Let $\mathcal{M}(\mathbb R)$ be the space of Radon measures, equipped with topology $\tau$ generated by the following "weak convergence":

$$ \mu_n \rightarrow \mu \quad \text{iff} \quad \int f \, d\mu_n \rightarrow \int f \, d\mu \quad $$ for all continuous function $f$ with quadratic growth: $|f(x)|\leq C(1+|x|^2)$ for some $C>0$. Let $\mathcal{M}_2(\mathbb R)$ be the subspace of $\mathcal{M}(\mathbb R)$ that contains all Radon measures with finite second moment.

I would like to know if there is a description of the topological dual of $(\mathcal{M}(\mathbb R),\tau)$ and $(\mathcal{M}_2(\mathbb R),\tau)$.

I know $\mathcal{M}(\mathbb R)$ is the dual of $C_0(\mathbb R)$, so we have $$ (\mathcal{M}(\mathbb R),\sigma(\mathcal{M}(\mathbb R),C_0(\mathbb R)))^*=C_0(\mathbb R) $$ where $\sigma(\mathcal{M}(\mathbb R)$ is the weak star topology. It is also obvious that convergence $\tau$ implies convergence in the weak star topology. So I was hopping the dual of dual of $(\mathcal{M}(\mathbb R),\tau)$ or $(\mathcal{M}_2(\mathbb R),\tau)$ would just be the family of continuous functions wit quadratic growth.

I also notice that $\tau$ convergence is the same as convergence in Wasserstein 2 distance, when restricted to probability measures with finite second moment. I will also be interested to see if there is any connection.

I hope my question makes sense and looking forward to any hints and ideas!


In general, if $V$ is a vector space, $W$ is a vector space of functionals on $V$, and we give $V$ the weak topology with respect to $W$, then the topological dual of $V$ will be just $W$. The proof is pretty much exactly the same as the case of the weak* topology, which you seem to be familiar with. See this post for more details.

So, in your case, the dual of either $\mathcal{M}(\mathbb{R})$ or $\mathcal{M}_2(\mathbb{R})$ with your weak topology would be the space of continuous functions of quadratic growth.

  • $\begingroup$ Thanks a lot. I overlook the obvious fact that for any dual pair $(X,Y)$, it always holds that $(X,\sigma(X,Y))^*=Y$. $\endgroup$
    – Ryan
    May 24 '18 at 15:33

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