# Equivalent definition of Schwartz space

Definition of Schwartz space is the following. $$f(x) \in \mathcal{S} \overset {\mathrm{def}} {\Leftrightarrow} \displaystyle \sup_{x \in \mathbb{R^d} } \left|x^\alpha\partial^\beta_x f(x)\right| < \infty$$

$$\forall\alpha,\forall\beta$$ $$\in$$ $$\mathbb{Z^d_+}$$ ($$\alpha,\beta$$ is multi-index notation)

My textbook is written the following statement.

$$\displaystyle \sup_{x \in \mathbb{R^d} } \left|x^\alpha\partial^\beta_x f(x)\right| < \infty\Leftrightarrow \displaystyle \sup_{x \in \mathbb{R^d} } \left|\partial^\alpha_x (x^\beta f(x))\right| < \infty$$

I have proved $$\Rightarrow$$ by using Leibniz's rule. But I haven't proved $$\Leftarrow$$. Please tell me proof $$\Leftarrow$$.

1) With $$\alpha = 0$$ you have $$\sup_{x \in \mathbb R} |x^\beta f(x)| < \infty$$ for all $$\beta$$.
2) With $$\alpha = 1$$ you have $$\sup_{x \in \mathbb R} |\beta x^{\beta - 1} f(x) + x^\beta f'(x)| < \infty$$ for all $$\beta$$. You already know from 1) that $$\sup_{x \in \mathbb R} |x^{\beta - 1}f(x)| < \infty$$, so it follows that $$\sup_{x \in \mathbb R} |x^\beta f'(x)| < \infty.$$
3) Do the same with $$\alpha = 2$$, etc.