Inverse transform

Remarks

Example: the Fourier transform $\hat f(\xi) = \int_\R f(x) e^{-2\pi i x \xi}\, dx$ is a bijection of the Schwartz space onto itself, and its inverse is $f(x) = \int_\R \hat f(\xi) e^{2\pi i x \xi}\, d\xi$. So here the inverse transform is again an integral transform, with kernel $e^{2\pi i x \xi}$. Non-example: the transform $Tf(u) = \int_0^1 f(t)\, dt$ with kernel $K(t, u) = 1$ on $[0,1] \times \R$ has no inverse transform on the continuous functions, since $f(t) = 0$ and $f(t) = 2t - 1$ both have $Tf = 0$. When the inverse is an integral transform with kernel $H$, one writes $f(t) = \int F(u)\, H(u, t)\, du$, and $(f, F)$ is a transform pair.