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). $\square$