Lecture 6. Equicontinuity and Series of Functions

We have already encountered power series, which can be thought of as series of functions. When solving equations involving functions as unknowns, one often can produce a sequence of approximate solutions. To prove that at least some subsequence converges requires one to obtain some compactness. For families of functions defined on a compact set, equicontinuity (along with boundedness) turns out to provide a characterization of compactness. Let $X$ be a compact metric space and denote by $$ \operatorname{C}(X,\mathbb{R}) $$ the space of real-valued continuous functions defined on it. By taking $X=[0,1]$ and $$ f_k(x)=x^k,\: x\in [0,1], $$ we obtain a sequence continuous functions where $f_1,\dots,f_n$ are linearly independent no matter how large $n\in\mathbb{N}$ is. This shows that the space $\operatorname{C}(X,\mathbb{R})$ cannot be finite dimensional. The Arzéla-Ascoli Theorem, the main result of this lecture, contains a characterization of compactness for subsets of $\operatorname{C}(X,\mathbb{R})$ and shows that the necessary conditions of boundedness and closure, which are also sufficient in finite dimensions, are not enough in this case. Let us dwell on this a little more. On $\operatorname{C}(X,\mathbb{R})$ we introduce the distance $d_\infty$ via $$ d_\infty(f,g)=\sup _{x\in X}|f(x)-g(x)|,\: f,g\in\operatorname{C}(X,\mathbb{R}). $$ It is induced by the norm $\|\cdot\| _\infty$ defined through $\| f\|_\infty=\sup_{x\in X}|f(x)|$. Notice that $$ \| f_k\| _\infty\leq 1,\: k\in \mathbb{N}, $$ so that the set $S=\{ f_k\, :\, k\in \mathbb{N}\}$ is bounded in $\operatorname{C}(X,\mathbb{R})$. Next we show that $S$ is also closed. Towards this goal, take a convergent sequence $(g_j)_{j\in\mathbb{N}}$ in $S$, which is is therefore of the form $g_j=f_{k_j}$, $j\in \mathbb{N}$. Notice that the indeces $k_j$ are not necessarily ordered as they would be for a subsequence. Convergence amounts to the existence of $g_\infty\in \operatorname{C}(X,\mathbb{R})$ such that $$ \lim _{j\to\infty}d_\infty(g_j,g_\infty)=\lim _{j\to\infty}\sup _{x\in X}|g_j(x)-g_\infty(x)|=0. $$ First we show that there can only be finitely many distinct indeces in the set $I=\{ k_j\, :\, j\in \mathbb{N}\}$. Assume this is not the case. Then we find a monotone increasing subsequence $(k_{j_l})_{l\in \mathbb{N}}$ of indeces in $I$. Without loss of generality let it be the original sequence of indeces $(k_j)_{j\in\mathbb{N}}$. In this case we have that $$ g_j(x)=f_{k_j}(x)\to 0,\: x\in [0,1),\text{ and } g_j(1)=f_{k_j}(1)=1\to 1, $$ as $j\to\infty$, but this is impossible since $g_\infty$ is continuous (as the uniform limit of a sequence of continuous functions) and cannot possibly simultaneously satisfy $$ g_\infty(x)=0\text{ for }x\in [0,1)\text{ and }g_\infty(1)=1. $$ Thus we indeed have that $I$ is finite and thus that $$ \min _{l\neq\tilde l\in I} d_\infty(f_{l},f_{\tilde l})\geq\delta>0, $$ for some $\delta$. The convergent sequence $(g_j)_{j\in\mathbb{N}}$ must therefore become stationary, i.e. $g_j=f_l$ for all $j\geq N$, some $l\in I$, and some $N\in \mathbb{N}$. This shows $g_\infty\in S$ and concludes the proof.
The set is therefore closed and bounded but not compact, since it cannot contain a convergent subsequence. If it did the limit would have to be both continous and discontinous which is clearly not possible. Additional conditions are therefore required for compactness of subsets in the metric space $\bigl( \operatorname{C}(X,\mathbb{R}),d_\infty\bigr)$. Equicontinuity (see below) happens to be the right condition. Finally observe that the family $S$ is not equicontinous.

Equicontinuity

Definition

Let $(X,d)$ be a metric space and $K \subset X$ compact. We use the notation $$ \operatorname{C}(K)=\{ f:K\to \mathbb{R}\, |\, f \text{ is continuous}\} $$ and consider a subset $F\subset C(K)$. The we say that
(i) $F$ is pointwise bounded iff, for any given $x\in K$ there is $M_x>0$ such that $$ |f(x)| \le M_x\text{ for all }f\in F\, . $$ (ii) $F$ is equi-bounded (or simply bounded) iff there is $M>0$ such that $$ |f(x)| \le M\text{ for all }x\in K\text{ and }f \in F\, . $$ (iii) $F$ is equi-continuous iff, for any $\epsilon>0$, there is $\delta>0$ such that $$ |f(x) - f(y) |\leq\epsilon\text{ whenever }d(x,y)\leq\delta\text{ for }x,y \in K\text{ and for all }f\in F. $$

Examples

The first two examples are straightforward. As for the last take $\epsilon>0$ and let $\delta=\epsilon/M$. For $x,y \in [a,b]$ with $|x-y|\leq\delta$ we have that $$ |f_\lambda(x) - f_\lambda(y)| =| f'_\lambda(\xi)| |x-y|\leq M\, |x-y | \le \delta = \epsilon $$ Since $\delta$ is independent of $x\in[a,b]$ as well as of $\lambda\in\Lambda$, $F$ is equicontinuous on $[a,b]$.

Lemma

Let $(X,d)$ be a compact metric space. Then it contains a countable dense subset.

For any $k\in \mathbb{N}$, the balls $B(x,1/k)$ for $x\in X$ yield an open cover of $X$. By compactness, finitely many suffice. For each fixed $k\in \mathbb{N}$, we can therefore find $x^k_j\in X$ for $j=1,\dots, n_k$ and some $n_k\in \mathbb{N}$ such that $$ X\subset\bigcup _{j=1}^{n_k}B(x^k_j,1/k)\, . $$ The set $$ \bigcup_{k\in \mathbb{N}}\{ x^k_j\, |\, j=1,\dots, n_k\}\subset X $$ is then dense and countable as desired.

Theorem (Arzéla-Ascoli)

Let $F\subset \operatorname{C}(K)$ for a compact metric space $K$. Then the following two statements are equivalent
(i) $F$ is equi-continuous and pointwise bounded.
(ii) $F$ is bounded and every sequence $(f_n)_{n\in\mathbb{N}}$ in $F$ has a uniformly convergent subsequence.

(ii)$\Rightarrow$(i)
By (ii) $F$ is bounded pointwise. Suppose that $F$ is not equicontinuous. Then there exists $\epsilon_0>0$ such that, for any $k\in \mathbb{N}$ ($\delta=\frac{1}{k}$), there are $x_k, y_k \in K$ and $f_k\in F$ with $$ |x_k - y_k |\leq\frac{1}{k}\text{ and }|f_k(x_k) - f_k(y_k) |\geq\epsilon_0. $$ By (ii) there are a subsequence $(f_{k_j})_{j\in \mathbb{N}}$ and $f_\infty\in \operatorname{C}(K)$ such that $$ f_{k_j}\to f_\infty\text{ uniformly as }j\to\infty. $$ We may also choose a convergent subsequence $(x_{k_j})_{j\in \mathbb{N}}$ (without loss of generality we take it to be same subsequence as above), so that $$ x_{k_j} \to x\text{ for some }x\in K\, . $$ Since $|x_{k_j}-y_{k_j}|<\frac{1}{k_j}\to 0$ as $j\to\infty$, the sequence $(y_{k_j})_{j\in\mathbb{N} }$ has the same limit $x$. Therefore \begin{equation*} 0<\epsilon_0 \le|f_{k_j}(x_{k_j}) - f_{k_j} (y_{k_j}) | \le |f_{k_j}(x_{k_j}) - f_\infty(x_{k_j}) |+|f _\infty(x_{k_j}) - f_\infty(y_{k_j}) | + |f_\infty(y_{k_j}) - f_{k_j} (y_{k_j}) |\longrightarrow 0\text{ as }j\to\infty. \end{equation*} This is impossible and so $F$ has to be equicontinuous.
(i) $\Rightarrow$ (ii)
It is easily to prove that (i) implies boundedness. We therefore only prove that (i) implies relative compactness. Given a sequence $(f_n)_{n\in\mathbb{N}}$ in $F$, we need to find a (uniformly) convergent subsequence.
Step 1: Let $D=\{ x_1,x_2,\dots\}$ be a dense subset of $K$ which exists thanks to the lemma above. Since $\bigl(f_n(x_1)\bigr)_{n\in \mathbb{N}}$ is a bounded sequence in $\mathbb{R}$, it admits a convergent subsequence, i.e. $$ \lim_{j\to\infty}f_{n^1_j}(x_1)=f(x_1), $$ for some sequence of indices $(n^1_j)_{j\in\mathbb{N}}$ and some $f(x_1)\in\mathbb{R}$. We can then extract a further subsequence $(n^2_j)_{j\in\mathbb{N}}$ of $(n^1_j)_{j\in\mathbb{N}}$ for which $$ \lim_{j\to\infty}f_{n^2_j}(x_2)=f(x_2), $$ for some $f(x_2)\in \mathbb{R}$. Continuing in this fashion we obtain limits $f(x_k)$ along subsequences $(n^k_j)_{j\in\mathbb{N}}$ for $k\in\mathbb{N}$. Considering the sequence of indeces $(n^j_j)_{j\in\mathbb{N}}$ we obtain that $$ \lim _{j\to\infty}f_{n^j_j}(x_k)=f(x_k)\, ,\: k\in \mathbb{N}. $$ Step 2: To simplify notation, define $g_j:=f_{n^j_j}$. In Step 1 it was proved that $$ g_j(x)\to f(x)\text{ as }j\to\infty\text{ for any }x\in D\, , $$ but the convergence is not necessarily uniform in $x\in D$. We claim that it actually is. To see this, fix $\epsilon>0$ and choose $m\in \mathbb{N}$ so that $$ |g_i(x)-g_i(y)|\leq \epsilon/3\text{ provided }d(x,y)\leq 1/m, $$ which is possible by the equi-continuity assumption. By using the construction of a set $D$ described in the proof of the above density lemma and given any $x\in D$ we can always find $x^m_{l_x}\in D$ with $l_x\in\{ 1,\dots, n_m\}$ such that $d(x,x^m_{l_x})\leq 1/m$. Then \begin{equation*} |g_i(x)-g_j(x)|\leq|g_i(x)-g_i(x^m_{l_x})|+|g_i(x^m_{l_x})- g_j(x^m_{l_x})|+|g_j(x^m_{l_x})-g_j(x)|\leq 2\epsilon/3 + \max_{1\leq l\leq n_m}|g_i(x^m_{l})-g_j(x^m_{l})|\qquad \forall x\in D\, . \end{equation*} Then it suffices to choose $N\in \mathbb{N}$ with $\max_{1\leq l\leq n_m}|g_i(x^m_{l})-g_j(x^m_{l})|\leq \epsilon/3$ for $i,j\geq N$ (possible since we are taking the maximum over only a finite number of points at which convergence occurs) to conclude that $$ |g_i(x)-g_j(x)|\leq \epsilon\: \forall x\in D\, ,\:\forall\, i,j\geq N. $$ Letting $i\to\infty$ we see that $$ |f(x)-g_j(x)|\leq\epsilon\:\forall x\in D\, ,\:\forall j\geq N, $$ which is the desired uniform convergence on $D$.
Step 3: The function $f:D\to \mathbb{R}$ is uniformly continuous. Indeed, fix $\epsilon>0$ and choose $N\in \mathbb{N}$ such that \begin{equation*} |f(x)-f(y)|\leq |f(x)-g_j(x)|+|g_j(x)-g_j(y)|+|g_j(y)-f(y)|\leq 2\epsilon/3+|g_j(x)-g_j(y)|\qquad \forall x,y\in D\, ,\:\forall j\geq N, \end{equation*} by Step 2. Now equi-continuity yields $\delta>0$ such that $|g_j(x)-g_j(y)|\leq\epsilon/3$ whenever $x,y\in K$ satisfy $d(x,y)\leq \delta$ regardless of $j$. It then follows that $$ |f(x)-f(y)|\leq\epsilon\, ,\:\forall x,y\in D\text{ provided }d(x,y)\leq\delta, $$ which is the desired uniform continuity.
Step 4: We now apply the extension theorem of Lecture 5 to extend $f:D\to\mathbb{R}$ to a uniformly continuous function $f:K=\overline{D}\to\mathbb{R}$. We leave it as an excercise to conclude the proof by showing that $g_j(x)=f_{n^j_j}(x)\to f(x)$ for all $x\in K$.

Remark

Series of functions

Definitions

Let $K\subset X$.
(i) We say that $\sum_{n=1}^\infty f_n$ converges pointwise on $K$ iff $\lim_{n \to \infty} s_n(x)$ exists for each $x \in K$.
(ii) We say that $\sum_{n=1}^\infty f_n$ converges absolutely on $K$ iff $\sum_{n=1}^\infty |f_n(x)|$ converges on $K$, that is, for each $x\in K$.
(iii) We say that $\sum_{n=1}^\infty f_n(x)$ converges uniformly on $K$ iff it converges pointwise on $K$ but the convergence is independent of $x\in K$, i.e. iff, for every $\epsilon>0$, there is $N\in \mathbb{N}$ such that $$ \sup _{x\in K}|s_n(x)-s_\infty(x)|\leq\epsilon\, ,\: n\geq N\, , $$ for some function $s_\infty:K\to \mathbb{R}$.

Theorem (Weierstrass M-Test)

Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions on $K$ such that $$ \sup _{x\in K}|f_n(x) | \le M_n\, ,\: n\in \mathbb{N}. $$ If $\sum_{n=1}^\infty M_n < \infty$, then $\sum_{n=1}^\infty f_n$ converges absolutely and uniformly on $K$.

Given $\epsilon>0$, we need to find $N\in \mathbb{N}$ such that $$ \big | \sum_{k=n}^m f_k(x) \big |\leq\sup _{y\in K}\sum_{k=n}^m |f_k(y) |\leq\epsilon\text{ if }m>n\ge N\, . $$ Indeed, if this is the case, the Cauchy test will imply pointwise absolute convergence and, since $N$ does not depend on $x\in K$, the convergence is uniform. By assumption there is $N\in \mathbb{N}$ with $$ \sum_{k=n}^m M_k\leq\epsilon\text{ if }m>n\ge N\, . $$ It follows that $\sum_{k=n}^m |f_k(x)| \le\sum_{k=n}^m M_k\leq\epsilon\, ,\: x \in K$, when $m>n\ge N$.

Example

Observe that $$ \left | \frac{\sin (nx)}{n^2} \right | \leq \frac{1}{n^2}\text{ for } x \in\mathbb{R}\text{ and for all } n\in \mathbb{N}, $$ and that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges (it is a "$p$-series'' with $p=2$). Therefore we apply the Weierstrass M-test to see that the series converges absolutely and uniformly in $\mathbb{R}$.

Example

Clearly $f(0) = \sum_{n=1}^\infty \frac{0^2}{(1+0)^n} = 0$, whereas for any $x\ne 0$ \begin{equation*} f(x)=\sum_{n=1}^\infty \frac{x^2}{(1+x^2)^n}=x^2 \sum_{n=1}^\infty \left( \frac{1}{1+x^2} \right)^n=x^2 \frac{\frac{1}{1+x^2}}{1- \frac{1}{1+x^2}}=\frac{x^2}{1+x^2} \frac{1+x^2}{x^2}=1\, . \end{equation*} Therefore $ f(x)=\begin{cases} 0\, ,& x=0,\\ 1\, , & x\ne 0. \end{cases} $

Example

$f_n$ is well-defined on $$ K=\mathbb{R}\setminus\{-1/n^2 : n\in \mathbb{N}\}\, , $$ where it is also continuous.
(a) As $f_n(0) =1\not\to 0$, the series does not converge at $x=0$ and $f(0)=+\infty$. Let $K_0 = K \setminus\{0\}$. Take first $K_0\ni x_0 >0$ and observe that $$ f_n(x_0) = \frac{1}{1+n^2x_0} \le \frac{1}{x_0} \cdot \frac{1}{n^2}. $$ Thus $\sum_{n=1}^\infty f_n(x_0)$ converges by the comparison test. Next take $x_0<0$. There is $N$ such that $1+n^2 x_0<0$ for $n\ge N$ and thus $$ |f_n(x_0)| = \Big|\frac{1}{1+n^2x_0}\Big| \le \frac{1}{n^2 |x_0|-1} \le \frac{2}{n^2|x_0| } = \frac{2}{|x_0|} \cdot \frac{1}{n^2}\, ,\: n\ge N. $$ Again it follows that $\sum_{n=1}^\infty f_n(x_0)$ converges absolutely by the comparison test.
(b) We prove that $\sum_{n=1}^\infty \frac{1}{1+n^2 x}$ converges uniformly on $\{x\in \mathbb{R}\, |\, |x| \ge \delta\}\cap K_0$ for any fixed $\delta>0$. Indeed, let $\delta>0$. Then there is $N\in \mathbb{N}$ such that $$ n^2 \delta \ge \frac{n^2 \delta}{2} +1\text{ if }n\ge N, $$ so that $$ |f_n(x) |= \left | \frac{1}{1+n^2 x} \right | \le \frac{1}{n^2x} \leq \frac{1}{n^2\delta} $$ for any $n\ge N$ and $x\in K_0$ with $x\ge \delta$. If $x<-\delta$ and $n>1/\delta$ we have that $$ |f_n(x) | = \Big| \frac{1}{1+n^2 x} \Big|\le \frac{1}{n^2 |x| -1}\le\frac{1}{\frac{n^2 \delta}{2}+1-1}\leq\frac{2}{n^2 \delta}. $$ and so $\sum_{n=1}^\infty |f_n| $ converges uniformly on $\{|x| \ge \delta \}$.
(c) The series $\sum_{n=1}^\infty f_n$ does not converge uniformly on $(-\delta,\delta) \cap K_0$ for any $\delta>0$. To see this, let $\epsilon_0 = 1/2$. Choose $m=n+1$, $n=N$, and set $x_N= \frac{1}{(1+N)^2} \in (-\delta, \delta)$ for $N\in\mathbb{N}$ large enough. Then $$ \left | \sum_{k=n+1}^m f_k(x_N) \right | = |f_{N+1} (x_N) | = \frac{1}{1+(N+1)^2\frac{1}{(N+1)^2} }= \frac{1}{2} = \epsilon _0 $$ Uniform convergence on $(-\delta, \delta) \cap K_0$ is therefore not possible.
(d) We know by part (a) that $f(x)$ is well-defined on $K_0$. We know that $f_n$ is continuous on $K_0$ for $n\in \mathbb{N}$ and, by (b), that $\sum_{n=1}^\infty f_n$ converges uniformly to a function $f$ on $K_0 \cap \{x \in \mathbb{R}\, |\, |x| \geq\delta \}$ for any given $\delta>0$. It follows that the limiting function $f$ is continuous on $K_0 \cap \{|x| \ge \delta\}$ for any $\delta >0$, or that continuity holds on $$ K_0 \cap \cup_{\delta>0} \{|x| \ge \delta \} = K_0 \cap \mathbb{R} \setminus \{0\} = K_0. $$

Example

First observe that $f_n(x) = \frac{(nx)}{n^2} = \frac{nx - [nx]}{n^2} $ is continuous on $$ K=\mathbb{R}\setminus\{x=\frac{k}{n}\, |\, k \in \mathbb{Z}\}\subset\mathbb{R}\setminus\mathbb{Q} $$ for all $n\ge 1$. Next notice that $$ \frac{|(nx)|}{n^2} \le \frac{1}{n^2}\, ,\: n\in \mathbb{N}. $$ As we know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, the Weierstrass M-Test shows that $f(x)$ is well-defined for $x\in \mathbb{R}$ and $f$ is continuous on $K=\mathbb{R}\setminus \mathbb{Q}$.

Theorem (Stone-Weierstrass)

Let $f$ be continuous on $[a,b]$. Then there is a sequence $(p_n)_{n\in\mathbb{N}}$ of polynomials such that $$ p_n(x) \to f(x)\text{ as }n\to\infty $$ uniformly on $[a,b]$.