Unsure how $(\ell^1)^*\cong\ell^\infty$ I do not really understand how the dual space of $\ell^1$ is isomorphic $\ell^\infty$ in that $\ell^\infty$ is a set of sequences and a dual space is a set of functions. I know I am most likely not understanding something relatively simple but I cannot seem to wrap my head around why this is the case.

Let $L$ be a linear functional on $\ell^1$, we have $L:\ell^1 \rightarrow \mathbb{R}$. $L$ is defined by its action on a basis, $\{L(e_i)\}$ and can be written as $L(a) = L(\sum_i a_i e_i) = \sum_i a_i L(e_i)$.  So lets form the isomorphism $\phi(L) = \{L(e_i)\}$. If $sup_i |L(e_i)| < \infty$ then $|\sum_i a_i L(e_i)| \leq \sum_i |a_i L(e_i)| \leq \sup_i |L(e_i)| \times \sum_i |a_i|  < \infty$. Hence $\ell^{\infty} \subset \phi(\ell^1)$, Now if $sup_i |L(e_i)| = \infty$, then there exits a subsequence $\{L(e_{n_i}) : |L(e_{n_i})| > N_M, \forall i \geq M \}$. Let $M_N = \inf\{n_p: |L(e_{n_i})| > N, \forall i \geq n_p\} + 1$. Now choose the $\ell_1$ sequence $x_{M_N} = sign(L(e_{M_N})) \times \frac{1}{N^2}$  and other $x_i$ are equal to $0$.  Now $L(x) = \sum_{N} x_{M_N} L(e_{M_N}) \geq \sum_{N} \frac{1}{N} $ which is not in $\ell^1$. Hence $\phi(\ell^1) = \ell^{\infty}$.
EDIT: In the above I assumed $(\ell^1)^*$ is a space of linear functionals on $\ell^1$ of bounded norm i.e., $\sup_x |Lx|/||x||_{\ell^1} < \infty $ for $L \in (\ell^1)^*$. The insight is linear functional which is a linear map of bounded norm is defined by its action on the basis elements and hence can be mapped to a fixed sequence and since the linear functional maps a sequence to its inner product by a fixed sequence and since we want this inner product to converge in norm this forces the sequence to which the linear functional is mapped to have certain  properties w.r.t norm. Here the norm forced upon is $\ell^{\infty}$. In general it's the holder's inequality which dictates what norm is forced upon in the sequence to which linear functional is mapped to so that the inner product converges. Goes to say about beauty of holder's inequality.
