Tempered distribution

Remarks

The inequality says that $T$ is a continuous linear map for the Fréchet topology of $\mathcal{S}(\R^n)$ given by the seminorms $p_{\alpha,\beta}$. Equivalently, $T(\varphi_k) \to 0$ whenever $p_{\alpha,\beta}(\varphi_k) \to 0$ for all $\alpha, \beta$.

Example: the Dirac delta $\delta(\varphi) = \varphi(0)$ is tempered. It is linear, and $\abs{\delta(\varphi)} \le p_{0,0}(\varphi)$, so the condition holds with $C = 1$ and $N = 0$. More generally, if $f$ is measurable and $\int \abs{f(x)} (1 + \abs{x})^{-m} \, dx < \infty$ for some $m \ge 0$, then $\varphi \mapsto \int f \varphi \, dx$ is tempered.

Non-example: on $\R$, the function $\varphi(x) = e^{-\sqrt{1 + x^2}}$ lies in $\mathcal{S}(\R)$, but $\int e^x \varphi(x) \, dx = \infty$. So the formula $\varphi \mapsto \int e^x \varphi \, dx$ does not define a linear map on all of $\mathcal{S}(\R)$, and $e^x$ gives no tempered distribution. It still defines a distribution on compactly supported test functions.

Derivatives act by $(\partial^\beta T)(\varphi) = (-1)^{\abs{\beta}} T(\partial^\beta \varphi)$, and the Fourier transform acts by $\hat{T}(\varphi) = T(\hat{\varphi})$.