I have a question which maybe so naive but I want to know the result about it.

Let $\mathcal{M}=\mathcal{M}(\mathbb{R})$ be the space of bounded measures. Then by some materiau such as Multidimensional diffusion processes and Large deviations, we know that the dual space of $\mathcal{M}$ is $\mathcal{C}_b^0(\mathbb{R})$ which is the space of continuous bounded functions defined on $\mathbb{R}$. Here the topology of $\mathcal{M}$ is induced by weak convergence.

Now we consider a subspace $\mathcal{M}_p$ of $\mathcal{M}$ such that:

$$\mathcal{M}_p=\{\mu\in\mathcal{M}: \int_{\mathbb{R}} x^2\mu(dx)<\infty\}$$

I would like to know the dual space of $\mathcal{M}_p$, I guess it is the space of continuous functions $f$ satisfying

$$|f(x)|\leq C(1+|x|^2)$$

for some constant $C$. But I don't know how to prove it. If someone knows it please let me know. Thanks a lot!

  • 4
    $\begingroup$ The notion of dual is usually associated to a vector space. The space of probability measures is not a vector space. $\endgroup$ Dec 5, 2013 at 21:28
  • $\begingroup$ There is probably a sensible question to be asked, but this should be "put on hold" until CodeGolf repairs it. $\endgroup$ Dec 5, 2013 at 22:21
  • $\begingroup$ Since every (Borel) probability measure on $\mathbb{R}$ can be written as a weak limit of linear combinations of Dirac measures (which all have finite second moment), isn't the closure of $\mathcal{M}_p$ in the topology of weak convergence $\mathcal{M}$? $\endgroup$ Dec 6, 2013 at 0:12
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    $\begingroup$ These are signed measures? So we can have $\int x^2 \mu(dx) = 0$ for nontrivial $\mu$? Perhaps you want a "norm" defined as $\int x^2 |\mu|(dx)$ ? $\endgroup$ Dec 6, 2013 at 1:13
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    $\begingroup$ Well in a Banach space, it is easy to show that a bounded linear functional on a dense set extends to the entire space, so your dual space would have to be the same as the dual space of $\mathcal{M}$. This is because continuity of a linear functional in a Banach space is the same as Lipschitz continuity and Lipschitz maps take Cauchy sequences to Cauchy sequences. I honestly do not even know the definition of the dual space of something other than a vector space, so maybe it's more complicated there. $\endgroup$ Dec 6, 2013 at 1:58

1 Answer 1


If $X$ is any locally convex space and $L$ is a subspace (endowed with the relative topology) then the (continuous) dual $L'$ of $L$ is a quotient of $X'$ by the subspace $L^\perp=\lbrace f\in X': f|_L=0\rbrace$ (this follows from Hahn-Banach: the restriction map $X' \to L'$ is surjective).

Even if you modify the definition of $\mathcal M_p$ as suggested by Gerald Edgar there is thus no reason to expect $\mathcal M_p'$ to be a subspace of $\mathcal M'$.

  • $\begingroup$ Would it be a quotient space then? $\endgroup$
    – ABIM
    Aug 11, 2017 at 13:22

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