If $f\in S(\mathbb R^n)$ (schwarz space), why $f\in L^p(\mathbb R^n)$?

Question

Let $$\mathcal S(\mathbb R^n)=\left\{f\in \mathcal C^\infty (\mathbb R^n)\mid \forall N\in\mathbb N,\forall \alpha \in\mathbb N^n, \sup_{x\in\mathbb R^n}|(1+|x|^N)\partial _x^\alpha f(x)|<\infty \right\},$$ the schwarz space where $\partial _x^\alpha f(x)=\frac{\partial ^{|\alpha |}}{\partial x_1^{\alpha _1}\cdot ...\cdot \partial x_n^{\alpha _n}}f(x)$ and $|\alpha |=\alpha _1+...+\alpha _n$. Why do we have that $f\in L^p(\mathbb R^n)$ for all $p\in [1,+\infty ]$ whenever $f\in \mathcal S(\mathbb R^n)$. It's often used in my course, but I don't understand the reason (and can't prove it as well).

Answer 1

If $f$ is a Schwartz function then it is infinitely differentiable, hence is certainly measurable.

And from the definition of the Schwartz space, for each $N\geq 1$ there is a constant $M_N$ such that $$|f(x)|\leq M_N(1+|x|^N)^{-1}$$ for all $x$. Choosing $N$ large enough that $(1+|x|^N)^{-1}\in L^p(\mathbb{R}^n)$ shows that $\int_{\mathbb{R}^n}|f(x)|^p\;dx$ is finite. For $p=\infty$, taking $N=1$ shows that $f$ is bounded.