[Math] Is there dual space of the distributions $\mathcal{D}'(R)$

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Dear MOs,

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

Best Answer

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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