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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.

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    The point not whether $A$ is continuous but whether it maps $\mathcal S$ to $\mathcal S$. If so, then it is automatically continuous by the closed graph theorem. Anyway, your suspicion is right, The function $f(x)=\sin(e^x)$ is an example. – 2017-02-10
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    @Jochen, your comment and Paul's answer have actually confused me. While you say that A wouldn't be continuous, Paul says that tempered distribution is closed under differentiation. Which should mean that A is continuous. – 2017-02-10
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    My bad answer (now deleted) was misguided, as I had misread the question... So, yes, while differentiation of the tempered distribution $\sin(e^x)$ is still a tempered distribution despite its exponentially growing pointwise values, that's not your question. My apologies for not paying better attention! – 2017-02-10

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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).

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    Aah so the range is not contained in $\mathcal{S}(\mathbb{R})$. So we can't talk about continuity from $\mathcal{S}(\mathbb{R})$ to $\mathcal{S}(\mathbb{R})$. – 2017-02-10
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    @SahibaArora Yes, for a closed linear operator $T : X \to X$, continuity is equivalent to $\forall \varphi \in X, T(\varphi) \in X$ (or something like that) – 2017-02-10
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    @user1952009, but we can't talk about A being continuous or not if the range is not in the codomain, right? A is not an operator from $\mathcal{S}(\mathbb{R})$ to $\mathcal{S}(\mathbb{R})$. – 2017-02-10
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    @SahibaArora Your operator is densely defined $U \subset S(\mathbb{R}) \to S(\mathbb{R})$ and we want to know if we can extend it to the whole space $S(\mathbb{R})$ or not. Jochen proved it is not possible, so it can't be continuous. – 2017-02-10
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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 $n=0,1,2,...$.

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    hmm ? you meant functions with all derivatives polynomially bounded – 2017-08-16
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    @AK1977 What are polynomially bounded functions? – 2017-08-16
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    $O_{M}$ denote the set of infinitely differentiable functions on $\mathbb{R}$ which together with their derivatives are polynomially bounded, that is $f \in O_{M} $ is $C^{\infty}$ and for each $k\in \{0,1,2,3,....\}$ there are $N(k)$ and $C(k)$ with $|{\partial}^{\,k}_{x}f(x)| \le C\,{[1+x^2]}^{N}$ – 2017-08-16
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    The proof is in 3 steps (a, b, c) written in the problem 23 of page 176 (problem at the end of chapter "Locality of Convex Spaces") of the book "Methods of Mathematical Physics (Vol 1 functional analysis) Simon&Reed 1980" – 2017-08-17
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    a) Let $F\in O_{M}$ then prove $ f \to F f$ is continuous map on $S$. b) Let $F$ be a measurable function so that $Ff \in S $ for all $f \in S $ then prove $F$ is $C^{\infty}.$ c) If $f\to F f$ is continuous, prove that $F\in O_{M}$ – 2017-08-17
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    @SahibaArora **It is obvious** : if $\forall n, \forall x, |f^{(n)}(x)| \le 1+C_n |x|^{k_n}$ then $\forall \varphi \in S(\mathbb{R}), f \varphi \in S(\mathbb{R})$. Conversely if $f^{(n)}$ isn't bounded by any polynomial then you can construct a Schwartz function such that $f \varphi$ isn't Schwartz. – 2017-08-20