In general, if $E_i$ is a $\mathbb R$-vector space and $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing between $E_1$ and $E_2$ and $$p_{x_2}(x_1):=q_{x_1}(x_2):=\left|\langle x_1,x_2\rangle\right|\;\;\;\text{for }(x_1,x_2)\in E_1\times E_2,$$ then $p_{x_2}$ is a seminorm on $E_1$ for all $x_2\in E_2$ and $q_{x_1}$ is a seminorm on $E_2$ for all $x_1\in E_1$.
Hence, $\left\{p_{x_2}:x_2\in E_2\right\}$ is inducing a locally convex topology $\sigma(E_1,E_2)$ on $E_1$ and $\left\{q_{x_1}:x_1\in E_1\right\}$ is inducing a locally convex topology $\sigma(E_2,E_1)$ on $E_2$.
If $\varphi\in E_1^\ast$, we can show that $\varphi$ is $\sigma(E_1,E_2)$-continuous iff $\varphi=\langle\;\cdot\;x_2\rangle$.
By the aforementioned result, we are able to identify $$E_1':=\left\{\varphi\in E_1^\ast:\varphi\text{ is }\sigma(E_1,E_2)\text{-continuous}\right\}$$ with $E_2$. Does that mean $E_1'$ is again a locally convex topological vector space when endowed with the topology $\sigma(E_2,E_1)$?
I'm mainly interested in the following instance: Let $E_1:=C_b(E)$ be the set of real-valued bounded continuous functions on a metric space $E$, $E_2:=\mathcal M(E)$ denote the set of finite signed measures on $\mathcal B(E)$ and $$\langle f,\mu\rangle:=\int f\:{\rm d}\mu\;\;\;\text{for }(f,\mu)\in C_b(E)\times\mathcal M(E).$$
If $E_1=C_b(E)$ is equipped with the supremum norm, we may identify $\mathcal M(E)$ with a subset of $C_b(E)'$. How is $\sigma(E_2,E_1)$ related the the subspace topology on $\mathcal M(E)$ induced from the weak* topology on $C_b(E)'$?
Best Answer
Q1: Yes, $E_1'$ is a locally convex space when endowed with the $\sigma(E_1',E_1)$-topology. (It is just $E_2$ with the $\sigma(E_2,E_1)$-topology.)
Q2: The weak-$*$ topology respects subspaces. In general, if $\langle E, G \rangle$ is a dual pair and if $F \subseteq G$ is a subspace, then the relative $\sigma(G,E)$-topology on $F$ coincides with the $\sigma(F,E/F^\perp)$-topology.¹ If $F$ separates points on $E$, so that $\langle E,F\rangle$ is again a dual pair, then one has $F^\perp = \{0\}$, so now the $\sigma(F,E/F^\perp)$-topology is simply the $\sigma(F,E)$-topology.
¹: For a textbook reference, see [Sch99, §IV.4.1, Corollary 1 (p. 135)] or [Köt83, Proposition 22.2.(1) (p. 276)], among others. This is basically just a special case of the transitive law of initial topologies (see halfway through this very long answer), since weak topologies and subspace topologies are examples of initial topologies.
In your example, we would have $E = C_b(\Omega)$, $F = \mathcal M(\Omega)$ and $G = C_b(\Omega)'$. It is easy to see that $\mathcal M(\Omega)$ separates points on $C_b(\Omega)$, so $F^\perp = \{0\}$. Furthermore, we know from this other question of yours that $C_b(\Omega)$ separates points on $\mathcal M(\Omega)$, so $\mathcal M(\Omega)$ can be viewed as a subspace of $C_b(\Omega)'$. Hence we are in the setting described above.
References
[Köt83] Gottfried Köthe, Topological Vector Spaces I, Second revised printing, Grundlehren der mathematischen Wissenschaften 159, Springer, 1983.
[Sch99] H.H. Schaefer, with M.P. Wolff, Topological Vector Spaces, Second Edition, Graduate Texts in Mathematics 3, Springer, 1999.