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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?

user310374
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    In fact the spaces $(\ell^1)^*$ and $\ell^\infty$ are not equal but isomorphic. – mathcounterexamples.net Feb 25 '17 at 19:17
  • @mathcounterexamples.net so the equality does not mean set equality, but rather that there exists some isomorphism? So what can I say about that specific $T$ I defined above? That there is some isometric function that takes it to a corresponding element in $\ell^\infty$? Is that the meaning of saying they are isomorphic? – user310374 Feb 25 '17 at 19:27
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    The idea is that the mapping $\ell^\infty\to(\ell^1)^*$ given by $\{a_i\}\mapsto T$ is an isometric isomorphism onto $(\ell^1)^*$. – Aweygan Feb 25 '17 at 19:43
  • @Aweygan that is helpful. However, for $g=(a_1,a_2,...) \in \ell^\infty$ as above, $\|g\| = \sup_i a_i$. On the other hand, if $T \in (\ell^1)^*$ is defined such that for $f=(f1,f2,...)\in \ell^1$, $Tf = \sum a_i f_i$, then $\|T\| = \sum_i |a_i|$, which is not in general equal to $\|g\|$, and so $\{a_i\} \rightarrow T$ is not an isometric isomorphism, right? So, perhaps $T$ should instead be defined as $T = \sup_i\{a_i f_i\}$? – user310374 Feb 25 '17 at 20:01
  • I guess I am just trying to understand, what is the $T\in (\ell^1)^*$ that corresponds to $\{a_i\} \in \ell^\infty$? – user310374 Feb 25 '17 at 20:03

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