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If $f \in C^\infty$ and $g$ is a real valued function can we say anything about their product? In particular is $fg \in C^\infty$ or maybe if we stipulate $g$ has compact support can we make the claim?

What's a good way to look at this problem? Im looking to understand this as part of a derivation for a weak (FEM) formulation for Stokes Flow.

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  • $\begingroup$ I guess I dont understand the downvotes/close votes. What can I do to improve my question? $\endgroup$ – still_learning May 28 '15 at 20:43
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    $\begingroup$ Apparantly someone is downvoting the question and all answers :(. $\endgroup$ – Eff May 28 '15 at 20:50
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    $\begingroup$ I saw this question while reviewing close votes and it looks totally reasonable. I'm usually kind of quick to vote to close, but your post is on topic and shows a reasonable amount of effort. Someone is just being a jerk. I voted to leave open and upvoted your post. :) $\endgroup$ – user223391 May 29 '15 at 0:49
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The only thing you can say that $fg$ has the same smoothness as $g$. That is, if $g$ is in $C^k$, then $fg$ is in $C^k$ as well.

If $f$ in addition has compact support, then $g\in C^k$ implies that $fg$ is in $C^k$ with uniformly continuous derivatives up to order $k$. Similarly, if $g$ is in some Sobolev space $W^{m,p}$, then $fg\in W^{m,p}$ as well.

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I don't think you can say anything without further assumption on $g$. The constant function $f=1$ is $\mathcal{C}^\infty$, and $fg=g$ can be arbitrary.

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    $\begingroup$ (Care to explain the downvote? I am slightly baffled.) $\endgroup$ – Clement C. May 28 '15 at 20:46

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