A relation between $L_p$ and $L_\infty$ spaces for any $1\le p < \infty$ for Schwartz functions

Question

Let us consider only Lebesgue measure on $\mathbb{R}^n$. The norms are defined by $$ \| f \|_p := \left(\int_{\mathbb{R}^n} |f|^p dx \right)^{1/p} \qquad \text{and} \qquad \|f\|_\infty := \inf\{C \ge 0 : |f(x)|\le C \text{ for almost every $x$}\}. $$ and define Schwartz space $\mathcal{S}$, for every $\alpha, \beta \in \mathbb{N}^n$, $$f\in\mathcal{S} \quad \iff \quad \sup_{x\in\mathbb{R}^n} |x^\alpha D^\beta f(x)|<\infty \; \text{for every}\; f \in C^{\infty}(\mathbb{R}^n).$$ My question is here.

For every $1\le p < \infty$, is the following inequality holds: $$ \|f\|_p \le \|f\|_\infty$$ for every $f \in \mathcal{S}$ ?

Answer 1

Not true. Take $n=1,p=1,f(x)=e^{-cx^{2}}$ where $c>0$. The inequality gives $\int_{\mathbb R} e^{-cx^{2}} dx \leq 1$. Letting $c \to 0$ and using Fatou's Lemma we get $\infty=\int_{\mathbb R} 1 dx\leq 1$, a contradiction.

Answer 2

The other answer is fine. It may be of interest to see a "softer" answer. Here we don't need to know any specific Schwartz functions, we just need $\newcommand\S{\mathcal S}$ $\S\ne0$: Say $f\in\S$, $f\ne0$, and for $t>0$ define $$f_t(x)=f(tx).$$Then $||f_t||_\infty=||f||_\infty$, $\lim_{t\to0}||f_t||_p=\infty$ and $\lim_{t\to\infty}||f_t||_p=0$, so neither norm is bounded by a constant times the other.