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Let $H$ be a Hilbert space, and let $X$ be a topological vector space.

  • Under what conditions on the topologies of $X$ and $H$ does there exist an injective, continuous linear map $f : H \to X$?

Supposing that such an injective embedding $f : H \to X$ exists, consider the topological closure $F := \overline{fH}$ in $X$.

  • When does the closure $F$ admit the structure of a Fréchet space, with $f : H \to F$ an isometric embedding?
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    $\begingroup$ The obvious idea to embed the separable Hilbert space into a sequentially complete locally convex space is to find a sequence $x_n$ which converges fast to $0$ and to define $f:\ell_2 \to X$ by $f((a_n)_{n\in\mathbb N})= \sum\limits_{n=1}^\infty a_n x_n$. This map will be injective if the sequence is topologically linearly m-independent$. This is discussed on page 37 in the book *Barrelled locally convex spaces of Bonet and Perez-Carreras. The second question is not clear to me: If there is no norm on the Frechet space $F$, what do you mean by isometric? On the other hand, if $F$ is ... $\endgroup$
    – Jochen Wengenroth
    Commented Mar 15, 2013 at 14:45
  • $\begingroup$ ...normed and $f$ is an isometry then $f(H)$ is complete and hence closed in $X$ (if you assume that $X$ is Hausdorff). $\endgroup$
    – Jochen Wengenroth
    Commented Mar 15, 2013 at 14:46
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    $\begingroup$ Since you can always embed $H$ isometrically into $H \times Y$ for any topological vector space $Y$, there would seem to be plenty of scope. $\endgroup$
    – jbc
    Commented Mar 15, 2013 at 16:18
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    $\begingroup$ Are the inclusions of Levi-Sobolev spaces $H^s(T^n)\rightarrow H^t(T^n)$ on products $T^n$ of circles, for $s>t$, the kind of example you want? If so, it illustrates that a Hilbert-space structure can get weaker... $\endgroup$
    – paul garrett
    Commented Mar 15, 2013 at 19:52
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    $\begingroup$ @Tom: Even if there is any metric on $F$ inducing the vector space topology such that $f$ is an isometry then $f(H)$ will be already a completely metrizable topological vector space. This not obvious (since the uniformities may be a priori different) but true because of a theorem of Victor Klee (solving a problem of Banach). Therefore, $F=f(H)$ still holds. $\endgroup$
    – Jochen Wengenroth
    Commented Mar 16, 2013 at 10:52

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