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.

arbitrarynets need not be contained in any limitand of this colimit, butboundedCauchy netsare. (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). $\endgroup$ – paul garrett Sep 17 '17 at 19:142more comments