5
$\begingroup$

I would like to work with a slightly loser definition of an LF-space but am unsure what niceties I'm throwing away in the process. Let me provide a comparison of the conventional definition and my own slightly broader one.


Generalized LF-Topology: Let $E$ be a vector space over $K = \mathbb{R}$ or $\mathbb{C}$. Let $\{E_i : i \in I\}$ be a countable family of $K$-linear subspaces of $E$ such that

  1. The relation $i \leq j \iff E_i \subseteq E_j$ defines a partial order on $I$.
  2. We have $$E = \varinjlim_{i \in I}E_i = \bigoplus_{i \in I} E_i \bigg/ \langle g_i - g_j \circ g_{j,i} \mid i \leq j\rangle,$$ where for each $i \in I$ we define the canonical map $g_i: E_i \to \oplus_{j \in I}E_j$ and for each $j \geq i$ we define the inclusion map $g_{j,i}: E_i \to E_j$. (Here the topology on $\oplus_{i \in I}E_i$ is defined to be the finest locally convex topology such that $g_i$ is continuous for all $i \in I$.
  3. For each $i \in I$ we have fixed a topology $\tau_i$ on $E_i$ with respect to which $E_i$ is a Fréchet space.
  4. The directed family of LCSs $\{(E_i,\tau_i) : i \in I\}$ is strict, meaning that if $i \leq j$ the topology on $E_i$ induced by $\tau_j$ is exactly $\tau_i$. (If we only required that the imbedding $E_i \to E_j$ was continuous, then the induced topology could be coarser than $\tau_i$.

After potentially adding more restrictions to $(I,\leq)$ and the $E_i$, the LF-topology on $E$ may be defined as follows: A convex subset $U$ of $E$ is an open neighborhood of $0$ in $E$ if and only if $U \cap E_i$ is open in $E_i$ for all $i \in I$.

Conventional Definition: My definition above, but where it is assumed that $I = \mathbb N$ under the usual total order.


My question is: if I permit partially ordered countable families of Fréchet spaces, but still require the inductive limit to be strict (as defined above), do I wind up with the same class of LF-spaces, or is this definition actually more general? Are there any modern references (say, post early 1970's) that have studied broader definitions of LF-spaces than those that were investigated classically by Dieudonné-Schwartz and Köthe?

Thanks in advance for any answers or references!

Edit: Thinking about this a bit more, it seems at least convenient (and probably necessary) to assume further that the poset $I$ is well-founded.

$\endgroup$
  • $\begingroup$ If the family is directed in the sense that for all $i,j$ there is a $k$ with $E_i \cup E_j \subseteq E_k$, then you get nothing new. If there are $i,j$ such that $(E_i \cup E_j)\setminus E_k \neq \varnothing$ for all $k$, I'm not sure whether $\bigcup E_k$ can be a vector space, maybe you then need more than countably many spaces to get a vector space as the union. $\endgroup$ – Daniel Fischer Jun 13 '15 at 21:27
  • $\begingroup$ @DanielFischer: Whoops! I was being too sloppy with my assumptions in the definition. I was thinking about the more general case you mention, but the definition of the limit as a union doesn't make sense here. Thanks for the reply! I'll update my question. $\endgroup$ – Dan Jun 13 '15 at 21:29
3
$\begingroup$

My question is: if I permit partially ordered countable families of Fréchet spaces, but still require the inductive limit to be strict (as defined above), do I wind up with the same class of LF-spaces, or is this definition actually more general?

Fact: The limit $E$ of a countable inductive system $(E_i)_{i \in I}$ of Fréchet spaces is already an (LF)-space in the conventional sense, i.e. $E$ is the inductive limit of a sequence of Fréchet spaces $(E_n)_{n \in \mathbb{N}}$.

Thus, you do not get anything new when considering general countable directed index sets $I$. You do not need the system to be strict. For a reference see Meise, Vogt: "Introduction to Functional Analysis" (2004), p. 291 (Remark). The construction of the sequence can be found in Lemma 24.34: Forget about the partial order on $I$ (induced by $\subseteq$ on $E$) and identify $I = \mathbb{N}$ (hereby inducing a linear order on $I$). Then the sequence $(E_n)_{n \in \mathbb{N}}$ has an increasing subsequence $(E_{n_k})_{k \in \mathbb{N}}$ (with $E_\nu \subseteq E_{n_{k+1}}$ for all $1 \leq \nu \leq n_k$ and all $k$) which produces the same inductive limit topology on $E$:

$$\displaystyle \lim_{\stackrel{\longrightarrow}{i \in I}} E_i = \lim_{\stackrel{\longrightarrow}{k \in \mathbb{N}}} E_{n_k}.$$

$\endgroup$
  • $\begingroup$ Thank you for the reference, and for answering my old inquiry! I actually came up with more-or-less the same justification in my own work after posing the question, and it's good to see that it appears somewhere in the literature! I really appreciate the response and the suggestion of Meise and Vogt. I'd originally learned from Trèves's and Schaefer's texts, but this looks like a great alternative/reference. $\endgroup$ – Dan May 6 '16 at 18:59

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.