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I'm looking for a proof of $\mathcal{S}(\mathbb{R}) \subset L^p(\mathbb{R})$ for $1 \leq p \leq \infty$.

My informal probe follow like this: For any function $f \in L^p(\mathbb{R})$ exists a piecewise function $h_n$ such that $||h_n(x) - f(x)||_{L^p} \rightarrow 0 \ \text{with} \ n\rightarrow \infty \ (\forall x)$, and any $h_n$ could be approximated by compactly supported smooth functions ($C_{c}^{\infty}(\mathbb{R})$). And Since $C_{c}^{\infty}(\mathbb{R})$ is dense in $\mathcal{S}(\mathbb{R})$, then can be concluded that $\mathcal{S}(\mathbb{R}) \subset L^p(\mathbb{R})$.

Any help to formalize that, or any different proof will be helpful.

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    $\begingroup$ By "probe", do you mean "proof"? You attempt uses too sophisticated tools, and seems to be an attempt of proof of the density of the Schwartz space in $L^p$ rather than the inclusion. For the inclusion, just note that $(1+x^2)|f(x)|$ is bounded independently of $x$ if $f\in\mathcal S(\mathbb R)$ and $\int_{\mathbb R}1/(1+x^2)^p\mathrm dx$ is finite for $p=1$, hence for $p\geqslant 1$. The case $p=+\infty$ follows by definition of the Schwartz space. $\endgroup$ Commented May 27, 2016 at 15:43
  • $\begingroup$ I don't understand; it is obvious that Schwartz functions are in $L^p$ for every $p$.. and for the density, one can prove that $L^1$ is dense in $L^p$ (simply remove the tail of a $L^p$ function to get a $L^p([a,b]) \subset L^1$ function) and that $S$ is dense in $L^1$ (by convolution with $\frac{1}{n}e^{-n^2 x^2}$) $\endgroup$
    – reuns
    Commented May 27, 2016 at 19:27

2 Answers 2

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To make Davide Giraudo's comment more clear, here are the steps:

$$ \int |f(x)|^p = \int \left((1+x^2)|f(x)|\right)^p\frac{1}{(1+x^2)^p} $$

From here, since $(1+x^2)f(x)$ is bounded (because $f\in \mathcal{S}$), and $\frac{1}{(1+x^2)^p}\leq \frac{1}{1+x^2}\in L^1$ (for $p\geq 1$), one gets

$$ \int |f(x)|^p \leq \|(1+x^2)f(x)\|^p_\infty \int \frac{1}{1+x^2}\leq \pi (\|f\|_{0,0}+\|f\|_{2,0})^p $$

where

$$ \|f\|_{\alpha,\beta} = \underset{x\in\mathbb{R}^n}{\sup} |x^\alpha D^\beta f(x)| $$

Since $f\in \mathcal{S}$, $\|f\|_{\alpha,\beta}$ is finite for all $\alpha,\beta\geq 0$.

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  • $\begingroup$ Pretty clear, thanks! $\endgroup$
    – mavillan
    Commented May 27, 2016 at 18:08
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I think that a simpler proof comes from

$$\left|f(x)\right|\leq\frac{K}{1+x^2}\quad \Longrightarrow\quad \int_{\mathbb{R}}\left|f(x)\right|^p\,dx \leq K^p\cdot\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(p-\frac{1}{2}\right)}{\Gamma(p)}.$$

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  • $\begingroup$ I think you want $K^p$ on the right. $\endgroup$
    – zhw.
    Commented May 27, 2016 at 16:23
  • $\begingroup$ @zhw.: right, fixed. $\endgroup$ Commented May 27, 2016 at 17:12

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