such that for all $z\in \theta\text{-}P_M(O_\delta(x))$ $(\subset \theta\text{-}P_M^\delta(x))$ and $x\in X$, $x'\in O_\delta(x)$. Remark 8. The proof of Theorem 2 shows that the conclusion of this theorem remains valid if one assumes the condition not for all points $x\in X$ with $r(x):=\varrho_\theta(x,M)>a$, but only for the points from the ball $B(x,R+\varepsilon)$, where $\varepsilon>0$ is an arbitrary small fixed number. Definition 2. An asymmetric space $(X,\|\,{\cdot}\,|)$ is a left-Hausdorff space (a right- Hausdorff space) if, for all distinct points $a,b\in X$, there exists $\varepsilon>0$ such that $B^-(a,\varepsilon)\cap B^-(b,\varepsilon)=\varnothing$ $(B(a,\varepsilon)\cap B(b,\varepsilon)\,{=}\,\varnothing)$. Any left- (right-)Hausdorff space is right- (left-)Hausdorff). This claim is verified by the transformation $x\mapsto -x$. So, in the setting of asymmetric spaces, we will simply speak of Hausdorff spaces. Definition 3. A sequence $\{x_n\}$ in an asymmetric space $(X,\|\,{\cdot}\,|)$ is called a Cauchy sequence (an inverse Cauchy sequence) if, for each $\varepsilon>0$, there exists $N\in \mathbb{N}$ such that $\|x_m-x_n|<\varepsilon$ $(\|x_n-x_m|<\varepsilon)$ for all $m\geqslant n\geqslant N$. A space $(X,\|\,{\cdot}\,|)$ is called right-complete (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 $n\to \infty$. A right-complete space will be simply referred to as a complete space. A space $(X,\|\,{\cdot}\,|)$ is called inversely right-complete (inversely left-complete) if, for any inverse 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 $n\to\infty$. Theorem 3. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric Hausdorff right-complete space, let $\varphi\colon X\to \mathbb{R}$, let $\psi$ be a right-uniformly upper semicontinuous function on $X$, $f\uparrow$, and let for all points $x\in X$ with $r(x):=\varrho_\theta^-(x,M)>a$. Then for each $R>a$ and each $x\in X $ with $R>r(x)\geqslant a$. In addition, if the restriction of the function $r(\,{\cdot}\,)$ to any interval is continuous on this interval, then we may assume that $r(x_R)= R$. Proof. The proof is similar to that of Theorem 1. Assume that, for any point $y\in X$, there exist a vector $\Delta x$ and a number $\lambda\in (0,\mu)$ such that $\|\Delta x|\in (\delta(y)/2,\delta(y)]$ and $r(y+\Delta x)-r(y)> (f(r(y))- \lambda)\|\Delta x | $. We fix an arbitrary $\lambda'\in (\lambda,\mu)$. For a sufficiently small $\eta>0$, an arbitrary point $x'\in O_\eta^-(y):=\{z\mid \|y-z|<\eta\}$, and any $\varepsilon=(\lambda'-\lambda)\|\Delta x|/3$, there exists $z\in M$ such that
$$
\begin{equation*}
r(y)-r(x')=\varrho_\theta(y)-\varrho_\theta(x')\geqslant\psi(z-y)-\psi(z-x')-\varepsilon> -2\varepsilon.
\end{equation*}
\notag
$$
Hence, for the vector $\Delta x':=y+\Delta x-x'$, we have Here, it can be assumed that $\eta$ is so small that $f(r(y))\geqslant f(r(x')) - \varepsilon$ (since $f(\,{\cdot}\,)$ is continuous and monotone increasing and since $r(\,{\cdot}\,)$ is left-upper semicontinuous). Hence
As in the proof of Theorem 1, for vectors $\Delta x':=y+\Delta x-x'$ and some number $\lambda''\in (\lambda',\mu)$, for sufficiently small $\eta$, we have
$$
\begin{equation*}
(f(r(x'))-\lambda')\|\Delta x| >(f(r(x'))-\lambda'')\|\Delta x'|.
\end{equation*}
\notag
$$
Here, as $x'$ we can take the points $x_n$ starting from some number (the points $x_n$ are constructed as in the proof of Theorem 1).
Since $X$ is Hausdorff, there exists $\sigma\in (0,\eta)$ such that $B^-(y,\sigma)\cap B^-(y+\Delta x,\sigma)= \varnothing$. Hence, if $x'\in O^-_\sigma(y)$, then $\lim_{n\to \infty}\|\Delta x_n|\geqslant \|\Delta x'|/2 \geqslant \sigma/2$, which is impossible. The rest of the proof repeats that of Theorem 2. Theorem 3 is proved. Remark 9. An application of Theorem 3 to the space $(X,\| - {\cdot}\,|)$ produces the following assertion. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric inversely left-complete Hausdorff space, $\varphi\colon X\to \mathbb{R}$, let $\psi$ be a left-uniformly upper semicontinuous function on $X$, $f\uparrow$, and let for all $x\in X$ with $r(x):=\varrho_\theta(x,M)>a$. Then, for each $R>a$ and each $x\in X$ with $R>r(x)\geqslant a$, In addition, if the restriction of the function $r(\,{\cdot}\,)$ to any interval is continuous on this interval, then we can assume that $r(x_R)= R$. Remark 11. In Theorem 3, as $\psi$ one may take the asymmetric norm $\|\,{\cdot}\,|$ itself (without the additional assumption that the unit ball is closed; cf. Remark 5). Remark 12. The assumption that the space should be Hausdorff in Theorem 3 can be relaxed by assuming instead that, for an arbitrary point $y\in X$, there exists a vector $\Delta x$ such that and there exists $\sigma>0$ such that $B^-(y,\sigma)\cap B^-(y+\Delta x,\sigma)=\varnothing$. § 3. The rate of variation of the $\theta$-metric function as a function of approximative properties of setsNow let $\psi(x):=\|x|$, $\varphi$ be a bounded function on $M$, and $|\varphi|\leqslant \tau<+\infty$. Given any point $a\neq 0$, we set For an arbitrary vector $\ell$ with $\|-\ell|=1$ and a point $a\neq 0$, we setGiven an arbitrary $\varepsilon>0$, we define
$$
\begin{equation*}
R=R(\varepsilon,x):=\inf\{\psi(y-x)\mid y\in \theta\text{-}P_M^\varepsilon(x)\}.
\end{equation*}
\notag
$$
We will assume that $x\in X$ is such that $R(\varepsilon,x)>\tau$. In this case, we set
$$
\begin{equation*}
\upsilon(x,\varepsilon):=\inf_{a\in M\cap B(x,R+2\tau)}\alpha(\ell,a-x).
\end{equation*}
\notag
$$
The regularity of $M$ at a point $x\in X$ along a vector $\ell$ is defined by
$$
\begin{equation*}
\omega_x(\ell):=\lim_{\varepsilon\to 0+}\upsilon(x,\varepsilon).
\end{equation*}
\notag
$$
Theorem 4. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric space, $M\subset X$, $\psi(\,{\cdot}\,):=\|\,{\cdot}\,|$, let $|\varphi|< \tau<+\infty$ on $M$, $x\in X$, be such that $\inf\{\psi(y-x)\mid y\in M\}>\tau$, and let $\ell\in X$, $\|-\ell|=1$. Then Proof. Consider an arbitrary $\varepsilon\in (0,\tau/2)$. Let $u\in (0,\delta)$ be arbitrary, where $\delta>0$ is such that $\|\pm\delta \ell|<\varepsilon$ (that is, $\delta'=\max\{\|\delta\ell|,\|-\delta\ell|\}<\varepsilon$). Note that $\theta\text{-}P_M(x)\subset \theta\text{-}P_M^{\varepsilon}(x)\subset M\cap B_\theta(x,R+\tau+\varepsilon) $. Indeed, the condition $b\in P_M^{\varepsilon}(x)$ implies that
$\|b-x|+\varphi(b)\leqslant \inf_{c\in M}(\|c-x|+\varphi(c))+\varepsilon$
${}\leqslant \inf_{c\in P_M^{\varepsilon}(x)}(\|c-x|+\varphi(c))+\varepsilon$
${}\leqslant R+\tau+\varepsilon$, that is, $\theta\text{-}P_M^{\varepsilon}(x)\subset B_\theta(x,R+\tau+\varepsilon)$. Since
$\varrho_\theta(x-u\ell,M)$
${}=\inf_{z\in M}(\|z-(x-u\ell)|+\varphi(z))$
${}\leqslant \inf_{z\in M}(\|z-x|+\varphi(z)+\|u\ell|)$
${}\leqslant \varrho_\theta(x,M)+\delta'<\varrho_\theta(x,M)+\varepsilon$, we have $\theta\text{-}P_M(x-u\ell) \subset\theta\text{-}P_M^{\delta'}(x)\subset\theta\text{-}P_M^{\varepsilon}(x)\subset M\cap B_\theta(x,R+\tau+\varepsilon)$ for all $u\in (0,\delta)$. Let $\Delta x=-u\cdot \ell$. For each point $a\in M\cap B_\theta(x,R+2\tau)$ and any $y^*\in A^*(a-x)$, we have
$$
\begin{equation*}
\|a-(x+\Delta x)|\geqslant y^*(a-x-\Delta x)=y^*(a-x)+uy^*(\ell).
\end{equation*}
\notag
$$
Choosing $y^*$ so that $y^*(\ell)$ would be arbitrarily close to $\alpha(\ell,a-x)$, we get the estimate $\|a-(x+\Delta x)|\geqslant \|a-x|+u\upsilon(x,\varepsilon)$. Hence
$(\psi(a-(x+\Delta x))+\varphi(a))-\varrho_\theta(x,M)$
${}\geqslant(\psi(a-(x+\Delta x))+\varphi(a))-(\psi( a-x)+\varphi(a))$
${}= \|a-(x+\Delta x)|-\|a-x|\geqslant u\upsilon(x,\varepsilon)$. We can choose a point $a$ so that $\|a-(x+\Delta x)|+\varphi(a)$ would be arbitrarily close to $\varrho_\theta(x+\Delta x,M)$ (in this case, of course, we have $\|a-x|+\varphi(a)\geqslant \varrho_\theta(x,M)$). Hence $\varrho_\theta(x+\Delta x,M)-\varrho_\theta(x,M)\geqslant u\upsilon(x,\varepsilon)=\|\Delta x|\upsilon(x,\varepsilon)$. Therefore,
$$
\begin{equation*}
\varlimsup_{\|\Delta x|\to 0}\frac{r(x+\Delta x)-r(x)}{\|\Delta x|}\geqslant \omega_x(\ell),
\end{equation*}
\notag
$$
which proves Theorem 4. Definition 4. Let $X=(X,\|\,{\cdot}\,|)$ be an asymmetric space. Given $A,B\subset X$, we set A mapping $\Phi\colon O(x_0)\to 2^X\setminus\{\varnothing\}$ (where $O(x_0)\subset X$ is some neighbourhood of $x_0$) is said to be stable at $x_0\in X$ if the condition $\{x_n\}\subset O(x_0)$, $\|x_n-x_0|\to 0$, $n\to\infty$, implies that Remark 13. In particular, a mapping $\Phi$ is stable at a point $x_0$ if it is upper semicontinuous at $x_0$ (both the image and pre-image are equipped with the topology generated by open balls of $X)$. This conditions holds a fortiori for points of $\theta$-approximative compactness $x_0$, that is, for points at which $\bigcap_{n}\theta\text{-}P_M^{\delta_n}(x_0)=\theta\text{-}P_M(x_0)$ for any positive null sequence $\{\delta_n\}$, and $\theta\text{-}P_M(x_0)$ is a compact set. For arbitrary $x\in X\setminus M$ and $\varepsilon>0$, we set
$$
\begin{equation*}
\vartheta(x,\varepsilon):=\inf_{a\in \theta\text{-}P_M^\varepsilon(x)}\alpha(\ell,a-x)
\end{equation*}
\notag
$$
and define
$$
\begin{equation*}
\varpi_x(\ell):=\lim_{\varepsilon\to 0+}\vartheta(x,\varepsilon).
\end{equation*}
\notag
$$
Theorem 5. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric space, $M\subset X$, $\psi(\,{\cdot}\,):=\|\,{\cdot}\,|$, a function $\varphi\colon M\to \mathbb{R}$ be uniformly upper semicontinuous on $M$, the $\theta$-metric projection be stable at a point $x\in X$, and let $\ell\in X$, $\|-\ell|=1$. Then Proof. Consider an arbitrary $\varepsilon>0$. There exists a number $\eta\in(0,\varepsilon/2)$ such that $\varphi(a)-\varphi(b)<\varepsilon/2$ for all $a,b\in M$ such that $\|a-b|<\eta$. We set $\Phi:=\theta\text{-}P_M$ and define $\Delta x=-u\cdot \ell$. There exists a number $\delta_0>0$ such that, for all $u\in (0,\delta_0)$, There exist $a\in \Phi(x+\Delta x)$, $b\in \Phi(x)$ such that $\|a-b|<\varepsilon/2$. Since $\psi(a-x)-\psi(b-x)=\|a-x|-\|b-x|\leqslant \|a-b|$, we have Hence $a\in \theta\text{-}P_M^\varepsilon(x)$. Let $y^*\in S^*$ be such that $y^*(\ell)$ is arbitrarily close to $\alpha(\ell,a-x)$. Since $\|a-(x+\Delta x)|\geqslant y^*(a-x-\Delta x)=y^*(a- x)+uy^*(\ell)$, we have
$\|a-(x+\Delta x)|\geqslant \|a-x|+u\vartheta(x,\varepsilon)$ and
$\varrho_\theta(x+ \Delta x,M) -\varrho_\theta(x,M)$
${}\geqslant(\psi(a-(x+\Delta x))+\varphi(a))-(\psi( a-x)+\varphi(a))$
${}=\|a-(x+\Delta x)|-\|a-x|\geqslant u\vartheta(x,\varepsilon)$. Therefore, This proves the theorem. Definition 5. Let $X=(X,\|\,{\cdot}\,|)$ be an asymmetric space, $O(x_0)\subset X$ be a neighbourhood of a point $x_0$. A mapping $\Phi\colon O(x_0)\to 2^X\setminus\{\varnothing\}$ is called lower semicontinuous at a point $x_0$ if $\nu(\Phi(x_n),\{y_0\})\to 0$ for each point $y_0\in \Phi(x_0)$ as $\|x_n-x_0|\to 0$, $n\to\infty$. Remark 14. In Theorem 5, the condition that the function $\varphi$ is uniformly upper semicontinuous on $M$ can be replaced by the assumption that $\varphi$ is upper semicontinuous on $M$ (in the usual sense) together with the assumption that the $\theta$-metric projection is lower semicontinuous at $x\in X$. In this case, the conclusion of this theorem remains valid. In the same way, the upper semicontinuity of $\varphi$ on $M$ can be replaced by the upper semicontinuity on the set $\theta\text{-}P_M^\varepsilon(x)$ for some $\varepsilon>0$. Remark 15. Let the function $\varphi$ in Theorem 5 be defined on $X\times X$ and let uniformly with respect to $a\in M$. Then We set
$$
\begin{equation*}
\begin{gathered} \, \varsigma(x,\varepsilon,\ell) :=\inf\{\alpha(\ell,a-x)\mid a\in\theta\text{-}P_M( O_\varepsilon(x)) \cap O_\varepsilon(\theta\text{-}P_M(x))\}, \\ \zeta_x(\ell) :=\lim_{\varepsilon\to 0+}\varsigma(x,\varepsilon,\ell). \end{gathered}
\end{equation*}
\notag
$$
Theorem 6. Let $(X,\|\,{\cdot}\,|)$ be an asymmetric space, let $M\subset X$, ${\psi(\,{\cdot}\,):=\|\,{\cdot}\,|}$, let $\varphi\colon X\times X\to \mathbb{R}$ be upper semicontinuous on $M\times \{x\}$ with respect to the first variable (that is, for all $b\in M$ and $\varepsilon>0$, there exists $\eta>0$ such that $|\varphi(a,x)-\varphi(b,x)|<\varepsilon$ for all $a\in M$ such that $\|a-b|<\eta$), and let the limit be uniform with respect to $a\in M$. Next, let the $\theta$-metric projection be lower semicontinuous at $x\in X$, and let $\ell\in X$, $\|-\ell|=1$. Then In addition, if $ \theta\text{-}P_M(x)$ is a singleton $\{b\}$, $\ell:=(b-x)/\|x-b|$, and $\|\,{\cdot}\,|$ is a uniformly smooth (symmetric) norm, then Proof. Consider an arbitrary $\varepsilon > 0$. There exists $\eta \in (0,\varepsilon/2)$ such that $\varphi(a,x)-\varphi(b,x)<\varepsilon/2$ for all $a,b\in M$ such that $\|a-b|<\eta$. We set $\Phi:=\theta\text{-}P_M$ and define $\Delta x=-u\cdot \ell$. There exists $\delta_0>0$ such that, for all $u\in (0,\delta_0)$ and $b\in \Phi(x)$,
$$
\begin{equation*}
\|\Delta x|,\ \nu(\Phi(x+\Delta x),\{b\})<\frac{\varepsilon}2.
\end{equation*}
\notag
$$
For each point $b\in \Phi(x)$, there exists a point $a\in \Phi(x+\Delta x)$ such that $\|a-b|<\varepsilon/2$. Hence $a\in \theta\text{-}P_M(O_\varepsilon(x)) \cap O_\varepsilon(\theta\text{-}P_M(x))$ and, choosing $y^*\in S^*$ so that $y^*(\ell)$ would be arbitrarily close to $ \alpha(\ell,a-x)$, and using the inequality
$\|a-(x+\Delta x)|\geqslant y^*(a-x-\Delta x)=y^*(a-x)+uy^*(\ell)$, we get the estimate
$\|a-(x+\Delta x)| \geqslant \|a-x|+u\,{\cdot}\,\varsigma(x,\varepsilon,\ell)$.
There exists $u_0>0$ such that
$\varphi(a,x+\Delta x)-\varphi(a,x)\geqslant (-s-\varepsilon)\|\Delta x|$ for all
$u\in (0,u_0)$, where $\Delta x=-u\ell$. Hence
$\varrho_\theta(x+\Delta x,M) -\varrho_\theta(x,M)\geqslant(\psi(a-(x+\Delta x))+\varphi(a,x+\Delta x))-(\psi( a-x)+\varphi(a,x))$
${}=\|a-(x+\Delta x)|-\|a-x|+(-s-\varepsilon)\|\Delta x|$
${}\geqslant u\cdot\varsigma(x,\varepsilon,\ell)+(s-\varepsilon)u$. Therefore,
$$
\begin{equation*}
\varlimsup_{\|\Delta x|\to 0}\frac{r(x+\Delta x)-r(x)}{\|\Delta x|}\geqslant \zeta_x(\ell)-s.
\end{equation*}
\notag
$$
The second assertion is trivial. Theorem 6 is proved. Remark 16. The condition that $\varphi$ is upper semicontinuous on $M\times \{x\}$ can be replaced by the assumption that $\theta\text{-}P_M^\varepsilon(x)\times \{x\}$ is upper semicontinuous (for some $\varepsilon>0)$ (or on an arbitrary $\theta$-bounded subset $N\subset M)$; similarly, the assumption that the limit Theorem 7. Let $(X,\|\,{\cdot}\,\|)$ be a normed linear space, let $M\subset X$, ${\psi(\,{\cdot}\,):=\|\,{\cdot}\,\|}$, $\varphi\colon M\to \mathbb{R}$, $\theta=(\|\,{\cdot}\,\|,\varphi)$, and let the $\theta$-metric projection be lower semicontinuous at a point $x_0\in X$ with $\varrho_\theta(x_0,M)>0$. Then Proof. We may assume without loss of generality that $\varrho_\theta(x_0,M)=1$. For each $\varepsilon>0$, there exists $\delta>0$ such that, for all $x$ with $\|x-x_0\|=\delta$ and an arbitrary point $y_0\in \theta\text{-}P_M(x_0)$, there exists a point $y\in \theta\text{-}P_M(x)$ such that $\|y-y_0\|<\varepsilon/2$. We fix a point $y_0$ and consider $x$ such that $x_0 \in (y_0,x)$, $\|x- x_0\|=\delta$. Let $z\in (y,x)$ be a point such that the intervals $[y,y_0]$ and $[z,x_0]$ are parallel. Then $\|z-x_0\|=(\delta/(1+\delta))\|y-y_0\|< \delta\varepsilon/(2(1+\delta))$. We have
$$
\begin{equation*}
\begin{aligned} \, \|y-x\|+\varphi(y) &=\|y-z\|+\varphi(y)+\|z-x\| \\ &\geqslant \|y-x_0\|-\|z-x_0\|+\varphi(y)+\|x-x_0\|-\|z-x_0\| \\ &=\|y-x_0\|+\varphi(y)+\delta-\frac{\delta\varepsilon}{1+\delta}\geqslant \|y_0-x_0\|+\varphi(y_0)+\delta-{\delta\varepsilon}. \end{aligned}
\end{equation*}
\notag
$$
Therefore, Hence, for $\Delta x:=x-x_0$, we have proving the theorem. Definition 6. Let $\varnothing \ne M \subset X$. We say that $x \in X\setminus M$ is a $\theta$-solar point for $M$ if there exists a point $y\in \theta\text{-}P_Mx\ne \varnothing$ (called a $\theta$-luminosity point) such that (geometrically, this means that there is a “solar” ray emanating from $y$ and passing through $x$ such that $y$ is a nearest point from $M$ for each point from this ray). We say that $x\in X\setminus M$ is a strict $\theta$-solar point for $M$ if $\theta\text{-}P_Mx\ne\varnothing$ and condition (3.1) is met for each point $y\in \theta\text{-}P_Mx$. A set $M\subset X$ is said to be a $\theta$-sun (respectively, a strict $\theta$-sun) if each point $x\in X\setminus M$ is a $\theta$-solar point (respectively, a strict $\theta$-solar point for $M$). Theorem 8. Let $(X,\|\,{\cdot}\,\|)$ be a normed linear space, $M\subset X$ be a boundedly compact set, ${\psi(\,{\cdot}\,):=\|\,{\cdot}\,\|}+x^*(\,{\cdot}\,)$, $x^*\in X^*$, $\varphi\colon M\to \mathbb{R}$, $\theta=(\psi,\varphi)$, and let the $\theta$-metric projection be single-valued and continuous on $ X$. Then $M$ is a $\theta$-sun. Proof. Consider an arbitrary point $x_0$ such that $\varrho_\theta(x_0,M)>0$. By continuity of $\theta\text{-}P_M$ on $X$, there exists $\varepsilon>0$ such that $M_0 :=\theta\text{-}P_M(B(x_0,\varepsilon))$ is a precompact set (this follows from local boundedness of the continuous function $\theta\text{-}P_M$ and bounded compactness of $M$). Consider the mapping $\phi\colon B(x_0,\varepsilon)\to B(x_0,\varepsilon)$ defined by
$$
\begin{equation*}
\phi(x):=x_0+\varepsilon\frac{x_0 -\theta\text{-}P_Mx}{\|x_0 -\theta\text{-}P_Mx\|}.
\end{equation*}
\notag
$$
This mapping carries continuously the ball $B(x_0,\varepsilon)$ into its precompact subset. Therefore, there exists a fixed point $z\in S(x_0,\varepsilon) $ such that $z=\phi(z)$. Hence we have $x_0\in (z,\theta\text{-}P_Mz)$. We claim that $\theta\text{-}P_Mx_0=\theta\text{-}P_Mz=:w$. Assume that there exists a point $\widehat{w}\in M$, $\widehat{w}\neq w$, which is a nearest point in $M$ for $x_0$. Since
$$
\begin{equation*}
\begin{aligned} \, &\|w-z\|+x^*(w-z)+\varphi(w) <\|\widehat{w}-z\|+x^*(\widehat{w}-z)+\varphi(\widehat{w}) \\ &\qquad\leqslant \|x_0-z\|+\|\widehat{w}-x_0\|+x^*(x_0-z)+x^*(\widehat{w}-x_0) +\varphi(\widehat{w}), \end{aligned}
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
\begin{aligned} \, &\|w-x_0\|+x^*(w-x_0)+\varphi(w) \\ &\qquad=\|w-z\|- \|x_0-z\|+x^*(w-z)-x^*(x_0-z)+\varphi(w) \\ &\qquad<\|\widehat{w}-x_0\|+x^*(\widehat{w}-x_0)+\varphi(\widehat{w}). \end{aligned}
\end{equation*}
\notag
$$
But this contradicts the assumption that $\widehat{w}$ is a nearest point for $x_0$. Hence $\theta\text{-}P_Mx_0=\theta\text{-}P_Mz$. A similar argument shows that $\theta\text{-}P_Mz$ is a nearest point for each point from $ [z,\theta\text{-}P_Mz]$. Repeating this argument for the point $z$ (in place of $x_0$), we conclude that $w=\theta\text{-}P_Mx_0$ is a nearest point for each point from the ray $\ell:=\{w+t(x_0-w)\mid t\geqslant 0\}$. Since $x_0$ is arbitrary, the set $M$ is a $\theta$-sun. Theorem 7 is proved. The following example illustrates the potency of the above results for establishing non-unique solvability in Dirichlet problems. For similar approaches to this problem, see Tsar’kov [19], [20]. Example 1. Let $H$ be a real Hilbert space with norm $|\,{\cdot}\,|$ and $M\subset H$ be a non- empty closed subset. Let $\varphi\colon M\to \mathbb{R}$ be a uniformly continuous bounded function such that $|\varphi|\leqslant \tau$. Consider the minimization problem of the functional $G(a,x):=|a-x|^2+\varphi(a-x)$, $a\in M$, on the set $M\times H$. We will assume that, for all $x\in H$, this problem has a unique solution and the $\theta$-metric projection is continuous. Therefore, for the functional $G_0:=\sqrt{G}$, a solution for $x\in H$ also exists and is unique. The functional $G_0$ can be represented as $|a-x|+g(a,x)$, where
$$
\begin{equation*}
g(a,x):= |a-x|\Biggl(\sqrt{1+\frac{\varphi(a-x)}{|a-x|^2}}-1\Biggr) =\frac{\varphi(a-x)}{|a-x|+\sqrt{|a-x|^2+\varphi(a-x)}}.
\end{equation*}
\notag
$$
We claim that this condition implies that $\operatorname{\overline{conv}}M$ is contained in a uniform neighbourhood of the set $M$ of order $\sqrt{\tau}$. In addition, we will assume that that each $x \in H$ is a point of $\theta$-approximative compactness relative to $\theta:=(|\,{\cdot}\,|,g)$. In this case, the $\theta$-metric projection is single-valued and continuous on $H$. It is also easily seen that it suffices to assume that $\varphi$ is locally uniformly continuous on $M$ rather than uniformly continuous on $H$. For an arbitrary point $x\in H\setminus M$, let $a$ be a solution of the minimization problem for $x$. Following Theorem 6, we take as $\ell$ the vector $(a-x)/|a-x|$. Consider the space $H:=\mathring{W}^1_2(\Omega)$, where $ \Omega\subset \mathbb{R}^n $ is a bounded domain with Lipschitz boundary, and the norm on $H$ is defined by $|u|:=\bigl(\int_\Omega |\nabla u(x)|^2\, dx\bigr)^{1/2}$. As an example, as $\varphi$ we consider the function
$$
\begin{equation*}
\varphi(u):=\int_{\Omega}\alpha\arctan u^2(t)\, dt,\qquad \alpha>0.
\end{equation*}
\notag
$$
For the derivative in the direction $\ell$, we have
$$
\begin{equation*}
\frac{\partial}{\partial\ell}(\varphi(a-x))= -\int_\Omega\frac{2\alpha (a(t)-x(t))\ell(t)}{1+(a(t)-x(t))^4}\, dt\leqslant 0,
\end{equation*}
\notag
$$
and the modulus of this derivative is majorized by $\alpha \varkappa$, where the constant $\varkappa>0$ depends only on $\Omega$. We will assume that $\tau>(4\alpha\varkappa)^2$. We have
$$
\begin{equation*}
\begin{aligned} \, &\frac{\partial g}{\partial\ell}(a,x) =\frac{\partial (\varphi(a-x))/\partial\ell}{|a-x|+\sqrt{|a-x|^2+\varphi(a-x)}} \\ &\ \ + \frac{\varphi(a-x)}{(|a-x|+\sqrt{|a-x|^2+\varphi(a-x)})^2} \biggl\{\frac{(a-x,\ell)}{|a-x|}+\frac{(a-x,\ell)- \partial(\varphi(a-x))/\partial\ell}{\sqrt{|a-x|^2+\varphi(a-x)}}\biggr\} \\ &\qquad\leqslant \frac{|\varphi(a-x)|}{(|a-x|+\sqrt{|a-x|^2+\varphi(a-x)})^2} \biggl\{\frac{(a-x,\ell)}{|a-x|}+\frac{(a-x,\ell)+ \alpha\varkappa}{\sqrt{|a-x|^2+\varphi(a-x)}}\biggr\} \\ &\qquad\leqslant\frac{2\tau}{\sqrt{|a-x|^2+(\varphi(a-x))}\,(|a-x|+\sqrt{|a-x|^2+\varphi(a-x)})} \leqslant \frac{2\tau}{|a-x|^2-\tau} \end{aligned}
\end{equation*}
\notag
$$
if $|a-x|>\sqrt{\tau}$. In addition, $|g|\leqslant \tau/|a-x|$ ($\leqslant\sqrt{\tau}$). So, taking as $\theta$ the functional $|a-x|+g(a,x)$ (here $\psi=|\,{\cdot}\,|$), it follows from Theorem 6 that, for $r(x)>2\sqrt{\tau}$,
$$
\begin{equation*}
\varlimsup_{|\Delta x|\to 0}\frac{r(x+\Delta x)-r(x)}{|\Delta x|}\geqslant 1- \frac{2\tau}{|a-x|^2-\tau}\geqslant 1-\frac{2\tau}{(r(x)-\sqrt{\tau}\,)^2-\tau}.
\end{equation*}
\notag
$$
By Theorem 1, for all $R$ and $x\in H$ with $R>r(x)=\varrho_\theta(x,M)>(\sqrt{3}+1)\sqrt{\tau}$,
$$
\begin{equation*}
\inf\{|x_R-x|\mid r(x_R)=R \}\leqslant \int_{r(x)}^R\frac{((r-\sqrt{\tau}\,)^2-\tau )\, dr}{(r-\sqrt{\tau}\,)^2-3\tau}.
\end{equation*}
\notag
$$
For an arbitrary point $x\in H$ such that $r(x)=a=2(\sqrt{3}+1)\sqrt{\tau}$, we have
$$
\begin{equation*}
\int_{a}^{R}\frac{(r-\sqrt{\tau}\,)^2-\tau}{(r-\sqrt{\tau}\,)^2-3\tau}\, dr =R-a+\frac{2\tau}{2\sqrt{3\tau}} \ln\frac{r-\sqrt{\tau}-\sqrt{3\tau}}{r-\sqrt{\tau}+\sqrt{3\tau}}\bigg|_{a}^{R}.
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
\biggl|\frac{1}{2}\ln\frac{1-x}{1+x}+x\biggr|<\frac{1}{6},
\end{equation*}
\notag
$$
for $0<x\leqslant 1/2$, we have where as $b$ we can take the number
$$
\begin{equation*}
\begin{aligned} \, &2\tau\biggl(-\frac{1}{R-\sqrt{\tau}}+\frac{1}{a-\sqrt{\tau}}\biggr) +\frac{2\sqrt{\tau}}{3\sqrt{3}}=2\tau\frac{R-a}{(R-\sqrt{\tau}\,)(a-\sqrt{\tau}\,)}+ \frac{2\sqrt{\tau}}{3\sqrt{3}} \\ &\qquad\leqslant \frac{2\tau}{a-\sqrt{\tau}} +\frac{2\sqrt{\tau}}{3\sqrt{3}} =\sqrt{\tau}\biggl(\frac{2}{2\sqrt{3}+1}+\frac{2}{3\sqrt{3}}\biggr)< \frac{2}{\sqrt{3}}\sqrt{\tau}=:b_0. \end{aligned}
\end{equation*}
\notag
$$
Since $R_0:=\varrho(x_R,M)\geqslant R-\sqrt{\tau}$, there exists a point $x_R$ such that $r(x_R)=R\leqslant R_0+\sqrt{\tau}$ and $|x_r-x|<R-a+b_0\leqslant R_0-a+b_0+\sqrt{\tau}\leqslant R_0-a/5$, $r(x)=a$. As a result, the ball $B(x_R,R_0)$ is disjoint from the set $M$ and contains the ball $B(x,a/5)$. Making $R\to \infty$ (hence $R_0\to\infty$), proceeding as in [15], and using the condition that the space $H$ is uniformly smooth, we find that any point $x$ with $r(x)=a$ can be separated from $M$ by some hyperplane. Therefore, $\operatorname{\overline{conv}}M$ is disjoint from the set of points $x$ at which $r(x)\geqslant a$. Hence $\operatorname{\overline{conv}}M$ is contained in the $(a+\sqrt{\tau}\,)$-neighbourhood of $M$. Corollary 1. If, under the conditions of the above example, the set $M$ is such that there exist points $\alpha,\beta\in M$ and a point $s\in (\alpha,\beta)$ such that $M\cap B_\theta(s,a+\sqrt{\tau}\,)=\varnothing$, then $\theta$-$P_Mv$ is contains at least two points for some point $v\in H$. In particular, this is so if $M$ is a non-convex cone (that is, $tM=M$ for all $t\geqslant 0$). A similar assertion can also be proved under the condition $|\varphi(a)|\leqslant C+C_0|a|^\gamma$, where $C,C_0>0$ are some constants, and $\gamma\in [0,1)$. So, on the space $H:=\mathring{W}^1_2(\Omega)$, where $ \Omega\subset \mathbb{R}^n $ is a bounded domain with Lipschitz boundary, the norm is defined by $|u|:=\bigl(\int_\Omega |\nabla u(x)|^2\, dx\bigr)^{1/2}$. As $\varphi(u)$ we take the functional
$$
\begin{equation*}
\varphi(u):=\int_{\Omega}\alpha\arctan u^2(t)\, dt,\qquad \alpha>0,
\end{equation*}
\notag
$$
and as the set $M$ we consider the set
$$
\begin{equation*}
\biggl\{u\in \mathring{W}^1_2(\Omega)\biggm| -\frac{1}{2}\int_\Omega |u(x)|^2\, dx\leqslant -C\biggr\}, \qquad C>0,
\end{equation*}
\notag
$$
where $1<p<(n+2)/(n-2)$ for $n\geqslant 3$ and $p>1$ for $n<3$. Here, we choose $C>0$ to be sufficiently large to satisfy the conditions of Corollary 1. A standard argument easily shows that the set $M$ is a $\theta$-approximatively compact in the space $H$. So, if the $\theta$-projection is single-valued, then it is continuous. Solving the minimization problem of the functional $G(u,v):=|u-v|^2+\varphi(u-v)$ on the non-convex set $M$ for all $v\in H$, and using Lemma 2 in [20], we find that, for some function $v_0\in \mathring{W}^1_2(\Omega)$, there exist two different critical points (functions) $u_1,u_2\in \mathring{W}^1_2(\Omega)$ for the functional
$$
\begin{equation*}
G(u,v_0)-\frac{\lambda}{2} \int_\Omega | u(t)|^2\, dt\quad \text{for some }\lambda>0.
\end{equation*}
\notag
$$
Substituting $w=u-v_0$, we see that there exist two different functions $w_1,w_2\in \mathring{W}^1_2(\Omega)$ which are critical points for the functional
$$
\begin{equation*}
|w|^2+\varphi(w)-\frac{\lambda}{2} \int_\Omega | w(t)+v_0(t)|^2\, dt.
\end{equation*}
\notag
$$
Hence the functions $w_1,w_2\in \mathring{W}^1_2(\Omega)$ are generalized solutions of the problem where $f=\lambda v_0\in W^{1}_0(\Omega)$, for some $\lambda>0$.
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\Bibitem{Tsa24} |
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