Why is multiplication on the space of smooth functions with compact support continuous?

Question

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.

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. — Hugo, Mar 03 '16 at 04:08
Btw. If you don't like sequences, just don't use them. The proof with neighborhoods is basically the same. — user251257, Mar 03 '16 at 04:27
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. — , Mar 03 '16 at 04:30
The reason why it's not Fréchet space is because it's not First Countable. This fact is proven here: http://math.stackexchange.com/questions/982556/why-is-the-topology-of-compactly-supported-smooth-function-in-mathbb-rd-not — Hugo, Mar 03 '16 at 05:30
There is another way of seeing that the space is not metrizable. Suppose it has a metric $d$. For each $N\in \mathbb N$ we have a bump function $\phi_N$ that is positive on $B[0,N]$ and $0$ elsewhere. We know $d(0,\phi_N)>0$ and for $\epsilon>0$ small enough we have $d(0,\epsilon \phi_N)<1/N$ because $\lim_{t\to 0} d(0,t\phi_N)=0.$ So we define $f_N = \epsilon \phi_N$. Notice that $f_N \to 0$. Since $f_N$ is a convergent sequence on $C_c^\infty(\mathbb R^d)$ there is a compact $K\subset R^d$ such that the support of $f_n$ is contained on $K$, that is a contradiction. — Hugo, Mar 03 '16 at 05:46
@MikeMiller , i don't understand your comment. Can you give me some detail? — Hugo, Mar 03 '16 at 05:50
Nope, my comment is nonsense, since it would prove a false conclusion. Leaving it up in case anyone else makes the same mistake I did. — , Mar 03 '16 at 05:51
The space of test functions is not sequential. The comment of federico on the answer of the link you send show that. On the post
http://math.stackexchange.com/questions/706061/topologies-of-test-functions-and-distributions#comment1483406_706463

he posted the article http://www.ams.org/journals/proc/1971-027-03/S0002-9939-1971-0270145-X/S0002-9939-1971-0270145-X.pdf and on the second paragraph is said that it was proved that $\mathcal D = C_c^\infty(\mathbb R^d)$ is not sequential (and the paper proves \mathcal D' is not sequential). — Hugo, Mar 03 '16 at 13:25
But it's true that if $\mathfrak{X}$ is a locally convex vector space and $F:C_c^\infty(\mathbb R^d)\mapsto \mathfrak{X}$ is linear then $F$ is continuous if and only if $F$ is sequentially continuous. — Hugo, Mar 03 '16 at 13:34
isn't $C_c^\infty(\mathbb R^d \times \mathbb R^d)$ continuously isomorph to $C_c^\infty(\mathbb R^d) \times C_c^\infty(\mathbb R^d)$? If they are, then this is finally settle... — user251257, Mar 03 '16 at 18:24

score 4 · Accepted Answer · answered Mar 23 '16 at 20:18

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 $$ \psi_i\chi_j = (\psi_i \varphi_i) \chi_j =\psi_i (\varphi_i\chi_j)\in V_i\cdot V_i \subseteq U\cap K_i \subseteq U.$$

If you wish give your answer here too: http://mathoverflow.net/questions/234025/why-is-multiplication-on-the-space-of-smooth-functions-with-compact-support-cont — Hugo, Mar 24 '16 at 01:08
@HugoC.Botós Put a link on mathoverflow, I lost my login data and feel no need to recover it somehow. — Vobo, Mar 25 '16 at 13:02

Why is multiplication on the space of smooth functions with compact support continuous?

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