Since I am reading some stuff about weak convergence of probability measures, I started to wonder what is the dual space of the space consisting of all the finite (signed) measures (which is well known to be a Banach space with the norm being total variation). Is there any characterization of it? We may impose extra assumptions on the underlying space if necessary.
Measure Theory – Dual Space of the Space of Finite Measures
banach-spacesfunctional-analysismeasure-theoryprobability theory
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Let $E$ be a locally convex space with topological dual $E'$. Equip $E'$ with the weak*-topology.
Every continuous linear functional $\varphi \colon E' \to \mathbb C$ is of the form $f \mapsto f(e)$ for some $e \in E$, so $(E',\text{weak*})' = E$. In particular, the dual space of $M(X)$ with the weak*-topology is $C(X)$ again.
Since $\varphi$ is continuous, the set $U = \left\{f \in E' \mid \lvert \varphi(f) \rvert \lt 1\right\}$ is an open neighborhood of $0$. Thus, recalling how the basis of neighborhoods at a point in $(E',\text{weak*})$ is defined, there exist $e_1,\dots,e_n \in E$ and $\varepsilon \gt 0$ such that $V = \left\{f \in E' \mid \max\left\{\lvert f(e_i)\rvert \mid i = 1,\dots,n\right\} \lt \varepsilon\right\}$ is contained in $U$. Let $\varphi_i(f) = f(e_i)$.
From $V \subseteq U$ it follows that $\ker{\varphi} \supseteq \bigcap_{i=1}^n \ker{\varphi_i}$ and from linear algebra (see here) we deduce that $\varphi = \sum_{i=1}^n \lambda_i \varphi_i$ for some $\lambda_i \in \mathbb{C}$. In other words, $$ \varphi(f) = \sum_{i=1}^n \lambda_i \varphi_i(f) = \sum_{i=1}^n \lambda_i f(e_i)= f\left(\sum_{i=1}^n \lambda_i e_i \right)$$ and we've shown that $\varphi$ is evaluation at $e = \lambda_1 e_1 + \cdots + \lambda_n e_n \in E$.
The answer is only partially YES. However $\mathcal{M}^+(\mathbb{\mathbb R})$ obviously cannot be a vector space due to the positivity constraint. So this rules out both questions as currently written. What is true, though, is that the metric space $(\mathcal{M}^+(\mathbb{R}),d_{BL})$ is complete and metrizes the weak convergence. I will only prove rigorously the completeness, see my final remark for how to get the "metrization". The proof below actually works in any dimension, and also in any domain $\Omega\subset \mathbb R^d$.
Let $\{\mu_n\}$ be a sequence of positive measures, let me denote the mass $m_n:=\mu_n(\mathbb R)\geq 0$, and assume that the sequence is Cauchy $$ d_{BL}(\mu_p,\mu_q)\to 0 \qquad \mbox{as }p,q\to\infty. $$
As a first step, it is easy to see that $\{m_n\}$ is a (real) Cauchy sequence: for, testing $ f \equiv 1$ in the definition of $d_{BL}$, we get $$ |m_p-m_q|=\left|\int 1 d\mu_p -\int 1 d\mu_q \right|\leq d_{BL}(\mu_p,\mu_q). $$ Since the real line is complete, there is $m\geq 0$ such that $m_n\to m$. If $m=0$ then it is immediate to see that, for any $f\in \mathcal C_b$, there holds $|\int f d \mu_n|\leq ||f||_\infty m_n\to 0$, which proves that $\mu_n\to 0$ weakly (narrowly).
If $m>0$ then we can assume that $m/2\leq m_n\leq 2m$ for $n$ large enough, and the renormalized sequence $\tilde \mu_n:=\frac {\mu_n}{m_n}\in\mathcal P(\mathbb R)$ is well-defined. I claim that $\{\tilde\mu_n\}$ is $d_{BL}$-Cauchy as well. Indeed, for $p,q$ large enough we have by triangular inequality \begin{multline*} \left| \int f d\tilde\mu_p -\int f d\tilde\mu_q\right| = \left| \int f \frac{1}{m_p}d\mu_p -\int f \frac{1}{m_q}d\mu_q \right| \\ \leq \frac 1m \left| \int f d\mu_p -\int f d\mu_q \right| \\ + \left|\left(\frac 1{m_p}-\frac 1m \right)\int f d\mu_p\right| + \left|\left(\frac 1{m_q}-\frac 1m \right)\int f d\mu_q\right| \\ \leq \frac 1m \left| \int f d\mu_p -\int f d\mu_q \right| \\ + \left|\frac 1{m_p}-\frac 1m \right| \|f\|_\infty 2m + \left|\frac 1{m_q}-\frac 1m \right| \|f\|_\infty 2m. \end{multline*} Taking the supremum over $f$ such that $\|f\|,Lip(f)\leq 1$ gives $$ d_{BL}(\tilde\mu_p,\tilde\mu_q)\leq \frac 1md_{BL}(\mu_p,\mu_q) + 2m\left|\frac 1{m_p}-\frac 1m \right| +2m\left|\frac 1{m_q}-\frac 1m \right| $$ and entails my claim.
Since $(\mathcal P(\mathbb R),d_{BL})$ is complete there is a proabability measure $\tilde \mu\in \mathcal P(\mathbb R)$ such that $d_{BL}(\tilde\mu_n,\tilde \mu)\to 0$. Because we already proved that $m_n\to m$, it is then easy to check that $ \mu_n=m_n\tilde\mu_n$ converges (in the Bounded-Lipschitz distance) to the limit $\mu:=m\tilde\mu$. Indeed for fixed $f$ \begin{multline*} \left|\int f d\mu_n- \int f d\mu \right| =\left|m_n\int f d\tilde\mu_n- m\int f d\tilde\mu \right| \\ \leq |m_n-m|\cdot \left|\int f d\tilde \mu_n\right| + m\left|\int f d\tilde\mu_n-\int f d\tilde\mu \right| \\ \leq |m_n-m|\cdot\|f\|_\infty+ m\left|\int f d\tilde\mu_n-\int f d\tilde\mu \right|. \end{multline*} Taking one last time the supremum over $f$'s gives $d_{BL}(\mu_n,\mu)\leq |m_n-m| + md_{BL}(\tilde\mu_n,\tilde\mu)\to 0$ and the proof is complete.
Final remark: following the same lines it is easy to see that $d_{BL}$ does indeed metrize the weak convergence. The strategy of proof is identical: show that the masses converge, use this to suitably renormalize $\tilde\mu_n:=\frac{1}{m_n}\mu$, and exploit that the statement is already known for probability measures. (The case of vanishing mass $m_n\to 0$ must be treated separately.)
Best Answer
Well, your space of measures is isometric to $L^1(\mu)$ for some (probably very big, non-sigma-finite) measure $\mu$. So it is enough to know what is the dual of an $L^1$ space.