I read from my note that $$\mathcal{S}(\mathbb{R})\subset L^2(\mathbb{R})\subset\mathcal{S}'(\mathbb{R}).$$
Where $\mathcal{S}$ is the space of rapidly decreasing function on $\mathbb{R}$, $\mathcal{S}'(\mathbb{R})$ is the space of tempered distribution on $\mathbb{R}.$
I could not identify the proof of above in my functional analysis text. Could anyone give me a proof of it?
If $\phi \in \def\S{\mathcal S}\def\R{\mathbf R}\S(\R)$ is given, note that by definition of rapidly decreasing, we have $\phi \in C(\R)$, that is $\phi$ is continuous - hence measurable - and moreover for some $M \in \R$: $$ \def\abs#1{\left|#1\right|}\abs{\phi(x)}(1 + \abs x^2) \le M $$ Hence $$ \int_\R \abs{\phi}^2 \,d\lambda \le M^2 \int_\R (1 + \abs x^2)^{-2}\, d\lambda(x) = \frac{M^2\pi}2 < \infty \tag 1 $$ so $\phi \in L^2(\R)$. Thus we have shown $\S(\R) \subseteq L^2(\R)$.
To prove $L^2(\R)\subseteq \S'(\R)$, let $\phi \in L^2(\R)$ be an arbitrary function, note that for any $\psi \in \S(\R)$ we have by the above, that $$ \abs{\int_\R \phi\psi\,d\lambda} \le \def\norm#1{\left\|#1\right\|}\norm{\phi}_2 \norm{\psi}_2 \le \norm{\phi}_2 \frac{\sqrt\pi}{\sqrt 2}\cdot \sup_x \abs{(1 + \abs x^2)\psi(x)} $$ As the latter is a continuous seminorm on $\S(\R)$, this proves that $\psi \mapsto \int_\R \phi\psi\, d\lambda$ is continuous on $\S(\R)$, therefore $\phi \in \S'(\R)$.