It's a duplicate of this question, since I really want to get an explanation.

Let $\left(V_{n},\phi_{n,n+1}\right)_{n\in\mathbb{N}}$ be an inductive sequence of LCTV spaces. A locally convex inductive limit of $\left(V_{n},\phi_{n,n+1}\right)_{n\in\mathbb{N}}$ is it's algebraic inductive limit $V=\lim_{\to}V_{n}$, equipped with final locally convex topology w.r.t. $\phi_{n}:V_{n}\to V$.

$V$ can also be described as a quotient of the locally convex direct sum $\bigoplus V_{n}$ by the subspace $D$ , generated by vectors $\left\{ \left(\dots,v_{n},-\phi_{n,n+1}\left(v_{n}\right),\dots\right)|v_{n}\in V_{n}\right\} $. We denote by $\pi:\bigoplus V_{n}\to V\simeq\bigoplus V_{n}/D$ the quotient map. The universal property of V follows from the universal properties of $\bigoplus V_{n}$ and of the quotient.

A map $f:V\to U$ in LCTV is continuous iff maps $f\circ\phi_{n}$ are continuous. Taking $f=\mathrm{id}_V$ one get that if subspace $F\subset V$ is closed then $F_n=\phi_{n}^{-1}\left(F\right)$ is closed. Clearly $F\cong\lim_\to F_n$ as linear space.

The question I am trying to understand is the following: if $F$ is a closed subspace, why it does not have to be the locally convex inductive limit of $F_n$ (why topologies can be different)?

We clearly have $\pi^{-1}(F)=\bigoplus F_n$, which is a closed subspace of $\bigoplus V_n$. Doesn't this force the quotient topology on $\pi^{-1}(F)/\pi^{-1}(F)\cap D$ (which is the locally convex inductive limit topology) be the subspace topology of $F$?

Would it matter if $V_n$ are Banach spaces?

  • $\begingroup$ "...if subspace $F\subset V$ is open (closed) then..." -- this must be a mistake? An open subspace $F$ in a locally convex space $V$? $\endgroup$ Sep 6, 2013 at 7:39
  • $\begingroup$ @SergeiAkbarov: I edited to avoid confusion. $\endgroup$
    – user35953
    Sep 6, 2013 at 8:34
  • 1
    $\begingroup$ In this science if you can't prove something immediately, this means that there must be a counterexample... Your words - "why topologies can be different" - do they mean that you saw this counterexample before? $\endgroup$ Sep 6, 2013 at 9:48
  • $\begingroup$ @SergeiAkbarov: there is an exercise in Bourbaki's TVS book, but it is pretty involved and in more general situation than LB(limits of Banach)-spaces. $\endgroup$
    – user35953
    Sep 6, 2013 at 11:56

1 Answer 1


Here is my favorite example which is very natural and simple: Let $\mathscr E (\Omega)$ be the Frechet space of smooth functions on an open set in $\Omega\subseteq\mathbb R^d$ so that it dual $\mathscr E'(\Omega)$ is the space of compactly supported distributions. This is a nice inductive limit of Banach spaces. Hence $V= \mathscr E(\Omega)\times \mathscr E'(\Omega)$ is an inductive limit of Frechet spaces. Now, the "diagonal" $F$ is closed and algebraically isomorphic to the space of test functions $\mathscr D(\Omega)$ but the subspace topology of $V$ on $F$ is much coarser than the topology of $\mathscr D(\Omega)$: the dual of the former is the space of distributions with compact singular support whereas the dual of the latter is the space of all distributions.


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