Dual of bounded uniformly continuous functions - MathOverflow

Dual of bounded uniformly continuous functions

2010-10-29T21:20:17Z

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.

Answer by Greg Kuperberg for Dual of bounded uniformly continuous functions

2010-11-01T23:03:10Z

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.

Answer by anonymous for Dual of bounded uniformly continuous functions

2010-11-02T02:24:55Z

$C_u(\mathbb R)^*$ 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.

Answer by jbc for Dual of bounded uniformly continuous functions

2012-08-29T17:00:43Z

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".