Inverse Fourier transform

Remarks

This matches the convention $\hat f(\xi) = \int f(x) e^{-2\pi i x \cdot \xi} \, dx$ for the Fourier transform. So $\check g(x) = \hat g(-x)$, and $\check g$ is continuous, bounded by $\int \abs{g}$, and tends to $0$ at infinity. With the convention $\hat f(\xi) = \int f(x) e^{-i x \cdot \xi} \, dx$ the inverse transform carries the factor $(2\pi)^{-n}$.

Example: for $n = 1$ and $g = \mathbf{1}_{[-1/2, 1/2]}$, $$\check g(x) = \int_{-1/2}^{1/2} e^{2\pi i x \xi} \, d\xi = \frac{\sin(\pi x)}{\pi x} \quad (x \ne 0), \qquad \check g(0) = 1.$$

Non-example: the constant function $g = 1$ on $\R$ is not integrable, and the integral $\int_\R e^{2\pi i x \xi} \, d\xi$ diverges for every $x$. Its inverse Fourier transform exists only in the sense of tempered distributions, where it is the Dirac delta at $0$.

The name is justified by the Fourier inversion theorem: if $f$ and $\hat f$ are integrable, then $(\hat f)^\vee = f$ almost everywhere.