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Definition of rapidly decreasing function

$$\sup_{x\in\mathbb{R}} |x|^k |f^{(l)}(x)| < \infty$$ for every $k,l\ge 0$.

Given the Gaussian function $f(x) = e^{-x^2}$, I know that its derivatives will always be in form of $P(x)e^{-x^2}$ where $P(x)$ is a polynomial of degree, say, $n$. Then $|x|^k |f^{(l)}(x)|$ will be $Q(x) e^{-x^2}$ where $Q(x)$ is of degree $n+k$. $e^{-x^2}$ is bounded apparently. But how could I "immediately" argue this whole thing is bounded?

3x89g2
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  • I guess that you never proved that $x e^{-x} \to 0$ when $x \to \infty$ because the proof is exactly the same here – reuns Apr 22 '16 at 05:03

2 Answers2

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Suppose $(e^{-x^2})^{(n)} =p_n(x) e^{-x^2} $. Then

$\begin{array}\\ (e^{-x^2})^{(n+1)} &=(p_n(x) e^{-x^2})'\\ &=p_n'(x) e^{-x^2}-p_n(x)(2x) e^{-x^2}\\ &=e^{-x^2}(p_n'(x) -2xp_n(x))\\ \end{array} $

so if we define $p_0(x) = 1$ and $p_{n+1}(x) =p_n'(x) -2xp_n(x) $, then $(e^{-x^2})^{(n)} =p_n(x) e^{-x^2} $.

Looking at this recurrence, we see that $p_n(x)$ is a polynomial of degree $n$ with leading coefficient $(-2)^n$.

We could derive a recurrence for the coefficients, but this is enough to show that $|p_n(x)| \le C_n x^n $ for $x \ge 1$, where $C_n$ is the sum of the absolute value of the coefficients of $p_n(x)$.

Since $x^ne^{-x^2} \to 0$ as $x \to \infty$, $p_n(x)e^{-x^2} \to 0$ as $x \to \infty$.

Easy proof that $x^ne^{-x^2} \to 0$ as $x \to \infty$:

From the power series, $e^{x^2} \gt \dfrac{x^{2n}}{n!} $ so $x^n e^{-x^2} < n! x^{-n} \to 0 $ as $x \to \infty$.

marty cohen
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An exponential always beats a polynomial in the end...

$|x| \le e^{|x|}$ so $|x|^n < e^{n |x|} < e^{x^2}$ if $|x| > n$, therefore $\left|x^n e^{-x^2}\right| < 1$ there.

Since the continuous function $x^n e^{-x^2}$ is also bounded on the finite interval $[-n,n]$, we conclude that $x^n e^{-x^2}$ is bounded on $\mathbb R$.

Take a linear combination of those and you have your $Q(x) e^{-x^2}$.

Robert Israel
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