
# 4.12: Sequences and Series of Functions


I. Let

$f_{1}, f_{2}, \ldots, f_{m}, \dots$

be a sequence of mappings from a common domain $$A$$ into a metric space $$\left(T, \rho^{\prime}\right) .$$ For each (fixed) $$x \in A,$$ the function values

$f_{1}(x), f_{2}(x), \ldots, f_{m}(x), \ldots$

form a sequence of points in the range space $$\left(T, \rho^{\prime}\right).$$ Suppose this sequence converges for each $$x$$ in a set $$B \subseteq A.$$ Then we can define a function $$f : B \rightarrow T$$ by setting

$f(x)=\lim _{m \rightarrow \infty} f_{m}(x) \text { for all } x \in B.$

This means that

$(\forall \varepsilon>0)(\forall x \in B)(\exists k)(\forall m>k) \quad \rho^{\prime}\left(f_{m}(x), f(x)\right)<\varepsilon.$

Here $$k$$ depends not only on $$\varepsilon$$ but also on $$x,$$ since each $$x$$ yields a different sequence $$\left\{f_{m}(x)\right\}.$$ However, in some cases (resembling uniform continuity), $$k$$ depends on $$\varepsilon$$ only; i.e., given $$\varepsilon>0,$$ one and the same $$k$$ fits all $$x$$ in $$B.$$ In symbols, this is indicated by changing the order of quantifiers, namely,

$(\forall \varepsilon>0)(\exists k)(\forall x \in B)(\forall m>k) \quad \rho^{\prime}\left(f_{m}(x), f(x)\right)<\varepsilon.$

Of course, (2) implies (1), but the converse fails (see examples below). This suggests the following definitions.

Definition 1

With the above notation, we call $$f$$ the pointwise limit of a sequence of functions $$f_{m}$$ on a set $$B(B \subseteq A)$$ iff

$f(x)=\lim _{m \rightarrow \infty} f_{m}(x) \text { for all } x \text { in } B;$

i.e., formula (1) holds. We then write

$f_{m} \rightarrow f(\text {pointwise}) \text { on } B.$

In case (2), we call the limit uniform (on $$B )$$ and write

$f_{m} \rightarrow f(\text {uniformly}) \text { on } B.$

II. If the $$f_{m}$$ are real, complex, or vector valued (§3), we can also define $$s_{m}=\sum_{k=1}^{m} f_{k}$$ (= sum of the first $$m$$ functions) for each $$m$$, so

$(\forall x \in A)(\forall m) \quad s_{m}(x)=\sum_{k=1}^{m} f_{k}(x).$

The $$s_{m}$$ form a new sequence of functions on $$A.$$ The pair of sequences

$\left(\left\{f_{m}\right\},\left\{s_{m}\right\}\right)$

is called the (infinite) series with general term $$f_{m} ; s_{m}$$ is called its $$m$$ th partial sum. The series is often denoted by symbols like $$\sum f_{m}, \sum f_{m}(x),$$ etc.

Definition 2

The series $$\sum f_{m}$$ on $$A$$ is said to converge (pointwise or uniformly) to a function $$f$$ on a set $$B \subseteq A$$ iff the sequence $$\left\{s_{m}\right\}$$ of its partial sums does as well.

We then call $$f$$ the sum of the series and write

$f(x)=\sum_{k=1}^{\infty} f_{k}(x) \text { or } f=\sum_{m=1}^{\infty} f_{m}=\lim s_{m}$

(pointwise or uniformly) on $$B$$.

Note that series of constants, $$\sum c_{m},$$ may be treated as series of constant functions $$f_{m},$$ with $$f_{m}(x)=c_{m}$$ for $$x \in A.$$

If the range space is $$E^{1}$$ or $$E^{*},$$ we also consider infinite limits,

$\lim _{m \rightarrow \infty} f_{m}(x)=\pm \infty.$

However, a series for which

$\sum_{m=1}^{\infty} f_{m}=\lim s_{m}$

is infinite for some $$x$$ is regarded as divergent (i.e., not convergent) at that $$x$$.

III. Since convergence of series reduces to that of sequences $$\left\{s_{m}\right\},$$ we shall first of all consider sequences. The following is a simple and useful test for uniform convergence of sequences $$f_{m} : A \rightarrow\left(T, \rho^{\prime}\right).$$

Theorem $$\PageIndex{1}$$

Given a sequence of functions $$f_{m} : A \rightarrow\left(T, \rho^{\prime}\right),$$ let $$B \subseteq A$$ and

$Q_{m}=\sup _{x \in B} \rho^{\prime}\left(f_{m}(x), f(x)\right).$

Then $$f_{m} \rightarrow f(\text {uniformly on } B)$$ iff $$Q_{m} \rightarrow 0$$.

Proof

If $$Q_{m} \rightarrow 0,$$ then by definition

$(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad Q_{m}<\varepsilon.$

However, $$Q_{m}$$ is an upper bound of all distances $$\rho^{\prime}\left(f_{m}(x), f(x)\right), x \in B.$$ Hence (2) follows.

Conversely, if

$(\forall x \in B) \quad \rho^{\prime}\left(f_{m}(x), f(x)\right)<\varepsilon,$

then

$\varepsilon \geq \sup _{x \in B} \rho^{\prime}\left(f_{m}(x), f(x)\right),$

i.e., $$Q_{m} \leq \varepsilon.$$ Thus (2) implies

$(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad Q_{m} \leq \varepsilon$

and $$Q_{m} \rightarrow 0.$$ $$\square$$

Examples

(a) We have

$\lim _{n \rightarrow \infty} x^{n}=0 \text { if }|x|<1 \text { and } \lim _{n \rightarrow \infty} x^{n}=1 \text { if } x=1.$

Thus, setting $$f_{n}(x)=x^{n},$$ consider $$B=[0,1]$$ and $$C=[0,1)$$.

We have $$f_{n} \rightarrow 0$$ (pointwise) on $$C$$ and $$f_{n} \rightarrow f(\text { pointwise })$$ on $$B,$$ with $$f(x)=0$$ for $$x \in C$$ and $$f(1)=1.$$ However, the limit is not uniform on $$C,$$ let alone on $$B .$$ Indeed,

$Q_{n}=\sup _{x \in C}\left|f_{n}(x)-f(x)\right|=1 \text { for each } n.$

Thus $$Q_{n}$$ does not tend to $$0,$$ and uniform convergence fails by Theorem 1.

(b) In Example (a), let $$D=[0, a], 0<a<1 .$$ Then $$f_{n} \rightarrow f$$ (uniformly) on $$D$$ because, in this case,

$Q_{n}=\sup _{x \in D}\left|f_{n}(x)-f(x)\right|=\sup _{x \in D}\left|x^{n}-0\right|=a^{n} \rightarrow 0.$

(c) Let

$f_{n}(x)=x^{2}+\frac{\sin n x}{n}, \quad x \in E^{1}.$

For a fixed $$x$$,

$\lim _{n \rightarrow \infty} f_{n}(x)=x^{2} \quad \text { since }\left|\frac{\sin n x}{n}\right| \leq \frac{1}{n} \rightarrow 0.$

Thus, setting $$f(x)=x^{2},$$ we have $$f_{n} \rightarrow f$$ (pointwise) on $$E^{1}.$$ Also,

$\left|f_{n}(x)-f(x)\right|=\left|\frac{\sin n x}{n}\right| \leq \frac{1}{n}.$

Thus $$(\forall n) Q_{n} \leq \frac{1}{n} \rightarrow 0.$$ By Theorem 1, the limit is uniform on all of $$E^{1}.$$

Theorem $$\PageIndex{2}$$

Let $$f_{m} : A \rightarrow\left(T, \rho^{\prime}\right)$$ be a sequence of functions on $$A \subseteq(S, \rho).$$ If $$f_{m} \rightarrow f\left(\text {uniformly } \text { on a set } B \subseteq A, \text { and if the } f_{m} \text { are relatively (or uniformly) }\right.$$ continuous on $$B$$ , then the limit function $$f$$ has the same property.

Proof

Fix $$\varepsilon>0.$$ As $$f_{m} \rightarrow f$$ (uniformly) on $$B,$$ there is a $$k$$ such that

$(\forall x \in B)(\forall m \geq k) \quad \rho^{\prime}\left(f_{m}(x), f(x)\right)<\frac{\varepsilon}{4}.$

Take any $$f_{m}$$ with $$m>k,$$ and take any $$p \in B.$$ By continuity, there is $$\delta>0,$$ with

$\left(\forall x \in B \cap G_{p}(\delta)\right) \quad \rho^{\prime}\left(f_{m}(x), f_{m}(p)\right)<\frac{\varepsilon}{4}.$

Also, setting $$x=p$$ in (3) gives $$\rho^{\prime}\left(f_{m}(p), f(p)\right)<\frac{\varepsilon}{4}.$$ Combining this with (4) and (3), we obtain $$\left(\forall x \in B \cap G_{p}(\delta)\right)$$

\begin{aligned} \rho^{\prime}(f(x), f(p)) & \leq \rho^{\prime}\left(f(x), f_{m}(x)\right)+\rho^{\prime}\left(f_{m}(x), f_{m}(p)\right)+\rho^{\prime}\left(f_{m}(p), f(p)\right) \\ &<\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}<\varepsilon. \end{aligned}

We thus see that for $$p \in B$$,

$(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in B \cap G_{p}(\delta)\right) \quad \rho^{\prime}(f(x), f(p))<\varepsilon,$

i.e., $$f$$ is relatively continuous at $$p(\text { over } B),$$ as claimed.

Quite similarly, the reader will show that $$f$$ is uniformly continuous if the $$f_{n}$$ are. $$\square$$

Note 2. A similar proof also shows that if $$f_{m} \rightarrow f$$ (uniformly) on $$B,$$ and if the $$f_{m}$$ are relatively continuous at a point $$p \in B,$$ so also is $$f.$$

Theorem $$\PageIndex{3}$$ (Cauchy criterion for uniform convergence)

Let $$\left(T, \rho^{\prime}\right)$$ be complete. Then a sequence $$f_{m} : A \rightarrow T, A \subseteq(S, \rho),$$ converges uniformly on a set $$B \subseteq A$$ iff

$(\forall \varepsilon>0)(\exists k)(\forall x \in B)(\forall m, n>k) \quad \rho^{\prime}\left(f_{m}(x), f_{n}(x)\right)<\varepsilon.$

Proof

If (5) holds then, for any (fixed) $$x \in B,\left\{f_{m}(x)\right\}$$ is a Cauchy sequence of points in $$T,$$ so by the assumed completeness of $$T,$$ it has a limit $$f(x).$$ Thus we can define a function $$f : B \rightarrow T$$ with

$f(x)=\lim _{m \rightarrow \infty} f_{m}(x) \text { on } B.$

To show that $$f_{m} \rightarrow f$$ (uniformly) on $$B,$$ we use (5) again. Keeping $$\varepsilon, k,$$ $$x,$$ and $$m$$ temporarily fixed, we let $$n \rightarrow \infty$$ so that $$f_{n}(x) \rightarrow f(x)$$. Then by Theorem 4 of Chapter 3, §15, $$\rho^{\prime}\left(f_{m}(x), f_{n}(x)\right) \rightarrow p^{\prime}\left(f(x), f_{m}(x)\right).$$ Passing to the limit in (5), we thus obtain (2).

The easy proof of the converse is left to the reader (cf. Chapter 3, §17, Theorem 1). $$\square$$

IV. If the range space $$\left(T, \rho^{\prime}\right)$$ is $$E^{1}, C,$$ or $$E^{n}$$ (*or another normed space), the standard metric applies. In particular, for series we have

\begin{aligned} \rho^{\prime}\left(s_{m}(x), s_{n}(x)\right) &=\left|s_{n}(x)-s_{m}(x)\right| \\ &=\left|\sum_{k=1}^{n} f_{k}(x)-\sum_{k=1}^{m} f_{k}(x)\right| \\ &=\left|\sum_{k=m+1}^{n} f_{k}(x)\right| \quad \text { for } m<n. \end{aligned}

Replacing here $$m$$ by $$m-1$$ and applying Theorem 3 to the sequence $$\left\{s_{m}\right\},$$ we obtain the following result.

Theorem $$\PageIndex{3'}$$

Let the range space of $$f_{m}, m=1,2, \ldots,$$ be $$E^{1}, C,$$ or $$E^{n}$$ (*or another complete normed space). Then the series $$\sum f_{m}$$ converges uniformly on $$B$$ iff

$(\forall \varepsilon>0)(\exists q)(\forall n>m>q)(\forall x \in B) \quad\left|\sum_{k=m}^{n} f_{k}(x)\right|<\varepsilon.$

Similarly, via $$\left\{s_{m}\right\},$$ Theorem 2 extends to series of functions. (Observe that the $$s_{m}$$ are continuous if the $$f_{m}$$ are.) Formulate it!

V. If $$\sum_{m=1}^{\infty} f_{m}$$ exists on $$B,$$ one may arbitrarily "group" the terms, i.e., replace every several consecutive terms by their sum. This property is stated more precisely in the following theorem.

Theorem $$\PageIndex{4}$$

Let

$f=\sum_{m=1}^{\infty} f_{m}(\text {pointwise}) \text { on } B.$

Let $$m_{1}<m_{2}<\cdots<m_{n}<\cdots$$ in $$N,$$ and define

$g_{1}=s_{m_{1}}, \quad g_{n}=s_{m_{n}}-s_{m_{n-1}}, \quad n>1.$

(Thus $$g_{n+1}=f_{m_{n}+1}+\cdots+f_{m_{n+1}}.)$$ Then

$f=\sum_{n=1}^{\infty} g_{n}(\text {pointwise}) \text { on } B \text { as well; }$

similarly for uniform convergence.

Proof

Let

$s_{n}^{\prime}=\sum_{k=1}^{n} g_{k}, \quad n=1,2, \ldots$

Then $$s_{n}^{\prime}=s_{m_{n}}$$ (verify!), so $$\left\{s_{n}^{\prime}\right\}$$ is a subsequence, $$\left\{s_{m_{n}}\right\},$$ of $$\left\{s_{m}\right\} .$$ Hence $$s_{m} \rightarrow f(\text { pointwise })$$ implies $$s_{n}^{\prime} \rightarrow f$$ (pointwise); i.e.,

$f=\sum_{n=1}^{\infty} g_{n} \text { (pointwise). }$

For uniform convergence, see Problem 13 (cf. also Problem 19). $$\square$$