### Proposition 3.1

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

- 1.
\(\lim_{j}\mu _{ij}=0 \)*for every*
\(i\in N\).

- 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 in**Y**that verifies*

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

(1)

*for every**F**in*
\(Y^{*}\)*and every**x**in**X*. *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*.

### 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

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

- 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.
\(\sum_{j=1}^{\infty } \vert F(z_{ij}) \vert ^{2}< \Vert F \Vert ^{2}\)*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

*Let**X**and**Y**be 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*
\(X^{*}\)*and**Y*, *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*

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

*where**K**is any subset of the index set**N**with*
\(\operatorname{card} K \leq n\).

### 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

*Let**X**and**Y**be 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 *3.1*are verified*, *and let*
\(\{f_{ij}\}_{j=1}^{\infty }\), \(\{z_{ij}\}_{j=1}^{\infty }\)*for*
\(i=1,2,\ldots \) , *be sequences in*
\(X^{*}\)*and**Y*, *respectively*, *such that conditions of Theorem *3.4*are 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)

*where**K**is a subset of the index set**N**with*
\(\operatorname{card} K \leq n_{i}\).

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

- 2.
$$\begin{aligned} \nu (D)=\sum_{i=1}^{\infty }\sup _{j} \vert \mu _{ij} \vert = \frac{\pi ^{2}}{6}< \infty, \end{aligned}$$

then by using Lemma 2.1*D* is a nuclear operator.

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

- 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.