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Recall Lemma 0.5.1 from the Helemskii's monograph "The homology of Banach and Topological Algebras":

$\textbf{Lemma}$ Let $\phi\colon X\to Y$ be an injective map between Banach spaces with dense range. If the dual map $\phi^*$ is surjective then so is $\phi$ itself.

My question is: what if $X,Y$ are complete DF-spaces? Is this lemma still true? The situation I am considering is even more specific. Namely, $X$ is a quotient of a complete DF-space (by a closed subspace) and $Y$ is a closed subspace of a complete DF-space.

The only thing I can see is that $\phi^*$ is a topological isomorphism $\textit{onto}$.

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  • $\begingroup$ I don't see anything ''more specific'': $X$ is a quotient of itself and $Y$ is a closed subspace of itself. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 30, 2017 at 6:20
  • $\begingroup$ The closed range theorem implies that the bitransposed map $\phi^{\ast\ast}$ has closed range, if you knew that $X$ and $Y$ were reflexive this would be enough. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 30, 2017 at 6:24
  • $\begingroup$ The closed range theorem is perhaps overkill: $\phi^*$ is injective since $\phi$ has dense range and if you assume that it is surjective then it is an isomorphism for the strong duals. This means that for every (closed absolutely convex) bounded set $B\subseteq Y$ there is $C\subseteq X$ with the same properties such that $C^\circ\subseteq \phi^*(B^\circ)$ which yields $\phi(C)^\circ = (\phi^*)^{-1}(C^\circ) \subseteq B^\circ$ and hence $B\subseteq \phi(C)^{\circ\circ}=\overline{\phi(C)}$. I don't think that, in general, this implies surjectivity of $\phi$ but I don't have an example. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 30, 2017 at 6:59
  • $\begingroup$ If you don't require completeness, a counterexample would be any non-complete DF space $X$ embedded into its completion $Y$. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 30, 2017 at 12:47
  • $\begingroup$ In fact I am assuming completeness. As for reflexivity, I would like to omit this assumption, if possible. I guess barrelledness can help but this is again something I would like to assume only if there is a non-barrelled counterexample. $\endgroup$
    – Krzysztof
    Commented Jul 3, 2017 at 12:35

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Here is a counterexample: For every countable inductive limit $E=\lim E_n$ of Banach (or locally convex) spaces one has (either by definition or a simple construction) an algebraically exact sequence $0\to \bigoplus E_n \to \bigoplus E_n \to E\to 0$ where the map $\phi$ on the direct sum is defined by $(e_n)_{n\in\mathbb N} \mapsto (e_n-j_{n-1}(e_{n-1}))_{n\in\mathbb N}$ (where $j_n:E_n\to E_{n+1}$ are the linking maps of the inductive spectrum and $j_0(e_0)=0$).

There are inductive limits $E$ of Banach spaces such that the limit topology is trivial (only the empty set and the whole space are open). In this case $X=Y=\bigoplus E_n$ are (very nice, in particular, barrelled) DF-spaces and $\phi$ is continuous with dense range (because the associated Hausdorff space of $E$ is $\lbrace 0\rbrace =\bigoplus E_n/\overline{\text{range}(\phi)}$. Then $\phi^*$ is an isomorphism but $\phi$ is not surjective.

As mentioned in the comments, the answer to your question is positive if $X$ is a reflexive DF-space or if $\phi$ is the transposed of an operator between Frechet spaces (in that case the closed range theorem or the arguments in my third comment apply).

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  • $\begingroup$ That's a very nice construction. I did not mention that explicitly but I consider only Hausdorff spaces. Can you see a counterexample with $E$ Hausdorff? $\endgroup$
    – Krzysztof
    Commented Jul 11, 2017 at 12:22
  • $\begingroup$ I don't understand your comment. $X=Y$ is clearly Hausdorff. The quotient $Y/\phi(X)$ is Hausdorff if and only if $\phi(X)$ is closed. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 11, 2017 at 14:27
  • $\begingroup$ I'm sorry. I have just noticed that $E$ is (obviously) different from $\oplus E_n$. Now everything is clear for me. $\endgroup$
    – Krzysztof
    Commented Jul 13, 2017 at 8:13

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