# Equivalent definitions of Schwartz Space

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

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 $$|x^{\alpha}\partial^{\beta}f(x)|\leq C_{\alpha,n}(1+|x|)^{N}|\partial^{\beta}f(x)|\leq C_{\alpha,n}C_{N,\beta}$$.