# 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.

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Can you do it in one dimension? – GEdgar Aug 21 '12 at 18:32
If $f$ satisfies the condition and you are given $\alpha$ and $\beta$, take $P=(1+x_1^2+\cdots+x_n^2)x^\alpha$ and $L=D^\beta$: by hypothesis, $PLf$ is bounded. Show that implies that $x^\alpha D^\beta f\to 0$ for large $x$. – Mariano Suárez-Alvarez Aug 21 '12 at 18:41
Now I feel like a fool. :) – user38404 Aug 21 '12 at 18:56
Well, don't! :-) – Mariano Suárez-Alvarez Aug 21 '12 at 19:52

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}

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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)

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