Chapter 16: Series of Functions

Theorem (Cauchy Criterion)

The function series \(\sum_{n=1}^{\infty}f_n  \) converges uniformly on \(D \)  if and only if for any given \(\varepsilon>0   \), there exists \(N\in \mathbb{N}  \) such that
\begin{equation}\label{195}
   |f_{n+1}(t)+\cdots+f_m(t)|<\varepsilon
\end{equation}
for all \(m,n\in \mathbb{N}  \) satisfying \(m>n  \) and every point \(t\in D   \).

Proof

1.  Necessity.

Suppose that the function series \(\sum_{n=1}^{\infty}f_n \)  converges uniformly on \(D  \) and its sum function is \(F   \). Then, for any given \(\varepsilon>0   \), there exists \(N\in \mathbb{N}  \) such that \(n>N  \) implies
 \[\left|F(t)-\sum_{k=1}^n f_k(t)\right|<\frac{\varepsilon}{2}  \quad\mbox{for all}\quad t\in D.     \notag\]
Hence, for all \(m>n>N  \) and all \(t\in D   \), we have
 \[ \begin{aligned}
   |f_{n+1}(t)+\cdots+f_m(t)|
   &=& \left|\sum_{k=1}^mf_k(t)-\sum_{k=1}^nf_k(t)\right|  \\ \\
   &=& \left|\sum_{k=1}^mf_k(t)-F(t)-\sum_{k=1}^nf_k(t)+F(t)\right|  \\ \\
   &\leq& \left|\sum_{k=1}^mf_k(t)-F(t)\right|
         +\left|F(t)-\sum_{k=1}^nf_k(t)\right|  \\ \\
   &<& \frac{\varepsilon}{2}+\frac{\varepsilon}{2}  \\ \\
   &=& \varepsilon.
  \end{aligned}   \notag\]

2.  Sufficiency.

Suppose that for any given \(\varepsilon>0   \), there exists \(N\in \mathbb{N}  \) such that \eqref{195} holds for all \(m>n>N  \) and all \(t\in D   \). Fix \(t_0\in D   \). Then the numerical series \(\sum_{n=1}^{\infty}f_n(t_0)  \) satisfies the Cauchy criterion for convergence of a numerical series. Thus, the numerical series \(\sum_{n=1}^{\infty}f_n(t_0)  \) is convergent. Because \(t_0\in D  \) was arbitrarily chosen, we conclude that the series \(\sum_{n=1}^{\infty}f_n  \) is pointwise convergent on \(D   \). Let \(F  \) be its sum function. Choose \(n  \) for
 \[\left|\sum_{k=1}^mf_k(t)-\sum_{k=1}^nf_k(t)\right|<\frac{\varepsilon}{2}
    \quad\mbox{for all}\quad t\in \mathbb{T}.     \notag\]
If \(m  \to\infty   \), then we get
 \[\left|F(t)-\sum_{k=1}^nf_k(t)\right|
   \leq\frac{\varepsilon}{2}<\varepsilon  \quad\mbox{for all}\quad t\in D.     \notag\]
Therefore, \(\sum_{k=1}^{\infty}f_k  \) converges uniformly to \(F  \) on \(D   \).

This completes the proof.

Corollary 

A necessary condition for the series \(\sum_{n=1}^{\infty}f_n  \) to converge uniformly on \(D  \) is that \(f_n  \to 0  \) uniformly on \(D  \) as \(n  \to\infty   \).

Theorem (The Abel Test)

Suppose that every term \(f_n  \) of the function series
\(\sum_{n=1}^{\infty} f_n  \) satisfies
 \[|f_n(t)|\leq a_n  \quad\mbox{for all}\quad n\in \mathbb{N}  \quad\mbox{and all}\quad t\in D,     \notag\]
and the numerical series \(\sum_{n=1}^{\infty}a_n  \) is convergent. Then the function series \(\sum_{n=1}^{\infty}f_n \)  is uniformly convergent on \(D   \).

Proof

Since the numerical series \(\sum_{k=1}^{\infty}a_k  \) is convergent, for any \(\varepsilon>0   \), there exists \(N\in \mathbb{N}  \) such that \(m>n>N  \) imply
 \[a_{n+1}+a_{n+2}+\cdots+a_m<\varepsilon.     \notag\]
Hence,
 \[ \begin{aligned}
   \left|\sum_{k=n+1}^mf_k(t)\right|
   &\leq& \sum_{k=n+1}^m|f_k(t)|  \\ \\
   &\leq& \sum_{k=n+1}^ma_k  \\ \\
   &<& \varepsilon.
  \end{aligned}   \notag\]
Then, we conclude that \(\sum_{n=1}^{\infty}f_n  \) is uniformly convergent on \(D   \).

Theorem (Dirichlet Test)

Suppose that the function sequence \(\{f_n(t)\}_{n\in \mathbb{N}}  \) is monotone for each \(t\in D  \) with respect to \(n \)  and \(\{f_n\}_{n\in \mathbb{N}}  \) is uniformly convergent to \(0  \) on \(D   \). In addition, assume that the partial sum sequence of \(\sum_{n=1}^{\infty}g_n(t)  \) is uniformly bounded on \(D   \), i.e.,
 \[\left|\sum_{k=1}^n g_k(t)\right|\leq M
    \quad\mbox{for all}\quad t\in D  \quad\mbox{and}\quad n\in \mathbb{N}.     \notag\]
Then \(\sum_{n=1}^{\infty}f_ng_n  \) is uniformly convergent on \(D   \).

Proof

Since \(\{f_n\}_{n\in \mathbb{N}}  \) is uniformly convergent to \(0  \) on \(D   \), for any given \(\varepsilon>0   \), there exists \(N\in \mathbb{N}  \) such that \(n>N  \) implies
 \[|f_n(t)|<\varepsilon  \quad\mbox{for all}\quad t\in D.     \notag\]
Moreover, for all \(m>n>N   \), we have
 \[ \begin{aligned}
   \left|\sum_{k=n+1}^mg_k(t)\right|
   &=& \left|\sum_{k=1}^mg_k(t)-\sum_{k=1}^{n}g_k(t)\right|  \\ \\
   &\leq& \left|\sum_{k=1}^mg_k(t)\right|+\left|\sum_{k=1}^ng_k(t)\right|  \\ \\
   &\leq& 2M.
  \end{aligned}   \notag\]
Using the Abel lemma, we get
 \[ \begin{aligned}
   \left|\sum_{k=n+1}^mf_k(t)g_k(t)\right|
   &\leq& 2M\left(|f_{n+1}(t)|+2|f_m(t)|\right)  \\ \\
   &<& 6M\varepsilon.
  \end{aligned}   \notag\]
Hence, it follows that \(\sum_{n=1}^{\infty}f_ng_n  \) is uniformly convergent on \(D   \).