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Let a sequence of test functions ${\phi_{m(x)}}, m = 1, 2, . . .$ converge in D to zero, and let $\psi(x)$ be an infinitely differentiable function. Show that $\psi\phi_{m(x)}$ also converges to zero.

Any help/hints would be greatly appreciated.

A sequence of ${\phi_{m(x)}}, m = 1, 2, . . .$ converges to $\phi \; $if - all ${\phi_{m(x)}}, m = 1, 2, . . .$ as well as $\phi$ vanishes outside the region. and $D^{k}{\phi_{m(x)}} \rightarrow D^{k}\phi$ uniformly over $R^{n}$ as $m$ goes to infinity for all multiindices $k$ and also we have $\psi \phi_{m}$ in D since $\psi$ is a function of $C^{\infty}$.

mathsstudent
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  • What is your definition of "test function"? – Math1000 Dec 09 '19 at 19:27
  • A sequence of ${\phi_{m(x)}}, m = 1, 2, . . .$ converges to $\phi ; $if
    • all ${\phi_{m(x)}}, m = 1, 2, . . .$ as well as $\phi$ vanishes outside the region.

    and $D^{k}{\phi_{m(x)}} \rightarrow D^{k}\phi$ uniformly over $R^{n}$ as $m$ goes to infinity for all multiindices $k$ and also we have $\psi \phi_{m}$ in D since $\psi$ is a function of $C^{\infty}$.

     – mathsstudent Dec 09 '19 at 20:00
  • That is not quite what Math1000 asked, but it gives the general idea. Test functions are smooth functions of compact support. Also it tells us you are talking about a topology where convergence requires all derivatives to converge, which limits the topology on $D$. But it still doesn't tell us for sure that $D$ is. In the post, you just throw it out without any explanation. So what exactly is $D$? Obviously a set of smooth functions, but on what domain? is it all of $C^\infty(R^n)$ or is it more restrictive? Is there anything more known of its topology? What do you know about it? – Paul Sinclair Dec 10 '19 at 02:57

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