Showing that $l^p(\mathbb{N})^* \cong l^q(\mathbb{N})$ I'm reading functional analysis in the summer, and have come to this exercise, asking to show that the two spaces $l^p(\mathbb{N})^*,l^q(\mathbb{N})$ are isomorphic, that is, by showing that every $l \in l^p(\mathbb{N})^*$ can be written as
$l_y(x)=\sum y_nx_n$
for some $y$ in $l^q(\mathbb N)$.
The exercise has a hint. Paraphrased:
"To see $y \in l^q(\mathbb N)$ consider $x^N$ defined such that $x_ny_n=|y_n|^q$ for $n \leq N$ and $x_n=0$ for $n > N$. Now look at $|l(x^N)| \leq ||l|| ||x^N||_p$."
I can't say I understand the first part of the hint. To prove the statement I need to find a $y$ such that $l=l_y$ for some $y$. How then can I define $x$ in terms of $y$ when it is $y$ I'm supposed to find. Isn't there something circular going on?
The exercise is found on page 68 in Gerald Teschls notes at http://www.mat.univie.ac.at/~gerald/ftp/book-fa/index.html
Thanks for all answers.
 A: We know that the $(e_n)$ with a $1$ at the $n$-th position and $0$s elsewhere is a Schauder basis for $\ell^p$ (which has some nice alternative equivalent definitions, I recommend Topics in Banach Space Theory by Albiac and Kalton as a nice reference about this.
So, every $x \in \ell^p$ has a unique representation by
$$x = \sum_{k = 1}^\infty y_k e_k.$$
Now consider $l \in \ell^q$. Because $l$ is bounded we also have that
$$l(x) = \sum_{k = 1}^\infty y_k l(e_k).$$
Now set $z_k = f(e_k)$. Consider the following $x_n = (y_k^{(n)})$ where
$$y_k^{(n)} = \begin{cases} \frac{|z_k|^q}{z_k} &\text{when $k \leq n$ and $z_k \neq 0$,}\\
0 &\text{otherwise.}
\end{cases}.$$
We have that
$$\begin{align}l(x_n) &= \sum_{k = 1}^\infty y_k^{(n)} z_k\\
&= \sum_{k = 1}^n |z_k|^q\\
&\leq \|l\|\|x_n\|\\
&= \|l\| \left ( \sum |x_k^{(n)}|^p \right )^{\frac1p}\\
&= \|l\| \left ( \sum |x_k^{(n)}|^p \right )^{\frac1p}\\
&= \|l\| \left ( \sum |z_k|^q \right )^{\frac1q}.
\end{align}$$
Hence we have that 
$$\sum |z_k|^q = \|l\| \left (\sum |z_k|^q \right )^{\frac1q}.$$
Now we divide and get
$$\left ( \sum_{k = 1}^n |z_k|^q \right )^{\frac1q} \leq \|l\|.$$
Take the limit to obtain
$$\left ( \sum_{k \geq 1} |z_k|^q \right )^{\frac1q} \leq \|l\|.$$
We conclude that $(z_k) \in \ell^q$.
So, now you could try doing the same for $L^p(\mathbf R^d)$ with a $\sigma$-finite measure. A small hint: Using the $\sigma$-finiteness you can reduce to the finite measure case.
