what is the definition of the function u?? From book
I. Singer, Bases in Banach Spaces I, Grundlehren der mathema-
tischen Wissenschaften, Springer Berlin Heidelberg, 2012.
Let $1 \leq p < \infty$ and let $\{E_n\}$ be a sequence of Banach spaces. We denote by $(E_1 \times E_2 \times ...)_{l^p}$ the space of all sequences $\{x_n\}$ with $x_n \in E_n (n = 1,2,...),$ for which $\{\|x_n\|\} \in l^p,$ endowed with the norm
$$\|\{x_n\}\| = (\sum_{i=1}^\infty \|x_i\|^p)^{1/p}.$$
lemma: Let $1  \leq p < \infty.$ Then we have $$(l^p \times l^p \times ...)_{l^P} \equiv l^p.$$
Proof
Let $\{e_n\}$ be the unit vector basis of $l^p$ and let $u$ be an arbitrary one to one mapping of the set of elements
$$z_{ij} = \{\underbrace{0, ..., 0}_{j-1}, e_i, 0, 0,...\} \in (l^p \times l^p \times ...)_{l^p} (i,j = 1,2,...)$$
onto the set $\{e_n\},$ extended by linearity to the set of all finite linear combinations. Then we have, for any scalars $\alpha_{ij} (i = 1, ...,n; j = 1,...,m),$
\begin{equation}
\begin{split}
\|u(\sum_{i=1}^n \sum_{j=1}^m \alpha_{ij} z_{ij})\| &= \|\sum_{i=1}^n \sum_{j=1}^m \alpha_{ij} u(z_{ij})\| = (\sum_{i=1}^n \sum_{j=1}^m |\alpha_{ij}|^p)^{1/p} = (\sum_{j=1}^m \sum_{i=1}^n |\alpha_{ij}|^p)^{1/p}\\
\nonumber &= (\sum_{i=1}^n |\alpha_{i1}|^p + \sum_{i=1}^n |\alpha_{i2}|^p + ... + \sum_{i=1}^n |\alpha_{im}|^p)^{1/p} \\
\nonumber &= (\|(\alpha_{11}, \alpha_{21},..., \alpha_{n1})\|_p^p + \|(\alpha_{12}, \alpha_{22}, ..., \alpha_{n2})\|_p^p\\
\nonumber &\quad + ... + \|(\alpha_{1m}, \alpha_{2m},..., \alpha_{nm})\|_p^p)^{1/p}\\
\nonumber &= (\|\alpha_{11} e_1 + \alpha_{21} e_2+ ... + \alpha_{n1} e_n\|_p^p + \|\alpha_{12}e_1 + \alpha_{22}e_2 + ... +\alpha_{n2}e_n\|_p^p\\
\nonumber &\quad + ... + \|\alpha_{1m} e_1 + \alpha_{2m} e_2 + ... + \alpha_{nm}e_n\|_p^p)^{1/p}\\
\nonumber &= (\|\sum_{i=1}^n \alpha_{i1}e_i\|_p^p + \|\sum_{i=1}^n \alpha_{i2}e_i\|_p^p + ... + \|\sum_{i=1}^n \alpha_{im}e_i\|_p^p)^{1/p}\\
\nonumber &= (\sum_{j=1}^m \|\sum_{i=1}^n \alpha_{ij} e_i\|_p^p)^{1/p} \\
\nonumber &= \|\{\sum_{i=1}^n \alpha_{i1} e_i, ..., \sum_{i=1}^n \alpha_{im} e_i, 0, 0, ...\}\|_p \\
\nonumber &= \|\sum_{j=1}^m \{\underbrace{0, ..., 0}_{j-1} , \sum_{i=1}^n \alpha_{ij} e_i, 0, 0, ...\}\|_p \\
\nonumber &= \|\{\sum_{i=1}^n \alpha_{i1} e_i, 0, 0, ...\} + \{0, \sum_{i=1}^n \alpha_{i2}e_i, 0, 0, ...\}\\
 \nonumber &\quad + ... + \{\underbrace{0,..., 0}_{m-1}, \sum_{i=1}^n \alpha_{im}e_i , 0, 0, ...\} \|_p\\
\nonumber &= \|\sum_{j=1}^m \sum_{i=1}^n  \{\underbrace{0, ..., 0}_{j-1},  \alpha_{ij} e_i, 0, 0,...\}\|_p\\
\nonumber &= \|\sum_{i=1}^n \sum_{j=1}^m \alpha_{ij} z_{ij}\|_p,
\end{split}
\end{equation}
whence the assertion follows, since $\{z_{ij}\}$ is linearly independent and $[z_{ij}] = (l^p \times l^p \times ...)_{l^p}, [e_n] = l^p.$
Question
I don't know what's the definition of the function u??
So i don't know how $$\|\sum_{i=1}^n \sum_{j=1}^m \alpha_{ij} u(z_{ij})\| = (\sum_{i=1}^n \sum_{j=1}^m |\alpha_{ij}|^p)^{1/p}$$ in the first equality in the proof of the lemma??
 A: Idea
$z_{ij}$ is an infinite sequence of infinite sequences that is $1$ at the $i^{\text{th}}$-row $j^{\text{th}}$-column and $0$ elsewhere.
$$\begin{pmatrix}
\overbrace{0} & \overbrace{0} & \dots & \overbrace{0} & \overbrace{0} & \dots\\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots\\
0 & 0 & \dots & 0 & 0 & \dots\\
0 & 0 & \dots & 1 & 0 & \dots \\
0 & 0 & \dots & 0 & 0 & \dots\\
\underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \ddots
\end{pmatrix}$$
Intuitively, we can think of it as a "standard basis vector for infinite matrix".
$u$ is then an arbitary one to one mapping of these "standard basis vectors" onto $\{e_i\}$ extended linearly. It is analogous to how we define linear functional $\phi: \mathbb{R}^n \to \mathbb{R}$ by specifying what is $\phi(e_i)$.
Definition of $u$
We can express the above idea more precisely as followed.
Write $(\ell^p)^{\mathbb{N}} = (\ell^p \times \ell^p \times \dots)_{\ell^p}$ and
$$(\ell^p)^{\mathbb{N}}_{00} = \\
\left\{\begin{pmatrix}
\overbrace{c_{11}} & \overbrace{c_{12}} & \overbrace{c_{13}} & \dots\\
c_{21} & c_{22} & c_{23} & \dots \\
c_{31} & c_{32} & c_{33} & \dots \\
\underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \ddots
\end{pmatrix}
\in (\ell^p)^{\mathbb{N}} \middle | \text{ all but finitely many } c_{ij} = 0\right\}$$
Define
$$z_{ij} =
\begin{pmatrix}
\overbrace{0} & \overbrace{0} & \dots & \overbrace{0} & \overbrace{0} & \dots\\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots\\
0 & 0 & \dots & 0 & 0 & \dots\\
0 & 0 & \dots & 1 & 0 & \dots \\
0 & 0 & \dots & 0 & 0 & \dots\\
\underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \ddots
\end{pmatrix}$$
Let $E_{\text{std}} = \{e_i \mid i \in \mathbb{N}\} \subseteq \ell^p$, $Z_{\text{std}} = \{z_{ij} \mid i, j \in \mathbb{N}\} \subseteq (\ell^p)^{\mathbb{N}}_{00}$ and $u|_{Z_{\text{std}}}$ to be an arbitrary one to one function from $Z_{\text{std}}$ onto $E_{\text{std}}$.
We can now extend $u|_{Z_{\text{std}}}$ linearly to $u: (\ell^p)^{\mathbb{N}}_{00} \to \ell^p$ by defining
$$u(c) =
\begin{cases}
u|_{Z_{\text{std}}}(c) \text{ if } c \in Z_{\text{std}} \\
\sum_\limits{i, j \in \mathbb{N}} c_{ij} u|_{Z_{\text{std}}}(z_{ij}) \text{ if } c \notin Z_{\text{std}}
\end{cases}$$
where $$c =
\begin{pmatrix}
\overbrace{c_{11}} & \overbrace{c_{12}} & \overbrace{c_{13}} & \dots\\
c_{21} & c_{22} & c_{23} & \dots \\
c_{31} & c_{32} & c_{33} & \dots \\
\underbrace{\vdots} & \underbrace{\vdots} & \underbrace{\vdots} & \ddots
\end{pmatrix}
$$
with all but finitely many $c_{ij} = 0$
Proof of equality
Write $e_r = u(z_{ij})$ and $e_{r'} = u(z_{i'j'})$. Since $u|_{Z_{\text{std}}}$ is a one to one function from $Z_{\text{std}}$ onto $E_{\text{std}}$, we must have $$(i, j) \neq (i', j') \implies z_{ij} \neq z_{i'j'} \implies u(z_{ij}) \neq u(z_{ij}) \implies e_r \neq e_{r'} \implies r \neq r'$$
Therefore, distinct $(i, j)$ index must map to distinct $r$ index.
Let $K$ be the number of $a_{ij}$ such that $a_{ij} \neq 0$. Here, we require $K$ to be finite. We can then write
$$\sum_{i, j \in \mathbb{N}} a_{ij} u(z_{ij}) = \sum_{k = 1, 2, \dots, K} a'_k e_{r_k} = (0, \dots, 0, a'_1, 0, \dots, 0, a'_2, \dots \dots, 0, \dots, 0, a'_K, 0, 0, \dots)$$ after a suitable re-labelling of $a_{ij}$ into $a'_k$.
Taking $p$-norm
$$\left|\left|\sum_{i, j \in \mathbb{N}} a_{ij} u(z_{ij})\right|\right| = ||(0, \dots, 0, a'_1, 0, \dots, 0, a'_2, \dots \dots, 0, \dots, 0, a'_K, 0, 0, \dots)|| = \left(\sum_{k = 1, 2, \dots, K} |a'_k|^p\right)^{\frac{1}{p}}$$
Undoing the re-labelling
$$\left(\sum_{k = 1, 2, \dots, K} |a'_k|^p\right)^{\frac{1}{p}} = \left(\sum_{i, j \in \mathbb{N}} |a_{ij}|^p\right)^{\frac{1}{p}}$$
We obtain
$$\left|\left|\sum_{i, j \in \mathbb{N}} a_{ij} u(z_{ij})\right|\right| = \left(\sum_{i, j \in \mathbb{N}} |a_{ij}|^p\right)^{\frac{1}{p}}$$
which is the same as your equality because we require all but finitely many $a_{ij} = 0$.
