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There are many representation theorems which state that the dual space of a Banach space $X$ has a particularly concrete form. For example, if $X = C([0,1],\mathbb R)$ is the space of real-valued continuous functions on $[0,1]$, then $X^*$ is the space of Radon measures on $[0,1]$. When you are confronted with some Banach space, where do you go to figure out a representation of its dual space? Is there a book or survey article with a rich set of examples?

Here is the particular example which motivates this question. Let $\operatorname{Sym}$ be the space of symmetric $n \times n$ real matrices with a usual matrix norm. Let $U \subseteq \mathbb R^n$ be compact, and let $X = C^{2+\alpha}(U, \operatorname{Sym})$ with the obvious norm. What is the dual space of $X$?

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The $\operatorname{Sym}$ is a red herring here. Since $X=C^{2+\alpha}(U)\otimes\operatorname{Sym}$, you get $X^*=C^{2+\alpha}(U)^*\otimes\operatorname{Sym}^*$. The same goes if $\operatorname{Sym}$ is replaced by any finite-dimensional space. (The tensor product of two infinite dimensional Banach space is an entirely different kettle of fish, of course.) – Harald Hanche-Olsen Apr 16 '10 at 2:30
Just to add to Harald's comment: of course this doesn't determine the dual space up to isometry, but up to (linear, bicontinuous) isomorphism. In many contexts one only cares about the Banach space up to isomorphism, but occasionally one might wish to determine the norm more precisely. – Yemon Choi Apr 16 '10 at 2:36
Note ... you mean Radon measures on [0,1], not on X . – Gerald Edgar Apr 16 '10 at 12:28
up vote 1 down vote accepted

In wikipedia, there is a list of banach spaces with its dual space and it also tells you if it is reflexive, for example.

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Hi Dan, welcome to MathOverflow. Thanks for the reference! – Tom LaGatta Aug 9 '11 at 1:44

Find many worked-out examples, all the important ones, in Dunford & Schwartz, volume I. See the top row of the table beginning on page 374.

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Great reference. Thanks, Gerald. – Tom LaGatta Apr 16 '10 at 16:50

In many cases, you simply work through or unwind the definition of a dual Banach space, namely the vector space of bounded linear functionals. In the specific case you're interested in, you simply get a space of $Sym^*$-valued distributions (in the sense of Laurent Schwartz).

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Since $U$ is compact and we're dealing with a fixed degree of differentiability, I don't think we get distributions per se. My guess would be `$Sym*$'-valued complex measures on $U$. – Yemon Choi Apr 16 '10 at 2:37
Well, they're not smooth distributions. But they're not just measures, either, because if $x_0 \in U$, the dual space contains the functional $f \mapsto \langle A,\partial_i\partial_jf(x_0)\rangle$, where $A \in Sym^*$. – Deane Yang Apr 16 '10 at 11:07

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