Proposition 12.4.4 (Fourier Transform on Schwartz Functions).label Suppose that $f,g\in \cali{S}(\R^{n})$ then
- (1)
$\ha{f}\in C^{\infty}$ and $\pd^{\be}(\ha{f})=\ha{c f}$ where $c=(-2\pi ix)^{\be}$ [1]
- (2)
$\ha{\pd^\be f}(\xi)=(2\pi i\xi)^{\be} \ha{f}(\xi)$
- (3)
$\cali{S}$ is stable under Fourier Transform
- (4)
$\integral{}{}(f\ha{g})dx=\integral{}{}(\ha{f}g)dx$
Proof. See Folland
- (1)
Take $\pd^{\be}$ under $\integral{}{}$ which generate the multiplicative term
- (2)
Integration by parts with boundary terms vanishing by $f\in\cali{S}$
- (3)
Follows from (1) and (2)
- (4)
An application of Fubini shows
\begin{align*}\integral{}{}f(x)\ha{g}(x)dx=\integral{}{}f(x)\integral{}{}e^{-2\pi ix\cd y}g(y)dydx=\integral{}{}g(y)\integral{}{}e^{-2\pi ix\cd y}f(x)dxdy=\integral{}{}\ha{f}gdx\end{align*}
$\square$
- under multi-index notationkeyboard_return
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