4.12: Sequences and Series of Functions
This page is a draft and is under active development.
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.
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.
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 \(f_{m} : A \rightarrow\left(T, \rho^{\prime}\right).\)
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 \(Q_{m} \rightarrow 0.\) \(\square\)
(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 \(E^{1}.\)
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 \(f.\)
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 l imit 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.
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.
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). \(\square\)