description of dual space of space of Radon measure equipped with topology of weak convergence

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.
• Thanks a lot. I overlook the obvious fact that for any dual pair $(X,Y)$, it always holds that $(X,\sigma(X,Y))^*=Y$.