Prove the dual space of $l^p$ is isomorphic to $l^q$ if $\frac{1}{q}+\frac{1}{p}=1$ 
Prove the dual space of $\ell^p$ is isomorphic to $\ell^q$ if $\frac{1}{q}+\frac{1}{p}=1$ ($1<p<\infty$)

Define a map $J:\ell^q \to (\ell^p)'$ such that $Jy(x)=\sum_{k=1}^\infty x_ky_k,x\in \ell^p,y\in \ell^q$
I have verified that $Jy\in (\ell^p)'$, $J$ is linear and $\lVert Jy \rVert\leq \lVert y \rVert_q$. 
How to show $\lVert Jy \rVert \geq \lVert y \rVert_q$ and $J$ is surjective?
 A: Given $y=\{y_k\}\in\ell^q$ nonzero, consider the sequence $x=\{e^{-i\theta_k}\|y\|_q^{1-q}|y_k|^{q/p}\}$, where $\theta_k=\arg(y_k)$ (define $\theta_k$ arbitrarily if $y_k=0$).  Then $x\in\ell^p$ with $\|x\|_p=1$ (it's not too hard to see), and 
\begin{align}
Jy(x)&=\sum_{k=1}^\infty y_ke^{-i\theta_k}\|y\|_q^{1-q}|y_k|^{q/p}\\
&=\|y\|_q^{1-q}\sum_{k=1}^\infty|y_k|^{1+q/p}\\
&=\|y\|_q^{1-q}\sum_{k=1}^\infty|y_k|^q \\
&=\|y\|_q.
\end{align}
Thus $\|Jy\|\geq|Jy(x)|=\|y_p\|$.
To show that $J$ is surjective, fix $f\in(\ell^p)^*$ and look at what it does to the (Schauder) basis vectors $e_n=\{\delta_{k,n}\}$ for each $n\in\mathbb{N}$.  This gives you a sequence $y=\{y_k\}$, and show that this sequence is in $\ell^q$ and $Jy=f$.
A: Surjectivity of the map $J$.
Let $\varphi\in (\ell^p)'$, with $\|\varphi\|_*=1$, and set $y_n=\varphi(e_n)$, where $e_n=(0,\ldots,0,1,0,\ldots)$, with exactly one $1$ in the $n$th place. We shall show that $\{y_n\}\in \ell^q$, $\|\{y_n\}\|_q=1$ and $\varphi(\{x_n\})=\sum x_ny_n$.
I. $\{y_n\}\in \ell^q$. For every
$x_1,\ldots,x_n$, with $|x_1|^p+\cdots+|x_n|^p\le  1$, we have
$\|(x_1,\ldots,x_n,0,0,\ldots)\|_p\le 1$ and hence
$$
1\ge \varphi(x_1e_1+\cdots+x_ne_n)=x_1y_1+\cdots+x_ny_n.
$$
But
$$
\sup_{|x_1|^p+\cdots+|x_n|^p\le  1}x_1y_1+\cdots+x_ny_n=\big(|y_1|^q+\cdots+|y_n|^q\big)^{1/q}, \tag{1}
$$
and hence
$$
|y_1|^q+\cdots+|y_n|^q\le 1,
$$
and since this holds for every $n$, then $\|\{y_n\}\|_q\le 1$ and $\{y_n\}\in \ell^q$.
II. $\varphi(\{x_n\})=\sum x_ny_n$. Let $x=\{x_n\}\in\ell^p$ and set $x^n=(x_1,x_2,\ldots,x_n,0,0,\ldots)$. Then $\|x^n-x\|\to 0$ and hence
$\varphi(x^n-x)\to0$. But
$$
\varphi(x^n)=\sum_{k=1}^n x_ky_k,
$$
and as the series $\sum_{n=1}^\infty x_ny_n$, converges then
$$
\varphi(x)=\lim_{n\to\infty}\varphi(x^n)=\sum_{n=1}^\infty x_ny_n=(x,y).
$$
The fact that $\|y\|_q=1$ can be readily shown.
Proof of (1). Clearly, (Hölder)
$$
|x_1y_1+\cdots+x_ny_n|\le \left(|x_1|^p+\cdots+|x_n|^p\right)^{1/p}
\left(|y_1|^q+\cdots+|y_n|^q\right)^{1/q}\le \left(|y_1|^q+\cdots+|y_n|^q\right)^{1/q}
$$
It is easily shown that equality is obtained for
$$
x_i=\frac{|y_i|^{q/p}\mathrm{sgn}(y_i)}{\left(|y_1|^q+\cdots+|y_n|^q\right)^{1-1/q}}, \quad i=1,\ldots,n.
$$
