Timeline for What is the "dual" of the space of currents?
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| Apr 25, 2018 at 14:20 | comment | added | Jochen Wengenroth | Well, in nuclear (more generally: Montel) spaces the strong and weak topologies have the same convergent sequences (or bounded nets or filters with basis in a bounded set -- just because bounded sets are strongly compact and there is no strictly coarser Hausdorff topology on a compact space). But even if two locally convex topologies have the same convergent sequences they need not have the same dual. | |
| Apr 25, 2018 at 13:29 | comment | added | Liviu Nicolaescu | Let me rephrase: In a perfect space strong and weak convergence coincide: Gelfand & Shilov, vol.2, Section I.6.3. The theorem in Sec. I.6.2. of the same book shows that $C^\infty(M)$ is perfect if $M$ compact. | |
| Apr 25, 2018 at 12:33 | comment | added | Jochen Wengenroth | No. Whenever an $\infty$-dimensional space $X$ has a continuous norm (i.e., $p(x)=0$ implies $x=0$, seminorms need not satisfy this) then its topology is strictly finer than $\sigma(X,X^*)$ (just because the latter topology doesn't have continuous norms). The space $C^\infty(M)$ you mentioned clearly has continuous norms. | |
| Apr 25, 2018 at 10:01 | comment | added | Liviu Nicolaescu | In the terminology of Gelfand and Shilov, if $X$ is a perfect locally convex space than its topology coincides with the $\sigma(X,X^*)$ topology. The nuclear spaces are perfect. | |
| Apr 25, 2018 at 7:28 | comment | added | Jochen Wengenroth | What is the role of nuclearity? Of course, the dual of $(X^\ast,\sigma(X^\ast,X))$ is $X$ (more precisely, $\{L_x:x\in X\}$) for every locally convex space $X$. If you ask for a topological isomorphism between $(X^\ast,\sigma^\ast)^\ast$ you have to specify the topology on that bidual. For the topology of uniform convergence on equicontinuous sets you have again topological isomorphy for all locally convex spaces. If you consider uniform convergence on bounded sets you need a barrelldness condition for $X$. For the weak$^\ast$-topology on the bidual you need that $X$ has its weak topology... | |
| Apr 25, 2018 at 4:58 | vote | accept | Li Yutong | ||
| Apr 24, 2018 at 17:04 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
added 2 characters in body
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| Apr 24, 2018 at 16:29 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |