$L^2$ implies tempered distribution

Question

I am reading a book and I don't understand some of the statements in the proof. It says $1_{B(0,1)}f\in L^2(\mathbb{R}^d;(1+|\xi|)^sd\xi))$ so it also belongs to $S'(\mathbb{R^d})$ thus it is a tempered distribution. I don't quite know why the function belongs to $L^2$ space of some measure implies it is a tempered distribution. Thanks for any help!

Answer 1

An easy way to see that $L^2(\mathbb{R}^d) \subset \mathcal{S}'(\mathbb R^d)$: $\mathcal{S}'(\mathbb R^d)$ is defined to be the dual space of $\mathcal{S}(\mathbb R^d)$, the Schwartz space, which is contained in $L^2$ (that is easy to see). From $\mathcal{S} \subset L^2$, taking the duals we get $(L^2)' \subset \mathcal{S}$, and the result follows because $(L^2)'=L^2$.