We have the following definition of Schwartz space $S(\mathbb{R}^k)$: we tell that the function $\psi: \mathbb{R}^k \to \mathbb{C}$ is in $S(\mathbb{R}^k)$ iff 1. $\psi \in C^{\infty}(\mathbb{R}^k)$ and 2. for all $m \in \mathbb{N}$ and for all $\alpha \in \mathbb{N}^k$, there exists $C_{m, \alpha}(\psi) >0$ such as $\forall x \in \mathbb{R}^k, (1+|x|^2)^m |D^\alpha \psi(x)| \leq C_{m,\alpha}(\psi)$.
Please can you help me to prove the following proposition: we tell that $\psi: \mathbb{R}^k \to \mathbb{C}$ is in $S(\mathbb{R}^k)$ iff 1. $\psi \in C^\infty(\mathbb{R}^k)$ and 2. forall $l \in\mathbb{N}$ and for all $\alpha \in \mathbb{N}^k$ the function $x \to |x|^l D^{\alpha} \psi(x)$ is bounded.
Thanks in advance for the help.
Definition 1 implies definition 2 because for each $l\in\mathbb N$ and $\alpha\in \mathbb N^k$, $$ \left\lvert |x|^l D^{\alpha} \psi(x)\right\rvert\leqslant \left(1+\left\lvert x\right\rvert^2\right)^{l}\left\lvert D^{\alpha} \psi(x)\right\rvert\leqslant C_{l,\alpha}\left(\psi\right). $$ For the opposite direction, let $m\in\mathbb N$ and $\alpha\in \mathbb N^k$. Using the binomial formula, the following equality holds $$ (1+|x|^2)^m |D^\alpha \psi(x)| =\sum_{j=0}^m\binom mj|x|^{2j}|D^\alpha \psi(x)| $$ hence $$\sup_{x\in\mathbb R^k} (1+|x|^2)^m |D^\alpha \psi(x)| \leqslant \sum_{j=0}^m\binom mj\sup_{t\in\mathbb R^k}|t|^{2j}|D^\alpha \psi(t)| \lt+\infty. $$