Continuity of a linear operator in Schwartz Space

Question

Let $f: \mathbb{R}\rightarrow\mathbb{R}$ be a $\mathbb{C}^{\infty}$ function which is bounded. Define $A:\mathcal{S}(\mathbb{R})\rightarrow\mathcal{S}(\mathbb{R})$ as $A(\phi)=f\phi$. Is $A$ continuous?

Intuitively, I think this should not be true. If we take a function $f$ which is bounded but has an unbounded derivative then for a sequence $\{\phi_n\}\rightarrow0$ in $\mathcal{S}(\mathbb{R})$, $\{f\phi_n\}\nrightarrow0$ in $\mathcal{S}(\mathbb{R})$. I thought of taking, $f=\sin(x^2)$. However, I'm unable to find a suitable $\{\phi_n\}$.

Note: $\mathcal{S}(\mathbb{R})$ is the Schwartz space.

Answer 1

Consider the bounded smooth function $f(x)=\sin(\exp(x^2))$ and $g(x)=\exp(-x^2/2)$. Then $g$ belongs to $\mathcal S(\mathbb R)$ but $fg$ does not because $$(fg)'(x)= 2x \exp(x^2/2)\cos(\exp(x^2)) - x\exp(-x^2/2)f(x)$$ does not converge to $0$.

The "multiplier space" of $\mathcal S(\mathbb R)$ is calculated in Laurent Schwartz' book on distribution theory (which I do not have at hand, right now).

Answer 2

The biggest set for function $f$ of your operator $A$ for which $A$ is continuous on $S(R)$ is the subspace of $C^{\infty}$ for which $\partial^{n}_{x} f(x)$ are polynomially bounded functions for $n=0,1,2,...$.