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Can someone explain the notion of a rapidly decreasing function? Namely, a function in the Schwartz space:

$$\mathscr{S}(\mathbb{R}^n):= \{ f \in C^{\infty} (\mathbb{R}^n) : ||f||_{\alpha, \beta} < \infty \, \, \forall \alpha, \beta\}$$

with $$ ||f||_{\alpha,\beta}:= \sup_{x \in \mathbb{R}^n} |x^{\alpha} D^{\beta} f(x)|$$

What exactly is this trying to say? I'm not quite familiar with the notation $x^\alpha$. Is this the same $x \in \mathbb{R}^n$ that we take the sup over? I understand that $\alpha, \beta$ are multi-indices, and I understand the notation $$D^\beta : = \frac{\partial^{ |\beta|}}{\partial x_1^{\beta_1} \cdots \partial x_n^{\beta_n}}$$, but I just don't quite understand why these functions are "rapidly decreasing". So, an explanation of the definition and notation would be appreciated.

Pedro
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Anthony Peter
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    $x^{\alpha}$ commonly refers to $x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ where the $\alpha_i$ are integers. This is the beauty and annoyance of multi-index notation. It's great for people who are well-acquainted with the material and awful for people who aren't. I'm of the opinion that Fourier transform theory should be taught in one variable and then a brief word or two said about several variables (since the ideas are very similar). Schwartz functions are called rapidly decreasing since they decay faster than any polynomial. Think exponential decay. The usual example is $e^{-x^2}$ – Cameron Williams Jul 21 '14 at 23:54
  • Naively, $f \in C^{\infty}(\mathbb{R^n})$ will be in $\mathscr{S}(\mathbb{R}^n)$ if, and only if, every derivative of $f$ goes to $0$ faster than every $\frac{1}{x^{\alpha}}$ when $x \to \infty$. – Amateur Jul 21 '14 at 23:54
  • @CameronWilliams so why is it that we multiply the derivative by this product of $x$'s components? – Anthony Peter Jul 21 '14 at 23:58
  • Well the problem is that it is perhaps not obvious how to quantify the phrase "decays faster than any exponential". Intuitively what that would mean is that if you multiply your function $f$ by any polynomial $p$, that $\lim_{x\to\pm\infty} p(x)f(x) = 0$. This isn't a true characterization though since lots of horribly-behaved functions obey just a relationship. An example is $f(x) = e^{-|x|}/x$. So perhaps we should require that $f$ be infinitely differentiable since we want to take arbitrarily many derivatives in Fourier theory. – Cameron Williams Jul 22 '14 at 00:01
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    Particularly, this says that $f$ must be continuous. Well as soon as we make $f$ continuous, it's not hard to see that the condition $\lim_{x\to\pm\infty} p(x)f(x) = 0$ for any polynomial $p$ is equivalent to saying $\sup_{x\in\Bbb R}|x^nf(x)| <\infty$ for any $n\in\Bbb N$. We would like such a condition to hold for our derivatives as well since we would like our vector space to be closed under differentiation - otherwise really bad things might happen in our integrals! – Cameron Williams Jul 22 '14 at 00:03
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    "really bad things"? Also, can you explain the equivalence of the definitions? @CameronWilliams – Anthony Peter Jul 22 '14 at 00:09
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    I was being pretty colloquial but what I mean is that all bets would be off as to how the derivatives of our function decayed which spells disaster for Fourier transform theory (which you are building up to). In Fourier transform theory, $x$ and $\frac{d}{dx}$ play a dual role and the whole point (in my opinion) of distribution theory is to make this duality explicit. You'll be multiplying functions by powers and differentiating them all over the place. It is best that the derivatives also decay exponentially or you could end up with divergent integrals. – Cameron Williams Jul 22 '14 at 00:13
  • @CameronWilliams Can you explain the equivalence of definitions? – Anthony Peter Jul 22 '14 at 05:00
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    If a function is continuous and goes to zero at infinity, it has to have a maximum. That's the general idea. – Cameron Williams Jul 22 '14 at 05:21

1 Answers1

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I just don't quite understand why these functions are "rapidly decreasing".

The reason for the terminology is that, if $\phi\in\mathcal{S}(\mathbb R^n)$, then each $D^\alpha \phi$ tends to $0$ faster than $|x|^{-N}$ for all $N\geq 0$ as $|x|\to\infty$. (Introduction to the Theory of Distributions, page 93)

See the proof here.

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Pedro
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