# multiplying by a $C^\infty$ function

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.

• I guess I dont understand the downvotes/close votes. What can I do to improve my question? – still_learning May 28 '15 at 20:43
• Apparantly someone is downvoting the question and all answers :(. – Eff May 28 '15 at 20:50
• 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. :) – user223391 May 29 '15 at 0:49

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.
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.