One important question about Schwartz space I have one problem with Schwartz space. I remind you which definitions which I am using.
Multi-index: $\alpha=(\alpha_{1}, \ldots , \alpha_{n}) \in \mathbb{Z}_{+}^{n}$, length of multi-index: $| \alpha | = \sum_{i=1}^{n} \alpha_{i}$, partial differential operator: $$D^{\alpha} = \frac{\partial ^{|\alpha|}}{\partial x_{1}^{\alpha_{1}} \ldots \partial x_{n}^{\alpha_{n}}}.$$ Moreover $|x|=\sqrt{x_1^2+ \ldots + x_n^2}$, Schwartz space: $$\mathcal{S}(\mathbb{R}^n) = \{f \in C^{\infty}(\mathbb{R}^n) \colon \sup_{x \in \mathbb{R}^{n}} (1+|x|)^{m}|D^{\beta}f(x)| < \infty, \: m \in \mathbb{N}, \: \beta \in \mathbb{Z}_{+}^{n} \}.$$
It is easy to prove that if $f \in \mathcal{S}(\mathbb{R}^n)$ then $D^{\beta}f \in \mathcal{S}(\mathbb{R}^n)$ for every multi-index $\beta$. I need to prove that if $f \in \mathcal{S}(\mathbb{R}^n)$ then $(1+|x|)^{m}f(x) \in \mathcal{S}(\mathbb{R}^n)$ for every integer $m$. Has anyone idea how to prove it?
I found theorem: If $a(x) \in C^{\infty}(\mathbb{R}^n)$ is such that $$\forall_{\beta \in \mathbb{Z}_{+}^{n}} \; \exists_{c_{\beta}>0,\; n_{\beta} \in \mathbb{N}} \colon |D^{\beta} a(x)| \leq {c_{\beta}} \cdot (1+|x|)^{n_{\beta}}$$
then $a(x)\cdot f(x) \in \mathcal{S}(\mathbb{R}^n)$ for all $f \in \mathcal{S}(\mathbb{R}^n)$. So, if we show that $$|D^{\beta} (1+|x|)^{m}| \leq c_{\beta} \cdot (1+|x|)^{n_{\beta}} $$ then it will be proved. I got only $$|D^{\beta} (1+|x|)^{m}| \leq c_{\beta} \cdot (1+|x|)^{k}\cdot |x|^{-|\beta|}$$
for all $k\geq m$, but it doesn't help. I will be gratful for all tips and solutions.
Greetings from Poland,
Marcin.
 A: The function $x \mapsto |x|$ is not smooth, so we cannot expect that $x \mapsto (1+|x|) \cdot f(x)$ is differentiable. Let's prove instead that
$$x \mapsto (1+|x|^2)^m \cdot f(x) \in \mathcal{S}(\mathbb{R}^n) \tag{1}$$
By the Binomial theorem, we have
$$(1+|x|^2)^m = \sum_{k=0}^m {m \choose k} \cdot |x|^{2k}$$
Thus, by the product rule,
$$D^{\beta} \big( (1+|x|^2)^m \cdot f(x) \big) = \sum_{\alpha+\gamma=\beta} C_{\alpha,\gamma} \sum_{k=0}^m {m \choose k} \cdot D^{\alpha}(|x|^{2k}) \cdot D^{\gamma}(f(x))$$
Since the Schwartz space is a linear space and since $x \mapsto D^{\alpha} (|x|^{2k})$ is a polynomial (in $x$) for fixed $\alpha \in \mathbb{N}_0^n$, $k \in \mathbb{N}_0$, it suffices to show that functions of the form
$$x \mapsto g(x) := x^{\alpha} \cdot D^{\gamma} f(x))$$
satisfy
$$\sup_{x \in \mathbb{R}^n} (1+|x|)^{\ell} \cdot |g(x)| < \infty \tag{2}$$
for arbitrary $\ell \in \mathbb{N}$. This follows from the fact that
$$|x_j|^{\alpha_j} \leq (1+|x_j|)^{\alpha_j} \leq (1+|x|)^{\alpha_j}$$
i.e.
$$|x^\alpha| \leq (1+|x|)^{\sum_j \alpha_j} = (1+|x|)^{|\alpha|}$$
Consequently,
$$\begin{align*} (1+|x|)^{\ell} \cdot |g(x)| &\leq (1+|x|)^{|\alpha|} \cdot (1+|x|)^{\ell} \cdot |D^{\gamma}f(x)|  \\ &= (1+|x|)^{\ell+|\alpha|} \cdot |D^{\gamma} f(x)|. \end{align*}$$
Thus, $(2)$ follows from the definition of the Schwartz space.
