1 Introduction and basic definitions

Banach spaces, which are separable and reflexive, can exist without a Schauder basis as proved by Enflo in 1973 [11]. However, in 1972, Morrell and Retherford [8] showed that in each infinite-dimensional Banach space and for any sequence of positive numbers, that is, monotonically convergent to zero \((\lambda _{i})_{i\in N}\), where \(N=\{1,2,3,\ldots \}\), one can construct a weakly square-summable basic sequence whose norms equal to \((\lambda _{i})_{i \in N}\).

In 1977, Makarov and Faried [7] showed how to construct compact operators of the form \(\sum_{i\in N} \mu _{i}f_{i}\otimes x_{i}\) between arbitrary infinite-dimensional Banach spaces such that its sequence of approximation numbers has a specific rate of convergence to zero. It was also proved that the operator ideal, whose sequence of approximation numbers are p-summable, is a small ideal; see [4, 10, 11].

In this work, we show how to construct compact operators between arbitrary infinite-dimensional Banach spaces using a countable number of basic sequences and nuclear operators, represented in the form of an infinite-dimensional matrix \((\mu _{ij})_{i,j\in N}\) defined over the space \(\ell _{1}\) of all absolutely summable sequences, which verifies

$$\begin{aligned} \lim_{j}\mu _{ij}=0 \end{aligned}$$

for every \(i\in N\). For such double-summation operators, a choice of matrix elements is more convenient than choosing sequence elements in the case of single-summation operators. Such a construction will help give counterexamples of operators between Banach spaces without a Schauder basis. An upper estimate of the sequence of approximation numbers is given for such double-summation operators. For basic notions and some related results, one can see [1, 6, 9, 13].

The following notations are used throughout this study. The normed space of bounded linear operators from a normed space X into a normed space Y is denoted by \(L(X, Y)\), while the dual space of the normed space X is denoted by \(X^{*}=L(X, R)\), where R is the set of real numbers.

Also as mentioned before, the space \(\{x=(x_{i})_{i=1}^{\infty }:\sum_{i}|x_{i}|^{p} <\infty \}\) of all sequences of real numbers that are p-absolutely summable, is denoted by \(\ell _{p}\), which is equipped with the norm \(\|x\|=(\sum_{i\in N}|x_{i}|^{p})^{\frac{1}{p}}\). The space \(\{x=(x_{i})_{i=1}^{\infty }: \lim x_{i}=0\}\) of all sequences of real numbers that are convergent to zero, is denoted by \(c_{o}\), which is equipped with the norm \(\|x\|=\sup_{i\in N}|x_{i}|\).

Definition 1.1

([12])

A map s, which assigns a unique sequence \(\{s_{r}(T)\}_{r=0}^{ \infty }\) of real numbers to every operator \(T\in {L(X,Y)}\), is called an s-number sequence if the following conditions are verified:

  1. 1.

    \(\|T\|=s_{0}(T)\geq s_{1}(T)\geq \cdots \geq 0\) for \(T\in L(X,Y)\).

  2. 2.

    \(s_{r+m}(U+V)\leq s_{r}(U)+s_{m}(V)\) for \(U,V\in L(X,Y)\).

  3. 3.

    \(s_{r}(UTV)\leq \|U\|s_{r}(T)\|V\|\) for \(V\in L(X_{0},X), T \in L(X,Y)\) and

    \(U\in L(Y,Y_{0})\).

  4. 4.

    \(s_{r}(T)=0\) if and only if \(\operatorname{rank}(T)\leq r\) for \(T\in L(X,Y)\).

  5. 5.

    \(s_{r}(I_{k})=\bigl\{ \begin{array}{l@{\quad}l} 1, & \text{for }r< k; \\ 0, & \text{for }r\geq k, \end{array} \)

where \(I_{k}\) is the identity operator on Euclidean space \(\ell _{2} ^{k}\).

As an examples of s-numbers, we mention the approximation numbers \(\alpha _{r}(T)\), Gelfand numbers \(c_{r}(T)\), Kolmogorov numbers \(d_{r}(T)\), and Tikhomirov numbers \(d_{r}^{*}(T)\), defined by

  1. 1.

    \(\alpha _{r}(T)=\inf \{\|T-A\|: A\in L(X,Y)\) and \(\operatorname{rank}(A)\leq r\}\). Clearly, we always have \(\|T\|=\alpha _{0}(T)\geq \alpha _{1}(T)\geq \alpha _{2}(T)\geq \cdots \geq 0\).

  2. 2.

    \(c_{r}(T)=\alpha _{r}(J_{Y}T)\), where \(J_{Y}\) is a metric injection from the space Y into a higher space \(\ell ^{\infty }( \varLambda )\) of all bounded-real functions for a suitable index set Λ.

  3. 3.
    $$\begin{aligned} d_{r}(T)=\inf_{\operatorname{dim}K\leq r} \sup_{ \Vert x \Vert \leq 1} \inf_{y\in K} \Vert Tx-y \Vert , \end{aligned}$$

    where \(K\subseteq Y\).

  4. 4.

    \(d_{r}^{*}(T)=d_{r}(J_{Y}T)\).

Definition 1.2

([11])

An operator \(T\in L(X,Y)\) is nuclear if and only if it can be represented in the form

$$\begin{aligned} T(x)=\sum_{i=1}^{\infty }a_{i}(x)y_{i}, \end{aligned}$$

with \(a_{1}, a_{2},\ldots \in X^{*}\) and \(y_{1}, y_{2}, \ldots \in Y\), such that

$$\begin{aligned} \sum_{i=1}^{\infty } \Vert a_{i} \Vert \Vert y_{i} \Vert < \infty. \end{aligned}$$

On the class \(N(X,Y)\) of all nuclear operators from X into Y, a norm \(\nu (T)\) is defined by

$$\begin{aligned} \nu (T)=\inf \biggl\{ \sum_{i} \Vert a_{i} \Vert \Vert y_{i} \Vert \biggr\} , \end{aligned}$$

where the inf is taken over all possible representations of the operator T.

2 Basic theorems and technical lemmas

It is well known that an infinite matrix defines a linear continuous operator from the space \(\ell _{1}\) into itself if its columns are absolutely uniformly-summable; see [3, 4, 10].

Lemma 2.1

([11], 6.3.6)

An operator \(T\in L(\ell _{1},\ell _{1})\) is nuclear if and only if there is an infinite matrix \((\sigma _{ik})_{i,k\in N}\) such that

$$\begin{aligned} T(x)= \Biggl(\sum_{k=1}^{\infty }\sigma _{ik}x_{k} \Biggr)_{i=1}^{\infty } \quad\textit{for } x=(x_{k})_{k=1}^{\infty }\in \ell _{1} \end{aligned}$$

and

$$\begin{aligned} \sum_{i=1}^{\infty }\sup_{k} \vert \sigma _{ik} \vert < \infty. \end{aligned}$$

In this case

$$\begin{aligned} \nu (T)=\sum_{i=1}^{\infty }\sup _{k} \vert \sigma _{ik} \vert . \end{aligned}$$

Lemma 2.2

([3])

If \((T_{i})_{i=1}^{\infty }\)is an absolutely summable sequence of bounded linear operators then

$$\begin{aligned} \alpha _{n} \Biggl(\sum_{i=1}^{\infty }T_{i} \Biggr)\leq \inf \Biggl\{ \sum_{i=1}^{\infty } \alpha _{n_{i}}(T_{i}):\sum_{i=1}^{\infty }n_{i}=n \Biggr\} , \end{aligned}$$

where the inf is taken over all possible representations for

$$\begin{aligned} \sum_{i=1}^{\infty }n_{i}=n. \end{aligned}$$

The following is a consequence of Lemma 2 in [2].

Theorem 2.3

Let \((x_{i})_{i=1}^{\infty }\)be a sequence in a Banach spaceXsuch that

$$\begin{aligned} \sum_{i=1}^{\infty } \bigl\vert f(x_{i}) \bigr\vert < \infty\quad \textit{for every } f\in X^{*}, \end{aligned}$$

then the series \(\sum_{i=1}^{\infty }\lambda _{i}x_{i}\)converges unconditionally inXfor every sequence \((\lambda _{i})_{i=1}^{ \infty }\in c_{o}\).

Theorem 2.4

(Morrell and Retherford [8])

LetXbe an infinite-dimensional Banach space and let \((\lambda _{i})_{i=1} ^{\infty }\in c_{o}\)with \(0<\lambda _{i}<1\), then there is a basic sequence \((x_{i})_{i=1}^{\infty }\)inXsuch that \(\|x_{i}\|=\lambda _{i}\)for all \(i=1,2,\ldots \)that verifies

$$\begin{aligned} \sum_{i=1}^{\infty } \bigl\vert f(x_{i}) \bigr\vert ^{2}\leq \Vert f \Vert ^{2} \quad\textit{for every } f\in X^{*}. \end{aligned}$$

Remark 2.5

Theorem 2.4 is valuable in the case of sequences that are slowly convergent to zero \((\lambda _{i})_{i=1}^{\infty }\). Indeed, if \((\lambda _{i})_{i=1}^{\infty }\) converges rapidly to zero then \(\sum_{i=1}^{\infty }\|x_{i}\|<\infty \) and hence, one can write

$$\begin{aligned} \sum_{i=1}^{\infty } \bigl\vert f(x_{i}) \bigr\vert ^{2}\leq \sum _{i=1}^{\infty } \Vert f \Vert ^{2} \Vert x_{i} \Vert ^{2}\leq C \Vert f \Vert ^{2} \quad\text{for every } f\in X^{*}. \end{aligned}$$

Theorem 2.6

(Dini’s theorem [5])

For a convergent series \(\sum_{i=1}^{\infty }a_{i}\)of positive real numbers, the series

$$\begin{aligned} \sum_{i=1}^{\infty }\frac{a_{i}}{R_{i}^{m}} \quad\textit{is } \textstyle\begin{cases} \textit{convergent} & \textit{for }m< 1; \\ \textit{divergent} & \textit{for }m\geq 1, \end{cases}\displaystyle \end{aligned}$$

where \(R_{i}=\sum_{j=i}^{\infty }a_{j}\)is the remainder of the series \(\sum_{i=1}^{\infty }a_{i}\).

Theorem 2.7

([7])

LetXandYbe infinite-dimensional Banach spaces and let \((\lambda _{r})_{r=1}^{\infty }\)be a monotonically decreasing sequence of positive real numbers, then there is a completely continuous operator \(A\in L(X,Y)\)verifying

$$\begin{aligned} 2^{-4}\lambda _{3r}\leq d_{r}^{*}(A) \leq \alpha _{r}(A)\leq 8\lambda _{r} \quad\textit{for every } r \in \{1,2,\ldots \}. \end{aligned}$$

Lemma 2.8

([3])

Let \(\{\xi _{i}\}_{i\in N}\)be a bounded family of real numbers and let \(K\subseteq N\)be an arbitrary subset of indices, such that cardKis the number of elements inK. Then

$$\begin{aligned} \sup_{\operatorname{card} K=r+1} \inf_{i\in K}\xi _{i} = \inf_{\operatorname{card} K=r} \sup_{i\notin K} \xi _{i}. \end{aligned}$$

3 Main results

Proposition 3.1

LetXandYbe infinite-dimensional Banach spaces and let \(M=(\mu _{ij})_{i,j\in N}\)be an infinite matrix verifying that:

  1. 1.

    \(\lim_{j}\mu _{ij}=0 \)for every \(i\in N\).

  2. 2

    \(\sum_{i=1}^{\infty }\sup_{j=1}^{\infty } \vert \mu _{ij} \vert <\infty\).

Let \((f_{ij})_{i,j\in N}\)be a matrix of functionals in \(X^{*}\)and \((z_{ij})_{i,j\in N}\)be a matrix of elements inYthat verifies

$$ \sup_{i=1}^{\infty }\sum _{j=1}^{\infty } \bigl\vert f_{ij}(x)F(z_{ij}) \bigr\vert < \infty $$
(1)

for everyFin \(Y^{*}\)and everyxinX. Then the expression

$$\begin{aligned} T(x)=\sum_{i=1}^{\infty }\sum _{j=1}^{\infty }\mu _{ij} f_{ij}(x) z _{ij} \end{aligned}$$

defines a linear continuous operator fromXintoY.

Proof

Let

$$\begin{aligned} \lambda _{n}=\sum_{i\geq n}\sup _{j=1}^{\infty } \vert \mu _{ij} \vert , \end{aligned}$$

then from Dini’s theorem 2.6 we get

$$\begin{aligned} \sum_{i=1}^{\infty }\frac{\sup_{j=1}^{\infty } \vert \mu _{ij} \vert }{\sqrt{ \lambda _{i}}}< \infty. \end{aligned}$$

From condition (1) and Theorem 2.3, the formula

$$\begin{aligned} T_{i}(x)=\sum_{j=1}^{\infty } \frac{\mu _{ij}}{\sqrt{\lambda _{i}}} f _{ij}(x) z_{ij} \end{aligned}$$
(2)

defines a linear continuous operator \(T_{i}\in L(X,Y)\) for every \(i=1,2,\ldots \) .

Now we need to prove the unconditional convergence of the series

$$\begin{aligned} T(x)=\sum_{i=1}^{\infty }\sqrt{\lambda _{i}} T_{i}(x). \end{aligned}$$

In order to do so, it is enough to apply again Theorem 2.3, noting that \(\lambda _{n}\rightarrow 0\) and we only have to verify that

$$\begin{aligned} \sum_{i=1}^{\infty } \bigl\vert g T_{i}(x) \bigr\vert < \infty, \quad\text{for every } g\in Y^{*}. \end{aligned}$$

In fact,

$$\begin{aligned} \sum_{i=1}^{\infty }\sum _{j=1}^{\infty } \biggl\vert \frac{\mu _{ij}}{\sqrt{ \lambda _{i}}} f_{ij}(x) g(z_{ij}) \biggr\vert &\leq \sum _{i=1}^{\infty }\sup_{j=1}^{\infty } \frac{ \vert \mu _{ij} \vert }{\sqrt{\lambda _{i}}}\sum_{j=1}^{ \infty } \bigl\vert f_{ij}(x) g(z_{ij}) \bigr\vert \\ &\leq \sum_{i=1}^{\infty }\sup _{j=1}^{\infty }\frac{ \vert \mu _{ij} \vert }{\sqrt{ \lambda _{i}}} \Biggl[ \sup _{i=1}^{\infty } \sum_{j=1}^{\infty } \bigl\vert f_{ij}(x) g(z_{ij}) \bigr\vert \Biggr]< \infty. \end{aligned}$$

Then the expression

$$\begin{aligned} T(x)=\sum_{i=1}^{\infty }\sum _{j=1}^{\infty }\mu _{ij} f_{ij}(x) z _{ij} \end{aligned}$$

defines a linear continuous operator from X into Y. □

Remark 3.2

From Theorem 2.4 and for every \(i=1,2,\ldots \) , there exist a basic sequence of functionals \(\{f_{ij}\}_{j=1}^{\infty }\) in \(X^{*}\) and a basic sequence of elements \(\{z_{ij}\}_{j=1}^{\infty }\) in Y such that

$$\begin{aligned} \sum_{j=1}^{\infty } \bigl\vert f_{ij}(x) \bigr\vert ^{2}\leq \Vert x \Vert ^{2} \quad\text{for every } x\in X \end{aligned}$$

and

$$\begin{aligned} \sum_{j=1}^{\infty } \bigl\vert F(z_{ij}) \bigr\vert ^{2}\leq \Vert F \Vert ^{2} \quad\text{for every } F\in Y^{*}. \end{aligned}$$

Basic sequences can be found by choosing different convergent to zero sequences \((\lambda _{i})_{i=1}^{\infty }\in c_{o}\), as mentioned in Theorem 2.4, according to their rate of convergence.

As a consequence of Proposition 3.1 and Remark 3.2 we get the following result.

Theorem 3.3

LetXandYbe Banach spaces and let \(\{f_{ij}\}_{j=1}^{\infty }\)and \(\{z_{ij}\}_{j=1}^{\infty }\), where \(i\in N\), be basic sequences in \(X^{*}\)andY, respectively. Verifying the following,

  1. 1.

    \(\sum_{j=1}^{\infty } \vert f_{ij}(x) \vert ^{2}< \Vert x \Vert ^{2}\)for every \(x\in X\), and \(i\in N\).

  2. 2.

    \(\sum_{j=1}^{\infty } \vert F(z_{ij}) \vert ^{2}< \Vert F \Vert ^{2}\)for every \(F\in Y^{*}\)and \(i\in N\), then every nuclear operator

    $$\begin{aligned} M=\{\mu _{ij}\}:\ell _{1}\rightarrow \ell _{1}, \quad\textit{with } \lim_{j}\mu _{ij}=0, \end{aligned}$$

    defines an operator \(T:X\rightarrow Y\)of the form

    $$\begin{aligned} T(x)=\sum_{i=1}^{\infty }\sum _{j=1}^{\infty }\mu _{ij} f_{ij}(x) z _{ij}. \end{aligned}$$

Proof

The proof follows directly from Proposition 3.1 and Remark 3.2. □

Theorem 3.4

LetXandYbe infinite-dimensional Banach spaces and let \(\{\mu _{i}\}_{i=1}^{\infty }\)be a sequence of real numbers that is convergent to zero and \(\{f_{i}\}_{i=1}^{\infty }\), \(\{z_{i}\}_{i=1} ^{\infty }\)be sequences in \(X^{*}\)andY, respectively. Verifying that

$$\begin{aligned} \sum_{i=1}^{\infty } \bigl\vert f_{i}(x) \bigr\vert ^{2}\leq \Vert x \Vert ^{2} \quad\textit{for every } x\in X, \end{aligned}$$

and

$$\begin{aligned} \sum_{i=1}^{\infty } \bigl\vert F(z_{i}) \bigr\vert ^{2}\leq \Vert F \Vert ^{2}\quad \textit{for every } F\in Y^{*}. \end{aligned}$$

Then for the operator

$$\begin{aligned} T=\sum_{i=1}^{\infty }\mu _{i} f_{i} \otimes z_{i} \end{aligned}$$

we have

$$\begin{aligned} \alpha _{n}(T)\leq \inf_{\operatorname{card} K\leq n} \sup _{i\notin K} \vert \mu _{i} \vert , \end{aligned}$$

whereKis any subset of the index setNwith \(\operatorname{card} K \leq n\).

Proof

For every operator \(T\in L(X,Y)\) and every subset of indices \(K\subset N\) with \(\operatorname{card} K\leq n\), we define a finite rank operator

$$\begin{aligned} A_{K}=\sum_{i\in K}\mu _{i} f_{i} \otimes z_{i} \end{aligned}$$

with \(\operatorname{rank}(A_{K})\leq n\). From the definition of approximation numbers we get

$$\begin{aligned} \alpha _{n}(T) &\leq \Vert T-A_{K} \Vert = \biggl\Vert \sum_{i\notin K}\mu _{i} f_{i} \otimes z_{i} \biggr\Vert \\ &= \sup_{ \Vert x \Vert =1} \sup_{ \Vert F \Vert =1} \biggl\vert \sum_{i\notin K}\mu _{i} f_{i}(x) F(z_{i}) \biggr\vert \\ &\leq \sup_{ \Vert x \Vert =1} \sup_{ \Vert F \Vert =1} \sum _{i\notin K} \bigl\vert \mu _{i} f _{i}(x) F(z_{i}) \bigr\vert \\ &\leq \sup_{i\notin K} \vert \mu _{i} \vert \sup_{ \Vert x \Vert =1} \sup_{ \Vert F \Vert =1} \sum _{i\notin K} \bigl\vert f_{i}(x) F(z_{i}) \bigr\vert \\ &\leq \sup_{i\notin K} \vert \mu _{i} \vert . \end{aligned}$$

Since this relation is true for every index subset K with \(\operatorname{card} K\leq n\),

$$\begin{aligned} \alpha _{n}(T)\leq \inf_{\operatorname{card} K\leq n} \sup _{i\notin K} \vert \mu _{i} \vert . \end{aligned}$$

 □

Remark 3.5

As a consequence of Theorem 3.4 and by using Lemma 2.8, we can get the following similar result:

$$\begin{aligned} \alpha _{n}(T)\leq \sup_{\operatorname{card} K=n+1} \inf _{i\in K} \vert \mu _{i} \vert . \end{aligned}$$

Theorem 3.6

LetXandYbe infinite-dimensional Banach spaces and let \((\mu _{ij})_{i,j\in N}\)be an infinite matrix with linearly independent rows such that conditions of Proposition 3.1are verified, and let \(\{f_{ij}\}_{j=1}^{\infty }\), \(\{z_{ij}\}_{j=1}^{\infty }\)for \(i=1,2,\ldots \) , be sequences in \(X^{*}\)andY, respectively, such that conditions of Theorem 3.4are fulfilled for all \(i=1,2,\ldots \) . Then for the operator

$$\begin{aligned} T=\sum_{i=1}^{\infty }\sum _{j=1}^{\infty }\mu _{ij} f_{ij} \otimes z _{ij} \end{aligned}$$

we have

$$\begin{aligned} \alpha _{n}(T)\leq \inf_{\varSigma n_{i}=n} \sum_{i=1}^{\infty } \Bigl\{ \inf _{\operatorname{card} K\leq n_{i}} \sup_{j\notin K} \vert \mu _{ij} \vert \Bigr\} , \end{aligned}$$
(3)

whereKis a subset of the index setNwith \(\operatorname{card} K \leq n_{i}\).

Proof

From Lemma 2.2, Theorem 3.4 and by using the same operator \(T_{i}\) defined by Eq. (2) throughout the proof of Proposition 3.1, we get

$$\begin{aligned} \alpha _{n}(T)=\alpha _{n}\Biggl(\sum _{i=1}^{\infty }T_{i}\Biggr)\leq \sum _{i=1} ^{\infty }\alpha _{n_{i}}(T_{i}) \leq \sum_{i=1}^{\infty } \inf _{\operatorname{card} K\leq n_{i}} \sup_{j\notin K} \vert \mu _{ij} \vert . \end{aligned}$$

This relation is true for every \(\varSigma n_{i}=n\), then we get the proof.

In the following, we are going to give two examples of nuclear operators over \(\ell _{1}\) and use them to construct operators over general Banach spaces with specific approximation numbers. □

Example 3.7

Consider the operator \(A\in L(c_{0},\ell _{1})\) such that \(A=(a_{ij})_{i,j=1} ^{\infty }\), where

$$\begin{aligned} &a_{ij} =0 \quad\text{for } i\neq j, \\ &a_{ii} =\frac{1}{2^{k}(k+1)^{2}} \quad\text{for } 2^{k} \leq i< 2^{k+1}. \end{aligned}$$

Also, consider \(B\in L(\ell _{1},c_{0})\), such that

$$\begin{aligned} B= \begin{pmatrix} B_{0}&0&0&\cdots \\ 0&B_{1}&0&\cdots \\ 0&0&B_{2}&\cdots \\ \cdot &\cdot &\cdot \\ \cdot &\cdot &\cdot \\ \cdot &\cdot &\cdot \end{pmatrix}, \end{aligned}$$

where

$$\begin{aligned} &B_{0} =(1), \\ &B_{k} = \begin{pmatrix} B_{k-1}&B_{k-1} \\ B_{k-1}&-B_{k-1} \end{pmatrix} \quad\text{is a } 2^{k}\times 2^{k} \text{ matrix for } k=1,2,3,\ldots. \end{aligned}$$

Thus we have \(D=AB\in L(\ell _{1},\ell _{1})\), such that

$$\begin{aligned} D= \begin{pmatrix} D_{0}&0&0&\cdots \\ 0&D_{1}&0&\cdots \\ 0&0&D_{2}&\cdots \\ \cdot &\cdot &\cdot \\ \cdot &\cdot &\cdot \\ \cdot &\cdot &\cdot \end{pmatrix}, \end{aligned}$$

where

$$\begin{aligned} &D_{0} =(1), \\ &D_{k} =\frac{k^{2}}{2(1+k)^{2}} \begin{pmatrix} D_{k-1}&D_{k-1} \\ D_{k-1}&-D_{k-1} \end{pmatrix} \quad\text{is a } 2^{k}\times 2^{k} \text{ matrix for } k=1,2,3,\ldots. \end{aligned}$$

Let \(D=(\mu _{ij})_{i,j=1}^{\infty }\), then this operator has the following properties:

  1. 1.
    $$\begin{aligned} \sum_{i=1}^{\infty } \vert \mu _{ii} \vert &=1+\biggl(\frac{1}{8}+\frac{1}{8} \biggr)+\biggl( \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+ \frac{1}{36}\biggr)+\biggl(\frac{1}{128}+ \frac{1}{128}+ \cdots \biggr)+\cdots \\ &=\sum_{i=1}^{\infty }\frac{1}{i^{2}}= \frac{\pi ^{2}}{6}. \end{aligned}$$
  2. 2.
    $$\begin{aligned} \nu (D)=\sum_{i=1}^{\infty }\sup _{j} \vert \mu _{ij} \vert = \frac{\pi ^{2}}{6}< \infty, \end{aligned}$$

    then by using Lemma 2.1D is a nuclear operator.

  3. 3.

    \(\operatorname{Trac}(D)=1+(\frac{1}{8}-\frac{1}{8})+( \frac{1}{36}-\frac{1}{36}+\frac{1}{36}-\frac{1}{36})+(\frac{1}{128}- \frac{1}{128}+\cdots )+\cdots =1\).

  4. 4.

    \(D=(\mu _{ij})_{i,j=1}^{\infty }\) is having linearly independent rows.

Now, for \(D=(\mu _{ij})_{i,j=1}^{\infty }\) and by using Proposition 3.1 and Theorem 3.6 one can construct an operator \(T\in L(X,Y)\) for any Banach spaces \(X,Y\) of the form

$$\begin{aligned} T=\sum_{i=1}^{\infty }\sum _{j=1}^{\infty }\mu _{ij} f_{ij} \otimes z _{ij}, \end{aligned}$$

where \(\{f_{ij}\}_{i,j=1}^{\infty }\), \(\{z_{ij}\}_{i,j=1}^{\infty }\), are basic sequences in \(X^{*}\) and Y, respectively, such that conditions of Theorem 3.4 are fulfilled for all \(i=1,2,\ldots \) .

Now by applying Eq. (3), one can get

$$\begin{aligned} \alpha _{n}(T)\leq \frac{\pi ^{2}}{6}-\sum _{i=1}^{k+1}\frac{1}{i^{2}} \quad\text{for } n=1,2,3,\ldots \text{ where } 2^{k}\leq n< 2^{k+1}. \end{aligned}$$

Hence, we have

$$\begin{aligned} \lim_{n\rightarrow \infty }\alpha _{n}(T)\leq \frac{\pi ^{2}}{6}-\sum_{i=1}^{\infty } \frac{1}{i^{2}}=0, \end{aligned}$$

which is consistent with the properties of the approximation numbers.

By applying Eq. (3) in the case of \(n=0\), we get

$$\begin{aligned} \alpha _{0}(T)&= \Vert T \Vert \leq 1+\biggl( \frac{1}{8}+\frac{1}{8}\biggr)+\biggl(\frac{1}{36}+ \frac{1}{36}+\frac{1}{36}+\frac{1}{36}\biggr)+\biggl( \frac{1}{128}+\frac{1}{128}+ \cdots \biggr)+\cdots \\ &=\sum_{i=1}^{\infty }\frac{1}{i^{2}}= \frac{\pi ^{2}}{6}. \end{aligned}$$

Example 3.8

Consider the operator \(J\in L(\ell _{1},\ell _{1})\) such that \(J=(\lambda _{ij})_{i,j=1}^{\infty }\) where \(\lambda _{ij}= \frac{ij}{2^{i+j}}\), then this operator has the following properties:

  1. 1.

    \(\nu (J)=\sum_{i=1}^{\infty }\sup _{j} \vert \lambda _{ij} \vert =\sum _{i=1}^{\infty }\frac{i}{2^{i}}\sup _{j}(\frac{j}{2^{j}})=1<\infty,\) then by using Lemma 2.1J is a nuclear operator.

  2. 2.

    \(J=(\lambda _{ij})_{i,j=1}^{\infty }\) has linearly independent rows.

Now for \(J=(\lambda _{ij})_{i,j=1}^{\infty }\) and by using Proposition 3.1 and Theorem 3.6, one can construct an operator \(T\in L(X,Y)\) for any Banach spaces \(X,Y\) on the form,

$$\begin{aligned} T=\sum_{i=1}^{\infty }\sum _{j=1}^{\infty }\lambda _{ij} f_{ij} \otimes z_{ij}, \end{aligned}$$

where \(\{f_{ij}\}_{i,j=1}^{\infty }\) and \(\{z_{ij}\}_{i,j=1}^{\infty }\) are basic sequences in \(X^{*}\) and Y, respectively, such that conditions of Theorem 3.4 are fulfilled for all \(i=1,2,\ldots \) .

Applying Eq. (3) yields

$$\begin{aligned} \alpha _{n}(T)\leq \frac{n+1}{2^{n}}\quad \text{for } n=1,2,3, \ldots. \end{aligned}$$

Thus, we have \((\alpha _{n}(T))_{n=1}^{\infty }\in \ell _{1}\) because

$$\begin{aligned} \sum_{n=1}^{\infty }\alpha _{n}(T)\leq \sum_{n=1}^{\infty } \frac{n+1}{2^{n}}=3< \infty. \end{aligned}$$

Applying Eq. (3) in the case of \(n=0\) yields

$$\begin{aligned} \alpha _{0}(T)= \Vert T \Vert \leq \frac{1}{2}\sum _{i=1}^{\infty } \frac{i}{2^{i}}= \frac{1}{2}\times 2=1, \end{aligned}$$

noting that this is independent of the selection of \(\{f_{ij}\}_{i,j=1} ^{\infty }\) and \(\{z_{ij}\}_{i,j=1}^{\infty }\).

If we choose \(\{f_{ij}\}_{i,j=1}^{\infty }\) and \(\{z_{ij}\}_{i,j=1} ^{\infty }\) such that

$$\begin{aligned} \Vert f_{ij} \Vert = \Vert z_{ij} \Vert = \frac{1}{\sqrt{ij}}, \end{aligned}$$

then we get

$$\begin{aligned} \nu (T)\leq \sum_{i,j=1}^{\infty }\lambda _{ij} \Vert f_{ij} \Vert \Vert z_{ij} \Vert = \sum_{i,j=1}^{\infty }\biggl( \frac{ij}{2^{i+j}}\biggr) \biggl(\frac{1}{ij}\biggr)=1< \infty, \end{aligned}$$

which means that T, in this case, is a nuclear operator.

4 Conclusion

By using nuclear operators defined over \(\ell _{1}\) with particular representation, one can construct compact operators over general Banach spaces with specific approximation numbers. Such compact operators are been constructed using a countable number of basic sequences and nuclear operators. For such nuclear operators, its construction in a matrix form will yield to double-summation operators. This double-summation gives more freedom rather than choosing sequence elements in the case of single-summation operators. Such a construction will help give counterexamples of operators between Banach spaces without a Schauder basis.