# 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

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