Fourier transform of tempered distributions

Remarks

The definition makes sense because $\mathcal{F}$ is a continuous linear bijection of $\mathcal{S}(\R^n)$ onto itself. For an integrable function $f$, viewed as a tempered distribution, $\mathcal{F}f$ is the usual Fourier transform; this follows from Fubini's theorem, $\int (\mathcal{F}f)\varphi = \int f (\mathcal{F}\varphi)$. On $\mathcal{S}'(\R^n)$, $\mathcal{F}$ is a linear bijection, and $\mathcal{F}(\partial_j T) = 2\pi i \xi_j\, \mathcal{F}T$.

Example: for the Dirac delta $\delta$ at $0$, $\langle \mathcal{F}\delta, \varphi \rangle = \mathcal{F}\varphi(0) = \int \varphi(x)\, dx$. So $\mathcal{F}\delta$ is the constant function $1$. By the Fourier inversion formula, $\mathcal{F}1 = \delta$ in the same way.

Non-example: on $\R$, the function $e^{\abs{x}}$ is locally integrable but does not define a tempered distribution. For the Schwartz function $\varphi(x) = 1/\cosh(x/2)$, the integral $\int e^{\abs{x}} \varphi(x)\, dx$ diverges. So the formula above does not define a Fourier transform of $e^{\abs{x}}$ as a tempered distribution.