Why is $x^\alpha \partial^\beta f$ a linear combination of terms of the form $\partial ^\alpha (x^\beta f)$? Let $f:\mathbb{R}^m \to \mathbb{R}$ be a smooth function.
Why is $x^\alpha \partial^\beta f$ a finite linear combination of terms of the form $\partial ^\alpha (x^\beta f)$? Here, $\alpha$ and $\beta$ denote multi-indices, that is, they are elements in $\mathbb{Z}_{\geq 0} \times \cdots \times \mathbb{Z}_{\geq 0}$.
(Motivation: I want to prove a smooth function $f$ with bounded $\partial ^\alpha (x^\beta f)$ for all $\alpha, \beta$ is in the Schwartz class.)
It is clear that the converse is true by the product rule for derivatives. But I don't think it is easy to prove that such a procedure is invertible.
Any help will be fully appreciated. Thank you.
 A: Solution 1. Since
$$ x^{\alpha}\partial^{\beta} f = (x_1^{\alpha_1}\partial_1^{\beta_1}) \cdots (x_m^{\alpha_m}\partial_m^{\beta_m}) f, $$
it suffices to prove when $m = 1$. In such case, write $\partial = \frac{d}{dx}$. Then we can prove by induction that
$$ g \partial^n f = \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} \partial^{k} \left( (\partial^{n-k}g) f \right) $$
Then the desired claim follows by plugging $g(x) = x^a$.

Solution 2. For a concrete induction argument, write $|\alpha|=\sum_{i=1}^{m}\alpha_i$ and set 
$$ W_n = \operatorname{span}\left\{ \partial^{\alpha}(x^{\beta}f) : |\alpha| \leq n \right\}. $$
Then we make the following observation:

Claim. If $g \in W_n$, then $x_i g \in W_n$.

Indeed, the claim is obviously true for $n = 0$. Next, assume that the claim is true for $n$. Let $g \in W_{n+1}$ be arbitrary and write $g = h_0 + \sum_{j=1}^{m} \partial_j h_j$ for some $h_0 \in W_0$ and $h_1,\cdots,h_m \in W_n$. Then
\begin{align*}
x_i g
&= x_ih_0 + \sum_{j=1}^{m} x_i \partial_j h_j \\
&= x_ih_0 + \sum_{j=1}^{m} \big(\partial_j(x_i h_j) - \delta_{ij}h_j\big)
\end{align*}
By the induction hypothesis, we know that $x_i h_j \in W_n$, and so, $\partial_j(x_i h_j) \in W_{n+1}$. All the rest terms are obviously in $W_{n+1}$, and therefore $x_i g \in W_{n+1}$. This completes the induction step, and we are done. $\square$
Returning back to the original problem, note that $\partial^{\beta} f \in W_{|\beta|}$. So the claim shows that $x^{\alpha}\partial^{\beta}f$ also lies in $W_{|\beta|}$.
A: The basic identity that you need is $\partial(x\cdot f) - f = x\partial f$. You can build up to multiindices step by step. First it is sufficient to consider the 1-d case $m=1$ because you do the derivates one variable at a time. Next $\partial(x^n\cdot f) - nx^{n-1}f = x^n\partial f$. Finally you can iterate this equation by sticking in $\partial f$ instead of $f$ to get higher order derivatives.
A: Let $\prec$ denotes the graded lexicograhpic order on $(\alpha,\beta)$, and $(x^{\alpha'}\partial^{\beta'}f)_{(\alpha',\beta')\preceq(\alpha,\beta)}$, $(\partial^{\beta'}(x^{\alpha'}f))_{(\alpha',\beta')\preceq(\alpha,\beta)}$ be horizontal vectors (ordered decreasingly by $\prec$). Then we have
$$ (\partial^{\beta'}(x^{\alpha'}f))_{(\alpha',\beta')\preceq(\alpha,\beta)}=(x^{\alpha'}\partial^{\beta'}f)_{(\alpha',\beta')\preceq(\alpha,\beta)}M $$
where $M$ is a lower triangular square matrix whose every diagonal component is $1$. Since such $M$ is invertible, we can obtain a desired representation by multiplying $M^{-1}$ from the right.
A: Remember the following properties of the Fourier Transform:
$$\widehat{D_z^{\alpha}\varphi}=\xi^{\alpha}\widehat{\varphi} \hbox{ and } D_\xi^\alpha(\widehat{\varphi})=\widehat{z^\alpha \varphi}.$$
Therefore,
\begin{eqnarray*}
z^{\beta}D_z^\alpha f(z) &=& z^\beta \{D_z^\alpha f(z) \}^{\land \lor}\\
&=& z^\beta\{\xi^\alpha \hat{f}(\xi) \}^{\lor}(z)\\
&=& \{z^\beta[\xi^\alpha \hat{f}(\xi)]^\lor(z) \}^{\land \lor}\\
&=&\{D_\xi^\beta[\xi^\alpha\hat{f}(\xi)] \}^\lor(z)\\
&=&\sum_{\delta \leq \beta} \left(\begin{array}{c} \beta \\ \delta\end{array}\right) \{D_\xi^\delta \xi^\alpha D_\xi^{\beta-\delta}\hat{f}(\xi) \}^\lor(z)\\
&=&\sum_{\delta \leq \beta} \left(\begin{array}{c} \beta \\ \delta\end{array}\right) \frac{\alpha!}{(\alpha-\beta)!}\{ \xi^{\alpha-\delta} D_\xi^{\beta-\delta}\hat{f}(\xi) \}^\lor(z)\\
&=&\sum_{\delta \leq \beta} \left(\begin{array}{c} \beta \\ \delta\end{array}\right) \frac{\alpha!}{(\alpha-\beta)!}\{ \xi^{\alpha-\delta} [D_\xi^{\beta-\delta}\hat{f}(\xi)]^{\lor \land} \}^\lor(z)\\
&=&\sum_{\delta \leq \beta} \left(\begin{array}{c} \beta \\ \delta\end{array}\right) \frac{\alpha!}{(\alpha-\beta)!}D_z^{\alpha-\delta} \{D_\xi^{\beta-\delta}\hat{f}(\xi)\}^{\lor}(z)\\
&=&\sum_{\delta \leq \beta} \left(\begin{array}{c} \beta \\ \delta\end{array}\right) \frac{\alpha!}{(\alpha-\beta)!}D_z^{\alpha-\delta} \{z^{\beta-\delta} f(z)\}.\end{eqnarray*}
