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 $\Leftarrow$.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.