Exp (-x^2) is Schwartz Test Function

Theorem
$\map \exp {-x^2}$ is a Schwartz test function.

Proof
Let:


 * $\dfrac {\d^n}{\d x^n} \map \exp {-x^2} = \map {p_n} x \map \exp {-x^2}$

We have that $\map {p_0} x = 1$ and $\map {p_1} x = -2 x$.

Then:

where


 * $\map {p_{n + 1}} x := \map {p_n'} x - 2 x \map {p_n} x$

Note that each subsequent $\map {p_n} x$ is constructed from derivatives, products and differences of polynomials.

Hence, $\map {p_n} x$ is a polynomial.

Consider the Maclaurin series of $\map \exp {x^2}$.

Suppose $\size x \ge 1$.

Then:

Suppose $\size x \le 1$.

We have that $x^k \map \exp {-x^2}$ is a continuous function.

Hence, it is bounded:


 * $\forall x \in \closedint {-1} 1 \exists M \in \R : \size {x^k \map \exp {-x^2}} \le M$

Altogether:


 * $\sup \size {x^k \map \exp {-x^2}} \le \map \max {M, k!} < \infty$

Thus:


 * $\map \exp {-x^2} \in \map \SS \R$.