Inequality between $L^p$ norm and Schwartz seminorm

Question

Let $u\in \mathscr{S}$ (Schwartz space). I can prove easily that $u$ is in $L^p$ for every $p$. But I cannot prove the following. Let $\{p_k\}$ be the family of seminorms that defines the topology of the Schwartz space

$$p_k(\psi)=\max_{|\alpha|,|\beta|\leq k} ||x^\alpha D^\beta \psi||_{L^{\infty}}$$

Then there is a sufficiently large $N$ such that $$||u||_{L^p}\leq Cp_N(u)$$ for an opportune costant $C$.

Answer 1

Hint:

The topology on $\mathcal{S}$ is generated by an equivalent sequence of pseudonorms:

$$ \rho_m(\phi):=\sup_{\substack{x\in\mathbb{R}^n\\|\beta|\leq m}} |(1+|x|^2)^m\partial^\beta\phi(x)|,\qquad m\in\mathbb{Z}_+ $$

For $\phi\in\mathcal{S}$ and $1\leq p<\infty$, $$ |\phi(x)|=(1+|x|^2)^n|\phi(x)|\frac{1}{(1+|x|^2)^n} \leq \frac{\rho_n(\phi)}{(1+|x|^2)^n}, $$ Thus \begin{align*} \|\phi\|_p&\leq \Big(\int^\infty_0\frac{1}{(1+|x|^2)^{np}}\,dx\Big)^{1/p}\rho_n(\phi) \end{align*}