"Elementary" proof for the dual space of $\ell^p$ ($1Let $1<p<\infty$. Given a measure space $X$ one typically proves that $(L^p)^* =L^q$ using (essentially) two components: (i) the Holder inequality shows that $L^q\subset (L^p)^*$ and (ii) the Radon-Nikodym derivative establishes the reverse inclusion.
In particular, I find part (ii) of this proof somewhat inaccessible for first time learners, and so I am wondering if there is a more ``elementary'' proof to show that $(L^p)^*\subset L^q$ in the case that we have the counting measure.  Many questions on this site have established that $(\ell^1)^*=\ell^\infty$, so my question lies specifically with $p>1.$
So far, I have the idea to represent $\phi \in (\ell^p)^*$ through projections: $a_n := (\phi, e_n)$. Then, assuming that the sequence $a_n\not\in \ell^q$ I hope to construct some particular $b_n\in \ell^p$ so that $(a_n,b_n) = \infty$, contradicting the boundedness of $\phi$.
Any reference or direction would be very helpful. Thank you!
 A: In the following discussion, assume that $1<p<\infty$.
Recall that the space $\ell^p=\ell^p(\mathbb{N})$ is defined as 
the set of infinite (complex) sequences $x=(x_k)$ with $\sum |x_k|^p<\infty$ together with the vector space structure and the norm
$$
\|x\|_p = (\sum |x_k|^p)^{1/p}\;.
$$
One can prove the fact that $(\ell^p)^* = \ell^q$ ($1/p+1/q = 1$) in a fairy "elementary" (without measure theory) way as follows. 
Proof. Suppose $f\in (\ell^p)^*$. We show that there is a $z\in \ell_q$ such that for all $x\in\ell_p$,
$$
f(x) = \sum x_iz_i\;,
$$
and $\|f\| = \|z\|_q\;$.
Let $e_1 = (1,0,\cdots)$, $e_2=(0,1,0,\cdots)$, and in general $e_k$ the vector having the $k$-th entry equal to one and all other entries equal to zero. One can follow the following steps to complete the proof:


*

*the partial sum $s_n(x) := \sum_{k=1}^nx_ke_k$ converges to $x$ in the $\ell^p$ norm;

*by linearity and continuity of $f$, one has $|f(x)-f(s_n)|\to 0$ as $n\to\infty$;

*thus, 
$$
f(x) = \sum_{k=1}^\infty x_kf(e_k)\;.
$$

*show $z = (z_k)\in\ell^q$ with $z_k:=f(e_k)$ by showing that $\|z\|_q\leq\|f\|$;

*show that $\|f\| \leq \|z\|_q$.

To show $\|z\|_q\leq\|f\|$, it suffices to show that 
$$
(\sum_{k=1}^n|z_k|^q)^{1/q}\leq \|f\|\tag{1}
$$
for all positive integer $n$. Consider $w\in\ell^p$ defined as
$$
w_k = \begin{cases}
|z_k|^{q-2}z_k,&\textrm{when $z_k\neq 0$};\\
0,&\textrm{when $z_k=0$}.
\end{cases}
$$
Then it is very straightforward to check the following:


*

*$\|s_n(w)\|_p^p = \sum_{k=1}^n|z_k|^q$;

*$$
\sum_{k=1}^n|z_k|^q=f(s_n(w))\leq |f(s_n(w))|
\leq \|f\|(\sum_{k=1}^n|z_k|^q)^{1/p}
$$
The estimate (1) follows by $1/p+1/q = 1$.
To show $\|f\|\leq\|z\|_q$, observe that
$$
|f(x)| \leq \sum |x_kz_k|\leq \|x\|_p\|z\|_q
$$
by Holder's inequality.
