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Mathematics LibreTexts

4.12: Sequences and Series of Functions

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    21178
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    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). \(\square\)