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Geometric progressions in distance spaces; applications to fixed points and coincidence points E. S. Zhukovskiy Derzhavin Tambov State University, Tambov, Russia
Bibliographic databases:
     § 1. IntroductionMany results in nonlinear analysis depend on a construction, in a complete metric space $(X,\rho_X)$, of a special (in particular, iterative) sequence $\{x_i\}$ such that, for some $\gamma < 1$,
$$
\begin{equation}
\rho_X(x_{i+1},x_i)\leqslant \gamma \rho_X(x_i,x_{i-1}), \quad i=1,2,\dots\,.
\end{equation}
\tag{1.1}
$$
Since the metric satisfies the triangle inequality, such a sequence is always Cauchy and therefore converges. The limit point of this sequence is shown (under certain conditions) to be the required solution of the problem under consideration (namely, a fixed point, a coincidence point of mappings, a minimum point or a zero of a functional, and so on). This idea underlies Banach’s fixed point theorem (see [1]), Nadler’s theorem (a set-valued analogue of Banach’s theorem; see [2]), Arutyunov’s coincidence point theorem for single- and set-valued mappings (see [3], Theorems 1 and 2) and their numerous refinements and extensions (see [4]–[9]). Using this approach, Fomenko (see [10] and [11]) constructed iteration algorithms for approximations of the required sets, in particular, sets of simultaneous fixed points or coincidence points of mappings, and proved several versions of the principle of the existence of zeros for $(\alpha,\beta)$-search functionals (see [12] and [13]), from which the classical theorems [1]–[3], as well as some known and new conditions for the existence of coincidence and fixed points can be derived. If the distance $\rho_X$ in the space $X$ is not a metric and, in place of the ‘ordinary’ triangle inequality, satisfies a generalized $f$-triangle inequality (that is, $\rho_X (x,v)\leqslant f\bigl(\rho_X (x,u), \rho_X (u,v)\bigr)$ for all $ x,u,v\in X$, where the fixed function $f\colon {\mathbb{R}_+}^{2} \to \mathbb{R}_+$ is continuous at $(0,0)$ and $f(0,0) = 0$), then property (1.1) does not imply in general that the sequence $\{x_i\}$ is convergent in $(X,\rho_X)$. In such spaces, which are called $f$-quasimetric spaces, the problem of the convergence of iterations requires different approaches, which involve relations for the distances between all elements of the sequence $\{x_i\}$, rather than only between neighbouring elements. Such approaches were developed in [14], where an analogue of the Krasnosel’skii-Browder fixed point theorem was established for generalized contractions in a $f$-quasimetric space. A similar idea was used in [15] to obtain, for an $f$-quasimetric space, the principle of the existence of zeros for $\lambda$-generalized search functionals, and to derive the fixed-point theorem for generalized contractions save for uniqueness results (see [14]) and an analogue of Nadler’s theorem from this principle. Note that if a function $f$ is linear (that is, $f(r_1,r_2)=q_1 r_1 + q_2 r_2$, $r_1,r_2\in \mathbb{R}_+$), then any sequence with property (1.1) is a Cauchy sequence. This was established by Arutyunov and Greshnov [16]–[19] and Fomenko [20] in the proofs of theorems on fixed points, coincidence points of (single- or set-valued) mappings and theorems on zeros of $(\alpha, \beta)$-search functionals in the corresponding spaces (which are known as ($q_1, q_2)$-quasimetric spaces). The above suggests the following natural questions. First, does any space with generalized distance (not necessarily satisfying even the $f$-triangle inequality for some function $f$) in which any ‘geometric progression’ (a sequence of the form (1.1)) is convergent admit analogues of the above results on fixed and coincidence points, and zeros of search functionals? Second, what conditions on the distance $\rho_X$ (similar to the generalized triangle inequality) imply the convergence in $X$ of sequences of the form (1.1) in the case when $\gamma<1$? These two questions are considered in our work. The paper is organized as follows. In § 2 we consider spaces with general distances and look at important particular cases of such spaces. In § 3 we examine geometric progressions and study their properties required for results on fixed and coincidence points. In § 3, given a space $(X,\rho_X)$, we define the set $\mathcal{B}_X $ of numbers $\gamma \geqslant 0$ such that any geometric progression $\{ x_i\}_{i=0}^{\infty}\subset X$ with common ratio $\gamma$ is a Cauchy sequence. We also define the function $\overline{\mathcal{P}}_{X,\gamma}$ associating with $r_0 \geqslant 0$ the number $\overline{\mathcal{P}}_{X,\gamma}(r_0):=\sup\bigl\{\varlimsup_{i\to \infty}\rho(x_0, x_i)\bigr\}$, where the supremum is taken over all geometric progressions with ratio $\gamma$ such that $r_0=\rho_X(x_0,x_1)$. In § 4 an answer to the first question is given. Namely, we show that a coincidence point of an $\alpha$-covering and a $\beta$-Lipschitz set-valued mappings exists for $\gamma:=\beta^{-1}\alpha \in \mathcal{B}_X$. We also obtain estimates for a coincidence point in terms of the function $\overline{\mathcal{P}}_{X,\gamma}$. From these results a fixed point theorem for set-valued mappings is derived. The second question is addressed in the concluding two sections, where we find conditions on the distance that ensure the convergence of geometric progressions. In § 5 we investigate the properties of the set $\mathcal{B}_{X}$ for $f$-quasimetric spaces $X$. On this basis, in § 6 we find the set $\mathcal{B}_X$ for $f$-quasimetric spaces with concrete functions $f$ (in particular, linear and power functions) and estimate the values of the function $\overline{\mathcal{P}}_{X,\gamma}$ (where $\gamma \in \mathcal{B}_{X}$). In addition, in that section we prove that, for a sufficiently wide class of functions $f$, the corresponding $f$-quasimetric spaces satisfy the ‘zero-one law’, that is, in such spaces the set $\mathcal{B}_X$ is either the minimal possible (contains only $0$) or the maximal possible (coincides with $[0,1)$). § 2. Some simple properties of spaces with distanceLet $X\neq \varnothing$. A mapping $\rho_X\colon X^2\to \mathbb{R}_+ $ is said to be a distance in $X$ if it satisfies the identity axiom
$$
\begin{equation}
\forall\, x,u \in X \quad \rho_X (x,u)=0 \quad\Longleftrightarrow\quad x=u.
\end{equation}
\tag{2.1}
$$
The pair $X:=(X,\rho_X)$ will be called a distance space (or just a space). A distance $\rho_X$ is a metric (correspondingly, $X$ is a metric space) if $\rho_X$ is symmetric, and satisfies the triangle inequality
$$
\begin{equation}
\forall\, x,u,v\in X \quad \rho_X(x,v)\leqslant \rho_X(x,u)+\rho_X(u,v).
\end{equation}
\tag{2.3}
$$
We recall some known generalizations of a metric. A distance satisfying the triangle inequality (2.3) (but not symmetric in general) is a quasimetric (the corresponding space is a quasimetric space). The studies of quasimetric spaces go back to Alexandroff, Wilson and Nemytskii (see [21] and [22]). Let $f\colon {\mathbb{R}_+}^{2} \to \mathbb{R}_+$ be such that We say that the $f$-triangle inequality holds if
$$
\begin{equation}
\exists\, \sigma >0 \quad \forall\, x,u,v \in X \left. \begin{array}{l} \rho_X (x,u)<\sigma, \\ \rho_X (u,v)<\sigma \end{array}\right\}\quad \Longrightarrow \quad \rho_X (x,v)\leqslant f\bigl(\rho_X (x,u), \rho_X (u,v)\bigr).
\end{equation}
\tag{2.5}
$$
In this case the distance $\rho_X$ is called an $f$-quasimetric and $X$ is an $f$-quasimetric space (see [23] and [14]). Such spaces were introduced by Fréchet (see [24], p. 18). A necessary and sufficient condition ensuring that a function $f$ satisfying (2.5) and (2.4) exists is that the asymptotic triangle inequality holds (see Theorem 2 in [25]):
$$
\begin{equation}
\forall\, \{x_i\}, \{u_i\}, \{v_i\}\subset X \left. \begin{array}{l} \rho_X(x_i,u_i)\to 0, \\ \rho_X(u_i,v_i)\to 0 \end{array}\right\}\quad \Longrightarrow\quad \rho_X(x_i,v_i)\to 0.
\end{equation}
\tag{2.6}
$$
If (2.5) holds for $\sigma=\infty$ and a linear function $f$, that is, if
$$
\begin{equation}
\forall\, x,u,v \in X \quad \rho_X (x,v)\leqslant q_1 \rho_X (x,u)+q_2\rho_X (u,v)
\end{equation}
\tag{2.7}
$$
for $q_1,q_2 \geqslant 1$, then $X$ is said to be a $(q_1,q_2)$-quasimetric space (the distance $\rho_X $ is a $(q_1,q_2)$-quasimetric). The properties of $(q_1,q_2)$-quasimetric spaces were investigated in [16]–[18]. It was shown in [26] that for an arbitrary $(q_1,q_2)$-quasimetric space there exists a function $f$ for which the $f$-triangle inequality (2.5) is stronger than (2.7). If, in addition, the distance $\rho_X $ is symmetric, then we will drop the prefix ‘quasi’ in constructions like ‘quasimetric’, ‘quasimetric space’, ‘$f$-quasimetric’, ‘$f$-quasimetric space’, ‘$(q_1,q_2)$-quasimetric’, and ‘$(q_1,q_2)$-quasimetric space’. Again, consider spaces $(X,\rho_X)$ without additional constraints on the distance $\rho_X$. We equip such spaces with the ‘natural’ topology, which was proposed and investigated by Nedev [27]. For such spaces we introduce the requisite definitions similar to those given in [14], [23] and [28] for concrete classes of $f$-quasimetric spaces. Let $x_0\in X$ and $r\geqslant 0$. Consider the balls
$$
\begin{equation*}
O_X (x_0, r)=\{x\in X\colon \rho_X(x_0, x)<r \}\quad\text{and} \quad B_X (x_0, r)=\{x\in X\colon \rho_X(x_0, x)\leqslant r \}
\end{equation*}
\notag
$$
(for $r=0$, we have $O_X (x_0, 0)=\varnothing$ and $B_X (x_0, 0)=\{x_0\}$). A set $V\subset X$ is said to be open if, for each $v\in V$, there exists $\delta>0$ such that $O_X (v, \delta) \subset V$. The set of all such open sets defines a topology on $X$. Note that a ball $O_X (x_0, r)$ is not necessarily a member of this topology, that is, $O_X (x_0, r)$ can fail to be an open set even for a $(q_1,q_2)$-metric space, where $q_1 > 1$ and $q_2 > 1$ (see Example 3.4 in [18]). A set is said to be closed if its complement is open. Clearly, each singleton is a closed set (without any additional assumptions on the distance $\rho_X$), that is, this topology satisfies the $T_1$ separation axiom. However, this topology can fail to be Hausdorff (even for quasimetric spaces). We say that a sequence $\{ x_i\}_{i=1}^{\infty}\subset X$ converges to a point $x\in X $ (and write $\lim_{i\to \infty}x_i=x$ or $x_i \to x$) if $\lim_{i\to \infty}\rho_X(x,x_i)=0$, that is, if any of the two equivalent relations holds:
$$
\begin{equation*}
\forall\, \varepsilon\,{>}\, 0 \quad \exists\, I \ \ \forall\, i \,{>}\,I \quad \rho_X(x,x_i)\,{<}\,\varepsilon \quad \Longleftrightarrow \quad \forall\, \varepsilon\,{>}\, 0 \quad \exists\, I \ \ \forall\, i \,{>}\,I \quad x_i \in O_X (x, \varepsilon).
\end{equation*}
\notag
$$
Of course, the limit point of a sequence is not unique in general (this is so even in $(q_1,q_2)$-quasimetric spaces, where the limit point is not unique in general and has some other ‘unusual’ properties; for the corresponding examples, see [14], [18], [29]). If, in a space $X$, the convergence $\rho_X(x,x_i) \to 0$ implies that $\rho_X(x_i,x) \to 0$, then the distance $\rho_X$ is said to be weakly symmetric ($X$ is a weakly symmetric space). As in the case of ‘usual’ metric spaces, the closeness of sets can be expressed in terms of the convergence of sequences. Namely, the following result holds. Proposition 2.1. A set $U\subset X$ is closed if and only if any limit point of any convergent sequence in $U$ lies in $U$. Proof. Let $U$ be a closed set and let the sequence $\{ x_i\}_{i=1}^{\infty}\subset U$ converge to $x\in X$. If $x\in X\setminus U$, then, since the complement $X\setminus U$ is open, there exists $\delta>0$ such that $O_X (x, \delta) \subset X\setminus U$. But this contradicts the fact that $x_i \in O_X (x, \delta)$ for all $i$ starting from some index, which follows from the convergence $x_i \to x$.
Assume that any limit point of any converging sequence from $U$ lies in $U$. If $U$ were not closed, then the complement $ X\setminus U$ would not be open, that is, there would exist $x\in X\setminus U$ such that, for any $i$, there exists $x_i\in O_X (x, i^{-1})$ such that $x_i \notin X \setminus U$. The sequence $\{ x_i\}_{i=1}^{\infty}\subset U$ thus defined converges to $x\in X\setminus U$, which contradicts the assumption. Proposition 2.1 is proved. We also weaken the definition of a closed set as follows. A set $U\subset X$ is said to be quasiclosed if, for an arbitrary convergent sequence of points in $U$, at least one of its limit points lies in $U$. We say that $\{ x_i\}_{i=1}^{\infty}$ is a Cauchy sequence if
$$
\begin{equation*}
\forall\, \varepsilon >0 \quad \exists\, I \quad \forall\, j >i>I \quad \rho_X(x_i,x_j)<\varepsilon \quad(\text{that is, } x_j \in O_X(x_i,\varepsilon)).
\end{equation*}
\notag
$$
This definition of a Cauchy sequence was used in [18] and [23], while in [14] the inequality $\rho_X(x_i,x_j)<\varepsilon$ was required to hold for all $j$ and $i$ exceeding $I$, rather than for $j >i$ only. Note that a convergent sequence in a space $(X,\rho_X)$ can fail to be a Cauchy sequence (for conditions necessary for the Cauchy property of convergent sequences in $f$-quasimetric spaces to hold, see [14]). A space $(X,\rho_X)$ is said to be complete if each Cauchy sequence in it is convergent. A set $U\subset X$ is complete if any Cauchy sequence in $U$ is convergent and all of its limit points lie in $U$. A set $U\subset X$ is said to be quasicomplete if any Cauchy sequence in $U$ is convergent and at least one of its limit points lies in $U$. § 3. Geometric progression in spaces with distanceA sequence $\{ x_i\}_{i=0}^{\infty}$ in a space $(X,\rho_X)$ is called a geometric progression with ratio $\gamma \geqslant 0$ if (1.1) holds. We let $\mathrm{GP}_X [\gamma]$ denote the set of all geometric progressions of $X$ with common ratio $\gamma $. The subset of $\mathrm{GP}_X [\gamma]$ consisting of all convergent sequences is denoted by $\mathrm{GP}^{\mathrm{con}}_X [\gamma]$. Both sets are nonempty, since, for any $x_0 \in X$, the constant sequence $x_i=x_0$, $i\in \mathbb{N}$, is a convergent geometric progression with ratio $\gamma$. It is easily seen that, for all $\gamma',\gamma \in \mathbb{R}_+$,
$$
\begin{equation*}
\gamma' < \gamma \quad\Longrightarrow\quad \mathrm{GP}_X [\gamma'] \subset \mathrm{GP}_X [\gamma], \quad \mathrm{GP}^{\mathrm{con}}_X [\gamma'] \subset \mathrm{GP}^{\mathrm{con}}_X [\gamma].
\end{equation*}
\notag
$$
Next, let $\mathcal{B}_X$ be the set of all $\gamma \geqslant 0$ such that any $\{ x_i\}_{i=0}^{\infty}\in \mathrm{GP}_X [\gamma]$ is a Cauchy sequence. Since $0 \in \mathcal{B}_X$, the set $\mathcal{B}_X$ is nonempty. Note that if a space $X$ is complete, then $\mathrm{GP}^{\mathrm{con}}_X [\gamma]=\mathrm{GP}_X [\gamma]$ for any $\gamma \in \mathcal{B}_X$. The next property of the set $\mathcal{B}_X$ follows easily from the definition:
$$
\begin{equation}
\forall\, \gamma',\gamma \in \mathbb{R}_+ \quad \gamma' < \gamma, \quad \gamma \in\mathcal{B}_X \quad\Longrightarrow\quad \gamma' \in\mathcal{B}_X.
\end{equation}
\tag{3.1}
$$
We set $\Lambda_X :=\sup \mathcal{B}_X$. By (3.1), for any $\gamma \in [0, \Lambda_X )$ each geometric progression in $X$ with ratio $\gamma$ is a Cauchy sequence. Next, if $\gamma > \Lambda_X$, then there is a non-Cauchy sequence $\{ x_i\}_{i=0}^{\infty}\in \mathrm{GP}_X [\gamma]$. Clearly, in any space $X$, an example of a non-Cauchy sequence is provided by the sequence $\{ x_i\}_{i=0}^{\infty}$ for which the distance from each element to the next is constant, that is, $\rho_X(x_i,x_{i+1})=\rho_X(x_{i-1},x_{i})>0$ for any $i$. But this equality means that $\{ x_i\}_{i=0}^{\infty}$ is a geometric progression with ratio $\gamma=1$. Hence, if a space $X$ contains such a sequence, then $\Lambda_X \leqslant 1$. Proposition 5.2 in § 5 shows that, for any function $f$ there exists an $f$-quasimetric space $X$ which contains such a geometric progression, that is, $\Lambda_X \leqslant 1$ for the class of $f$-quasimetric spaces. From the results of [16]–[18] it readily follows that in $(q_1,q_2)$-quasimetric spaces, just as in ‘usual’ metric spaces, any geometric progression with ratio $\gamma\in [0,1)$ is a Cauchy sequence, that is, $\Lambda_X=1$ for the class of $(q_1,q_2)$-quasimetric spaces. In § 6 we obtain a condition on a function $f$ under which the ‘zero-one law’, which means that $\Lambda_X $ is either $0$ or $1$, holds for the class of $f$-quasimetric spaces. In general distance spaces (in particular, in some concrete $f$-quasimetric spaces) this law fails to hold, and in this case $\Lambda_X $ can take any intermediate value. Example 3.1. On the set $X:=\{ x_i\}_{i\in \mathbb{Z}}$ define a symmetric distance by (2.2) and the equalities We claim that if a distance is defined by (3.2), then it satisfies the asymptotic triangle inequality (2.6), that is, the space under consideration is an $f$-metric space. Now we consider various sequences consisting of elements $x_i$, $u_i$, $v_i$ of $X$ such that $\rho_X(x_i,u_i)\to 0$ and $ \rho_X(u_i,v_i)\to 0$. 1. Let $u_i=x_{i+1}$ and $v_i=x_{i+2}$. Then $\rho_X(x_i,v_i)=(i+1)^{-1}+(i+2)^{-1} \to 0$ as ${i\to \infty}$. 2. Let $u_i=x_{i+j_i}$ and $v_i=x_{i+j_i+1}$, where $j_i\geqslant 2$. If $\rho_X(x_i,u_i)=(i+1)^{-1}+\dots +(i+j_i)^{-1} \to 0$ as $i\to \infty$, then, clearly, $ (i+1)^{-1}+\dots +(i+j_i+1)^{-1} \to 0$, which implies that $ \rho_X(x_i,v_i)\to 0$. 3. Let $u_i=x_{i+j_i}$ and $v_i=x_{i+j_i+m_i}$, where $j_i,m_i\geqslant 2$. In this case, since $ \rho_X(x_i,u_i)=(i+1)^{-1}+\dots +(i+j_i)^{-1} \to 0$ and $ \rho_X(u_i,v_i)=(i+j_i+1)^{-1}+\dots +(i+j_i+m_i)^{-1} \to 0$, we have $ \rho_X(x_i,v_i)=(i+1)^{-1}+\dots +(i+j_i+m_i)^{-1}=\rho_X(x_i,u_i)+ \rho_X(u_i,v_i) \to 0$. That the sequence $\{x_{-i} \}_{i=0}^{\infty}$ satisfies (2.6) is clear. We introduce some other characteristics of spaces $(X,\rho_X)$ which are relevant for the study of geometric progressions in such spaces. We set
$$
\begin{equation}
{\mathcal{D}}_{X}:=\bigl\{\rho_X(x_0,x_1)\colon x_0,x_1\in X \bigr\}.
\end{equation}
\tag{3.3}
$$
Let $\gamma \in \mathcal{B}_X$. For any $r_0 \in {\mathcal{D}}_{X}$ there exists a geometric progression $\{ x_i\}_{i=0}^{\infty}$ with ratio $\gamma$ such that $\rho_X(x_0,x_1)=r_0$ (to define such a progression it suffices to consider any pair $(x_0,x_1)$ of points in $ X$ ‘lying’ at the required distance, and define the progression $x_0,x_1,x_1,\dots$ all of whose elements, starting from the second, are equal). Consider the function $\overline{\mathcal{P}}_{X,\gamma}\colon {\mathcal{D}}_{X} \to \overline{\mathbb{R}}_+ $, $\overline{\mathbb{R}}_+:=\mathbb{R}_+\cup \{+\infty \}$ which associates with any $r_0 \in {\mathcal{D}}_{X}$ the value
$$
\begin{equation*}
\overline{\mathcal{P}}_{X,\gamma} (r_0) :=\sup \Bigl\{\varlimsup_{i\to \infty}\rho(x_0, x_i) \colon \{x_i\}_{i=0}^{\infty}\in \mathrm{GP}_X [\gamma],\, \rho_X(x_0, x_1)=r_0 \Bigr\}.
\end{equation*}
\notag
$$
Next, consider an arbitrary convergent sequence $\{ x_i\}_{i=0}^{\infty}\in \mathrm{GP}^{\mathrm{con}}_X [\gamma]$ such that $\rho_X(x_0,x_1)\in {\mathcal{D}}_{X}$ (it is clear that such a sequence exists). Let $\operatorname{Lim}x_i$ be the set of all limit points of this sequence. The distance from $x_0$ to this set is defined by
$$
\begin{equation*}
\rho_X(x_0,\operatorname{Lim}x_i):=\sup_{x\in\operatorname{Lim}x_i} \rho_X(x_0, x).
\end{equation*}
\notag
$$
We also consider the function $\mathcal{P}_{X,\gamma}\colon \mathcal{D}_{X} \to \overline{\mathbb{R}}_+ $, which associates with each $r_0 \in \mathcal{D}_{X}$ the value
$$
\begin{equation*}
\mathcal{P}_{X,\gamma} (r_0) :=\sup \bigl\{\rho_X(x_0,\operatorname{Lim}x_i)\colon \{x_i\}_{i=0}^{\infty}\in \mathrm{GP}^{\mathrm{con}}_X [\gamma],\, \rho_X(x_0, x_1)=r_0 \bigr\} .
\end{equation*}
\notag
$$
The above characteristics of the space $X$ will be used in the theorem on coincidence points of set-valued mappings. § 4. Coincidence points of covering and Lipschitz set-valued mappingsLet $\rho_X$ and $\rho_Y$ be distances in spaces $X$ and $Y$, respectively. We equip the product $X \times Y$ with the distance $\rho_{X \times Y}$ defined by
$$
\begin{equation*}
\rho_{X \times Y}((x,y), (u,z))=\rho_{X}(x,u) + \rho_{Y}(y,z) \qquad \forall\, (x,y), (u,z)\in X \times Y.
\end{equation*}
\notag
$$
Given arbitrary nonempty sets $U,V \subset Y$, we define the distance from $U$ to $V$ in the ‘standard’ way:
$$
\begin{equation*}
\operatorname{dist}_Y(U,V):=\inf \bigl\{ \rho_Y(u,v)\colon u\in U,\, v\in V \bigr\}.
\end{equation*}
\notag
$$
Let $\Psi,\Phi\colon X\rightrightarrows Y$ be set-valued mappings with nonempty values $\Psi(x),\Phi(x)\subset Y$ for all $x\in X$. Consider the problem of the existence of a coincidence point of these mappings — by definition, this a point $\xi \in X$ such that Let $\mathrm{Coin}(\Psi,\Phi):=\{\xi \in X\colon \Psi(\xi)\cap\Phi(\xi) \neq \varnothing \}$ be the set of coincidence points of $\Psi$ and $\Phi$. We are interested in conditions under which this set is nonempty.In the particular case of $Y=X$ and the identity mapping $\Psi\colon X\to X$ (that is, $\Psi(x)=x$ for any $x\in X$; this mapping can be looked upon as set-valued, with singleton values) a coincidence point is a classical fixed point, which is a solution of the inclusion We define covering and Lipschitz mappings similarly to the case of metric spaces (see [3]). For $(q_1,q_2)$-quasimetric spaces the analogous definitions were given in [18]. The graph of a mapping $\Psi$ is denoted by $\operatorname{gph}(\Psi):=\bigl\{(x,y)\in X\times Y\colon y\in \Psi(x) \bigr\}$. Let $\alpha>0$. By definition, a mapping $\Psi\colon X\rightrightarrows Y$ is $\alpha$-covering if
$$
\begin{equation*}
\forall\, x\in X \quad\forall\, r \geqslant 0 \quad \bigcup_{y\in \Psi(x)} B_Y (y,\alpha r) \subset \Psi (B_X (x, r)).
\end{equation*}
\notag
$$
Clearly, $\Psi$ is an $\alpha$-covering mapping if and only if
$$
\begin{equation*}
\forall\, x \in X \quad\forall\, y \in \Psi(x) \quad\forall\, y' \in Y \quad\exists\, x' \in X \quad y' \in \Psi(x'), \quad\rho_X(x,x')\,{\leqslant}\, \frac{1}{\alpha}\rho_Y(y,y').
\end{equation*}
\notag
$$
The ‘standard’ definition of a Lipschitz set-valued mapping involves the Hausdorff distance. In the case of $(q_1,q_2)$-quasimetric spaces this definition was given in [18]. We can extend the definition of Hausdorff distance to the distance spaces under consideration here and then give a definition of Lipschitz set-valued mappings. However, we give an equivalent definition, which does not involve the Hausdorff distance. It will be used in the proof of our results on coincidence points. Let $\beta \geqslant 0$. A mapping $\Phi\colon X\rightrightarrows Y$ will be called $\beta$-Lipschitz if
$$
\begin{equation*}
\begin{aligned} \, & \forall\, x\in X \quad\forall\, x'\in X \quad\forall\, y\in \Psi(x) \quad\forall\, \varepsilon>0 \\ & \exists\, y'\in Y \quad y'\in \Phi(x'), \quad \rho_Y(y,y')\leqslant (\beta+\varepsilon)\rho_X(x,x'). \end{aligned}
\end{equation*}
\notag
$$
Lemma 4.1. Let $\alpha>0$ and $\beta\geqslant 0$, let $\Psi\colon X\rightrightarrows Y$ be an $\alpha$-covering mapping, and let $\Phi\colon X\rightrightarrows Y$ be a $\beta$-Lipschitz mapping. Then for an arbitrary $x_0 \in X$ and any $\varepsilon > 0$, there exist geometric progressions $\{x_i\}_{i=0}^{\infty}\in \mathrm{GP}_{X}[\gamma]$ and $\{y_i\}_{i=0}^{\infty}\in \mathrm{GP}_Y[\gamma]$ with ratio $\gamma=\varepsilon+{\alpha}^{-1}{\beta}$ satisfying Proof. Let $x_0 \in X$ and $\varepsilon > 0$, and let $\varepsilon_0 >0$ be such that
$$
\begin{equation*}
\varepsilon_0 \leqslant \varepsilon, \qquad \varepsilon_1:=\frac{\varepsilon_0}{\alpha}(\varepsilon_0 + \beta + \operatorname{dist}_Y (\Psi(x_0),\Phi(x_0))) \leqslant \varepsilon\quad\text{and} \quad \frac{\varepsilon_0}{\alpha} \leqslant \varepsilon.
\end{equation*}
\notag
$$
Next, let $y\in \Psi(x_0)$, $y_0\in \Phi(x_0)$ be such that $\rho_Y(y,y_0)< \varepsilon_0+ \operatorname{dist}_Y (\Psi(x_0),\Phi(x_0))$. Since $\Psi$ is an $\alpha$-covering mapping, there exists $x_1\in X$ satisfying
$$
\begin{equation*}
y_0\in \Psi(x_1)\quad\text{and} \quad \rho_X(x_0,x_1)\leqslant \frac{1}{\alpha} \rho_Y(y,y_0) \leqslant\frac{\varepsilon_0}{\alpha}+\frac{1}{\alpha} \operatorname{dist}_Y (\Psi(x_0),\Phi(x_0)).
\end{equation*}
\notag
$$
This proves inequality (4.1). Next, since $\Phi$ is a $\beta$-Lipschitz mapping, there exists $y_1\in Y$ such that
$$
\begin{equation*}
\begin{gathered} \, y_1 \in \Phi(x_1), \\ \begin{split} \rho_Y(y_0,y_1) &\leqslant (\varepsilon_0 + \beta)\rho_X(x_0,x_1) \leqslant (\varepsilon_0 + \beta) \biggl(\frac{\varepsilon_0}{\alpha}+\frac{1}{\alpha} \operatorname{dist}_Y (\Psi(x_0),\Phi(x_0))\biggr) \\ &=\varepsilon_1+\frac{\beta}{\alpha}\operatorname{dist}_Y (\Psi(x_0),\Phi(x_0)), \end{split} \end{gathered}
\end{equation*}
\notag
$$
which proves (4.2).
Let us show using induction that, for any $i=1,2,\dots$, there exist $x_i \in X$ and $y_i \in Y $ satisfying
$$
\begin{equation}
y_i\in \Psi(x_{i+1})\cap \Phi(x_i), \qquad y_{i+1} \in \Phi(x_{i+1}),
\end{equation}
\tag{4.4}
$$
$$
\begin{equation}
\rho_X(x_i,x_{i+1})\leqslant\biggl(\frac{\varepsilon_0}{\alpha} + \frac{\beta}{\alpha}\biggr) \rho_X(x_{i-1},x_i),
\end{equation}
\tag{4.5}
$$
$$
\begin{equation}
\rho_Y(y_i,y_{i+1})\leqslant\biggl(\frac{\varepsilon_0}{\alpha} + \frac{\beta}{\alpha}\biggr) \rho_Y(y_{i-1},y_i).
\end{equation}
\tag{4.6}
$$
Since $\Psi$ is an $\alpha$-covering mapping, there exists $x_2\in X$ such that
$$
\begin{equation*}
y_1\in \Psi(x_2) \quad\Longrightarrow\quad y_1\in \Psi(x_2)\cap \Phi(x_1)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\rho_X(x_1,x_2)\leqslant \frac{1}{\alpha} \rho_Y(y_0,y_1) \leqslant \frac{1}{\alpha} (\varepsilon_0 + \beta)\rho_X(x_0,x_1) .
\end{equation*}
\notag
$$
Since $\Phi$ is a $\beta$-Lipschitz mapping, for some $y_2\in Y$ we have $y_2 \in \Phi(x_2)$ and
$$
\begin{equation*}
\rho_Y(y_1,y_2)\leqslant (\varepsilon_0 + \beta)\rho_X(x_1,x_2) \leqslant (\varepsilon_0 + \beta)\frac{1}{\alpha} \rho_Y(y_0,y_1).
\end{equation*}
\notag
$$
This proves (4.4)–(4.6) for $i=1$.
Assume that for all $i\leqslant j$ we have found $x_i$ and $y_i$ satisfying (4.4)–(4.6). Since $\Psi$ is an $\alpha$-covering mapping, there exists $x_{j+1}\in X$ such that
$$
\begin{equation*}
y_j\in \Psi(x_{j+1}) \quad\Longrightarrow\quad y_j\in \Psi(x_{j+1})\cap \Phi(x_j)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\rho_X(x_j,x_{j+1})\leqslant \frac{1}{\alpha} \rho_Y(y_{j-1},y_j) \leqslant \frac{1}{\alpha} (\varepsilon_0 + \beta)\rho_X(x_{j-1},x_j).
\end{equation*}
\notag
$$
Since $\Phi$ is an $\beta$-Lipschitz mapping, for some $y_{j+1}\in Y$ we have $y_{j+1} \in \Phi(x_{j+1})$ and
$$
\begin{equation*}
\rho_Y(y_j,y_{j+1})\leqslant (\varepsilon_0 + \beta)\rho_X(x_j,x_{j+1}) \leqslant (\varepsilon_0 + \beta)\frac{1}{\alpha} \rho_Y(y_{j-1},y_j) .
\end{equation*}
\notag
$$
This proves (4.4)–(4.6) for $i=j+1$.
It remains to note that the resulting sequences $\{x_i\}_{i=0}^{\infty}$ and $\{y_i\}_{i=0}^{\infty}$ are geometric progressions with ratio $\gamma={\alpha}^{-1}{\beta}+\varepsilon$ which obey conditions (4.1)–(4.3). This proves Lemma 4.1. Theorem 4.1. Let $\alpha> 0$ and $\beta \geqslant 0$ be such that ${\alpha}^{-1}{\beta}<\Lambda_X$ and ${\alpha}^{-1}{\beta}<\Lambda_Y$. Next, let $\Psi\colon X\rightrightarrows Y$ be an $\alpha$-covering mapping and $\Phi\colon X\rightrightarrows Y$ be a $\beta$-Lipschitz mapping. Assume that either one of the graphs of these mappings, $\operatorname{gph}(\Psi)$ and $\operatorname{gph}(\Phi)$, is complete and the other is quasiclosed in $X \times Y$, or one of the graphs is quasicomplete and the other is closed in $X \times Y$. Then $\mathrm{Coin}(\Psi,\Phi)\neq \varnothing $ and, moreover, for an arbitrary $x_0 \in X$ and any $\varepsilon > 0$, the geometric progression $\{x_i\}_{i=0}^{\infty} \in \mathrm{GP}_{X}[\gamma]$ with ratio $\gamma={\alpha}^{-1}{\beta}+\varepsilon$ in Lemma 4.1 converges to a coincidence point $\xi\in X$ of $\Psi$ and $\Phi$, and and Proof. Let $x_0 \in X$ and $\varepsilon > 0$, and let $\varepsilon_0 \in (0, \varepsilon]$ satisfy
First we consider the case when the graph $\operatorname{gph}(\Psi)$ is complete and the graph $\operatorname{gph}(\Phi)$ is quasiclosed. Let $\{x_i\}_{i=0}^{\infty}\in \mathrm{GP}_{X}[\gamma]$ and $\{y_i\}_{i=0}^{\infty}\in \mathrm{GP}_Y[\gamma]$, where ${\gamma=\varepsilon_0+{\alpha}^{-1}{\beta}}$, be sequences from Lemma 4.1 satisfying (4.1)–(4.3). By (4.9) these are Cauchy sequences. Hence $\{(x_{i+1},y_i)\}_{i=0}^{\infty}\subset X \times Y$ is also a Cauchy sequence. By (4.3) this sequence lies in the complete set $\operatorname{gph}(\Psi)\subset X \times Y$, and so, for the set of its limit points we have
$$
\begin{equation}
\operatorname{Lim} (x_{i+1},y_i) \neq \varnothing\quad\text{and} \quad \operatorname{Lim} (x_{i+1},y_i)\subset \operatorname{gph}(\Psi).
\end{equation}
\tag{4.10}
$$
It is clear that
$$
\begin{equation}
\operatorname{Lim} (x_{i+1},y_i)=\operatorname{Lim} (x_{i},y_i)
\end{equation}
\tag{4.11}
$$
and the sequence $\{(x_{i},y_i)\}_{i=0}^{\infty}$ lies in the graph $\operatorname{gph}(\Phi)$. Since $\Phi$ is quasiclosed, there exists a pair $(\xi,\mathfrak{y})\in \operatorname{Lim} (x_{i},y_i)$ such that $(\xi,\mathfrak{y})\in \operatorname{gph}(\Phi)$. By (4.10), $(\xi,\mathfrak{y})\in \operatorname{gph}(\Psi)$. So $\xi$ is a coincidence point of $\Psi$ and $\Phi$. Note that $\rho_X(x_0,\xi)\leqslant \mathcal{P}_{X,\gamma}(\rho_X(x_0,x_1))$, which implies (4.7) by Lemma 4.1. We also have $\lim_{i \to \infty}\rho_X(x_0,x_i) \leqslant\overline{\mathcal{P}}_{X,\gamma} ( \rho_X(x_0,x_1) )$, from which (4.8) follows by Lemma 4.1.
Now let $\operatorname{gph}(\Psi)$ be quasicomplete and $\operatorname{gph}(\Phi)$ be closed. In this case the Cauchy sequence $\{(x_{i+1},y_i)\}_{i=0}^{\infty}$ lies in the quasicomplete graph $\operatorname{gph}(\Psi)$. Therefore, it converges, and, among its limit points, there exists a point $(\xi,\mathfrak{y})\in \operatorname{gph}(\Psi)$. The sequence $\{(x_{i},y_i)\}_{i=0}^{\infty}$ lies in the closed set $\operatorname{gph}(\Phi)$, and for the set of its limit points we have
$$
\begin{equation*}
\operatorname{Lim} (x_{i},y_i) \subset \operatorname{gph}(\Phi)
\end{equation*}
\notag
$$
and (4.11) holds. Hence $(\xi,\mathfrak{y})\in \operatorname{gph}(\Phi)$. So $\xi$ is a coincidence point of $\Psi$ and $\Phi$ which satisfies inequalities (4.7) and (4.8).
In the cases when $\operatorname{gph}(\Psi)$ is closed (quasiclosed) and the graph $\operatorname{gph}(\Phi)$ is quasicomplete (complete) the argument goes along the same lines, with $\Psi$ and $\Phi$ interchanged. This proves Theorem 4.1. Remark 4.1. In Theorem 4.1 we proved that, for all $\varepsilon > 0$ and $x_0 \in X$, there exists a coincidence point $\xi$ of $ \Psi$ and $\Phi$ satisfying (4.7) and (4.8). For this point $\xi$ we can obtain a similar estimate for the distance of $y_0$ to a point $\mathfrak{y}\in \Psi(\xi) \cap \Phi(\xi)$, which is a limit point of the sequence $\{y_i\}_{i=0}^{\infty}\in \mathrm{GP}_Y[\gamma]$, $\gamma=\varepsilon+{\alpha}^{-1}{\beta}$. Hence Remark 4.2. Theorem 4.1 relaxes the assumptions of Theorem 2 in [3] and Theorem 5.7 in [18], which require that $\Phi$ should be closed and at least one of the graphs of $\Psi$ and $\Phi$ should be complete. In Theorem 4.1 this condition is replaced by quasiclosedness and quasicompleteness. This relaxation is essential. For example, if at least one sequence in $X$ has a nonunique limit point, then the graph of the identity mapping $I_X\colon X\to X$, $x\mapsto I_X(x)=x$, is neither complete, nor closed. At the same time, the graph of the identity mapping is quasiclosed and quasicomplete. This, in particular, allows us to employ Theorem 4.1 in the analysis of fixed points of set-valued mappings. It is worth pointing out that, notwithstanding the fact that the assumptions of Theorem 4.1 refine and relax substantially those of Theorem 2 in [3] and Theorem 5.7 in [18], and in spite of the utmost generality of the distance spaces under consideration, the arguments used here in the proof of Theorem 4.1 are fairly close to those used in Theorem 2 of [3] and Theorem 5.7 of [18]. Corollary 4.1. For $\gamma \in [0,\Lambda_X)$ let $\Phi\colon X \rightrightarrows X$ be a $\gamma$-Lipschitz mapping with closed graph $\operatorname{gph}(\Phi)$ in $X \times X$. Then, for an arbitrary $x_0 \in X$ and any $\varepsilon > 0$, $\Phi$ has a fixed point $\xi\in X$ satisfying and § 5. Geometric progressions in $f$-quasimetric spacesIn order to apply Theorem 4.1 to coincidence points of set-valued mappings, given a space $X$, we need to know the quantity $\Lambda_X$ and the functions $\mathcal{P}_{X,\gamma}$ and $\overline{\mathcal{P}}_{X,\gamma}$, where $\gamma \in [0,\Lambda_X)$. In this section we discuss the properties of the set $\mathcal{B}_{X}$ that facilitate the calculation of these characteristics. Here we assume that $X$ is an $f$-quasimetric space. Below, in § 6, the results of this section will be used to evaluate these characteristics for concrete $f$-quasimetric spaces. Let $\mathbf{F}$ be the class of functions $f\colon \mathbb{R}_+^{2} \to \mathbb{R}_+$ satisfying (2.4). Two functions $f,\varphi\in \mathbf{F}$ are be said to be equivalent (we write $f\sim \varphi$) if there exists $\delta > 0$ such that $f(r_1,r_2)=\varphi(r_1,r_2)$ for all $r_1,r_2\in (0,\delta)$. By $\mathcal{F}$ we denote the quotient space $\mathbf{F}/{\sim}$ . Below $f\colon \mathbb{R}_+^{2} \to \mathbb{R}_+$ will denote both a function satisfying (2.4) and its equivalence class. We define an order relation in $\mathcal{F}$ as follows: given $f,\varphi\in \mathcal{F}$, we say that $f\leqslant \varphi$ if there exists $\delta > 0$ such that $f(r_1,r_2)\leqslant \varphi(r_1,r_2)$ for all $r_1,r_2\in (0,\delta)$. We also set
$$
\begin{equation}
\varphi_{\infty}(r_1,r_2)=\max\{r_1,r_2\}\quad\text{and} \quad \varphi_1(r_1,r_2)=r_1+ r_2, \qquad \varphi_{\infty}, \varphi_1\in \mathcal{F}.
\end{equation}
\tag{5.1}
$$
For each $f \in \mathcal{F}$ let $\mathrm{Sp}[f]$ denote the class of all spaces in which the distance satisfies inequality (2.5) with this function $f$. Note that any three relatively close points in any $\mathrm{Sp}[\varphi_1]$-space satisfy the ‘ordinary’ triangle inequality (2.3). We set
$$
\begin{equation*}
\mathcal{B}_{\mathrm{Sp}[f]} :=\bigcap_{X \in \mathrm{Sp}[f]} \mathcal{B}_{X}\quad\text{and} \quad \Lambda_{\mathrm{Sp}[f]} :=\inf_{X \in \mathrm{Sp}[f]} \Lambda_{X} .
\end{equation*}
\notag
$$
In view of this definition, given a function $f$ and $\gamma\in \mathcal{B}_{\mathrm{Sp}[f]}$ (then, in particular, $\gamma\in [0,\Lambda_{\mathrm{Sp}[f]})$), each geometric progression with ratio $\gamma$ in any $f$-quasimetric space is a Cauchy sequence. If $\gamma \notin \mathcal{B}_{\mathrm{Sp}[f]}$ (in particular, if $\gamma >\Lambda_{\mathrm{Sp}[f]}$), then there exists an $f$-quasimetric space containing a geometric progression with ratio $\gamma$ that is not a Cauchy sequence. For any $\gamma \in \mathcal{B}_X$ and all $r_0 \geqslant 0$ there exists a space $X\in \mathrm{Sp}[f]$ containing a geometric progression $\{x_i\}_{i=0}^{\infty}\in \mathrm{GP}_X [\gamma]$ such that $\rho_X(x_0,x_1)=r_0$. A necessary and sufficient condition for a space $X$ to contain such a geometric progression is that $r_0 \in \mathcal{D}_X $, where the set $\mathcal{D}_X $ is defined by (3.3), that is, $X$ should contain at least one pair of points $x_0,x_1$ such that $\rho_X(x_0,x_1)=r_0$. Consider the function $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}\colon \mathbb{R}_+ \to \overline{\mathbb{R}}_+$ that associates, with each $r_0 \geqslant 0$, the quantity
$$
\begin{equation*}
\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma} (r_0) := \sup \bigl\{ \overline{\mathcal{P}}_{X,\gamma} (r_0) \colon X\in \mathrm{Sp}[f], \mathcal{D}_X \ni r_0\bigr\}
\end{equation*}
\notag
$$
(we show in Corollary 5.1 below that the function $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma} $, defined on the class of $\mathrm{Sp}[f]$-spaces, cannot be equal to $+\infty$, unlike $\overline{\mathcal{P}}_{X,\gamma}$, which is defined for an arbitrary space $X$). Similarly, for all $\gamma \in \mathcal{B}_{\mathrm{Sp}[f]}$ and $r_0 \geqslant 0$ there exists a space $X\in \mathrm{Sp}[f]$ containing a convergent geometric progression $\{x_i\}_{i=0}^{\infty}\in \mathrm{GP}_X^{\mathrm{con}} [\gamma]$ such that $\rho_X(x_0,x_1)=r_0$. For such a space $X$ we always have $r_0 \in \mathcal{D}_X $. Hence we can consider the function $\mathcal{P}_{\mathrm{Sp}[f],\gamma}\colon \mathbb{R}_+ \to \overline{\mathbb{R}}_+$ defined by
$$
\begin{equation*}
\forall\, r_0 \in \mathbb{R}_+ \quad {\mathcal{P}}_{\mathrm{Sp}[f],\gamma} (r_0) :=\sup \bigl\{ {\mathcal{P}}_{X,\gamma} (r_0) \colon X\in \mathrm{Sp}[f], \mathcal{D}_X \ni r_0\bigr\}.
\end{equation*}
\notag
$$
We also define the class $\mathrm{Sp}^s[f]$ of weakly symmetric spaces in which the distance satisfies inequality (2.5) for a function $f\in \mathcal{F}$. Proceeding as above, for the class $\mathrm{Sp}^s[f]$ we define the sets $\mathcal{B}_{Sp^s}[f]$, the numbers $\Lambda_{\mathrm{Sp^s}[f]}$ and the functions ${\mathcal{P}}_{\mathrm{Sp}^s[f],\gamma}$ and $\overline{\mathcal{P}}_{\mathrm{Sp}^s[f],\gamma}$ (in particular, $\mathcal{B}_{Sp^s}[f]:=\bigcap_{X \in \mathrm{Sp^s}[f]} \mathcal{B}_{X}$). The following relations are clear:
$$
\begin{equation*}
\begin{gathered} \, \mathrm{Sp}^s[f] \subset \mathrm{Sp}[f], \qquad \mathcal{B}_{Sp^s}[f]\supset \mathcal{B}_{Sp}[f], \qquad \Lambda_{\mathrm{Sp^s}[f]} \geqslant \Lambda_{\mathrm{Sp}[f]}, \\ \forall\, r_0 \geqslant 0 \quad {\mathcal{P}}_{\mathrm{Sp}^s[f],\gamma}(r_0) \leqslant {\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0), \qquad \overline{\mathcal{P}}_{\mathrm{Sp}^s[f],\gamma} (r_0) \leqslant \overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0). \end{gathered}
\end{equation*}
\notag
$$
Proof. Let $\gamma \in \mathcal{B}_{\mathrm{Sp}[\varphi]} $. Consider an arbitrary space $X\in \mathrm{Sp}[f]$ and a geometric progression $\{x_i\}_{i=0}^{\infty}$ with ratio $\gamma$ in it. Inequality (2.5) is satisfied for the distance in $X$ and the function $f$, and since $f \leqslant \varphi$, the same inequality holds for $X$ and $\varphi$. Therefore, $X\in \mathrm{Sp}[\varphi]$, and $\{x_i\}_{i=0}^{\infty}$ is a Cauchy sequence. So $\gamma \in \mathcal{B}_{ X }$. Since this inclusion holds for an arbitrary space $X\in \mathrm{Sp}[f]$, we have $\gamma \in \mathcal{B}_{\mathrm{Sp}[f]}$.
The proof for weakly symmetric spaces goes along the same lines. Proposition 5.1 is proved. Below we consider $f$-quasimetric spaces under the assumption that the following condition is met: $\mathbf{(\mathfrak{F})}$ the function $f$ satisfies (2.4), is (nonstrictly) increasing with respect to both arguments, and
$$
\begin{equation}
f(r_1,r_2)\geqslant \varphi_{\infty}(r_1,r_2) \quad\textit{for all } (r_1,r_2)\in \mathbb{R}_+^{2},
\end{equation}
\tag{5.2}
$$
where the function $\varphi_{\infty}$ is defined by (5.1). Remark 5.1. Inequality (5.2) is quite natural — if it fails, then an $f$-metric space $X$ (with symmetric distance $\rho_X$) cannot contain more than two points. Indeed, let there exist $d_1\geqslant d_2 > 0$ such that $f(d_1,d_2)< \varphi_{\infty}(d_1,d_2)=d_2$, and let there exist $x,u,v\in X$ such that $\rho_X(x,u)=\rho_X(u,x)=d_1$ and $\rho_X(u,v)=\rho_X(v,u)=d_2$. Then but the latter inequality implies that $d_1<d_1$, which is impossible. Proposition 5.2. If a function $f$ satisfies condition $\mathbf{(\mathfrak{F})}$, then $1\notin \mathcal{B}_{\mathrm{Sp}[f]}$ and $1\notin \mathcal{B}_{\mathrm{Sp^s}[f]}$ (so that $\Lambda_{\mathrm{Sp}[f]}\leqslant 1$ and $\Lambda_{\mathrm{Sp^s}[f]}\leqslant 1$). Proof. Given a countable set $X$, we equip it with the discrete distance (the distance between any two distinct points is 1). Any sequence of points in this space is a geometric progression with ratio $\gamma=1$; hence it is not a Cauchy sequence. Therefore, $1\notin \mathcal{B}_{X}$. But this $X$ is a $\varphi_{\infty}$-metric space, and therefore, if $f$ satisfies $\mathbf{(\mathfrak{F})}$, then $X\in \mathrm{Sp}[f]$ and $X\in \mathrm{Sp^s}[f]$. Now the required result follows. Remark 5.2. In the proof of Proposition 5.2 we defined a space $X$ such that $X\in \mathrm{Sp}[f]$ and $X\in\mathrm{Sp^s}[f]$ for any function $f$ satisfying $\mathbf{(\mathfrak{F})}$. Therefore, $\mathrm{Sp}[f]\neq\varnothing$ and $\mathrm{Sp^s}[f]\neq \varnothing$. Note also that $\mathrm{Sp}[f]$ and $\mathrm{Sp^s}[f]$ contain countable spaces. Other ‘not so trivial’ countable $f$-metric spaces will be considered below. Given a function $f\colon \mathbb{R}_+^{2} \to \mathbb{R}_+$ and numbers $\gamma\geqslant 0$ and $r_0>0$, we define a space which will used below in testing the inclusion $\gamma\in \mathcal{B}_{\mathrm{Sp}[f]}$ for concrete $f$ and $\gamma $. Consider a countable set $X=\{x_i\}_{i=0}^{\infty}$. The distance $\rho_X$ of $x_i$ to $x_{i+k}$ is denoted by $d_{i,k}$ ($i=0,1,\dots$, $k=1,2,\dots$), and the distance of $x_i$ to $x_{i-j}$ will be denoted by $d_{i,-j}$ ($i=1,2,\dots$, $j=1,\dots,i$). Given $i=0,1,\dots$, we set
$$
\begin{equation}
\begin{gathered} \, d_{i,1}:=r_0 \gamma^i, \qquad d_{i,2}:=f(d_{i,1}, d_{i+1,1}), \\ d_{i,k}:=\min\bigl\{ f(d_{i,1}, d_{i+1,k-1}),f(d_{i,2}, d_{i+2,k-2}),\dots, f(d_{i,k-1}, d_{i+k-1,1}) \bigr\}, \\ k=3,4,\dots\,. \end{gathered}
\end{equation}
\tag{5.3}
$$
Similarly, we define
$$
\begin{equation}
\begin{gathered} \, d_{i,-1}:=r_0 \gamma^i, \quad i=1,2,\dots, \qquad d_{i,-2}:=f(d_{i,-1}, d_{i-1,-1}), \quad i=2,3,\dots, \\ d_{i,-k}:=\min\bigl\{ f(d_{i,-1}, d_{i-1,-k+1}), f(d_{i,-2}, d_{i-2,-k+2}), \dots,f(d_{i,-k+1}, d_{i-k+1,-1}) \bigr\}, \\ i=3,4,\dots, \quad k=-1,-2,\dots, -i. \end{gathered}
\end{equation}
\tag{5.4}
$$
By $\mathcal{X}[f,\gamma,r_0]$ we denote the space $(X,\rho_X)$ equipped with the distance
$$
\begin{equation}
\begin{gathered} \, \rho_X (x_i,x_{i+k}):=d_{i,k}, \qquad i=0,1,\dots, \quad k=1,2,\dots, \\ \rho_X (x_i,x_{i-k}):=d_{i,-k}, \qquad i=1,2,\dots, \quad k=1,\dots,i , \end{gathered}
\end{equation}
\tag{5.5}
$$
where $d_{i,k}$ and $d_{i,-k}$ are given by (5.3) and (5.4). We show below that the inclusion $\gamma\in \mathcal{B}_{\mathrm{Sp}[f]}$ is equivalent to the relation $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,r_0]}$ for all $r_0\in \mathbb{R}_+ $. This will allow us to use the space $\mathcal{X}[f,\gamma,r_0]$ as a model space for the whole class $\mathrm{Sp}[f]$ in the study of the Cauchy property of geometric progressions. The following property of sequences defined by (5.3) and (5.4) will be required in the proof of results concerning the set $\mathcal{B}_{\mathrm{Sp}[f]}$ Lemma 5.1. Let the function $f$ satisfy $\mathbf{(\mathfrak{F})}$. Then, for any $i=0,1,\dots$, the sequence $\{d_{i,k}\}_{k=1}^{\infty}$ is increasing; for any $i=1,2,\dots$, the finite sequences $\{d_{i,-k}\}_{k=1}^{i}$ and $\{d_{i-k,k}\}_{k=1}^{i}$ are also increasing. Proof. First we show that, for any fixed $i=0,1,\dots$, the sequence $\{d_{i,k}\}_{k=1}^{\infty}$ is increasing. We use induction on $k$.
By the definition of $d_{i,2}$ we have
$$
\begin{equation*}
d_{i,2}=f(d_{i,1}, d_{i+1,1})\geqslant \varphi_{\infty}(d_{i,1}, d_{i+1,1})\geqslant d_{i,1}.
\end{equation*}
\notag
$$
Assume that for any $i$ we have
$$
\begin{equation*}
d_{i,1}\leqslant d_{i,2} \leqslant \dots \leqslant d_{i,k}
\end{equation*}
\notag
$$
for some integer $k>2$. We claim that $d_{i,k}\leqslant d_{i,k+1}$. We have
$$
\begin{equation}
\begin{aligned} \, \notag d_{i,k+1} &=\min\bigl\{ f(d_{i,1}, d_{i+1,k}), f(d_{i,2}, d_{i+2,k-1}), \\ &\qquad\qquad \dots, f(d_{i,k-1}, d_{i+k-1,2}), f(d_{i,k}, d_{i+k,1}) \bigr\}. \end{aligned}
\end{equation}
\tag{5.6}
$$
If the minimum in (5.6) is attained at $f(d_{i,k}, d_{i+k,1})$, then
$$
\begin{equation*}
d_{i,k+1}=f(d_{i,k}, d_{i+k,1})\geqslant \varphi_{\infty}(d_{i,k}, d_{i+k,1})\geqslant d_{i,k}.
\end{equation*}
\notag
$$
Now assume that the minimum in (5.6) is attained at some $f(d_{i,j}, d_{i+j,k+1-j})$, where $j<k$. We compare (5.6) with $d_{i,k}$ given by (5.3). Since the function $f$ is increasing, by the induction assumption we have
$$
\begin{equation*}
\begin{gathered} \, f(d_{i,1}, d_{i+1,k})\geqslant f(d_{i,1}, d_{i+1,k-1}), \\ f(d_{i,2}, d_{i+2,k-1}) \geqslant f(d_{i,2}, d_{i+2,k-2}), \\ \dots, \\ f(d_{i,k-1}, d_{i+k-1,2}) \geqslant f(d_{i, k-1}, d_{i+k-1,1}). \end{gathered}
\end{equation*}
\notag
$$
So, in this case the required inequality $d_{i,k+1} \geqslant d_{i,i+k}$ also holds.
Let us now show that, for any $i=1,2, \dots $, the finite sequence $\{d_{i,-k}\}_{k=1}^i$ is increasing. We argue by induction on $k$. We have
$$
\begin{equation*}
d_{i,-2}=f(d_{i,-1}, d_{i-1,-1}) \geqslant \varphi_{\infty} (d_{i,-1}, d_{i-1,-1})\geqslant d_{i,-1}.
\end{equation*}
\notag
$$
Assume that, for any $i> 2$, for some integer $k\in[2,i-1]$,
$$
\begin{equation*}
d_{i, -1}\leqslant d_{i, -2} \leqslant \dots \leqslant d_{i, -k}.
\end{equation*}
\notag
$$
We claim that in this case $d_{i, -k-1} \geqslant d_{i,-k}$. By (5.4),
$$
\begin{equation}
d_{i,-k-1}=\min\bigl\{ f(d_{i,-1}, d_{i-1,-k}), f(d_{i,-2}, d_{i-2,-k+1}),\dots, f(d_{i,-k}, d_{i-k,-1}) \bigr\}.
\end{equation}
\tag{5.7}
$$
If the minimum in (5.7) is attained at $f(d_{i,-k}, d_{i-k,-1}) $, then this inequality is obvious:
$$
\begin{equation*}
d_{i, -k-1}=f(d_{i,-k}, d_{i-k,-1}) \geqslant \varphi_{\infty}(d_{i,-k}, d_{i-k,-1})\geqslant d_{i,-k}.
\end{equation*}
\notag
$$
Now let the minimum in (5.7) be delivered by $f(d_{i,-j}, d_{i-j,-k-1+j})$ for some $j<k$. Let us compare $d_{i,-k-1}$ with $d_{i,-k}$ (recall that $d_{i,-k}$ is given by (5.4)). By the induction assumption and since $f$ is increasing, we have
$$
\begin{equation*}
\begin{gathered} \, f(d_{i,-1}, d_{i-1,-k}) \geqslant f(d_{i,-1}, d_{i-1,-k+1}), \\ f(d_{i,-2}, d_{i-2,-k+1}) \geqslant f(d_{i,-2}, d_{i-2,-k+2}), \\ \dots , \\ f(d_{i,-k+1}, d_{i-k+1,-2}) \geqslant f(d_{i,-k+1}, d_{i-k+1,-1}). \end{gathered}
\end{equation*}
\notag
$$
So the required inequality $d_{i, -k-1}\geqslant d_{i,-k}$ also holds in this case. Hence the finite sequence $\{d_{i,-k}\}_{k=1}^i$ is increasing.
To conclude the proof we show that the finite sequence $\{d_{i-k,k}\}_{k=1}^{i}$ is also increasing. We use induction on $k$. We have
$$
\begin{equation*}
d_{i-2,2}=f(d_{i-2,1}, d_{i-1,1}) \geqslant \varphi_{\infty} (d_{i-2,1}, d_{i-1,1})\geqslant d_{i-1,1}.
\end{equation*}
\notag
$$
Assume that, for any $i> 2$ and some integer $k\in[2,i-1]$,
$$
\begin{equation*}
d_{i-1, 1}\leqslant d_{i-2,2} \leqslant \dots \leqslant d_{i-k, k}.
\end{equation*}
\notag
$$
We claim that $d_{i-k-1, k+1} \geqslant d_{i-k,k}$. Recall that
$$
\begin{equation}
\begin{aligned} \, d_{i-k ,k} &=\min\bigl\{ f(d_{i-k,1}, d_{i-k+1,k-1}), f(d_{i-k,2}, d_{i-k+2,k-2}), \notag \\ &\qquad\qquad\dots, f(d_{i-k,k-1}, d_{i-1,1}) \bigr\} \end{aligned}
\end{equation}
\tag{5.8}
$$
and
$$
\begin{equation}
\begin{aligned} \, d_{i-k-1,k+1} &=\min\bigl\{ f(d_{i-k-1,1}, d_{i-k,k}), f(d_{i-k-1,2}, d_{i-k+1,k-1}), \notag \\ &\qquad\qquad \dots, f(d_{i-k-1,k}, d_{i-1,1}) \bigr\}. \end{aligned}
\end{equation}
\tag{5.9}
$$
Let us show that $d_{i-k-1, k+1} \geqslant d_{i-k,k}$. If the minimum in (5.9) is attained at $f(d_{i-k-1,1}, d_{i-k,k}) $, then this inequality is secured by the relation
$$
\begin{equation*}
d_{i-k-1, k+1}=f(d_{i-k-1,1}, d_{i-k,k}) \geqslant \varphi_{\infty} (d_{i-k-1,1}, d_{i-k,k})\geqslant d_{i-k,k}.
\end{equation*}
\notag
$$
Now assume that the minimum in (5.9) is delivered by $f(d_{i-k-1,j}, d_{i-k-1+j,k+1-j})$ for some $j<k$. By the induction assumption and since $f$ is increasing, we have
$$
\begin{equation*}
\begin{gathered} \, f(d_{i-k-1,2}, d_{i-k+1,k-1}) \geqslant f(d_{i-k,1}, d_{i-k+1,k-1}), \\ f(d_{i-k-1,3}, d_{i-k+2,k-2}) \geqslant f(d_{i-k,2}, d_{i-k+2,k-2}), \\ \dots , \\ f(d_{i-k-1,k}, d_{i-1,1}) \geqslant f(d_{i-k,k-1}, d_{i-1,1}) . \end{gathered}
\end{equation*}
\notag
$$
Hence, in view of (5.8) and (5.9) we obtain $d_{i-k-1, k+1} \geqslant d_{i-k,k}$. So the finite sequence $\{d_{i-k, k}\}_{k=1}^i$ is increasing. Lemma 5.1 is proved. Proposition 5.3. Let the function $f$ satisfy condition $\mathbf{(\mathfrak{F})}$. Then $\mathcal{X}[f,\gamma,r_0]\in {\mathrm{Sp}[f]}$ for all $\gamma \geqslant 0$ and $r_0>0$. Proof. We check that the distance (5.5) satisfies inequality (2.5) for the function $f$.
Given arbitrary points $x_i, x_{i+j}$ and $ x_{i+k}$, if $0 < j < k$, then
$$
\begin{equation*}
\rho_X(x_i,x_{i+k}) \leqslant f(\rho_X(x_i,x_{i+j}), \rho_X(x_{i+j},x_{i+k}))
\end{equation*}
\notag
$$
by the definition of $d_{i,i+k}$. If $j > k$, then, since for fixed $i$ the sequence $\{d_{i,k}\}_{k=1}^{\infty}$ is monotone (see Lemma 5.1), we have
$$
\begin{equation*}
\begin{aligned} \, f (\rho_X(x_i,x_{i+j}), \rho_X(x_{i+j},x_{i+k})) &=f(d_{i,j},d_{i+j,-j+k})\geqslant \varphi_{\infty}(d_{i,j},d_{i+j,-j+k}) \\ &\geqslant d_{i,j} \geqslant d_{i,k}=\rho_X(x_i,x_{i+k}). \end{aligned}
\end{equation*}
\notag
$$
Next, for $x_i, x_{i-j}$ and $ x_{i+k}$, $0<j\leqslant i$, by Lemma 5.1 we have
$$
\begin{equation*}
\begin{aligned} \, &f (\rho_X(x_i,x_{i-j}), \rho_X(x_{i-j},x_{i+k}))=f(d_{i,-j},d_{i-j,j+k})\geqslant \varphi_{\infty}(d_{i,-j},d_{i-j,j+k}) \\ &\qquad \geqslant d_{i-j,j+k}=d_{i+k-j-k, j+k} \geqslant d_{i+k-k,k}=d_{i,k}=\rho_X(x_i,x_{i+k}). \end{aligned}
\end{equation*}
\notag
$$
A similar argument involving Lemma 5.1 shows that, for all $i$, $j$ and $k$,
$$
\begin{equation*}
\begin{aligned} \, \rho_X(x_{i+k},x_{i}) &\leqslant f(\rho_X(x_{i+k},x_{i+j}), \rho_X(x_{i+j},x_{i})), \\ \rho_X(x_{i+k},x_{i}) &\leqslant f(\rho_X(x_{i+k},x_{i-j}), \rho_X(x_{i-j},x_{i})). \end{aligned}
\end{equation*}
\notag
$$
So $\mathcal{X}[f,\gamma,r_0]\in {\mathrm{Sp}[f]}$, which proves Proposition 5.3. The next result shows that the $f$-quasimetric (5.5) is maximal if the $d_{i,k}$ are given by (5.3), that is, the $f$-triangle inequality (2.5) fails to hold for larger $d_{i,k}$. To formulate this result, let us define an equivalence and an order relation for distances between points of a fixed set $X$. So assume that an arbitrary set $X$ is equipped with two distances $\rho_X,\varrho_X$: ${X^2 \to \mathbb{R}_+}$. We say that these distances are equivalent (and write $\varrho_X \sim \rho_X$) if there exists $\sigma>0$ such that $\varrho_X(x,u)=\rho_X(x,u)$ whenever $x,u \in X$ are such that $\varrho_X(x,u)<\sigma $ or $\rho_X(x,u)<\sigma$. The class of distances equivalent to a fixed distance $\rho_X$ will also be denoted by $\rho_X$. We write $\varrho_X \leqslant \rho_X $ if there exists $\sigma>0$ such that $\varrho_X(x,u)\leqslant \rho_X(x,u)$ whenever $x,u \in X$ are such that $\rho_X(x,u)<\sigma $. Proposition 5.4. Let the function $f$ satisfy condition $\mathbf{(\mathfrak{F})}$, and let $d_{i,k}$ and $d_{i,-k}$ be given by the recurrence relations (5.3) and (5.4). Let the $f$-quasimetric $\rho_X$ on the space $X=\{ x_i\}_{i=0}^{\infty}$ be defined by (5.5). If $X$ is equipped with another distance $\varrho_X\colon X^2 \to \mathbb{R}_+$ such that $\varrho_X \nleq \rho_X$, $\varrho_X(x_i,x_{i+1}) \leqslant d_{i,1}$ and $\varrho_X(x_{i+1},x_{i})\leqslant d_{i+1,-1}$ for all $i=0,1, \dots$, then this distance does not satisfy the $f$-triangle inequality (2.5). Proof. The inequality $\varrho_X \nleq \rho_X$ means that
$$
\begin{equation*}
\forall\, \sigma >0 \quad \exists\, x,u \in X \quad \rho_X(x, u)< \sigma, \quad \varrho_X(x, u) > \rho_X(x, u).
\end{equation*}
\notag
$$
Since $X=\{x_i\}_{i=0}^{\infty}$, there exist natural numbers $k$ and $i$ such that
$$
\begin{equation}
\rho_X(x_i, x_{i+k})< \sigma\quad\text{and} \quad \varrho_X(x_i, x_{i+k}) > \rho_X(x_i, x_{i+k})
\end{equation}
\tag{5.10}
$$
or
$$
\begin{equation*}
\rho_X(x_i, x_{i-k})< \sigma\quad\text{and} \quad \varrho_X(x_i, x_{i-k}) > \rho_X(x_i, x_{i-k}).
\end{equation*}
\notag
$$
For definiteness assume that (5.10) holds. Fix the least $k=k_0$ for which there exists a natural number $i$ satisfying (5.10). Note that $k_0\geqslant 2$ because, for all $i=0,1, \dots$, by the assumptions of the proposition we have $\varrho_X(x_i,x_{i+1})\leqslant \rho_X(x_i,x_{i+1})$.
Since $\{d_{i,k}\}_{k=1}^{\infty} $ is increasing (see Lemma 5.1), for all $k<k_0$ we have $d_{i,k}\leqslant d_{i,k_0}$, and therefore
$$
\begin{equation*}
\rho_X(x_{i}, x_{i+k}) \leqslant \rho_X(x_{i}, x_{i+k_0})< \sigma.
\end{equation*}
\notag
$$
Next, since $k_0$ is the smallest $k$ for which (5.10) holds, for all $k<k_0$ we have
$$
\begin{equation*}
\varrho_X(x_{i}, x_{i+k}) \leqslant \rho_X(x_{i}, x_{i+k}).
\end{equation*}
\notag
$$
From (5.3) we have
$$
\begin{equation*}
\begin{aligned} \, &\rho_X(x_{i}, x_{i+k_0})=d_{i, k_0} \\ &\qquad =\min\bigl\{ f(d_{i,1}, d_{i+1,k_0-1}), f(d_{i,2}, d_{i+2,k_0-2}),\dots,f(d_{i,k_0-1}, d_{i+k_0-1,1}) \bigr\}, \end{aligned}
\end{equation*}
\notag
$$
and so there exists $k\in [1, k_0-1]$ such that $ d_{i, k_0}=f(d_{i,k}, d_{i+k,k_0-k})$. Therefore, by (5.10) we have
$$
\begin{equation*}
\varrho_X (x_{i}, x_{i+k_0}) > d_{i, k_0}=f(d_{i,k}, d_{i+k,k_0-k}) \geqslant f(\varrho_X(x_i, x_{i+k}), \varrho_X(x_{i+k},x_{i+k_0})).
\end{equation*}
\notag
$$
Thus, for any $\sigma>0$ there exist $x_{i},x_{i+k}$ and $ x_{i+k_0}$ such that
$$
\begin{equation*}
\begin{gathered} \, \varrho_X(x_{i}, x_{i+k_0})> f( \varrho_X(x_{i}, x_{i+k}),\varrho_X(x_{i+k}, x_{i+k_0}), \\ \varrho_X(x_{i}, x_{i+k})\leqslant \rho_X(x_{i}, x_{i+k})\leqslant \rho_X(x_{i}, x_{i+k_0}) <\sigma \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, \varrho_X(x_{i+k}, x_{i+k_0}) &\leqslant \rho_X(x_{i+k}, x_{i+k_0}) =d_{(i+k_0)-(k_0 -k),k_0 -k} \\ &\leqslant d_{(i+k_0)-k_0,k_0}=\rho_X(x_{i}, x_{i+k_0}) <\sigma \end{aligned}
\end{equation*}
\notag
$$
(the third relation follows since the finite sequence $\{d_{i-n,n}\}_{n=1}^{i}$ is increasing for any $i$: see Lemma 5.1). Hence the distance $\varrho_X $ does not satisfy the $f$-triangle inequality (2.5). Proposition 5.4 is proved. Remark 5.3. The proof of Proposition 5.4 depends essentially on the following property of the distance $\varrho_X$: For otherwise it would be possible to find a mapping $\varrho_X\colon X^2 \to \mathbb{R}_+$, $\varrho_X \nleq \rho_X$, satisfying the $f$-triangle inequality. In fact, to do this it suffices to set $\varrho_X (x_0,x_0)=1$ for $x_0 \in X$ and $\varrho_X (x,u)=\rho_X(x,u)$ for any pair $(x,u)\in X^2$, $(x,u)\ne (x_0,x_0)$. However, the proof of Proposition 5.4 does not use the second part of the distance axiom, which claims that if $\varrho_X (x,u)=0$, then $x=u$. Thus, the conclusion of Proposition 5.4 remains true if $\varrho_X\colon X^2 \to \mathbb{R}_+$ satisfies only equality (5.11) in the identity axiom (2.1) (so that $\varrho_X$ is a ‘semidistance’, rather than a distance). Corollary 5.1. If $f$ satisfies $\mathbf{(\mathfrak{F})}$, then $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0)\!<\! \infty$ for all $\gamma \!\in\! \mathcal{B}_{\mathrm{Sp}[f]}$ and ${r_0 \!\geqslant\! 0}$. Proof. For $\gamma \in \mathcal{B}_{\mathrm{Sp}[f]}$ and $r_0 \geqslant 0$ consider a space $(U,\varrho_U)\in \mathrm{Sp}[f]$ containing a sequence $\{u_i\}_{i=0}^{\infty}\subset \mathrm{GP}_U [\gamma]$ such that $\varrho_U(u_0,u_1)=r_0$. Let $X=\mathcal{X}[f,\gamma,r_0]=\{x_i\}_{i=0}^{\infty} $ be the space with the $f$-quasimetric $\rho_X$ defined by (5.5). Since $\{x_i\}_{i=0}^{\infty}$ is a Cauchy sequence relative to $\rho_X$, there exists a natural number $I$ such that $\rho_X(x_I, x_{I+k})\leqslant 1$ for any natural number $k$.
Let $\varrho_X\colon X^2 \to \mathbb{R}_+$ be defined by $\varrho_X(x_i,x_j)=\varrho_U(u_i,u_j)$, $i,j=0,1,\dots$ . This mapping is a ‘semidistance’ in $X$, that is, it obeys (5.11) and the $f$-triangle inequality (2.5). By Proposition 5.4 and in view of Remark 5.3 we have
$$
\begin{equation*}
\varrho_U(u_0, u_{I+k})=\varrho_X(x_0, x_{I+k}) \leqslant \rho_X(x_0, x_{I+k})\leqslant f(\rho_X(x_0, x_{I}), 1 ), \qquad k=1,2,\dots\,.
\end{equation*}
\notag
$$
So $\overline{\mathcal{P}}_{U}(r_0)\leqslant f(\rho_X(x_0, x_{I}), 1 )< \infty$ for any space $U\in \mathrm{Sp}[f]$ that contains a geometric progression with the required properties. Remark 5.4. The above proof of Corollary 5.1 does not just show that $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0)$ is finite, but it also gives us a formula for the evaluation of this quantity: Since the distance in the space $\mathcal{X}[f,\gamma,r_0]$ is maximal, as established in Proposition 5.4, we can use this space as a model space not only for the evaluation of $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0)$, but also for finding $\mathcal{B}_{\mathrm{Sp}[f]}$. To formulate the corresponding result, we require the following property of the set $\mathcal{B}_{\mathcal{X}[f,\gamma,r_0]}$. Lemma 5.2. Let $f$ satisfy condition $\mathbf{(\mathfrak{F})}$. Then $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,r_0]}$ for all $r_0 > 0 $ if and only if this inclusion holds for some $r_0> 0 $. Proof. Given $r_0 > 0 $, we claim that the inclusion $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,r_0]}$ holds if and only if $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,1]}$ (of course, this implies the required result).
First assume that $r_0 \geqslant 1$. Consider the spaces
$$
\begin{equation*}
X=\mathcal{X}[f,\gamma,r_0]=\{x_i\}_{i=0}^{\infty}\quad\text{and} \quad \overline{X}=\mathcal{X}[f,\gamma,1]=\{\overline{x}_i\}_{i=0}^{\infty}.
\end{equation*}
\notag
$$
Assume that $\{x_i\}_{i=0}^{\infty}$ is a Cauchy sequence (this is so if $\gamma < 1$, see Proposition 5.2). For any $i=0,1,\dots$ we have
$$
\begin{equation*}
\rho_{X}(x_i,x_{i+1})=r_0 \gamma^i \geqslant \gamma^i=\rho_{\overline{X}}(\overline{x}_i,\overline{x}_{i+1}).
\end{equation*}
\notag
$$
Since $f$ is increasing, from the definitions of the distances $\rho_{X}$ and $\rho_{\overline{X}}$ via (5.3)–(5.5) it follows that, for all $i=0,1,\dots $ and $k=1,2,\dots$,
$$
\begin{equation*}
\rho_X (x_i,x_{i+k}) \geqslant \rho_{\overline{X}} (\overline{x}_i,\overline{x}_{i+k}),
\end{equation*}
\notag
$$
and therefore $\{\overline{x}_i\}_{i=0}^{\infty}$ is a Cauchy sequence.
Now let $\{\overline{x}_i\}_{i=0}^{\infty}$ be a Cauchy sequence. Then (see Proposition 5.2) $\gamma < 1$, and there exists a natural number $I$ such that $r_0 \gamma^{I}\leqslant 1$. For any $i=0,1,\dots$ we have
$$
\begin{equation*}
\rho_{X}(x_{i+I},x_{i+I+1})=r_0 \gamma^{i+I} \leqslant \gamma^i=\rho_{\overline{X}}(\overline{x}_i,\overline{x}_{i+1}).
\end{equation*}
\notag
$$
Since $f$ is increasing, from the definition of the distances $\rho_{X}$ and $\rho_{\overline{X}}$ we have
$$
\begin{equation*}
\rho_X (x_{i+I},x_{i+I+k})\leqslant \rho_{\overline{X}} (\overline{x}_i,\overline{x}_{i+k}), \qquad i=0,1,\dots, \quad k=1,2,\dots\,.
\end{equation*}
\notag
$$
Hence $\{x_i\}_{i=0}^{\infty}$ is a Cauchy sequence.
The proof for $r_0 < 1$ is similar. Lemma 5.2 is proved. Theorem 5.1. Let the function $f$ satisfy condition $\mathbf{(\mathfrak{F})}$. Then $\gamma\in \mathcal{B}_{\mathrm{Sp}[f]}$ if and only if $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,1]}$. Proof. Let $\gamma\in \mathcal{B}_{\mathrm{Sp}[f]}$. By the definition of the set $\mathcal{B}_{\mathrm{Sp}[f]}$, for any space $X \in \mathrm{Sp}[f]$ we have $\mathcal{B}_{\mathrm{Sp}[f]} \subset \mathcal{B}_{X}$ and, in particular, $\mathcal{B}_{\mathrm{Sp}[f]} \subset \mathcal{B}_{\mathcal{X}[f,\gamma,1]}$ (recall that $\mathcal{X}[f,\gamma,1] \in \mathrm{Sp}[f]$ by Proposition 5.3). Therefore, $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,1]}$.
Now let $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,1]}$. Consider an arbitrary space $(U,\varrho_U)\in \mathrm{Sp}[f]$ and choose a geometric progression $\{u_i\}_{i=0}^{\infty}\subset U$ with ratio $\gamma $ in this space; let $r_0=\varrho_U(u_0,u_1)$. On the subspace $V=\{u_i\}_{i=0}^{\infty} \subset U$ consider the induced distance $\varrho_V$. Since ${V\in \mathrm{Sp}[f]}$, for this induced distance we have $\varrho_V \leqslant \rho_V$ by Proposition 5.4, where $\rho_V$ is the $f$-quasimetric defined by (5.5) on $V$. The space $(V,\rho_V)$ is isometric to $\mathcal{X}[f,\gamma,r_0]$. By Lemma 5.2, $\gamma\in \mathcal{B}_{\mathcal{X}[f,\gamma,1]}$ and so $\gamma\in \mathcal{B}_{(V,\rho_V)}$. Since the geometric progression $\{u_i\}_{i=0}^{\infty}$ is a Cauchy sequence relative to $\rho_V$, the sequence $\{u_i\}_{i=0}^{\infty}$ is also a Cauchy sequence relative to the distance $\varrho_V$. Thus, $\gamma\in \mathcal{B}_{U}$ for any space $U\in \mathrm{Sp}[f]$, which shows that $\gamma\in \mathcal{B}_{\mathrm{Sp}[f]}$. Theorem 5.1 is proved. The following equivalent reformulation of Theorem 5.1 provides a convenient tool for testing the Cauchy property of geometric progressions in $f$-quasimetric spaces. Theorem 5.1'. Let the function $f$ satisfy condition $\mathbf{(\mathfrak{F})}$. Then all geometric progressions with ratio $\leqslant \gamma$ in any $f$-quasimetric space are Cauchy sequences if and only if $\{x_i\}_{i=0}^{\infty}$ is a Cauchy sequence in the space $\mathcal{X}[f,\gamma,1]= \{x_i\}_{i=0}^{\infty}$. § 6. Calculating $\Lambda_{\mathrm{Sp}[f]}$, $\mathcal{P}_{\mathrm{Sp}[f],\gamma}$ and $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}$ for concrete classes of $f$-quasimetric spacesUsing the results of the previous section, let us provide examples of the calculation of $\Lambda_{\mathrm{Sp}[f]}$ and $\mathcal{B}_{\mathrm{Sp}[f]}$ for concrete $f\in \mathcal{F}$ (and, in particular, for linear and power functions). In the examples that follow we also give estimates for the functions $\mathcal{P}_{\mathrm{Sp}[f],\gamma} $ and $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}$ for $\gamma \in \mathcal{B}_{\mathrm{Sp}[f]}$. In addition, we show that for quite a wide class of functions $f \in \mathcal{F}$ the ‘zero-one law’ holds for the corresponding $f$-quasimetric spaces, that is, $\Lambda_X $ is equal to either 0 or 1. Example 6.1. Let the linear function $f\colon \mathbb{R}_+^{2} \to \mathbb{R}_+$ with coefficients $q_1\!\geqslant\! 1$ and ${q_2\!\geqslant\! 1}$ be defined by In addition, from the results of [18] it readily follows that for any $(q_1,q_2)$-quasimetric space
$$
\begin{equation*}
\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0)\leqslant \frac{q_1^2 S(q_2\gamma, m_0 -1)+ q_1 (q_2\gamma)^{m_0 -1}}{1-q_2 \gamma^{m_0}}r_0,
\end{equation*}
\notag
$$
and for any weakly symmetric $(q_1,q_2)$-quasimetric space where Example 6.2. Let $f_1\colon \mathbb{R}_+ \to \mathbb{R}_+$ be an increasing function such that $f_1(r)\geqslant r$ for all $r\in \mathbb{R}_+$ and $f_1(r)\to 0$ as $r\to 0$. Let the function $f\colon \mathbb{R}_+^{2} \to \mathbb{R}_+$ be given by From inequality (6.1) for $(X,\rho_X)\in \mathrm{Sp}[f]$ we have
$$
\begin{equation*}
\forall\, n \in \mathbb{N} \quad \rho_X (x_{0},x_{n})\leqslant f_1(r_0) \quad\Longrightarrow\quad \overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma }(r_0)\leqslant f_1(r_0).
\end{equation*}
\notag
$$
In the case of a weakly symmetric space $X \in \mathrm{Sp}^s[f]$, if the set $\operatorname{Lim} x_i$ of limit points of this Cauchy sequence is nonempty, then for $\xi\in \operatorname{Lim} x_i$ we have $\rho_X(\xi,x_i)\to 0$ and $\rho_X(x_i,\xi)\to 0$. Now by the $f$-triangle inequality Therefore, Example 6.3. Let $f_2\colon \mathbb{R}_+ \to \mathbb{R}_+$ be an increasing function and let $f_2(r)\geqslant r$ for all $r\in \mathbb{R}_+$ and $f_2(r)\to 0$ as $r\to 0$. Arguing as in Example 6.2, let us show that, for the function Next, for $i=3,4,\dots$ we have
$$
\begin{equation*}
\begin{aligned} \, \rho_X (x_{0},x_{i}) &\leqslant \max\bigl\{ \dots \max\{ \max\{ r_0,f_2(r_0\gamma)\}, f_2(r_0\gamma^2)\}, \dots , f_2(r_0\gamma^{i-1})\bigr\} \\ &=\max \bigl\{r_0, f_2(r_0\gamma), f_2(r_0\gamma^2) ,\dots, f_2(r_0\gamma^{i-1})\bigr\} =\max \{ r_0, f_2(r_0\gamma)\}. \end{aligned}
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation*}
\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0)\leqslant \max\{ r_0, f_2(r_0\gamma)\}.
\end{equation*}
\notag
$$
For an arbitrary $\xi\in \operatorname{Lim} x_i$, using the $f$-triangle inequality we obtain Therefore, for $X \in {Sp^s}[f]$ we have $\rho_X(x_0,\xi)\leqslant \max\{ r_0, f_2(r_0\gamma)\}$, and therefore Example 6.4. We claim that, for any $\eta\in (0,1]$, for the function Another appeal to (6.4) shows that, in any space $(X,\rho_X) \in {Sp}[f]$,
$$
\begin{equation*}
\forall\, n \in \mathbb{N} \quad \rho_X (x_{0},x_{n})\leqslant \frac{r_0^{\eta}}{1-\gamma^{\eta}} + r_0\gamma^{ n-1}
\end{equation*}
\notag
$$
and Therefore, Example 6.5. Let $\eta\in (0,1]$ and let the function $f$ be given by By inequality (6.5), for any $X\in \mathrm{Sp}[f]$ we have
$$
\begin{equation*}
\forall\, i=3,4,\dots \quad \rho_X (x_{0},x_{i}) \leqslant r_0 + \frac{r_0^{\eta} \gamma^{\eta }}{1-\gamma^{\eta}}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}(r_0) \leqslant r_0 + \frac{r_0^{\eta} \gamma^{\eta }}{1-\gamma^{\eta}}.
\end{equation*}
\notag
$$
Next, from the $f$-triangle inequality, for an arbitrary $\xi\in X$ we have If $\xi\in \operatorname{Lim} x_i$, then for $X \in \mathrm{Sp}^s[f]$ we have $\rho_X(x_i,\xi) \to 0$, and therefore Table 1 summarizes the estimates in Examples 6.1–6.5 for $\overline{\mathcal{P}}_{\mathrm{Sp}[f],\gamma}^{}(r_0)$ and $\mathcal{P}_{\mathrm{Sp^s}[f], \gamma}(r_0)$ for $f$-quasimetric spaces with various functions $f$. Table 1
For the $f$-quasimetric spaces in Examples 6.1–6.5 we have ${\mathcal{B}_{\mathrm{Sp}[f]}\,{=}\,\mathcal{B}_{\mathrm{Sp^s}[f]}\,{=}\,[0,1)}$. Let us now give an example of a function $f$ such that, for the corresponding $f$-quasimetric space, the set $\mathcal{B}_{\mathrm{Sp}[f]}$ is different from $[0,1)$. Example 6.6. Consider the function $f$ given by where $\eta\in (0,2^{-1}]$. We claim that $\mathcal{B}_{\mathrm{Sp}[f]}=\{0\}$. Given an arbitrary $\gamma \in (0,1)$ and $r_0>0$, let us show that the geometric progression $\{x_i\}_{i=0}^{\infty}$ in $\mathcal{X}[f,\gamma,r_0]$ is not a Cauchy sequence. Since $\gamma < 1$, we have $\rho_X(x_i,x_{i+1})<1$ for all $i$ starting from some index. Assume that the progression starts with this index, that is, assume without loss of generality that $r_0 <1$. We claim that, for all $i=0,1,\dots$, $m=1,2,\dots$, $k=2^{m-1}+1,\dots, 2^m$,
$$
\begin{equation}
\rho(x_i,x_{i+k})\geqslant r_0^{\eta^m}\gamma^{(i+k)\eta^{m}}.
\end{equation}
\tag{6.7}
$$
We argue by induction on $m$. For any $i$ and $m=1$ we have $k=2$ and
$$
\begin{equation*}
\rho(x_i,x_{i+2})=\max\bigl\{r_0^{\eta}\gamma^{i\eta},r_0^{\eta}\gamma^{(i+1)\eta} \bigr\} =r_0^{\eta}\gamma^{i\eta}>r_0^{\eta}\gamma^{(i+2)\eta},
\end{equation*}
\notag
$$
that is, inequality (6.7) holds. Assume that (6.7) holds for some natural number $m=\widetilde{m}$, all $i$ and all ${k=2^{\widetilde{m}-1}+1, \dots, 2^{\widetilde{m}}}$. Let us show that this inequality holds for $m=\widetilde{m}+1$, all $i$ and all $k=2^{\widetilde{m}}+1,\dots, 2^{\widetilde{m}+1}$. For $k=2^{\widetilde{m}}+1$, by the definition of the distance in $\mathcal{X}[f,\gamma,r_0]$ there exists a natural number $j\in [1, 2^{\widetilde{m}}]$ such that
$$
\begin{equation*}
\rho(x_i,x_{i+2^{\widetilde{m}}+1})=\bigl(\max\{ \rho(x_i,x_{i+j}), \rho(x_{i+j},x_{i+2^{\widetilde{m}}+1})\}\bigr)^\eta.
\end{equation*}
\notag
$$
One of the integers $j$ and $2^{\widetilde{m}}+1-j$ is greater than $2^{\widetilde{m}-1}$. Hence, by the induction assumption
$$
\begin{equation*}
\rho(x_i,x_{i+2^{\widetilde{m}}+1})\geqslant\bigl(r_0^{\eta^{\widetilde{m}}} \gamma^{(i+2^{\widetilde{m}}+1)\eta^{\widetilde{m}}}\bigr)^\eta =r_0^{\eta^{\widetilde{m}+1}} \gamma^{(i+2^{\widetilde{m}}+1)\eta^{\widetilde{m}+1}}.
\end{equation*}
\notag
$$
For $k=2^{\widetilde{m}}+1$ inequality (6.7) is proved. Next, proceeding as above we see that for $k=2^{\widetilde{m}}+2$ there exists a natural number $j\in [1, 2^{\widetilde{m}}+1]$ such that
$$
\begin{equation*}
\rho(x_i,x_{i+2^{\widetilde{m}}+2})=\bigl(\max\{ \rho(x_i,x_{i+j}), \rho(x_{i+j},x_{i+2^{\widetilde{m}}+2})\}\bigr)^\eta.
\end{equation*}
\notag
$$
One of the integers $j$ and $2^{\widetilde{m}}+2-j$ is greater than $2^{\widetilde{m}-1}$. If it is no greater than $2^{\widetilde{m}}$, then by the induction assumption
$$
\begin{equation*}
\rho(x_i,x_{i+2^{\widetilde{m}}+2})\geqslant\bigl(r_0^{\eta^{\widetilde{m}}} \gamma^{(i+2^{\widetilde{m}}+2)\eta^{\widetilde{m}}}\bigr)^\eta =r_0^{\eta^{\widetilde{m}+1}} \gamma^{(i+2^{\widetilde{m}}+2)\eta^{\widetilde{m}+1}}.
\end{equation*}
\notag
$$
If this number is $2^{\widetilde{m}}+1$, then
$$
\begin{equation*}
\begin{aligned} \, \rho(x_i,x_{i+2^{\widetilde{m}}+2}) & \geqslant \bigl(\max\{\rho(x_i,x_{i+2^{\widetilde{m}}+1}), \rho(x_{i+1},x_{i+2^{\widetilde{m}}+2})\}\bigr)^{\eta} \\ & \geqslant \bigl(r_0^{\eta^{\widetilde{m}+1}} \gamma^{(i+2^{\widetilde{m}}+2)\eta^{\widetilde{m}+1}}\bigr)^\eta> r_0^{\eta^{\widetilde{m}+1}} \gamma^{(i+2^{\widetilde{m}}+2)\eta^{\widetilde{m}+1}}. \end{aligned}
\end{equation*}
\notag
$$
This proves inequality (6.7) for $k=2^{\widetilde{m}}+1$. A similar argument establishes inequality (6.7) for all $k$ up to $k=2^{\widetilde{m}+1}$. From (6.7), for $k=2^{m}$ we obtain
$$
\begin{equation*}
\rho(x_i,x_{i+2^{m}})\geqslant r_0^{\eta^m}\gamma^{(i+2^m)\eta^{m}}.
\end{equation*}
\notag
$$
In the case when $\eta=2^{-1}$, as $k=2^{m}\to \infty$ we have the convergences Therefore, $\{x_i\}_{i=0}^{\infty}$ is not a Cauchy sequence. In the case when $\eta <2^{-1}$ we have and therefore $\{x_i\}_{i=0}^{\infty}$ is not a Cauchy sequence either. For a function $f \in \mathcal{F}$ of the form (6.6) with exponent $\eta \in (2^{-1},1)$ we have not succeeded here to find the set $\mathcal{B}_{\mathrm{Sp}[f]}$ of common ratios of geometric progressions that are Cauchy sequences. Nevertheless, we can assert that this set is either minimal (the singleton $\{0\}$) or is the maximal interval $[0,1)$, and, respectively, $\Lambda_{\mathrm{Sp}[f]}=0$ or $\Lambda_{\mathrm{Sp}[f]}=1$. This ‘zero-one law’ is a direct consequence of the following result, which holds for the function (6.6) with exponent $\eta \in (2^{-1},1)$ (and for many functions in the examples considered in this section). Let $\varphi_{\eta}$ be the function defined by (6.6), that is,
$$
\begin{equation*}
\varphi_{\eta}\colon \mathbb{R}_+^{2} \to \mathbb{R}_+, \qquad \varphi_{\eta}(r_1,r_2)=\max \{r_1^{\eta},r_2^{\eta}\}, \quad r_1,r_2 \geqslant 0.
\end{equation*}
\notag
$$
Proposition 6.1. Let $f\in \mathcal{F}$ satisfy condition $\mathbf{(\mathfrak{F})}$ and the inequality $f \leqslant \varphi_{\eta}$ for some $\eta\in (2^{-1},1)$. Then $\Lambda_{\mathrm{Sp}[f]}$ can assume only two values, $0$ and $1$. Proof. Let $f\in \mathcal{F}$ satisfy the hypotheses of Proposition 6.1. Assume that $\Lambda_{\mathrm{Sp}[f]}> 0$. We claim that in this case $\Lambda_{\mathrm{Sp}[f]}=1$.
Since $\Lambda_{\mathrm{Sp}[f]}> 0$, there exists positive $\gamma \in \mathcal{B}_{\mathrm{Sp}[f]}$. For this $\gamma$, by Lemma 5.2, $\{x_i\}$ is a Cauchy sequence in the space $X:=\mathcal{X}[f,\gamma,1],$ $X=(\{x_i\},\rho_X)$. Let $\nu \in (2^{-1}\eta^{-1},1 )$ be arbitrary. We show that the sequence $\{u_i\}$ is also a Cauchy sequence in $U:=\mathcal{X}[f,{\gamma}^{\nu},1]$. First we show that, for all $i=0,1,\dots$ and $k=1,2,\dots$,
$$
\begin{equation}
\rho_U (u_{2i}, u_{2i+2k}) \leqslant \rho_X (x_{i}, x_{i+k}).
\end{equation}
\tag{6.8}
$$
This inequality will be proved by induction on $k$ (for all $i$ simultaneously).
For $k=1$ we have
$$
\begin{equation*}
\begin{aligned} \, \rho_U (u_{2i}, u_{2i+2}) &=f(\rho_U (u_{2i}, u_{2i+1}),\rho_U (u_{2i+1}, u_{2i+2})) =f ({\gamma}^{\nu 2 i}, {\gamma}^{\nu(2i+1)} ) \\ &\leqslant \varphi_{\eta} ({\gamma}^{\nu 2i}, {\gamma}^{\nu(2i+1)} )={\gamma}^{\nu 2 i \eta} \leqslant {\gamma}^i=\rho_X (x_{i}, x_{i+1}). \end{aligned}
\end{equation*}
\notag
$$
Assume that inequality (6.8) holds for all $i=0,1,\dots$ and all natural numbers $k<k_0$. For $k=k_0$, by the definition of the distance in $\mathcal{X}[f,{\gamma},r_0]$ there exists $l<k_0$ such that
$$
\begin{equation*}
\rho_X (x_{i}, x_{i+k_0})=f(\rho_X (x_{i},x_{i+l}),\rho_X(x_{i+l}, x_{i+k_0})).
\end{equation*}
\notag
$$
Hence, by the induction assumption
$$
\begin{equation*}
\rho_X (x_{i}, x_{i+k_0}) \geqslant f(\rho_U (u_{2i},x_{2i+2l}), \rho_U(u_{2i+ 2l}, x_{2i+2 k_0})).
\end{equation*}
\notag
$$
Now, by the definition of the distance in $\mathcal{X}[f,{\nu}^{\eta},1]$,
$$
\begin{equation*}
\rho_X (x_{i}, x_{i+k_0}) \geqslant \rho_U (u_{2i}, x_{2i+2k_0}).
\end{equation*}
\notag
$$
Let us use inequality (6.8) to show that $\{u_i\}_{i=0}^{\infty}\subset U$ is a Cauchy sequence. As concerns the sequence $\{x_i\}_{i=0}^{\infty}\subset X$, for any $\varepsilon > 0$ there exists $I$ such that $\rho_X(x_i, x_{i+k}) < \varepsilon$ for all integers $i>I$ and all natural numbers $k$. Therefore, $\rho_U(x_{2i}, x_{2i+2k})<\varepsilon$. By Lemma 5.1,
$$
\begin{equation*}
\rho_U(x_{2i+1}, x_{2i+2k}) < \varepsilon, \qquad \rho_U(x_{2i}, x_{2i+2k-1}) < \varepsilon\quad\text{and} \quad \rho_U(x_{2i+1}, x_{2i+2k-1}) < \varepsilon.
\end{equation*}
\notag
$$
So $\rho_X(x_j, x_{j+m}) < \varepsilon$ for all integers $j>2I$ and all natural numbers $m$.
Thus, we have shown that $\gamma^\nu \in \mathcal{B}_{\mathrm{Sp}[f]}$. Repeating the above arguments, we find that $\gamma^{\nu^2} \in \mathcal{B}_{\mathrm{Sp}[f]}$, $\gamma^{\nu^4} \in \mathcal{B}_{\mathrm{Sp}[f]}$ and so on, that is, $\gamma^{\nu^{2^l}} \in \mathcal{B}_{\mathrm{Sp}[f]}$, $l=1,2,\dots$ . Since $\gamma^{\nu^{2^l}}\to 1$ as $l\to \infty$, we have $\Lambda_{\mathrm{Sp}[f]}=1$. This proves Proposition 6.1. To conclude, we note that if a function $f \in \mathcal{F}$ does not satisfy the hypotheses of Proposition 6.1, then it is unknown whether or not the ‘zero-one law’ holds for the corresponding class of spaces $\mathrm{Sp}[f]$. 
Citation in format AMSBIB
\Bibitem{Zhu23} |
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