Adding Infinitely Many Functions

For this post, let $K$ be a set and $\mathbb K$ be an ordered field.

Given a sequence $\{f_n\}$ of functions $f_n : K \to \mathbb K$ , define $F_n : K \to \mathbb K$ by

$\displaystyle F_n(x) := \sum_{i=1}^n f_i(x).$

How can we guarantee that the sequence of sums $\{F_n\}$ converge uniformly? Just like the case with real numbers, the Cauchy criterion gives us some useful handles on this question.

To abbreviate, we will define the supremum norm and characterise uniform convergence in terms of the supremum norm.

Definition 1. For any function $g : K \to \mathbb R$ , define

$\displaystyle \|g\|_\infty := \sup_{x \in K}|g(x)|,$

whenever the right-hand side exists. Then for any sequence $\{f_n\}$ of real-valued functions and $f : K \to \mathbb R$ , $f_n \to f$ uniformly if and only if

$\forall \epsilon > 0 \quad \exists N \in \mathbb N : \quad (n > N \quad \Rightarrow \quad \| f_n - f \|_\infty < \epsilon).$

Lemma 1. Let $f, g : K \to \mathbb R$ be functions such that $\| f \|_\infty, \| g \|_\infty$ exists. The following properties hold:

$\| f \|_\infty \geq 0$ .
$\| f \|_\infty = 0 \Rightarrow f = 0$ .
$\| kf \|_\infty = |k| \cdot \| f \|_\infty$ , $k \in \mathbb R$ .
$\| f + g \|_\infty \leq \| f \|_\infty + \| g \|_\infty$ .

This justifies calling $\| \cdot \|_\infty$ a norm.

Proof. Exercise.

We say that $\{f_n\}$ converges uniformly if there exists $f : K \to \mathbb R$ such that $f_n \to f$ uniformly.

Theorem 1. Let $K$ and $\{f_n\}$ be a sequence of real-valued functions $f_n : K \to \mathbb R$ . Then $\{F_n\}$ converges uniformly if and only if

$\displaystyle \forall \epsilon > 0 \quad \exists N \in \mathbb N: \quad m > n > N\quad \Rightarrow \quad \left\| \sum_{i = n+1}^m f_i \right \|_\infty < \epsilon.$

Proof. We will prove in two directions, starting with $(\Rightarrow)$ . Fix $\epsilon > 0$ . Since $\{ F_n \}$ converges uniformly, for any $k > 0$ , there exists $F : K \to \mathbb R$ and $N \in \mathbb N$ such that for any $x \in K$ ,

$n > N \quad \Rightarrow \quad \| F_n - F \|_\infty < k \cdot \epsilon.$

Hence, for $m > n > N$ ,

$\displaystyle \begin{aligned} \left\| \sum_{i = n+1}^m f_i \right \|_\infty = \| F_m - F_n \|_\infty &\leq \| F_m - F \|_\infty + \| F - F_n \|_\infty \\ &< k \cdot \epsilon + k \cdot \epsilon \\ &= (2k) \cdot \epsilon. \end{aligned}$

Setting $k =1/2$ yields the result. In fact, this result holds for any sequence in a metric space. But we will reserve that discussion for another post.

For the direction $(\Leftarrow)$ , we will need to take advantage of the Cauchy-completeness of $\mathbb R$ . In particular, we first show that for each $x \in K$ , $\{F_n(x)\}$ is Cauchy, and thus converges in $\mathbb R$ to some unique limit $F(x)$ , after which we strengthen the convergence $F_n \to F$ to uniform convergence.

Fix $x \in K$ and $\epsilon > 0$ . By the hypothesis, for any $k > 0$ , there exists $N \in \mathbb N$ such that

$\displaystyle m > n > N\quad \Rightarrow \quad \| F_m - F_n \|_\infty < k \cdot \epsilon.$

Unraveling the definition of $\|\cdot\|_\infty$ ,

$\displaystyle \forall u \in K \quad \left| \sum_{i = n+1}^m f_i(u) \right| < k \cdot \epsilon.$

Particularising to $u = x$ ,

$\displaystyle |F_m(x) - F_n(x)| = \left| \sum_{i = n+1}^m f_i(x) \right| < k \cdot \epsilon.$

Setting $k = 1$ , the sequence $\{F_n(x)\}$ is Cauchy and converges to some unique limit $F(x)$ , i.e. $F_n \to F$ point-wise. To strengthen the result to uniform convergence, we recall that for any $u \in K$ ,

$|F_n(u) - F_m(u)| = |F_m(u) - F_n(u)| < k \cdot \epsilon.$

Taking $m \to \infty$ ,

$\displaystyle |F_n(u) - F(u)| = \lim_{m \to \infty} |F_n(u) - F_m(u)| \leq \lim_{m \to \infty} (k \cdot \epsilon) = k \cdot \epsilon.$

Setting $k = 1/2$ , $| F_n(u) - F(u) | \leq \epsilon/2 < \epsilon$ . Since this estimate holds for any $u \in K$ , $F_n \to F$ uniformly.

This result helps us determine when sequences of sums $\{F_n\}$ converge uniformly or not. But what’s a useful test to establish such convergence? One ridiculously useful test is the Weierstrass M-test.

Theorem 2. Let $\{f_n\}$ be a sequence of functions $f_n : K \to \mathbb R$ . If $\displaystyle \sum_{n=1}^\infty \|f_n\|_\infty$ converges, then $\{F_n\}$ converges uniformly.

Proof. Fix $\epsilon > 0$ . Since $\displaystyle \sum_{n=1}^\infty \|f_n\|_\infty$ converges, for any $k > 0$ , there exists $N \in \mathbb N$ such that for $m > n > N$ ,

$\displaystyle \left\| \sum_{i = n+1}^m f_i \right \|_\infty \leq \sum_{i=n+1}^m \|f_i\|_\infty < k \cdot \epsilon.$

Setting $k = 1$ , by Theorem 1, $\{F_n\}$ converges uniformly.

Corollary 1 (Weierstrass M-test). Let $\{f_n\}$ be a sequence of functions $f_n : K \to \mathbb R$ . Suppose for each $n$ , there exists $M_n \geq 0$ such that $\|f_n\|_\infty \leq M_n$ . Suppose furthermore that $\displaystyle \sum_{n=1}^\infty M_n$ converges. Then $\{F_n\}$ converges uniformly.

Proof. Employ the comparison test.

Corollary 2. Let $\{a_n\}_{n = 0}^\infty \subseteq \mathbb R$ be a sequence of real numbers. If $\displaystyle \sum_{n=0}^\infty |a_n| < \infty$ , then the power series $x \mapsto \displaystyle \sum_{n=0}^\infty a_nx^n$ converges uniformly on $[-1, 1]$ and thus is continuous on $[-1, 1]$ .