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Let $\mathcal{D}(R):=C_c^\infty(R)$ be the smooth functions with compact support. Its dual space is the space $\mathcal{D}'(R)$ of distributions. This space $\mathcal{D}(R)$ has its weak *-topology induced by $\mathcal{D}(R)$, which makes it into a locally convex space (See Rudin Functional Analysis P.160).

The question is: Does the double dual space $\mathcal{D}''(R)$, i.e., the dual space of the distributions $\mathcal{D}'(R)$, exist? For Banach space $X$, we know that double dual $X^{**}\supseteq X$. They are equal when $X$ is reflective. A first guess is that $\mathcal{D}''(R)\supseteq \mathcal{D}(R)$.

Thanks for any comments! :-)

Anand

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I just find a reference. The answer is yes. See Yoshida's Functional analysis (6th Ed.) P. 140. –  Anand Feb 7 '12 at 17:30
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Of course the dual space exists. The question is not whether or not it exists but if we can find a nice description of it. –  Johannes Hahn Feb 7 '12 at 18:16
    
I don't understand the question: as Johannes notes, the dual space of course exists... Anand, can you edit the question so that it asks what you really want to ask? –  Mariano Suárez-Alvarez Feb 7 '12 at 18:46
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up vote 4 down vote accepted

Well, that depends on what topology you want to put on the space of distributions. The weak$^*$ is probably not really the one you would like to take. Instead, the strong dual might be more useful. The seminorms of this topology are given by $$p_B(\varphi) = \sup_{f \in B} |\varphi(f)|$$ where $B \subseteq \mathcal{D}(\mathbb{R})$ runs through the bounded subsets of the LF space $\mathcal{D}(\mathbb{R})$. The it is a theorem that (since the test functions are Montel etc) the dual with respect to this is again the space of test functions, i.e. the test functions are reflexive...

I guess for the weak$^\ast$ topology this is not true and one gets a different dual of the dual. Your inclusion is correct, any test function gives a linear functional on the distribtions (by evaluation) which is continuous in the weak$^*$-topology. But you probably get more...

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The dual of the weak*-space can is also isomorphic (via the natural map) to the space of test functions. –  Johannes Hahn Feb 7 '12 at 18:19
    
Hi Johannes: Ok, that I didn't know. But the weak$^*$ topology is still different from the strong one, right? I should be strictly weaker... –  Stefan Waldmann Feb 7 '12 at 18:35
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For any topological vector space $X$ the dual of $(X',\sigma(X',X))$ is $X$ (more precisely, given by an evaluation in some point of $X$). This is rather linear algebra than topological vector spaces: If $F$ is a continuous linear functional on $X'$ there are finitely many $x_n \in X$ and a constant $C$ such that [ |F(\phi)|\le C \max\{\phi(x_n)\} ] which implies that $F$ is a linear combination of the evaluations in $x_n$ and (by linearity) an evaluation in the linear combination of the $x_n$. –  Jochen Wengenroth Feb 8 '12 at 9:01
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The weak$^*$ topology hardly ever coincides with the topology of uniform convergence on bounded sets as described in Stefan's answer. This is the case if and only if all bounded sets are finite-dimensional. –  Jochen Wengenroth Feb 8 '12 at 9:03
    
Thanks for the clarification, Jochen! –  Stefan Waldmann Feb 8 '12 at 9:32
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Check Francois Treves' book (now available in Dover)

Topological Vector Spaces, Distributions and Kernels

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Thanks Liviu Nicolaescu. I will have a look. My main references are Rudin and Yoshida's book. :-) –  Anand Feb 8 '12 at 9:50
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