Regarding the definition of Schwartz Space of functions I came across a definition of Schwartz Space where they were defined as functions $f$ such that $\mathrm{lim}_{|x|\to \infty} |x^{\alpha}D^{\beta}f(x)|=0$ for any pair of multiindices $\alpha,\ \beta$. Now this clearly implies that $P(x)Lf(x)$ is bounded for any polynomial $P$ and differential operator with constant coefficients $L$. How do I prove the converse? If $P(x)Lf(x)$ is bounded for any polynomial $P$ and differential operator with constant coefficients $L$, would it still hold that $\mathrm{lim}_{|x|\to \infty} |x^{\alpha}D^{\beta}f(x)|=0$ for any pair of multiindices $\alpha,\ \beta$?
If not, then $\forall N>0, \exists x_N $ and $  M_N>0, $ so that $ |x_N|>N$ and $|x^{\alpha}D^{\beta}f(x)|>M_N$. How does this contradict boundedness of the same? If $M_N$ were not to depend on $N$, and were fixed, we could contradict boundedness, but as it is I can't see a way out. 
 A: This definition is problematic because it only dictates the behavior of $f$ outside of some arbitrarily sized sphere $\left(\raise{3pt}{\lim\limits_{|x|\to\infty}}\right)$. This definition also requires that $f\in C^\infty$.
The definition I usually see is
$$
\left|x^\alpha\partial^\beta f(x)\right|\le C_{\alpha,\beta}\tag{1}
$$
for all $x\in\mathbb{R}^n$. This definition implies that $f\in C^\infty$.
However, if we assume that $P(x)Lf(x)$ is bounded for any polynomial $P$ and constant coefficient differential operator $L$, which is easily equivalent to $(1)$, then because $P(x)=|x|^2x^{\alpha}$  is a polynomial, we have
$$
\begin{align}
|x|^2\left|x^{\alpha}\partial^{\beta}f(x)\right|
&=\left||x|^2x^{\alpha}\partial^{\beta}f(x)\right|\\
&\le C\tag{2}
\end{align}
$$
Then, from $(2)$ we get
$$
\begin{align}
\lim_{|x|\to\infty}\left|x^{\alpha}\partial^{\beta}f(x)\right|
&\le\lim_{|x|\to\infty}\frac{C}{|x|^2}\\
&= 0\tag{3}
\end{align}
$$
A: Fix $\alpha$ and $\beta$ multi-indices. We define the polynomial $P(x):=x^{\alpha}=\prod_{j=1}^dx_j^{\alpha_j}$ and $L=\partial^{\beta}$. 
(actually, you managed to solve the more difficult part of the problem)
