Denote the dual of $\ell^1$ as $(\ell^1)^*$. I have learned that $(\ell^1)^* = \ell^\infty$. I am trying to better understand this statement. I found this question to be helpful, but a question still remains.
For concreteness, $\ell^1$ is the set of all functions $f : \mathbb{N} \rightarrow \mathbb{F}$ such that $\sum_{i \in \mathscr{N}} |f(i)| <\infty$ and $\ell^\infty$ is the set of all functions $f : \mathbb{N} \rightarrow \mathbb{F}$ such that $\sup_{i \in \mathbb{N}} |(i)| <\infty$. In other words, $\ell^1$ is the set of absolutely convergent sequences, where $\ell^\infty$ is the set of sequences where each element is bounded.
Consider a sequence of scalars $\{\alpha_i\}_{i=0}^\infty$, such that $\sup_i|a_i| < \infty$. Then, for $f=(f_1, f_2,...) \in \ell^1$, define $T$ as $Tf = \sum_{i\in \mathbb{N}} a_if_i$. Noting that $\| Tf \| \leq \|a\| \|T\| $, $T$ is clearly a bounded linear functional on $\ell^1$ and hence it is in $(\ell^1)^*$.
The set equality above must mean that if $T \in (\ell^1)^*$, then $T \in \ell^\infty$. But how could this be? Every element of $\ell^\infty$ is a function from $\mathbb{N} \rightarrow \mathbb{F}$. $T$ is a function $\ell^1 \rightarrow \mathbb{F}$. They just seem fundamentally different.
I am clearly missing something obvious--and I suspect it is related to the fact that I defined the $T$ by using a sequence. Perhaps the equality is saying something about this sequence?
