Edit:
Let $X$ be a strict LB-space described by $\lim X_n$ and suppose that $\{x_n\}_{n \in \mathbb{N}}$ converges in $X$. I'm looking for a reference showing that $x_n$ must converge in some $X_N$.
$\begingroup$
$\endgroup$
4
-
$\begingroup$ The lc case is taken care of by the fact that this is a strict $LF$ inductive limit in the sense of Dieudonné and Schwartz. The top case follows immediately. $\endgroup$– user131781Commented Apr 20, 2020 at 10:19
-
$\begingroup$ But what do convergence sequences look like in the $LF$ space setting? I'm only familiar with the universal property of the construction but I've never seen a discussion on convergence in that topology. $\endgroup$– ABIMCommented Apr 20, 2020 at 10:23
-
$\begingroup$ It is a great difference if you ask for convergence of sequences or convergence of nets in locally convex inductive limits: If all $X_n$ are closed (I think user131781 just forgot this assumption) then a sequence in $Y$ converges if and only if the sequence is contained in some $X_n$ and converges there. For nets nothing like this is true. $\endgroup$– Jochen WengenrothCommented Apr 20, 2020 at 14:03
-
$\begingroup$ @JochenWengenroth I updated the question to reflect a reference to precisely this point. $\endgroup$– ABIMCommented May 12, 2020 at 21:56
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
The result (even for LF-spaces) is due to J. Dieudonné and L. Schwartz La dualité dans les espaces (F) et (LF), Annales de l’institut Fourier, tome 1 (1949), p. 61-101, propositions 2 and 4.
(Proposition 2 says that the inductive limit topology induces on the ,,steps'' their original topologies, and proposition 4 says that bounded subsets of the inductive Limit are contained in steps.)
answered May 13, 2020 at 11:29
-
$\begingroup$ I gave this some thought and I wonder, this means that eventually the sequence is stuck on one of the steps and not that it must always lie on the same step. No? $\endgroup$– ABIMCommented May 28, 2020 at 21:17
-
1$\begingroup$ If you mean by eventually stuck on one of the steps that there is a step which contains all but finitely many terms of the sequence then there is another step containing all terms of the sequence. $\endgroup$ Commented May 29, 2020 at 6:10
-
$\begingroup$ But why do we need Prop. 4? Isn't the fact that $f \in X_n$, the fact that $X_n$ is closed, and Prop. 2 enough? $\endgroup$– ABIMCommented May 29, 2020 at 19:40