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 $C_b(E)'$?

  • $\begingroup$ Hm, the bounty does not seem to have increased the number of views by much. I think this is for the most part a functional analysis question, so maybe you should add the functional-analysis tag, seeing as that's the tag most people are following. $\endgroup$ Dec 3 '20 at 21:00
  • $\begingroup$ Too bad you let the bounty expire. In any case, I hope my answer was useful to you. Have a nice day! $\endgroup$ Dec 8 '20 at 12:56

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.


[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.

  • $\begingroup$ Thank you for your answer. (a) When you say what we exactly need to show to conclude that "$E_1'$ is just $E_2$ with the $\sigma(E_2,E_1)$-topology". In fact, I don't know whether we need to show anything at all or whether this is simply how we define the topology on $E_1'$. In the case of a normed space $E$, we usually equip $E'=\mathfrak L(E,\mathbb R)$ with the operator norm. And if we have a "representation result" as in the question of the form $E\cong F$ for some other normed space $F$, it usually means that there is an isometric isomorphism from $E$ to $F$. $\endgroup$
    – 0xbadf00d
    Dec 5 '20 at 18:13
  • $\begingroup$ (b) There's still an unanswered question in the comments below this answer, but if this answer is correct, the bilinear pairing should always be non-degenerate, since every finite measure on the Borel $\sigma$-algebra of a metric space has a certain regularity property. What do you think? $\endgroup$
    – 0xbadf00d
    Dec 5 '20 at 18:15
  • $\begingroup$ @0xbadf00d (a) we don't have to show anything at all, unless I misunderstood your original question. However, you make a good point that the weak-$*$ topology on $E_1'$ is not the usual norm topology. The norm dual is generalized by the so-called strong topology on the dual of an arbitrary TVS. (b) Thanks for pointing that out; I wasn't aware of that. $\endgroup$ Dec 7 '20 at 7:29

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