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?

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 $Jy=f$.
