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Let $X$ be a locally compact Hausdorff space. As far as I understand, the space $C_c(X) = C_c(X; \mathbb{C})$ of compactly supported continuous complex-valued functions on $X$ is (most?) often topologized as the limit of the spaces $C_K(X)$, where $$C_K(X) := \{ f\,\colon X \to \mathbb{C} \text{ continuous}, \, \mathrm{supp}(f) \subseteq K \}$$ and $K$ ranges over the compact subsets of $X$. However, it would also seem somewhat natural to consider $C_c(X)$ as a subspace of the space $C(X) = C(X; \mathbb{C})$ of all continuous complex-valued functions on $X$, which carries its own natural topology, namely the compact-open topology. My question is: are these two topologies secretly the same? (A reference would be great, but a sketch of an argument would of course also do.)

Also on this note, let $\mu$ be a nonzero Radon measure on $X$ (by which I mean a countably additive positive measure defined on the Borel-$\sigma$-algebra of $X$, finite on compact sets, inner regular on open sets and outer regular on Borel sets; the case I'm ultimately interested in is: $X = G$ is a locally compact group and $\mu$ is Haar measure). Then I'm given to understand that $C_c(X)$ is a dense subspace of $L^2(X) = L^2(X, \mu; \mathbb{C})$. Does the subspace topology coming from the inclusion $C_c(X) \hookrightarrow L^2(X)$ agree with either of the two topologies mentioned above? I cannot seem to find an answer, either by myself or with the help of a search engine, but this sounds like something which should be well-known (or maybe it's trivially true/false and I'm just not seeing it).

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    $\begingroup$ All the spaces you mention are complete in their natural topologies. But a proper subspace of a complete space can never be both dense and complete in the induced topology. $\endgroup$ May 27, 2021 at 20:06
  • $\begingroup$ @bathalf15320 thank you for your comment. As far as I can see, we know e.g. that $C([0, 1])$ is complete in the sup-norm and that $C_c([0, 1])$ is a proper dense subspace. (Also I'm aware that $L^2(X)$ is complete and that $C_c(X)$ is a proper subspace in general.) But I wasn't aware that $C_c(X)$ would generally/always be complete in the limit topology (would you happen to have a reference?) nor that the compact-open topology on $C(X)$ would generally/always come from a norm—isn't this (only) applicable to the space of bounded continuous functions? Sorry if I'm missing something obvious $\endgroup$
    – epitaph
    May 27, 2021 at 21:02
  • $\begingroup$ Sorry, I wrote nonsense in the first line (because $C_c([0, 1]) = C([0, 1])$) but I can't edit the comment anymore. I originally meant to write that $C_b(\mathbb{R})$ (bounded continuous functions) is complete in the sup-norm and that $C_c(\mathbb{R})$ is then a proper dense subspace. Still I'm missing why the inductive limit topology makes $C_c$ complete and whether $C(X)$ is complete (in the compact-open topology) $\endgroup$
    – epitaph
    May 27, 2021 at 21:26
  • $\begingroup$ Incidentally I just came across mathoverflow.net/questions/359019/… which seems to definitively show that the inductive limit topology and the compact-open topology on $C_c(X)$ differ even for $X = \mathbb{R}$. (The second question about agreement with the "$L^2$ topology" remains open.) $\endgroup$
    – epitaph
    May 27, 2021 at 21:30
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    $\begingroup$ @epitaph: The example which will tell you everything you need to know is $X=\mathbb{N}$ with the discrete topology. $\endgroup$ May 27, 2021 at 21:37

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The example $X = \mathbb N$ (with Radon measure $\mu =$ counting measure) suggested by Abdelmalek Abdesselam indeed shows that the three topologies on $C_c(X)$ are not the same in general. Here are the details I have worked out (hopefully correctly):$\newcommand{\abs}[1]{|#1|}$ $\newcommand{\norm}[1]{\|#1\|}$

  • the topology $\tau_2$ induced by the inclusion $C_c(\mathbb N) \hookrightarrow L^2(\mathbb N) = \ell^2(\mathbb N)$ is finer than the topology $\tau_{\rm co}$ coming from the inclusion $C_c(\mathbb N) \hookrightarrow C(\mathbb N)$. Indeed, a basic neighbourhood of $0$ in the latter topology is of the form $$V = V(n_1, \dotsc, n_r ; \varepsilon_1, \dotsc, \varepsilon_r) = \{f : \abs{f(n_i)} < \varepsilon_i, i = 1, \dotsc, r\}$$ where the $n_i$ are some natural numbers and the $\varepsilon_i$ are positive real numbers. If $\varepsilon > 0$ is smaller than the finitely many $\varepsilon_i$'s, then $V$ contains $\{f \in C_c(\mathbb N) : \norm{f}_2 < \varepsilon\}$ because each $f$ in the latter set satisfies $\abs{f(n)} < \varepsilon$ for all $n \in \mathbb N$.
    To see that the two topologies are not the same, note that on $V$ (for any choice of $n_i$ and $\varepsilon_i$), the $L^2$-norm is unbounded (because there is no control over all the infinitely many natural numbers outside of $\{n_1, \dotsc, n_r\}$), i.e., no $\tau_2$-open ball around $0$ fully contains a $\tau_{\rm co}$-neighbourhood of $0$.

  • the inductive limit topology $\tau_{\rm lim}$ on $C_c(\mathbb N)$ is finer than $\tau_2$. Indeed, a fundamental neighbourhood of $0$ in the former topology is of the form $$W = W((\varepsilon_n)_{n \in \mathbb N}) = \{f : \abs{f(n)} < \varepsilon_n ~ \forall n\}$$ for some sequence $(\varepsilon_n)$ of positive real numbers. So for instance, for any $\varepsilon > 0$, the fundamental $\tau_2$-neighbourhood $\{f \in C_c(\mathbb N) : \norm{f}_2 < \varepsilon\}$ of $0$ contains $W((\varepsilon_n))$ with $\varepsilon_n = \varepsilon a_n$ where $(a_n)$ is any square-summable sequence of (strictly) positive real numbers with $\sum a_n^2 \le 1$.
    To see that the two topologies are not the same, observe that, whenever $(\varepsilon_n)$ is a sequence s.t. $\varepsilon_n \to 0$ for $n \to \infty$, the $\tau_{\rm lim}$-neighbourhood $W((\varepsilon_n))$ cannot contain any $\tau_2$-open ball $B_\varepsilon(0)$: indeed, such a ball always contains all the functions $$n \mapsto \begin{cases} \varepsilon, & \text{if } n = m, \\ 0, & \text{else} \end{cases} \qquad (m \in \mathbb N),$$ but $W((\varepsilon_n))$ will only contain finitely many of these functions because $\varepsilon_n$ becomes smaller than $\varepsilon$ eventually.

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    $\begingroup$ Good job, but I think you are still missing the most important topology: the locally convex topology defined by the seminorms $f\mapsto\sum_n \varepsilon_n|f(n)|$ indexed by arbitrary sequences of nonnegative reals (or nonnegative integers to make the collection a bit smaller). In any case you will need uncountably many seminorms, since this topology, which is the most natural for $C_c(\mathbb{N})$, is nonmetrizable. $\endgroup$ Jul 15, 2021 at 18:35
  • $\begingroup$ @AbdelmalekAbdesselam thank you for your comment, I don't think I was actually aware of this topology. Does it have a name? Also could you explain or give a reference on where it pops up and what it can be used for? So far I would've considered the inductive limit topology to be the most natural/suitable since it is e.g. the one used by Bourbaki to define (complex) measures on a locally compact Hausdorff space. $\endgroup$
    – epitaph
    Jul 15, 2021 at 19:24
  • $\begingroup$ I retagged your question so hopefuly an expert on TVSs can explain to you the difference between inductive limits in the category of topological spaces and that of topological vector spaces. The above topology is the simplest example of LF space, it is also the finest locally convex topology. I don't know about good references, but would recommend the book by Osborne link.springer.com/book/10.1007/978-3-319-02045-7 $\endgroup$ Jul 15, 2021 at 20:13
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    $\begingroup$ Before all that, read and do the exercises mentioned in math.stackexchange.com/questions/3510982/… $\endgroup$ Jul 15, 2021 at 20:15
  • $\begingroup$ @AbdelmalekAbdesselam thank you for the links and for retagging the question. Actually I'm aware of the difference you mention, maybe I should've specified that I always meant the inductive limit topology in the category of locally convex TVSs. $\endgroup$
    – epitaph
    Jul 16, 2021 at 9:46
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A bit long for a comment so I will post it as an answer.

I did not think of $C_c(\mathbb{N})$ as $\mathscr{D}(M)$ for some zero-dimensional manifold $M$ (with countably many connected components), but now that you mention it, yes that is exactly it. I like to think of $C_c(\mathbb{N})$ as a "bare bones" toy model for $\mathscr{D}(\Omega)=C_{c}^{\infty}(\Omega)$. A key difficulty that beginniners in the subject have to overcome is to understand why convergent sequences of test functions must be supported in a common compact set. Once this hurdle is cleared, you will become more comfortable with these funny spaces and they will become your friends. The easiest way to understand the issue is to treat the elementary case $C_c(\mathbb{N})$ first.

A convergent sequence form a bounded set and this last property is why you must have a common compact support. Recall that $A\subset C_c(\mathbb{N})$ is bounded iff for all continuous seminorm $\rho:C_c(\mathbb{N})\rightarrow [0,\infty)$, $$ \sup_{f\in A}\rho(f)<\infty\ . $$ Arguing by contradiction, suppose there is no common compact (i.e., here finite) support for the elements of $A$. Then one can construct a sequences $f_k\in A$ and $n_1<n_2<\cdots$ in $\mathbb{N}$, such that $f_k(n_k)\neq 0$ for all $k$. Recall that the natural topology of $C_c(\mathbb{N})$ is defined by the seminorms $$ \|f\|_{\varepsilon}=\sum_{n\in \mathbb{N}}\varepsilon_n|f(n)| $$ indexed by all sequences $\varepsilon\in [0,\infty)^{\mathbb{N}}$. Now let us construct such a sequence by letting $\varepsilon_n=k\times |f_k(n_k)|^{-1}$ if $n=n_k$ and otherwise let $\varepsilon_n=0$. Clearly $\|f_k\|_{\varepsilon}\ge k$, and so $\sup_{f\in A}\|f\|_{\varepsilon}=\infty$.

Now for $\mathscr{D}(\Omega)$, take a sequence of points $x_n$ that escape to the boundary of $\Omega$. Build some small disjoint balls around these points, and construct some bum functions $\phi_n$ supported inside these balls. The map $f\mapsto \sum_{n}f(n)\phi_n$ gives you a way to embed $C_c(\mathbb{N})$ inside $\mathscr{D}(\Omega)$ and also transfer counterexamples from one setting to the other.

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