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Everybody says that the Schwartz space is a space of rapidly decreasing functions, or of functions that rapidly vanish, but I am baffled with proving it formally. I can come up with nice reasons why it is true, but not with a solid formal proof.

What I want to prove is $$f\in S(\mathbb{R}^n)\implies\lim_{|x|\to\infty}f(x)=0$$ or more precisely $\forall\varepsilon>0,\exists K\subset\mathbb{R}^n$ compact, that $x\in(\mathbb{R}^n\setminus K)\implies|f(x)|<\varepsilon$.

The Schwartz space is $$S(\mathbb{R}^n)=\{f:\mathbb{R}^n\to\mathbb{R}^n\mid f\in C^\infty(\mathbb{R}^n),\; \|f\|_{\alpha,\beta}<\infty\}$$ where $\alpha,\beta\in\mathbb{N}_0^n$ are multi-indexes and $$\|f\|_{\alpha,\beta}=\sup_{x\in\mathbb{R}^n}|x^\alpha \partial_\beta f(x)|.$$

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    $\begingroup$ You should first clarify what your characterization/definition of the Schwartz space is, and also what you mean by those other phrases, ... and then see what gap, if any, you find between the two. That is, depending on what you mean, it's a tautology... $\endgroup$ – paul garrett Oct 24 '14 at 22:59
  • $\begingroup$ I don't see $\forall\varepsilon>0,\exists K\subset\mathbb{R}^n$ compact, that $x\in(\mathbb{R}\setminus K)\implies|f(x)|<\varepsilon$ there $\endgroup$ – user74200 Oct 24 '14 at 23:06
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Proposition. Let $f\in C^\infty(\mathbb R^n)$. The following statements are equivalent.

(a) $\displaystyle\|f\|_{\alpha,\beta}:=\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|<\infty$ for all $\alpha,\beta\in\mathbb{N}^n$.

(b) $\displaystyle\sup_{x\in\mathbb{R}^n}\|x\|^k|D^\beta f(x)|<\infty$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.

(c) $\displaystyle\lim_{|x|\to\infty}\|x\|^kD^\beta f(x)=0$ for all $k\in \mathbb N$ and $\beta\in\mathbb{N}^n$.

(d) $\displaystyle\lim_{|x|\to\infty} x^\alpha D^\beta f(x)=0$ for all $\alpha,\beta\in\mathbb{N}^n$.

As a consequence we have the following

Corollary. Let $f\in C^\infty(\mathbb R^n)$. Then, $f$ is a Schwartz function if and only if $f$ is a rapidly decreasing function.

Proof of the Corollary: To say that $f$ is in the Schwartz space means that $f$ satisfies the condition $(a)$ of the proposition. To say that $f$ is rapidly decreasing means that each $D^\beta f$ tends to $0$ faster than $\|x\|^{-k}$ for all $k\geq 0$ as $|x|\to\infty$ which is precisely the condition $(c)$ of the proposition. $\blacksquare$

Proof of the Proposition:

$\text{(a)}\Rightarrow\text{(b)}$ Let $e_1,...,e_n$ be the canonical basis of $\mathbb{R}^n$. Then $$\|x\|^k=\left(\sum_{j=1}^n x_j^2 \right)^{k/2}\leq C_{k,n}\sum_{j=1}^n(x_j^2)^{k/2}=C_{k,n}\sum_{j=1}^n|x_j|^k=C_{k,n}\sum_{j=1}^n|x^{ke_j}|$$ for all $x=(x_1,...,x_n)\in\mathbb{R}^n$, where $C_{k,n}$ is a positive constant that depends on $k$ and $n$. Thus, $$\begin{align} \sup_{x\in\mathbb R^n}\|x\|^k|D^\beta f(x)|&\leq\sup_{x\in\mathbb R^n}\left(C_{k,n}\sum_{j=1}^n|x^{ke_j}|\right)|D^\beta f(x)|=C_{k,n}\sup_{x\in\mathbb R^n}\sum_{j=1}^n|x^{ke_j}D^\beta f(x)|\\ &\leq C_{k,n}\sum_{j=1}^n\|f\|_{ke_j,\beta}<\infty. \end{align}$$

$\text{(b)}\Rightarrow\text{(c)}$ There exists a constant $M$ such that $\|x\|^{k+1}|D^{\beta}f(x)|\leq M$ for all $x\in\mathbb R^n$ and thus $$\left|\|x\|^{k}D^{\beta}f(x)\right|\leq\frac{M}{|x|},\qquad\forall\ x \neq 0.$$ Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.

$\text{(c)}\Rightarrow\text{(d)}$ Let $\alpha=(\alpha_1,...,\alpha_n)$ be a multi-index and $k=\alpha_1+\cdots+\alpha_n$. Then $$|x^\alpha|=\prod_{j=1}^n|x_j^{\alpha_j}|=\prod_{j=1}^n(x_j^{2})^{\alpha_j/2}\leq \prod_{j=1}^n\left(\sum_{r=1}^nx_r^2\right)^{\alpha_j/2}=\|x\|^{k}$$ for all $x=(x_1,...,x_n)\in\mathbb{R}^n$. Thus, $$|x^\alpha D^\beta f(x)|\leq\left|\|x\|^kD^\beta f(x)\right|,\qquad\forall \ x \in\mathbb R^n.$$ Since the right hand side of the last inequality goes to zero as $|x|\to\infty$, we get the result.

$\text{(d)}\Rightarrow\text{(a)}$ There exists a constant $M$ such that $$|x^\alpha D^\beta f(x)|\leq 1,\quad\forall\ x\notin \overline{B(0,M)}$$ and thus $$\sup_{x\in\mathbb{R}^n}|x^\alpha D^\beta f(x)|\leq\left\{1,\max_{x\in \overline{B(0,M)}}|x^\alpha D^\beta f(x)|\right\}<\infty.\blacksquare$$

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  • $\begingroup$ Awesome! Thanks a lot! $\endgroup$ – user74200 Dec 29 '16 at 22:20
  • $\begingroup$ What does $x^{ke_j}$ mean? $\endgroup$ – Bob Mar 8 '20 at 11:27
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    $\begingroup$ @BobOakley For example, if $n=3$ then $x=(x_1,x_2,x_3)$ and $ke_2=k(0,1,0)=(0,k,0)$. Thus $x^{ke_2}=x_1^0 x_2^k x_3^0=x_2^k$. $\endgroup$ – Pedro Mar 8 '20 at 11:36
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My "informal" proof: It must hold for all $\alpha,\beta\in\mathbb{N}_0^n$. So if $\sup |x^\alpha f(x)|$ has to be bounded, then the limit of $f(x)$ in infinity has to be zero, because $|x^\alpha|$ in infinity is infinity. So if $f(x)$ is schwartz and $x^\alpha f(x)$ shall be bounded, then the only way to achieve this is to "put $f(x)$ down" to zero in infinity(because if $\lim_{x\to\infty} f(x)=c$, then $\lim_{x\to\infty}f(x)x^\alpha=\lim_{x\to\infty}f(x)\lim_{x\to\infty}x^\alpha=c.\infty=\infty$). If $f(x)$ is zero in infinity and $f(x)$ is smooth enough, then we can use L'hospital's rule $|\alpha|$ times and the resulting limit will be zero.

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    $\begingroup$ It is not true that $|x^\alpha|$ in infinity is infinity. The behaviour of $|x^\alpha|$ depends on $\alpha$ and the path taken. For example, if $x$ goes to infinity on the $x_1$-axis and $\alpha=(0,1,...,1)$ then $|x^\alpha|=0$. $\endgroup$ – Pedro Aug 17 '16 at 14:05

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