It is known that the Schwartz space is dense in $L^p$. And I was told that Schwartz functions are bounded in $L^p$. Could anyone show me

"Every Schwartz function is bounded in $L^p$" by explicitly computing the $L^p$ norm?


  • $\begingroup$ Their p-th power goes to zero faster than any inverse power of x. $\endgroup$ Oct 31, 2015 at 5:37

1 Answer 1


You could use the equivalent definition of the Schwartz space: $$f\in\mathcal{S}(\mathbb{R})\iff \sup_{x\in\mathbb{R}}|x^iD^jf(x)|<\infty\iff\sup_{x\in\mathbb{R}}|(1+|x|)^nD^kf(x)|<\infty$$For $i,j,k\in\mathbb{N}_0$, $n\geq k$. See here or Folland's Real Analysis. From this it follows readily. Alternatively, \begin{align*} \lVert f\rVert_p^p&=\int_{\mathbb{R}}\lvert{f(x)}\rvert^pdx=\int_{\lvert{x}\rvert\leq1}\lvert{f(x)}\rvert^pdx+\int_{\lvert{x}\rvert>1}\lvert{f(x)}\rvert^pdx\\ &\leq \biggl(\sup_{x\in\mathbb{R}}{\lvert{f(x)}\rvert}\biggr)^p\cdot2+\sup_{x\in\mathbb{R}}\bigl(\lvert{x}\rvert^N\lvert{f(x)}\rvert\bigr)^p\int_{\lvert{x}\rvert>1}\frac{1}{\lvert{x}\rvert^{Np}}dx\\ &=C_1+C_2 \end{align*} For $N$ sufficiently large such that the second integral in the inequality is finite.

  • $\begingroup$ Why $2^p$? You can simply bound $\int_{|x|\le1}|f|^p\le(\sup f)^p\cdot2$. $\endgroup$
    – FShrike
    Apr 12, 2023 at 8:54
  • 1
    $\begingroup$ You are correct, I mixed up dimension of the space (in this case 1) and the norm $L^p$. $\endgroup$
    – Moya
    Apr 12, 2023 at 18:49

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