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Let $D\subset X$ be a dense subset of a complete separable locally convex space $X$ over $\mathbb{R}$. Though the question seems simple enough, I can't seem to find the answer in the literature:

  • If $D$ is a dense subspace of $X$, then do the continuous duals $D'$ and $X'$ coincide?

  • More generally, if $D$ is not a linear subspace, can we give any meaning to its "dual" $D'$ and if so, how does this object relate to $X$?

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    $\begingroup$ For $D$ a subspace this seems immediate to me, for $X$ an arbitrary topological vector space. Use Cauchy nets (and completeness of $\mathbf{R}$). $\endgroup$
    – YCor
    Commented Jun 4, 2021 at 8:36
  • $\begingroup$ The second question ($D$ not a subspace) is quite speculative. For instance, in an infinite-dimensional Hilbert space, $D$ might be a subset that is linearly independent over $\mathbf{R}$. $\endgroup$
    – YCor
    Commented Jun 4, 2021 at 8:40
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    $\begingroup$ As YCor remarks, the first statement is well-known and quite simple (one does not need completeness or separability of $X$). Instead of the extension by uniform continuity one can also use Hahn-Banach for the extension which is unique by density. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 4, 2021 at 8:56
  • $\begingroup$ @YCor Interesting point, in that case I guess I can reduce to the first and just consider the span in my setting. Let me ask, is there a good reference where I can find the first result? $\endgroup$
    – John_Algorithm
    Commented Jun 4, 2021 at 11:25
  • $\begingroup$ But Hahn-Banach is false for general (Hausdorff) topological vector spaces. And in the locally convex setting anyway it sounds curious to use a non-constructive theorem to get a unique extension. In any case, the use of Zorn's lemma in the proof has a step which is that the maximal subspace on which the extension is defined, has to be closed. This eventually makes use of this immediate Cauchy argument. $\endgroup$
    – YCor
    Commented Jun 4, 2021 at 12:35

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