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I was reading Terence Tao post https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/ 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.

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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  • $\begingroup$ Yes, but $C_c^{\infty}( \mathbb R^d)$ is not Fréchet. It is not First-countable space too. On the other hand F restricted to $C_c^{\infty}( K)\times C_c^{\infty}( K)$ is continuous for any compact $K\subset R^d$ because $C_c^{\infty}(K)$ is metrizable. $\endgroup$ – Hugo C Botós Mar 3 '16 at 4:08
  • $\begingroup$ Btw. If you don't like sequences, just don't use them. The proof with neighborhoods is basically the same. $\endgroup$ – user251257 Mar 3 '16 at 4:27
  • $\begingroup$ You can make your large family of semi norms countable with no big change. The overpowered way of saying this would be to pick a countable dense subset of the space of compact subsets of $\Bbb R^d$ with the Hausdorff metric. This is a more or less standard trick. $\endgroup$ – user98602 Mar 3 '16 at 4:30
  • $\begingroup$ The reason why it's not Fréchet space is because it's not First Countable. This fact is proven here: math.stackexchange.com/questions/982556/… $\endgroup$ – Hugo C Botós Mar 3 '16 at 5:30
  • $\begingroup$ @HugoC.Botós sorry. My bad. I should go to bed. $\endgroup$ – user251257 Mar 3 '16 at 5:45
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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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  • $\begingroup$ OH! That's awesome! I loved your argument. Thanks! $\endgroup$ – Hugo C Botós Mar 24 '16 at 1:06
  • $\begingroup$ If you wish give your answer here too: mathoverflow.net/questions/234025/… $\endgroup$ – Hugo C Botós Mar 24 '16 at 1:08
  • $\begingroup$ @HugoC.Botós Put a link on mathoverflow, I lost my login data and feel no need to recover it somehow. $\endgroup$ – Vobo Mar 25 '16 at 13:02

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