Please tell me about the 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 $\Leftarrow$.
To keep notation simple I will sketch the idea in 1D.
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.