Integral transforms of tempered distributions

Remarks

The definition agrees with the integral on Schwartz functions: for $f, \varphi \in \mathcal{S}(\R^n)$, Fubini's theorem gives $\int (Tf)\varphi = \iint K(x, y) f(y) \varphi(x)\, dy\, dx = \int f\, (T^t\varphi)$, when the double integral converges absolutely. Example: the Fourier transform has the symmetric kernel $K(x, y) = e^{-2\pi i x \cdot y}$, so $T^t = T$ maps $\mathcal{S}$ onto $\mathcal{S}$. For the Dirac delta $\delta$ at $0$, $\langle T\delta, \varphi \rangle = (T\varphi)(0) = \int \varphi$, so $T\delta = 1$. Non-example: the kernel $K(x, y) = e^{xy}$ on $\R$ does not qualify, since for the Schwartz function $\varphi(y) = e^{-\sqrt{1 + y^2}}$ the integral $\int_\R e^{2y} \varphi(y)\, dy$ diverges, so $T\varphi$ is not defined at $x = 2$.