The dual of $\ell^p(\textbf{r})$ is $\ell^q(\textbf{r})$, where $\textbf{r}$ is a weighted vector I'm trying to show that the dual space of $\ell^p$ is $\ell^q$ with the typical conditions, only that we will include a weight to our space. The proof that I want to imitate is the Kreyszig, but I have two problems. So I star with
For each $k\in\mathbb{N}$, we consider the canonical sequence in $\ell^p(\textbf{r})$ defined by  $e_k = \left(\delta_{kj}\right)_{j\in\mathbb{N}}$, where  $\delta_{kj}$ is the known Kronecker delta, that is,  $\delta_{kj} = 1$ if $k=j$ y $0$ otherwise for $k, j \in \mathbb{N}$. Then, for each  $\textbf{x}=\left(x_k\right) \in \ell^p(\textbf{r})$ it is true that
$$
  \lim_{N\to\infty}
  \left\|\textbf{x}-  \displaystyle\sum^{N}_{k=1} x_k e_k \right\|_r^p
  = \lim_{N\to\infty}\displaystyle\sum^{\infty}_{k=N+1} \left|x_k\right|^p  r_k = 0
 $$
and then all $\textbf{x} \in \ell^p(\textbf{r})$ has a single representation of the form
$$
 \textbf{x} = \displaystyle\sum^{\infty}_{k=1} x_k e_k.
 $$
which means that $\left\{e_k\right\}_{k\in\mathbb{N}}$ is a Schauder basis for $\ell^p(\textbf{r})$.
Consider some $f \in \left(\ell^p(\textbf{r})\right)'$,  the dual space of $\ell^p(\textbf{r})$, and define the sequence $\textbf{y}=\left(y_k\right)$ by
\begin{equation}\label{def-yk}
 y_k = f\left(e_k\right). \hspace{10cm} (1)
\end{equation}
Since  $f$ is linear and continuous, for any $\textbf{x}=\left(x_k\right) \in \ell^p(\textbf{r})$ it is true that
\begin{eqnarray*}
 f\left(\textbf{x}\right) &= & f\left( \displaystyle\sum^{\infty}_{k=1} x_k e_k\right)\\
 &=&  f\left(\lim_{N\to\infty} \displaystyle\sum^{N}_{k=1} x_k e_k\right)\\
 &=& \lim_{N\to\infty} f\left( \displaystyle\sum^{N}_{k=1} x_k e_k\right)\hspace{1cm}\text{(by continuity)}\\
 &=&  \lim_{N\to\infty} \displaystyle\sum^{N}_{k=1} x_k f\left(e_k\right)\hspace{1.3cm}\text{(by linearity)}\\
  &=&  \lim_{N\to\infty} \displaystyle\sum^{N}_{k=1} x_k y_k =
  \displaystyle\sum^{\infty}_{k=1} x_k y_k 
\end{eqnarray*}
and the formula $f\left(\textbf{x}\right) =\displaystyle\sum^{\infty}_{k=1} x_k y_k$ holds; so now we have to show that the sequence $\textbf{y}=\left(y_k\right)$ defined in (1) is in $\ell^q(\textbf{r})$.
Indeed, for each $n\in\mathbb{N}$ the sequence  $\textbf{x}_n = (\xi^{(n)}_k)$ is considered with ( THE FIRST PROBLEM IS ADDING WEIGHT TO THIS SUCCESSION, WHICH SEEMS NATURAL, BUT LATER IT AFFECTS ME PROOF (or so I think))
\begin{equation}\label{d2}
\xi^{(n)}_k =
    \begin{cases}
\frac{|y_k|^q}{y_k},        & \mbox{si } k \le n \hspace{2mm} \mbox{y }  y_k \neq 0 \\
0,                         & \mbox{si } k > n \hspace{2mm} \mbox{o }  y_k = 0.
    \end{cases}
\end{equation}
Then $\textbf{x}_n\in\ell^p\left(\textbf{r}\right)$  since it has a finite amount of non-null elements; so by the formula $f\left(\textbf{x}\right) =\displaystyle\sum^{\infty}_{k=1} x_k y_k$ it is allowed to write
$$
f(\textbf{x}_n) = \displaystyle\sum^{\infty}_{k=1} \xi^{(n)}_k y_k = \displaystyle\sum^{n}_{k=1} |y_k|^q.
$$
Now using the definition of $\xi^{(n)}_k$ and the fact that $(q - 1)p = q$,
$$
\begin{aligned}
\left|f(\textbf{x}_n)\right| 
&\le \left\| f \right\| \left\| \textbf{x}_n \right\|_r\\ 
& = \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |\xi^{(n)}_k|^p r_k    \right)^{1/p}\\ 
& = \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^{(q-1)p}    r_k \right)^{1/p}\\ 
& = \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^q r_k\right)^{1/p}
\end{aligned}
$$
and when joining the ends you have to
$$
\left|f(\textbf{x}_n)\right| = \displaystyle\sum^{n}_{k=1} |y_k|^q  
\le \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^q r_k \right)^{1/p}.
$$
(HERE IS THE OTHER PROBLEM, BECAUSE ONE PART HAS THE WEIGHTED VECTOR, BUT AND THE OTHER DOES NOT)
Thanks for the help
 A: Actually your proof is already there, because you only need to define two terms (you can use Riesz's representation theorem, as an analogy, since the integral is seen as the sum, the functions as the sequences and the weighted vector as the discrete measure).
The functional needs three terms,
$$
f\left(\textbf{x}\right) = \sum^{\infty}_{k=1} x_k y_k r_k
$$
then you must define the sequence $(y_k)$ as follows
$$
y_k = \frac{f\left(e_k\right)}{r_k}
$$
thus, you no longer need to multiply by the weighted vector in the sequence $\xi^{(n)}$.
Finally, with that substitution, you should easily arrive at what
$$
    f(\textbf{x}_n) = \displaystyle\sum^{\infty}_{k=1} \xi^{(n)}_k y_k r_k = \displaystyle\sum^{n}_{k=1} |y_k|^q r_k.
    $$
Achieving the expected result
$$
    \left|f(\textbf{x}_n)\right| = \displaystyle\sum^{n}_{k=1} |y_k|^q r_k 
    \le \left\| f \right\| \left( \displaystyle\sum^{n}_{k=1} |y_k|^q r_k \right)^{1/p}.
    $$
although you still haven't shown that $\ell^q$ is its dual space, already the proof is natural
