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Let $(X,d)$ be a metric space, and let $C_u(X)$ be the Banach space of bounded uniformly continuous functions on $X$ (with the uniform norm). How can I characterize its dual space $C_u(X)^*$?

I would guess it can be described as some space of measures. I would even be interested in the case $X=\mathbb{R}$.

Obviously if $X$ is compact this is just the signed (or complex) Radon measures on the Borel $\sigma$-algebra of $X$. If $d$ is a discrete metric then we have all finitely additive measures on $X$. But more generally I do not know.

Edit: If $C_b(X)$ is the Banach space of all bounded continuous functions on $X$, we of course have $C_u(X) \subset C_b(X)$ as a closed subspace, and we know that $C_b(X)^*$ can be identified with the space of finite, regular, finitely additive signed Borel measures on $X$. Certainly each such measure gives us a continuous linear functional on $C_u(X)$, so we have a map $C_b(X)^* \to C_u(X)^*$ which is just the restriction map, but it is not injective. Conversely, by Hahn-Banach each bounded linear functional on $C_u(X)$ extends to a bounded linear functional on $C_b(X)$, but not in a canonical way.

Also, it is clear that in general $C_u(X)^*$ contains more than just the countably additive measures, since e.g. if $X=\mathbb{R}$ it contains some Banach limits. So we have all the countably additive finite Radon measures, but not all the finitely additive finite regular measures. This is why I would imagine that $C_u(X)^*$ consists of all finitely additive measures satisfying some condition that is more than "regular" but less than "countably additive". But I have no idea what it might be.

As mentioned in comments, I am happy to know about nontrivial special cases: $X$ locally compact, $X$ complete and separable, etc.

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Do you want to restrict to the case where X is locally compact? –  Yemon Choi Oct 29 '10 at 23:11
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My impression is that the best one can do in general is to say $C_u(X) = C(UX)$ where $UX$ is the spectrum of the unital C*-algebra $C_u(X)$, and then by Riesz representation say that the dual space is the space of measures on the compactification UX. The problem of what you can say about $UX$ is, I think, both much studied by others and little understood by me. –  Yemon Choi Oct 29 '10 at 23:14
    
@Yemon Choi: Sure, anything known in the locally compact case or other special cases would be interesting. –  Nate Eldredge Oct 30 '10 at 3:07
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To elaborate a little on what Yemon says: $C_u(G)$ is well-studied for a locally compact group $G$ (using the uniformity definition of uniformly continuous, but if the group if metric, that's the same). I don't think I've ever seen a "nice" description of $C_u(G)^*$ of the type you seem to be after... –  Matthew Daws Nov 1 '10 at 20:09

3 Answers 3

up vote 15 down vote accepted

$C_u(\mathbb R)^*$ is essentially the space of complex measures on $\beta \mathbb Z\coprod (\beta\mathbb Z\times(0,1)).$ Here $\beta \mathbb Z$ is the Stone-Čech compactification of $\mathbb Z,$ and the $\coprod$ denotes disjoint union.

One can identify $C_u(\mathbb R)$ with $C_0(\beta \mathbb Z \coprod (\beta \mathbb Z\times (0,1)))$ in the following way: for $f\in C_u(\mathbb R),$ and write $f=g+h$, where $g(n)=0$ for all $n\in \mathbb Z$ and $h$ is continuous and linear on each interval $[n,n+1].$ We will identify $g$ with a function $\tilde g:\beta \mathbb Z\times [0,1]\to \mathbb C$ in the following way: since $f:\mathbb R\to \mathbb C$ is uniformly continuous, the functions $g|_{[n,n+1]}, n\in \mathbb Z$ form an equicontinuous family, considered as functions $g_n\in C([0,1]).$ By Arzelà-Ascoli, the set $\{g_n:n\in \mathbb Z\}$ is precompact in the uniform topology. By the universal property of $\beta \mathbb Z$, there is a unique continuous function $\varphi: \beta \mathbb Z\to C([0,1])$ such that $\varphi(n)=g_n$ for $n\in \mathbb Z.$ Now the function $\tilde g(x,y):=\varphi(x)(y)$ is a continuous function from $\beta \mathbb Z\times [0,1]$ to $\mathbb C$; the joint continuity is obtained by again applying equicontinuity of the family $\{\varphi(x):x\in \beta\mathbb Z\}.$

We have identified $f$ with a pair $(\tilde g, h),$ where $\tilde g: \beta \mathbb Z\times [0,1]\to \mathbb C$, $\tilde g(x,0)=\tilde g(x,1)=0$ for all $x\in \beta \mathbb Z,$ and $h:\mathbb R\to \mathbb Z$ is determined by the sequence of values $h(n),n\in \mathbb Z.$ It's easy to check that every such pair $(\tilde g, h)$ uniquely determines a function $f\in C_u(\mathbb R)$ by $f(n+x)=\tilde g(n,x)+h(n+x), x\in [0,1), n\in \mathbb Z.$

If we use the norm $|(\tilde g, h)|:=|\tilde g|+|h|$ (these are sup norms), the identification $f\leftrightarrow (\tilde g,h)$ is an identification of $C_u(\mathbb R)$ with $C_0(\beta \mathbb Z \coprod (\beta \mathbb Z\times (0,1))),$ as Banach spaces.

So it appears that finding a non-obvious element of $C_u(\mathbb R)^*$ is more or less equivalent to finding a non-obvious element of $C_b(\mathbb Z)^*,$ as Greg predicted.

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Okay, fair enough. I was hoping for something that lives on $\mathbb{R}$ itself, like the finitely additive measures that are the dual of $C_b(\mathbb{R})$. I know that "non-obvious" examples are going to be nonconstructive no matter how they're expressed. I'll keep the bounty up for a few more days, but I could be satisfied with this. –  Nate Eldredge Nov 2 '10 at 5:49
    
I think this description of $C_u(\mathbb R)$ says the extreme points of the unit ball of $C_u(\mathbb R)^*$ are finitely additive $\{0,1\}$-valued measures supported on cosets of $\mathbb Z.$ This makes a bit of sense, since an atomless $\{0,1\}$-valued finitely additive measure supported on $\{n+\frac{1}{n}:n\in \mathbb N\}$ induces an element of $C_u(\mathbb R)^*$ which comes from such a measure supported on $\mathbb N,$ thanks to uniform continuity. –  John Griesmer Nov 2 '10 at 19:32

This is a belated reply to your query, more a comment but too long for that and, anyway, I don't have that option.

Firstly, there is a concept of a compactification which plays for metric spaces the same role as the Stone-Cech compactification for completely regular spaces (it works even for uniform spaces). It is called the Samuel compactification and the measures you are looking for are the Radon measures on this compact set (which contains the original space in a canonical way). In functional analytic terms it is the spectrum of the Banach algebra you are using---the bounded, uniformly continuous functions on the metric space (again, more generally, uniform space---in fact, I think that the category of uniform spaces is the more natural one for your question).

I kind of suspect that you are looking for ways to get nice spaces of measures (or spaces of nice measures) as duals of spaces of bounded continuous or uniformly continuous functions. If so, this has been investigated in some detail. In the case of continuous functions this is accomplished by the so-called strict topology (introduced originally by Buck for locally compact spaces, then generalised to completely regular spaces by several mathematicians in the 70's). Pachl (in Pac. J. Math. paper "Measures as functionals on uniformly continuous functions"---available online) did something similar for measures on uniform spaces---he called them uniform measures. In particular, he proved a deep theorem on compactness in this space---a theorem that is not as well known as it should be.

A unified approach to these subjects can be found in the book "Saks Spaces and Applications to Functional Analysis".

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Part of $C_u(X)^*$ is well understood. Every uniformly continuous function on $X$ uniqely extends to the completion $\bar{X}$, so certainly any signed Borel measure on $\bar{X}$ is a continuous functional on $C_u(X)$. If $\bar{X}$ is compact, then you're done. Beyond that, I don't think that much can be said. For example, if $X = \mathbb{Z}$, then every bounded function on $X$ is uniformly continuous. So $C_u(\mathbb{Z})^*$ is the set of measures on the Stone-Cech compactification $\beta\mathbb{Z}$ of $\mathbb{Z}$. It is well known that you cannot construct points in $\beta\mathbb{Z} \setminus \mathbb{Z}$ without the axiom of choice, or some other extension of ZF. That does not by itself imply that you cannot explicitly construct functionals in $C_u(\mathbb{Z})^*$ other than linear combinations of values, but I think that that's not possible either.

Moreover, $C_u(\mathbb{Z})$ embeds as closed Banach subspace of $C_u(\mathbb{R})$, for instance by taking the piecewise linear extension of a bounded function on $\mathbb{Z}$. So there must be many non-obvious functionals in $C_u(\mathbb{R})^*$ that restrict to non-obvious functionals in $C_u(\mathbb{Z})^*$. This type of argument applies to lots of metric spaces. It applies to any metric space that has an unbounded uniformly continuous map to $\mathbb{R}$, which is to say, any unbounded metric space. In detail, let $X$ be an unbounded metric space with a base point $x$, and let $S$ be an infinite set of distances from $x$ to other points, such that any two elements of $S$ are at least 1 apart. Then a bounded function on $S$ extends to $\mathbb{R}$ by linear interpolation (and constant extrapolation below the minimum of $S$). Then this function $f(t)$ pulls back to the function $f(y) = f(d(x,y))$ on $X$. This is a closed embedding of $C_u(S)$ into $C_u(X)$, and $C_u(S)^* = C(S)^*$ is wild except for the functionals that are linear combinations of values on $S$.

This is not a conclusive proof that $C_u(\mathbb{R})^*$ (say) does not have any non-obvious functionals in it whatsoever that can be constructed within ZF. But I would bet that this is true.

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The non-constructibility of such objects is addressed in econ.kuleuven.be/ces/discussionpapers/Dps10/DPS1010.pdf –  Nate Eldredge Nov 2 '10 at 5:40
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I think what I am most interested in is a characterization of $C_u(X)^*$ as a space of measures. Certainly it is going to contain weird things. $C_b(\mathbb{R})^*$ contains Banach limits and the like, which "need" AC. But they can be realized as finitely additive measures with some regularity properties (see, e.g. Dunford and Schwartz). This is deceptively simple because the existence of finitely additive measures that are not countably additive also "needs" AC. But I think it's a pretty fact, and I was hoping there was something analogous for $C_u(\mathbb{R})^*$ or $C_u(X)^*$. –  Nate Eldredge Nov 2 '10 at 5:44
    
@Nate I see --- I understand the question better now. What you really want is a natural interpretation of these dual vectors, whether or not they can be constructed. –  Greg Kuperberg Nov 2 '10 at 21:19
    
@Greg: Yes, exactly. –  Nate Eldredge Nov 2 '10 at 22:46

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