Generalized Fourier Transform: Schwartz Space¶
The Schwartz space \(\mathcal{S}\) is the topological vector space of functions \(f : \mathbb{R}_n \to \mathbb{C}\) such that:
\(f(x)\) infinitely differentiable
any derivitive tends to zero fast than any power of x:
\[\forall m, n \ge 0 \implies \lim_{x \to \infty} |x|^n | f^{(m)} (x) | = 0\]
Properties without proof:
the Schwartz space is closed under differentiation and multiplication by polynomials.
Schwartz class functions are bounded and decay faster than any polynomial
Schwartz class functions are integrable
Lemma¶
Fourier transform of Schwartz class is bounded
Proof:
\[\begin{split}\because
| \int_{-\infty}^{+\infty} e^{-2 \pi i s x} f(x) dx| & \le
\int_{-\infty}^{+\infty} | e^{-2 \pi i s x} | \cdot | f(x) | dx
\\ & =
\int_{-\infty}^{+\infty} | f(x) | dx
\\ & =
\| f \|_1\end{split}\]
\[\therefore
| \mathcal{F} f (s) | \le \| f \|_1\]
Differentiation Formulas¶
First statement:
\[f \in \mathcal{S}
\implies
\mathcal{F} f^{(n)} =
(2 \pi i s)^n \cdot \mathcal{F} f\]
Second statement
\[f \in \mathcal{S}
\implies
\frac{\partial^n \mathcal{F} f}{\partial s^n} =
\mathcal{F} ((-2 \pi i x)^n f(x))\]
Proof of first statement:
\[ \begin{align}\begin{aligned}& \because
f \in \mathcal{S}\\& \therefore
f' \in \mathcal{S}\end{aligned}\end{align} \]
\[\begin{split}\therefore
(\mathcal{F} f') (s) & =
\int_{-\infty}^{+\infty}
e^{-2 \pi i s x} f' (x) dx
\\ & =
e^{-2 \pi i s x} \cdot f(x) |_{-\infty}^{+\infty} -
\int_{-\infty}^{+\infty} f(x) d e^{-2 \pi i s x}
\\ & =
- \int_{-\infty}^{+\infty}
(-2 \pi i s) \cdot e^{-2 \pi i s x} f(x) dx
\\ & =
2 \pi i s \cdot \mathcal{F} f (s)\end{split}\]
By induction it follows that:
\[\mathcal{F} f^{(n)} =
(2 \pi i s)^n \cdot \mathcal{F} f\]
Proof of first statement:
\[ \begin{align}\begin{aligned}& \because
f \in \mathcal{S}\\& \therefore
x f(x) \in \mathcal{S}\end{aligned}\end{align} \]
\[\begin{split}\therefore
-2 \pi i \cdot (\mathcal{F} (x f(x))) (s) & =
\int_{-\infty}^{+\infty}
-2 \pi i \cdot e^{-2 \pi i s x} \cdot x f(x) dx
\\ & =
\int_{-\infty}^{+\infty}
\frac{\partial}{\partial s} e^{-2 \pi i s x} \cdot f(x) dx
\\ & =
\frac{\partial \mathcal{F} f (s)}{\partial s}\end{split}\]
By induction it follows that:
\[\frac{\partial^n \mathcal{F} f}{\partial s^n} =
\mathcal{F} ((-2 \pi i x)^n f(x))\]
\[\tag*{$\blacksquare$}\]
Closure under Fourier Transform¶
\[f \in \mathcal{S} \implies \mathcal{F} f \in \mathcal{S}\]
Proof:
\[\begin{split}(2 \pi i s)^n \cdot \frac{\partial^m}{\partial s^m} \mathcal{F} f(s) & =
(2 \pi i s)^n \cdot \mathcal{F} ((-2 \pi i x)^m f(x)) (s)
\\ & =
\mathcal{F} (\frac{\partial^n}{\partial x^n} (-2 \pi i x)^m f(x)) (s)
\\ & =
(-2 \pi i)^m \cdot \mathcal{F} (\frac{\partial^n}{\partial x^n} x^m f(x)) (s)\end{split}\]
\[ \begin{align}\begin{aligned}\therefore
|(2 \pi s)^n| \cdot |\frac{\partial^m}{\partial s^m} \mathcal{F} f(s)| =
|(-2 \pi)^m| \cdot |\mathcal{F} (\frac{\partial^n}{\partial x^n} x^m f(x)) (s)|\\\therefore
|s^n| \cdot |\frac{\partial^m}{\partial s^m} \mathcal{F} f(s)| =
(2 \pi)^{m - n} \cdot |\mathcal{F} (\frac{\partial^n}{\partial x^n} x^m f(x)) (s)|\end{aligned}\end{align} \]
\[\because
f \in \mathcal{S}\]
\[\therefore
\frac{\partial^n}{\partial x^n} x^m f(x) \in \mathcal{S}\]
\[\therefore
|s^n| \cdot |\frac{\partial^m}{\partial s^m} \mathcal{F} f(s)| \le
(2 \pi)^{m - n} \cdot \| \frac{\partial^n}{\partial x^n} x^m f \|\]
\[\therefore
\mathcal{F} f \in \mathcal{S}\]
also:
\[\mathcal{F}^{-1} f \in \mathcal{S}\]
Parseval’s Theorem of Fourier Transform¶
\[f, g \in \mathcal{S}
\implies
\int_{-\infty}^{+\infty} F (s) \bar{G} (s) ds =
\int_{-\infty}^{+\infty} f (x) \bar{g} (x) dx =\]
where
\[F = \mathcal{F} f, G = \mathcal{F} g\]
Proof:
\[ \begin{align}\begin{aligned}\because
g(x) = \int_{-\infty}^{+\infty} e^{2 \pi i s x} G(s) ds\\\therefore
\bar{g}(x) = \int_{-\infty}^{+\infty} e^{-2 \pi i s x} \bar{G}(s) ds\end{aligned}\end{align} \]
\[\begin{split}\therefore
\int_{-\infty}^{+\infty} f (x) \bar{g} (x) dx & =
\int_{-\infty}^{+\infty} \left(
f (x)
\int_{-\infty}^{+\infty} e^{-2 \pi i s x} \bar{G}(s) ds \right) dx
\\ & =
\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
e^{-2 \pi i s x} f(x) \bar{G} (s) ds dx\end{split}\]
Since everything converges, integrations above are interchangeable.
\[\begin{split}\therefore
\int_{-\infty}^{+\infty} f (x) \bar{g} (x) dx & =
\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
e^{-2 \pi i s x} f(x) \bar{G} (s) dx ds
\\ & =
\int_{-\infty}^{+\infty} \left(
\int_{-\infty}^{+\infty}
e^{-2 \pi i s x} f(x) dx
\right)
\bar{G} (s) ds
\\ & =
\int_{-\infty}^{+\infty} F(x) \bar{G} (s) ds\end{split}\]
Inference:
\[f, g \in \mathcal{S}
\implies
\int_{-\infty}^{+\infty} | F (s) |^2 ds =
\int_{-\infty}^{+\infty} | f (x) |^2 dx\]
Back to Fourier Transform.