Fourier Transform of Tempered Distribution on 1-Lebesgue Space equals Tempered Distribution of Fourier Transform of defining Function

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Theorem

Let $\map {L^1} \R$ be the Lebesgue $1$-space.

Let $f \in \map {L^1} \R$.

Let $T_f \in \map {\SS'} \R$ be a tempered distribution associated with $f$.


Then:

$\hat T_f = T_{\hat f}$

where the hat denotes the Fourier transform of the tempered distribution $T_f$ and of the real function $f$ respectively.


Proof

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

\(\ds \map { {\hat T}_f} \phi\) \(=\) \(\ds \map {T_f} {\hat \phi}\) Definition of Fourier Transform of Tempered Distribution
\(\ds \) \(=\) \(\ds \int_\R \map f x \map {\hat \phi} x \rd x\) Definition of Tempered Distribution
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty \map f x \paren {\int_{-\infty}^\infty \map \phi \xi e^{-2\pi i x \xi} \rd \xi } \rd x\) Definition of Fourier Transform
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty \int_{-\infty}^\infty \map f x \map \phi \xi e^{-2\pi i x \xi} \rd \xi \rd x\)
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty \map \phi \xi \paren {\int_{-\infty}^\infty \map f x e^{-2\pi i x \xi} \rd x} \rd \xi\)
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi\) Definition of Fourier Transform

We have that the Fourier Transform of 1-Lebesgue Space Function is Bounded.

Then, from Lebesgue Infinity-Space is Subset of Tempered Distribution Space it follows that $\ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi$ is a tempered distribution.

That is, by definition:

$\ds \ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi = T_{\hat f}$

$\blacksquare$


Sources

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