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

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.

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:


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