Metric Compactness Criteria Involving Sequences of Mappings and a Proof of the Ascoli–Arzelà Theorem with the use of Bernstein Polynomials

Abstract

We establish inter alia a compactness criterion in metric spaces involving a sequence of completely continuous mappings, which is continuously convergent, in the sense of H. Hahn, to the identity mapping. For Banach spaces, the linear version of that result coincides with the compactness theorem due to Mazur. We also present a new proof of the Ascoli–Arzelà theorem, in which we use the above compactness criterion applied to the sequence of Bernstein operators.

1 Introduction and Preliminaries

Let (X, d) be a metric space, $x\in X$ and $r>0$. By B(x, r), we denote the open ball of radius r centered at x. Recall that a set $A\subseteq X$ is totally bounded, if for any $\varepsilon >0$, there exists a finite $\varepsilon$-net $C\subseteq X$ for A, i.e., C is finite and $A\subseteq \bigcup _{x\in C}B(x,\varepsilon )$. A set A is said to be relatively compact, if it is contained in a compact set; equivalently, if its closure is compact. If a set is relatively compact, then it is totally bounded. On the other hand, if (X, d) is complete, then every totally bounded set is relatively compact, so the two notions are then equivalent.

In Sect. 2 we first give a simple general characterization of totally bounded sets in metric spaces involving sequences of continuous mappings acting on these sets; see Proposition 2.1. For Banach spaces, we modify that result and with the help of Krasnosel’skiĭ’s [12] extension theorem, we obtain a characterization of relatively compact sets using this time sequences of completely continuous mappings acting on the whole space; see Proposition 2.2. Recall that a mapping $F:X\rightarrow Y$, where X and Y are metric spaces, is completely continuous, if F is continuous and the image F(A) of any bounded set $A\subseteq X$ is relatively compact in Y; see, e.g., [6, p. 112]. The latter result has an interesting consequence: we may say that every Banach space X has the nonlinear compact approximation property, i.e., for any compact set $A\subseteq X$ and every $\varepsilon >0$, there exists a completely continuous mapping $F:X\rightarrow X$ such that $\Vert x-Fx\Vert <\varepsilon$ for any $x\in A$. Both the above results deal with sequences of mappings, which depend on considered sets. However, in a reasonable class of spaces it is possible to test the compactness of a set by using one universal sequence of completely continuous mappings. That is shown in our Theorem 2.1 which gives the compactness criterion in a complete metric space having the property that there exists a sequence of completely continuous mappings, which is continuously convergent to the identity mapping. Recall that a sequence of mappings $F_n:X\rightarrow Y$, where X and Y are metric spaces, is continuously convergent to a mapping $F:X\rightarrow Y$ (we use then the denotation $F_n\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,F$), if for any $x\in X$ and any sequence $(x_n)$, $x_n\rightarrow x$ implies that $F_n(x_n)\rightarrow F(x)$. Clearly, if $F_n\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,F$, then $(F_n)$ is pointwise convergent to F. On the other hand, if $F_n\rightrightarrows F$, i.e., $(F_n)$ is uniformly convergent to F, then $F_n\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,F$ if, and only if, F is continuous (cf. [13, Theorem 3, p. 220]). The notion of continuous convergence was introduced by Hahn [7, p. 238]; see also [13, p. 197].

The restriction of Theorem 2.1 to Banach spaces and sequences of linear operators, coincides with Mazur’s compactness criterion, which was first mentioned, without a proof, in Banach’s French monograph [1, p. 237]. However, the boundedness assumption of a set has been inadvertently omitted in Banach’s book as well as in its 1987 English edition. As noticed by one of the referees, a correct statement of this compactness criterion has been given and proved by Phillips [17, Theorem 3.7]. In fact, we suppose that Phillips did not know Mazur’s result because he noted in a footnote [17, p. 526] that the compactness criterion was suggested to him by T.H. Hildebrandt, the supervisor of his Ph.D. thesis. Nevertheless, the need of using the boundedness condition is well known to specialists and some later articles discussing Mazur’s criterion, like the paper [9], assumed the boundedness of a set and mentioned Banach’s book as reference; others refer to Phillips [17].

As an application, we present in Sect. 3 a new proof of the Ascoli–Arzelà, in which we use the compactness criterion from the previous section applied to the space C([0, 1]) and the sequence of Bernstein operators. In fact, a linear version of Theorem 2.1 is sufficient for that proof, so we would welcome other applications, where Theorem 2.1 would be used in its full generality. Let us notice that in the literature we may find various proofs of the Ascoli–Arzelà theorem. In many classical textbooks on functional analysis the authors either construct a finite $\varepsilon$-net for a suitable family of functions (see, e.g., [4] or [18, p. 394]), or use a diagonalization argument as done, e.g., in [10, p. 154]. There are also other lesser-known approaches that we want to mention now. In particular, the proof given by Ullrich [19] is based on the Tychonoff compactness theorem. Nagy [16] presented a functional analytic proof with the help of the Banach–Alaoglu theorem. Beer [2] derived the Ascoli–Arzelà theorem from the Zarankiewicz compactness theorem for sequences of closed sets in a separable metric space. Hanche-Olsen and Holden [8] established a clever simple lemma on metric compactness, which implies the Ascoli–Arzelà theorem as well as the Kolmogorov–Riesz compactness theorem for $L^p({\mathbb R}^n)$. At last, recently, Wójtowicz [21] provided yet another proof that uses the Stone–Čech compactification technique.

For our proof of the Ascoli–Arzelà theorem, it is crucial that the sequence of Bernstein operators is uniformly convergent to the identity mapping on any bounded and equicontinuous family of functions on [0, 1]; see Proposition 3.1. Actually, in our proof of this fact we almost repeat an argument from Bernstein’s proof of the Weierstrass approximation theorem (see, e.g., [5, pp. 158–159]).

Finally, in the appendix, we consider the pointwise version of the notion of continuous convergence and we give its characterization, which let us also characterize a (global) continuous convergence.

2 Compactness Criteria in Metric and Banach Spaces Involving a Sequence of Mappings

For $A,B\subseteq X$, we set $D(A,B):=\sup _{x\in A}d(x,B)$, where $d(x,B):=\inf _{y\in B} d(x,y)$. By I, we denote the identity mapping on X, and by $I|_A$, its restriction to set A.

We start with the following simple characterization of totally bounded sets in metric spaces.

Proposition 2.1

Let (X, d) be a metric space and $A\subseteq X$. The following statements are equivalent:

• (i) A is totally bounded;

• (ii) there exists a sequence of continuous mappings $F_n:A\rightarrow X$ such that $F_n(A)$ is totally bounded for each $n\in {\mathbb N}$, and $(F_n)$ is uniformly convergent to $I|_A$;

• (iii) there exists a sequence $(A_n)$ of totally bounded subsets of X such that $D(A,A_n)\rightarrow 0$.

Proof

(i)$\Rightarrow$(ii): It suffices to set $F_n:=I|_A$ for $n\in {\mathbb N}$.

(ii)$\Rightarrow$(iii): Set $A_n:=F_n(A)$ and $\alpha _n:=\sup _{x\in A}d(x,F_nx)$ for $n\in {\mathbb N}$. By (ii), each $A_n$ is totally bounded and $\alpha _n\rightarrow 0$. For any $x\in A$, we have that $F_nx\in A_n$, so

$$\begin{aligned} d(x,A_n)\le d(x,F_nx)\le \alpha _n, \end{aligned}$$

which yields that $D(A,A_n)\le \alpha _n$. Since $(\alpha _n)$ converges to 0, so does $(D(A,A_n))$.

(iii)$\Rightarrow$(i): Fix $\varepsilon >0$. By (iii), there exists $k\in {\mathbb N}$ such that $D(A,A_k)<\varepsilon /2$. Since $A_k$ is totally bounded, there exists a finite $\frac{\varepsilon }{2}$-net $\{x_1,\ldots ,x_p\}$ for $A_k$. Now, fix $x\in A$. Then

$$\begin{aligned} d(x,A_k)\le D(A,A_k)<\frac{\varepsilon }{2}, \end{aligned}$$

so there exists $y\in A_k$ such that $d(x,y)<\varepsilon /2$. Next, there is $i\in \{1,\ldots ,p\}$ such that $d(y,x_i)<\varepsilon /2$. Hence we infer that $d(x,x_i)<\varepsilon$. Thus $\{x_1,\ldots ,x_p\}$ is an $\varepsilon$-net for A. $\square$

We pass to the case of Banach spaces, in which we may somewhat modify Proposition 2.1 as follows. In the proof we use Krasnosel’skiĭ’s extension theorem in the form given in [6, p. 118].

Proposition 2.2

Let X be a Banach space and $A\subseteq X$. The following statements are equivalent:

• (i) A is relatively compact;

• (ii) A is bounded and there exists a sequence of completely continuous mappings $F_n:X\rightarrow X$ such that $(F_n|_A)$ is uniformly convergent to $I|_A$.

Proof

(i)$\Rightarrow$(ii): Clearly, A is bounded and $\overline{A}$ is compact. By Krasnosel’skiĭ’s theorem, $I|_{\overline{A}}$ is extendable to a continuous mapping $F:X\rightarrow X$ such that F(X) is relatively compact. So the more F is completely continuous and $F|_A=I|_A$. Thus it is enough to set $F_n:=F$ for $n\in {\mathbb N}$.

(ii)$\Rightarrow$(i): Since each $F_n$ is completely continuous and A is bounded, we get that $F_n(A)$ is relatively compact, hence totally bounded for any $n\in {\mathbb N}$. By Proposition 2.1 applied to the sequence $(F_n|_A)$, we obtain that A is totally bounded. Since X is complete, we infer that A is relatively compact. $\square$

Remark 2.1

Recall that a Banach space X is said to have the compact approximation property (C.A.P. in short), if for any compact set $A\subseteq X$ and any $\varepsilon >0$, there exists a compact linear operator $T:X\rightarrow X$ (i.e., T(B) is relatively compact whenever B is a bounded subset of X) such that $\Vert x-Tx\Vert <\varepsilon$ for any $x\in A$; see, e.g., [14, p. 94]. Equivalently, for any compact set $A\subseteq X$, there exists a sequence of linear compact operators $T_n:X\rightarrow X$ such that $(T_n|_A)$ converges uniformly to $I|_A$. It is known that there exist Banach spaces which fail to have the C.A.P. [14, p. 94]. On the other hand, we may say that Proposition 2.2 ((i)$\Rightarrow$(ii)) states that every Banach space has the nonlinear C.A.P.

Let us notice that a sequence $(F_n)$ appearing in Proposition 2.2depends on set A. However, in a reasonable class of spaces there exist universal sequences of completely continuous mappings, which detect every relatively compact set. Our next result deals with such a case.

Theorem 2.1

Let (X, d) be a complete metric space such that there exists a sequence of completely continuous mappings $F_n:X\rightarrow X$, which is continuously convergent to the identity mapping I. Then, for any set $A\subseteq X$, A is relatively compact if, and only if, A is bounded and the sequence $(F_n|_A)$ is uniformly convergent to $I|_A$.

Proof

($\Rightarrow$): Clearly, A is bounded since it is relatively compact. Since $\overline{A}$ is compact and also $F_n|_{\overline{A}}\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,I|_{\overline{A}}$, Kuratowski’s theorem [13, Theorem 5, p. 221] implies that $F_n|_{\overline{A}}\rightrightarrows I|_{\overline{A}}$; in particular, $F_n|_A\rightrightarrows I|_A$.

($\Leftarrow$): By Proposition 2.1, we may infer that A is totally bounded. Since (X, d) is complete, A is relatively compact. $\square$

The following example shows that in general we cannot weaken the assumptions of Theorem 2.1 by assuming only that $(F_n)$ is pointwise convergent to I.

Example 2.1

Set $X:={\mathbb R}$ and endow X with the Euclidean metric. For $n\in {\mathbb N}$ and $x\in X$, define

$$\begin{aligned} F_n(x):=\left\{ \begin{array}{ll} (2n+1)x &{} { \text{ if } }x\in \left[ 0,\frac{1}{2n}\right] ,\\ (-2n+1)x+2 &{} { \text{ if } } x\in \left( \frac{1}{2n},\frac{1}{n}\right) ,\\ x &{} { \text{ if } } x\in (-\infty ,0)\cup \left[ \frac{1}{n},\infty \right) . \end{array}\right. \end{aligned}$$

It is easily seen that $(F_n)$ is pointwise convergent to I. Moreover, each $F_n$ is completely continuous since it is continuous and for any bounded set $B\subseteq {\mathbb R}$, $F_n(B)$ is contained in the compact set $F_n(\overline{B})$. Set $A:=[0,1]$. Clearly, A is compact, but $(F_n|_A)$ does not converge uniformly to $I|_A$ since

$$\begin{aligned} \sup _{x\in [0,1]}|x-F_n(x)|\ge \left| \frac{1}{2n}-F_n\left( \frac{1}{2n}\right) \right| =1. \end{aligned}$$

However, if X is a Banach space and mappings $F_n$ are linear, then both types of convergence are equivalent in view of the following

Lemma 2.1

Let X be a Banach space, Y be a normed linear space, $(T_n)$ be a sequence of linear bounded operators from X to Y, and $T:X\rightarrow Y$. Then $(T_n)$ is continuously convergent to T if, and only if, $(T_n)$ is pointwise convergent to T.

Proof

Part ‘only if’ is trivial. So assume that $(T_n)$ is pointwise convergent to T. By the uniform boundedness principle, $M:=\sup _{n\in {\mathbb N}}\Vert T_n\Vert <\infty$. Let $(x_n)$ be convergent to a point $x\in X$. Then we have that

$$\begin{aligned} \Vert T_nx_n-Tx\Vert \le \Vert T_nx_n-T_nx\Vert +\Vert T_nx-Tx\Vert . \end{aligned}$$

By hypothesis, $\Vert T_nx-Tx\Vert \rightarrow 0$, whereas

$$\begin{aligned} \Vert T_nx_n-T_nx\Vert \le \Vert T_n\Vert \Vert x_n-x\Vert \le M\Vert x_n-x\Vert \rightarrow 0. \end{aligned}$$

Consequently, we obtain that $T_nx_n\rightarrow Tx$, so $T_n\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,T$. $\square$

As an immediate consequence of Theorem 2.1 and Lemma 2.1, we get the following compactness criterion in Banach spaces due to Mazur [1, p. 237]. Notice that for a linear operator, its compactness means the same as its complete continuity in the sense of the definition used here from [6, p. 112]. However, as pointed out by one of the referees, a different terminology is used in linear functional analysis: here compact operators and completely continuous operators are in general different objects; see, e.g., [15, p. 336].

Corollary 2.1

Let X be a Banach space such that there exists a sequence of linear compact operators $T_n:X\rightarrow X$, which is pointwise convergent to the identity operator I. Then, for any set $A\subseteq X$, A is relatively compact if, and only if, A is bounded and the sequence $(T_n|_A)$ uniformly converges to $I|_A$.

Let us also notice that a particular version of Corollary 2.1 is given (without references) in Kadets’ monograph [11, p. 293]; here operators $T_n$ are assumed to be of finite rank. Then, a Banach space admitting such a sequence $(T_n)$ is said by Kadets to have the pointwise approximation property (P.A.P. in short). So adhering to this terminology, the property of a Banach space which possesses a sequence of linear compact operators pointwise convergent to the identity operator, could be called the pointwise compact approximation property (P.C.A.P. in short). It is clear that the P.A.P. implies the P.C.A.P. However, the reverse implication is not true. That can be deduced, though not directly, from the result of Willis [20], who constructed a Banach space X that does not have the approximation property (A.P. in short), but which does have the C.A.P. (Recall that the A.P. is defined as the C.A.P. by replacing compact operators with operators of finite rank; see, e.g., [14, p. 30].) In fact, it follows from Willis’ construction that the space X has the stronger property, namely—in our terminology—the P.C.A.P. as observed by Benyamini [3].

We close this section with a result, which shows in particular that separability is a necessary condition for a metric space to have the property described in Theorem 2.1.

Proposition 2.3

Let (X, d) be a metric space such that there exists a sequence of completely continuous mappings $F_n:X\rightarrow X$, which is pointwise convergent to the identity mapping. Then (X, d) is separable.

Proof

Fix $x_0\in X$ and $m\in {\mathbb N}$, and set $B_m:=B(x_0,m)$. By hypothesis, for any $n\in {\mathbb N}$, $\overline{F_n(B_m)}$ is compact, hence separable, so there exists a countable set $A_{nm}\subseteq \overline{F_n(B_m)}$ such that $\overline{A_{nm}}=\overline{F_n(B_m)}$. Fix $x\in B_m$ and $\varepsilon >0$. Since $F_nx\rightarrow x$, there exists $k\in {\mathbb N}$ such that $d(x,F_kx)<\varepsilon /2$. Since

$$\begin{aligned} F_kx\in F_k(B_m)\subseteq \overline{A_{km}}, \end{aligned}$$

there is $a\in A_{km}$ such that $d(F_kx,a)<\varepsilon /2$. Thus we may infer that $d(x,a)<\varepsilon$. Set $A_m:=\bigcup _{i\in {\mathbb N}}A_{im}$. Clearly, $a\in A_m$, so we have shown that $B_m\subseteq \overline{ A_m}$. Set $A:=\bigcup _{m\in {\mathbb N}}A_m$. Since each $A_m$ is countable, so is A. Moreover, we have that

$$\begin{aligned} X=\bigcup _{m\in {\mathbb N}}B_m\subseteq \bigcup _{m\in {\mathbb N}}\overline{A_m}\subseteq \overline{\bigcup _{m\in {\mathbb N}}A_m}=\overline{A}, \end{aligned}$$

which shows that A is dense in X. Thus (X, d) is separable. $\square$

3 A Proof of the Ascoli–Arzelà Theorem  with the Use of Bernstein Polynomials

First, let us notice that even the weaker version of Corollary 2.1—with operators of finite rank—yields relatively easy compactness criteria in Banach spaces $c_0$, $l_p$ and $L_1([0,1])$ as presented by Kadets [11, pp. 295–297]. On the other hand, Corollary 2.1 was used by Hanche-Olsen, Holden and Mallinnikova [9] to give a nice proof of the Kolmogorov–Riesz–Sudakov compactness criterion in $L_p({\mathbb R}^n)$ with $1\le p<\infty$. Our purpose here is to present another application of Corollary 2.1 to obtain a new proof of the Ascoli–Arzelà theorem.

Recall that for any $n\in {\mathbb N}$, the Bernstein operator ${{\mathcal {B}}}_n$ is defined on C([0, 1]) by the following formula: for $f\in C([0,1])$ and $x\in [0,1]$,

$$\begin{aligned} ({{\mathcal {B}}}_nf)(x):=\sum _{k=0}^n{n\atopwithdelims ()k}f\left( \frac{k}{n}\right) x^k(1-x)^{n-k}. \end{aligned}$$

Clearly, all ${{\mathcal {B}}}_n$ are linear bounded operators of finite rank and their values are in $C^{\infty }([0,1])$. In particular, each ${{\mathcal {B}}}_n$ satisfies the assumptions of the following lemma. We denote by $B_1$ the closed unit ball in C([0, 1]) centered at 0.

Lemma 3.1

Let $T:C([0,1])\rightarrow C([0,1])$ be a linear bounded operator of finite rank such that $Tf\in C^1([0,1])$ for every $f\in C([0,1])$. Then the family $T(B_1)$ is equilipschitzian.

Proof

Set $X:=C([0,1])$ and for $f\in T(X)$, $Df:=f'$. Clearly, D is linear and bounded since T(X) is finite dimensional. Thus operator $D\circ T:X\rightarrow X$ is also linear and bounded. Hence, for any $f\in B_1$,

$$\begin{aligned} \Vert (D\circ T)(f)\Vert \le \Vert D\circ T\Vert =:L, \end{aligned}$$

i.e., $\sup _{x\in [0,1]}|(Tf)'(x)|\le L$. By Lagrange’s theorem, we obtain that for any $x,y\in [0,1]$ and $f\in B_1$,

$$\begin{aligned} |(Tf)(x)-(Tf)(y)|\le L |x-y| \end{aligned}$$

which means that $T(B_1)$ is equilipschitzian. $\square$

Proposition 3.1

Let ${{\mathfrak F}}$ be a bounded and equicontinuous family of functions from C([0, 1]) endowed with the supremum norm. Then the sequence $({{\mathcal {B}}}_n|_{{{\mathfrak F}}})$ converges uniformly to $I|_{{{\mathfrak F}}}$.

Proof

Since ${{\mathfrak F}}$ is bounded,

$$\begin{aligned} M:=\sup \{|f(x)|: f\in {{\mathfrak F}},\ x\in [0,1]\} \end{aligned}$$

is finite. Since [0, 1] is compact and ${{\mathfrak F}}$ is equicontinuous, it is also uniformly equicontinuous. Hence, for any $\varepsilon >0$, there exists $\delta >0$ such that for any $x,y\in [0,1]$, $|x-y|<\delta$ implies that $|f(x)-f(y)|<\varepsilon /2$ for any $f\in {{\mathfrak F}}$. For $n\in {\mathbb N}$, $k=0,1,\ldots ,n$ and $x\in [0,1]$, set

$$\begin{aligned} a_{kn}:=\frac{k}{n}\ { \text{ and } }\ p_{kn}(x):={n\atopwithdelims ()k}x^k(1-x)^{n-k}. \end{aligned}$$

Further, define $e_j(x):=x^j$ for $j=0,1,2$ and $x\in [0,1]$. Now, fix $n\in {\mathbb N}$, $f\in {{\mathfrak F}}$ and $x\in [0,1]$. Since ${{\mathcal {B}}}_ne_0=e_0$, we have that

$$\begin{aligned} \begin{array}{ll} |({{\mathcal {B}}}_nf)(x)-f(x)|\!&{}=|\sum \limits _{k=0}^nf(a_{kn})p_{kn}(x)-\sum \limits _{k=0}^nf(x)p_{kn}(x)| \\ &{} \le \sum \limits _{k=0}^n|f(a_{kn})-f(x)|p_{kn}(x). \end{array} \end{aligned}$$

(3.1)

Define $I_1:=\{k\in \{0,1,\ldots ,n\}:|a_{kn}-x|<\delta \}$ and $I_2:=\{0,1,\ldots ,n\}\setminus I_1$. Then we get that

$$\begin{aligned} \begin{array}{ll} \sum \limits _{k=0}^n|f(a_{kn})-f(x)|p_{kn}(x)\!&{}=\sum \limits _{k\in I_1}|f(a_{kn})-f(x)|p_{kn}(x)\\ &{} +\sum \limits _{k\in I_2}|f(a_{kn})-f(x)|p_{kn}(x). \end{array} \end{aligned}$$

(3.2)

Denote the components of the above sum by $S_1$ and $S_2$, respectively. Then we have that

$$\begin{aligned} S_1\le \frac{\varepsilon }{2}\sum \limits _{k\in I_1}p_{kn}(x)\le \frac{\varepsilon }{2}\sum \limits _{k=0}^np_{kn}(x)=\frac{\varepsilon }{2}. \end{aligned}$$

(3.3)

Now, for $k\in I_2$, $|a_{kn}-x|\ge \delta$, so $|a_{kn}-x|/\delta \ge 1$, and hence $(a_{kn}-x)^2/\delta ^2\ge 1$. Therefore, since simultaneously $|f(a_{kn})-f(x)|\le 2M$, we obtain that

$$\begin{aligned} \begin{array}{ll} S_2&{}\le \sum \limits _{k\in I_2}\frac{(a_{kn}-x)^2}{\delta ^2}2Mp_{kn}(x)\le \frac{2M}{\delta ^2}\sum \limits _{k=0}^n(a_{kn}-x)^2p_{kn}(x)\\ &{}=\frac{2M}{\delta ^2}\left( \sum \limits _{k=0}^na_{kn}^2p_{kn}(x)-2x\sum \limits _{k=0}^na_{kn}p_{kn}(x)+x^2\sum \limits _{k=0}^np_{kn}(x)\right) \\ &{}=\frac{2M}{\delta ^2}\left( ({{\mathcal {B}}}_ne_2)(x)-2x({{\mathcal {B}}}_ne_1)(x)+x^2\right) . \end{array} \end{aligned}$$

It is well known that ${{\mathcal {B}}}_ne_1=e_1$ and ${{\mathcal {B}}}_ne_2=\left( 1-\frac{1}{n}\right) e_2+\frac{1}{n}e_1$ (see, e.g., [5, p. 159]). Thus we may infer that

$$\begin{aligned} \begin{array}{ll} S_2\! &{}\le \frac{2M}{\delta ^2}\left( \left( 1-\frac{1}{n}\right) x^2+\frac{1}{n}x-x^2\right) \\ &{}=\frac{2M}{\delta ^2n}x(1-x)\le \frac{M}{2\delta ^2n}. \end{array} \end{aligned}$$

(3.4)

By (3.1)–(3.4), we conclude that for any $n\in {\mathbb N}$ and $f\in {{\mathfrak F}}$,

$$\begin{aligned} \Vert {{\mathcal {B}}}_nf-f\Vert \le \frac{\varepsilon }{2}+\frac{M}{2\delta ^2n}. \end{aligned}$$

Thus, for sufficiently large n, we obtain that $\Vert {{\mathcal {B}}}_nf-f\Vert <\varepsilon$ for any $f\in {{\mathfrak F}}$, which completes the proof. $\square$

Theorem 3.1

(Ascoli–Arzelà). Let ${{\mathfrak F}}$ be a family of functions from C([0, 1]) endowed with the supremum norm. Then ${{\mathfrak F}}$ is relatively compact if, and only if, ${{\mathfrak F}}$ is bounded and equicontinuous.

Proof

Applying Proposition 3.1 to a one-element family, we get that $({{\mathcal {B}}}_n)$ is pointwise convergent to I.

($\Rightarrow$): Now, assume that ${{\mathfrak F}}\subseteq C([0,1])$ is relatively compact. By Corollary 2.1, $({{\mathcal {B}}}_n|_{{{\mathfrak F}}})$ converges uniformly to $I|_{{{\mathfrak F}}}$, so for any $\varepsilon >0$, there exists $k\in {\mathbb N}$ such that

$$\begin{aligned} \Vert {{\mathcal {B}}}_kf-f\Vert <\frac{\varepsilon }{3}\ { \text{ for } \text{ any } }\ f\in {{\mathfrak F}}. \end{aligned}$$

(3.5)

Since ${{\mathfrak F}}$ is bounded and ${{\mathcal {B}}}_k$ satisfies the assumptions of Lemma 3.1, we may conclude that the family ${{\mathcal {B}}}_k({{\mathfrak F}})$ is equilipschitzian with a constant $L>0$, i.e., for any $f\in {{\mathfrak F}}$ and $x,y\in [0,1]$,

$$\begin{aligned} |({{\mathcal {B}}}_kf)(x)-({{\mathcal {B}}}_kf)(y)|\le L|x-y|. \end{aligned}$$

(3.6)

Now, fix $f\in {{\mathfrak F}}$ and $x,y\in [0,1]$. By (3.5) and (3.6), we get that

$$\begin{aligned} \begin{array}{ll} |f(x)-f(y)|\!&{}\le |f(x)-({{\mathcal {B}}}_kf)(x)|+|({{\mathcal {B}}}_kf)(x)-({{\mathcal {B}}}_kf)(y)|+|({{\mathcal {B}}}_kf)(y)-f(y)|\\ &{}<\frac{2}{3}\varepsilon +L|x-y|. \end{array} \end{aligned}$$

Set $\delta :=\frac{\varepsilon }{3L}$. Then, for any $x,y\in [0,1]$ with $|x-y|<\delta$, we obtain that

$$\begin{aligned} |f(x)-f(y)|<\varepsilon \ { \text{ for } \text{ any } }\ f\in {{\mathfrak F}}, \end{aligned}$$

so ${{\mathfrak F}}$ is uniformly equicontinuous.

($\Leftarrow$): This follows immediately from Corollary 2.1 and Proposition 3.1. $\square$

4 Appendix: Another Look at the Notion of Continuous Convergence

We start with discussing the pointwise version of the notion of continuous convergence.

Proposition 4.1

Let (X, d) and $(Y,\rho )$ be metric spaces, $x_0\in X$ and for $n\in {\mathbb N}$, $F_n:X\rightarrow Y$ be continuous at $x_0$. The following statements are equivalent:

• (i) For any sequence $(x_n)$, $x_n\rightarrow x_0$ implies the convergence of sequence $(F_n(x_n))$;

• (ii) The limit $y_0:=\lim _{n\rightarrow \infty }F_n(x_0)$ exists and for any sequence $(x_n)$, $x_n\rightarrow x_0$ implies the convergence of $(F_n(x_n))$ to $y_0$;

• (iii) The sequence $(F_n(x_0))$ is convergent and the family $\{F_n:n\in {\mathbb N}\}$ is equicontinuous at $x_0$.

Proof

(i)$\Rightarrow$(ii): By (i), setting $x_n:=x_0$ we obtain that $(F_n(x_0))$ converges to some point $y_0\in Y$. Now, let $x_n\rightarrow x_0$. For $n\in {\mathbb N}$, define $x_{2n-1}':=x_0$ and $x_{2n}':=x_{2n}$. Since $x_n'\rightarrow x_0$, we get by (i) that $(F_n(x_n'))$ is convergent, so $\lim _{n\rightarrow \infty }F_{2n-1}(x_{2n-1}')=\lim _{n\rightarrow \infty }F_{2n}(x_{2n}')$. By (i), $(F_n(x_n))$ is also convergent, so we may infer that

$$\begin{aligned} \lim _{n\rightarrow \infty }F_n(x_n)=\lim _{n\rightarrow \infty }F_{2n}(x_{2n})=\lim _{n\rightarrow \infty }F_{2n}(x_{2n}')=\lim _{n\rightarrow \infty }F_{2n-1}(x_{2n-1}')=y_0. \end{aligned}$$

(ii)$\Rightarrow$(iii): Suppose, to the contrary, $\{F_n:n\in {\mathbb N}\}$ is not equicontinuous at $x_0$. Then there exist $\varepsilon _0>0$ and sequences $(x_n)$ and $(k_n)$ such that $x_n\rightarrow x_0$ and $\rho (F_{k_n}(x_n),F_{k_n}(x_0))\ge \varepsilon _0$ for each $n\in {\mathbb N}$. Consider the set

$$\begin{aligned} N:=\{k_n:n\in {\mathbb N}\}. \end{aligned}$$

Suppose that N is finite. Then, for some $p\in {\mathbb N}$, the set $\{n\in {\mathbb N}:k_n=p\}$ is infinite. Hence there exists a strictly increasing sequence $(m_n)$ of positive integers such that $k_{m_n}=p$, and then

$$\begin{aligned} \rho (F_p(x_{m_n}),F_p(x_0))\ge \varepsilon _0\ { \text{ for } \text{ each } }\ n\in {\mathbb N}, \end{aligned}$$

which contradicts the continuity of $F_p$ at $x_0$. Thus N is infinite and without loss of generality, by passing to a subsequence if necessary, we may assume that $k_n<k_{n+1}$ for each $n\in {\mathbb N}$. We show that $F_{k_n}(x_n)\rightarrow y_0$. Define a sequence $(x_n')$ by setting

$$\begin{aligned} x_{k_n}':=x_n\ { \text{ and } }\ x_i:=x_0\ { \text{ for } }\ i\in {\mathbb N}\setminus N. \end{aligned}$$

Clearly, $x_n'\rightarrow x_0$, so by (ii), $F_n(x_n')\rightarrow y_0$ which yields that $F_{k_n}(x_{k_n}')\rightarrow y_0$, i.e., $F_{k_n}(x_n)\rightarrow y_0$. Simultaneously, $F_{k_n}(x_0)\rightarrow y_0$, so we obtain that

$$\begin{aligned} 0<\varepsilon _0\le \rho (F_{k_n}(x_n),F_{k_n}(x_0))\rightarrow 0, \end{aligned}$$

which gives a contradiction. Thus (iii) holds.

(iii)$\Rightarrow$(i): Denote $y_0:=\lim _{n\rightarrow \infty }F_n(x_0)$ and assume that $x_n\rightarrow x_0$. Then we have that

$$\begin{aligned} \rho (F_n(x_n),y_0)\le \rho (F_n(x_n),F_n(x_0))+\rho (F_n(x_0),y_0). \end{aligned}$$

The equicontinuity of $\{F_n:n\in {\mathbb N}\}$ implies easily that $\rho (F_n(x_n),F_n(x_0))\rightarrow 0$, so the above inequality let us infer that $F_n(x_n)\rightarrow y_0$. $\square$

As an immediate consequence, we obtain the following characterization of continuous convergence.

Corollary 4.1

Let (X, d) and $(Y,\rho )$ be metric spaces. For $n\in {\mathbb N}$, let $F_n:X\rightarrow Y$ be arbitrary functions and $F:X\rightarrow Y$. The following statements are equivalent:

• (i) $(F_n)$ is continuously convergent to F and each $F_n$ is continuous;

• (ii) $(F_n)$ is pointwise convergent to F and the family $\{F_n:n\in {\mathbb N}\}$ is equicontinuous.

Proof

Assume that (i) holds and fix $x_0\in X$. Since $F_n\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,F$, we get taking a constant sequence $(x_0)$ in the definition of continuous convergence that $F_nx_0\rightarrow Fx_0$. Since each $F_n$ is continuous at $x_0$, Proposition 4.1 ((i)$\Rightarrow$(iii)) yields that the family $\{F_n:n\in {\mathbb N}\}$ is equicontinuous at $x_0$. Thus (ii) holds.

Now, assume that (ii) is satisfied and fix $x_0\in X$. Then, obviously, each $F_n$ is continuous at $x_0$ and condition (iii) of Proposition 4.1 holds. Thus, by Proposition 4.1 ((iii)$\Rightarrow$(ii)), we obtain that for any sequence $(x_n)$, $x_n\rightarrow x_0$ implies the convergence of $(F_n(x_n))$ to $\lim _{n\rightarrow \infty }F_nx_0$, i.e., $F_n(x_n)\rightarrow Fx_0$. That means $F_n\longrightarrow ^{\!\!\!\!\!\!\!\!c}\; \,F$. $\square$

This result highlights better the phenomenon observed in Lemma 2.1 that both types of convergence are equivalent for any sequence of linear bounded operators on a Banach space. Indeed, by the uniform boundedness principle, if such a sequence is pointwise convergent, then the family of these operators is automatically equicontinuous.