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.



Dual of bounded uniformly continuous functions
$C\_u(\mathbb R)^\*$ 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.


