Show that $f$ not define tempered distribution

Question

Let $f(x)=\exp(|x|²)$.

How can show that $f\in \mathcal{D}'(\mathbb{R}^n)$ but $f\notin \mathcal{S}'(\mathbb{R}^n)$

Answer 1

The function $f$ defines a distribution because it is locally integrable: $$\forall R>0\quad \int_{-R}^R|f(x)|dx<+\infty.$$

After that, consider a positive Schwartz function $\phi\in S$ with non-compact support. Without losing generality we can suppose that $\forall x\in[-1,1]\,\phi(x)>1$.

After that consider a sequence of Schwartz functions $g_n(x) = \frac 1n \phi(x-n)$. Clearly, all seminorms for the function $g_n$ defining the Schwartz space vanish as $n\to \infty$, therefore, $g_n\to0$ in the sense of $S$.

However, $$\langle f,g_n\rangle=\int_{\Bbb R}e^{x^2}g_n(x)dx \ge \int_{n-1}^{n+1}e^{x^2}g_n(x)dx\ge \frac 1n\int_{n-1}^{n+1}e^{x^2} dx\ge \frac 2n e^{(n-1)^2}\to +\infty\text{ as }n\to \infty.$$

Therefore, $f$ can not be a tempered distribution.