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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!

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The notion of dual is usually associated to a vector space. The space of probability measures is not a vector space. –  Liviu Nicolaescu Dec 5 '13 at 21:28
    
There is probably a sensible question to be asked, but this should be "put on hold" until CodeGolf repairs it. –  Gerald Edgar Dec 5 '13 at 22:21
    
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 $\mathcal{M}$? –  Chris Janjigian Dec 6 '13 at 0:12
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These are signed measures? So we can have $\int x^2 \mu(dx) = 0$ for nontrivial $\mu$? Perhaps you want a "norm" defined as $\int x^2 |\mu|(dx)$ ? –  Gerald Edgar Dec 6 '13 at 1:13
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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. –  Chris Janjigian Dec 6 '13 at 1:58
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closed as unclear what you're asking by Gerald Edgar, Stefan Kohl, Daniel Moskovich, Jack Huizenga, Andrey Rekalo Dec 6 '13 at 8:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

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 $\mathcal M'$.

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