dual space of a subspace of the space of bounded measures Let $\mathcal{M}=\mathcal{M}(\mathbb{R})$ be the space of bounded measures. Equipped with the weak convergence, the dual space of $\mathcal{M}$ is $\mathcal{C}_b(\mathbb{R})$ consisting of continuous bounded functions defined on $\mathbb{R}$.
Now we consider a subspace $\mathcal{M}_1$ of $\mathcal{M}$ such that:
$$\mathcal{M}_1=\{\mu\in\mathcal{M}: \int_{\mathbb{R}} |x|\mu(dx)<\infty\}$$
I would like to know the dual space of $\mathcal{M}_1$. In other words, which topology $\mathcal{M}_1$ should be used and for this topology which is the dual space of $\mathcal{M}_1$.
For example, we can take the topology of Wasserstein metric:
http://en.wikipedia.org/wiki/Wasserstein_metric
Which is the dual space associated to Wasserstein metric?
Does someone know the related results? Thanks a lot!
 A: The proper setting for your problem is that of weighted norms.  The basic situation is the symmetric duality between $C^b$ and $M_b$, the bounded continuous functions and the bounded measures on the real line.  Note that the former is not provided with the sup norm topology, but with the strict topology which was introduced in the 50's by R.C. Buck precisely for this duality.  Supose now that $ \phi$ is a weight, i.e. a positive, continuous function.  One then has a corresponding duality between $C^b_\phi$ and the space of measures $ \mu$ for which $ \mu \phi$ is bounded, where $C^b_\phi$ is the space of continuous functions with $f \phi$ bounded.  The former is provided with the corresponding weighted  sup-norm and strict topology.  The case you are interested in is where the weight is $1/(1+ |x|) $.
This is a brief sketch but I can provide details if desired.
A: A measure belongs to $\mathcal{M}_1$ if and only if every Lipschitz function is integrable (has finite integral); if and only if every uniformly continuous function is integrable. So you have at least two choices for what you want the dual of $\mathcal{M}_1$ to be: The space of all Lipschitz functions, and the space of all uniformly continuous functions. And you can get either of the two as dual of $\mathcal{M}_1$ by putting the appropriate "weak" topology on $\mathcal{M}_1$. The space of all Lipschitz functions is also the dual of $\mathcal{M}_1$ if $\mathcal{M}_1$ has the Vasershtein norm.
For more about duality with Lipschitz functions, see Weaver's 1999 book "Lipschitz algebras".
A: Thanks so much for 7891user's help. But it is not complet clear for me. So I summarize what I understand according to 7891user and please let me know if I am right.
Let $\mathcal{C}:=\mathcal{C}(\mathbb{R})$ be the space of continuous functions on $\mathbb{R}$. 
Define by $\mathcal{C}_b\subset\mathcal{C}$ consisting of bounded elements. Define for $\phi(x):=1+|x|$, 
$$\mathcal{C}_{\phi}=\{f\in\mathcal{C}: \frac{f}{\phi}\in\mathcal{C}_b\}$$
If we define for any $f, g\in\mathcal{C}_{\phi}$
$$d(f,g)=\sup_{x\in\mathbb{R}}|f(x)-g(x)|$$
then $(\mathcal{C}_{\phi}, d)$ is a complet metric space.
If we define for any $f\in\mathcal{C}_{\phi}$
$$||f||=\sup_{x\in\mathbb{R}}|\frac{f(x)}{\phi(x)}|$$
then $(\mathcal{C}_{\phi}, ||\cdot||)$ is a Banach space. 
Let $\mathcal{M}(\mathbb{R})$ be the space of bounded measures on $\mathbb{R}$ and denote
$$\mathcal{M}_1(\mathbb{R})=\{\mu\in\mathcal{M}(\mathbb{R}): \int_{\mathbb{R}} |x|d|\mu|<\infty\}$$
It is still not complet clear for me, $\mathcal{M}_1(\mathbb{R})$ is the dual space of $(\mathcal{C}_{\phi}, d)$ or $(\mathcal{C}_{\phi}, ||\cdot||)$? 
If $\mathcal{M}_1(\mathbb{R})$ is the dual space of some space, then does the convergence of $(\mu_n)\subset\mathcal{M}_1(\mathbb{R})$ imply  the convergence of $\int fd\mu_n$ for any $f\in\mathcal{C}_{\phi}$?
