# 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!

• I very much doubt your first claim that the dual space of $M(R)$ is $C_b(R)$. We know $C_b(R)^*=M(R)$ -- but that the double dual is recoverable seems perverse.
– JHM
Jan 14, 2014 at 19:41
• The Wasserstein metric is a metric between probability measures, not arbitrary measures. You can turn this metric into a norm, but then the set of measures with finite first moment isn't complete for it. Jan 14, 2014 at 20:20
• The "weak" topology of the question is known as the "weak-star" topology to functional analysts. If $X^*=Y$, then the dual of $Y$ (with the weak-star topology) is $X$ again. Jan 14, 2014 at 22:29
• @GeraldEdgar Yes, but then one should also point out that the dual of $C_b(R)$ is not $\mathcal M(R)$, but rather the measures on the Stone-Cech compactification of $R$. (As far as I remember...) Jan 15, 2014 at 10:09
• There is a pairing between the two spaces $C_b(R)$ and $M(R)$. That is all that is needed for the weak dual of one to be the other, "weak" in the sense of the pairing. Jan 15, 2014 at 16:29

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.

• I would appreciate some details.
– JHM
Jan 14, 2014 at 21:17
• Thanks so much for 7891user and I would like to know the details Jan 14, 2014 at 21:23
• Thanks for J. Martel pointing that and I will reformulate my question. Jan 14, 2014 at 21:23
• Forget $\mathcal{M}$ and we focus on $\mathcal{M}_1$, which topology we can use for the space $\mathcal{M}_1$ and which is its dual space associated to the topology? Jan 14, 2014 at 21:24
• I answered that---it is the corresponding space of weightrd continuous functions. The topology is the weak one generated by it on your space of measures. Jan 15, 2014 at 7:02

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".

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 $f\in\mathcal{C}_{\phi}$?

• Could 7891user make me clear? And I need the associated reference for this duality. Thanks a lot! Jan 15, 2014 at 11:03
• It s important that you use the strict topogy on the bounded, continuous functions, not the norm (the dual in this case would be the FINITELY additive, bounded measures, not what you want I imagine. The strict topology was defined by Buck, confusingly in this context by using weighted seminorms. A more modern description would be: the finest locally convex topology which agrees with compact convergence on the unit ball. The topology on $C^b_\phi$ that you want is described in an analogous fashion. Jan 15, 2014 at 11:43
• A suitable reference is (declaration of interest) my book "Saks spaces and applications to functional analysis". Jan 15, 2014 at 11:45
• Thanks a lot! My last question is given $\mathcal{M}_1(\mathbb{R})$, I need some type of convergence: $(\mu_n)\subset\mathcal{M}_1(\mathbb{R})\rightarrow\mu\in\mathcal{M}_1(\mathbb{R})$ implies either $\int fd\mu_n\rightarrow\int fd\mu$ for every $f\in\mathcal{C}_b$ and $\int |x|d|\mu_n|\rightarrow\int |x|d|\mu|$ either $\int fd\mu_n\rightarrow\int fd\mu$ for every $f\in\mathcal{C}_{\phi}$ I would like to know the description of the dual space of $\mathcal{M}_1(\mathbb{R})$ associated to either of these two convergence. If you know the result please let me know, thanks again! Jan 15, 2014 at 13:24