I have a stupid question about a topology on $C_c(X)$. Here $X$ is locally compact Hausdorff. Can assume $\sigma$-compact if it helps.

Definition (topology on $C_c(X)$): For each compact $K \subset X$, $C_K(X)$ is the set of functions in $C_c(X)$ with support in $K$. $C_K(X)$ is given a Banach space structure with the sup norm. Call this topology $\tau_K$.

Let $\beta$ be the set of $V \subset C_c(X)$ which are convex, balanced, and which have $V\cap C_K(X) \in \tau_K$ for each compact $K$.

Define $\tau$ to be the collection of sets in $C_c(X)$ given by $\bigcup (\phi_i + W_i)$ where $W_i \in \beta$. This gives a topology on $C_c(X)$ making it a locally convex topological vector space with $\beta$ as a local basis. Moreover, the topology $\tau_K$ coincides with the subspace topology $\tau|_{C_K(X)}$.

For all the above facts see Rudin's functional analysis $\S 6.3$ onwards.

Question: Take the product topology $\tau \times \tau$ on $C_c(X) \times C_c(X)$. Say $V\subset C_c(X)\times C_c(X)$ is convex and balanced. Suppose for every pair of compact sets $K, F \subset X$, we have that $V \cap C_K(X)\times C_F(X) \in \tau_K \times \tau_F$. Is $V$ open in the product topology $\tau\times \tau$?

I came across this question while studying pseudomeasures and induced representations from Folland's Harmonic analysis.

Attempt: Say $(\phi_1, \phi_2) \in V$. It suffices to find $W_1, W_2 \in \beta$ such that $(\phi_1,\phi_2) + W_1\times W_2 \subset V$...Pretty lost at this point.

Also posted here.


2 Answers 2


The topology $\tau$ you describe makes $(C_c(X),\tau)$ the colimit (or inductive or direct limit) in the category LCS of locally convex spaces of the system $C_K(X)$ with inclusions $i_{K,L}:C_K(X) \hookrightarrow C_L(X)$ for compact subsets $K\subseteq L$ of $X$, i.e., the inclusions $i_K:C_K(X)\to C_c(X)$ have the universal property that, for every family of continuous linear maps (the morphisms of the category) $f_K:C_K(X)\to Y$ with $f_L\circ i_{K,L}=f_K$, there is a unique morphism $f:C_c(X)\to Y$ with $f_K=f\circ i_K$. A big advantage of the categorical viewpoint is that you easily get permanence properties like colimits commute with colimits.

Since the category LCS is additive (you can add two morphisms $(f+g)(x)=f(x)+g(x)$ and this distributes over the composition -- sometimes the term pre-additive category is used) general abstract nonsense tells you that the product is also a coproduct and hence a colimit which commutes with colimits.

Therefore and without any calculations, the product topology $\tau\times \tau$ is the colimit topology of the system $C_K(X)\times C_L(X)$.

EDIT. I hope that the following helps. Whenever you have colimits $E=$colim$_I E_i$ and $F=$colim$_J F_j$ in a category having codropucts (denoted by $\oplus$) it is always true that $E\oplus F =$colim$_{I\times J} E_i\oplus F_j$. Since LCS is additive, coproducts and products are "the same" so that you can replace $\oplus$ by $\times$. (You see that I am quite reluctant to replace the abstract argument by concrete calculations.)

  • $\begingroup$ Thank you very much for this interesting answer! Please allow me some time to read. $\endgroup$
    – Latimer
    Dec 7, 2021 at 9:56
  • $\begingroup$ Sir, I might be misunderstanding, but when I try to show that $(C_c(X) \times C_c(X), \tau\times\tau)$ is the inductive limit of the system $\{C_K(X)\times C_L(X)\}$, I run into exactly the same problem - I want to show the universal property and it's easy enough to define the map. But when I try to show continuity, I end up with a set $V \subset C_c(X)\times C_c(X)$ whose intersection with every $C_K(X) \times C_L(X)$ is open...how do I then show this $V$ is in $\tau \times \tau$? $\endgroup$
    – Latimer
    Dec 7, 2021 at 14:59
  • $\begingroup$ It might be helpful to spell out the exact meaning of my claim that the product is the coproduct. If you want to avoid the categorical viewpoint, note first that it is enough to take $V$ an absolutely convex neighbourhood of $0$ (those determine the topology in every locally convex space) which thus contains the union of neighbourhoods $U_K\times U_K$ with $0$-neighbourhoods $U_K$ in $C_K(X)$. Then "calculate" the absolutely convex hull of this union. $\endgroup$ Dec 7, 2021 at 16:02
  • $\begingroup$ 'my claim that the product is the coproduct' - sorry can you please elaborate? My interpretation is that you want me to think about two objects in the category; the first being $(C_c(X) \times C_c(X), \tau\times\tau)$ and the second being $(C_c(X)\times C_c(X), \omega)$ where $\omega$ is the topology defined by declaring all convex balanced $V$ that intersect every $C_K(X)\times C_L(X)$ openly as a local base. You then want me to show that each of these is the direct limit of the product system $\{C_K(X)\times C_L(X)\}$ thereby answering my question. $\endgroup$
    – Latimer
    Dec 8, 2021 at 5:33
  • $\begingroup$ I quite like the categorical viewpoint but it seems that I'm left with the same question as above when I try to show that $(C_c(X)\times C_c(X),\tau\times\tau)$ is the limit of the system. $\endgroup$
    – Latimer
    Dec 8, 2021 at 5:35

Ok, I think I understand Jochen's wonderful answer. To see if it is so, I try to write some details of what he says.

Category theory (boo!): Define a local base $\alpha$ on $C_c(X) \times C_c(X)$ by the property $V\in \alpha$ if and only if $V$ is convex, balanced, and $V$ intersects every $C_K(X)\times C_L(X)$ openly. One can show that this indeed gives a topology $\omega$ on $C_c(X)\times C_c(X)$ making it a TVS (same argument as in Rudin).

Using the family of continuous maps $C_K(X) \to (C_c(X)\times C_c(X),\omega); f\mapsto(f,0)$, we get a continuous map $\iota_1:(C_c(X),\tau) \to (C_c(X)\times C_c(X), \omega)$. Similarly, we also get a continuous map $\iota_2(g):= (0,g)$.

This gives a continuous map (the identity) $$\iota:=(i_1,i_2):(C_c(X)\times C_c(X), \tau \times \tau) \to (C_c(X)\times C_c(X), \omega).$$ Thus we're done?

Element wise (yay!): Let $\pi_1$ and $\pi_2$ denote the projections from $C_c(X)\times C_c(X)$ to each of the factors. If $V$ is as in the question, we see that $\pi_1\left(\frac{1}{2}V \cap \left(C_K(X)\times \{0\}\right)\right) \subset C_K(X)$ is open. This shows that $V_1:=\pi_1\left(\frac{1}{2}V \cap \left(C_c(X)\times \{0\}\right)\right)$ is open. That is, it belongs to $\tau$. Similarly $V_2:=\pi_2\left(\frac{1}{2}V \cap \left(C_c(X)\times \{0\}\right)\right)$ is in $\tau$. We have that $V_1\times V_2 = V_1\times\{0\}+ \{0\}\times V_2 \subset V$. So we're done?

Not sure if I'm misunderstanding again...


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