Characterization of $\mathscr{S}(\mathbb R^n)$?

Question

Consider the vector space $$\displaystyle\mathscr{S}(\mathbb R^n)=\{f\in C^\infty(\mathbb R^n): \lim_{|x|\to \infty} |x^\alpha \partial^\beta \phi(x)|=0, \forall \alpha, \beta\in\mathbb N_0^n\}.$$

I'd like to show $f\in C^\infty(\mathbb R^n)$ is an element of $\mathscr{S}(\mathbb R^n)$ if and only if $$\displaystyle \sup_{x\in\mathbb R^n} |x^\alpha \partial^\beta f(x)|<\infty.$$

I proved the implication $(\Rightarrow)$ as follows:

Let $\varepsilon>0$. There exists $M>0$ such that $|x|>M\Rightarrow |x^\alpha \partial^\beta f(x)|<\varepsilon$. Now suppose $x\in\mathbb R^n$ is such that $|x|\leq M$. Then, $$|x^\alpha \partial^\beta f(x)|\leq |x|^{|\alpha|}|\partial^\beta f(x)|\leq M^{|\alpha|}\sup_{x\in \overline{B(0, M)}}|\partial^\beta f(x)|.$$ Setting $C_{\alpha, \beta}=\max\{\varepsilon, M^{|\alpha|} \displaystyle\sup_{x\in \overline{B(0, M)}}|\partial^\beta f(x)|\}$ we find $|x^\alpha \partial^\beta f(x)|\leq C_{\alpha, \beta}$ for all $x\in\mathbb R^n$ what implies $\displaystyle\sup_{x\in\mathbb R^n} |x^\alpha \partial^\beta f(x)|<\infty$.

However, I wasn't manage to prove the implication $(\Leftarrow)$. Any hint?

Answer 1

Hint. The point for $(\Leftarrow)$ is, that you have boundedness for all $\alpha$. Let $\alpha, \beta$ be given, we have $$ |x^\alpha \partial^\beta f(x)| \le |x|^{|\alpha|}|\partial^\beta f(x)| = \frac{|x|^{|\alpha| + 1}|\partial^\beta f(x)|}{|x|} $$ If we can bound the numerator, we are done.

Answer 2

Since it's bounded for any multiindex add $1$ to $\alpha_i$ to show that $|x^{\alpha} \partial^\beta f(x)|\leq \frac{M_i}{|x_i|}$ for every $i$. Selecting the largest $M_i$ to be $M$ we get $|x^{\alpha} \partial^\beta f(x)|\leq \min_i\frac{M_i}{|x_i|}\leq\frac{M}{\max_i|x_i|}$ for all $x\neq0$. But if $|x|\to\infty$ then $\max_i|x_i|\to\infty$.

Answer 3

Well, using @martini 's idea I wrote a proof:

Let $\varepsilon>0$ be given and fix $\alpha, \beta\in\mathbb N_0^n$. If $p\in\mathbb N$ then using the multinomial theorem:

$$ \begin{align} \displaystyle |x|^{2p} |\partial^\beta f(x)|^2&=(x_1^2+\ldots+x_n^2)^p |\partial^\beta f(x)|^2\\ &=\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}x^{2\upsilon}|\partial^\beta f(x)|^2\\ &=\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}|x^\upsilon|^2|\partial^\beta f(x)|^2\\ &=\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}|x^\upsilon \partial^\beta f(x)|^2\\ &\leq \sum_{|\upsilon|=p} \frac{p!}{\upsilon!}C_{\upsilon, \beta}^2\\ &\leq (\max_{|\upsilon|=p} C_{\upsilon, \beta})^2\sum_{|\upsilon|=p} \frac{p!}{\upsilon!}\\ &=n^p (\max_{|\upsilon|=p} C_{\upsilon, \beta})^2\\ &=C_{p, \beta}^2, \end{align}$$ Consequently: $$|x|^p |\partial^\beta f(x)|\leq C_{p, \beta}.$$

Applying this with $p=|\alpha|+1\in\mathbb N$ we find:

$$\begin{align*} \displaystyle |x^\alpha \partial^\beta f(x)|&\leq \frac{|x|^{|\alpha|+1}}{|x|}|\partial^\beta f(x)|\\ &\leq \frac{C_{\alpha, \beta}}{|x|} \end{align*}$$ Now choosing $\displaystyle M=\frac{C_{\alpha, \beta}}{\varepsilon}$ we see that $|x|>M\Rightarrow |x^\alpha \partial^\beta f(x)|<\varepsilon$. Therefore, $$\lim_{|x|\to +\infty} |x^\alpha \partial^\beta f(x)|=0.$$ Obs: Above I also used the general fact $$\sum_{|\upsilon|=p}\frac{p!}{\upsilon!}=n^p,$$ whenever $\upsilon\in\mathbb N_0^n$. This follows expanding $(1+\ldots+1)^p$ with the multinomial theorem.