Fourier Transform of Dirac Delta Distribution
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Theorem
Let $\delta \in \map {\SS'} \R$ be the Dirac delta distribution.
Let $\mathbf 1 : \map \SS \R \to \R$ be the constant tempered distribution such that for all $\phi \in \map \SS \R$ we have:
- $\ds \map {\mathbf 1} \phi = \int_{-\infty}^\infty 1 \cdot \map \phi x \rd x$
Then in the distributional sense it holds that:
- $\hat \delta = \mathbf 1$
where the hat denotes the Fourier transform of a tempered distribution.
Theorem
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Then:
\(\ds \map {\hat \delta} \phi\) | \(=\) | \(\ds \map \delta {\hat \phi}\) | Definition of Fourier Transform of Tempered Distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\hat \phi} 0\) | Definition of Tempered Dirac Delta Distribution | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{- \infty}^\infty \map \phi x e^{-2 \pi i 0 x} \rd x\) | Definition of Fourier Transform of Real Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_{- \infty}^\infty \map \phi x \cdot 1 \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\mathbf 1} \phi\) |
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions