Fourier Transform of Derivative of Tempered Distribution

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Theorem

Let $T \in \map {\SS'} \R$ be a tempered distribution.

Let $\xi \in \R$ be a real number.

Let the hat denote the Fourier transform.


Then in the distributional sense it holds that:

$\hat {\paren{T'} } = 2 \pi i \xi \hat T$


Proof

Let $\phi \in \map \SS \R$ be a Schwartz test function.

Then:

\(\ds \map {\hat {\paren {T'} } } {\map \phi x}\) \(=\) \(\ds \map {T'} {\map {\hat \phi} x}\) Definition of Fourier Transform of Tempered Distribution
\(\ds \) \(=\) \(\ds \map {T'} {\int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }\) Definition of Fourier Transform of Real Function
\(\ds \) \(=\) \(\ds -\map T {\dfrac \d {\d x} \int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }\) Definition of Derivative of Tempered Distribution
\(\ds \) \(=\) \(\ds -\map T {-2\pi i \xi \int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }\)
\(\ds \) \(=\) \(\ds 2\pi i \xi \map T {\int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }\) Definition of Tempered Distribution
\(\ds \) \(=\) \(\ds 2\pi i \xi \map T {\map {\hat \phi} x}\) Definition of Fourier Transform of Real Function
\(\ds \) \(=\) \(\ds 2\pi i \xi \map {\hat T} {\map \phi x}\) Definition of Fourier Transform of Tempered Distribution


$\blacksquare$


Sources

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