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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 $\Leftarrow$.

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1 Answer 1

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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.

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