I'm trying to prove the properties of convergence on the set of test functions, $\mathscr{D}$, but the following is giving me some problems.
Let $\phi_n\to\phi$ and $\psi_n\to\psi$ on $\mathscr{D}$. Then $f\phi_n\to f\phi$ on $\mathscr{D}$ for any smooth function $f$ defined on $\mathbb{R}^N$.
Any help/hints would be greatly appreciated.
Convergence of the $\phi_n$ in $\mathscr{D}$ implies that the $\phi_n$ are convergent in $\mathscr{D}_K$ where $K$ is a compact subset of $\Bbb{R}^N$. Rudin Functional Analysis 6.5. In $\mathscr{D}_K$, the topology is given by a collection of sup norms, see Rudin 6.2, and so convergence of $\phi_n$ is the same as convergence in all those norms, i.e., uniform convergence of $D^\alpha (\phi_n)$ for all multiindices $\alpha$. Likewise, convergence of $f\phi_n$ is the same as convergence in all those norms, i.e., uniform convergence of $D^\alpha (f\phi_n)$ for all multiindices $\alpha$.
The latter can be shown by the rule for derivatives of a product. For example for $N=1$, and for first derivatives, we have $D(f\phi_n-f\phi)=(Df)(\phi_n-\phi)+fD(\phi_n-\phi)$, so that $||D(f\phi_n-f\phi)||\le ||(Df)(\phi_n-\phi)||+||fD(\phi_n-\phi)||\le ||Df||\cdot ||\phi_n-\phi||+$ $||f||\cdot||D\phi_n-D\phi||\to 0$ as $n\to\infty$. The norms here are sup norms on $K$. We use here the fact that $f$ is smooth so that all the sup norms $||D^\alpha f||$ exist (i.e., are $<\infty$).