relations between some double sequences and series of functions(SEQUENCE AND SERIES OF FUNCTIONS)

8.3 Theorem Given a double sequence $\{a_{i,j}\}$, $i=1,2,3,\ldots, j=1,2,3,\ldots,$ suppose that

$$ \begin{equation} \sum_{j=1}^{\infty}\vert a_{i,j}\vert=b_i\hspace{1cm}(i=1,2,3,\ldots)\end{equation} $$

and $\sum b_i$ converges. Then

$$ \begin{equation} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$

Proof Let $E$ be a countable set, consisting of the points $x_0, x_1, x_2, \ldots,$ and suppose $x_{n} \rightarrow x_{0}$ as $n \rightarrow \infty$. Define

$$ \begin{equation} \begin{aligned} f_i(x_n)&=\sum_{j=1}^{n}a_{i,j} \hspace{1cm}(i,n=1,2,3,\ldots),\\ f_i(x_0)&=\sum_{j=1}^{\infty}a_{i,j} \hspace{1cm}(i=1,2,3,\ldots),\\ g(x)&=\sum_{i=1}^{\infty}f_{i}(x) \hspace{1cm}(x\in E). \end{aligned}\end{equation} $$

And they are pretty well-defined. Since

$$ \begin{equation} \lim_{n\rightarrow\infty} f_i(x_n)=f_i(x_0)\end{equation} $$

and

$$ \begin{equation} \lim_{n \rightarrow \infty}x_n=x_0,\end{equation} $$

$f_i$ is continuous at $x_{0}$. In addition, since

$$ \begin{equation} f_i(x_0)\leq \sum_{j=1}^{\infty}\vert a_{i,j}\vert=b_i\hspace{1cm}(i=1,2,3,\ldots),\\ f_i(x_n)\leq \sum_{j=1}^{\infty}\vert a_{i,j}\vert=b_i\hspace{1cm}(i=1,2,3,\ldots),\end{equation} $$

and $\sum b_i$ converges, we know that $g(x)=\sum f_i$ converges uniformly, thus $g$ is continuous at $x_{0}$ too. The continuity of $g$ implies that

$$ \begin{equation} \begin{aligned} \lim_{n \rightarrow \infty}g(x_n)&=g(x_0), \\ g(x_0)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}&=\lim_{n \rightarrow \infty}g(x_n)=\lim_{n \rightarrow \infty}\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j}. \end{aligned}\end{equation} $$

  Note that

$$ \begin{equation} g(x_n)=\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j} =\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j},\end{equation} $$

which can be proved by using the mathematical induction.

  Firstly, it’s obivious that the equation is correct when $n=1$. Assume that it holds for $n=k(k\in N^{*})$. i.e.,

$$ \begin{equation} g(x_k)=\sum_{i=1}^{\infty}\sum_{j=1}^{k}a_{i,j}= \sum_{j=1}^{k}\sum_{i=1}^{\infty}a_{i,j},\end{equation} $$

and then consider the situation where $n=k+1$. Since

$$ \begin{equation} \begin{aligned} g(x_{k+1})&=\sum_{i=1}^{\infty}\sum_{j=1}^{k+1}a_{i,j}= \sum_{i=1}^{\infty}\sum_{j=1}^{k}a_{i,j}+\sum_{i=1}^{\infty}a_{i,k+1}\\ &=\sum_{j=1}^{k}\sum_{i=1}^{\infty}a_{i,j}+\sum_{i=1}^{\infty}a_{i,k+1}\\ &=\sum_{j=1}^{k+1}\sum_{i=1}^{\infty}a_{i,j}, \end{aligned}\end{equation} $$

hence we conclude that

$$ \begin{equation} g(x_n)=\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j} =\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$

  Thus,

$$ \begin{equation} \begin{aligned} g(x_0)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}&=\lim_{n \rightarrow \infty}g(x_n)=\lim_{n \rightarrow \infty}\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{i,j}\\ &=\lim_{n \rightarrow \infty}\sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{i,j}\\ &=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}, \end{aligned}\end{equation} $$

i.e.,

$$ \begin{equation} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{i,j}.\end{equation} $$