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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.

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    This is not true for general $g \in C^\infty $. Für example, if $s=0$, then your inequality holds if and only if $g \in L^\infty $, which can fail for $g \in C^\infty $ (for example $g (x) = |x|^2$). – 2017-02-28
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    Oh, here $C^{\infty}(\mathbb R^{d})$ means the space of functions with bounded derivatives, also all higher orders. – 2017-02-28
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    Since you know the claim is true for $s \in \Bbb{Z} $, you can use interpolation to derive the general result. – 2017-02-28
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    Oh yes it works. Thanks! – 2017-03-03

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