I know that in Sobolev spaces $H^{s}(\mathbb R^{d})$, $s\in \mathbb R$, the multiplication by $g\in S(\mathbb R^{d})$ is continuous, namely the following inequality holds:
$$||fg||_{H^{s}(\mathbb R^{d})}\leq C||f||_{H^{s}(\mathbb R^{d})}$$
for all $f\in H^{s}(\mathbb R^{d})$ with constant $C$ independent to $f$.
I am asking that is it also true for $g\in C^{\infty}(\mathbb R^{d})$?
(Here $C^{\infty}(\mathbb R^{d})$ means the space of smooth functions $f$ with derivatives $\bigtriangledown ^{k} f$ bounded, $k\geq0$.)
What I have done is to check when $s$ is a nonnegative integer, and in turn a negative integer by duality. If the coefficient function $g$ is a Schwartz function I can apply the convolution and Peetre inequality, but in this case I do not see how to prove.
How can we prove this when $g\in C^{\infty}(\mathbb R^{d})$?
Thanks.