Exp (-x^2) is Schwartz Test Function

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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:

\(\ds \dfrac {\d^{n + 1} } {\d x^{n + 1} } \map \exp {-x^2}\) \(=\) \(\ds \dfrac \d {\d x} \dfrac {\d^n}{\d x^n} \map \exp {-x^2}\)
\(\ds \) \(=\) \(\ds \dfrac \d {\d x} \paren {\map {p_n} x \map \exp {-x^2} }\)
\(\ds \) \(=\) \(\ds \map {p_n'} x \map \exp {-x^2} - 2 x \map {p_n} x \map \exp {-x^2}\)
\(\ds \) \(=\) \(\ds \paren{\map {p_n'} x - 2 x \map {p_n} x} \map \exp {-x^2}\)

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 $\map \exp {x^2}$ .

\(\ds \map \exp {x^2}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {x^2}^k} {k!}\)
\(\ds \) \(\ge\) \(\ds \frac {\paren {x^2}^k} {k!}\) $k \in \N$
\(\ds \) \(=\) \(\ds \frac {\size x^k} {k!} \size x^k\) Absolute Value Function is Completely Multiplicative
\(\ds \leadsto \ \ \) \(\ds \size x^{-k} k!\) \(\ge\) \(\ds \size x^k \map \exp{-x^2}\)

Suppose $\size x \ge 1$.

Then:

\(\ds \size {x^k \map \exp {-x^2} }\) \(=\) \(\ds \size x^k \map \exp {-x^2}\) Absolute Value Function is Completely Multiplicative
\(\ds \) \(\le\) \(\ds \frac {k!} {\size x^k}\)
\(\ds \) \(\le\) \(\ds k!\) $\size x \ge 1$

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:

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

$\blacksquare$


Sources

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