It seems to me I saw a counterexample somewhere, but I can't find it, can anybody help me?

Let $\varphi:X\to Y$ be a linear continuous map of locally convex spaces, and $\widetilde{\varphi}:\widetilde{X}\to\widetilde{Y}$ the corresponding linear continuous map of their completions.

If $\varphi:X\to Y$ is injective, is $\widetilde{\varphi}:\widetilde{X}\to\widetilde{Y}$ injective as well?

I think, the answer must be "no", so another question is

Under which conditions the injectivity of $\varphi:X\to Y$ implies the injectivity of $\widetilde{\varphi}:\widetilde{X}\to\widetilde{Y}$?

  • 1
    $\begingroup$ I don't have any references at hand, by I believe that there is a criterion of W. Robertson ensuring the injectivity of the extension to the compltions: $X$ should have a $0$-neighbourhood basis of absolutely convex sets $U$ such that $\varphi(U)$ is closed in $Y$. Probably, this can be found in Köthe's book on topological vector spaces. $\endgroup$ – Jochen Apr 16 '16 at 12:57
  • $\begingroup$ @Jochen you know, I don't see this criterion in Köthe's book. $\endgroup$ – Sergei Akbarov Apr 29 '20 at 15:25
  • 1
    $\begingroup$ This is §18(4) on page 210 of Köthe's Topological Vector Spaces. $\endgroup$ – Jochen Apr 29 '20 at 16:24
  • $\begingroup$ @Jochen, once again I see that you know everything. :) I feel like a peasant from a Siberian village near you. :) Thank you! $\endgroup$ – Sergei Akbarov Apr 29 '20 at 16:51
  • 1
    $\begingroup$ You are welcome, Sergei. I have a searchable copy of Köthe's book on my computer. That helps a lot since in Köthe's books there is everything -- but very hard to find. $\endgroup$ – Jochen Apr 29 '20 at 18:24

Here is a way of getting lots counterexamples. Let $\widetilde{X}$ be your favorite infinite-dimensional complete space, let $X\subset \widetilde{X}$ be a dense subspace, let $v\in\widetilde{X}\setminus X$, and let $L$ be the span of $v$. Then we can take $Y=\widetilde{X}/L$, and the quotient map $\widetilde{\varphi}:\widetilde{X}\to Y$ is not injective but its restriction to $X$ is.

I don't know of any useful conditions under which the answer is yes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.