$T: C_b(X, \mathbb R)\to C_b(X,\mathbb R)$ a linear map where $X$ is complete, separable, locally compact metric space. $C_b(X,\mathbb R)$ is vector space of all bounded continuous functions. Can I say that $T^*: M_1(X)\to M_1(X)$ where $M(X)$ is space of all probability measures on $X$? Thanks!
Dual space of space of space of bounded continuous maps
measure-theorymetric-spacesprobability theory
Related Solutions
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.)
The other answer is correct, but a little brief so I decided to add a few words.
Let's begin with the Riesz representation: For a locally compact Hausdorff space $X$, the map taking the finite signed regular Borel measure $\mu$ to the functional $f\mapsto \langle f,\mu\rangle:= \int f\,d\mu$ is an isometric isomorphism of the Banach space $M_r(X)$ onto the dual of the Banach space $C_0(X)$. Here $C_0(X)$ is the space of continuous functions that vanish at infinity, and the norm on $M_r(X)$ is the total variation norm.
Now Banach-Alaoglu guarantees that the unit ball in $C_0(X)^*$ is weak${}^*$ compact, which corresponds to vague compactness for the measures. However, generally, the unit sphere in $C_0(X)^*$ is not weak${}^*$ closed and hence not compact. Note however that the cone $\cal P$ of positive elements in $M_r(X)$ is vaguely closed: $${\cal P}=\bigcap_{f\in C_0(X)_+} \left\{\mu: \langle f,\mu\rangle \geq 0\right\},$$ so its intersection with the unit ball is compact.
The intuition here is that the positive part of the unit sphere is tiny, so could well be compact even though the whole sphere itself is not.
So how does the compactness of $X$ enter into the picture? If $X$ is compact, then $C_0(X)=C_b(X)$ and the vague and weak topologies of measures coincide. In particular, the constant function "1" belongs to $C_0(X)$ so the space of probability measures is the compact set $$ {\cal P}\cap \{\mu: \|\mu\|\leq 1\}\cap \{\mu: \langle 1,\mu\rangle =1\}.$$
Best Answer
No. Even for locally compact $X$, the dual of $C_b(X)$ is larger than the space of (regular Borel) measures.
Say $X=\Bbb N$, so $C_b(X)=\ell_\infty$. Let $\Lambda\in\ell_\infty^*$ be a Banach limit, which is to say that $\Lambda x=\lim_{j\to\infty}x_j$ for all $x\in\ell_\infty$ such that the limit exists. Then $\Lambda$ is not given by a measure on $X$.
Note $\Bbb N$ is certainly locally compact and separable; it's also complete in the standard metric $|n-m|$.
In fact one can give an analogous example if $X$ is any non-compact locally compact Hausdorff space. A bit of Banach algebra stuff shows that $C_b(X)$ is isometrically isomorphic to $C(K)$, where $K$ is a certain compact Hausdorff space containing $X$ as a dense subset. Any measure on $K$ which is not supported on $X$ gives an example of a bounded linear functional on $C_b(X)$ which does not arise from a measure on $X$.