Show that the dual space of $\ell^1$ is isomorphic to $\ell^{\infty}$ I've started taking my first course in multivariable analysis and in our notes there is an exercise about the dual space of the sequence space $\ell_1$ and I'm unsure how to prove that it is isomorphic to $\ell_{\infty}$ as in class we haven't gone over the dual space in detail.
If somebody could also give me some extra intuition about dual spaces that would be much appreciated :)
 A: The dual space consists of all continuous linear functionals on the space. What continuity buys you is that, if you can determine the functionals on a dense subspace $M$, you can bootstrap to the full space by continuity. Continuity of a linear function $F$ on the normed space assures that $F$ is determined by its values on the dense subspace $M$.
For example, because $\ell^1$ has a basis of sequences
$$
              e_0 = \{ 1, 0, 0, 0, \cdots \} \\
              e_1 = \{ 0, 1, 0, 0, \cdots \} \\
              e_2 = \{ 0, 0, 1, 0, \cdots \} \\
                    \vdots
$$
then there is a natural way to bootstrap. Specifically, if $x=\{ \alpha_k \}\in\ell^1$, then
$$
     \left\| x - \sum_{n=0}^{N}\alpha_n e_n \right\|_{\ell^1} \le \sum_{k=N+1}^{\infty}|\alpha_n|\rightarrow 0 \mbox{ as } N\rightarrow\infty.
$$
Therefore, by continuity, if $F$ is a continuous linear functional on $\ell^1$, then
$$
        F(x) = \lim_{N\rightarrow\infty}F\left(\sum_{k=0}^{N}\alpha_k e_k \right)
    = \lim_{N\rightarrow\infty} \sum_{k=0}^{N}\alpha_k F(e_k) = \sum_{k=0}^{\infty}\alpha_k F(e_k).
$$
So, already you have a representation of any such linear functional. The sum on the right is guaranteed to converge. In fact it must converge absolutely because you can choose unimodular constants $u_k$ such that $u_k\alpha_k F(e_k)=|\alpha_k||F(e_k)|$ and apply the above to $\{u_k\alpha_k\}$ instead. Continuity of $F$ is equivalent to boundedness, meaning the existence of a smallest non-negative constant $\|F\|$ such that $|F(x)| \le \|F\|\|x\|_{\ell^1}$ for all $x\in\ell^1$. From this, you get $\{ F(e_k) \}\in\ell^{\infty}$ and $\|\{ F(e_k)\}\|_{\ell^{\infty}} \le \|F\|$. Conversely, 
$$|F(\{\alpha_k\})| \le \|\{F(e_k)\}\|_{\infty}\|\{\alpha_k\}\|_{\ell^1}$$
gives $\|F\|\le\|\{F(e_k)\}\|_{\infty}$ because $\|F\|$ is the smallest such constant. Hence,
$$\|F\|=\|\{F(e_k)\}\|_{\ell^{\infty}}.$$
So there is an isometric linear correspondence between $(\ell^{1})^*$ and $\ell^{\infty}$.
