# How is it useful to know what the dual of a space is isomorphic to?

Wikipedia has a neat spaces and their duals. For example, it lists that the dual of $c_0$ is $l_1$.

But how can I use that knowledge? I'm trying to prove something for all functions of the dual of $c_0$, and figured that I could use my familiarity of $l_1$ to get some properties of those functions. But I don't know how.

For example, which element of $l_1$ corresponds to which functional of $c_0$? I suppose their "scalar product" (multiply $x\in c_0$ with $y\in l_1$ element-wise and sum the products) would be such a functional. If that's true, I could probably prove it. But I only came up with it because that was my intuition. Is there a more mechanical way?

I realize that it's hard to answer the question without knowing exactly which problem I'm trying to solve, but perhaps there are some general strategies?

Sorry if the question seems confused and "ranty"; I'm having a hard time sorting this out.

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Indeed, the typical "representation theorem" for those dual spaces gives you an explicit isomorphism. It is not enough to tell which space the dual is isomorphic to. And your conjecture about the ismorphism between $\ell^1$ and $c_0^\star$ is true. –  Giuseppe Negro Dec 6 '12 at 0:42

When working with infinite dimensional spaces (such as spaces of functions or of sequences) it is not enough to consider the topology induced by the norm. The norm topology has so many open sets that compact sets are hard to find. And we need compactness for all kinds of proofs in analysis.

Knowing that your space $X$ is the dual of some other Banach space $Y$ gives you weak* topology on $X$. (I don't like how this subject is treated in Wikipedia, so I link to these lecture notes instead). For this topology we have a fantastic theorem of Banach and Alaoglu: the closed balls $\{x\in X\colon \|x\|\le R\}$ are compact. All of a sudden, we are able to pick convergent subsequences from bounded sequences, as we do all the time in finite dimensions. Convergence must be understood in the weak* sense, but that is better than nothing.

Another tool that requires compactness is Krein-Milman theorem: every compact convex set is the closed convex hull of its extreme points. Say, we have a nonempty bounded closed convex set $K\in \ell_1$. Since $\ell_1$ is the dual space of $c_0$, we have weak* topology on $\ell_1$, which makes $K$ compact. Applying Krein-Milman theorem, we conclude that $K$ has extreme points. In contrast, none of this works in $L^1[0,1]$, where the closed unit ball has no extreme points. The problem is that there is no Banach space whose dual is isomorphic to $L^1[0,1]$.

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The problems that you are trying to solve (in sequence-spaces such as $c_0, l^p$, etc) are meant to give you practice to attack the harder problems that show up in function spaces such as $C_0(X), L^p(X,\mu)$, etc. Identifying the dual spaces of these function spaces is a long and difficult story in measure theory - a fact that is not obvious unless you have spent some time struggling through their counterparts in these sequence-spaces.

More philosophically, the interplay between a vector space and its dual is an example of a phenomenon that is quite widely used in mathematics. For instance, a similar duality for abelian topological groups is the reason the Fourier transform is so interesting.

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