Let $f : \mathbb R \to \mathbb R$ and $a > 0$ given by $f(x) = e^{-a|x|}$. Show that $f$ is rapidly decreasing and belongs to $L_1(\mathbb R)$, but not to $\mathcal S(\mathbb R)$.
I had shown that it is rapidly decreasing and in $L_1(\mathbb R)$, but I am unsure about $\mathcal S(\mathbb R)$. In my opinion it belongs to $\mathcal S(\mathbb R)$...
To get this off the unanswered list (all credits go to Bryan): $f$ is not differentiable at $0$ since $$\lim_{h\downarrow 0}\frac{1}{h}(f(h)-f(0))\neq\lim_{h\uparrow 0}\frac{1}{h}(f(h)-f(0)).$$ You cannot show it to be rapidly decreasing if it is not differentiable (rapidly decreasing usually means to be an element of Schwartz space, but perhaps you are using a different use of the term?). But it will be a pleasure for you to show $\exp(-ax^2)$ to lie in Schwartz space.