If $\phi\in \mathcal{S}(\mathbb R) $ then $\phi_{t}(x)=\frac{1}{t} \phi(x/t)\in\mathcal{S}(\mathbb R)$?

Question

Let $\phi:\mathbb R \to \mathbb C$ be a function; and define $$\phi_{t}(x):=\frac{1}{t}\phi (x/t), (t>0).$$

We note that, if $\phi\in L^{1}(\mathbb R),$ then $\phi_{t}\in L^{1}(\mathbb R);$ in fact, we have, $\|\phi \|_{L^{1}}= \|\phi_{t}\|_{L^{1}(\mathbb R)}.$

Consider the Schwartz space, $\mathcal{S}(\mathbb R)= \{f\in C^{\infty}(\mathbb R): \sup_{x\in \mathbb R} (1+|x|)^{m} |\partial^{n} f(x) |< \infty, \forall m, n \in \mathbb N \} $

My Question is: (1) Suppose $\phi \in \mathcal{S}(\mathbb R).$ Is it true that, $\phi_{t}\in \mathcal{S}(\mathbb R)$ (Take any fix $t>0$)? (2) If answer in (1) is negative, then is it possible to find some non-zero $\phi \in \mathcal{S}(\mathbb R)$ such that $\phi_{t}$ is also in $\mathcal{S}(\mathbb R)$?

My Little attempt: I think, If $\phi \in C^{\infty}(\mathbb R),$ then $\phi_{t}\in C^{\infty}(\mathbb R);$ as composition of differentiable functions is differentiable and $\partial^{n}\phi_{t}(x)= \frac{1}{t}\partial^{n}\phi(x/t)= \frac{1}{t^{n+1}} \partial^{n} \phi(x/t),$...

Thanks,

Answer 1

For a fixed $x$, we have $$(1+|x|)^m|\partial^n\phi_t(x)|=t^{-(n+1)}(1+|x|^m)(\partial^n\phi)(x/t) =t^{-(n+1)}(1+t^m|x/t|^m)|(\partial^n\phi)(x/t)|\\ \leqslant t^{-(n+1)}\sup_{y\in\mathbb R}(1+t^m|y|^m)|\partial^n\phi(y)|.$$ To conclude, notice that $1+t^my^m\leqslant (1+t^m)(1+|y|^m)$, hence $$\sup_{x\in\mathbb R}(1+|x|)^m|\partial^n\phi_t(x)|\leqslant (1+t^m)t^{-(n+1)}\sup_{y\in\mathbb R}(1+t^m|y|^m)|\partial^n\phi(y)|<+\infty.$$