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I asked the question Why is multiplication on the space of smooth functions with compact support continuous? on M.SE sometime ago but I didn't receive a satisfactory answer.

I was reading this post of Terence Tao and I'm not able to prove the last item of exercise 4.

I have a map $F:C_c^{\infty}(\mathbb R^d)\times C_c^{\infty}(\mathbb R^d)\to C_c^{\infty}(\mathbb R^d)$ given by $F(f,g) = fg$.

The question is: Why is $F$ continuous?

I proved that if a sequence $(f_n,g_n)$ converges to $(f,g)$ then $F(f_n,g_n) \to F(f,g)$, that is, $F$ is sequentially continuous. But, as far as i know, this does not implies that $F$ is continuous because $C_c^\infty (\mathbb R^d)$ is not first countable.

The topology of $C_c^{\infty}(\mathbb R^d)$ is given by seminorms $p:C_c^{\infty}(\mathbb R^d) \to \mathbb R_{\geq 0}$ such that $p\big|_{C_c^{\infty}( K)}:{C_c^{\infty}( K)} \to \mathbb R_{\geq 0}$ is continuous for every $K\subset \mathbb R^d$ compact; the topology of ${C_c^{\infty}( K)}$ is given by the seminorms $ f\mapsto \sup_{x\in K} |\partial^{\alpha} f(x)|$, $\alpha \in \mathbb N^d,$ and $C_c^{\infty}( K)$ is a Fréchet space.

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    $\begingroup$ Did you try generalizing your argument from sequences to nets? If you can do that, you're done. Often arguments generalize pretty easily from sequences to nets (at least when the results they lead to are true). $\endgroup$
    – John Baez
    Commented Mar 19, 2016 at 8:17
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    $\begingroup$ Have you tried simply showing that the $F$-pre-image of any open set open? $\endgroup$ Commented Mar 19, 2016 at 12:44
  • $\begingroup$ @John Baez: To make the sequential argument i needed to prove the following result: A sequence $f_n$ converges to $0$ on $C_c^\infty(\mathbb R^d)$ if, and only if, there is a $K\subset \mathbb R^d$ compact such that the support of each $f_n$ is contained in $K$ and $f_n \to 0$ in $C_c^\infty( K)$. (I'll write the proof bellow). $\endgroup$
    – Hugo
    Commented Mar 19, 2016 at 16:11
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    $\begingroup$ You are correct that arbitrary nets need not be contained in any limitand of this colimit, but bounded Cauchy nets are. (This is part of a proof that such strict colimits are quasi-complete, although not complete in general.) But we can also give a straightforward non-sequence/net argument on the Frechet spaces $C^\infty_c(K)$, as @Bazin writes below (and there your sequence-oriented argument is sufficient, although not really necessary). $\endgroup$ Commented Sep 17, 2017 at 19:14
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    $\begingroup$ My problem with the @Bazin argument is that showing that $C^\infty_c(\mathbb R^d )\times C^\infty_c(\mathbb R^d )$ is limit of the diagram given by $C^\infty_c(K)\times C^\infty_c(L)$, with $K,L$ compact, does not proof that $F$ is continuous because the limit universal property is about linear maps and not bilinear ones (as far as i know). $\endgroup$
    – Hugo
    Commented Sep 18, 2017 at 0:38

3 Answers 3

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You can spare yourself the functional analytic abstract nonsense by using an explicit set of seminorms on $\mathcal{D}(\mathbb{R}^d)=C_{c}^{\infty}(\mathbb{R}^d)$ which, unfortunately, are not well-known but can be found in the excellent book "Topological Vector Spaces and Distributions" by Horváth on p.171.

Let $\mathbb{N}=\{0,1,\ldots\}$, and denote the set of multiindices by $\mathbb{N}^d$. A locally finite family $\theta=(\theta_{\alpha})_{\alpha\in\mathbb{N}^d}$ of continuous functions $\mathbb{R}^d\rightarrow \mathbb{R}$ is one such that for all $x\in\mathbb{R}^d$ there is a neighborhood $V$ such that $V\cap {\rm Supp}\ \theta_{\alpha}=\varnothing$ for all but finitely many $\alpha$'s. Let $$ \|f\|_{\theta}=\sup_{\alpha\in\mathbb{N}^d}\sup_{x\in\mathbb{R}^d} |\theta_{\alpha}(x)D^{\alpha}f(x)|\ , $$ then the seminorms $\|\cdot\|_{\theta}$ where $\theta$ runs over all such locally finite families define the topology of $\mathcal{D}(\mathbb{R}^d)$.

Continuity of the pointwise product follows once you show that for every $\theta$, there exists $\theta'$ and $\theta''$ such that $$ \|fg\|_{\theta}\le \|f\|_{\theta'}\|g\|_{\theta''} $$ for all test functions $f$ and $g$, which one can do by hand.

For instance, you can use the Leibniz or product rule $$ D^{\alpha}(fg)=\sum_{\beta+\gamma=\alpha}\frac{\alpha!}{\beta!\gamma!} D^{\beta}f D^{\gamma}g\ , $$ and the brutal $l^1$-$l^{\infty}$ estimate $$ |D^{\alpha}(fg)|\le \prod_{i=1}^{d}(\alpha_i+1) \times\max_{\beta+\gamma=\alpha} \frac{\alpha!}{\beta!\gamma!} |D^{\beta}f| |D^{\gamma}g|\ , $$ in order to see that $\theta'=\theta''$ works if it is defined by $$ \theta'_{\beta}(x):=\frac{1}{\beta!} \sup_{\alpha\ge \beta} \sqrt{\prod_{i=1}^{d}(\alpha_i+1)!}\times\sqrt{|\theta_{\alpha}(x)|}\ . $$


Brief Feb 2020 addendum:

@Martin Sleziak: Thank you for the edit. I didn't know one could link to a specific page as you did for the reference to Horváth. That's great!

Request for references: I attribute these explicit seminorms to Horváth because I only saw them in the book I mentioned. If you are aware of an earlier reference where these seminorms appeared, please let me know.

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    $\begingroup$ About the seminorms $\|\cdot\|_\theta$: Hörmander (The Analysis of Lin. Par. Diff. Op. I, notes of Chap. 2) writes "the convenient explicit form of the seminorms (..) has been adopted from Gårding and Lions", and quotes: L. Gårding et J.L. Lions: Functional Analysis. Nuovo Cimento N. 1 del Suppl. al Vol. 10(14) 1959 (9-66). $\endgroup$ Commented Dec 12, 2022 at 20:00
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    $\begingroup$ @PietroMajer: Thank you for the reference. Indeed, I see the seminorms being described on p. 21 of link.springer.com/article/10.1007/BF02724838 $\endgroup$ Commented Dec 12, 2022 at 21:53
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The spaces $C^\infty_c(\mathbb R^d)$ and $C^\infty_c(\mathbb R^d)\times C^\infty_c(\mathbb R^d)$ are $LF$ spaces (inductive limit of Frechet spaces) and their standard topologies are not metrizable. We need only to check that for $K,L$ given compact subsets of $\mathbb R^d$, the restriction of $F$ to the Frechet space $C^\infty_K(\mathbb R^d)\times C^\infty_L(\mathbb R^d)$ is continuous (here $C^\infty_K(\mathbb R^d)$ stands for the $C^\infty_c(\mathbb R^d)$ functions with support included in $K$). Calling $F_{K,L}$ this restriction (valued in $C^\infty_{K\cap L}(\mathbb R^d)$), it is enough to check its sequential continuity, which is what you have already done.

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  • $\begingroup$ I agree that $C_c^\infty(\mathbb R^d)$ is the inductive limit of $C_c^\infty(K)$ with $K\subset \mathbb R^d$ compact. I also know that any linear map $\phi:C_c^\infty(\mathbb R^d) \to \mathfrak{X} $, where $\mathfrak{X}$ is a locally convex vector space, is continuous if, and only if, for every $K$ compact the map $\phi\big|_{C_c^\infty(K)}$ is continuous. From the characterization of sequence i gave above, i obtain that a linear map $\phi$ is continuous if and only if it's sequentially continuous. $\endgroup$
    – Hugo
    Commented Mar 19, 2016 at 18:38
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    $\begingroup$ I believe that your argument would show an analogous result for a linear map $C_c^\infty(\mathbb R^d)\times C_c^\infty(\mathbb R^d) \to \mathfrak{X} $. The problem is that $F$ is bilinear map. If i show that a bilinear map is continuous $C_c(\mathbb R^d)\times C_c(\mathbb R^d) \to \mathfrak{X}$ if, and only if, is continuous to restrictions to $C_c(K)\times C_c(L)$ then the problem is solved. $\endgroup$
    – Hugo
    Commented Mar 19, 2016 at 18:40
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    $\begingroup$ @Hugocito This is hopefully doable passing through the (projective) tensor product? I.e. proving $C^\infty_c(\mathbb{R}^d)\otimes C^\infty_c(\mathbb{R}^d)=C^\infty_c(\mathbb{R}^{2d})$ even topologically - which sounds true, as it is a matter of a commutation of two colimits (but needs a careful proof) $\endgroup$ Commented Dec 9, 2022 at 8:37
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The problem is solved here: https://math.stackexchange.com/a/1710723/41494

The solution given by the user Vobo is the following:

Let $B_n$ be the ball with radius $n$, $K_n=C_c^\infty(B_n)$ with its metrizable topology, $\varphi_n\in K_n$ a function with support contained in $B_{n}$ and $\varphi_n(x)=1$ for $x\in B_{n-1}$. First observe that $$ F_n\colon K_n\times K_n \to K_n $$ is a continuous map, which can be easily seen by the defining seminorms for these metric spaces.

Now let $U$ be a convex neighbourhood of $0$, i.e. $U\cap K_n$ is a convex neighbourhood of $0$ in $K_n$ for each $n$. Inductively for each $n$, you can find a $0$-neighbourhood $V_n$ of $K_n$ such that $$ F[V_n,V_n] \subseteq U\cap K_n $$ (by the continuity of $F_n$) and $$ \varphi_k V_n \subseteq V_k\,\,\,\,\, (1\leq k < n).$$ Set $W_n:=V_n\cap K_{n-1}$ and $W$ as the convex hull of $\bigcup_n W_n$. Observe that for each $n$, $W_n$ is neigbourhood of $0$ in $K_{n-1}$, so $W\cap K_{n-1}\supseteq W_n$ is one too, hence $W$ is a neighbourhood of $0$ in $C_c^\infty(\mathbb{R}^d)$. Now $F[W,W]\subseteq U$ would establish the continuity of $F$.

Let $\psi, \chi\in W$, i.e. $\psi=\alpha_1\psi_1+\cdots + \alpha_m\psi_m$ and $\chi=\beta_1 \chi_1 + \cdots + \beta_m \chi_m$ with $\alpha_i, \beta_i\geq 0$, $\sum \alpha_i = \sum \beta_i =1$ and $\psi_i,\chi_i\in V_i$. As $$ F(\psi,\chi)=\psi\cdot \chi = \sum_{i,j} \alpha_i\beta_j \cdot \psi_i\chi_j $$ and $\sum_{i,j} \alpha_i\beta_j = 1$, it it sufficient to verify $\psi_i\chi_j\in U$. Now if $i=j$, $$ \psi_i\chi_i = F(\psi_i,\chi_i)\in F[V_i,V_i]\subseteq U\cap K_i \subseteq U.$$ If $i\neq j$, e.g. $i<j$, then $\psi_i\in V_i$ and $\chi_j\in V_j$ and so $$ \psi_i\chi_j = (\psi_i \varphi_i) \chi_j =\psi_i (\varphi_i\chi_j)\in V_i\cdot V_i \subseteq U\cap K_i \subseteq U.$$

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