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I'm trying to show the following but have no idea how to begin. I'm quite new to analysis and multi-index notation.

$$ f \in \mathcal{S} \quad \Longleftrightarrow \quad \forall N \in \mathbb{N}, \alpha \in \mathbb N_0 \text{ a multi-index} \ \exists C_{N,\alpha} > 0: \ \vert \partial^{\alpha} f(x) \vert \leq \frac{C_{N,\alpha}}{(1+|x|)^{N}} $$ where $\mathcal{S}$ denotes the Schwartz space.

For the forward direction I've written down the definition of being in Schwartz space i.e. $$ \sup_{x \in \mathbb{R}^{n}}|x^{\alpha}\partial^{\beta}f(x)| \leq C_{\alpha,\beta} $$ where $ \alpha, \beta \in \mathbb N_0$ are multi-indices

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First note that $|x^{\alpha}|\leq C_{\alpha,n}|x|^{|\alpha|}$ and $|x|^{|\alpha|}\leq C_{\alpha,n}\displaystyle\sum_{|\beta|\leq|\alpha|}|x^{\beta}|$. See this qeustion for a proof.

"$\implies$": Note that using binomial expansion, we can get $(1+|x|)^{N}\leq C_{N}(1+|x|+\cdots+|x|^{N})$ and each $|x|^{i}\leq C_{i,n}\displaystyle\sum_{|\beta|\leq i}|x^{\beta}|$ and we have $|x|^{i}|\partial^{\alpha}f(x)|\leq C_{i,n}\displaystyle\sum_{|\beta|\leq i}|x^{\beta}\partial^{\alpha}f(x)|\leq C_{i,n}\sum_{|\beta|\leq i}\sup_{x\in{\bf{R}}^{n}}|x^{\beta}\partial^{\alpha}f(x)|$.

"$\impliedby$": Let $N=|\alpha|$, then $|x^{\alpha}|\leq C_{\alpha,n}|x|^{N}\leq C_{\alpha,n}(1+|x|)^{N}$ and hence $|x^{\alpha}\partial^{\beta}f(x)|\leq C_{\alpha,n}(1+|x|)^{N}|\partial^{\beta}f(x)|\leq C_{\alpha,n}C_{N,\beta}$.

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  • $\begingroup$ Thank you, very clear! $\endgroup$
    – VBACODER
    Commented Mar 26, 2018 at 8:31

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