Folland Real Analysis -- Schwartz Space example

Question

In Folland's Real Analysis 8.1, they introduce the Schwartz space $S = \left\lbrace f \in C^{\infty} : \|f\|_{N,\alpha} < \infty \forall N,\alpha\right\rbrace$ where $\alpha$ is a multi-index, $N$ is a nonnegative integer, and $\|f\|_{N,\alpha} = \sup_{x \in \mathbb{R}^n}(1+|x|)^N|\partial^\alpha f(x)|$. They then say that $f_\alpha(x) = x^{\alpha}e^{-|x|^2}$ is a function in $S$.

A couple things:

1) I don't understand the notation $x^{\alpha}e^{-|x|^2}$. The book says $x^\alpha = \Pi x_j^{\alpha_j}$, so it looks like I have: $f(x) = x_1^{\alpha_1}e^{-\sqrt{x_1^2 + \cdots + x_n^2}} \times \cdots \times x_n^{\alpha_n}e^{-\sqrt{x_1^2 + \cdots + x_n^2}}$... this doesn't seem right.

2) Why is $x^{\alpha}e^{-|x|^2}$ in $S$? Intuitively, the exponential suggests the function is rapidly decreasing, but the partial derivatives get really messy really quickly, so it seems like it's not obvious how to show this function is in $S$ (I would think if I can show the partial derivatives are bounded, I'd be done)

Answer 1

For 1), what you wrote is wrong, the exponential is present only one time in the function. The correct expression would be :

$$f_\alpha(x) = \left( x_1^{\alpha_1} x_2^{\alpha_2}\cdots x_n^{\alpha_n}\right) e^{-(x_1^2+x_2^2+\cdots +x_n^2)}$$

Or, in another form

$$f_\alpha(x) = \prod_{i=1}^n x_i^{\alpha_i}e^{-x_i^2}$$

For 2), it suffice to show by reccurence that

$$\partial^m x^\alpha e^{-|x|^2} = P_{m,\alpha}(x) e^{-|x|^2}$$

where $P_{m,\alpha}$ is a polynomial. No need to compute explicitely this polynomial

Answer 2

1. Let $f_\beta(x)=x^\beta e^{-|x|^2}$, if we are in $d$ dimensions, $$f_\gamma(x) = x_1^{\gamma_1}x_2^{\gamma_2}\dots x_d^{\gamma_d} \exp( - x_1^2-x_2^2-\cdots-x_d^2).$$

2. The definition means that the derivatives up to any order decay faster than the inverse of any polynomial in $x$. The derivatives need to be more than just bounded, e.g. functions that are infinitely continuously differentiable with compact support are not necessarily in Schwartz space.

The Leibniz rule gives $\partial^\alpha(x^\gamma e^{-|x|^2})=\sum_{|\beta|\leq \alpha} { \alpha \choose \beta} ( \partial^\beta x^\gamma )(\partial^{\alpha-\beta} e^{-|x|^2})$.

$$\begin{split} \partial^\beta x^\gamma =& {\partial \over \partial_1^{\beta_1}} x_1^{\gamma_1} \cdots {\partial \over \partial_d^{\beta_d}} x_d^{\gamma_d} \\ =& {\gamma_1! \over (\gamma_1-\beta_1)!}x_1^{\gamma_1-\beta_1} \dots {\gamma_d! \over (\gamma_d-\beta_d)!}x_d^{\gamma_d-\beta_d} \\ =& {\gamma! \over (\beta-\gamma)!} x^{\gamma-\beta} \end{split} $$

the last line follows by definition of multi-index factorials, we assume that $\beta \leq \gamma$ else the derivative is zero. It is straight forward to show that in one dimension, any order derivative in $x$ of $e^{-x^2}$ is $P(x)e^{-x^2}$ for $P$ some polynomial of $x$. Skipping the details this time we find that

$$ \partial^{\alpha-\beta} e^{-|x|^2} = \prod_{i=1}^d P_i(x_i) e^{-x_i^2} = e^{-|x|^2 } Q_{\alpha,\beta}(x), $$

where $Q(x) = P_1(x_1) \cdots P_d(x_d)$, so far this shows that

$$\begin{split} \partial^{\alpha} f_\gamma(x) =& \sum_{|\beta|\leq \alpha, \gamma \geq \beta} { \alpha \choose \beta} {\gamma! \over (\beta-\gamma)!} x^{\gamma-\beta} e^{-|x|^2 } Q_{\alpha,\beta}(x) \\ =& e^{-|x|^2} R_{\alpha, \gamma}(x) \end{split} $$

and $f_\gamma\in C^\infty( \Bbb{R}^d)$, as required. The exponential grows faster than any polynomial in $x$, so for any $N \in \Bbb{N}$ and $\alpha \in \Bbb{R}^d$, $$ \begin{split} \lVert f_\gamma \rVert_{N,\alpha} =& \sup_{x \in \Bbb{R^d} } |\partial^\alpha f_\gamma(x) |\\ =& \sup_{x \in \Bbb{R^d} } e^{-|x|^2} R_{\alpha, \gamma}(x) (1+|x|)^{N} < \infty \end{split}$$ so that $f_\gamma \in \mathcal{S}(\Bbb{R}^d)$.