5
$\begingroup$

The Schwartz space $\mathcal{S}(\Bbb{R})$ consists of all indefinitely differentiable functions $f$ such that for all $\ell, k \geq 0$, we have $$ \sup_{x \in \Bbb{R}} |x|^k |f^{(\ell)}(x)| \lt \infty $$ i.e. $f$ and all of its derivatives are rapidly decreasing in that sense.

How do I show that this is true of $f(x) = e^{-x^2}$?

I know that $\mathcal{S}(\Bbb{R})$ is closed under multiplication by polynomials in $x$ over $\Bbb{C}$ and that it is also a vector space over $\Bbb{C}$. Obviously it's closed under taking derivatives too.

We have that $f^{(\ell)}(x) = P(x) e^{-x^2}$ for some polynomial $P$.

$\endgroup$
5
  • 1
    $\begingroup$ Since you already know the structure of $f^{(\ell)}$, all that remains is to see that $P(x)e^{-x^2}$ is bounded for every polynomial $P$. $\endgroup$ Nov 8, 2013 at 18:12
  • $\begingroup$ How do I prove that it's bounded, without looking at the graph? $\endgroup$ Nov 8, 2013 at 18:15
  • 1
    $\begingroup$ Brutal estimate. Say the polynomial has degree $d$, then you know that $\lvert P(x)\rvert \leqslant C\lvert x\rvert^d$ for some $C > 0$ and $\lvert x\rvert \geqslant 1$. You know $e^{x^2} \geqslant 1 + \frac{x^{2d}}{d!}$, hence $\lvert P(x)e^{-x^2}\rvert \leqslant\: ?$ $\endgroup$ Nov 8, 2013 at 18:19
  • $\begingroup$ Thanks @DanielFischer I will try to answer this with your help in the comments. $\endgroup$ Nov 8, 2013 at 18:27
  • 1
    $\begingroup$ If you face difficulties, feel free to ping me. $\endgroup$ Nov 8, 2013 at 18:27

1 Answer 1

7
$\begingroup$

$f^{(\ell)}(x) = P_{\ell}(x) e^{-x^2}$ and $|P_{\ell}(x)|$ is bounded by $C_{\ell}|x|^d$, where $C_{\ell}$ is some constant and $d \geq$ degree of $P_{\ell}$. Let $d$ be even for convenience (so we can divide by $2$).

We can write $e^{x^2} = 1 + x^2 + (x^2)^2/2! + \dots$ so that $e^{x^2} \geq 1 + (x^2)^{d/2}/(d/2)!$ (all terms are positive) and $e^{-x^2} \leq \frac{1}{1+x^d/(d/2)!} \implies |e^{-x^2}| \leq |\frac{1}{1+x^d/(d/2)!}|$.

So we have $\sup_{x \in \Bbb{R}}|x|^k|f^{\ell}(x)| \leq \sup C_{\ell}\left | \frac{x^d}{1 + \frac{x^{d}}{(d/2)!}} \right | \leq C_{\ell}\left | \frac{x^d}{x^d / (d/2)!} \right | \leq C_{\ell}(d/2)!$

Thus $f(x) = e^{-x^2}$ and all of its derivatives are rapidly decreasing in the Schwartz space sense and so $f$ belongs to $\mathcal{S}(\Bbb{R})$.

$\endgroup$
3
  • 4
    $\begingroup$ Pedantically, you only have $\lvert P_\ell(x)\rvert \leqslant C_\ell \lvert x\rvert^d$ for $\lvert x\rvert \geqslant a > 0$, since $\lvert x\rvert^d$ vanishes in $0$ [for $d > 0$], but $P_\ell$ need not. You can restrict to $\lvert x\rvert \geqslant 1$, and conclude the boundedness on $[-1,1]$ from the continuity, or you could bound by $C_\ell(1 + \lvert x\rvert^d)$. But in principle, that's it. $\endgroup$ Nov 8, 2013 at 19:05
  • 2
    $\begingroup$ In the first line of the post, shouldn't "$|P_{\ell}(x)|$ is bounded" be replaced by "$|x|^k|P_{\ell}(x)|$ is bounded by"? $\endgroup$
    – Pedro
    Mar 14, 2016 at 13:36
  • $\begingroup$ @Pedro good question. This was too long ago for me to answer though. Feel free to edit the post. $\endgroup$ Feb 5, 2021 at 8:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .