\section{Introduction}

The importance of the spaces of Lipschitz and H\"older continuous functions in mathematics and other sciences cannot be overestimated. This mere introduction is too short to even list all the theories where such functions play a key role let aside describe them in detail. Therefore, we will limit ourselves to mentioning only those instances which are closely related to compactness, i.e., the main object of our study. One of the oldest examples of an application of H\"older continuous functions dates back to 1930s and to Leray and Schauder who used functions satisfying the H\"older condition and estimates for such functions in the theory of classical solutions to partial differential equations. (For a detailed discussions of classical solutions to partial differential equations in the space of H\"older continuous functions we refer the reader to the well-known monographs, e.g., \cites{evans, gilbarg_trudinger}.) When talking about applications of the spaces of H\"older continuous functions one should not forget singular operators appearing in the fractional calculus. It turns out that such spaces are natural domains for the Riemann—Liouville integral operator of fractional order or one of its most notable representatives, that is, the Abel operator (for more detail see, e.g.,~\cite{GV} and~\cite{samko}*{Chapters 3.1 and 3.2}). The other noteworthy situation where Lipschitz spaces emerge naturally is the Kantorovich–Rubinstein metric defined in the space of Borel measures on a compact metric space. These metrics are used to define the weak sequential convergence of measures and applied in the treatment of the Monge--Kantorovich mass transport problem. (The introductory discussion on the subject, from the perspective of the spaces of Lipschitz continuous functions, may be found in~\cite{cobzas_book}*{Section~8.4}.)

Because of their applications -- in the previous paragraph we barely touched this topic -- Lipschitz and H\"older continuous functions constantly enjoy great interest among researchers. Each year numerous scientific articles and chapters are devoted just to studying their properties. Even whole books on Lipschitz and H\"older functions are written. Among the recent ones let us mention two monographs: one by Weaver, whose second, extended, edition was published in 2018 (see~\cite{weaver}) and the other by Cobza{\c{s}} \emph{et al.}s published in 2019 (see~\cite{cobzas_book}). In those monographs one may find plenty of profound ideas, perspectives and applications related to the spaces of Lipschitz and H\"older continuous functions. What they lack, however, is a (strong) compactness criterion in such spaces. No matter how hard we tried, we could not find one in the literature, although, as it turned out, in the case of real-valued functions it had stayed hidden in a plain sight for more than half a century. In a paper of Cobza{\c{s}} from 2001 it is even stated: ,,\emph{\textup[\ldots\textup] apparently there is no compactness criterion in spaces of H\"older functions, and some criteria given in the literature turned to be false}’’ (see~\cite{cobzas2001}*{p.~9}).  

Over the last years several attempts to find a compactness criterion in the spaces of Lipschitz and/or H\"older continuous mappings have been made. For example, in the already mentioned paper~\cite{cobzas2001} Cobza{\c{s}} worked in the space of those vector-valued Lipschitz maps defined on a compact and convex subset $K$ of a normed space which can be extended over a given open set containing $K$ to a continuously differentiable function and he gave necessary and sufficient conditions for a non-empty and bounded subset of such a space to be precompact. In~\cite{banas_nalepa} the authors gave a sufficient condition for relative compactness of subsets of the space of H\"older continuous real-valued functions; this result was later repeated in~\cite{banas_nalepa2}.  It should be underlined here that this condition, which we will refer to as \emph{uniform local flatness}, is far from being necessary, as many natural Lipschitz and H\"older continuous functions (like, for example, $f \colon [0,1] \to \mathbb R$ given by $f(x)=x$) do not satisfy it. It turns out, however, that uniform local flatness can be used to characterize precompactness, but one need to work in the so-called \emph{little Lipschitz \textup(H\"older\textup) space} rather than the classical Lipschitz space. (For the appropriate definitions and the compactness criterion in little Lipschitz spaces see Sections~\ref{sec:52} and~\ref{sec:55} below.) This was already known to Johnson in the real-valued setting (see~\cite{johnson}*{Theorem~3.2}) and to Garc\'{\i}a-Lirola \emph{et al.} in a slight different context of little Lipschitz functions that are continuous with respect to a topology which needs not to come from the metric of the underlying space (see~\cite{GPR}*{Lemma~2.7}). Finally, based on the ideas of Bana\'s and Nalepa, in~\cite{Saiedinezhad} Saiedinezhad used the uniform local flatness to provide a sufficient condition for relative compactness of a non-empty subset of the space $C^{k,\alpha}(X,\mathbb R)$ of real-valued multivariate functions that are $k$-times continuously differentiable and whose partial derivative of order $k$ is H\"older continuous with exponent $\alpha$; here $X$ is a compact subset of $\mathbb R^n$. Clearly, this result cannot be necessary. However, Saiedinezhad used it to introduce and study a measure of non-compactness in the space $C^{k,\alpha}(X,\mathbb R)$.

When thinking about possible approaches to proving a compactness criterion in a given normed space, two obvious methods come to mind: direct and indirect one. The former one is self-explanatory. The latter one often consists of two steps: finding another -- simpler -- space (linearly) isomorphic with our given space and proving a compactness criterion in this simpler space. It turns out that in the case of Lipschitz and H\"older continuous mappings defined on a metric space $(X,d)$ and with values in a normed space $E$ as the simpler space we can take the product of $E$ and the space of bounded (or bounded and continuous) functions. The isomorphism can be constructed using the so-called de Leeuw’s map. There is only one ,,little'' detail that needs to be taken care of. To a given H\"older continuous function $f \colon X \to E$ the de Leeuw’s map assigns the bounded continuous mapping $\Phi(f)$, which for distinct $x,y \in X$ is defined by the formula $\Phi(f)(x,y)=(f(x)-f(y))/\varphi(d(x,y))$; here $\varphi$ is the so-called comparison function. (More detail on comparison functions and de Leeuw’s map can be found in Section~\ref{sec:compact_in_lip} and in the references given therein.) So, as we can see, the function $\Phi(f)$ is not defined on the whole product $X \times X$, but on its subset $(X\times X) \setminus \dset{(x,x)}{x \in X}$, which does not need to be compact even if $X$ is. And the compactness criteria for $B(Y,E)$ with $Y$ being an arbitrary non-empty set or $C(Y,E)$ with $Y$ being a non-compact topological space, in general, are not easy to come by, especially if $E$ is infinite-dimensional. 

One example of such a criterion for real-valued functions can be found in the celebrated monograph of Dunford and Schwarz. Using an approach of Phillips (see~\cite{phillips} as well as~\cite{eveson}) it is shown that a non-empty and bounded subset $A$ of $B(Y,\mathbb R)$ or $C(Y,\mathbb R)$ is relatively compact if and only if for every $\varepsilon>0$ there is a finite cover $U_1,\ldots,U_N$ of $Y$ and points $t_i \in U_i$ such that $\sup_{f \in A}\sup_{x \in U_i}\abs{f(x)-f(t_i)}\leq \varepsilon$ for $i=1,\ldots,N$ (cf.~\cite{DS}*{Theorem~IV.5.6, p.~260 and Theorem~IV.6.5, p.~266}); here $Y$ is either an arbitrary (non-empty) set in the case of $B(Y,\mathbb R)$, or a topological space in the case of $C(Y,\mathbb R)$. (For completeness let us also add that the condition appearing in the above mentioned result is equivalent to the condition (DS) discussed in the present article and because of that we will refer to it also as the (DS) condition.)

In our study we have been interested in a compactness criterion for the space of H\"older continuous mappings taking values in an arbitrary normed space $E$. What may come as a surprise (having in mind, for example, the Arzel\`a--Ascoli theorem) is that the mere replacement of absolute values with norms in the results of Dunford and Schwarz simply does not work. It is not difficult to provide examples of compact (or even finite) subsets of the spaces $B(Y, E)$ or $C(Y,E)$, with $Y$ being a metric space, which do not satisfy the condition (DS) – for more detail and examples see Section~\ref{sec:discussion} below. Because of that we needed a completely new approach to establishing compactness criteria in $B(Y,E)$ and/or $C(Y,E)$ (not utilizing Phillip's results) and a completely new condition equivalent to precompactness in those spaces (note that we do not assume that $E$ is complete) which is more general than the condition (DS). Our method of proof is based on a version of a very recent abstract compactness criterion proved in~\cite{GKM} which, roughly speaking, allows to translate precompactness/relative compactness to a certain inequality involving the norm of the considered space. We should also mention that the ideas of Ambrosetti  (see~\cite{Ambrosetti}*{Section~2} and~\cite{BBK}*{Lemma~1.2.8}) were of much help in our study. It is worth underlining here that our approach although more elementary than the one presented in the book of Dunford and Schwarz (in a sense that it does not use advanced concepts from functional analysis or topology) allows not only to prove the compactness criteria in the spaces $B(Y, E)$ or $C(Y,E)$ in the general case when $E$ is an arbitrary normed space (and thus also in the space $\Lip_\varphi(Y,E)$ of H\"older continuous mappings), but also to draw some additional conclusion about the regularity of sets $U_i$ covering the space $Y$. (It turns out that those sets might be chosen either closed or open in the case of the space $C(Y,E)$). We refer the reader to Section~\ref{sec:discussion} for a detailed discussion of our condition and the condition (DS) and of differences and similarities between the case when $E$ is finite and infinite dimensional. 

As already mentioned, using the results for $B(Y, E)$ and $C(Y,E)$ in the second part of the paper we prove a compactness criterion for the space $\Lip_\varphi(X,E)$ of H\"older continuous mappings taking values in a normed space $E$ (see Theorem~\ref{thm:compact_in_Lip}). However, we do not stop there. We analyse the condition appearing in that result and show that what really matters is certain uniform behaviour of functions for arguments which are close to each other. This allows us to formulate an equivalent -- localized -- version of the compactness criterion in $\Lip_\varphi(X,E)$, which we later use to obtain a compactness criterion in the little Lipschitz space $\lip_\varphi(X,E)$.               


\section{Preliminaries}
The main goal of this section is to introduce the notation and conventions used throughout the paper and to recall some basic facts. We will use default symbols to denote the classical spaces and norms. In particular, if $(E, \norm{\cdot}_E)$ is a normed space, then by $B(X,E)$ we will denote the normed space of all bounded $E$-valued maps defined on a non-empty set $X$ endowed with the supremum norm $\norm{f}_\infty:=\sup_{x \in X}\norm{f(x)}_E$. Assuming that $X$ is a topological space, by $C(X,E)$ we will denote the normed subspace of $B(X,E)$ consisting of all continuous maps $f \colon X \to E$. Given a set $A\subseteq B(X,E)$ and a point $x \in X$ let $A(x):=\dset{f(x)}{f \in A}$; we will call such a set a \emph{section} of $A$ at $x$. We also set $A(X):=\bigcup_{x \in X}A(x)$. The open and closed balls in a metric space $X$ with center at $x\in X$ and radius $r>0$ will be denoted by $B_X(x,r)$ and $\ball_X(x,r)$, respectively. Finally, if $A, B$ are two non-empty subsets of a linear space, then $A+B:=\dset{a+b}{a\in A,\ b\in B}$ and $A-B:=\dset{a-b}{a\in A,\ b\in B}$. For functions $f,g \colon X \to E$ mapping a set $X$ into a normed space $E$ the difference $f-g$ will be always defined pointwise.

\subsection{Precompactness and relative compactness} 

The definition of precompactness is well-known. However, it is sometimes mistaken with relative compactness. So, to avoid any ambiguity, let us state the definition explicitly. A metric space $X$ is \emph{precompact} if its completion is compact. Equivalently, $X$ is precompact if every sequence in $X$ contains a Cauchy subsequence, or if it is \emph{totally bounded}, that is, if for each $\varepsilon>0$ there is a finite collection of points $x_1,\ldots,x_n \in X$ such that $X= \bigcup_{i=1}^n\ball_X(x_i,\varepsilon)$; such a collection is often called an $\varepsilon$-net. Moreover, a subset $A$ of a metric space $X$ is \emph{precompact} if it is precompact as a metric space itself (within the metric inherited from $X$). A non-empty subset $A$ of a metric space $X$ is called \emph{relatively compact in $X$} if the closure of $A$ with respect to $X$ is compact, or, equivalently, if each sequence in $A$ contains a subsequence convergent to an element in $X$. In complete metric spaces these two notions connected with compactness coincide. For more information concerning precompact and relatively compact metric spaces see~\cite{BBK}*{Section~1.1} or~\cite{MV}*{Chapter~4}.

\subsection{Abstract compactness criterion}
The proofs of the main results of the paper will be based on a modification of a new abstract compactness criterion that was presented in~\cite{GKM}. 

Let $(E,\norm{\cdot}_E)$ be a normed space endowed with a family $\bigl\{\norm{\cdot}_i\bigr\}_{i \in I}$ of semi-norms on $E$ satisfying the following two conditions:
\begin{enumerate}[label=\textbf{\textup{(A\arabic*)}}]
 \item\label{i} $\norm{x}_E=\sup_{i \in I}\norm{x}_i$ for $x \in E$,
 \item\label{ii} for every $i,j \in I$ there is an index $k \in I$ such that $\norm{x}_i \leq \norm{x}_k$ and $\norm{x}_j \leq \norm{x}_k$ for $x \in E$; in other words, the family $\bigl\{\norm{\cdot}_i\bigr\}_{i \in I}$ forms a directed set. 
\end{enumerate}
A non-empty subset $A$ of $E$ is called \emph{equinormed} (\emph{with respect to the family $\bigl\{\norm{\cdot}_i\bigr\}_{i \in I}$}) if for every $\varepsilon>0$ there exists $k \in I$ such that $\norm{x}_E\leq \varepsilon + \norm{x}_k$ for all $x \in A$.

Now, we are in position to state the afore-mentioned abstract compactness criterion. Although this is a slight modification of~\cite{GKM}*{Theorem~19} we will skip the proof completely, as it is identical to the original one. 

\begin{theorem}\label{thm:compactness_ver2}
Let $(E,\norm{\cdot}_E)$ be a normed space equipped with a family $\bigl\{\norm{\cdot}_i\bigr\}_{i \in I}$ of semi-norms satisfying conditions~\ref{i} and~\ref{ii}. A non-empty subset $A$ of $E$ is precompact if and only if the set $A-A$ is equinormed and for each sequence $(x_n)_{n \in \mathbb N}$ of elements of $A$ there is a subsequence $(x_{n_k})_{k \in \mathbb N}$ which is Cauchy with respect to each semi-norm $\norm{\cdot}_i$.
\end{theorem}

\begin{remark}\label{rem:boundedness}
In Theorem~\ref{thm:compactness_ver2} we need not require the set $A$ to be bounded, as its boundedness follows easily from other assumptions. It is clear that if $A$ is precompact, it is also bounded. In the other case suppose on the contrary that the set $A-A$ is equinormed and each sequence $(x_n)_{n \in \mathbb N}$ in $A$ has a  subsequence $(x_{n_k})_{k \in \mathbb N}$ which is Cauchy with respect to each semi-norm $\norm{\cdot}_i$, but $A$ is not bounded in $E$. Then, there is a sequence $(y_n)_{n \in \mathbb N}$ in $A$ such that $\norm{y_n}_E \to +\infty$ as $n \to +\infty$. Let $(y_{n_k})_{k \in \mathbb N}$ be its subsequence which is Cauchy with respect to each semi-norm $\norm{\cdot}_i$. In particular, $(y_{n_k})_{k \in \mathbb N}$ is bounded in each semi-norm, that is, $M_i:=\sup_{k \in \mathbb N}\norm{y_{n_k}}_i<+\infty$ for every $i \in I$. As the set $A-A$ is equinormed, there is an index $j \in I$ such that $\norm{\xi-\eta}_E \leq 1+\norm{\xi-\eta}_j$ for all $\xi,\eta \in A$. And then we have $\norm{y_{n_k}-y_{n_1}}_E \leq 1+ \norm{y_{n_k}-y_{n_1}}_j \leq 1+2M_j$ for $k \in \mathbb N$, meaning that $\sup_{k \in \mathbb N}\norm{y_{n_k}}_E<+\infty$ and leading to a contradiction.
\end{remark}


\section{Compactness in $B(X,E)$}

In Section~\ref{sec:compactness_B_X_E} we provide a full characterization of precompact subsets of the space of bounded mappings with values in a normed space. The proof of this result will not be a mere rewriting of the argument for real-valued functions which can be found in e.g.~\cite{DS}*{Theorem~6, p.~260} (see also~Theorem~\ref{thm:DS} below). A completely new approach is needed, partly because the condition imposed on a non-empty set $A\subseteq B(X,E)$ guaranteeing its precompactness in the vector-valued case is different from the ``classical'' condition considered in~\cite{DS} (cf. condition (DS) below). Our proof will be based on Theorem~\ref{thm:compactness_ver2}. We will therefore begin Section~\ref{sec:compactness_B_X_E} by defining an appropriate family of semi-norms on $B(X,E)$.

In Section~\ref{sec:discussion_of_B} we discuss the condition (B) guaranteeing precompactness of a non-empty subset $A$ of $B(X,E)$ little bit more. We will provide examples illustrating that it cannot be easily simplified. We will also show that in the case of real-valued functions it agrees with the condition presented in~\cite{DS}*{Theorem~6, p.~260}.

\subsection{Compactness criterion in $B(X,E)$}\label{sec:compactness_B_X_E} 
Let us fix a non-empty set $X$ and a normed space $(E,\norm{\cdot}_E)$. Furhtermore, let $\mathcal F$ be the family of all non-empty and finite subsets of $X$, and for any $Y \in \mathcal F$ set
\begin{equation}\label{eq:norms}
\norm{f}_Y:=\sup_{x \in Y}\norm{f(x)}_E. 
\end{equation}
It is clear that the family of semi-norms $\set[\big]{\norm{\cdot}_Y}_{Y \in \mathcal F}$ satisfies the conditions~\ref{i} and \ref{ii}.

Let us introduce a condition which plays a key role in our considerations. 

\begin{definition}
A non-empty set $A$ in $B(X,E)$ is said to satisfy the condition (B) if for every $\varepsilon>0$ there is a finite cover $U_1,\ldots,U_N$ of $X$ such that for every $i \in \{1,\ldots,N\}$ and every pair $x, y \in U_i$ for all $f \in A$ we have $\abs[\big]{\norm{f(x)}_E - \norm{f(y)}_E} \leq \varepsilon$.
\end{definition}

The proof of the compactness criterion in $B(X,E)$ will be based on Theorem~\ref{thm:compactness_ver2}. We will use the following two lemmas to show that one of the implications of that theorem is satisfied.  

\begin{lemma}\label{lem:5}
Let $A$ be a non-empty subset of $B(X,E)$ such that the algebraic difference $A-A$ satisfies the condition \textup{(B)} and the sections $A(x)$ are precompact for every $x \in X$. Then, each sequence $(f_n)_{n \in \mathbb N}$ of elements of the set $A$ contains a subsequence $(f_{n_k})_{k\in \mathbb N}$ which is Cauchy with respect to each semi-norm $\norm{\cdot}_Y$ given by~\eqref{eq:norms}. 
\end{lemma}

\begin{proof}
For each $i \in \mathbb N$ choose finitely many sets $U_1^i,\ldots,U_{N_i}^i$ satisfying the condition (B) with $\varepsilon=\frac{1}{i}$. In each set $U_j^i$ let us also fix a point $t_j^i$ and define $T:=\dset{t_j^i \in X}{\text{$i \in \mathbb N$ and $j=1,\ldots,N_i$}}$.

Now, let $(f_n)_{n \in \mathbb N}$ be a given sequence of elements of $A$. By the classical diagonal argument and the fact that all the sections $A(x)$ are precompact, we may select a subsequence  $(f_{n_k})_{k \in \mathbb N}$ which is Cauchy at every point $t \in T$. We will show that this subsequence is Cauchy at every $x \in X$. Take an arbitrary $x \in X$, fix $\varepsilon>0$ and choose $i \in \mathbb N$ such that $\frac{1}{i}\leq \frac{1}{2}\varepsilon$. As the sets $U_1^i,\ldots,U_{N_i}^i$ cover $X$ the point $x$ must belong to at least one of them; denote this set by $U_j^i$. The sequence $(f_{n_k}(t_j^i))_{k \in \mathbb N}$ is Cauchy in $E$, so there is an index $K$ such that $\norm{f_{n_k}(t_j^i) - f_{n_l}(t_j^i)}_E \leq \frac{1}{2}\varepsilon$ for all $k,l \geq K$. Then, in view of the condition (B), for all such indices $k,l$ we have $\norm{f_{n_k}(x)-f_{n_l}(x)}_E=\norm{(f_{n_k}-f_{n_l})(x)}_E - \norm{(f_{n_k}-f_{n_l})(t_j^i)}_E + \norm{f_{n_k}(t_j^i) - f_{n_l}(t_j^i)}_E \leq \varepsilon$. This proves that the sequence  $(f_{n_k}(x))_{k \in \mathbb N}$ is Cauchy.
\end{proof}


\begin{lemma}
Let $A$ be a non-empty subset of $B(X,E)$ such that $A-A$ satisfies the condition \textup{(B)}. Then, the algebraic difference $A-A$ is equinormed \textup(with respect to the family of semi-norms $\{\norm{\cdot}_Y\}_{Y \in \mathcal F}$ defined by~\eqref{eq:norms}\textup).
\end{lemma}

\begin{proof}
Similar to what we did in the proof of Lemma~\ref{lem:5} for each $i \in \mathbb N$ we choose finitely many sets $U_1^i,\ldots,U_{N_i}^i$ satisfying the condition (B) with $\varepsilon=\frac{1}{i}$, in each set $U_j^i$ we fix a point $t_j^i$ and, finally, we define $T:=\dset{t_j^i \in X}{\text{$i \in \mathbb N$ and $j=1,\ldots,N_i$}}$.

Now, let us fix $\varepsilon>0$ and let $i \in \mathbb N$ be such that $\frac{1}{i}\leq \frac{1}{2}\varepsilon$. Moreover, let $Y:=\{t_1^i,\ldots,t_{N_i}^i\}$. If $f,g \in A$, then there exists a point $x_\ast \in X$ such that $\norm{f-g}_{\infty} \leq \frac{1}{2}\varepsilon + \norm{(f-g)(x_\ast)}_E$. But $x_\ast$ belongs to one of the sets $U_1^i,\ldots,U_{N_i}^i$, say $U_j^i$, and so
\begin{align*}
 \norm{f-g}_{\infty} &\leq \tfrac{1}{2}\varepsilon + \norm{(f-g)(x_\ast)}_E\\
  & = \tfrac{1}{2}\varepsilon + \norm{(f-g)(x_\ast)}_E - \norm{(f-g)(t_j^i)}_E + \norm{(f-g)(t_j^i)}_E\\
	& \leq \varepsilon + \norm{f-g}_Y.
	\end{align*}
Therefore, the set $A-A$ is equinormed.	
\end{proof}

Now, we turn our attention to proving the opposite implication in Theorem~\ref{thm:compactness_ver2}.

\begin{lemma}\label{lem:7}
Let $A$ be a non-empty subset of $B(X,E)$ such that $A-A$ is equinormed \textup(with respect to the family of semi-norms $\{\norm{\cdot}_Y\}_{Y \in \mathcal F}$ defined by~\eqref{eq:norms}\textup) and each sequence $(f_n)_{n \in \mathbb N}$ of elements of the set $A$ contains a subsequence $(f_{n_k})_{k\in \mathbb N}$ which is Cauchy with respect to each semi-norm $\norm{\cdot}_Y$, then $A-A$ satisfies the condition \textup{(B)} and each section $A(x)$, where $x \in X$, is precompact. 
\end{lemma}

\begin{proof}
Note that for any $x\in E$ the sequence $(f_n)_{n \in \mathbb N}$ in $B(X,E)$ is Cauchy with respect to the semi-norm induced by the singleton $Y:=\{x\}$ if and only if the sequence $(f_n(x))_{n \in \mathbb N}$ is Cauchy in $E$. Thus, we can easily conclude that each section $A(x)$, where $x\in X$, is precompact, and so we only need to show that the algebraic difference $A-A$ satisfies the condition (B).  

Let $\varepsilon>0$ be fixed. Because the algebraic difference $A-A$ is equinormed there is a finite set $Y:=\{t_1,\ldots,t_m\} \subseteq X$ such that $\norm{f-g}_\infty \leq \frac{1}{16}\varepsilon + \norm{f-g}_Y$ for any $f,g \in A$. We already know that each section $A(t_i)$, where $i=1,\ldots,m$, is precompact, and so is the union $\bigcup_{i=1}^m A(t_i)$. Hence, there exists a finite collection of points $e_1,\ldots,e_n \in E$ such that $\bigcup_{i=1}^m A(t_i) \subseteq \bigcup_{i=1}^n B_E(e_i,\frac{1}{32}\varepsilon)$.\break By $\Psi$ denote the set of all functions $\psi\colon \{1,\ldots,m\}\to \{1,\ldots,n\}$ and for each $\psi \in \Psi$ let\break $A_\psi = \dset[\big]{f\in A}{\text{$f(t_i) \in B_E(e_{\psi(i)},\tfrac{1}{32}\varepsilon)$ for every $i=1,\ldots,m$}}$. Moreover, by $\Omega$ denote the set of all those $\psi\in\Psi$ for which $A_\psi\neq\emptyset$. It is obvious that for each $f\in A$ there exists $\psi_f\in \Omega$ such that $f\in A_{\psi_f}$. Let, additionally, for each $\psi\in\Omega$ a function $h_\psi\in A_\psi$ be fixed. There are only finitely many mappings in $\Omega$, say $h_1,\ldots,h_N$. Let us define the function $G \colon X \to \mathbb R^{N^2}$ by the formula
\begin{align*}
& G(x):=\bigl(\norm{(h_1-h_1)(x)}_E, \norm{(h_1-h_2)(x)}_E, \ldots, \norm{(h_1-h_N)(x)}_E, \norm{(h_2-h_1)(x)}_E,\\
&\hspace{4.5cm} \ldots, \norm{(h_2-h_N)(x)}_E, \ldots \norm{(h_N-h_1)(x)}_E,\ldots, \norm{(h_N-h_N)(x)}_E\bigr).
\end{align*}
In view of Remark~\ref{rem:boundedness} the set $A$ is bounded in $B(X,E)$. Let $M:=\sup_{f \in A}\norm{f}_{\infty}$. Then, $G(X)\subseteq [0,2M]^{N^2}$. Let $J_1,\ldots,J_k$ be a family of non-empty subsets of $\mathbb R$ covering $[0,2M]$ of diameters at most $\frac{1}{2}\varepsilon$. Moreover, let
\begin{align*}
& U_\lambda:=G^{-1}\bigl(J_{\lambda_{1,1}}\times J_{\lambda_{1,2}} \times \cdots \times J_{\lambda_{1,N}}\times J_{\lambda_{2,1}} \times \cdots \times J_{\lambda_{2,N}} \times \cdots \times J_{\lambda_{N,1}} \times \cdots \times J_{\lambda_{N,N}}\bigr), 
\end{align*}
where $\lambda:=(\lambda_{1,1}, \lambda_{1,2},\ldots, \lambda_{1,N}, \lambda_{2,1},\ldots, \lambda_{2,N}, \ldots, \lambda_{N,1}, \ldots, \lambda_{N,N}) \in \{1,\ldots,k\}^{N^2}$. The sets $U_\lambda$ form a finite cover of $X$. Let $\Lambda$ be the set of all those indices $\lambda$ for which $U_\lambda \neq \emptyset$.

We will show that for the sets $U_\lambda$, where $\lambda \in \Lambda$, the condition (B) holds. Let us fix $\lambda \in \Lambda$, mappings $f,g \in A$ and points $x,y \in U_\lambda$. Furthermore, let $h_f$ and $h_g$ be those maps among $h_1,\ldots,h_N$ for which we have $h_f \in A_{\psi_f}$ and $h_g \in A_{\psi_g}$. (Note that $h_f$ and $h_g$ may coincide.) By the definition of the sets $A_\psi$, we have $f(t_i), h_f(t_i) \in B_E(e_{{\psi_f}(i)},\frac{1}{32}\varepsilon)$ and $g(t_i), h_g(t_i) \in B_E(e_{{\psi_g}(i)},\frac{1}{32}\varepsilon)$ for $i=1,\ldots,N$. This implies that $\norm{f-h_f}_Y\leq \frac{1}{16}\varepsilon$ and $\norm{g-h_g}_Y\leq \frac{1}{16}\varepsilon$. Consequently, $\norm{f-h_f}_\infty\leq \frac{1}{16}\varepsilon + \norm{f-h_f}_Y \leq \frac{1}{8}\varepsilon$ and $\norm{g-h_g}_\infty \leq \frac{1}{8}\varepsilon$. Likewise, by the definition of the sets $U_\lambda$, we have $\norm{(h_f-h_g)(x)}_E, \norm{(h_f-h_g)(y)}_E \in J_{\lambda_{f,g}}$. And so, $\abs[\big]{\norm{(h_f-h_g)(x)}_E - \norm{(h_f-h_g)(y)}_E}\leq \frac{1}{2}\varepsilon$. Note that
\[
 \norm{(f-g)(x)}_E \leq  \norm{(f-h_f)(x)}_E + \norm{(h_f-h_g)(x)}_E + \norm{(h_g-g)(x)}_E
\]
and
\[
 \norm{(h_f-h_g)(y)}_E \leq \norm{(h_f-f)(y)}_E + \norm{(f-g)(y)}_E + \norm{(g-h_g)(y)}_E.
\]
Thus,
\begin{align*}
& \norm{(f-g)(x)}_E - \norm{(f-g)(y)}_E\\
& \quad \leq  \norm{(f-h_f)(x)}_E + \norm{(h_f-h_g)(x)}_E + \norm{(h_g-g)(x)}_E\\
&\qquad + \norm{(h_f-f)(y)}_E  - \norm{(h_f-h_g)(y)}_E + \norm{(g-h_g)(y)}_E\\
& \quad \leq 2\norm{f-h_f}_\infty + 2\norm{g-h_g}_\infty + \norm{(h_f-h_g)(x)}_E - \norm{(h_f-h_g)(y)}_E\\
& \quad \leq 2\cdot \tfrac{1}{8}\varepsilon + 2\cdot \tfrac{1}{8}\varepsilon + \tfrac{1}{2}\varepsilon = \varepsilon.
\end{align*}
Interchanging the roles of $x,y$ we get $\norm{(f-g)(y)}_E - \norm{(f-g)(x)}_E \leq\varepsilon$, which shows that the set $A-A$ satisfies the condition (B).
\end{proof}

From Theorem~\ref{thm:compactness_ver2} and Lemmas~\ref{lem:5}--\ref{lem:7} we finally get the following compactness criterion in $B(X,E)$.

\begin{theorem}\label{thm:compactness_BXE}
A non-empty subset $A$ of $B(X,E)$ is precompact if and only if the set $A-A$ satisfies the condition \textup{(B)} and for every $x \in X$ the sections $A(x)$ are precompact.
\end{theorem}

\begin{remark}\label{rem:relatively_compact}
If $E$ is a Banach space, then the space $B(X,E)$ is complete, and so the classes of relatively compact and precompact subsets coincide. Therefore, in such a case Theorem~\ref{thm:compactness_BXE} provides necessary and sufficient conditions for relative compactness of a non-empty subset of $B(X,E)$.
\end{remark}

\subsection{Discussion of the condition (B)}\label{sec:discussion_of_B}
\label{sec:discussion}

In this section we are going to study the condition (B) a little bit more. There are some questions that can (and should) be raised. One of the most natural ones is whether in the statement of the compactness criterion in $B(X,E)$, that is Theorem~\ref{thm:compactness_BXE}, the phrase ``the set $A-A$ satisfies the condition (B)'' can be replaced by ``the set $A$ satisfies the condition (B)''. It may come as a surprise that the answer is negative. Moreover, it cannot be done in any infinite-dimensional normed space $E$.

\begin{proposition}\label{prop:B:zero:one}
Let $X$ be the set consisting of all zero-one sequences with only finitely many non-zero terms. Moreover, let $E$ be an infinite-dimensional normed space. Then, there exists a non-empty set $A\subseteq B(X,E)$ with the following properties\textup:
\begin{enumerate}[label=\textup{(\alph*)}]
 \item $A$ satisfies the condition \textup{(B)},
 \item $A-A$ does not satisfy the condition \textup{(B)},
 \item for each $x \in X$ the section $A(x)$ is compact,
 \item $A(X)$ is not precompact,
 \item $A$ is not precompact.
\end{enumerate}
\end{proposition}

\begin{proof}
By the well-known Riesz lemma (see e.g.~\cite{kreyszig}*{Theorem~2.5-4}), there exists a sequence $(e_n)_{n \geq 0}$ of unit vectors in $E$ such that $\norm{e_i-e_j}_E \geq \frac{1}{2}$ for all distinct $i,j\in \N\cup\{0\}$. For each $n \in \mathbb N$ let us define a bounded mapping $f_n \colon X \to E$, by
\begin{equation}
f_n(x) = \begin{cases}
  e_0 & \text{if $\xi_n = 0$,}\\
	e_n & \text{if $\xi_n = 1$;}
\end{cases}
\end{equation}
here $x:=(\xi_n)_{n \in \mathbb N}$. Let also $A:=\dset{f_n}{n\in \mathbb N}$. 

Note that the set $A$ satisfies the condition (B) with the cover of $X$ consisting of a single set $U_1:=X$, because for every $x,y\in X$ and $n \in \mathbb N$ we have $\norm{f_n(x)}_E = \norm{f_n(y)}_E = 1$. However, the algebraic difference $A-A$ does not satisfy condition (B). To see this let $U_1,\ldots,U_N$ be any finite cover of $X$ and let $k$ be a positive integer such that $2^{k-1} > N$. Since there are exactly $2^{k-1}$ different zero-one sequences of length $k-1$, there exists at least one set $U_i$ which contains at least two distinct elements $x=(\xi_n)_{n \in \mathbb N}$ and $y=(\eta_n)_{n \in \mathbb N}$ such that $\xi_n = \eta_n = 0$ for $n\geq k$. As $x\neq y$, an index $l<k$ exists such that $\xi_l\neq \eta_l$. Of course, we may assume that $\xi_l=1$ and $\eta_l=0$. So, $f_k(x) = f_k(y) = f_l(y)= e_0$ and $f_l(x)=e_l$. Therefore, we have
\[
\abs[\big]{\norm{(f_k-f_l)(x)}_E - \norm{(f_k-f_l)(y)}_E} = \norm{e_0-e_l}_E \geq \tfrac{1}{2},
\]
which shows that the condition (B) for the set $A-A$ is not satisfied.

Now, let us fix $x=(\xi_n)_{n \in \mathbb N} \in X$ and let $m \in \mathbb N$ be such that $\xi_n=0$ for $n \geq m$. Then, $A(x)\subseteq \{e_0,e_1,\ldots,e_m\}$, meaning that the section $A(x)$ is compact. On the other hand, $A(X)=\dset{e_n}{n \in \mathbb N\cup\{0\}}$. This implies that $A(X)$ is not precompact, as the sequence $(e_n)_{n \in \mathbb N}$ does not contain a Cauchy subsequence. 

The fact that the set $A$ is not precompact follows from Theorem~\ref{thm:compactness_BXE}.
\end{proof}

Another natural question concerning the condition (B) is whether instead of the assumption ``the set $A-A$ satisfies the condition (B)'' we can assume in Theorem~\ref{thm:compactness_BXE} that the set $A \subseteq B(X,E)$ (or, eventually, $A-A$) satisfies the following stronger condition often encountered when dealing with compactness criteria for real-valued functions (cf.~\cite{alexiewicz1969analiza}*{pp. 370–371} or~\cite{DS}*{Section IV.4}):
\begin{enumerate}
 \item[(DS)] for every $\varepsilon>0$ there exist finitely many non-empty subsets $U_1,\ldots,U_N$ of $X$ whose union is $X$ such that for every $i \in \{1,\ldots,N\}$ and every pair $x, y \in U_i$ for all $f \in A$ we have $\norm{f(x) - f(y)}_E \leq \varepsilon$. 
\end{enumerate}
Once again the answer is negative as shown by the following example.

\begin{example}\label{ex:B:comp}
Let $X$ be the unit sphere in an infinite-dimensional normed space $E$. Consider the mappings $f,g \colon X \to E$ given by $f(x)=x$ and $g(x)=2x$, and let $A:=\{f,g\}$. It is easy to see that both the sets $A$ and $A-A$ satisfy the condition (B) with $U_1:=X$, as $\norm{f(x)}_E- \norm{f(y)}_E=0$, $\norm{g(x)}_E-\norm{g(y)}_E=0$ and $\norm{(f-g)(x)}_E-\norm{(f-g)(y)}_E=0$ for any $x,y \in X$. However, if either $A$ or $A-A$ satisfied the condition (DS), then for every $\varepsilon>0$ it would be possible to cover the unit sphere in an infinite-dimensional normed space with finitely many sets of diameter not greater than $\varepsilon$, which is absurd. Thus, neither the set $A$ nor the algebraic difference $A-A$ satisfy the condition (DS). Note also that for each $x \in X$ the section $A(x)$ is compact, but the set $A(X)$ is not.
\end{example}

A closer look at the above discussion reveals that the construction of the set $A$, showing that the assumptions of Theorem~\ref{thm:compactness_BXE} cannot be easily simplified, was possible because the normed space $E$ was infinite-dimensional, and thus the set $A(X)$ was not precompact. However, if we assume that the subset $A(X)$ of $E$ is precompact, the situation changes significantly.

\begin{proposition}\label{prop:DS_equivalent_B}
Let $X$ be a non-empty set and let $E$ be a normed space. For a non-empty subset $A$ of $B(X,E)$ the following conditions are equivalent\textup:
\begin{enumerate}[label=\textup{(\roman*)}]
 \item\label{it:DS_B_i}  $A$ satisfies the condition \textup{(DS)} and for every $x \in X$ the section $A(x)$ is precompact,
 \item\label{it:DS_B_ii} $A-A$  satisfies the condition \textup{(B)} and the set $A(X)$ is precompact.
\end{enumerate}
\end{proposition}

\begin{proof}
We begin with the proof of the implication~$\ref{it:DS_B_i}\Rightarrow\ref{it:DS_B_ii}$. The part concerning the condition (B) for $A-A$ is easy, as if the set $A$ satisfies the condition (DS), then so does $A-A$, and for any $x,y\in X$ and $f,g \in A$ we have $\abs[\big]{\norm{(f-g)(x)}_E-\norm{(f-g)(y)}_E}\leq \norm{(f-g)(x)-(f-g)(y)}_E$. Hence, we need only to show that the set $A(X)$ is precompact. Fix $\varepsilon>0$. By the condition (DS) the set $X$ can be covered by a finite collection of non-empty sets $U_1,\ldots,U_n$ such that $\sup_{f \in A}\norm{f(\xi)-f(\eta)}_E \leq \frac{1}{2}\varepsilon$ for every $i\in\{1,\ldots,n\}$ and every pair $\xi,\eta \in U_i$. In each set $U_i$ let us fix a point $t_i$. Then, for any $x \in U_i$ and $f \in A$ we have $f(x)=f(t_i)+f(x)-f(t_i) \in \{f(t_i)\}+\ball_E(0,\frac{1}{2}\varepsilon)$. Thus, $A(U_i):=\bigcup_{x\in U_i}A(x) \subseteq A(t_i) + \ball_E(0,\frac{1}{2}\varepsilon)$. But $A(t_i)$ is totally bounded by assumption, and so it can be covered with a finite family of balls $\ball_E(y_1^i,\frac{1}{2}\varepsilon)$, \ldots, $\ball_E(y_{m_i}^i,\frac{1}{2}\varepsilon)$, where $y_j^i \in A(t_i)$. This means that $A(X)\subseteq \bigcup_{i=1}^n A(U_i) \subseteq \bigcup_{i=1}^n \bigcup_{j=1}^{m_i} \ball_E(y_j^i,\varepsilon)$ and shows that $A(X)$ is totally bounded/precompact.

Now, we prove the implication~$\ref{it:DS_B_ii}\Rightarrow\ref{it:DS_B_i}$. The reasoning is similar to the one we used to establish Lemma~\ref{lem:7}. Let us assume that the algebraic difference $A-A$ satisfies the condition (B). Then, given $\varepsilon>0$ there exists a finite collection $W_1,\ldots,W_n$ of subsets of $X$ whose union is $X$ such that for every $i \in \{1,\ldots,n\}$ and all $x,y \in W_i$ and $f,g \in A$ we have $\abs[\big]{\norm{(f-g)(x)}_E - \norm{(f-g)(y)}_E} \leq \frac{1}{4}\varepsilon$. For each $i \in \{1,\ldots,n\}$ let us also fix $w_i\in W_i$. Furthermore, let $e_1,\ldots,e_m \in E$ be a finite $\frac{1}{8}\varepsilon$-net for $A(X)$.

By $\Psi$ denote the set of all functions $\psi\colon \{1,\ldots,n\}\to \{1,\ldots,m\}$ and for each $\psi \in \Psi$ let\break $A_\psi = \dset{f\in A}{\text{$f(w_i) \in B_E(e_{\psi(i)},\tfrac{1}{8}\varepsilon)$ for every $i=1,\ldots,n$}}$.
 Moreover, by $\Omega$ denote the set of all those $\psi\in\Psi$ for which $A_\psi\neq\emptyset$. 
 For each $\psi\in\Omega$ let us chose a function $g_\psi\in A_\psi$. Thus, we obtain a finite family $g_1,\ldots,g_k$ of functions, where $k$ is the cardinality of $\Omega$.  
  It is obvious that for each $f\in A$ there exists $\psi_f\in \Omega$ and a corresponding function $g_l \in \{g_1,\ldots,g_k\}$ such that $f,g_l\in A_{\psi_f}$.
 

Define the mapping $G \colon X \to E^k$ by  $G(x)=(g_1(x),\ldots,g_k(x))$ and let $V_\lambda:=G^{-1}(\prod_{i=1}^k \ball_E(e_{\lambda_i},\frac{1}{8}\varepsilon))$, where $\lambda:=(\lambda_1,\ldots,\lambda_k) \in \{1,\ldots,m\}^k$. Set $U_{i,\lambda}:=W_i \cap V_\lambda$ and by $I$ let us denote the set of those pairs $(i,\lambda)$, where $i \in \{1,\ldots,n\}$ and $\lambda \in \{1,\ldots,m\}^k$, for which the sets $U_{i,\lambda}$ are non-empty. Clearly, the finite family $\{U_{i,\lambda}\}_{(i,\lambda) \in I}$ is a covering of $X$. 

Let us now fix an index $(i,\lambda) \in I$ and take any points $x,y\in U_{i,\lambda}$ as well as any function $f \in A$ together with the corresponding function $g_l \in \{g_1,\ldots,g_k\}$ such that $f,g_l\in A_{\psi_f}$. Then, by condition (B) and the fact that $U_{i,\lambda}\subseteq W_i$, we can see that 
\[
\norm{(f-g_l)(x)}_E\leq \norm{(f-g_l)(w_i)}_E + \tfrac{1}{4}\varepsilon,
\]
and similarly
\[
\norm{(f-g_l)(y)}_E\leq \norm{(f-g_l)(w_i)}_E + \tfrac{1}{4}\varepsilon.
\]
As $f,g_l \in A_{\psi_f}$, we have $\norm{(f-g_l)(w_i)}_E \leq \frac{1}{4}\varepsilon$. Using the fact that the set $U_{i,\lambda}$ is included in $V_\lambda$ we obtain $\norm{g_l(x)-g_l(y)}_E \leq \frac{1}{4}\varepsilon$. Therefore, $\norm{f(x)-f(y)}_E \leq \norm{(f-g_l)(x)}_E + \norm{g_l(x)-g_l(y)}_E + \norm{(f-g_l)(y)}_E \leq \tfrac{1}{2}\varepsilon + \tfrac{1}{4}\varepsilon + \tfrac{1}{2}\varepsilon = \varepsilon$.


Since a subset of a precompact set is also precompact and $A(x)\subseteq A(X)$ for every $x \in X$, this ends the proof.
\end{proof}



As bounded sets in finite-dimensional Banach spaces are precompact, from Theorem~\ref{thm:compactness_BXE} and Proposition~\ref{prop:DS_equivalent_B} (cf. also Remark~\ref{rem:relatively_compact}) we immediately get the following result stated in a slightly different (but equivalent) form for $n=1$ as \cite{DS}*{Theorem~6, p.~260}. 

\begin{theorem}\label{thm:DS}
Let $X$ be a non-empty set. A non-empty set $A \subseteq B(X,\mathbb R^n)$ is relatively compact if and only if it is bounded and satisfies the condition \textup{(DS)}.
\end{theorem}


\section{Compactness in $C(X,E)$}
\label{sec:compactness_in_CXE}

In this section, as corollaries to our compactness results established previously, we will state compactness criteria for the space of bounded and continuous maps. The most interesting is the first one as it is probably not (widely) known. The other two (in the exact or similar version) are already known in the literature (see, e.g.~\cite{alexiewicz1969analiza}*{pp.~270--270}, \cite{BBK}*{Corollary~1.2.8}, \cite{DS}*{Section~IV.6} or~\cite{kreyszig}*{Theorem~8.7-4}). We have decided to state them, because we will need them when dealing with compactness in Lipschitz spaces and because, in contrast to, for example, the approach presented in~\cite{DS}, our method allows to say more about the regularity of the sets $U_1,\ldots,U_n$ appearing in the conditions (B) and (DS) (see Remarks~\ref{rem:open_closed} and~\ref{rem:open_closed2}).  

As an application of Theorem~\ref{thm:compactness_BXE} to a subspace of $B(X,E)$ consisting of continuous mappings we get the following result.

\begin{theorem}\label{thm:compactness_in_CXE_1}
Let $X$ be a topological space and let $E$ be a normed space. Then, a non-empty subset $A$ of $C(X,E)$ is precompact if and only if the algebraic difference $A-A$ satisfies the condition \textup{(B)} and for every $x \in X$ the sections $A(x)$ are precompact.  
\end{theorem}
 
\begin{remark}\label{rem:open_closed}
A closer look at the proof of Theorem~\ref{thm:compactness_BXE} allows us to say more about the regularity of the sets $U_1,\ldots,U_N$ appearing in the condition (B) in Theorem~\ref{thm:compactness_in_CXE_1}. If $A\subseteq C(X,E)$, then the function $G$ defined in the proof of Lemma~\ref{lem:7} is continuous. Thus, taking the open/closed covering $J_1,\ldots,J_k$ of the set $[0,2M]^{N^2}$, we may conclude that all the set $U_1,\ldots,U_N$ are open (closed) in $X$.
\end{remark}

We already know that in some situations both the conditions (B) and (DS) can be used to characterized precompact subsets of $B(X,E)$. Now, we prove a result similar to Proposition~\ref{prop:DS_equivalent_B} in the case when $A$ is a non-empty subset of the space $C(X,E)$.   

\begin{corollary}\label{rem:DS_B_3}
Let $X$ be a compact Hausdorff topological space and let $E$ be a normed space. Moreover, let $A$ be a non-empty subset of $C(X,E)$ such that each section $A(x)$, where $x \in X$, is precompact. Then, $A$ satisfies the condition \textup{(DS)} if and only if  $A-A$  satisfies the condition \textup{(B)}.
\end{corollary} 

\begin{proof}
In view of Proposition~\ref{prop:DS_equivalent_B} all we need to do is to show that if the algebraic difference $A-A$ satisfies the condition (B), then the set $A(X)$ is precompact. So let us fix $\varepsilon>0$. Then, by Theorem~\ref{thm:compactness_BXE} the set $A$ is precompact, and hence it has a finite $\varepsilon$-net $g_1,\ldots,g_n \in A$. As the functions $g_i$ are continuous and the topological space $X$ is compact, the images $g_i(X)$ as well as their union $\bigcup_{i=1}^n g_i(X)$ are compact. For every $f \in A$ and $x \in X$ we thus have $f(x)=f(x)-g_j(x) + g_j(x) \in \ball_E(0,\varepsilon)+g_j(X)$, where the function $g_j$ is an element of the $\varepsilon$-net $g_1,\ldots,g_n$ chosen so that $\norm{f-g_j}_\infty \leq \varepsilon$. Therefore, $A(X)\subseteq \ball_E(0,\varepsilon)+\bigcup_{i=1}^n g_i(X)$. This proves that the set $A(X)$ is precompact (cf. the first part of the proof of Proposition~\ref{prop:DS_equivalent_B}).
\end{proof}

\begin{remark}
It is worth underlining that using a similar approach to the one we used in the proof of Proposition~\ref{prop:DS_equivalent_B}, we can avoid applying Theorem~\ref{thm:compactness_BXE} in the proof of Corollary~\ref{rem:DS_B_3}. We decided, however, to proceed along this less elegant route to simplify the reasoning and not using the same arguments over and over again.     
\end{remark}

Proposition~\ref{prop:DS_equivalent_B} and Corollary~\ref{rem:DS_B_3} together with Theorem~\ref{thm:compactness_BXE} leads to another version of Theorem~\ref{thm:compactness_in_CXE_1} (cf. Theorem~\ref{thm:DS}). 

\begin{theorem}\label{thm:compactness_in_CXE_2}
Assume that either
\begin{enumerate}[label=\textup{(\alph*)}]
 \item\label{it:CXE_2_i} $X$ is an arbitrary topological space and $E=\mathbb R^n$, or
 \item $X$ is a compact Hausdorff topological space and $E$ is an arbitrary normed space.
\end{enumerate}
Then, a non-empty subset $A$ of $C(X,E)$ is precompact if and only if it satisfies the condition \textup{(DS)} and for every $x \in X$ the sections $A(x)$ are precompact.
\end{theorem}

\begin{remark}\label{rem:17}
Note that in Theorem~\ref{thm:compactness_in_CXE_2} in the case~\ref{it:CXE_2_i} when $E=\mathbb R^n$ the set $A(x)$, where $x \in X$, is precompact if and only if it is bounded. Furthermore, instead of assuming that for each $x \in X$ the sections $A(x)$ are bounded in $\mathbb R^n$, we may equivalently assume in Theorem~\ref{thm:compactness_in_CXE_2} that the set $A$ is bounded in $C(X,\mathbb R^n)$. When the normed space $E$ is infinite-dimensional the precompactness of the sections $A(x)$ cannot be replaced with their boundedness, or the boundedness of the whole set $A$. To see this take a separable infinite-dimensional Banach space $E$ and set $X:=\ball_{E^\ast}(0,1)$, where $E^\ast$ is the dual of $E$. From classical facts from functional analysis we know that $X$ is a metrizable compact Hausdorff topological space when endowed with the weak$^\ast$ topology (see~\cite{Conway}*{Theorem~V.3.1 and Theorem~V.5.1}). Given a point $y \in X$ consider the constant mapping $g_y \colon X \to (E^\ast,\norm{\cdot}_{E^\ast})$ given by $g_y(x)=y$ for $x \in X$ and let $A:=\dset{g_y}{y \in X}$. It is straightforward to check that the set $A$ satisfies the condition (DS) (with a family consisting of a single open set $U_1:=X$) and is bounded in $C(X,E^{\ast})$. However, $A(x)=\ball_{E^\ast}(0,1)$ for any $x \in X$. So no section is precompact in the norm of $E^\ast$.
\end{remark}

\begin{remark}\label{rem:open_closed2}
A remark similar to Remark~\ref{rem:open_closed} is also true in the case of Theorem~\ref{thm:compactness_in_CXE_2}, that is, for a set $A\subseteq C(X,E)$ in the condition (DS) we may require all the sets $U_1,\ldots,U_n$ to be either open or closed.
\end{remark}

Finally, from Theorem~\ref{thm:compactness_in_CXE_2}, Remark~\ref{rem:open_closed2} and the fact that every open covering of a compact metric space has a finite subcovering, the Arz\`ela--Ascoli theorem follows. For completeness, let us recall that that a non-empty set $A\subseteq C(X,E)$ is \emph{equicontinuous} if for every $\varepsilon>0$ there exists $\delta>0$ such that for every $x,y \in X$ with $d(x,y)\leq \delta$ and every $f \in A$ we have $\norm{f(x)-f(y)}_E\leq \varepsilon$. 

\begin{theorem}\label{thm:compactness_in_CXE_3}
Let $(X,d)$ be a compact metric space and let $E$ be a normed space. Then, a non-empty subset $A$ of $C(X,E)$ is precompact if and only if it is equicontinuous and for every $x \in X$ the sections $A(x)$ are precompact.
\end{theorem}


\section{Compactness in Lipschitz spaces}
\label{sec:compact_in_lip}

The goal of this section is to provide a full characterization of precompact subsets of both the Lipschitz and little Lipschitz spaces. Although we can define Lipschitz mappings on any metric space, to avoid unnecessary technicalities we will assume that the domains of such maps are compact. 

\begin{notation}
Throughout this section for a given metric space $X$ we will write $\tilde{X}$ for the metric space $(X\times X) \setminus \dset{(x,x)}{x \in X}$. We endow $\tilde{X}$ (and any other subset of $X\times X$) with the maximum metric inherited from $X \times X$.
\end{notation}

\subsection{Comparison function}
A key ingredient when defining classes of Lipschitz continuous mappings is the so-called \emph{comparison function}, that is, a non-zero function $\varphi\colon [0,+\infty)\to[0,+\infty)$ that is right-continuous at $0$, concave and such that $\varphi(0)=0$. A word of caution is in order here. The notion of a comparison function is ambiguous and its definition varies throughout the literature; see~\cite{AKORR}*{Section~2} for a detailed discussion of several types of comparison functions. In the sequel we will use several basic properties of such functions; we gathered them in the following lemma. We skip its proof, because it is straightforward and the arguments can be found scattered around various books on real functions or the internet (especially, on the Mathematics Stack Exchange forum).

\begin{lemma}\label{lem:comparision_function}
Let $\varphi\colon [0,+\infty)\to[0,+\infty)$ be a comparison function. Then, 
\begin{enumerate}[label=\textup{(\alph*)}]
 \item\label{lem:comparision_function_a} $\varphi$ is continuous on $[0,+\infty)$,
 \item\label{lem:comparision_function_b} $\varphi(t)/t$ is non-increasing on $(0,+\infty)$,
 \item\label{lem:comparision_function_c} $\varphi$ is sub-additive, that is, $\varphi(t+s)\leq \varphi(t)+\varphi(s)$ for $t,s\in [0,+\infty)$,
 \item\label{lem:comparision_function_d} $\varphi$ is non-decreasing on $[0,+\infty)$
 \item\label{lem:comparision_function_e} $\varphi(t)>0$ for every $t \in (0,+\infty)$,

 \item\label{lem:comparision_function_f} the limit $\lim_{t \to 0^+} \varphi(t)/t$ exists and is either a positive number, or $+\infty$.
\end{enumerate} 
\end{lemma}

\subsection{Lipschitz spaces}
\label{sec:52}
Fix a comparison function  $\varphi$. Moreover, let $(X,d)$ be a compact metric space and let $(E,\norm{\cdot}_E)$ be a normed space. By $\Lip_\varphi(X,E)$ we denote the class of all $\varphi$-Lipschitz continuous mappings, that is, mappings $f \colon X \to E$ which satisfy the condition
\[
 \abs{f}_{\varphi}:=\sup_{\substack{x,y \in X\\x\neq y}}\frac{\norm{f(x)-f(y)}_E}{\varphi(d(x,y))}<+\infty.
\]
Furthermore, by $\lip_\varphi(X,E)$ we denote the subclass of $\Lip_\varphi(X,E)$,  called the \emph{little Lipschitz space}, consisting of those $\varphi$-Lipschitz continuous mappings which are \emph{locally flat}, that is, satisfy the following condition
\begin{equation}\label{eq:little_lipschitz_def}
 \lim_{\delta\to 0^+} \sup\dset[\Bigg]{\frac{\norm{f(x)-f(y)}_E}{\varphi(d(x,y))}}{\text{$x,y \in X$ with $0<d(x,y) \leq \delta$}} = 0.
\end{equation}
It can be easily checked that $\Lip_\varphi(X,E)$ is a normed space spaces when endowed with the norm $\norm{\cdot}_\varphi:=\norm{\cdot}_{\infty}+\abs{\cdot}_{\varphi}$ (cf.~\cite{krepela}*{Theorem~24}) and $\lip_\varphi(X,E)$ is its closed subspace (in the norm inherited from $\Lip_\varphi(X,E)$). Of course, if $E$ is a Banach space, then both the spaces $\Lip_\varphi(X,E)$ and $\lip_\varphi(X,E)$ are complete.

In general $\lip_\varphi(X,E)\neq \Lip_\varphi(X,E)$; to see this it suffices to set $\varphi(t)=t$, $X:=[0,1]$, $E=\mathbb R$ and take a function $f(x)=x$. Actually, it turns out that in this setting the little Lipschitz space consists only of the constant functions (see~\cite{weaver}*{Example~4.8}). What is more surprising is that this phenomenon holds more generally, that is, if $X$ is a compact connected Riemannian manifold and $\varphi(t)=t$, then $\lip_\varphi(X,\mathbb R)$ is one-dimensional and consists of constant functions only (see~\cite{weaver}*{Example~4.9}).

On the other hand, for any comparison function $\varphi$ such that $\lim_{t \to 0^+} \varphi(t)/t=+\infty$, the class $\Lip(X,\mathbb R)$ of all functions $f \colon X \to \mathbb R$ satisfying the Lipschitz condition with the comparison function $\varphi(t)=t$ is dense in $\lip_\varphi(X, \mathbb R)$ (see~\cite{hanin}*{Proposition~4} and cf.~\cite{weaver2}*{Corollary~1.5}).

We end this part with stating a well-known result (with a classical proof) which describes equivalent ways of defining the little Lipschitz space.

\begin{proposition}\label{prop:equivalent_for_little_lip}
Let $\varphi$ be a comparison function. For a mapping $f \colon X \to E$ defined on a compact metric space $(X,d)$ with values in a normed space $(E,\norm{\cdot}_E)$ the following two conditions are equivalent\textup:
\begin{enumerate}[label=\textup{(\roman*)}]
 \item\label{ref:p1a} $ \lim_{\delta\to 0^+} \sup\dset[\Big]{\frac{\norm{f(x)-f(y)}_E}{\varphi(d(x,y))}}{\text{$x,y \in X$ with $0<d(x,y) \leq \delta$}} = 0$,
 
 \item\label{ref:p1c} for every $\varepsilon>0$ there exists $\delta>0$ such that $\norm{f(x)-f(y)}_E\leq \varepsilon \varphi(d(x,y))$ for all $x,y \in X$ with $d(x,y) \leq \delta$.
\end{enumerate}
\end{proposition} 

\subsection{De Leeuw's map}\label{sec:deLeeuw}
When dealing with Lipschitz continuous mappings, the de Leeuw's map $\Phi$ comes in handy. For each $f \in \Lip_\varphi(X,E)$ it assigns the mapping $\Phi(f) \colon \tilde X \to E$ given by  
\[
\Phi(f)(x,y) = \frac{f(x)-f(y)}{\varphi(d(x,y))}.
\]
The mapping $\Phi(f)$ is clearly continuous and bounded on $\tilde X$. Moreover, $\Phi(f)=\Phi(g)$ if and only if the difference $f-g$ is a constant function. (For more information on the de Leeuw's map see \cite{cobzas_book}*{Section 8.3.1} and \cite{weaver}*{Section 2.4}.) 

\subsection{Compactness criteria in $\Lip_\varphi(X,E)$}
Without further ado let us prove the following result.

\begin{theorem}\label{thm:compact_in_Lip}
Let $X$ be a compact metric space and let $E$ be a normed space. A non-empty subset $A$ of $\Lip_\varphi(X,E)$ is precompact if and only if the following two conditions hold\textup:
\begin{enumerate}[label=\textup{(\roman*)}]
 \item\label{it:compact_in_Lip_i} for every $\varepsilon>0$ there exist finitely many non-empty and open subsets $U_1,\ldots,U_N$ of $\tilde X$ whose union is $\tilde X$ such that for every $i \in \{1,\ldots,N\}$ and every pair $(x,y), (\xi,\eta) \in U_i$ for all $f,g \in A$ we have
 \[
 \abs[\Bigg]{ \norm[\bigg]{\frac{(f-g)(x) - (f-g)(y)}{\varphi(d(x,y))}}_E - \norm[\bigg]{\frac{(f-g)(\xi) - (f-g)(\eta)}{\varphi(d(\xi,\eta))}}_E  }\leq \varepsilon,
 \]

 \item\label{it:compact_in_Lip_ii} for every $x \in X$ the section $A(x)$ is precompact.
\end{enumerate}
\end{theorem}  

\begin{proof}
In the metric space $X$ choose a point $x_\ast$ and consider the linear operator $L\colon \Lip_\varphi(X,E)\to E\times C(\tilde{X},E)$ given by $L(f)=(f(x_\ast), \Phi(f))$, where $\Phi$ is the de Leeuw's map. It is easy to check that $L$ is an isomorphic embedding of $\Lip_\varphi(X,E)$ into $E\times C(\tilde{X},E)$ when the target space is endowed with the norm $\norm{(e,g)}:=\norm{e}_E + \norm{g}_\infty$ for $(e,g) \in E\times C(\tilde{X},E)$. Moreover, note that the condition~\ref{it:compact_in_Lip_i} states that the set $\Phi(A)-\Phi(A) \subseteq C(\tilde X,E)$ satisfies the condition (B) with open sets. The condition~\ref{it:compact_in_Lip_ii}, in turn, is equivalent with the fact that the sets $A(x_\ast)$ and $\Phi(A)(x,y):=\dset{\Phi(f)(x,y)}{ f\in A}$ for $(x,y) \in \tilde X$ are precompact.

Now, let us assume that the non-empty subset $A$ of $\Lip_\varphi(X,E)$ is precompact. Then, precompact are also the sets $\pi_1(L(A))$ and $\pi_2(L(A))$, where $\pi_1$ and $\pi_2$ are the projections onto the first and second factor of $E\times C(\tilde{X},E)$, respectively.  Note that $\pi_1(L(A))=A(x_\ast)$ and $\pi_2(L(A))=\Phi(A)$. So, by Theorem~\ref{thm:compactness_in_CXE_1}, Remark~\ref{rem:open_closed} and our preliminary observations we infer that the conditions~\ref{it:compact_in_Lip_i} and~\ref{it:compact_in_Lip_ii} above are satisfied. On the other hand, if the conditions~\ref{it:compact_in_Lip_i} and~\ref{it:compact_in_Lip_ii} are satisfied, then by~Theorem~\ref{thm:compactness_in_CXE_1} and Remark~\ref{rem:open_closed} the sets $A(x_\ast) \subseteq E$ and $\Phi(A) \subseteq C(\tilde X,E)$ are precompact. Hence, precompact is also their product $A(x_\ast) \times \Phi(A)$. As precompactness is hereditary, this implies the precompactness of $L(A) \subseteq A(x_\ast) \times \Phi(A)$. To end the proof it suffices to note that $A=L^{-1}(L(A))$. 
\end{proof}

In the finite-dimensional case, using Theorem~\ref{thm:compactness_in_CXE_2} instead of Theorem~\ref{thm:compactness_in_CXE_1} and reasoning as in the proof of the above result, we get the following criterion (cf. also Remark~\ref{rem:17}).

\begin{theorem}\label{thm:compact_in_Lip2}
Let $X$ be a compact metric space. A non-empty subset $A$ of $\Lip_\varphi(X,\mathbb R^n)$ is precompact if and only if it is bounded and for every $\varepsilon>0$ there exist finitely many non-empty and open subsets $U_1,\ldots,U_N$ of $\tilde X$ whose union is $\tilde X$ such that for every $i \in \{1,\ldots,N\}$ and every pair $(x,y), (\xi,\eta) \in U_i$ for all $f,g \in A$ we have
 \[
 \norm[\bigg]{\frac{f(x) - f(y)}{\varphi(d(x,y))} - \frac{f(\xi) - f(\eta)}{\varphi(d(\xi,\eta))}}\leq \varepsilon.
 \]
\end{theorem}

\begin{remark}
In Theorems~\ref{thm:compact_in_Lip} and~\ref{thm:compact_in_Lip2} we can clearly replace ``open'' with ``closed'' when talking about the sets $U_1,\ldots,U_N$ (cf.~Remarks~\ref{rem:open_closed} and~\ref{rem:open_closed2}). 
\end{remark}

We end this part with a very natural question whether in Theorem~\ref{thm:compact_in_Lip}~\ref{it:compact_in_Lip_i} and/or Theorem~\ref{thm:compact_in_Lip2} it is possible to replace the open sets $U_1,\ldots,U_N$ with open balls. The answer to this question is negative even  when $E$ is one-dimensional, as the following example shows.

\begin{example}\label{ex:balls_open_sets}
Set $X:=[0,2]$, $\varphi(t):=t$, and consider the function $f \colon X \to \mathbb R$ given by $f(x)=1-\abs{x-1}$, which clearly belongs to $\Lip_\varphi(X,\mathbb R)$. Now, take any finite family of open balls in $\tilde X$ that covers $\tilde X$ and let $B_{\tilde X}((\xi,\zeta),r)$ be a ball belonging to this family which contains infinitely many points of the form $(1-\frac{1}{n},1)$, where $n \in \mathbb N$. Then, $\abs{\xi-1+\frac{1}{n}}<r$ and $\abs{\zeta-1}<r$ for infinitely many $n \in \mathbb N$. Using the continuity of the absolute value, we infer that for all but finitely many $m \in \mathbb N$ we have $\abs{\zeta-1+\frac{1}{m}}<r$. This implies that the open ball 
$B_{\tilde X}((\xi,\zeta),r)$ contains (at least) two points $(1-\frac{1}{n_1},1-\frac{1}{m_1})$ and $(1-\frac{1}{n_2},1-\frac{1}{m_2})$ with $n_1<m_1$ and $n_2>m_2$. And then
\[
 \abs[\Bigg]{\frac{f\bigl(1-\frac{1}{n_1}\bigr) - f\bigl(1-\frac{1}{m_1}\bigr)}{\abs[\big]{\frac{1}{m_1}-\frac{1}{n_1}}} - \frac{f\bigl(1-\frac{1}{n_2}\bigr) - f\bigl(1-\frac{1}{m_2}\bigr)}{\abs[\big]{\frac{1}{m_2}-\frac{1}{n_2}}}}=2.
\]
This shows that replacing the open sets $U_1,\ldots,U_N$ with open balls in the statement of Theorem~\ref{thm:compact_in_Lip2} is not possible even if $E=\mathbb R$ and the set $A$ is a singleton.

A similar example works also in the case of Theorem~\ref{thm:compact_in_Lip}~\ref{it:compact_in_Lip_i}. It suffices to take the set $A:=\{f,g\}$ with the function $f$ defined above and $g(x)=0$ for $x \in X$ as well as the points $(1-\frac{1}{n},1)$ and $(1-\frac{1}{n}, 1+\frac{1}{n})$. (Note that for a sufficiently large $n \in \mathbb N$ those points must lie in the same open ball belonging to a given family of open balls covering $\tilde X$.) The details are left to the reader.
\end{example}

Before we formulate a compactness criterion in the little Lipschitz spaces, similarly to what we did in Section~\ref{sec:discussion} with the condition (B), we investigate the condition \ref{it:compact_in_Lip_i} of Theorem \ref{thm:compact_in_Lip}. We will show that all that really matters is the behaviour of the difference quotients appearing in that condition near the diagonal of $X\times X$. Let us start with two technical lemmas. In the sequel we will not need the full strength of the former one. We decided to state it, however, in a more general form, because it gives some additional insight into to relationship between various conditions considered in the paper.

\begin{lemma}\label{lem:introGH}
Let $X$ be a compact metric space and let $E$ be a normed space. Moreover, assume that $A$ is a non-empty and bounded subset of $\Lip_\varphi(X,E)$ such that the sections $A(x)$, where $x \in X$, are precompact. For a fixed $\delta>0$ let $\tilde X_\delta:=(X\times X)\setminus \bigcup_{x\in X}(B_X(x,\delta)\times B_X(x,\delta))$ and let $\tilde{A}_\delta$ consist of all the mappings $h$ which are the restrictions of $\Phi(f)$ to $\tilde X_\delta$, where $f \in A$ and $\Phi$ is the de Leeuw's map. Then, for any $\varepsilon>0$ there is a finite open cover of $\tilde X_\delta$ such that $\norm{h(x,y)-h(\xi,\eta)}_E\leq \varepsilon$ for any $h\in \tilde{A}_\delta$ and any $(x,y),(\xi,\eta)$ contained in the same element of the cover of $\tilde X_\delta$. 
\end{lemma}

\begin{proof} 
Clearly, $\tilde X_\delta$ is a compact metric space in the maximum metric $\tilde d_\delta$ inherited from $X\times X$ and $\tilde A_\delta \subseteq C(\tilde X_\delta,E)$. Furthermore, observe that precompactness of $A(x)$ for $x\in X$ implies the precompactness of $\tilde A_\delta (x,y):=\dset{h(x,y)}{h \in \tilde A_\delta}$ for $(x,y) \in \tilde X_\delta$. Thus, by Corollary~\ref{rem:DS_B_3} we only need to show that the algebraic difference $\tilde A_\delta - \tilde A_\delta$ satisfies the condition (B) with open sets (cf. Remarks~\ref{rem:open_closed} and~\ref{rem:open_closed2}).

By assumption, the set $A$ is bounded in $\Lip_\varphi(X,E)$. So, there is a positive constant $M$ such that $\norm{f}_\varphi:=\norm{f}_\infty+\abs{f}_\varphi \leq M$ for $f\in A$. We may also assume that $M\geq \sup\dset{\varphi(d(x,y))}{x,y\in X}$. Since the diagonal $\dset{(x,x)}{x\in X}$ is compact and disjoint from the compact set $\tilde X_\delta$, there is a positive constant $m$ such that $m\leq \inf\dset{\varphi(d(x,y))}{(x,y)\in \tilde X_\delta}$. 

Now, fix an arbitrary $\varepsilon>0$. The comparison function is continuous on the non-negative half-axis $[0,+\infty)$ and $\varphi(0)=0$, so there exists $r>0$ such that $\varphi(r) \leq \frac{m^2}{8M^2}\varepsilon$ and $\abs{\varphi(t)-\varphi(s)}\leq \frac{m^2}{8M}\varepsilon$ for any $t,s \in [0,\diam X]$ with $\abs{t-s}\leq 2r$. As the family $U_1,\ldots,U_N$ let us take any finite covering of $\tilde X_\delta$ with open balls in $\tilde X_\delta$ of radius $\frac{1}{2}r$. If the points $(x,y),(\xi,\eta) \in \tilde X_\delta$ belong to the same member of the covering and $h_1,h_2 \in \tilde A_\delta$, then, $h_1=\Phi(f)|_{\tilde X_\delta}$ and $h_2=\Phi(g)|_{\tilde X_\delta}$ for some $f,g \in A$ and 
\begin{align*}
&\abs[\big]{\norm{(h_1-h_2)(x,y)}_E - \norm{(h_1-h_2)(\xi,\eta)}_E}\\
&\ \ = \abs[\Bigg]{\frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}-\frac{\norm{(f-g)(\xi)-(f-g)(\eta)}_E}{\varphi(d(\xi,\eta))}}\\
&\ \ \leq m^{-2} \abs[\big]{ \varphi(d(\xi,\eta))\norm{(f-g)(x)-(f-g)(y)}_E - \varphi(d(x,y))\norm{(f-g)(\xi)-(f-g)(\eta)}_E}\\
&\ \ \leq m^{-2}\varphi(d(\xi,\eta))\norm{(f-g)(x)-(f-g)(\xi) + (f-g)(\eta)-(f-g)(y)}_E\\
&\qquad + m^{-2}\abs[\big]{\varphi(d(x,y))-\varphi(d(\xi,\eta))}\norm{(f-g)(\xi)-(f-g)(\eta)}_E\\
&\ \ \leq  m^{-2}\varphi(d(\xi,\eta))\norm{(f-g)(x)-(f-g)(\xi)}_E + m^{-2}\varphi(d(\xi,\eta))\norm{(f-g)(\eta)-(f-g)(y)}_E\\
&\qquad + m^{-2}\abs[\big]{\varphi(d(x,y))-\varphi(d(\xi,\eta))}\norm{(f-g)(\xi)}_E + m^{-2}\abs[\big]{\varphi(d(x,y))-\varphi(d(\xi,\eta))}\norm{(f-g)(\eta)}_E\\
&\ \ \leq 2m^{-2}M\varphi(d(\xi,\eta))\bigl[\varphi(d(x,\xi)) + \varphi(d(\eta,y))\bigr] + 4m^{-2}M\abs[\big]{\varphi(d(x,y))-\varphi(d(\xi,\eta))}\\
&\ \ \leq 2m^{-2}M^2\bigl[\varphi(d(x,\xi)) + \varphi(d(\eta,y))\bigr] + 4m^{-2}M\abs[\big]{\varphi(d(x,y))-\varphi(d(\xi,\eta))}.
\end{align*}
Recall that the metric space $\tilde X_\delta$ is endowed with the maximum metric $\tilde d_\delta$ inherited from $X \times X$. Hence, $\max\{d(x,\xi),d(\eta,y)\}=\tilde d_\delta((x,y),(\xi,\eta)) \leq r$ and $\abs{d(x,y)-d(\xi,\eta)}\leq d(x,\xi)+d(y,\eta)\leq 2r$. Thus, we obtain
\begin{align*}
 \abs[\big]{\norm{(h_1-h_2)(x,y)}_E - \norm{(h_1-h_2)(\xi,\eta)}_E} \leq 4\cdot \tfrac{1}{8}\varepsilon  + 4\cdot \tfrac{1}{8}\varepsilon =\varepsilon.
\end{align*}
Consequently, the algebraic difference $\tilde A_\delta - \tilde A_\delta$ satisfies the condition (B).
\end{proof}


\begin{lemma}\label{lem:tube}
Let $X$ be a compact metric space and for some $n \in \mathbb N$ let the diagonal of $X\times X$ be covered by a finite collection of open balls ${B}_X(x_i,\frac{1}{n})\times {B}_X(x_i,\frac{1}{n})$, where $i=1,\ldots, m$. Then, there is a number $\delta>0$ such that $\bigcup_{x\in X}({B}_X(x,\delta)\times {B}_X(x,\delta)) \subseteq \bigcup_{i=1}^m({B}_X(x_i,\frac{1}{n})\times{B}_X(x_i,\frac{1}{n}))$.
\end{lemma}

\begin{proof}
Let $n \in \mathbb N$ such that $\dset{(x,x)}{x \in X} \subseteq \bigcup_{i=1}^m ({B}_X(x_i,\frac{1}{n})\times {B}_X(x_i,\frac{1}{n}))$ be fixed. Suppose that the claim is not true, that is, for any $q\in \mathbb{N}$  there is a point $(\xi^q,\eta^q)\in {B}_X(\zeta^q,\frac{1}{q})\times {B}_X(\zeta^q,\frac{1}{q})$ such that $(\xi^q,\eta^q) \notin \bigcup_{i=1}^m({B}_X(x_i,\frac{1}{n})\times {B}_X(x_i,\frac{1}{n}))$. By compactness of $X$ we may assume that the sequence of centers $(\zeta^q)_{q \in \mathbb N}$ converges to a point $z\in X$. As the set $\bigcup_{i=1}^m (B_X(x_i,\frac{1}{n})\times B_X(x_i,\frac{1}{n}))$ is open in $X\times X$, there is some $\delta>0$ such that $B_X(z,\delta)\times B_X(z,\delta)\subseteq  \bigcup_{i=1}^m({B}_X(x_i,\frac{1}{n})\times {B}_X(x_i,\frac{1}{n}))$. But then, for all but finitely many $q \in \mathbb N$ we have ${B}_X(\zeta^q,\frac{1}{q})\times {B}_X(\zeta^q,\frac{1}{q})\subseteq B_X(z,\delta)\times B_X(z,\delta)$, which is impossible.
\end{proof}

Now, we are in position to prove an equivalent formulation of the condition~\ref{it:compact_in_Lip_i} of Theorem~\textup{\ref{thm:compact_in_Lip}}.

\begin{proposition}\label{lem:GH} 
Let $X$ be a compact metric space and let $E$ be a normed space. Assume that $A$ is a non-empty subset of $\Lip_\varphi(X,E)$ such that the sections $A(x)$, where $x \in X$, are precompact. Then, the condition~\ref{it:compact_in_Lip_i} of Theorem~\textup{\ref{thm:compact_in_Lip}} is equivalent to the following one\textup: 
\begin{enumerate}[label=\textup{(\roman*')}]

 \item\label{it:compact_in_Lip_i'} for every $\varepsilon>0$ and $n \in \mathbb N$ there is a radius $\delta>0$ and a finite number of open subsets $U_1,\ldots,U_k$ of $\tilde X$ with $\bigcup_{x\in X} (B_X(x,\delta)\times B_X(x,\delta))\cap\tilde X\subseteq \bigcup_{i=1}^k U_i\subseteq \bigcup_{x\in X} (B_X(x,\frac{1}{n})\times B_X(x,\frac{1}{n}))\cap\tilde X$ such that for any $f,g\in A$ and $i\in \{1,\ldots,k\}$ we have
\[
\sup_{(x,y)\in U_i} \frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}-\inf_{(x,y)\in U_i}\frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}\leq \varepsilon.
\]
\end{enumerate}
\end{proposition}

\begin{proof} 
First, assume that the condition \ref{it:compact_in_Lip_i} of Theorem~\ref{thm:compact_in_Lip} holds and fix any $\varepsilon>0$ and $n \in \mathbb N$. Then, there exist finitely many non-empty and open subsets $U_1,\ldots,U_N$ of $\tilde X$ covering $\tilde X$ such that for every $i \in\{1,\ldots,N\}$ and any $f,g \in A$ we have
\[ 
\sup_{(x,y)\in U_i}\frac{\norm{(f-g)(x) - (f-g)(y)}_E}{\varphi(d(x,y))} -\inf_{(x,y) \in U_i}\frac{\norm{(f-g)(x) - (f-g)(y)}_E}{\varphi(d(x,y))}\leq \varepsilon.
\]
As the set $\tilde X$ is open in $X\times X$, so are the members of the collection $U_1,\ldots,U_N$. Since the diagonal $\dset{(x,x)}{x\in X}$ is a compact subset of the metric space $X\times X$, there is a finite number of points $x_1,\ldots,x_k \in X$ such that $\dset{(x,x)}{x\in X}\subseteq \bigcup_{j=1}^k (B_X(x_j,{\frac{1}{n}})\times B_X(x_j,{\frac{1}{n}}))$. For each $i\in \{1,\ldots,N\}$ and $j\in \{1,\ldots,k\}$ set $V_{i,j}:=(B_X(x_j,{\frac{1}{n}})\times B_X(x_j,{\frac{1}{n}}))\cap U_i$. By Lemma~\ref{lem:tube} it is clear that the family consisting of all the non-empty sets $V_{i,j}$ satisfies all the requirements of the condition \ref{it:compact_in_Lip_i'}.

Now, we will move to the second part of the proof, which we will divide into two steps. We begin with showing that, under the assumption of precompactness of the sections $A(x)$, the condition~\ref{it:compact_in_Lip_i'} implies that the set $A$ is bounded in $\Lip_\varphi(X,E)$. Let $\varepsilon=n=1$ and let $U_1,\ldots, U_k$ be the sets appearing in the condition~\ref{it:compact_in_Lip_i'}; we do not mention the radius $\delta>0$ here, as we will not need it in this step. Suppose that $\sup_{f\in A}\sup_{x\in X}\norm{f(x)}_E=+\infty$. Then, there exist two sequences: $(x_m)_{m\in \mathbb{N}}$ in $X$ and $(f_m)_{m\in \mathbb{N}}$ in $A$ for which $\lim_{m\to\infty}\norm{f_m(x_m)}_E=+\infty$.   Since the metric space $X$ is compact we may assume that $\lim_{m\to\infty}x_m=x\in X$. If $x_{m_l}=x$ for infinitely many indices $l \in \mathbb N$, we would get $\lim_{l\to\infty}\norm{f_{m_l}(x)}_E=+\infty$, meaning that $A(x)$ is not precompact, which is absurd. Therefore, we may assume that $x_n \neq x$ for all $n \in \mathbb N$. Let $i \in \{1,\ldots,k\}$ be such an index that $(x,x_m)\in U_i$ for infinitely many $m$'s. Clearly, it is possible to find such $i$, because there are only finitely many sets $U_1,\ldots,U_k$. Passing to a subsequence yet another time, we may state that $(x,x_m)\in U_i$ for all $m \in \mathbb N$. Now, pick a function $f \in A$ and a point $(\xi,\eta) \in U_i$. Note that because of the compactness of the metric space $X$ and the continuity of the functions $\varphi$ and $f$ there is a constant $R>0$ with $\varphi(d(x_m,x))\leq R$ and $\norm{f(x_m)}_E\leq R$ for $m \in \mathbb N$. Observe also that, since the sections $A(\cdot)$ are precompact, (adjusting the constant $R$ if necessary) we have $\norm{(f_m-f)(y)}_E\leq R$ for $m \in \mathbb N$, where $y \in \{\xi,\eta,x\}$. Thus, for all $m \in \mathbb N$ we get
\begin{align*}
&\norm{f_m(x_m)}_E\\
 &\ \ \leq \norm{f_m(x_m)-f(x_m)}_E+\norm{f(x_m)}_E\\
 &\ \ \leq \norm{(f_m-f)(x_m)-(f_m-f)(x)}_E+\norm{(f_m-f)(x)}_E+\norm{f(x_m)}_E\\
 &\ \ = \varphi(d(x_m,x))\cdot \frac{\norm{(f_m-f)(x_m)-(f_m-f)(x))}_E}{\varphi(d(x_m,x))}+\norm{(f_m-f)(x)}_E+\norm{f(x_m)}_E\\
 &\ \ \leq  \varphi(d(x_m,x)) \cdot \sup_{(p,q)\in U_i}\frac{\norm{(f_m-f)(p)-(f_m-f)(q))}_E}{\varphi(d(p,q))}+\norm{(f_m-f)(x)}_E+\norm{f(x_m)}_E\\
 &\ \ \leq \varphi(d(x_m,x))\biggl(1+\frac{\norm{(f_m-f)(\xi)-(f_m-f)(\eta))}_E}{\varphi(d(\xi,\eta))}\biggr)+\norm{(f_m-f)(x)}_E+\norm{f(x_m)}_E\\
&\ \ \leq 3R + 2R^2/\varphi(d(\xi,\eta))<+\infty,
\end{align*}
which leads to a contradiction. Hence, $M:=\sup_{f\in A}\norm{f}_\infty<+\infty$.

Using a similar reasoning to the above one, we will show that the set $A$ is also bounded in the Lipschitz semi-norm $\abs{\cdot}_\varphi$. Suppose on the contrary that $\sup_{f\in A}\sup_{(x,y)\in \tilde X}\norm{f(x)-f(y)}_E/\varphi(d(x,y))=+\infty$. Hence, there are sequences $(x_m, y_m)_{m\in \mathbb{N}}$ in $\tilde X$ and $(f_m)_{m\in \mathbb{N}}$ in $A$ for which $\lim_{n\to\infty} \norm{f_m(x_m)-f_m(y_m)}_E/\varphi(d(x_m,y_m))=+\infty$. Let $i \in \{1,\ldots,k\}$ be such an index that $(x_m,y_m) \in U_i$ for infinitely many $m$'s. Passing to a subsequence if necessary, we may actually claim that the above relation holds for all $m \in \mathbb N$. If we fix $f \in A$ and $(\xi,\eta)\in U_i$, then in view of the condition~\ref{it:compact_in_Lip_i'} for all $m \in \mathbb N$ we have
\begin{align*}
 \frac{\norm{f_m(x_m)-f_m(y_m)}_E}{\varphi(d(x_m,y_m))}& \leq \frac{\norm{(f_m-f)(x_m)-(f_m-f)(y_m)}_E}{\varphi(d(x_m,y_m))}+\frac{\norm{f(x_m)-f(y_m)}_E}{\varphi(d(x_m,y_m))}\\
 & \leq \sup_{(p,q)\in U_i}\frac{\norm{(f_m-f)(p)-(f_m-f)(q)}_E}{\varphi(d(p,q))}+\frac{\norm{f(x_m)-f(y_m)}_E}{\varphi(d(x_m,y_m))}\\
 & \leq 1+\frac{\norm{(f_m-f)(\xi)-(f_m-f)(\eta)}_E}{\varphi(d(\xi,\eta))}+\frac{\norm{f(x_m)-f(y_m)}_E}{\varphi(d(x_m,y_m))}\\
 & \leq 1 + 4M/\varphi(d(\xi,\eta)) + \abs{f}_{\varphi}<+\infty,
\end{align*}
a contradiction.  Therefore, $\sup_{f\in A} \abs{f}_{\varphi}<+\infty$. Consequently, the set $A$ is bounded in $\Lip_\varphi(X,E)$. 

Now, we focus on the second -- main -- step of this part of the proof. If the condition \ref{it:compact_in_Lip_i'} holds, then for any fixed $\varepsilon>0$ and $n \in \mathbb N$ there is a radius $\delta>0$ and a finite number of open subsets $U_1,\ldots,U_k$ of $\tilde X$ with $\bigcup_{x\in X} (B_X(x,\delta)\times B_X(x,\delta))\cap\tilde X\subseteq \bigcup_{i=1}^k U_i\subseteq \bigcup_{x\in X} (B_X(x,\frac{1}{n})\times B_X(x,\frac{1}{n}))\cap\tilde X$ such that for any $f,g\in A$ and $i\in \{1,\ldots,k\}$ we have
\[
\sup_{(x,y)\in U_i} \frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}-\inf_{(x,y)\in U_i}\frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}\leq \varepsilon.
\]
From the previous step we know that the set $A$ is bounded in $\Lip_{\varphi}(X,E)$. Hence, we can apply Lemma~\ref{lem:introGH} with $\frac{1}{2}\varepsilon$ and $\frac{1}{4}\delta$ to obtain a finite family $W_{k+1},\ldots, W_N$ of open subsets of $\tilde X_{\frac{1}{4}\delta}$ such that 
\[
 \norm[\Bigg]{\frac{f(x)-f(y)}{\varphi(d(x,y))} - \frac{f(\xi)-f(\eta)}{\varphi(d(\xi,\eta))}}_E \leq \frac{1}{2}\varepsilon
\]
for any $f \in A$ and $(x,y),(\xi,\eta)$ belonging to the same member of the collection $W_{k+1},\ldots, W_N$. For every $i=k+1,\ldots,N$ let $U_i:=W_i\setminus \dset{(x,y) \in X\times X}{\inf_{z \in X}\max\{d(x,z),d(y,z)\}\leq \frac{1}{2}\delta}$. We may clearly assume that all the sets $U_i$ are non-empty; otherwise we just remove the empty ones from our collection.

We claim that the sets $U_i$, where  $i \in \{k+1,\ldots,N\}$, are open in $X \times X$ and hence in $\tilde X$. So, let us fix a set $U_j$ with $j \in \{k+1,\ldots,N\}$  and take any point $(a,b) \in U_j$. Since $(a,b) \in W_j$ and the set $W_j$ is open in $\tilde X_{\frac{1}{4}\delta}$, there is a radius $r>0$ such that $(B_X(a,r) \times B_X(b,r)) \cap \tilde X_{\frac{1}{4}\delta}\subseteq W_j$. Set $\rho:=\inf_{z \in X}\max\{d(a,z),d(b,z)\}$ and $R:=\min\{r,\frac{1}{4}\delta,\rho-\frac{1}{2}\delta\}$. Note that by definition we have $\rho>\frac{1}{2}\delta$, and so the radius $R$ is well-defined. We will show now that the open ball $B_{X}(a,R)\times B_X(b,R)$ in $X\times X$ is included in $U_j$. Clearly, it is included in $B_X(a,r) \times B_X(b,r)$. Furthermore,  $B_{X}(a,R)\times B_X(b,R) \subseteq \tilde X_{\frac{1}{4}\delta}$, since otherwise there would exist points $u,p,q \in X$ such that $d(a,p)<R$, $d(b,q)<R$, $d(p,u)<\frac{1}{4}\delta$ and $d(q,u)<\frac{1}{4}\delta$ and we would have $d(u,a)\leq d(a,p)+d(p,u)<\frac{1}{2}\delta$ and, similarly, $d(u,b)<\frac{1}{2}\delta$. But this is impossible, because $\rho= \inf_{z \in X}\max\{d(a,z),d(b,z)\}>\frac{1}{2}\delta$. Therefore, $B_{X}(a,R)\times B_X(b,R) \subseteq (B_X(a,r) \times B_X(b,r))\cap \tilde X_{\frac{1}{4}\delta} \subseteq W_j$. 

It remains to show that the sets $\dset{(x,y) \in X\times X}{\inf_{z \in X}\max\{d(x,z),d(y,z)\}\leq \frac{1}{2}\delta}$ and $B_{X}(a,R)\times B_X(b,R)$ are disjoint. Suppose on the contrary that there is a point $(p,q) \in  B_{X}(a,R)\times B_X(b,R)$ such that $\inf_{z \in X}\max\{d(p,z),d(q,z)\}\leq \frac{1}{2}\delta$. As the function $z \mapsto \max\{d(p,z),d(q,z)\}$ is continuous and the metric space $X$ is compact, there is a point $z_\ast\in X$ such that $\inf_{z \in X}\max\{d(p,z),d(q,z)\}=\max\{d(p,z_\ast),d(q,z_\ast)\}$. Now, we have
\begin{align*}
\rho&=\inf_{z \in X}\max\{d(a,z),d(b,z)\}\\
&\leq \max\{d(a,z_\ast),d(b,z_\ast)\}\\
&\leq \max\{d(a,p),d(b,q)\} + \max\{d(p,z_\ast),d(q,z_\ast)\}\\
&< \rho-\tfrac{1}{2}\delta + \tfrac{1}{2}\delta = \rho,
\end{align*}
a contradiction. Thus, the sets $\dset{(x,y) \in X\times X}{\inf_{z \in X}\max\{d(x,z),d(y,z)\}\leq \frac{1}{2}\delta}$ and $B_{X}(a,R)\times B_X(b,R)$ are disjoint, and $B_{X}(a,R)\times B_X(b,R) \subseteq U_j$.

The rest of the proof is straightforward, as it is not difficult to see that the family of sets $U_1,\ldots, U_k, U_{k+1},\ldots, U_N$ satisfies all the requirements of the condition~\ref{it:compact_in_Lip_i} of Theorem~\ref{thm:compact_in_Lip}.
\end{proof}

From Proposition~\ref{lem:GH} we immediately get the following second compactness criterion in the space of Lipschitz continuous functions $\Lip_\varphi(X,E)$.

\begin{theorem}\label{thm:compact_in_Lip'}
Let $X$ be a compact metric space and let $E$ be a normed space. A non-empty subset $A$ of $\Lip_\varphi(X,E)$ is precompact if and only if the condition~\ref{it:compact_in_Lip_i'} of Proposition~\textup{\ref{lem:GH}} holds and for every $x \in X$ the section $A(x)$ is precompact.
\end{theorem}  



\subsection{Compactness criterion in $\lip_\varphi(X,E)$}
\label{sec:55}
Finally, we characterize precompact subsets of the little Lipschitz space $\lip_\varphi(X,E)$. This time there is no need to distinguish between $E$ being finite- or infinite-dimensional. There is a very simple explanation of this phenomenon. Namely, if $f \in \lip_{\varphi}(X,E)$, then the mapping $\Phi(f)$, where $\Phi$ is the de Leeuw's map, can be extended continuously from $\tilde X$ over the whole (compact) product $X \times X$ by simply putting $\Phi(f)(x,x)=0$ for $x \in X$. 

\begin{theorem}\label{th:little:lip}
Let $X$ be a compact metric space. A non-empty subset $A$ of $\lip_\varphi(X,E)$ is precompact if and only if the set $A$ is \emph{uniformly locally flat}, that is, for every $\varepsilon>0$ there exists $\delta>0$ such that for any $x,y \in X$ with $d(x,y)\leq \delta$ we have $\norm{f(x)-f(y)}_E \leq \varepsilon \varphi(d(x,y))$, and for every $x \in X$ the section $A(x)$ is precompact.
\end{theorem}

\begin{proof}
If $A$ is a uniformly locally flat family of functions, then given $\varepsilon>0$ there exists $\eta>0$ such that $\norm{f(x)-f(y)}_E\leq \frac{1}{2}\varepsilon \varphi(d(x,y))$ 
for all $f \in A$ and all $x,y \in X$ with $d(x,y)\leq \eta$. (Note that to make it easier to refer to the previous results we have changed the notation a little bit here.) Let us fix $n \in \mathbb N$ and choose $m \in \mathbb N$ so that $\frac{1}{m}\leq \min\{\frac{1}{2}\eta,\frac{1}{n}\}$. Furthermore, pick a finite family of points $x_1,\ldots,x_k \in X$ such that the diagonal of $X\times X$ is covered by the union of balls  $B_X(x_i,\frac{1}{m})\times B_X(x_i,\frac{1}{m})$, where $i\in \{1,\ldots,k\}$.
 Define $U_i:=(B_X(x_i,\frac{1}{m})\times B_X(x_i,\frac{1}{m}))\cap \tilde X$ for $i\in \{1,\ldots,k\}$. Then, clearly $\bigcup_{i=1}^k U_i \subseteq \bigcup_{x\in X}(B_X(x,\frac{1}{n})\times B_X(x,\frac{1}{n}))\cap \tilde X$. Moreover, by Lemma~\ref{lem:tube} (applied with $m$) there is a radius $\delta>0$ such that $\bigcup_{x\in X}(B_X(x,\delta)\times B_X(x,\delta))\cap \tilde X \subseteq \bigcup_{i=1}^k U_i$. If we fix $f,g \in A$ and $i\in \{1,\ldots,k\}$, then by the uniform local flatness of $A$ we obtain
\begin{align*}
& \sup_{(x,y)\in U_i} \frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}-\inf_{(x,y)\in U_i}\frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))}\\
&\qquad \leq \sup_{(x,y)\in U_i} \frac{\norm{(f-g)(x)-(f-g)(y)}_E}{\varphi(d(x,y))} \leq \varepsilon.
\end{align*}
Thus, by Theorem~\ref{thm:compact_in_Lip'} we can conclude that $A$ is precompact in $\Lip_\varphi(X,E)$, and consequently in $\lip_\varphi(X,E)$.

To prove the opposite implication, let us assume that $A$ is a precompact subset of $\lip_\varphi(X,E)$ and $f_1,\ldots,f_m\in A$ is its finite $\frac{1}{2}\varepsilon$-net. Choose $\delta>0$ so that $\max_{1\leq i\leq m}\norm{f_i(x)-f_i(y)}_E\leq \frac{1}{2}\varepsilon\varphi(d(x,y))$. For any $f\in A$ there is some $j\in \{1,\ldots,m\}$ for which we have $\norm{f-f_j}_\varphi \leq \frac{1}{2}\varepsilon$, and thus
\begin{align*}
\norm{f(x)-f(y)}_E &\leq \norm{(f-f_j)(x)-(f-f_j)(y)}_E+\norm{f_j(x)-f_j(y)}_E\\
& \leq \tfrac{1}{2}\varepsilon\varphi(d(x,y))+ \tfrac{1}{2}\varepsilon\varphi(d(x,y))=\varepsilon\varphi(d(x,y)),
\end{align*}
whenever $x,y \in X$ are such that $d(x,y)\leq\delta$. This shows that the family $A$ is uniformly locally flat. To end the proof it suffices to observe that the Lipschitz norm $\norm{\cdot}_\varphi$ is stronger than the supremum norm $\norm{\cdot}_\infty$, and hence precompactness of $A$ implies precompactness of each section $A(x)$, where $x \in X$.
\end{proof}


\begin{remark}
Theorem~\ref{th:little:lip} in the case when $E=\mathbb R$ and $\varphi(t)=t$ was proven by Johnson using a different approach than ours (see~\cite{johnson}*{Theorem~3.2}). A similar result was also obtained by Garc\'{\i}a-Lirola \emph{et al.} who, motivated by some considerations of duality, studied compactness in the space of those functions $f \colon X \to \mathbb R$ defined on a bounded pointed metric space $(X,x_\ast, d)$ which are little Lipschitz, vanish at the designated point $x_\ast \in X$ and are continuous with respect to the topology $\tau$ on $X$ such that $(X,\tau)$ is compact and $d$ is $\tau$-lower semi-continuous (see~\cite{GPR}*{Lemma~2.7}).
\end{remark}


\subsection*{Acknowledgement}
We thank Luis Garc\'{\i}a-Lirola, Colin Petitjean and Abraham Rueda Zoca for bringing their paper~\cite{GPR} to our attention.


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