# Dual Pairs, topology of weak convergence and weak* topology

Edit for Bounty: I decided to put a bounty on this question because I would really like to get it properly. Thus, I would like to get feedbacks on my basic questions, and a detailed answer on my question (3) written in a language as accessible as possible, which is the basic source of confusion.

[Of course, any additional explanation regarding dual pairs is most welcome]

Here there is my question. To ask it, I will disseminate this text with few numbered questions.

Assume we have:

• a Polish space $$X$$;
• the set of continuous bounded functionals on $$X$$, denoted by $$C_b (X)$$;
• the Borel $$\sigma$$-algebra on $$\Omega$$, denoted by $$\mathcal{B} ( X)$$;
• the set of probability measures on $$\Omega$$, denoted by $$\Delta (X)$$.

Immagine that instead of simply endowing $$\Delta(X)$$ with the topology of weak convergence, we translate this all set up in terms of topological vector spaces, and consequently dual spaces, in order to get the same result by talking about the weak* topology.
Of course, we obtain the same result, but by doing so, I understand how this all thing works... :-)

1. Does what I have written here technically make sense?

If we do that, then we can say that we have a dual pair given by $$\langle C_b (X), \Delta (X) \rangle$$.

2. Is this literally (!) correct, or the correct expression of dual pair is on of the two:

1. $$\langle \mathcal{B} (X), \Delta (X) \rangle$$, or
2. $$\langle X, \Delta (X) \rangle$$?

I think (1) is wrong, because $$\mathcal{B}(X)$$ is not a vector space, thus it should not make sense to talk about $$\langle \mathcal{B} (X), \Delta (X) \rangle$$ as a dual pair. However, I am not sure about (2).

3. If one or the two, or both, are wrong and they cannot be dual pairs, why is the case?

Thus, we can say that we have a topological space $$( \Delta (X), \sigma (X^*, X))$$, where $$\sigma (X^*, X)$$ denotes the weak* topology, i.e.
$$\mu_n \overset{w^*}{\longrightarrow} \mu \in \Delta (X) \Longleftrightarrow \forall f \in C_b (X), \ \langle f, \mu_n \rangle \to \langle f, \mu \rangle \in \mathbb{R},$$

where $$\sigma (X^*, X) = w^*$$.

4. Is this correct?

Thus, we do have that $$( \Delta (X), w^*)$$ is a topological space, and is nothing more than $$\Delta (X)$$ endowed with the topology of weak convergence. Hence, this all process also shows why it should be more appropriate to write that $$\Delta (X)$$ is actually endowed with the topology of weak convergence* (even though the convention is another).

5. Again, is this sound?

Any feedback or answer is welcome, and it will be enormously appreciated.

• Dual pair usually is reserved for both being vector spaces. While the bounded continuous functions form one the probability measures don't. – C-Star-W-Star Mar 1 '15 at 20:13
• I'm not quite sure what may happen if the second domain of the dual pair lacks to be vector space and the bilinear form merely remains a linear form in its first argument but I guess it will still give a topological vector space as the crucial ingredient of a dual pair is the linearity of the evaluations. In your case you can consider: $\varepsilon_\mu(f):=\int_\Omega f\mathrm{d}\mu$ That will turn the continuous functions into a topological vector space: $\mathcal{C}(\Omega)$ So the dual pair would read: $\langle\mathcal{C}(\Omega),\mathcal{M}(\Omega)\rangle$ – C-Star-W-Star Mar 1 '15 at 20:50
• Besides, is it important for your study to consider positive measures? And did you have contact to complex measures already? – C-Star-W-Star Mar 1 '15 at 21:00
• @Freeze_S: I prefer not to enter yet into the realm of complex measures. For me, it is enough right now to get how it works in real terms. – Kolmin Mar 2 '15 at 11:56
• Ok fine enough. ;) – C-Star-W-Star Mar 2 '15 at 13:29

Disclaimer

This is just a formal treatment.
(It requires tedious verifications!)

Ground Space

Given a Borel space: $$\mathcal{B}(\Omega)=\sigma(\mathcal{T}(\Omega))$$ (Candidates: Polish Spaces)

Function Space

Consider the continuous functions: $$\mathcal{C}(\Omega):=\left\{f:\Omega\to\mathbb{C}:U\in\mathcal{T}(\mathbb{C}):f^{-1}U\in\mathcal{T}(\Omega)\right\}$$ They form a vector space by: $$(f+g)(\omega):=f(\omega)+g(\omega)\quad(\lambda f)(\omega):=\lambda f(\omega)$$ (Note: No topology yet!)

Measure Space

Consider the complex measures: $$\mathcal{M}(\Omega):=\left\{\mu:\Omega\to\mathbb{C}:A_k\in\mathcal{B}(\Omega):\mu\left(\bigsqcup_kA_k\right)=\sum_k\mu(A_k)\right\}$$

They form a vector space by: $$(\mu+\nu)(A):=\mu(A)+\nu(A)\quad(\lambda\mu)(A):=\lambda\mu(A)$$ (Note: No topology yet!)

Dual Pair

They become a dual pair by: $$\mathcal{C}(\Omega)\times\mathcal{M}(\Omega):\quad\langle f,\mu\rangle:=\int_\Omega f\mathrm{d}\mu$$ This endows the measures with a topology: $$\mathcal{T}(\mathcal{M}(\Omega)):=\sigma\{\mathcal{M}(\Omega);\mathcal{C}(\Omega)\}$$ (Hint: Reversing endows the functions with a topology.)

Modifications

Various function spaces: $$\mathcal{C}^\infty(\Omega)\quad\mathcal{C}_0(\Omega)\quad\mathcal{C}_\infty(\Omega)\quad\mathcal{C}^\infty_0(\Omega)\quad\ldots$$ Various measure spaces: $$\mathcal{M}_\mathbb{R}(\Omega)\quad\mathcal{M}_{\mathbb{R}_+}(\Omega)\quad\mathcal{M}_\mathbb{S}(\Omega)\quad\mathcal{M}_{\mathbb{S}_+}(\Omega)\quad\ldots$$ (Warning: Some don't form a vector space!)

• Thanks! However, I think there is a typo (not particularly serious) in the Dual Pair section, because first you write $\mathcal{C} (\Omega) \times \mathcal{M} (\Omega)$ and then you write $\langle \mu, f \rangle$. Anyway, summing up what you wrote, basically what I wrote is correct. – Kolmin Mar 3 '15 at 13:11
• Yes that's a typo ^^ (I was switching back and forth while writing.) – C-Star-W-Star Mar 3 '15 at 13:42
• However, your suggestion were not correct as you neither consider $\mathcal{B}(\Omega)\times\mathcal{M}(\Omega):\langle A,\mu\rangle:=\mu(A)$ nor do you $\Omega\times\mathcal{M}(\Omega):\langle\omega,\mu\rangle=??$. – C-Star-W-Star Mar 3 '15 at 13:47
• @Kolmin: I have no idea why mathjax doesn't want to compile. Hope it is readable anyway. (I guess it doesn't like me.) ;P – C-Star-W-Star Mar 3 '15 at 14:03
• It is the last "\mathcal{M)". You did not use the curly bracket. – Kolmin Mar 3 '15 at 14:05