© Wuhan University 2026

0 Introduction

It is a well-known fact that every absolutely convergent series in a scalar field $\mathbb{K}$ (where $\mathbb{K}$ is commonly $\mathbb{R}$ or $\mathbb{C}$) converges^[1]. To be specific,

$|{\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}{x}_k-{\displaystyle \sum _{k=\mathrm{1}}^n}{x}_k|=|{\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k|\le {\displaystyle \sum _{k=n+\mathrm{1}}^{\infty }}|x_k|$(1)

always holds by the continuity of $|\cdot |$ and the triangle inequality, where $x_k\in \mathbb{K}$. However,

$\Vert {\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k\Vert \le {\displaystyle \sum _{k=n+\mathrm{1}}^{\infty }}\Vert x_k\Vert$(2)

may not necessarily hold (see (19) for example) in a normed vector space $X$, where $x_k\in X$ may be a function, and ${\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k$ may indicate convergence almost everywhere or point wise. Put differently, this normed vector space is endowed with two topologies^[2]: one is induced by the norm, and the other by the point-wise convergence of vector sequences. When the two topologies in question coincide, it becomes a reasonable and intuitive step to interchange the order of taking limits and norms. This is because norms exhibit continuity with respect to this topology. However, as indicated in Ref. [3], these two topologies may be distinct. Thus, interchanging the order of limits and norms necessitates the employment of additional methods or tools.

In many practical applications, the question of whether limits and integrals commute frequently arises. For instance, when establishing limit distributions, if the sequence of probability density functions $\left\{p_n\left(x\right)\right\}$ corresponding to the sequence of random variables $\left\{X_n\right\}$ converges to a certain function, namely $p\left(x\right)$, the question arises as to whether $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\int }_S{p}_n(x)\mathrm{d}x={\int }_{S}p(x)\mathrm{d}x$ for all Borel sets $S$^[4].

In this article, several sufficient conditions that make equation (2) hold in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ space are studied. Generally, the Monotone Convergence Theorem (MCT), Fatou's lemma, and the Dominated Convergence Theorem (DCT) provide sufficient conditions for interchanging limits and integrals (or norms).

In fact, the MCT, Fatou's lemma and the DCT for abstract measure space have wide applications in many fields, such as for random elements in convex combination spaces^[5], in statistical convergence^[6]. Moreover, Fatou's lemma in several dimensions^[7], in a separable Banach space^[8], in infinite-dimensional spaces^[9] have also been studied. On the contrary, Tannery's theorem for series of numbers is really a special case of Weierstrass M-test or the DCT, which is why it probably doesn't get much attention^[10-11]. Meanwhile, the MCT and Fatou's lemma related to series of numbers have been rarely studied in literature, and their corresponding proofs and applications have not been systematically studied. The three theorems for series of numbers, serving as the theoretical basis for interchanging limits and ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ norms, deserve more attention.

Generally, the MCT is firstly proved, and then to prove Fatou's lemma by the MCT both for abstract measure space^[12] and for series of numbers^[13]. In the following text, we first show that the sum of the infimum is less than the infimum of the sum in real number field. If these elements involved in summing are non-negative, then Fatou's lemma for series of numbers can be obtained. If we replace the sum in Fatou's lemma and the MCT with supremum, we obtain Fatou's lemma and the MCT for ${\mathit{\ell }}^{\mathrm{\infty }}$ space. We also provide examples of strict inequality in Fatou's lemma. Furthermore, we obtain the manifestations of Fatou's lemma in ${\mathit{\ell }}^p(\mathrm{1}\le p\le +\mathrm{\infty })$ spaces. And then we prove the MCT for series of numbers and Tannery's theorem by Fatou's lemma for series of numbers. Then we prove the equivalent relationships among Fatou's lemma, the MCT and Tannery's theorem for series of numbers.

We present applications of these theorems. In ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ space, we have demonstrated that series that are absolutely convergent is also convergent. This proves the completeness of these function spaces. In the proof process, we have repeatedly utilized the technique of exchanging limits and norms. Specifically, we use the MCT to prove that the series which is absolutely convergence is in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$. Moreover, we use the MCT for $p=\mathrm{1}$ and $p=\mathrm{\infty }$, and Tannery's theorem for $\mathrm{1}\le p<\mathrm{\infty }$ to prove that the series which is absolutely convergence converges in the ${\mathit{\ell }}^p$ norm. The application of Fatou's lemma also involves proving the completeness of ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ spaces. For a Cauchy sequence in these function spaces, we first identify the potential limit sequence. Subsequently, by applying Fatou's lemma, we prove that this limit sequence is indeed an element of the space. Additionally, we demonstrate that the Cauchy sequence converges to this limit sequence in the ${\mathit{\ell }}^p$ norm.

The paper is organized as follows. In Section 1, Fatou's lemma for series of numbers and for ${\mathit{\ell }}^{\mathrm{\infty }}$, the MCT for series of numbers and for ${\mathit{\ell }}^{\mathrm{\infty }}$ are analyzed. We also give the equivalent relationships of Fatou's lemma, the MCT and Tannery's theorem for series of numbers. In Section 2.1, we provide the applications of the MCT for series of numbers and ${\mathit{\ell }}^{\mathrm{\infty }}$, along with those of Tannery's theorem. In Section 2.2, we provide the applications of Fatou's lemma for series of numbers and ${\mathit{\ell }}^{\mathrm{\infty }}$. In Section 3, we demonstrate that the DCT for ${\mathit{\ell }}^{\mathrm{\infty }}$ is not valid by providing a counterexample. In Section 4, the proofs of the three theorems for series of numbers based on the abstract measure and integration are also provided. In Section 5, we give conclusions of this paper.

1 Theorems for Series of Numbers and ${\mathit{\ell }}^{\mathbf{\infty }}$ Space

Definition 1   Let $\mathbb{R}$ denote the real field and $\mathbb{N}$ the set of natural numbers. The set ${\mathbb{R}}^{\mathbb{N}}$ of all sequences of elements of $\mathbb{R}$ is a vector space for component-wise addition and component-wise scalar multiplication. If $\mathrm{1}\le p<\mathrm{\infty }$, ${\mathit{\ell }}^p$ is the subspace of ${\mathbb{R}}^{\mathbb{N}}$ consisting of all sequences $x={\left\{x_j\right\}}_{j\in \mathbb{N}}$ satisfying ${\displaystyle \sum _{j=\mathrm{1}}^{\infty }}|x_j{|}^p<\mathrm{\infty }$. The real-valued function $\Vert \cdot {\Vert }_p$ is defined by

$\Vert x{\Vert }_p=(\sum _j|x_j{|}^p{)}^{\frac{\mathrm{1}}{p}}<\mathrm{\infty },\hspace{1em}\hspace{1em}\forall x\in {\mathit{\ell }}^p.$

If $p=\mathrm{\infty }$, then ${\mathit{\ell }}^{\mathrm{\infty }}$ is defined to be the space of all bounded sequences endowed with the norm

$\Vert x{\Vert }_{\mathrm{\infty }}=\underset{j}{\mathrm{s}\mathrm{u}\mathrm{p}}|x_j|<\mathrm{\infty },\hspace{1em}\hspace{1em}\forall x\in {\mathit{\ell }}^{\mathrm{\infty }}.$

Lemma 1   Suppose $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence of real numbers. Then for any $m,N\in \mathbb{N}$, we have

${\displaystyle \sum _{n=\mathrm{1}}^N}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^N}{x}_k(n),$(3)

and

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n).$(4)

Proof   It is obvious that

$\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le x_k(n),\forall k\ge m$

Thus

${\displaystyle \sum _{n=\mathrm{1}}^N}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le {\displaystyle \sum _{n=\mathrm{1}}^N}{x}_k(n),\forall k\ge m$

Thus, (3) is derived by taking the infimum with respect to $k\ge m$ on both side of the above inequality. (4) can be obtained in a similar manner.

Remark 1   If the condition of non-negative for $x_m(n)$ is added to this lemma, Fatou's lemma can be obtained.

Theorem 1   (Fatou's lemma for series of numbers^[13]). Suppose $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence of nonnegative real numbers (i.e., $x_m(n)\ge \mathrm{0}$ for each $m,n$). Then

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n).$(5)

Proof   If $x_k(n)\ge \mathrm{0},\forall k,n\in \mathbb{N}$, we have

$\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^N}{x}_k(n)\le \underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n),\forall m,N\in \mathbb{N},$(6)

by the non-decreasing property of the partial sum sequence of positive series with respect to the number of terms.

By virtue of equations (3) and (6), along with the transitive nature of inequalities, we can derive

${\displaystyle \sum _{n=\mathrm{1}}^N}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n),$(7)

We take the limit as $m\to \mathrm{\infty }$ on both sides of equation (7). By invoking the property that allows us to exchange the order of the limit and the finite-summation, we obtain

${\displaystyle \sum _{n=\mathrm{1}}^N}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n),$(8)

Letting $N\to \mathrm{\infty }$ on both sides of (8), and noting that $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}$ is just the $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}$, we have (5).

Theorem 2   (Fatou's lemma for ${\mathit{\ell }}^{\mathrm{\infty }}$ space). Suppose $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence of real numbers. Prove that

$\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n).$(9)

Proof   It is obvious that

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n),\forall n\in \mathbb{N}.$

Thus

$\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n).$

Remark 2   In Theorems 2 (Fatou's lemma for ${\mathit{\ell }}^{\mathrm{\infty }}$ space) and 4 (the MCT for ${\mathit{\ell }}^{\mathrm{\infty }}$ space), it's not necessary for $x_m(n)$ to be non-negative. But in applications of this paper, we only encounter non-negative situations.

The following Example 1 illustrates that equations (3), (5) and (9) can obtain strict inequality.

Example 1 Let

$x_k(n)=\{\begin{array}{l}\mathrm{1},k=n\\ \mathrm{0},k\ne n\end{array},$

where $n,k\in \mathbb{N}$. Then $\underset{\mathrm{1}\le k\le \mathrm{3}}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(\mathrm{1})+\underset{\mathrm{1}\le k\le \mathrm{3}}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(\mathrm{2})+\underset{\mathrm{1}\le k\le \mathrm{3}}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(\mathrm{3})=\mathrm{0}<$

$\mathrm{1}=\underset{\mathrm{1}\le k\le \mathrm{3}}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{3}}}{x}_k(n);{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{k\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)=\mathrm{0}<\mathrm{1}=\underset{k\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n);$

and $\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{k\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)=\mathrm{0}<\mathrm{1}=\underset{k\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_k(n).$

Corollary 1   (The manifestations of Fatou's lemma in ${\mathit{\ell }}^p(\mathrm{1}\le p\le +\mathrm{\infty })$ spaces). Suppose $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence, where $x_m(n)\in \mathbb{R}$. We have

$\Vert \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n){\Vert }_{\mathrm{\infty }}\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\Vert x_m(n){\Vert }_{\mathrm{\infty }},$(10)

and

$\Vert \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n){\Vert }_p^p\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\Vert x_m(n){\Vert }_p^p,\forall \mathrm{1}\le p<\mathrm{\infty },$(11)

provided all expressions that appeared in the above two inequalities are meaningful.

Proof   $\Vert \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n){\Vert }_{\mathrm{\infty }}=\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}|\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)|=\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}$

$|x_m(n)|=\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}|x_m(n)|\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}|x_m(n)|=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}$

$\Vert x_m(n){\Vert }_{\mathrm{\infty }}.$

The inequality in the above equation is due to (9).

$\mathrm{I}\mathrm{f}\mathrm{1}\le p<\mathrm{\infty },\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\Vert \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n){\Vert }_p^p={\displaystyle \sum _{n=\mathrm{1}}^{\infty }}|\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n){|}^p=$

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}|x_m(n){|}^p\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}|x_m(n){|}^p=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\Vert x_m(n){\Vert }_p^p$

The inequality in the above equation is due to (5).

Remark 3   If the condition

$x_m(n)\le x_{m+\mathrm{1}}(n),\forall n,m\in \mathbb{N}$

is added in Fatou's lemma, the inequality in Fatou's lemma will become the equality in the MCT.

Theorem 3   (the MCT for series of numbers^[13]). Suppose that $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence of nonnegative real numbers (i.e., $x_m(n)\ge \mathrm{0}$ for each $m,n$). Suppose that $x_m(n)\le x_{m+\mathrm{1}}(n)$ for each $m,n$. Set $x_{\mathrm{\infty },n}=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)$. Thus

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_{\mathrm{\infty },n}=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n).$(12)

Proof   By Fatou's lemma, we have

$\begin{array}{l}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_{\mathrm{\infty },n}={\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)={\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_m(n)\\ \le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n)=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n).\end{array}$(13)

Since $x_m(n)\le x_{m+\mathrm{1}}(n)\le x_{\mathrm{\infty }}(n)$, we have ${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n)\le {\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_{m+\mathrm{1}}(n)\le {\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_{\mathrm{\infty }}(n)$. Thus

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n)\le {\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_{\mathrm{\infty }}(n).$(14)

By (13) and (14), we have (12).

Corollary 2   The MCT and Fatou's lemma for series of numbers are equivalent to each other, that is, they can be proven to each other.

Proof   Theorem 3 illustrates that the MCT can be proved by Fatou's lemma. We direct the reader to Ref. [13] for the proof demonstrating that Fatou's lemma can be established through the MCT for numerical series.

Theorem 4   (the MCT for ${\mathit{\ell }}^{\mathrm{\infty }}$ space). Suppose that $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence of real numbers. Suppose that $x_m(n)\le x_{m+\mathrm{1}}(n),\forall m,n\in \mathbb{N}$. Set $x_{\mathrm{\infty },n}=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)$. Prove that

$\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)=\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n).$(15)

Proof   By (9), we have

$\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{x}_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n).$

Since $x_m(n)\le x_{m+\mathrm{1}}(n),\forall m,n\in \mathbb{N}$, the $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}$ in the above inequality can be written as $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}$. That is

$\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n).$

Conversely, we are required to establish inequality in the opposite direction. Since $x_m(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)$, we have $\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n)\le \underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)$. Thus

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{x}_m(n)\le \underset{n\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n).$

Then (15) is obtained.

Theorem 5   (The DCT for series of numbers known as Tannery's theorem^[10-11]). Suppose that $\{x_m{(n)\}}_{m,n\in \mathbb{N}}$ is a sequence of real numbers. $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)=x_{\mathrm{\infty }}(n),|x_m(n)|\le M(n)$, and ${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}M(n)<\mathrm{\infty }$, then

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}|x_m(n)-x_{\mathrm{\infty }}(n)|=\mathrm{0};$(16)

and

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_m(n)={\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_{\mathrm{\infty }}(n).$(17)

We will prove Tannery's theorem by using Fatou's lemma. The proof without Fatou's lemma and the MCT can be found in Refs. [10-11].

Proof   Since $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)=x_{\mathrm{\infty }}(n)$ and $|x_m(n)|\le M(n)$, we have that $|x_{\mathrm{\infty }}(n)-x_m(n)|\le \mathrm{2}M(n)$. Thus $\mathrm{2}M(n)-|x_{\mathrm{\infty }}(n)-x_m(n)|\ge \mathrm{0}$. By Theorem 4, we have

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}(\mathrm{2}M(n)-|x_{\mathrm{\infty }}(n)-x_m(n)|)$

$\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}(\mathrm{2}M(n)-|x_{\mathrm{\infty }}(n)-x_m(n)|).$

Subtracting ${\displaystyle \sum _{n=\mathrm{1}}^{\infty }}(\mathrm{2}M(n))$ from both sides of this inequality yields

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}(-|x_{\mathrm{\infty }}(n)-x_m(n)|)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}(-|x_{\infty }(n)-x_m(n)|).$

Thus

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}(|x_{\mathrm{\infty }}(n)-x_m(n)|)\ge \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}(|x_{\infty }(n)-x_m(n)|).$

Since $\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m(n)=x_{\mathrm{\infty }}(n)$, the left-hand side of the above-mentioned inequality equals $\mathrm{0}$. Thus

$\begin{array}{l}\mathrm{0}\ge \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{s}\mathrm{u}\mathrm{p}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}(|x_{\infty }(n)-x_m(n)|)\\ \ge \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\infty }}(|x_{\infty }(n)-x_m(n)|)\ge \mathrm{0}.\end{array}$

Thus, (16) is proved. (17) can be obtained by (16) immediately.

Corollary 3   Fatou's lemma for series of numbers can be proved by Tannery's theorem provided that

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)<\mathrm{\infty }.$(18)

Proof   It is obvious that $\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)$. By (18) and Tannery's theorem, we have

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)={\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n).$

By (4), we have

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n),$

and

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n),$

Thus,

${\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{x}_k(n)\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{k\ge m}{\mathrm{i}\mathrm{n}\mathrm{f}}{\displaystyle \sum _{n=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(n).$

Theorem 6   A normed vector space $X$ is complete if and only if every absolutely convergent series in $X$ converges^[12,14].

This theorem will be used to prove Example 2.

2 Applications

2.1 Applications of the MCT and the DCT

Example 2 Prove that every absolutely convergent series in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ converges. Thus ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ are complete.

Proof   We use Theorem 6. Suppose $\{x_k{\}}_{k=\mathrm{1}}^{\mathrm{\infty }}\subset {\mathit{\ell }}^p\mathrm{a}\mathrm{n}\mathrm{d}{\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_p=B<\mathrm{\infty }$where $x_k=\{x_k{(j)\}}_{j\in \mathbb{N}}.$ Let $\left|x_k\right|={\{\left|x_k(j)\right|\}}_{j\in \mathbb{N}},Y_n={\displaystyle \sum _{k=\mathrm{1}}^n}|x_k|,Y={\displaystyle \sum _{k=\mathrm{1}}^{\infty }}|x_k|,X={\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}{x}_k.\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{n}\Vert Y_n{\Vert }_p\le {\displaystyle \sum _{k=\mathrm{1}}^n}\Vert x_k{\Vert }_p\le {\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_p=B<\mathrm{\infty }$ by the triangle inequality of ${\mathit{\ell }}_p$ norm, and $|X(j)|=|{\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}{x}_k(j)|\le {\displaystyle \sum _{k=\mathrm{1}}^{\infty }}|x_k(j)|=Y(j),\forall j\in \mathbb{N}$. We claim $X\in {\mathit{\ell }}_p$ and $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\Vert X-{\displaystyle \sum _{k=\mathrm{1}}^n}{x}_k{\Vert }_p=\mathrm{0}$. Thus, the series $X={\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}{x}_k$ converges in the ${\mathit{\ell }}_p$ norm.

$\begin{array}{l}\mathrm{1})\mathrm{1}\le p<\mathrm{\infty }.\Vert X{\Vert }_p^p={\displaystyle \sum _{j=\mathrm{1}}^{\infty }}|X(j){|}^p\le {\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}Y{(j)}^p\\ ={\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{Y}_n{(j)}^p=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}{Y}_n{(j)}^p=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\Vert Y_n{\Vert }_p^p\le B^p<\mathrm{\infty },\end{array}$

where the third equality is due to the MCT for series of numbers. Thus $X\in {\mathit{\ell }}_p$. By the DCT for series of numbers and $|X-{\displaystyle \sum _{j=\mathrm{1}}^n}{x}_k{|}^p\le {(|X|+|{\displaystyle \sum _{j=\mathrm{1}}^n}{x}_k|)}^p\le {(\mathrm{2}Y)}^p\in {\mathit{\ell }}^{\mathrm{1}}$, we have

$\begin{array}{l}\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\Vert X-{\displaystyle \sum _{j=\mathrm{1}}^n}{x}_k{\Vert }_p^p=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{j=\mathrm{1}}^{\infty }}|X-{\displaystyle \sum _{j=\mathrm{1}}^n}{x}_k{|}^p={\displaystyle \sum _{j=\mathrm{1}}^{\infty }}\mathrm{0}=\mathrm{0}.\\ \mathrm{2})p=\mathrm{\infty }.\Vert X{\Vert }_{\mathrm{\infty }}=\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}|X(j)|\le \underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}Y(j)\end{array}$

$=\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{Y}_n(j)=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{Y}_n(j)=\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\Vert Y_n{\Vert }_{\mathrm{\infty }}\le B<\mathrm{\infty }$, where the third equality is due to the MCT for ${\mathit{\ell }}^{\mathrm{\infty }}$. Thus $X\in {\mathit{\ell }}^{\mathrm{\infty }}$. Since ${\displaystyle \sum _{k=\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_{\mathrm{\infty }}=B<\mathrm{\infty }$, we have $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_{\mathrm{\infty }}=\mathrm{0}$. If we can prove

$\Vert X-{\displaystyle \sum _{k=\mathrm{1}}^n}{x}_k{\Vert }_{\mathrm{\infty }}=\Vert {\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k{\Vert }_{\mathrm{\infty }}\le {\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_{\mathrm{\infty }},$(19)

then we can obtain $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\Vert X-{\displaystyle \sum _{k=\mathrm{1}}^n}{x}_k{\Vert }_{\mathrm{\infty }}=\mathrm{0}$. In fact,

$\begin{array}{l}\Vert X-{\displaystyle \sum _{k=\mathrm{1}}^n}{x}_k{\Vert }_{\mathrm{\infty }}=\Vert {\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k{\Vert }_{\mathrm{\infty }}=\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}|{\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k(j)|\\ \le \underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{\displaystyle \sum _{k=n+\mathrm{1}}^{\infty }}|x_k(j)|\\ =\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}\underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}|x_k(j)|=\underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}|x_k(j)|\\ \le \underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}\underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}|x_k(j)|=\underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}\Vert x_k{\Vert }_{\mathrm{\infty }}={\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_{\mathrm{\infty }},\end{array}$

where the fourth equality is due to the MCT for ${\mathit{\ell }}^{\mathrm{\infty }}$. Thus, (19) is proved.

In the above proof process, when $\mathrm{1}\le p<\mathrm{\infty }$, we use Tannery's theorem to prove that this series converges in the ${\mathit{\ell }}^p$ norm. When $p=\mathrm{\infty }$, we used the MCT to prove that this series converges in the ${\mathit{\ell }}^{\mathrm{\infty }}$ norm. It is worth noting that when $p=\mathrm{1}$, the MCT can also be used to prove that this series converges in the ${\mathit{\ell }}^{\mathrm{1}}$ norm. It suffices to prove (2). The proof process is as follows.

$\begin{array}{l}\Vert X-{\displaystyle \sum _{k=\mathrm{1}}^n}{x}_k{\Vert }_{\mathrm{1}}=\Vert {\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k{\Vert }_{\mathrm{1}}={\displaystyle \sum _{j=\mathrm{1}}^{\infty }}|{\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}{x}_k(j)|\\ \le {\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}{\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}|x_k(j)|={\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}\underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}|x_k(j)|\\ =\underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}|x_k(j)|=\underset{N\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\displaystyle \sum _{k=n+\mathrm{1}}^N}{\displaystyle \sum _{j=\mathrm{1}}^{\mathrm{\infty }}}|x_k(j)|={\displaystyle \sum _{k=n+\mathrm{1}}^{\mathrm{\infty }}}\Vert x_k{\Vert }_{\mathrm{1}},\\ \end{array}$

where the fourth equality is due to the MCT for series of numbers. Thus (2) is proved. Usually, the completeness of $L^p(\mathrm{1}\le p<\mathrm{\infty })$ can be proved by using the DCT ^[12]. However, as pointed out here, the completeness of $L^{\mathrm{1}}$ can also be proved without the DCT by using the MCT.

2.2 Applications of Fatou's Lemma

Example 3 Prove that every Cauchy sequence in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ converges. Thus ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ are complete.

Claim 1 Each Cauchy sequence in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$ converges point wise. In fact, if $\{x_n{\}}_{n=\mathrm{1}}^{\mathrm{\infty }}$ is a Cauchy sequence in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$, where $x_n=\{x_n{\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}$. According to the definition of Cauchy sequence, we have that $\forall \epsilon >\mathrm{0},\exists N\in \mathbb{N}$ such that

$\Vert x_n-x_m{\Vert }_p<\epsilon ,\hspace{1em}\forall n,m>N.$(20)

Since $|x_n\left(j\right)-x_m\left(j\right)|\le ({\displaystyle \sum _{j=\mathrm{1}}^{\infty }}|x_n\left(j\right)-x_m\left(j\right){|}^p{)}^{\frac{\mathrm{1}}{p}}=\Vert x_n-$

$\begin{array}{l}{x}_m{\Vert }_p,\forall \mathrm{1}\le p<\mathrm{\infty }\mathrm{a}\mathrm{n}\mathrm{d}|x_n\left(j\right)-x_m\left(j\right)|\le \underset{j\ge \mathrm{1}}{\mathrm{s}\mathrm{u}\mathrm{p}}|x_n\left(j\right)-x_m\left(j\right)|\\ =\Vert x_n-x_m{\Vert }_{\mathrm{\infty }},\mathrm{w}\mathrm{e}\mathrm{h}\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}|x_n\left(j\right)-x_m\left(j\right)|<\epsilon ,\forall n,m>N\end{array}$

for each $j\in \mathbb{N}$. That is for each fixed $j\in \mathbb{N}$, $\{x_m{\left(j\right)\}}_{m=\mathrm{1}}^{\mathrm{\infty }}$ is a Cauchy sequence. By the completeness of the field $\mathbb{R}$, there exists a $x(j)\in \mathbb{R}$ such that

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{x}_m\left(j\right)=x\left(j\right)$

for each $j\in \mathbb{N}$. We will prove that $x={\{x\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}\in {\mathit{\ell }}^p$ in the next step.

Claim 2 $x={\{x\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}$ is in ${\mathit{\ell }}^p(\mathrm{1}\le p\le \mathrm{\infty })$. For ${\epsilon }_{\mathrm{0}}=\mathrm{1}$, there exists $N_{\mathrm{0}}\in \mathbb{N}$ such that

$\Vert x_m-x_{N_{\mathrm{0}}}{\Vert }_p\le {\epsilon }_{\mathrm{0}}=\mathrm{1},\hspace{1em}\forall m\ge N_{\mathrm{0}}.$(21)

1) $\mathrm{1}\le p<\mathrm{\infty }$. By (21), we have

$\underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\Vert x_m-x_{N_{\mathrm{0}}}{\Vert }_p^p\le {\epsilon }_{\mathrm{0}}^p=\mathrm{1}.$

By (11), we have $\Vert x-x_{N_{\mathrm{0}}}{\Vert }_p^p=\Vert \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}(x_m-x_{N_{\mathrm{0}}}){\Vert }_p^p\le \underset{m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{f}}\Vert x_m-x_{N_{\mathrm{0}}}{\Vert }_p^p\le {\epsilon }_{\mathrm{0}}^p=\mathrm{1}.$ That is to say that $\{x\left(j\right)-x_{N_{\mathrm{0}}}{\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}\in {\mathit{\ell }}^p$. Thus ${\{x\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}=\{x\left(j\right)-x_{N_{\mathrm{0}}}{\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}+\{x_{N_{\mathrm{0}}}{\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}$ is in ${\mathit{\ell }}^p$ since $\{x_{N_{\mathrm{0}}}{\left(j\right)\}}_{j=\mathrm{1}}^{\mathrm{\infty }}\in {\mathit{\ell }}^p$ and ${\mathit{\ell }}^p$ is closed for addition.